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Is there an iron–iron bond in diiron nonacarbonyl?
Diiron nonacarbonyl, $\ce{[Fe2(CO)9]}$, is often depicted with an Fe–Fe bond as shown at the left-hand side:

The Fe—Fe bond is usually invoked in order to (1) explain the observed diamagnetism and (2) satisfy the 18-electron rule. However, I'm also aware that the presence of such a bond has been challenged numerous times in some textbooks and also by theoretical studies. Being untrained in computational chemistry, though, means I don't fully understand some of these papers.
So: is there a bond? If there isn't (which seems to be the case), can the bonding in this complex still be reconciled with the 18-electron rule?
I'd especially appreciate it if answers include some explanation of what is going on in the literature.

I'm also not a QC-savvy person, so I took the crystal structure [Cotton1974] and calculated Voronoi-Dirichlet solid angles (SA) for the iron atom. Even though the SA for Fe-C bonds (approx. 14) are 7-fold  higher than Fe-Fe (approx. 2),  C.N. of iron here is still considered 7 , involving a valence-type interaction between metals. The analysis is automated via AutoCN subroutine of ToposPro suit, which is used for topological crystal structure analysis. I'm not sure if it counts as an answer though.

tl;dr: There probably isn’t an Fe–Fe bond. We should probably stop drawing an Fe-Fe bond. Textbooks and lecturers should really stop teaching students that there is an Fe-Fe bond. But all that said, there isn’t one single satisfactory explanation for why there isn’t one…

Despite a lot of studies (and despite $\ce{[Fe2(CO)9]}$ being one of the earliest examples of a molecule with symmetrically bridging carbonyls) , the nature of bonding in $\ce{[Fe2(CO)9]}$ is still a little hard to get a clear picture of.
Whilst at a research level, most academics now accept that there is no significant bonding interaction between the two metal centres, this is not echoed at the undergraduate level where the Fe–Fe bonding model is still often taught.
Most textbooks either avoid the issue entirely (never providing a structure), or show the Fe–Fe bond (with some footnote that it might not be correct), and even those that do acknowledge the lack of the metal-metal bond (Elschenbroich being a good example) don’t really discuss how this works in practice, or how one would go about electron counting (a source of many a nasty tutorial problem).
Historical context
The simple answer is that there was never really any strong evidence for the presence of the Fe–Fe bond. As is often the case, it was adopted by several (well known) chemists at the time, and subsequent publications (and especially textbooks) propagated the idea, largely due to its convenience with respect to electron counting.
The earliest data that really looked at the nature of the Fe–Fe bond in $\ce{[Fe2(CO)9]}$  can be attributed to a crystal structure measured by Powell (2.46 Å)[1](later corrected by Cotton (2.53 Å)).[2]
Although these values didn’t necessarily prove the existence of an Fe–Fe bond (you can’t see a bond with X-ray, only the atoms, from which you can infer bonding), its presence was necessary to fulfil the 18-electron rule, and the values obtained were well in-keeping with other M–M bond lengths known.
The issue with the above argument (as convenient as it is) is that it’s impossible to tell whether the 2.53 Å represents a true bonding interaction or whether it’s simply a consequence of the bridging Fe–C(O)–Fe bonds pulling the metals close together. The iron itself is relatively contracted, so unlike in some other bridged systems, the steric problem of bringing the two metals together is relatively minimal.
The Beginning of Change
Although the Fe–Fe bond was fairly widely adopted, not all were in agreement.
Roald Hoffmann (of Woodward–Hoffmann fame) published several papers which appeared to contradict the presence of an Fe–Fe bond, showing that the interaction was in-fact “antibonding and repulsive”.[3]
Though not fully quantitative (simple Hückel calculations), Hoffmann was able to construct qualitative molecular orbital diagrams using molecular symmetry and isolobal/electronic fragments (from well behaved molecules) to piece together the likely bonding in $\ce{[Fe2(CO)9]}$. To date, his findings are largely still valid (though exact energies have now been calculated with far more accuracy).
Quantitative Computational Studies[#]
Modern computational studies agree that overall, the Fe–Fe interaction is net anti-bonding (though some bonding interactions have been found) with no genuine covalent Fe–Fe bond, as was initially proposed.
In Liddle’s Molecular Metal-Metal Bonds, McGrady provides a qualitative  molecular orbital diagram for the complex. The σ and σ* orbitals, as well as the δπ and δπ* orbitals (sandwiched between the former pair), are fully occupied and provide no net bonding character. However, the population of the Fe–Fe antibonding πδ* orbitals leads to a formal bond order of –2.[4]

This diagram correlates well to the results in one of the earliest computational studies on this system comparing various methods (DFT, MP2, SCF) by Reinhold.[5]
In the study, various properties of the molecules are calculated (M–M bond length, M–C bond lengths, bond angles etc.) and this data is compared to known data from crystal structures/other experimental work. They determine that only DFT gives accurate results (casting some doubt on earlier computational work using less rigorous methods).[##]
The molecular orbital diagram constructed (in agreement with the one from Ref [4] above) crucially shows a negative overlap of the Fe–Fe orbitals is observed, suggesting a direct repulsion. To quote directly from the paper:

in the iron systems, with more contracted metal orbitals, the direct metal–metal repulsion is relatively weak while the metal–bridge bonds are sufficiently strong

Other computational studies have looked to calculate properties of the complex, such as vibrational frequencies of any proposed Fe–Fe bond, or the NMR shieldings that would be present on the ligands with and without the presence of the Fe–Fe bond. This work is referenced in Comprehensive Organometallic Synthesis II (Vol. 6), but doesn’t seem to have lead to very useful results.
Conclusion
The take home message from this is that there really is unlikely to be an Fe–Fe bond in the way it is commonly taught to undergraduates, and certainly not for the purposes of electron counting.
A nice possible explanation is given by McGrady,[4] which provides some explanation for why the 18-electron rule isn’t important in this case (and at a level that an undergraduate could understand):

There are, in fact, many parallels between 34-electron $\ce{[Fe2(CO)9]}$ and square-planar 16-electron species, not least the tendency of both to react with nucleophiles. The breakdown of the 18-electron rule in square-planar complexes occurs because the vacant pz orbital fails to find an appropriate symmetry match among the linear combinations of ligand-based orbitals, and so is not used, either to form a bond or to accommodate metal-based lone pairs of electrons. [...]
In $\ce{[Fe2(CO)9]}$, there are two vacant $\mathrm{a_2''}$-symmetric metal-based combinations (primarily linear combinations of 4s and 4pz) but only one occupied ligand-based combination of the same symmetry. Thus, one of the two metal-based orbitals remains “unused” in the same sense that the pz orbital remains unused in square-planar species. This simple symmetry-based analysis suggests that there is no need to invoke an Fe–Fe bond in $\ce{[Fe2(CO)9]}$ simply because the metal centers do not conform to the 18-electron rule.

