text:
Molecular orbital theory and the conservation of molecular orbitals
I am not clear with the molecular orbital theory. I have read:

If the atomic orbitals are combined with the same phase they interfere constructively and a bonding orbital is formed.
  If the atomic orbitals are combined with different phases, they interfere destructively and an anti-bonding molecular orbital is formed.

So as per my understanding when two s-orbitals are combined either a bonding orbital can be formed or an anti-bonding orbital can be formed depending upon the type of interference.
This contradicts the following statement:

When two atomic orbitals combine they form two new molecular orbitals.

This means that when two s-orbitals combine they produce both bonding and anti-bonding orbitals. Where have I gone wrong?

The problem is that you interpret the notion of "orbital interference" literally. That is not your fault: many texts are still written in a way so that they give a false impression that orbitals indeed interfere with each other. But they do not. Strictly speaking, orbitals do not even exist, but that is a bit different story. Orbitals combine in a certain way which is just somewhat similar to wave interference, but is not wave interference.

I would agree with 'wildcat' that thinking of interference is not particularly helpful in this case. Thinking of combining orbitals of the same 'phase' is more useful as a Rule of Thumb. You are right in thinking that two orbitals produce a bonding and anti-bonding orbital. The sum of these orbital energies is the same as the sum of the original orbitals. Putting electrons in the bonding orbital lowers to total energy.

The two statements are not contradictory. There are indeed two ways to combine the two atomic orbitals to form new molecular orbitals. It is not either one way or the other--in fact both occur - one way creates a bonding molecular orbital and the other creates an antibonding molecular orbital.
So for example, if you have two ground state hydrogen atoms, each with an electron in the 1s orbital, you end up with a hydrogen molecule, with 2 electrons in the lower energy (sigma) bonding molecular orbital and 0 electrons in the (sigma*) antibonding molecular orbital. The same thing is true for formation of pi bonds from p orbitals.

But how can two waves interfere both constructively and destructively at the same time?

Here is a way to think about it: an interference pattern formed by combining two waves has regions of constructive interference and other regions of destructive interference.

So , do you mean to say that when two 1s orbitals combine , there are always two types of molecular orbital formed : bonding and anti bonding ?


text:
Migratory aptitude in Baeyer–Villiger reaction
I found a reaction in which we have to find the product:

I tried and got to know that it is Baeyer–Villiger reaction but I got confused after seeing the answer, I thought there would be a shift of methyl group according to migratory aptitude, but it is not like that the product is:

Why?

In the following image you can see the charge distribution in the transition state of the Baeyer–Villiger reaction (image taken from Clayden Organic Chemistry):

As you can see, there is partial positive charge at the migrating C atom, therefore a group will migrate easier if it contains cation stabilizing moieties. A benzene ring significantly stabilizes positive charge in its proximity through delocalization.

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text:
What causes mercury to expand equaly unregarded what the temperature increase is?
Mercury seemed to be a good liquid to use in thermometers. But why are most liquids expanding non-linearly with increase of temperature compared to other liquids?

related http://chemistry.stackexchange.com/questions/15944/why-is-mercury-used-in-thermometers http://chemistry.stackexchange.com/questions/13584/what-is-the-reason-for-anomalous-expansion-of-water/13588?s=5|0.0617#13588

Note that the thermal expansion of mercury is not purely linear, but is linear enough for an OK thermometer from 0 to 100C. NIST points to 'A Critical Evaluation of the Thermophysical Properties of Mercury, G.J.F. Holman and C.A. ten Seldam, in J. Phys. Chem. Ref. Data. 23(5) 807-827 (1994) for their preferred values.

The small but not zero thermal expansion is due to its mass and the fact that its liquid range is almost 400° C. Few other room temperature liquids have such large ranges.

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Can fungi metabolize pluronic polyol f127?
I am looking for a gelling agent that can be used in place of agar, where its carbon must not be bioavailable for fungi. I have read about pluronic polyol f127 as substitute for agar. Can fungi metabolize carbon from this polymer?

It is known that some species of fungi can use hydrocarbons as their carbon source. Polyethers like pluronic are less chemically inert than hydrocarbons. Indeed, degradation of polyethylene glycols by microorganisms is a well studied process (see DOI: 10.1007/s00253-001-0850-2 for example). Gloeophyllum trabeum is known to degrade poly(ethylene oxide) by a non-enzymatic, oxidative process (Proc. Natl. Acad. Sci. USA. 1998, 95(18):10373-7). Given those facts, there is a very good chance that some species of fungi can metabolize pluronics and similar compounds.

When you said that there is a very good chance that some fungal species can metabolize pluronics does that mean that most fungal species will not be able to metabolize pluronics? If that's the case then pluronics is ideal substitute for agarose as gelling agent for my study given that most fungi survive in agar. By the way, based from the structure, how can I tell whether carbon in a polymer is available for microorganisms?

Well I can't tell exactly how many species is able to metabolize polyethers, but I wouldn't be surprised if most of them can. About stability: anything that contains hydrolysable bonds (amides, esters and hemiacetal ethers) is potentially prone to biodegradation. Also chemical groups that can be oxidized - like alcohol groups in pluronics are good targets for enzymes. But that free-radical oxidation system described in the second reference is able to attack practically any organic molecule, so your only choice is to test several gelling agents with your particular species of fungi.


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What are typical runtimes for CASSCF calculations?
I'm doing single point energy calculations of FeS using ORCA. I originally used DFT with a variety of functionals, and these calculations all took less than a minute. Now I'm attempting to run CASSCF calculations using 8 active orbitals, 12 active electrons, and the aug-cc-pVTZ basis set. It's been running for 3 days. Is this typical, or should I be attempting to change tolerances and other settings? How long would one anticipate something like this to run? In addition, how can I (generally speaking) evaluate how close a given calculation is to finishing?

Would you consider remove diffuse function, which may save some time. In addition, molpro's casscf is typically faster than orca. Though dmrg can be faster, I heard it only becomes useful for beyond ~(12,12)

I can definitely try that! I'm replicating data from a paper, so it'll just be a matter of seeing how close it gets. Removing the diffuse function would be making my basis set cc-pVTZ instead of aug-cc-pVTZ, correct?
And that's good to know about Molpro, although since I just started and am starting with ORCA, I'm going to try to stick to that for the time being.

Yes. I mean cc-pVTZ..

Did you check the partial values of the ocupation numbers? Close to full or empty ocupation can give convergence problems

No, I haven't. I'll have to look into how to do that because I'm not sure how yet, but I'll give that a go. Thank you!

They are printed somewhere for each macro iteration. There must be as many as the number of orbitals in the active space. You can paste it in pastebin or similar if you prefer to show us.

Thank you. I'll take a look. In the meantime, a pastebin can be found here. Potentially unrelated issue, but I also submitted a job that was identical except the basis set was aug-cc-pVDZ (and increased the number of processors), but it had a fatal error regarding an incompatible matrix or vector ("BLAS-ERROR"). Looking into that now as well. I think it's because the CI-Iterations did not converge.

As shown in the Pastebin link above, it seems like the occupation number is okay (I think).

@AndrewR. I am not very experienced. I think that you probably can reduce the active space. The values 1.99865 1.99964 1.99868 1.99868 are very close to full occupation (Nocc=2).  I have particular interest in your question, as I started few days ago to learn about practical aspects of CASSCF by myself. I had not problems with my molecules but your case can be useful to learn about. Can you share the DOI of the paper? We can try to work together to solve this.

@user1420303. I will try reducing the active space and see if I can get a converged solution with that. I'd be happy to share the DOI. It can be found here 10.1016/j.cplett.2007.01.073.

IIRC your space has ${24 \choose 8} =735471$ Slater determinants and you're solving a dense square matrix eigenvalue problem of that size.

There is nothing trivial about MCSCF calculations because it is hard to predict a priori how long a calculation will take. There are well-defined equations for calculating how many determinants
$$
D(n,N,S) = \binom{n}{N/2+S} \binom{n}{N/2-S}
$$
or configuration state functions (CSFs)
$$
D(n,N,S) = \frac{2S+1}{n+1} \binom{n+1}{N/2-S} \binom{n+1}{N/2+S+1}
$$
will be required for your problem; $n$ is the number of active orbitals, $N$ is the number of active electrons, and $S$ the total spin. The calculation cost will scale with this number and can be further reduced with the use of point group symmetry. (Actually, depending on your system and the property of interest, calculations without symmetry can be meaningless, like for excited states and their energies.)
These equations should make it clear that the cost grows very rapidly, considering that for CASSCF, a full CI is being performed within your chosen active space. This is not what makes MCSCF calculations tricky in practice. Here is a plot of the total energy after the first 30 iterations.

This is because the occupation numbers are very close to 0/1/2, and they aren't changing with each iteration. Notice that the occupation numbers have inverted themselves, and repeated at every step is Rot=23,19. Orbital 19 is within the active space, but 23 is external. This is an indication that the orbitals in the active space are not the correct ones. From the paper:

the 12 or 13 valence electrons of the neutral or anion, respectively, were distributed among 14 valence active orbitals: 3p of sulphur plus 3d, 3d' and 4s orbitals of iron. With this active space we found that all important correlating orbitals are included in the active space.

MCSCF is not a black box; it's one of the few quantum chemical methods left that requires chemical intuition to even get started. This a (12,14) active space; the one in your calculations is (12,8). The procedure I use for running MCSCF calculations is generally:

Converge a single-reference calculation. This could be HF, DFT, MP2, etc. The usual recommendation is to form MP2 natural orbitals, then check the occupations numbers. If the occupation deviates more than 0.02 from an integer value, then it should probably be in the active space. This is in addition to considering the electrons and orbitals you wanted to correlate to begin with, along with antibonding partners. For example, in ethylene this is probably the pi bonding/antibonding pair, leading to a (2,2), and for 3d metal complexes this is whatever coordination (anti)bonds are present, with potentially the double d shell. I've had some luck with using DFT orbitals for metal complexes, where perturbation theory might be a bad idea due to near-degeneracies or cost. The goals here are to confirm your active space and provide the best possible starting orbitals for MCSCF.
Look at your orbitals. Of course, read the Mulliken or Lowdin population analysis, but from personal experience, I've caught many potential mistakes by doing this. It also helps when visualizing how the single-state, single-reference orbitals transform to the natural orbitals (potentially state-average) that result from MCSCF calculations.
Reorder your orbitals if necessary before running the MCSCF. Except for the most trivial cases, the orbitals that belong in the active space are not the orbitals that come straight out of your single-reference calculation. If they were, then you might not be running MCSCF to begin with. I also tend to look at the reordered orbitals one more time before doing MCSCF to confirm I haven't made any mistakes.
Read in the previous set of (reordered) orbitals into a separate MCSCF calculation and cross your fingers. Monitoring the occupation numbers of your active space is a good guide for seeing how the calculation is converging.
Do whatever post-processing you need for your project. For me, this is looking at final occupation numbers, the orbitals themselves, and then doing MR-CISD or some other multireference correlated calculation.