References & Notes
[1]: Powell, H.M.; Ewens, R.V.G J. Chem. Soc 1939, 286
[2]: Cotton, F.A.; Troup, J.M. J. Chem. Soc. Dalton Trans. 1974, 800
[3]: Lauher, J.W.; Elian, M.; Summerville, R.H.; Hoffman, R. J. Am. Chem. Soc., 1976, 98, 3219
[#]: Full disclosure, I’ve read most of the papers that will follow at some point whilst trying to answer this in the past, but (big but) I’m not a computational chemist. I’m hoping someone else may be able to shed light on how valid these computational results actually are.
[4]: McGrady, J.E. in Molecular Metal–Metal Bonds; Liddle, S., Eds.; Wiley-VCH: Weinheim, 2015
[5]: Hunstock, E.; Mealli, C.; Jose-Calhorda, M.; Reinhold, J. Inorg. Chem. 1999, 38, 5053. Note that this isn’t a complete geometry optimisation, they start with some known parameters. New J. Chem. 1994, 18, 465 is a paper supposedly attempting the complete optimisation from scratch, but I can’t find it anywhere right now to access it
[##]: Even now there are a lot of issues with accurately calculating properties of organometallic complexes… it takes a vast vast amount of computational time due to the need for basis sets far expanded compared to those developed for simple organic systems

As a note to anyone reading this, I wrote this more to satisfy myself of what I thought I knew. The stuff up until the computational section is all sound, but my understanding of how the computations discovered what they did is unknown to me (so basically, please don't comment "why did they use that basis set").  In any case, I'd be interested to see an answer from someone who does understand how the computation works.  Hopefully this serves as a useful starting point/has some useful references .

Actually, I'm not too concerned about the computational details - I agree it would be nice to know, but it's probably also easy to get lost in the details. Essentially you've covered all that I was looking for - the MO diagram, and the ending quote, are illuminating.

To supplement the accepted answer, here is something else I found.
Bonding in bridging carbonyls
An alternative explanation for the bonding in $\ce{[Fe2(CO)9]}$ is presented in a 2012 article by Green1 (which also contains numerous references to earlier computational studies indicating the absence of a bond). In it, the authors describe Fe–C(O)–Fe bonds as being of two "types". If there is backbonding from metal to carbonyl π*, then the bridging carbonyl is "ketonic" in nature; if there is no backbonding, then the interaction is best described as 3-centre-2-electron (3c2e).

(The half-arrow here is not related to any form of radical reactions - it is to "emphasise that the donor atom is not using two lone pairs, but rather that a single electron pair is being shared with both atoms", in the authors' words.)
Which of these two schemes is appropriate depends, crucially, on the occupation of the orbital(s) labelled $\lambda-$. This orbital is formed from Fe 3d and CO π* orbitals, and one can see that it corresponds to the $\mathrm{2e''}$ orbitals in the MO diagram provided by NotEvans.
The article then goes on to provide a qualitative MO diagram for the complex, using Hoffmann's earlier approach2 which invokes the isolobal analogy.

The thing to note here is that there are three bridging carbonyl ligands, but there are only two backbonding, $\lambda-$ type orbitals to fill (of symmetry $\mathrm{e''}$). Hence, two of the three bridging carbonyls are "ketonic"; the third bridging carbonyl is best described as being involved in a Fe–C(O)–Fe 3c2e bond.
Is there a bond?
According to the authors, the $\mathrm{a_1'}$ orbital, which is formed from iron $\mathrm{d}_{z^2}$ orbitals, cannot count as a Fe–Fe σ bond because these d orbitals are heavily involved in backbonding to terminal carbonyls.
The $\mathrm{e'}$ MOs (which correspond to $\lambda+$ orbitals in the first diagram, and are Fe–Fe π-bonding in character) and $\mathrm{e''}$ MOs ($\lambda-$ orbitals, Fe–Fe π-antibonding) are both occupied, and cannot lead to any net bonding. In fact, the bonding $\mathrm{e'}$ orbitals are mostly centred on CO, whereas the antibonding $\mathrm{e''}$ orbitals are mostly centred on Fe. So, there is a net repulsive interaction between the two iron atoms, in agreement with computational studies.
Relation to the 18-electron rule
Going back to the two types of bridging carbonyl ligands, the first "ketonic" carbonyl ligand counts as a one-electron donor to both metal atoms, as is usually taught in introductory organometallic chemistry. In electron counting notation, this is usually written as μ-X2, indicating that the carbonyl ligand is a one-electron donor (X) to the two metal atoms it bridges.
However, the 3c2e-type ligand counts as a two-electron donor to both metal atoms. It is not that the ligand donates two electrons to each metal atom (which is impossible here as CO can only donate one lone pair on carbon). Rather, the ligand donates two electrons to the entire bridge, which is shared between both metal atoms. This can be denoted μ-L2 (L indicating a two-electron donor).
Using this formalism, it is easy to see that the 18-electron rule is satisfied for both iron centres, if we assume the three bridging carbonyls to consist of two μ-X2 carbonyls and one μ-L2 carbonyl:

Of course, since the three bridging carbonyls are equivalent, we cannot single any one of them out to be the μ-L2 carbonyl; it is technically more accurate to consider it to be a hybrid of three resonance structures, each having a different bridging carbonyl being μ-L2.

References

Green, J. C.; Green, M. L. H.; Parkin, G. The occurrence and representation of three-centre two-electron bonds in covalent inorganic compounds. Chem. Commun. 2012, 48 (94), 11481–11503. DOI: 10.1039/C2CC35304K.
Summerville, R. H.; Hoffmann, R. M2L9 complexes. J. Am. Chem. Soc. 1979, 101 (14), 3821–3831. DOI: 10.1021/ja00508a019.