Doing an auto-occupation procedure without starting from previously-converged orbitals means that 1. you don't have any guarantees about the contents of your active space, 2. you start from orbitals that are close to Hartree-Fock in quality, and 3. the calculation has to do far more work in optimizing both sets of variational parameters (the MO coefficients $\{C\}$ and the CI coefficients $\{c\}$) since the internal and external orbitals (where the CI coefficients are frozen) are still far from convergence using a single-reference method.
To walk through part of this workflow, I ran a PBE0/aug-cc-pVTZ calculation (converges much faster than the SCF before MP2) to look at the Lowdin MO populations:
! pbe0 aug-cc-pvtz cc-pvtz/jk ri rijk tightscf usesym zora

%output
 print[p_orbpopmo_l] 1
 end

* xyz 0 5
Fe      0.000000      0.000000      0.000000
S       1.960000      0.000000      0.000000
*

The MO indices are S 3s: 14, Fe 4s: 20, Fe 3d: 15,16,17,18,19, and Fe 3d': 33,34,36-40. Automatic occupation would have gotten the occupieds right, but the virtuals wrong. I'm not sure which of the 7 virtual MOs that might be the Fe 3d' are the correct ones; there is probably some trial-and-error here, but these should be visualized. Another point is that I've only looked at the ordering for alpha-spin orbitals; for difficult cases such as transition metals, the spatial ordering for beta-spin orbitals can be very different. The reordering for alpha- and beta-spin orbitals might be different. Again, this is speaking from experience.
The final exhibit is the RI-MP2/aug-cc-pVTZ natural orbitals that might enter an MCSCF calculation based on their occupation numbers:
N[ 15]( A2) =   1.97761944
N[ 16]( A1) =   1.97753198
N[ 17]( B2) =   1.97009578
N[ 18]( B1) =   1.97009578
N[ 19]( A1) =   1.03828526
N[ 20]( A1) =   0.99215102
N[ 21]( B1) =   0.98317529
N[ 22]( B2) =   0.98317529
N[ 23]( A2) =   0.02013554
N[ 24]( B2) =   0.01947713
N[ 25]( B1) =   0.01947713
N[ 26]( A1) =   0.01816184

To be thorough, this should be combined with looking at the Lowdin MO populations of these, which might require reading them into another (SCF) calculation and not performing any iterations.
However, it is well-known that converging MCSCF equations is generally difficult. In my limited experience, DALTON, GAMESS, and Molcas have less trouble than ORCA. I haven't used Molpro for MCSCF. Which program you choose dictates what post-MCSCF calculations can be performed. The DIIS algorithm in ORCA, in particular, can stall even when it shouldn't. If you can afford it, the Newton-Raphson algorithm (switchstep nr), where the active space orbital Hessian is calculated at each step, works very well to force convergence once SuperCI is done. ORCA makes it easy to get started with CASSCF, and it has plenty of knobs, but it has a very slow integral engine, doesn't prevent you from doing stupid things like GAMESS, and more advanced features are poorly documented. The workflow described above (namely the plotting orbitals and rotating them) becomes very tedious as well, but that has more to do with MCSCF in general.
Here are some good resources that describe both practical and theoretical aspects of performing MCSCF calculations: 1, 2, 3, 4.

I can't thank you enough for such a thorough response. I am taking my time to digest everything you've said, and it's helping a lot, so I really appreciate it. I realize in the paper that they used a (12,14) active space. I thought it would be less costly to use a smaller active space, but that clearly wasn't the case. For some reason (12,10) ran extremely quickly and (12,14) is still running after about a week. Still, perhaps that means the orbitals chosen as the active orbitals by ORCA aren't correct. I will be sure to investigate further. Thanks for getting me started!

What I tried to convey is that everything is a balance. The (12,8) didn't run because there was no way to optimize the wavefunction. The (12,10) ran in reasonable time, but you could probably get a better energy from rotating your orbitals. The (12,14) is going to take a while (even with good orbitals) because the number of CSFs you're generating is huge, and this will require a lot of memory, plus your comment about the identity of the orbitals.

@AndrewR. Also, feel free to ping me in chat if you have any questions.

thank you for the help again. I've taken your comment into consideration and have a few (hopefully short) follow-up questions. I'm new to Stack Exchange so don't know how to ping you in chat. I created a room, but I'm not sure I was supposed to do that...

It's fine; to ping me in chat, go here (http://chat.stackexchange.com/rooms/3229/the-periodic-table), type '@pentavalentcarbon' and then some text, and I'll get a notification. I can't find the room you created on the chat page; if you'd rather chat there, you'll have to send me a link.

@pentavalentcarbon Thank you for the detailed answer. Could you please provide references for the equations which calculate the number of determinants and CSFs? The link that you have already provided in the answer is not working anymore. More specifically, a textbook or paper would be really helpful.

It seems increasing the active space helped make the CASSCF calculation converge. I've had success with increasing the number of active orbitals from 8 to 10. It only took an hour or so.


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Preference of locants in organic nomenclature
Why can't we write the compound 5-ethyl-4-methyloctane as 4-ethyl-5-methyloctane?

Are you sure that you don’t want to ask the other way round? “4-ethyl-5-methyloctane” is a correct systematic name while “5-ethyl-4-methyloctane” is not in accordance with IUPAC nomenclature.

As pointed out by Loong, the name you prefer is the correct name: 4-ethyl-5-methyloctane.
Generally we try to do the following:

Assign the smallest locant possible to the carbon bearing the highest priority functional gorup.
Assign locants to generate the lowest possible sum of the locants.

Often it is straightforward. In this case, both assignments of generate locants of 4 and 5, which sum to 9. Both place an alkyl substituent at the lower locant.
In this case, the locants are assigned so that the alphabetically first of the two substituents of equivalent priorty receives the smallest locant.

Correct - 4-ethyl-5-methyloxctane: ethyl is alphabetically before methyl and should get a smaller locant

Incorrent - 5-ethyl-4-methyloctane: ethyl is alphabetically before methyl and should get a smaller locant


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Does solute add to volume of solution?
I have to calculate the concentration of sucrose in water just by knowing their masses (and their temperature). Now, if the volume doesn't change when the sugar is being dissolved, I can do that quite simply over density. But I am not sure if it changes or not. Does it? And is that generally true (e.g. for electrolytes)?

That's completely incorrect.
When NaCl is dissolved in water there is a 2.5% reduction in the volume, as the break down of Hydrogen bonds to dissolve the salt results in water molecules being able to exist closer to Each Other reducing the volume.

@john a conflicting answer has appeared so do you have a reference?  Thanks!

Just open the Handbook of Chemistry and Physics, in the page "Specific Gravity of Sodium Chloride". Here everybody can read that by dissolving 315.5 g NaCl into 1,00 Liter water, one obtains a saturated salt solution having a volume of 1.115 Liter, and a density of 1.18 kg/L

Solute almost always changes the volume of final solution. The change (volume decrease or volume increase of final solution) of volume depends of solvent and solute chemicophysical properties, e.g. the full charge of each or the polar properties of each, or their molecular orbital structure and electron bond donors or acceptors. Two very non-interactive materials would not even mix, e.g. oil and water (H2O), though tiny tiny amounts of oil will "dissolve" in water as micelles, and tiny tiny amount of water will infiltrate oil, humidity of oil, just like humidity of breathable air.
If solute is not charged and does not have hydrogen-bonding-ready open orbitals or lone pairs of electrons, then the volume increases, both solute and solvent usually do not mix well.
If the solute is a ionic compound (e.g. salt, NaCl), then up to a certain amount of table salt the volume will decrease, the density will increase and reach the maximal amount of salt to reach the minimum amount of solvent, H2O, at a specific temperature and pressure.
If you mix a polar compound ready to accept or donate hydrogen bonds to H2O, then you get a similar picture as for NaCl in H2O, but not so drastic http://butane.chem.uiuc.edu/pshapley/genchem1/l21/vol.png , http://butane.chem.uiuc.edu/pshapley/genchem1/l21/1.html.

This can be seen from a very simple experiment.
Take water in a measuring cylinder and note the volume. Add 5 grams of NaCl and dissolve. Note the new volume.
You will observe that the final volume is greater than the initial volume. Hence the NaCl HAS added to the volume of the solution.
That being said, we generally make the assumption that the mass of the solute does not significantly alter the volume of the solution which is a reasonable approximation in the case of dilute solutions.
Your treatment will depend on how rigorous you want to be.

The calculation of concentration of sucrose in water is a comparison of masses and does not involve volume (or density) unless the volume of the final solution is specified, as in molarity or normality, both of which involve moles or equivalents of solute in a given volume of solution (1 L).
Combination of units like 20% w/v usually means a weight in grams over a volume in mL. This is done for convenience and does not usually give solutions which are used for further reactions or calculations. Such solutions would be used for washing, rinsing, etc. Sucrose solutions might be used for hummingbird feeding.

As you've described the situation, the concentration may be expressed as mmolality (or molality), (i.e. moles/kg) without making any assumptions likely to introduce errors.
Unless you are able to measure the density (as you noted), expressing concentration as moles per volume will err to the extent that the solution departs from ideality.
Binary solutions have volumes larger, smaller or the same as the sum of the volumes of the individual components. The differences - excess molar volumes - are negative when solutions contract (e.g. 2-propanol + water), positive when they expand.

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Quadrupole moment of a molecule
What is a quadrupole moment of a molecule and how does it arise? How is it measured for a particular molecule?
I've read the Wikipedia article on quadrupoles and understand that it has to do with the distribution of charge, but I'd like a more chemistry-oriented explanation.

Related: Can nonpolar molecules exhibit dipole-dipole forces? and Is the triiodide ion polar?

Chemically(?), quadruple moment is a quantity which describes the finer (fine than dipole) details of the charge distribution for a molecule

What is a quadrupole moment of a molecule and how does it arise?

This explanation is geared at someone with one year of chemistry. It captures the gist but is not rigorous:
To test for monopoles, dipoles or quadrupoles in a molecule or ion, assign charges or partial charges to each atoms. For partial charges, δ+ and δ- is fine (if you want estimated values for partial charges of simple molecules, use a tool like MolCalc).
To test for monopoles (ions), add up all the charges. If they are non-zero, it's an ion.
If the net charge is zero, test for presence of a dipole: If you can draw a line through the molecule such that more negative partial charges are on one side of the line and more positive partial charges are on the other side of the line, it has a dipole moment. An example for water is below.

If it does not have a dipole, you can check for a quadrupole moment: If you can draw an "X" through the molecule such that more negative partial charges are in opposite quadrants and more positive partial charges are in the other two quadrants, it has a quadrupole moment.

The animated GIF shows the model of carbon dioxide rotated in the presence of such an "X" (or "+" in this orientation if you will). In one orientation (where it is diagonal), the negative partial charges of oxygen are in opposite quadrants, while the positive partial charge is in the center (neither quadrant), showing that it has a quadrupole moment.
For a definition of the quadrupole moment (i.e. how to calculate it from a given charge distribution), take a look at this section of the wikipedia article on multipole expansions, which I found when reading the answer to Can nonpolar molecules exhibit dipole-dipole forces?.

How is it measured for a particular molecule?

The accepted answer for this question cites a paper describing one method.

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text:
How get a dilution at 150g/l of H2SO4 at 96%?
I've never done any chemistry before and I would like to get 5 liters of H2SO4 at a concentration of $150 [g/l]$.
One liter of $H_2SO_4$ at $96\% $W/W has a density of $1.84 [g/ml]$.
It means my bottle should weight $1840~[g]$ where I have $96/100\cdot1840=1766~[g]$ of pure acid.
5 liters of acid at $150~[g/l]$ means $5~[l]\cdot 150~[g/l] = 750~[g]$
The amount of acid at 96% is:
$$\frac{1840~[g]}{1766~[g]}\cdot 750~[g] = 781 [g] $$
Converted in liters I must have
$$\frac{781 [g]}{1840 [g/l]} = 424~[ml]$$
Eventually I need to poor 424 [ml] of $H_2SO_4$ at 96% into $ 5 [l] - 474 [ml] = 4.51~[l]$ of distilled water. Is that right?
I am wondering of the amount of heat will I get during this dilution. Is that ok to gently poor it (about 1 minute) into distilled water contained into a plastic jerrycan?

First of all: Never add water to sulphuric acid. The heat of the reaction may lead to an explosion. Use safety glasses, safety gloves, and wear protective clothing.
The densities at 20°C are

Water: $\rho_{water} = 0.9982~\mathrm{g~cm^{-3}}$
15% $\ce{H2SO4}$: $\rho_1 = 1.10~\mathrm{g~cm^{-3}}$
96% $\ce{H2SO4}$: $\rho_2 = 1.84~\mathrm{g~cm^{-3}}$

To get $5000~\mathrm{cm^3}$ 15% $\ce{H2SO4}$ with a mass of
$m = \rho_1 \cdot V = 1.10~\mathrm{g~cm^{-3}} \cdot 5000~\mathrm{cm^3} = 5500~\mathrm{g}$
you need
$m = 0.15 \cdot 5500~\mathrm{g} = 825~\mathrm{g}$ pure $\ce{H2SO4}$
These are contained in
$m = \frac{825~\mathrm{g}}{0.96} = 859.4~\mathrm{g}$ of 96% $\ce{H2SO4}$.
i.e.
$V = \frac{859.4~\mathrm{g}}{1.84~\mathrm{g~cm^{-3}}} = 467~\mathrm{cm^3}$ of 96% $\ce{H2SO4}$.
Finally, the amount of water you need to pour the acid in is
$m = 5500~\mathrm{g} - 859.4~\mathrm{g} = 4640.6~\mathrm{g}$ water, i.e. $4649.0~\mathrm{cm^3}$
In short: gently pour, while stirring, $467~\mathrm{cm^3}$ of 96% $\ce{H2SO4}$ into $4649~\mathrm{cm^3}$ distilled water to get $5000~\mathrm{cm^3}$ 15% $\ce{H2SO4}$. Break the procedure if the temperature gets too high.
Disclaimer: I do not take any responsibility for the correctness of the above calculation.

How did you get $\rho_1$?

I googled for it https://www.google.de/search?q=density+of+sulfuric+acid+at+20%C2%B0C

I actually made a mistake in calculating the amount of water by neglecting the water in the 96% acid. Corrected it from 4683 ccm to 4649 ccm.


text:
Suggestion for a Synthesis of Paracetamol — plausible?
I have seen a number of different plausible syntheses for paracetamol but they suffer from problems such poor regioselectivity of aromatic substitution (usually when nitrating phenol getting a mixture of undesired ortho and desired para). However, is the following not a plausible synthesis?

Take aniline (I believe it's commercially available but can be synthesised from mononitration then reduction).
React with acetic anhydride to make the amide — still ortho-, para-directing and mildly activating and the directing effects are not acid sensitive as in aniline.
Do a Friedel–Crafts acylation with the an acyl chloride.
React with hydrogen peroxide and sodium hydroxide to get paracetamol via a Dakin reaction.

Of course the acylation will suffer a little with regioselectivity issues but I feel like it will suffer less as the amide group is larger than a hydroxyl so favours para on steric grounds more strongly and is thus a better solution than nitration of phenol and subsequent reduction.
Nitration of phenol then reduction and reaction with acetic anhydride is the most common hit on google so I assume it's the most common method. But what's wrong with what I have suggested?

Amide group would react.

It certainly wouldn't react preferentially. I'm not even sure it would react much at all would it? Hydrolysis of amides requires quite forcing conditions. Are the conditions required to force a dakin reaction quite harsh?

Are you asking whether this synthesis would work at all or would it be viable on production scale?

I'd be keen to know about both

Your route sounds possible, yes. (Aniline is commercially available.) There would very likely be side reactions of the amide under Baeyer-Villiger conditions but nothing exceptionally out of control—except you lose a little bit of yield.
There seems to have been a little race concerning the shortest and most environmentally favourable route to paracetamol. To the best of my knowledge, and assuming that their NMR spectra are unaltered and directly from the reaction published, Joncour et al. have the shortest route in 1 (one) step starting from (also commercially available) hydroquinone and utilising ammonium acetate in acetic acid.[1]
However, for the industry to decide to use a non-established method, that new method must offer a significant advantage over what is presently used. ‘Significant advantage’ typically means much safer or much cheaper. The industrial paracetamol synthesis as employed by Hoechst–Celanese starts off from phenol (extremely inexpesive), does a Friedel–Crafts type acylation with acetic anhydride and $\ce{HF}$ (cheap materials), transforms the ketone into a ketoxime (cheap $\ce{NH2OH}$) and finally uses catalytic acids ($\ce{CF3COOH, SOCl2}$; the latter being rather cheap) to induce a Beckmann rearrangement.
Viewed zealously, this is technically already a more optimised synthesis than the one you propose—it is a step shorter and uses even cheaper reagents. There is probably not much room left for optimisation along these traditional lines.
Referene:
[1]: R. Joncour, N. Duguet, E. Métay, A. Ferreira, M. Lemaire, Green Chem. 2014, 16, 2997. DOI: 10.1039/C4GC00166D. The openly accessable supplementary information with experimental details and the NMR spectrum is available on the same page.

The synthesis proposed would not seem to offer much advantage as the question around selectivity is a matter of conjecture.  There are two major industrial syntheses of paracetamol,  mainly occurring in India and China. One is the nitration of phenol because it is so cheap. The other is cheaper but has technology barriers to technically inferior producers. This is the Bamberger rearrangement.  Starting with nitrobenzene, partial hydrogenation is carried out to phenylhydroxylamine. Being able to do this without full hydrogenation to aniline is the technical differentiator. Acid catalysed rearrangement of phenylhydroxylamine yields para-aminophenol. The hydrogenation and the rearrangement can be done in one step. The amino group then reacts preferentially to form the acetaminophen. Mallinckrodt use this process. This is extremely atom efficient as the most amount of waste is in the final step. On paper, the first two steps have no reagent byproducts.


text:
In Hückel's rule, can n be any integer?
Hückel's rule says that any planar compound with a ring of conjugated p orbitals with $4n+2$ electrons is aromatic. Here, can $n$ be any integer, or does $n$ have to be related to the number of p orbitals in the conjugated π system (as shown in the Frost circle)? For example, in benzene, there are 6 electrons in the pi system ($n=1$), which fill all 3 bonding orbitals. Thus, benzene is very stable. But say we remove 4 electrons so that there's only 2 electrons in the π system. Here, $n=0$, so is this cation considered aromatic?
On the other hand, if we add 4 electrons to benzene's π system, it has 10 electrons ($n=2$). Is this anion aromatic? It doesn't seem very stable to me, since it has now filled 2 antibonding orbitals as well.
In general, it is a requirement that the $4n+2$ p electrons exactly fill all the bonding orbitals of the pi system for the molecule to be aromatic?

Indeed too much charge disallows aromaticity, that's a very good question.

Agreed.  Even $\ce{C_8H_8^{2-}}$ is pushing it.  Some calculations suggest it's unstable as a free ion.  The thing we make when we combine potassium and COT in liquid ammonia is surely solvated by the ammonia and ion-paired with the $\ce{K^+}$.

As far as I can tell, n can be zero or any positive integer.

If the benzene tetraanion and tetracation are more stable than say the hexatriene tetraanion and tetracation, and I suspect they are, then the benzene ions could be said to be "aromatic".

This question is pointless, as the tetraionic compounds are completely hypothetic. Phantastic chemistry, that. ;-) Or putting it another way: It's Hückels rule, and you're bending it far beyond the point of breakage.