Beautiful answer. I read the same article and was confused about the weird-looking $\ce{^-C#O^+}$ bond. Doesn't the ketonic $\ce{C=O}$ structure already have a lone pair which it can donate, forming the 3c-2e bond? Why does it adopt this structure?

@AniruddhaDeb $\ce{C=O}$ and $\ce{^-C#O+}$ are resonance structures of each other; they aren't actually different things.


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Implication of completely non relativistic Hamiltonian
I am studying the
Application of perturbation theory to hydrogenic atoms subject to internal or external electromagnetic fields.
The very first equation it uses is the hamiltonian of hydrogenic atoms
$$H_0=\frac{p^2}{2\mu}-\frac{Ze^2}{r}$$
This equation is said to have no spin and is completely non-relativistic.

What does completely non-relativistic mean?

These are two (generally) unrelated questions - you should ask them separately. In this case you can [edit] your question to restrict the scope, if you wish.

@orthocresol removed the second question.

Your reduced equation is just the normal hydrogen Hamiltonian written in atomic units, that is with $\hbar=\frac{1}{4\pi\epsilon_{0}}=m_{e}=e=1$ (e is the charge of the electron, which they include for some reason despite it being 1 in atomic units). They also make the substitution $p=-i\hbar\nabla$ where, again to be consistent with atomic units, $\hbar=1$.
The form you have normally seen of the Hamiltonian is already spinless and non-relativistic (which are essentially synonymous for quantum mechanical operators). For something as light as Hydrogen, relativity (as in Einstein's theory of Special Relativity) is known to have very limited impact, and so it is neglected in forming the Hamiltonian.
There may be other questions on the Chem or Physics SE related to relativity in QM, but I would suggest unless you are particularly interested in relativity to simply focus on the QM side of things for now until you have that clear and can build on that knowledge.

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Reasons for solid or liquid soap
Soap is made by a saponification reaction, where a fat reacts with hydroxide ions to form a surfactant and glycerol.
To make a solid soap $\ce{NaOH}$ is used, while $\ce{KOH}$ is used for liquid soaps.
I don't understand why the alkali metal has such a great impact on the state of matter. Usually the argumentation is based on intramolecular interactions, such as Van-der-Waals forces or hydrogen bonds, but if the same fat is used once with $\ce{NaOH}$ and once with $\ce{KOH}$ the resulting surfactants are basically the same, so the interactions shouldn't differ too much.
The only reason I could think of is the size of the alkali metal. Potassium has an atomic radius of 231 pm which is quite a bit more than the radius of sodium, being 186 pm. But why the atomic radius should have an impact on the state of matter of the soap is still unclear to me. Maybe I'm also completely wrong with this assumption.
A while ago, this question was already asked here on this forum, but I wonder if the given explanation is the only reason for the different state of matter. It's absolutely true, that the reactivity of the alkali metals increases from top to bottom, but can this solely explain the phenomenon? @rch provides the solubility of $\ce{NaOH}$ and $\ce{KOH}$ to back up his answer, but I don't think that this is sufficient. You cannot simply change the hydroxide ion by a fatty acid and assume, that there are no substantial amendments in the reaction behavior, can you?
Does anyone can explain this in more detail?

related https://chemistry.stackexchange.com/questions/19663/soap-versus-shampoo

The answer to that older question is total nonsense. Reactivity of the metal has nothing to do with the physical properties of the finished product. Afaik the reason is mostly that potassium soaps are rather crude mixtures, and further the slightly larger ion seems to not fit well into any viable crystal structure of the carboxylate salt.

@Karl I thought so myself, that's why I reposted the question and am also asking for a better and more detailed answer. Could you explain, what you mean be crude mixture?

Well, most fats are wild mixtures of glycol triesters, so after saponification you have a dozen or so different carboxylate salts in your pot.

@Mithoron Although this is a duplicate of the given question, it would be very useful not to close this question. The thread 'Hard' soap vs 'Soft' soap: Why do they work this way? was already answered and has been marked as solved, even though the given answer is not meaningful. With my question I want to go into the phenomenon more closely. By your mark, other chemists can no longer answer and so the question will not be solved again.

You can't ping them like this https://chemistry.meta.stackexchange.com/questions/3889/pinging-in-comment-to-suggest-answering-question

I think it only superficially looked like duplicate, and may get reopened.

@Mithoron Let me express a well-meant criticism: If a question superficially looks like a duplicate, you should still take the time to check whether it is actually a simple duplicate of a lazy questioner, or whether it is a re-issue of an unanswered problem. I hope, that you can agree with this.

If it was reposting then situation would be obvious, if it was asking about the same problem then still one of questions should be closed (think Occam's Razor), but this case seems tricky.

https://chemistry.stackexchange.com/questions/82008/reasons-for-solid-or-liquid-soap

I don't know if anyone is still looking for the answer but here I go anyway because I spent 40 minutes researching this for an assignment
TL;DR - The better solubility of potassium salts is the key factor, but not in the way one would initially suspect. Industrial processes and the efficiency of large-scale soap making explains the choice.
The reason why potassium-based soaps are 'soft' or liquid is to do with the manufacturing process and the difficulty in separating them (R-COO-K) from the glycerin/glycerol after reaction with the soap lye (The alkaline salt solution used to make the soap e.g. KOH). The better solubility of the potassium salts contributes to this significantly.
Glycerin/glycerol has a lot of good properties for making a liquid soap too. It's colourless, viscous, non-toxic to humans and even has some antiviral and antimicrobial properties. It makes sense to utilise these and make the soap liquid instead of trying to get it out. Also it is often cost ineffective to dissolve out the glycerin with large amounts of expensive KCl or other K salt. Comparing this to sodium-based soaps, it takes smaller amounts of cheaper Na salts to separate the by-products.
It all comes down to money I guess.
Here are my sources, read them if you want to know more:

https://pubs.acs.org/doi/pdf/10.1021/ed025p379
http://www.madehow.com/Volume-2/Soap.html
https://pubchem.ncbi.nlm.nih.gov/compound/Glycerol

Not only money. Think of shaving soap (nowadays less common). Still, I guess if money wasn't involved we would have got shaving soap bars.