Or maybe not ... http://onlinelibrary.wiley.com/doi/10.1002/anie.197902291/abstract

No, you do not have to have as many π electrons as p-orbitals or any type of direct relationship. As long as you have a continuous, circular arrangement of p-orbitals and an electron count $4n+2$ (for monocycles), you can expect your molecule to be aromatic.

Pyrrole, furan and related molecules are aromatic. They have 6 π electrons like benzene, but only five p-orbitals.
The cyclobutadienyl dianion has been synthesised as a dilithium salt. The authors claim evidence for aromaticity.[1]
The dianion of cyclooctatetraene has also been synthesised and examined by ESR as early as the 1960’s. The authors state:

The assumption that both the dianion and the radical-anion are planar […] is consistend with [their equation] within experimental error.

Thus it can be assumed that $\ce{cot^2-}$ is also an aromatic system.[2]
Disubstituted dications of cyclooctatetraene have been synthesised a decade later by Olah et al. The published proton NMR data implies that the protons are more deshielded than a simple triene moiety would imply. The summary the authors give states the following:

The $\ce{^{1}H}$ and $\ce{^{13}C}$ NMR spectroscopic characterization of the reported new cyclooctatetraene dications indicates these ions possess those characteristics expected of a $\ce{C8}$ 6 π aromatic system and hence further extends the evidence for the great predictive utility of the $4n+2$ Hückel rule.

$\ce{cot^2+}$ is thus also an aromatic system.[3]

These last two bullet points must be the ones you are looking for. The cyclooctatetraene dianion, a cyclic $\ce{C8}$ species with 10 π electrons ($n=2$) is aromatic, as is also the cyclooctatetraene dication, the same $\ce{C8}$ species albeit with 6 π electrons ($n=1$).
For most systems, aromaticity is either already given (as they have e.g. six π electrons in their neutral state; compare benzene etc.) or they are only one electron away (cyclopentadienyl, cycloheptatrienyl or tropylium). The way these molecules obtain aromaticity is relatively fixed since greater charges destabilise molecules.
Only for those systems starting with $4n$ π electrons such as cyclooctatetraene, which are antiaromatic by the Hückel definition, can there be a choice. Both the dication and the dianion are generally unfavourable, but they are similarly unfavourable thus the system can be pushed into either direction.
It is highly unlikely that a tetraanion or tetracation of benzene exist outside of a $4~\mathrm{K}$ matrix, much like it is very unlikely for a cyclopentadienyl-trication or a cycloheptatrienyl-trianion to exist outside of similar conditions due to their great charge density.

References:
[1]: A. Sekiguchi, T. Matsuo, H. Watanabe, J. Am. Chem. Soc. 2000, 122, 5652. DOI: 10.1021/ja0004175.
[2]: H. L. Strauss, T. J. Katz, G. K. Fraenkel, J. Am. Chem. Soc. 1963, 85, 2360. DOI: 10.1021/ja00899a004.
[3]: G. A. Olah, J. S. Staral, G. Liang, L. A. Paquette, W. P. Melega, M. J. Carmody J. Am. Chem. Soc. 1977, 99, 3349. DOI: 10.1021/ja00452a027.

But all those systems have one specific value of $n$ for the given ring system.  The OP seems to ask whether a given ring system can show aromaticity for multiple values of $n$.  I would guess probably not unless you have a very large ring.

@OscarLanzi Agreed, I was trying to ask if a given ring system is aromatic for every value of $n$. The examples of pyrrole and furan have all the bonding MOs filled, so of course they're stable. Cyclobutadienyl and cyclooctatetraenyl dianions have all bonding and nonbonding MOs filled, so that would still be pretty stable. For more exotic values of $n$, I suppose it would still be aromatic because as ron mentioned in a comment, it's more stable than the isolated double bonds?

@swenger Okay, it seems I slightly misunderstood the question. But not to worry, I dug up the research paper claiming the cyclooctatetraenyl dication, too. Thus, $\ce{cot^2+}$ and $\ce{cot^2-}$ are both aromatic with $6$ and $10$ π electrons, respectively ($n = 1$ and $n=2$ in Hückel’s $4n+2$ rule).

Pyrrole, furan, etc. don't strictly obey Hückel's rule, since it was only derived for hydrocarbons, and they do not have the same symmetry restrictions.


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Chemical Biology - synthesizing/designing siRNA
I was reading an article on siRNA, and from what I read it sounded like we can edit genes using siRNA. First, did I even interpret this correctly? Second, I read that siRNA was being used to create drugs to help correct mutations. Does this mean that they could also give positive mutations to people? Finally, how does the structure of siRNA differ from that of normal RNA?

https://en.wikipedia.org/wiki/Small_interfering_RNA - I don't get your point

At this time, it it possible to completely chemically synthesize oligonucleotides (DNA/RNA) between 30-50 base pairs. siRNAs are a class of RNAs that, as the name suggests, bind to other RNAs in a sequence specific manner. Specifically, siRNAs preferentially target mRNAs, which blocks translation of those mRNAs into proteins. This regulation of gene expression has absolutely nothing to do with "creating genes for a human being".


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Do polarizability and electron delocalization have analogous effects on leaving groups?
With respect to bimolecular substitution and bimolecular elimination reactions, both electron delocalization and polarizability have an analogous effect on a species, making them good leaving groups.
For example I- is a good leaving group since its negative charge is spread out over a large area.
Also, the tosylate ion is a good leaving group because of electron delocalization where its electrons are spread out over a large area due to resonance stabilization.
However, the analogy seems to break down when considering their nucleophilicity in aprotic media, I-  is a good nucleophile and tosylate is not.
Is there a fundamental problem of the analogy between polarizability and electron delocalization?

Acetate is not a 'good' leaving group.  I think you have your nucleophilicity and leaving group ability backwards.

I replaced acetate with tosylate, but I think you're missing the point. Why does spreading charge out over a large space make resonance structures and polarizable species more stable -- but, increasing resonance decreases nucleophilicty, whereas increasing size and polarizability increases nucleophilicity? They're both spreading charge out over a large area.

TL;DR: Nucleophilicity is primarily a kinetic phenomena.  Leaving group ability is a thermodynamic phenomena.
The stabilization of an anion by delocalization of charge means that the resulting products will have lower energies than alternate situations. In accordance with the Hammond postulate (a few logical steps skipped), transition state barriers with low energy products will be smaller than those with higher energy products.
However, for a bimolecular reaction to occur, the molecule must have 'proper collision speed and orientation'.  Diffuse charges (over multiple atoms) generally make this more difficult.  This extra facet of the of the geometry of the transition state is why the equivalent stability argument does not apply to both leaving group ability and nucleophilicity.

Whoa, I was not expecting such a succinct answer. Thank you. Can you comment on bases, which are also said to be thermodynamic?

Explicitly not true, but a useful shortcut:  Basicity is the thermodynamic aspect of nucelophilicity.  Strong bases are higher in energy and thus, according to the Hammond postulate, transition state structures in bimolecular reactions will be close to the structure of the strong base, and hence activation energy barriers will be smaller.  I would encourage you to read the Wiki page on Hammond Postulate:  https://en.wikipedia.org/wiki/Hammond%27s_postulate

Nucleophilicity of a chemical species depends on the 'availability' of charge on the nucleophile. As you pointed out, acetates are bad nucleophiles. This is because the negative charge is delocalised over the 3 centres (O-C-O) and not readily available to any electrophile.
I- is a good nucleophile since it is spherical and polarisable, the charge being available.
The concepts of delocalisation and polarisability are independent from each other and any resemblance in a particular situation should be treated as coincidental and not read into too much.

While I appreciate your answer, it just seems like you're restating the premise. I understand that the two concepts are different, but they seem pretty analogous. Don't they both distribute negative charge over a large space, which contributes to stability? Is there a third concept that the "polarizability" and "electron delocalization" have in common, like "charge delocalization"?


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How to calculate the amount of substance of sodium that are in 10 mL of 30% sodium chloride?

How much of sodium is there in 10 mL of 30% sodium chloride?

$30\%$ of $\ce{10 ml}$ is $\pu{3 g}$ $\ce{NaCl}$. There is $\pu{3000 mg}$ of $\ce{NaCl}$ and would that be divided by the molecular weight of $\ce{NaCl}$ or of just $\ce{Na}$?
I divided $\pu{3000 mg}$ of $\ce{NaCl}$ by $\pu{58.44 mg/mol}$ and got $\pu{51.44 mol}$ $\ce{NaCl}$.
Since there is one mole of $\ce{Na}$ for every mole of $\ce{NaCl}$ there would be $\pu{51.44 mol}$ of \ce{Na}.
My teacher divided $\pu{3000 mg}$ (he didn't express what he had $\pu{3000 mg}$ of) by $\pu{22.99 mg/mol}$ of $\ce{Na}$ to get $\pu{130.5 mol}$ of $\ce{Na}$.  I have found other similar problems on the internet and feel I am correct.  My teachers only response was to read the question.  If I am wrong please help me understand why.

You are on the right track and with a few assumptions I'd say your calculation is correct. The one of your teacher is not.
Let's back up a moment and understand what is going on.
First of all we should make sure, that we are asking the right question. As you have given it, it seems incomplete. It should probably rather be:

What amount of substance of sodium (ions) can be found in 10 mL of a 30% aqueous solution of sodium chloride?

We have been given a mass fraction, which we need to convert to the amount of substance concentration first. The general formula for this is
$$w_i = \frac{\rho_i}{\rho} = \frac{c_i M_i}{\rho},$$
where $\rho$ is the density of the solution, $\rho_i$ is the mass concentration of component $i$, $M_i$ is the molar mass of component $i$, and $c_i$ is the amount of substance concentration of $i$.
In your calculation you made the assumption, that the density of a 30% sodium chloride solution is $\rho = 1~\mathrm{g/mL}$, which is not correct. You can look up a couple of values for brine solutions on wikipedia.
For the purpose of this example, let's assume $\rho(\ce{NaCl~30\%}) \approx 1.2~\mathrm{g/mL}$.
The molar mass of sodium chloride is $M(\ce{NaCl}) = 58.4~\mathrm{g/mol}$. Be careful with the units, as you used the incorrect ones.
The amount concentration is given through the formula
$$c_i = \frac{n_i}{V},$$
where $n_i$ is the amount of substance (not number of moles!) and $V$ is the volume of the solution.
You correctly identified that there is a one-to-one correlation between sodium ions and sodium chloride as expressed by
$$\ce{NaCl (s) ->C[H2O] Na+ (aq) + Cl- (aq)}.$$
Therefore we can write
$$n(\ce{Na+}) = n(\ce{NaCl}).$$
Putting everything together we arrive at the following formula:
\begin{align}
n(\ce{Na+}) &= \frac{w_{\ce{NaCl}}\cdot \rho(\ce{NaCl~30\%})}{M(\ce{NaCl})} \cdot V\\
            &= \frac{0.3\cdot 1.2~\mathrm{g/mL}}{58.4~\mathrm{g/mol}}\cdot 10~\mathrm{mL}\\
            &= 0.061~\mathrm{mol}
\end{align}
Basically you thought process was correct under the assumption of the inaccurate density. And it was off by $10^{-3}$ because you (and your teacher) used the wrong units.


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How does H's ionization energy relate to its transition energy (Bohr's Model)?
I am currently reviewing some material about orbital energy levels. In my review book there is a short snipet that reads:

The IE of $\ce{H}$ from its ground state ($n=1$) is $1312\ \mathrm{kJ/mol}$. Because of the squaring of the principle energy level, the ionization of an $\ce{e-}$ in $\ce{H}$ from the $n=2$ level is $1/4$ of that value ($328\ \mathrm{kJ/mol}$). The transition energy from the $n=1$ level to $n=2$ level is the difference between the two values, $984\ \mathrm{kJ/mol}$.

I am confused about how exactly an atom's ionization energy relate to its transition energy. If IE is the energy required to pluck off the first e- from an atom and TE is the energy required to change between different orbitals, does this mean that an atom with greater IE will need to absorb a greater amount of energy to jump orbitals?
Lastly, is there a equation that directly ties in both ionization energy with transition energy?

The visible  transitions in the $\ce{H}$ atom were observed experimentally to follow the equation
$$\overline{\nu} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$
where $R_\ce{H}$ is the Rydberg constant and $n_1$ and $n_2$ are integer constants, $1, 2, 3, \ldots, \infty$. The Bohr model of the atom produces the same formulae and a way to calculate the Rydberg constant; in wavenumber units this is  $109677\,\mathrm{cm^{-1}}$.
The ionisation energy of the $\ce{H}$ atom occurs when the electron is in its $n_1 = 1$ level and has just enough energy to be removed removed from the atom, this takes it to $n_2 = \infty$ and this energy is equal to the Rydberg. This is also $13.6$ electron volts which is often quoted. If the electron is in another energy level you can calculate how much energy is now needed to ionise it.  Have a look in a phys chem text book (such as McQuarrie & Simon or Atkins) for a proper description of this.

Ionization energy of some arbitrary atom is not related whatsoever to its transition energy. Also, there are many different transitions (even in a hydrogen atom), each with its own specific energy. It is just that this particular transition of this particular atom happens to have the energy of 3/4 its ionization energy, one of the few things the Bohr model managed to get right. You can't generalize that to other atoms and other transitions.

How exactly an atom's ionization energy relate to its transition energy?
I think it is best to clear up what the terminology of the question is so that we can address the correct answer.
The ionization energy is the direct measure of the electric interactions between the positively charged  nucleus center of an atom, and the negatively charged electrons around it. Ionization energy is like that invisible force you can feel between two magnets close together. If you were to measure how much effort is required to split the magnet apart till they are effectively not interacting, that would be like measuring the ionization energy and an electron from an atom. One way to supply the energy to pull off (knock-out) an electron,  can be by shining light (in the case of the first ionization energy for a hydrogen atom, $1312\ \mathrm{kJ\ mol^{-1}}$, where $\mathrm{kJ\ mol^{-1}}$ is amount of energy required = $15.60\ \mathrm{eV}$ (another unit of energy that express the amount of electric attraction the negative electron feels toward positive nucleus), and in terms of focused energy carried by light (electromagnetic waves, generated by moving charges or current) that is $= 91.18\ \mathrm{nm}$ electromagnetic waves, which we call  X-ray light.)
What is a transition energy?
The intuitive way to think about a transition state is to try and recall a time when you were in danger or had to make snap decision that brief instant before hell breaks lose is like a transition state. Something you have to build up eneough energy to get over. In the Bohr model, the transition state is when an electron moves from one orbital to another. The HUGE thing about this model that made it work is that Bohr assumed the energy between transition states in atoms (i.e the differences between the energies measured for each electron in an atom and the next) is quantized, meaning that it take it can be certain values only. Like having a light switch with 4 positions. It can have different levels but switches can just hover in between, either one or off .The transition energy is the energy required to move electrons between transition state. It is like when water is being heated. You are delivering energy into it by heating and nothing seems to happen until it reaches $100\mathrm{ºC}$ and then it boils.
Bohr related the model for the hydrogen atom and thus approximation for inonization energies for electron in hydrogen energy levels is give by.
Bohr model is given by $[ \Delta  E = \frac{m_ee^4}{8\varepsilon_0^2h^2} \lgroup\frac{1}{n_2} - \frac{1}{n_1} \rgroup$, where $n_i = 1,2,3,..]$
Where $n_1$ and $n_2$ are the quantum principle numbers between each quantized state. ie. $n_1 = 1$, means level 1, $n_2 = 2$, level 2 etc..
Even thought we know Bohrs model is incorrect, the quantization of energy as boundaries for electrons  is the basis of all modern models. It is not atoms that absorb energy in the case of ionization energy,
Misconception is that atoms absorb energy when its actually the electrons of nucleus particles that do. In the case of ionization energy, it its the electrons themselves that are being energized to the point of boiling off. Thus the valence electrons, which the ones furthest from the nucleus with lowest ionization energies, have lower ionization energies than core electrons (those closer to nucleus)

McQuarrie DA & Simon JD (2015) Physical Chemistry: A molecular approach . Viva Books


text:
Low solubility of propanal compared to propanol, propanone, propanoic acid and methyl ethanoate
I found on the internet that propanol, propanone, propanoic acid and methyl ethanoate where all completely soluble in water at room temperature. (23 degrees Celsius) But I found that propanal was, comparatively, was not completely soluble in water (Wikipedia said it was 20g/100ml). I was wondering why it was not completely soluble like all the others as small aldehydes can interact with hydrogen bonds and dipole-dipole forces with water (like all the others). Is there some other intermolecular force that influences solubility that I'm missing?

They're all on the verge of miscibility, methyl ethanoate (methyl acetate) is in fact not miscible according to wikipedia (soluble to 25%).
The acid is the most polar of all the C3 compounds you name. Butyric acid (C4) is also miscible with water, but valeric acid (c5) already not.
n-butanol is already rather badly soluble (8g in 100ml) in water. Same with butanone, etc.
Compared to the ketone, the aldehyde has a longer unpolar part, that is overall less influenced by the polar carbonyl bond. Both are clearly, i think, the least polar of the group, and so the aldehyde just drops beneath the line.
No mystery here. ;-)


text:
Why acetaldehyde and acetone have almost equal water solubility?
Given that boiling point of acetaldehyde and acetone are different due to H-bonding, why doesn't H-bond affect the solubility of these compounds in water?

They are both completely miscible with water.


text:
Are there stronger oxidizing agents than fluorine gas?
Are there stronger oxidizing agents than fluorine gas, so it could oxidize fluoride to fluorine?
Also, in case of oxygen, fluorine gas can oxidize oxygen gas to the exotic dioxygenyl ion. Can anything oxidize fluorine gas to $\ce{F+}$ ion?

Fluorine gas can't oxidize oxygen gas to dioxygenyl ion; $\ce{PtF6}$ can.

Fluorine gas by itself can't. Throw BF3 into the mix and it does.

Where do alpha particles fit in this scheme?  I think He++ ions would be pretty potent oxidizers too.

There are stronger oxidization agents than fluorine gas. I can name two in particular. But I don't think either would actually oxidize fluorine gas in it's pure state, for obvious reasons that you'll see shortly.
Dioxygen Difluoride. Affectionately named 'FOOF'. This is an extremely, extremely potent oxidizer. It also lights ice on fire. In fact, it lights essentially everything on fire or blows up on contact with said things.
Chlorine Trifluoride. Arguably, worse. This gas burns through practically everything, except for a few specific structural metals, but only because these form a very thin passivization layer. If this scratches, the tank burns. A notable incident with this gas, a 900kg spill of it burnt through 30cm of concrete, then 90cm of gravel underneath that concrete. It burnt over a meter into the ground.
As a foreword: I have no idea why you'd want to get within three miles of these things. Please don't light yourself on fire. It's not fun. (Trust me).

How are you quantifying stronger?

More readily performs a redox reaction.

Does FOOF or ClF3 has a higher standard electrode potential than fluorine gas? If yes, how much?

@Nissa They would dissolve your electrodes, so the question is sort of moot.

Note that many standard redox potentials are just formal, not really measured due kinetic or other reasons, but derived from the reaction $\Delta G_\mathrm{r} = nFE^{\circ}$

@Poutnik Yeah, but then again, FOOF can't be used anywhere close to standard conditions ;)

@Mithoron Well, it does not exclude each other.;-) The Wikipedia Standard_electrode_potential_(data_page) with "reasonable" compounds ends with KrF2.

@Poutnik Can help but wonder how they imagine aqueous KrF2, but ech, whatever.

As you say......

@Mithoron I have seen once tabelated potential for C(s)/CH4.