I can't answer your question. I thought I could, but once I looked into it, I couldn't find the data to support it.  That said, consider Na & K stearates (C18). BOTH have mp's well above 200°C. That is, neither of them are liquids. I believe these are reasonable models for fatty acid salts relevant to your question. Once you accept (tentatively) the idea that one isn't a liquid and the other a solid at room temperature, you're left trying to understand why the saponification reactions (apparently) result in products with different physical states (I'd have to see a bit more rigorous a study than anecdotal testimony, no offense.) I am unable to do more than speculate on this question. The reason which most suggests itself to me is that the degree of reaction (completion) is less with KOH than with NaOH. That is, the (anhydrous) products are less pure and so have less tendency to crystallize. (I suspect side-reactions aren't important, but I could be wrong.) The other "reasons" I can imagine (larger propensity of the K salt to hold onto water, less propensity of the K salt to continue to hydrolyze due to the lower level of KOH in the product (leading to a soap which is more stable (ages better)), etc.) all seem a bit far out. Finally, it is possible that the difference in size of the two ions results in the product with K+ to be substantially less ordered (more disordered) than with Na+. Believe it or not, it is unusual to have C12 -C18 fatty acids which are pure, the far most common situation is that there are distributions of both C number as well as saturation and branching. What this means as far as your question goes is that the data for the pure compounds is quite difficult to locate (if it has even been measured). (and so, without the "clean" data, everyone can have their pet hypothesis.)

Thanks for your answer. The statement is based not only on anecdotal  testimony but on experimental findings. I've tried the saponification with coconut fat using $\ce{KOH}$ and $\ce{NaOH}$ and the products were liquid respectively solid. Your objection with the sodium or potassium stearate is interesting, but according to PubChem potassium stearate is liquid.

The type of alkali metal used determines the kind of soap product. Sodium soaps, prepared from sodium hydroxide, are firm, whereas potassium soaps, derived from potassium hydroxide, are softer or often liquid. That's what Wikipedia says, but unfortunately there is not given any source to back up the statement. I can remember having read this in a textbook, but I don't know in which anymore.

I think we are thinking too chemically, i.e., too basically. From my (limited) experience, soft soaps contain more water. The correspondence of viscosity or some other measure of strength or hardness with water content is not exactly identical for the sodium and potassium soaps, but it is fairly close. If you make a potassium soap with low water content, it will be "hard"; if you make a sodium soap with more water it will be soft. All of the variables mentioned in previous answers are important, but there is (at least) one more...
The stability of the soaps is a different matter: a sodium or potassium soap with low water can be manufactured to be hard, but the potassium soap will be more expensive, and apparently potassium confers no great advantage. Hard soaps are stable.
However, a soft soap made can be made over a wide range of compositions and viscosities reproducibly with potassium, and will be stable and will not separate on storage. A soft soap made from sodium has much less stability, or a smaller stability range, and thus is more limited in what can be manufactured easily. Considering that soft soaps tend to be higher cost (cosmetic) soaps with higher cost (fragrances, colors, special feel, etc.), the extra potassium expense is a minimal burden.

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Is a spectrophotometric method of measuring the rate of a complex reaction valid?
A few months back I designed and carried out an experiment to measure the rate of the reaction between ferric chloride and copper.The dependent variable was the rate of the reaction; the independent variable was the the concentration of ferric chloride.
I searched up the reaction and got it from chemiday. There, it stated it as the following:
$$\ce{2FeCl3 + Cu -> 2FeCl2 + CuCl2}$$
So the method I used to measure the rate of this reaction was to use a spectrometer to find a rising absorbance peak for one of the reactants, which I selected to be: $\ce{CuCl2}$.
Then I got some anhydrous $\ce{CuCl2}$ and made a standard curve (Absorbance vs Concentration) for it, so that I could use that to convert the absorbance at it's $\lambda_\mathrm{max}$ to a concentration value.
I carried out the experiment taking the readings at that $\lambda_\mathrm{max}$ at set time intervals, and I reached the conclusion that the relationship between the concentration of ferric chloride and the rate of the reaction between it and copper was a linear one, at least for the concentrations I tested ($\pu{0.1 .. 0.6 M}$).
But now I'm doing more reading to write my report and I found out that this was a complex reaction, meaning that there were several steps to the reaction. After searching up some papers, the consensus seems to be:
\begin{align}
\ce{FeCl3 + Cu &-> FeCl2 + CuCl}\tag{1}\\
\ce{FeCl3 + CuCl &-> FeCl2 + CuCl2}\tag{2}\\
\ce{CuCl2 + Cu &-> 2CuCl}\tag{3}
\end{align}
As seen from above, it isn't as simple as the original single reaction that I mentioned earlier. First, $\ce{CuCl}$ is produced, then it becomes a reactant in the second reaction to produce $\ce{CuCl2}$, and lastly $\ce{CuCl2}$ becomes a reactant to produce $\ce{CuCl}$.
I'm seriously questioning my method after finding out that a series of reactions takes place, not a just single one. Although the data that I got and the conclusion that I reached matched nicely with other papers that I've read, I'm not sure if the explanation behing my method is correct.
Any input would be greatly appreciated. Thank you.

All this mechanism doesn't matter that much as you're reacting solid with liquid. Your experiment is probably okayish, at least linear dependence seems OK.

This is new for me; how does reacting solid with liquid make the steps irrelevant? Thank you for your input.

Redox itself is quite fast but only surface of metal can react; that's like you were slowly feeding your solution  with metal, and pace of this "feeding" determines change in conc. in solution.

The setup I had was to throw all the copper, which was very little, by the way (0.15g), into the solution. So, in terms of 'feeding' the metal, it was all fed at once.

The solid in liquid part doesn't matter. Having a solid in there means that the solution should be saturated in that reactant (unless the reaction is very fast) which is exactly what is desired in most circumstances, as the rate-dependence on copper concentration is being measured.

@HamzaQayyum sigh I tell you that inside of the solid is not reacting. If your metal was dissolved then everything would be different. System isn't isotropic so diffusion is very important.

@Mithoron Ah I see what you mean now. Thank you for clarifying.

@jheindel The rate dependence on Ferric Chloride is being measured, actually. The copper is a constant.