I'm not sure, but I think krypton difluoride is the strongest oxidizing agent I know of that's fairly stable at normal temperatures and pressures. (It decomposes at ~10% per hour at ~25°C, but is stable at dry ice temperature or below. Wikipedia claims it is "the most powerful known oxidizing agent known" and "more powerful even than elemental fluorine due to the even lower bond energy of Kr–F compared to F–F, with a redox potential of +3.5 V for the KrF2/Kr couple".) It is a colorless crystalline solid at room temperature that reacts with xenon to directly form XeF6 without any extra input of energy, and reacts with gold to form the hexafluoroaurate salt KrFAuF6, with gold in the +5 oxidation state (both releasing krypton gas). It can also be used to oxidize ClF5 to the ClF6+ cation. I don't know how it reacts with N2, Cl2, O2, or O3, but I feel like it must oxidize them, since things that oxidize xenon at all (such as platinum hexafluoride) can generally oxidize O2 to O2+ (dioxygenium). (I don't think O3+ is stable, but I imagine ozone could just be broken down into O2+ ions. In any case, it's probably easier to oxidize than XeF4, in which case KrF2 should be able to oxidize it.) Obviously, atomic fluorine would probably be stronger, but at normal pressures and temperatures it would spontaneously react with itself to form F2. (Atomic F and O can exist in the interstellar medium, though, with O being common even in the Earth's thermosphere.) I think that putting such an inert gas between the two fluorine atoms acts as kind of an approximation of atomic fluorine.
On the subject of ridiculously unstable things, positively charged things naturally attract electrons, so perhaps they are the real strongest oxidizing agents. (The only problem is that the compounds they form might break down as soon as the charge is balanced.) One might suppose that the F9+ ion (a bare fluorine nucleus) would be the best oxidizer, and I imagine it would be much better than F2, but if you're just interested in how good it is at stealing electrons from things in the short term, then charge matters much more than the electronegativity of the neutral atom, so a bigger nucleus would probably be a better "oxidizing agent". Thus, I might propose bare califonium or mendelevium nuclei as the best fairly stable particles for stealing electrons. However, these elements are, in fact, metals, so a lot of this work would be undone chemically whenever the charge went back to normal, which would happen quickly as it would basically just steal electrons from a as large an area as possible, so that positive charge would be spread out over a lot of only slightly positive, not-very electronegative, things, even if the electrons taken away to charge the ions were shot into deep space or something.
That being said, these considerations do shed some light on why electricity is often used in making strong oxidizing agents like krypton difluoride, as well as ozone, perchlorates, xenon fluorides, and even O2 by electrolysis of water. (It also makes strong reducing agents like alkali metals.) An F2+ ion is a much better oxidizing agent than neutral F2, and a Kr- ion is probably a much better reducing agent than caesium (it's certainly better than neutral rubidium, which has the same electron configuration). When you use electricity or UV light or extreme heat on a mixture of F2 and Kr you can get these ions and even more extreme ones, and that is how you can make KrF2. (Actually, the hot-wire synthesis method produces short-lived atomic F, which is also a better oxidizing agent than F2, as I mentioned.) The F2 doesn't have to oxidize the krypton: The F2+ or atomic F (or F+ or F3+ or F4+ or whatever*) does, and the Kr-F bonds just have to not immediately fall apart once the charge rebalances to make neutral molecules.
*My guess is that F2+ ions would sometimes bond to neutral F2 molecules to form ions like F3+ or F4+ (F3+ being isoelctronic with OF2), which is why I included them. So that might be an answer for what can oxidize fluorine.

Note: Kr- is probably irrelevant. I'm not sure if it can even exist, since noble gasses are thought to negative electron affinity, which means that it releases energy for Kr- to decay into neutral Kr and a free electron (e-). In any case, F2 or F oxidizing it would probably just result in Kr and F2(-) or F- and no chemical bond.

:/ You probably had better kept to redox potential - there's is estimated one for KrF2 when I've never seen it for ClF3 or FOOF.

Fluoride was inadvertently oxidized to fluorine in an attempt to oxidize oxygen to the +4 oxidation state.
Crawford and Klapötke[1] showed through quantum mechanical calculations that the $\ce{OF3^+}$ ion is vibrationally stable and potentially accessible. They therefore attempted the reaction
$\ce{OF2 +F2 +AsF5 -> OF3^+ +AsF6^-}$
but it did not work; instead they observed
$\ce{2OF2 +AsF5 -> O2^+ +AsF6^- +(3/2)F2}.$
Oxygen difluoride would not ordinarily oxidize fluoride to fluorine. Presumably it combines with the Lewis acid $\ce{AsF5}$ to generate a stronger oxidant.
Reference

Margaret-Jane Crawford, Thomas M. Klapötke (1999),
"The trifluorooxonium cation, OF3+", Journal of Fluorine Chemistry, 99(2), 151-156,
ISSN 0022-1139,
https://doi.org/10.1016/S0022-1139(99)00139-6.


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Can a more electropositive metal replaced by a less electropositive one?
I read from this reference that rubidium is extracted by reduction of its chloride with calcium or sodium. How is it possible?. Their electronegativities are much higher than rubidium. I think rubidium is far more reactive than either and can not be replaced from its salt by any other metal except cesium. While I have also read that cesium itself is replaced from its chloride by magnesium refer to this. How can a less electropositive metal replace a more electropositive metal from its salt? Are there some factors besides electronegativity, reduction potential and other reaction skills?

both reactions require extrernal source of heat. The process is actually an equilibrium, but at elevated temperatures rubidium and cesium relatively easily evaporate, which is used to shift the equilibrium.

But @permeakra how an equilibrium is established as extreme reactivity of such elements should not permit equilibrium at all. I think so

still, many processes are equilibriums. These ones included.

I tried to understand that @permeakra but I didn't even after much reading. Can you please explain it alongwith examples? I am not very good at chemistry.

And if equilibrium can be established between two metals wwith a considerable electronegativity difference then what's the use of electrolysis? Because almost every metal can be extracted by such equilibriums

While indeed equilibrium may be established for virtually all metals, it is still strongly shifted to more electronegative one. To shift it to less electronegative one, one have to be able to remove more electronegative one from the mixture. In the particular cases of cesium and rubidium it is achieved by vacuum distillation of the reaction mixture, because said metals vaporise relatively easily. That's not the case for most other metals.

@permeakra perhaps you meant to remove more electropositive?

yes, indeed more electropositive. To adress possible issue: quantum mechanincs states that every process is reversible, thus theoretically speking every reaction is reversible and it is possible to shift equilibrium in desirable direction. However, fragility of molecules involved or slow reaction rate often makes it impractical or plain impossible in realistic conditions.


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Acidic Strength of Nitrobenzoic Acid
The Correct acidic strength of the following :

But according to me pK value of 2-nitro benzoic acid should be more than 4-nitro benzoic acid because NO2 might have H-bonding  with COOH as NO2 have it in ortho-nitro phenol.
Correct me if i'm wrong and please do tell me how to identify the presence of H-Bonding.
Thanks !

Related: Ortho-effect in substituted aromatic acids and bases

When it comes to acidity, stability of the counter-ion is one of the main considerations. The nitro-group acts to stabilize the negative charge of the resulting nitrobenzoate by pulling the negative charge onto itself.
In the case of 2-nitrobenzoic acid, it is much nearer and therefore able to more effectively 'remove' the negative charge from the carboxylate group. In the case of 4-nitrobenzoic acid, it's still able to do so but less effectively as it is further away.

The hydrogen bonding effect that you describe does exist in 2-nitrobenzoic acid but this does not act to make it less acidic. While one might reason that the hydrogen bonding "holds the proton in place" and therefore makes it harder to leave, the presence of the hydrogen also means that both the nitro and the carboxylate group are highly unlikely to be simultaneously coplanar with the benzene ring (it's not obvious from the image because there's no bond between O and H). The energetics in this case are likely to favor the release of H+ in exchange for a coplanar configuration.


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Does 12 amu really equal the mass of one carbon-12 atom?
As I was studying chemistry, I encountered atomic mass units (amu) and based on some calculations, I realized that some numbers don't work out.
Apparently, 12 amu is the mass of a single carbon-12 atom. However, the mass of the atom is (mainly) composed of six protons and six neutrons. Since both protons and neutrons weigh a tad more than 1 amu, how is this possible?
Furthermore, if it were true, that would convey that 1 amu is more than $1.661 \times 10^{-24}$ grams. It doesn't work out even without considering electrons. Did we get the numbers wrong, or the mass of an atom of C-12 more than 12 amu?

When protons and neutrons are bound together by the strong nuclear force, some energy is released -- and, since $E=mc^2$, a little mass goes with it. https://en.wikipedia.org/wiki/Nuclear_binding_energy

Also note that the unified atomic mass unit $\mathrm u$ should not be confused with the obsolete atomic mass unit $\mathrm{amu}$ (which is deprecated since 1961). The atomic mass of C-12 is about $12.0005\ \mathrm{amu}$ (as used in chemistry until 1961) or $12.0000\ \mathrm u$ (by definition, since 1961). However, this small difference is not related to your actual question.


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Does dissociation happen in gases?
I understand that dissociation is when something dissolve. I heard this mostly only happen in liquids. But now I heard it can also happen in gases. Is this true? If so, when and why does this happen, and does it only happen to certain gases? I tried to do research but didn't find too much information. The only real information I found was on Wikipedia and it was confusing.

What information? Also are you asking about homo- or heterolytic dissociation?

Dissociation is a common way for things to be dissolved in liquids. In gases, things may dissociate but the mechanism is not about interaction with other things in the gas but because some molecules are unstable and may fall apart into other molecules depending on pressure and temperature. A famous example is dinitrogen tetroxide which dissociates at higher temperatures into nitrogen dioxide.

The information is confusing precisely because your question itself is confusing. There are different kinds of dissociation, and different degrees thereof. Moreover, there are different areas of study, each with its own concepts of what is significant and what is not. For example, if one molecule in a trillion splits into ions, this is nothing for chemistry, but probably quite a something for atmospheric physics.
As for the question you supposedly intended to ask: no, electrolytic dissociation of dissolved substances never happens in gases the way it happens in water.
Also, welcome to Chem.SE.

You know that compounds can dissociate in solution, for example copper sulphate as a solid , will dissociate in water into copper Cu$^{2+}$  and SO$_4^{2-} $ ions. This occurs because the water can solvate these ions and this turns out to be of lower energy that remaining in the solid. In the gas phase you have to add energy to dissociate a molecule, usually by absorbing a visible or ultra-violet photon. This has to have enough energy to break a bond, for example iodine vapour can be be dissociated with green-blue light (  < approx 500 nm) but a molecule like ozone O$_3$ needs ultra-violet light to break it apart. Dissociation in the gas phase usually produces two radicals such e.g. I$_2$ = I. + I. .