Ya hasty typing. Good catch.

This is a perfectly valid method of determining rates of chemical reactions. It is obviously limited to those reactions where either the product or the initial reactant can be monitored easily using spectroscopy as well as reactions which are slow enough to get an absorbance measurement at set intervals. I have used this method multiple times in undergraduate chemistry labs and it works, and in principle it does not matter if there are intermediates.
The intermediates should not matter because the rate of a reaction is defined as the rate at which product is formed. That is, the rate of a first-order reaction is,
$$
\text{rate}=k_{tot}[A]
$$
If the reaction involves only one elementary step then the prefactor $k_{tot}=k_1$, the rate constant of the one step (assuming no backwards reaction). Note that $[A]$ can be the concentration of either a product or reactant, but you might need a stoichiometric coefficient in there depending on the reaction. On the other hand, if there are intermediate steps, the prefactor can become arbitrarily messy. Maybe there's an equilibrium between two intermediates so you have a term in there $k_2/k_{-2}$ and whatever else you can imagine. This doesn't matter though because your experiment measures the change in concentration at either end, and the slope is then the effective rate constant for the reaction which has all of the intermediate stuff lumped together.
In fact, this method you use is probably the most direct means of measuring a rate in the sense that a rate is defined as a change in concentration over time. You are measuring a change in absorbance over time, so there is only step (and one assumption) in going from your data straight to the reaction rate. That assumption is simply the validity of Beer's law, which is usually not a problem for a large range of concentrations.

As a sanity check, look at this website which has a description of the same experimental setup and what appears to be real data. Note that they are measuring change in absorbance of a reactant while you say you are measuring the change in absorbance of a product, $\ce{CuCl2}$

Thank you very, very much for your input. I haven't studied rate equations as part of my studies yet, but I will certainly study them now so that I can understand that part of your answer. As for Beer's Law, I didn't think of that, but I see a clear pathway now as to how I can write about this. Thanks again.


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Can I get ride of dichlorobenzene in honey?
I have some honey that was stored near dichlorobenzene crystals & has taken on the smell. I'm wondering if I turn it into mead if it would remove the chemical?
The procedure would be to add 5-10 parts water to the honey. Then boil the solution for at least an hour. I would think this should rapidly remove the benzine component.
The second part of the process would to be to add a yeast culture that would not ferment until the chlorine has dissipated.
It would seem that I could tell the chemical is absent when the strong odor is gone.

I would get rid of that honey without a second thought.

Odor may be gone but it still may be dangerous.

Activated charcoal might work (after diluting the honey with water), but still, I wouldn't consume that.

Just theoretically, if you were to heat the honey (primarily fructose and glucose) to 373 K, the boiling point of water, it would speed up the evaporation of the dichlorobenzene (vapor pressure ~8,000 Pa). That might be enough to get rid of the odor, if kept at that temperature for a few days.
That said, the flavor of the honey would likely be destroyed, as well. Even though pasteurization of honey is not normally recommended, that only requires heating to 145° F (336 K) for 30 minutes, and it still changes the flavor.
So toss the honey, or use it as an embalming agent (see also National Geographic). The dichlorobenzene "moth flakes" should make this honey even more effective!


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Should the coupling between vibrational modes increase or decrease with increasing temperature/vibrational energy level?
At low temperatures, the amplitude of vibrations should be smaller, so the vibrations should be as harmonic as they are ever going to be. Thus, the normal mode description of vibrations should be at its best at low temperatures.
As this paper [1] describes, the normal mode description must be wrong when the vibrations reach an amplitude such that a bond is broken. Namely, if you consider the symmetric stretch of a molecule benzene, the normal mode coordinate moves all $\ce{C-H}$ bonds in unison so that they all pass through their minimum simultaneously. Yet, it is extremely unlikely that all of these bonds will break simultaneously. Much more likely is that only one bond breaks. Thus, the authors advocate the use of the local mode description of vibrations in high vibrational energy levels. Personally, I like the local mode description in most cases because it includes anharmonicity quite naturally. It has two shortcomings, however. It is only good for vibrations between a light and heavy atom ($\ce{X-H}$ generally). Second, it explicitly ignores coupling between modes by treating each vibration as that of a diatomic.
Both normal modes and local modes fail to capture any real coupling between vibrational modes, but it seems that the high-temperature limit is for the modes to behave as local modes. This seems to be in conflict with what I have heard on multiple occasions that the coupling between modes should increase as temperature increases. The temperature doesn't really matter, I'm just saying as you thermally populate higher vibrational states, what happens to the coupling.
Now, the final thing is that I believe there is ambiguity in what we mean by coupling between modes and this is probably the source of my confusion. When I say coupling, it could mean how well I am able to write each vibrationa using independent coordinates. That is, the potential energy of a particular oscillator is a function of one coordinate only. As temperature increases, does this approximation become better or worse?
The second meaning of coupling between modes could be how easily energy is redistributed between vibrational modes. I would think this and the previous point are related, but I believe the only condition for having energy redistributed between vibrational modes is that the vibrations are anharmonic. Not necessarily that they are coupled in the sense as above.
I believe that people mean each of these things when talking about coupling between modes even though they are distinct concepts. What is the right way of talking about coupling?
To summarize:
As higher vibrational energy levels are populated, does the assumption of each vibration depending on one coordinate only become better or worse?
As higher vibrational energy levels are populated, is energy redistributed between modes more easily or less easily, and does this relate to the first question at all?

Disclaimer:
I'm fairly certain energy redistribution should become faster at higher temperature as vibrations will become more anharmonic.