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Sulfuric acid from elemental sulfur and hydrogen peroxide
Will this reaction proceed under standard conditions? I don't really care if it is quite slow, as long as it takes less than a month to finish:
$$\ce{S8(s) + 24 H2O2(aq) -> 8 H2SO4(aq) + 16 H2O(l)}$$
My rationale for this happening is that the elemental sulfur can be oxidized by hydrogen peroxide to form sulfur dioxide, which is further oxidized by hydrogen peroxide to form sulfur trioxide. This then reacts with water to form sulfuric acid.

I expect it to be thermodynamically feasible, but quite slow.

You will likedly find it faster to perform the first oxidation by burning the elemental sulphur in a little tinfoil boat to produce gaseous sulphur dioxide. then set up some aperatus sucking the gasses produced to bubble up through the hydrogen peroxide to produce sulphuric acid. The peroxide does not oxidize sulphur dioxide to sulphur trioxide because it forms sulphuric acid directly.
Does require some aperatus but will be much faster than waiting for the peroxide. Calculate the gibs free energy if you are not sure if a reaction will be spontaneous.

To quote sulfur dioxide's Wikipedia page directly: "Inhaling sulfur dioxide is associated with increased respiratory symptoms and disease, difficulty in breathing, and premature death."

@ringo Of course, dealing with corrosive or toxic gases should be performed in a fume hood or with some kind of respirator.

Actually, here is a modification of your procedure involving some advanced radical chemistry with sulfur (see Scheme 1 graphics and pages 6 to 7) and the use of a photocatalyst with sunlight and oxygen from air.
My scheme is to burn a mix of Aluminum foil and sulfur. To the reaction product add dilute H2O2 (usually acidic for stability) in excess to capture all generated H2S and create a suspension of sulfur in dilute H2O2. Decant to remove Al(OH)3, but keep the sulfur suspension. Add a photocatalyst (like TiO2 or perhaps even Selenium sulfide sourced from SelSun shampoo,  see this source). Apply sunlight and pump in air. Unclear if periodic dosing of dilute H2O2 is even required.
Logic: Per the source paper, the one-electron reduction potential of H2S to create $\ce{.HS}$ requires a relatively strong one-electron oxidants such as hydroxyl radical (which is sourced from the action of the photocatalyst acting on H2O2). Some expected reactions assuming the photocatalyst has produced a solvated electron:
$\ce{H2O2 + e-(aq) -> •OH + OH-}$
$\ce{•OH + H2S -> H2O + •HS}$ (Source Page 7, Eq (9) )
$\ce{•HS/•S- + O2 -> •SO2- (+ H+)}$ (Source Page 7, Eq (7) )
$\ce{•SO2- + O2 -> SO2 + •O2-}$ (Source Page 7, Eq (8) )
$\ce{H+ + •O2- -> •HO2}$  (pH > 5)
$\ce{•HO2 + •HO2 -> H2O2 + O2}$
Or:
$\ce{•HO2 + e-(aq)  -> HO2-}$
and even water can apparently supply a needed H+ to form H2O2.
So in essence, an in situ formation of •HO2/H2O2 employing a starting solution of H2O2, colloidal sulfur, oxygen and a photocatalyst to eventually convert the colloidal sulfur to H2SO4.
Just came across a 2006 reference ('Spectrum of ·HS and its reactions with oxygen in aqueous solution) that appears to mirror the science I suggested above involving sulfur. To quote:

The 266 nm laser photolysis formation and chemical behavior of ·HS in aqueous solution have been studied by laser photolysis-transient absorption technique. The 266 nm laser photolysis of H2O2 leads to the formation of ·OH, which reacts with H2S to produce ·HS, while solutions of HS- can also be directly photolyzed with 266 nm laser pulses to produce ·HS radical. The transient UV-Vis absorption spectrum of ·HS has been attributed to 220-300 nm with a maximum absorption at 220 nm and a shoulder at 250-270 nm. The chemical behavior of ·HS is very active in oxygen saturated aqueous solution: ·HS will first react with oxygen to form ·SO2-, which further reacts with another oxygen molecule with the production of SO 2 and ·O2-. In acid solution, ·O2- immediately protonates to form ·HO 2


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Reactivity Towards Electrophilic Substitution
How do you decided which compound is the most or less reactive towards electrophilic substitution? For example, if the compounds are benzene, ethyl benzene, 2-phenylpropane and 2-methyl-2-phenylpropane?

Most reactive towards electrophilic substitution means that the compound itself needs to be an electrophile (so that they don't combine), this means there needs to be a +I effect, right? But +I effect increases with an increase in the number of alkyl groups, so that would indicate that 2-methyl-2-phenylpropane would be the most reactive, but benzene is the answer for the question. Thanks in advance!

Welcome to Chemistry! This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

You misunderstood question or there's a mistake. Steric effect is also important but not that much

Then how would I go about answering the question? Which factors are considered first?

see this earlier answer


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Diffusion-controlled reactions, activation-controlled reactions and rate equation
I am confused as to where the rate equation can be applied and where not to be applied.
https://en.wikipedia.org/wiki/Rate_equation
In diffusion-controlled reactions from what I understand reaction rate is analogous to substrate concentration and the rate determining step is the supply of substrate.
https://en.wikipedia.org/wiki/Diffusion-controlled_reaction
But in activation-controlled reactions does the same apply?
Does reaction rate also depend on substrate concentration?

I don't understand your confusion. Article about rate clearly says only zero order reactions don't depend on concentration.

To add a little to what @geo has written. In all types of bimolecular reactions in the liquid phase, the reactants diffuse together; they have to, there is no other way for them to meet. The reaction is diffusion controlled if the activation barrier is so small that the first time the reactants meet, reaction is effectively instantaneous. This means that the barrier is only a few times average thermal energy (3RT/2) and that reaction occurs before the species can diffuse apart again.
In an activation controlled reaction the species repeatedly meet by diffusion but only occasionally$^*$ do they acquire enough energy from the surrounding solvent molecules that, at that instant, have enough energy to surmount the activation barrier.  Once they have reached  the barrier top the energy fluctuations that caused the barrier to be surmounted in the first place must cease after the transition state is crossed and products are then formed. (Recall that molecules in solution are continually jostling about and, from the Boltzmann (exponential) distribution, there is always a small chance that a molecule has, by random fluctuations, and for a short time, many times the average energy ).
$^*$By 'occasionally' I mean perhaps 1 in a million or even tens of millions of encounters.

"In a diffusion-controlled reaction, the rate is controlled by the ability of the reactants to migrate through the solvent and encounter each other. In an activation-controlled reaction, the rate is controlled by the ability of the reactants that have met each other to acquire enough energy to react."
http://www.chemistryexplained.com/Hy-Kr/Kinetics.html#ixzz4Cgzm7LnN


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Gel electrophoresis separates proteins on the basis of what property?
In SDS-PAGE for separation of proteins, SDS is an ionic detergent that coats the unfolded protein to mask its native charge and give at a uniform charge to length ratio. Since the length of the protein is proportional to its molecular weight, the charge to mass ratio of all the loaded proteins is the same.
To quote this site:

Since the SDS-coated proteins have the same charge to mass ratio there will be no differential migration based on charge.

Similarly in a DNA agarose gel electrophoresis, the charge to mass ratio remains constant. So what is the separation based upon?
Proteins having larger size will have more SDS bind to it and thus more negative charge than to the smaller proteins. While proteins of higher molecular weight also have lower mobility in the gel. So is this some kind of a trade off between the charge and the size? If charge to mass is constant, what is the property that separation is based on?
What does the above quoted statement mean? Why not differential migration based on charge? Doesn't mobility through gel depend on the quantity of charge?

It is probably a good thing to compare protein electrophoresis with DNA electrophoresis here. For DNA, you know that it has a phosphosugar backbone that comes with a negative charge at ambient $\mathrm{pH}$ per phosphate group. This means, that the charge-to-mass ratio, as the quote implies, is similar, no matter how long the DNA fragment is. (Not identical, because different bases have different masses and DNA typically has a seemingly-random base distribution.) Yet they still travel at different speeds.
This is because the gel is essentially a spider’s web for the DNA fragment. It somehow needs to find the pores of the gel and fit through. These pores have different, randomly distributed sizes and are not all aligned perfectly as you would assume for a crystal structure. Therefore, this process requies some movement perpendicular to the direction of charge. The larger a DNA fragment is, the harder it is to constantly change directions to fit through the misaligned (and potentially small) pores, so the speed of diffusion is slower. Hence, low-molecular weight DNA fragments wander far through the gel while high-molecular weight fragments remain closer to the starting point. Since size and molecular weight nicely correspond to the length in bases, markers are often labelled in terms of $\mathrm{kb}$ — kilo-base pairs.
Similar concepts apply to SDS protein electrophoresis. By denaturing the proteins with the detergent SDS (sodium dodecyl sulphate), they are linearised and equipped with a strong negative charge from the sulphate groups. Much as is the case for DNA, the number of dodecyl sulphate molecules is approximately proportional to the length of the protein in amino acids. And since they are linearised, i.e. their tertiary and quaterneray structures (and to a certain extent also their secondary structures) are destroyed, they very much behave like a DNA fragment when forced through a gel: the shorter they are the better they fit through and the faster they travel.
In the protein case, however, the markers are typically not labelled as a certain number of amino acids but as corresponding to a certain molecular weight (often given in $\mathrm{kDa}$). This is because amino acids vary greatly in size from the very small glycine to the very large tryptophane or arginine. The vicinity of one of those larger amino acids may allow for an extra dodecyl sulphate to be added when compared to the vicinity of small amino acids.

But why don’t larger proteins/DNA fragments that carry a greater negative charge move through the gel faster? Isn’t a greater force being applied onto them?
I’m glad that you asked me that. Yes, a greater force is applied. But, as you may remember from physics classes, $F=m\cdot a$, so consequently $a = \frac{F}{m}$. Therefore, the larger molecule may experience a stronger force, but due to the molecule’s higher mass, the overall acceleration is equivalent (again: not identical but similar). This allows the size-discriminating gel network to do the sorting by size.

Thanks a lot! The 'identical but similar' just opened my eyes!! Never thought of it :)


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What makes DNA/Protein in a gel run in straight lines?
In a gel electrophoresis, let's assume an SDS-PAGE, I understand that the sample is loaded on a cuboidal well and the electric field is straight downward. It is easy to assume that the sample is going to run straight in the gel. Like it does here.