[1] Henry, B. R. (1977). Use of local modes in the description of highly vibrationally excited molecules. Accounts of Chemical Research, 10(6), 207-213.


text:
Derivation of an equality in Michaelis–Menten kinetics
Enzymatic action may be described as follows:
$$\ce{Enzyme + Substrate <=>[k_1] ES complex ->[k_\mathrm{2}] Enzyme + Product}$$
The initial rate of enzyme-catalyzed reactions can be described by the Michaelis-Menten equation:
$$\mathrm{rate} = \frac{V_\mathrm{max}[\ce{S}]}{K_\mathrm{M} + [\ce{S}]} = \frac{k_\mathrm{cat}[\ce{E}][\ce{S}]}{K_\mathrm{M} + [\ce{S}]}$$
where $V_\mathrm{max}$ is the maximum rate, $[\ce{S}]$ the substrate concentration, $[\ce{E}]$ is the enzyme concentration, $K_\mathrm{M}$ is the Michaelis constant and $k_\mathrm{cat}$ is the number of catalytic cycles per second.
It is known that
$$V_\mathrm{max} = k_2[E]_0$$
and by inspecting the equation above we can deduce that
\begin{align}
V_\mathrm{max} &= k_\mathrm{cat}[\ce{E}]\\
\implies k_2[\ce{E}]_0 &= k_\mathrm{cat}[\ce{E}]
\end{align}
How, and why, does this equality hold?
How do we know when to use ${[\ce{E}]_0}$ and $[\ce{E}]$ in rate equations, and what are the implications of using either?
${[\ce{E}]_0}$ refers to the enzyme concentration at the start of the reaction, and [E] refers to the concentration of enzyme at any point in time during the course of the reaction.

Can you define with $[\ce{E}]_0$ and $[\ce{E}]$ are, just to make sure everybody is on the same page

Something seems wrong, what is $\mathrm{k_2}$?

I think you are getting messed up by adding in extra constants. What you have listed in the reaction as $k_2$ is $k_{cat}$ and $V_{max}=k_{cat}[E]_0$ not $[E]$. On the Wikipedia page, they go through the derivation which explains why the equation depends only the initial enzyme concentration.

Your equation

$$\ce{Enzyme + Substrate <=>[k_1] ES complex ->[k_2] Enzyme + Product}$$

contains only two rate constants, $k_1$ and $k_2$, but not $k_{cat}$ you refer to.
The correct schema for the Michaelis-Menten kinetics would be
$$\ce{Enzyme + Substrate <=>[k_1][k_{-1}] ES complex ->[k_2 = k_{cat}] Enzyme + Product}$$
Note that $k_{cat} = k_2$ is the rate constant for the decay of the ES complex in the second step.
From this you can derive the Michaelis-Menten rate law as described in the Wikipedia article you refer to.
$[E]_0$ is the initial concentration of the enzyme. During the reaction the enzyme exists as free enzyme ($[E]$) and bound in the enzyme substrate complex $[ES]$. These concentrations are related by the equation $[E]_0 = [E] + [ES]$.

If k2 = kcat, then wouldn't that yield [E] nought = [E] ?

$[E] = [E_0]$ holds exactly only when $[ES] = 0$, e.g. at the (idealized) beginning of the reaction.

I'm slightly confused now. By that logic, at any point in time during the course of the reaction, [E] does not equal [E] nought, then k2 cannot equal to kcat, otherwise the equation $k_2[E]0 = k{cat}[E] $ will not hold.

$k_2$ and $k_{cat}$ are the same. And your equation only holds if $[E] = [E]_0$.

Then as the reaction proceeds, the equation $k_2 [E]o = k_{cat} [E] $ does not hold, and so the Michaelis-Menten equality I have described in my original post, does not hold either. Is it right to say that?

Your equation is wrong. One reason for this is your assumption $V_{max}=k_{cat} [E]$ which is wrong since $[E]$ usually is not constant over time. Another reason is that your $k_{cat}$ came from nowhere.

The assumption $V_{max} = k_{cat} [E] $ was obtained through inspection of the Michaelis-Menten equality I have described in my original post. It was an equation that was given in an assignment.

@Jonathan Smith that form of the equation is incorrect. If it were true, $V_{max}$, the maximum rate would vary over time, which wouldn't make sense. $V_{max}$ depends on $[E]_0$ because the max rate of the reaction is limited by how much enzyme you start out with, not how much you have at any given time.

Ah, so the error lies in me deriving the equality $V_{max} = k_{cat}[E] $ simply by comparing the 2 sides of the Michaelis-Menten equation in my original post?

@Jonathan Smith yes, in the rate equation you wrote, the [E] should be $[E]0$ and in your equation for $V{max}$ it should equal $k_{cat}[E]_0$.


text:
Comparing photocatalytic kinetic and reaction rate of PPCPs removal in water treatment
Can anyone explain to me what is the meaning the kinetic term in the question and how does it relate to reaction rate constant, k ? I have been reading a lot of articles but was unable to get an understanding on the kinetic term. In addition, is there a difference between degradation of PPCPs by photocatalyst and photo-oxidation ? Since photocatalyst process has oxidation as well, I am kinda confused between those two. Any help provided with be appreciated.


text:
Can it be said that on our planet we have roughly the same number of electrons as we have protons?
My question is quite simple I think. Is it true that we have approximately the same number of electrons as we have protons and the average electric charge of the molecules in our planet is roughly neutral?

Yes. You can extend your coverage to the universe and change the conditions to be "exactly" instead of "approximately" and "roughly".

Wow.... How was this conclusion drawn though?

This 1988 paper says it is still unknown whether or not Earth has a net electric charge: https://link.springer.com/article/10.1007/BF01054576

This more-recent paper puts upper limits on the charge: https://link.springer.com/article/10.1007%2Fs10714-012-1365-0

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

Obviously, the ideas expressed in the question are not (totally) alien to Chemistry. However, I think this is better off on the Physics.SE.

The surface of the earth is negatively charged, about $\pu{500,000C}$.

Yes
We can see that the charges in the observable world are close to balance by considering what we would observe if they were not.
Big imbalances of charges cause strong electrical fields. Those are measurable and their effects are observable. We can create such imbalances locally (that's what a Van de Graaff generator does) and the effects are spectacular: really big sparks as the strong fields ionise air and cause big electrical discharges. These separations also occur in nature (that's what thunderstorms are). In both cases, the consequence of significant temporary separations of charge are spectacular natural phenomena that act to restore the balance.
It could be true, however, that the earth as a whole had a net charge. We wouldn't notice this while on the surface as the net charge would be relative to interplanetary space. On the other hand any such charge would probably be neutralised by selective interaction with the charged particles in the solar wind which we could observe. Also, if the net charge were substantial, we would observe significant effects on planetary orbits (the electrical field is a lot stronger than the gravitational field.
These observations together suggest that both locally and globally we have approximately the same number of protons as electrons on the earth.


text:
Can cat urine corrode stainless steel?
My cat has a habit of urinating in the bathroom sink. It is unsanitary, but can it corrode the stainless steel plumbing hardware? My friend believes it will, but I disagree. I could believe that if the urine were left to sit over time it could possibly affect the electropolished layer, but it is generally cleaned up within 24 hours.

You do not have a stainless steel sink drain. It is 70:30 cartridge brass with nickle/chrome plating ( or equal; admiralty metal , muntz metal, etc) .

Well, infact yes, cat urine can corrode stainless steel plumbing.
Urine of any mammals contains quite a lot of ions, and the chloride ion concentration is quite high. Chlorides attack the passive layer of steel, exposing the reactive iron underneath.

Pitting corrosion - The passive layer on stainless steel can be attacked by certain chemical species. The chloride ion Cl- is the most common of these and is found in everyday materials such as salt and bleach. Pitting corrosion is avoided by making sure that stainless steel does not come into prolonged contact with harmful chemicals or by choosing a grade of steel which is more resistant to attack. The pitting corrosion resistance can be assessed using the Pitting Resistance Equivalent Number calculated from the alloy content.[1]

While you may clean up the urine quickly, if your cat does this frequently, it can add up to corrosion very quickly. I'd suggest doing whatever you can to make your cat from urinating in the sink.

“BRITISH STAINLESS STEEL ASSOCIATION Making the Most of Stainless Steel.” What Forms of Corrosion Can Occur in Stainless Steels?, www.bssa.org.uk/faq.php?id=9.

I would be surprised if you have stainless plumbing  ( cost >>> brass ). Any urine produces ammonia. Ammonia causes stress corrosion cracking in any brass and some other copper alloys. In the UK it may be called "season cracking" ( for the rain season in India ). So, if you have brass , I would keep it rinsed. Urine would most likely cause only cosmetic damage to stainless. These factors ( cracking of brass - no problem for stainless) have resulted in the flex tubing for connecting gas home appliances being made of stainless today when 30 + years ago they were brass. Footnote : Years ago, I had a gas space heater in a garage , a dog urinated on the brass flex tube and it cracked and leaked .


text:
Why does the addition of PCl5 increase the rate of dissociation of PCl5?

$\ce{PCl5 <=> PCl3 +Cl2} $

It's stated in my book that addition of $\ce{PCl5}$  to the equilibrium mixture increases the rate of forward reaction but no reason is mentioned for it.
However, before this the author derived $K_c = \dfrac{x^2}{(1-x)V}$ (using law of mass action at equilibrium ) where V is the volume of the container and $x$ is the number of $\ce{PCl5}$ molecules dissociated from $a$ number of moles initially.
I couldn't relate the "rate of forward reaction" statement to this equation while earlier in case of the reaction:
$\ce{H2 + I2 <=> HI}$
I was able to relate the equation I obtained through law of mass action with the effect of concentration on rate of forward or backward reaction.
Could someone please explain why addition of $\ce{PCl5}$ to the equilibrium mixture increases the rate of forward reaction?

Imagine a overpopulated kingdom suddenly gaining a large area of land (Start of Reaction). Immediately, groups of people start migrating to the area of land (Forward Reaction). However, as people migrate to this new land, some decide that, for whatever reason, they don't like this area (Reverse Reaction). As more and more people move there, more start coming back until there is an equal rate of people moving to the territory and people moving back (Dynamic Equilibrium).

To your question: To model this system using my example of immigration, we suddenly add an enourmous influx of refugees into the old kingdom (Increase in $\ce{PCl5}$).

So the land in the outer territory seems more valuable now (less crowding, more open space, better air, etc.). This causes a temporary increase in the forward direction until the reverse reaction increases to compensate. This is called Le Chatelier's principle in chemistry.
References: Le Chatelier's Principle, My Handy Old 2011 Introductory Chemistry Textbook.


text:
Is a 50/50 mix of Acetone and Water inert?
We are cleaning a gel out of a fuel tank that is dissolved by Acetone.  We want to neutralize the acetone by adding water to achieve a mix of 50/50 so that (PERHAPS) the acetone will not damage rubber / plastic compounds in the fuel system.
We have an electric fuel pump we are going to use to get the acetone and water out and we need to make sure we don't damage it. THEN there are the two brand new mercruiser marine fuel pumps, fuel filter and carburetor seals we don't want to damage.   We CAN  add more water than a 50/ mix as well.
We know we can't get all the contents of the tank out and so we want to make sure any remaining acetone will not be of a concentration to ruin the fuel system components.  By the way... this is for a famous movie director Roman Coppola.

Acetone is notoriously aggressive towards the plastic, rubber, and other non-metal components of internal combustion engines (and associated parts like fuel pumps). If the concentration of acetone is strong enough to dissolve your jell (assuming that is some organic compound) it is very likely strong enough to attack components of the fuel pump.
As a bit of anecdotal evidence regarding what I said above, note that of the dozen-odd oxygenated additives allowed by the US EPA for the oxygenated fuels program, neither acetone nor any other ketone is on the allowed list. Although this is in part due to the high volatility of acetone, it is also largely due to the fact that acetone would attack engine components as described above.
In short, what you've described is a bad idea for the health of your fuel pumps, even at very low concentrations of acetone.


text:
Understanding smoke chemistry of tobacco and/or marijuana
In an effort to understand the effects of tobacco vs marijuana smoke on health, the chemistry of smoking plants, and remembering some of the chemistry I took in college, I came across
a 2007 study comparing mainstream and sidestream marijuana and tobacco cigarette smoke,  according to which:

a number of chemicals were
  present in marijuana smoke at levels that were substantially higher
  than in tobacco smoke. For example, NO, NO x , hydrogen cyanide, and
  aromatic amines were present in marijuana smoke at levels 3-5 times
  higher than in mainstream tobacco smoke, while ammonia was present at
  levels 20 times higher than tobacco. Conversely, some compounds such
  as PAHs, formaldehyde, and acetaldehyde were found at moderately
  higher levels in tobacco.