But when I observed the apparatus in my lab, the anode was all throughout the base - looking something like this (cathode not shown here and the anode is insulated on the side - so effectively present at the base of the gel).

What makes the sample run in the gel only in a straight line? You may say that the forces cancel out on the either sides but what about the well that's at one end? It's being pulled downwards and also the other end.
So how does this work?

It's a very good question and I have no idea what the answer is. The only thing is, I am just not sure whether it belongs here... Fundamentally this is a physics question about a biological technique.

Should it be migrated? since cross posting isnt allowed right?

Please don’t add [tag:electrochemistry] to electrophoresis questions, since that tag is meant for chemical reactions occuring with the help of or to produce electricity.

What you are trying to achieve in the electrophoresis experiment is a homogenous electric field. If the electric field points downward homogenously, then the greatest force will be exercised in a downwards direction which will lead to a downwards movement of your charged sample.
While not being exactly equivalent, you can show a similar feature (homogenous field) with a horseshoe magnet. If you place a piece of paper or cardboard over the top of it and then sprinkle iron filings onto that, they should arrange themselves in a linear fashion between the magnet’s poles.

Oh right! Simple Gauss Law. Thanks :) High school physics. Now I understand why all that we learnt was important!

But still, wasn't that for an infinitely long charged wire? How do we manage to generate a homogenous field in this case?

@Polisetty It’s homogenous enough for the sake of the experiment ;)


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Reaction control on molecular level
Imagine I have a bottle with a random quantity of molecular hydrogen, and can apply exactly enough energy to break only one molecule of hydrogen. Is the reaction going to occur or the energy would be shared between other molecules in the bottle?

It depends on how you transfer this energy.

thank you , so you assume that if there is a way to optimise the transfer of the energie to only one molecule the reaction will occur

If you for example used single photon of appropriate energy.

Not necessarily. A molecule hit by that photon does not necessarily dissociate. The propability for that transition would need to be calculated, depending on the initial state (e.g. temperature of the gas).

The questioner asks that exactly enough energy is added to a molecule of hydrogen just to dissociate it and will energy be shared. The answer is both yes and no. It is better to take a step back and consider what happens when a photon is absorbed. In the case of hydrogen, dissociation must occur from an electronically excited state as it is a homo-nuclear diatomic and does not have a permanent or varying dipole. This means that the ground state cannot absorb a photon's energy from v=0,1,2 etc to any other vibrational level. (I'm assuming absorption into continuum just above v=$\infty$ is zero.) So excitation has to be with a UV photon. If the photon is just at dissociation, the H atoms will have effectively zero kinetic energy and so not depart from one another. In fact at this point life becomes complicated and interesting. The potential energy between two atoms increases very slowly with increasing separation as one reaches dissociation. In fact it is possible to form Rydberg atoms which may be a micron in diameter when excited to within fractions of wavenumbers of dissociation. This size is huge, about ten thousand times larger than the minimum H$_2$ atom separation, bigger than a protein and approaching the size of a bacterium! Clearly the two atoms hardly influence one another and consequently a small perturbation, a minute magnetic field for example, and certainly a nearby molecule may cause the Rydberg molecule either to break apart or recombine depending on whether some energy is added or subtracted in the interaction. If there is not enough energy then the H$_2$ will remain in a highly excited vibrational level until it suffers a collision with another molecule and some vibrational energy is transferred to this molecule. Alternatively the electronically excited H$_2$ could fluoresce and any vibrational energy left in the ground state shared on collision with another H$_2$.  If there is far too much energy than needed to dissociate, then the H atoms ballistically fly apart in opposite directions and transfer some of their energy at each collision with other H$_2$ molecules. Possibly reaction also occurs, but H$_3$ is almost certainly unstable.


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How to breakdown wax, preferably beeswax
I have a project that I'm working on at the moment, but I'm stuck on this detail. I'd like to find out what can I supplement beeswax with that is organic, non toxic, and safe if accidentally consumed, in order to speed up decomposition rate of beeswax to a time frame of about 1-6months. Any ideas?

As far as I know beeswax does not decompose. If you just want to melt it, use heat. If you want to break it down using a solvent, try turpentine (paint thinner).

So there's really no non toxic solvent to use?

I would try d-limonene or maybe perflubron if you can find it. Both are non polar non toxic solvents so they should work. Another thing that I have heard might work is vegetable oil or most other oils you can find in your kitchen. But otherwise d-limonene is the best bet because it's organic, nontoxic, and safe to consume.


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Why the intensity vs. frequency graph initially increases for black-body radiation?
Max Planck had said that $E = n h \nu$ in his quantum theory. Then according to it, as the frequency of the light emitted increases, so the light's intensity should decrease. But it is not in the initial observation of graph. Why?

Do you have the graph, or the link to it?

This equation is totally not about intensity.

As Ivan Neretin mentioned in comments, there is no any "intensity" in $E = n h \nu$ formula. Besides, this was the original supposition Planck made with the final formula being (essentially) the following,
$$
  B_\nu(\nu, T)
  \sim
  \nu^3 \frac{1}{\mathrm{e}^{\nu/T}-1} \, ,
$$
where $B_\nu$ is the spectral radiance per unit frequency, which is basically the "intensity" OP is looking for. A quick drawing of $\nu$-dependence shows that the $B_\nu$ curve has the desired shape:

Details can be found, for instance, in Wikipedia article.


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How to prevent dripping when using a Pasteur pipette with solvents like diethyl ether?
I'm trying to use a Pasteur pipette for pipetting diethyl ether and regardless of the angle I hold the pipette at, the diethyl ether will automatically start dispensing itself (I am unsure of what this phenomenon is called). I'm sure it's not the rubber bulb since I've been using the same one for pipetting other solvents that don't do this. I've also tried using a different rubber bulb, but to no avail.
Is there any way to prevent this from occurring? If not, is there anyway to mitigate the flow rate of the drips?

Low viscocity and high vapor pressure of such solvents makes it a pain indeed. Using an automatic pippet you can reverse pippet (push plunger button all the way down, and dispense only to first check. Also work quickly, deliberately and over short distances

DCM used to be my issue. Either work really quickly if your pipette is graduated or ditch the pipette and use a polypropylene plunger syringe (not the rubber ended ones, fine for bio, rubbish for wet organic chemistry ). I also used to overfill the (plastic) pipette and let it drip to roughly the correct volume, then transfer. Depends on your application.

Usually I avoid pipetting diethyl ether because it can be a bit annoying, but recently I've realized there is a very simple technique which markedly increases ease of transfer (though admittedly I've been using it mostly for DCM). All you have to do is saturate the interior of the pipette with the vapour of the low-boiling solvent before performing the transfer.
The problem arises because the low-boiling solvent will quickly (but not instantly) evaporate into the partial vacuum inside the pipette after pulling the solvent. If you try to transfer the liquid while it's still evaporating significantly, the large volume change from the phase transition will push the liquid beneath through the pipette exit.
So how do you get around this problem in practice? Using the pipette, simply pull and dispense the solvent a few times in succession over 5-10 seconds inside the outgoing vial. Then pull what you need and transfer. In my experience this greatly mitigates dripping. If you're not too slow between transfers, you can get away with doing this only once, at the start.
Other things which can be done to help:

As mentioned in the comments, avoid dragging the filled pipette over a  large distance. Keep the outgoing and ingoing vials right next to each other, at about the same level.
Pull a little bit of air into the pipette, keeping it between the exit and the solvent. This gives you a small time buffer before it drips out.
If you have a steady hand, keep very gently pulling air while you move the pipette between vials, matching the speed at which the liquid tries to drip out. This can be somewhat fiddly, so train to get a feel for it.
Don't fill the pipette to its maximum capacity on each transfer; do more transfers, each with less volume.
There's always a drip left on the very tip of the pipette, and that can easily fall. To avoid it, contact the remaining drop onto one of the outgoing vial's walls above the surface of the liquid, then pull a little bit of air into the pipette, sucking the last droplet along and leaving the tip dry.
Practice.


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Physical reason for the ideal lattice approximation
Why can we approximate a lattice as being infinite when we describe it as an ideal crystal? Is there any physical reason or is it just because the number of ions (/atoms) is so huge that we can treat an ion in the middle of the crystal as if it was surrounded by an infinite number of them, and just forget what happens on the surface?

Just like you said, ions in the middle feel pretty much the same way as if they were in an infinite crystal. There is no mystery and no deeper reason behind it.


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What is the hybridisation of hydrogen in methane?
I am trying to understand hybridisation. In methane carbon has $\mathrm{sp^3}$ hybridisation, but what is the hybridisation of hydrogen? Is it $\mathrm{sp^3}$? If yes then why?

What orbitals does hydrogen have available for bonding?

Orbitals of hydrogen atoms aren't hybridised in $\ce{CH4}$ at all:  1$\mathrm{s}$ orbital of each hydrogen atom forms a $\sigma$-bond with one of the four $\mathrm{sp^3}$-hybridised orbitals of the carbon atom.

Also, per calculations and experimental data, you could also argue for an unhybridised carbon since there are two different energies of the binding MOs.

You should answer your own question so that it may be useful to future users.

Hydrogen does not hybridise, as it only has one filled $s$ orbital. Hybridisation is the "mixing of $s$ and $p$ orbitals; here is a good explanation (for ethane, but it explains the general theory as well).

it is always possible to find the hybridization of the central atom i.e., Carbon. The atomic orbital of hydrogen does not undergo hybridization.


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Why do Bence-Jones proteins behave differently in urine heat test?
The preliminary test for proteinuria (proteins in urine) is precipitation/turbidity on heat due to denaturation. This turbidity should not disappear when 10% acetic acid is added - to differentiate from phosphates.
Bence-Jones proteinuria is found in multiple myeloma and they are light chains of IgG and are definitely proteins! They have this unusual behaviour, i quote,

Bence-Jones proteins are light chains of γ-globulin. These are excreted in multiple myeloma and other parapro- teinaemias. In heat and acetic acid test performed under temperature control, these proteins are precipitated at lower temperature ($\mathrm{56\ ^\circ C}$) and disappear on further heating above $\mathrm{90\ ^\circ C}$ but reappear on cooling to lower temperature again.

How is this behaviour explained?

What's the source of quote?