I have mostly understood the study but I do need some things clarified.
Question 1: Could the combustion of any plant material whether it is an edible herb or in my particular example, tobacco or marijuana, potentially produce similar products or by-products (e.g HCN, ammonia), because of the same functional groups present in the different compounds of the tobacco or marijuana plant?
Question 2: Would all of the molecules in the figure below which are found in the tobacco plant, be independently reacting with oxygen and what is the reaction of nicotine producing HCN and/or an aromatic amine which if I remember correctly is a cyclic molecule?

Picture source

Question 1 is easy to answer: Yes.
The ‘burning’ of whichever plant you’re (or somebody else is) smoking is an incomplete combustion on the ember end of the cigarette/joint/whatever. In any incomplete combustion of nitrogen-containing organic compounds, $\ce{HCN}$ and ammonia can (and likely will) be a side product. Aromatic amines are probably also always generateable as are $\ce{NO_x}$. And PAHs, aldehydes etc. can also result from the incomplete combustion of practically all plant or animal material.
Question 2 is slightly harder. To the best of my knowledge, however, the answer is: We do not know.
While we may be able to say that all the molecules in your question would be independently reacting, the mechanisms are so many and so complex that I very much doubt they have been the target of investigations. I think I recall somebody once studying the reaction mechanism of the complete combustion of methane and coming up with tens if not hundreds of partial reactions and steps. That was complete combustion (i.e. enough oxygen present), so it all boiled down to:
$$\ce{CH4 + 2O2 -> CO2 +  2H2O}$$
In incomplete combustion, we cannot even be fully sure of the reaction equation (and the stoichiometries involved therein), let alone a mechanism.
Side note: the complete combustion of a cigarette in pure oxygen takes about seven seconds ($\pu{7s}$) and oxidises it entirely to $\ce{CO2}$ and $\ce{H2O}$ (plus something from the nitrogens and likely $\ce{SO3}$ from sulphur/$\ce{P4O10}$ from phosphorus)

How would one go about ensuring the complete combustion of Marijuana? Asking for a friend.

@NonSecwitter Under high-oxygen conditions as shown in this video. Note: don’t try this at home.


text:
Calculate the number of atoms of oxygen present in 1.3 mol of sulfuric acid

Calculate the number of atoms of oxygen present in $\pu{1.3 mol}$ of $\ce{H_2SO_4}$

1 mol has $6.02\times 10^{23}$ atoms
So $\pu{1.3 mol}$ must have $1.3 \times 6.02 \times 10^{23} = 7.826 \times 10^{23}$ atoms

Since there are 4 oxygen atoms  out of 7 atoms in total in $\ce{H_2SO_4}$,
${4 \over 7} \times 7.826 \times10^{23}$
So there are $4.472 \times 10^{23}$ oxygen atoms present.
However the answer key says that $3.1 \times 10^{24}$ atoms is the answer.
What went wrong?

The key about the unit mole is that it does not reference which type of entity is specified. Using the mole only makes sense if you define what you are using the mole for. This could be:

protons
atoms
ions
molecules
formula units of an ionic or metallic solid
aggregates
apples
stars
…

The mole is basically just a large number that you can use instead of e.g. the number 1,000,000 or the number 0.5.
So in your question, $\pu{1.3 mol}$ of $\ce{H2SO4}$ means that you have $1.3 \times 6.022 \times 10^{23}$ molecules of $\ce{H2SO4}$, each of which contains one sulphur atom, two hydrogen atoms and four oxygen atoms. Thus, in one mole of $\ce{H2SO4}$ you can find:

$\pu{2mol}$ of hydrogen atoms
$\pu{1mol}$ of sulphur atoms
$\pu{50 mol}$ of protons
$\pu{50mol}$ of electrons
and $\pu{4mol}$ of oxygen atoms.

The correct solution is thus:
$$n(\ce{O}) = 4\times n(\ce{H2SO4}) = 4 \times 1.3 \times 6.022\times 10^{23} = 3.13 \times 10^{24}$$

Does 50 moles proton means 50 x 6,022 10^23 units of protons?

@Lifeforbetter Not units of protons, individual protons. Protons are countable and a mole is essentially just a shorthand for big number.

There are 1.3 mol of the entire acid molecule, so there are four as many moles of oxygen as there are of the acid, because one acid contains four oxygens.
$4×1.3=5.2$ and 5.2 times Avogadro's constant is (about) $3.1×10^{24}$ atoms.
Your error is in dividing by seven, which is neither conceptually nor mathematically correct.


text:
Structure of compound
The synthetic scheme below makes reference to a reagent $\ce{(CH_3)_3COCl}$  that is added at some points in the synthesis.
What kind of structure does $\ce{(CH_3)_3COCl}$  have? The closest structure I can imagine is that of $\ce{(CH_3)_3CCOCl}$ with an extra carbon for the 3 methyl groups to be attached to. $\ce{(CH_3)_3COCl}$ seems to imply that the 3 methyl groups are attached to the carbonyl carbon (which is impossible).

Image source: IChO 2015, Problem 6

$\ce{Me3COCl}$ is possible but would require a chlorine-oxygen bond. Seems like a typo to me …

Tert-butyl hypochlorite, cas no 507-40-4
Reagent for N-chlorination of amines and amides


text:
Which element has larger atomic radius -- beryllium or xenon?
I say that xenon radius is smaller comparing to beryllium, but my friend says otherwise.
What I think is that xenon is down in periodic table comparing to beryllium (3 periods down) and it's far toward the right (16 groups ahead). So, the nuclear pull dominate (16 by 3 which is very big) and therefore xenon's radius is smaller than beryllium's.
My friends said that xenon is down in periodic table compared to beryllium and hence it's bigger and the nuclear pull effect has only little effect.
So who is right ?

You cannot really effectively compare the atomic radii of xenon and beryllium empirically; the atoms are too far apart. Once you remove yourself from direct neighbours, it is generally not possible to estimate relative sizes of atoms.
The closest hint you can get are calculated atomic radii on Wikipedia’s data page. These show that xenon is slightly smaller ($\pu{108 pm}$ to $\pu{112pm}$) but the difference is smaller than the expected error. So, for all intents and purposes they have the same size.

How about if they are not far apart, such as Beryllium with Sulfur or Beryllium with Phosphorus ? Magnesium with Yttrium or Magnesium with Niobium ?

@JasonC Unless they are directly above/below/left/right of each other or top-right/bottom-left (in all cases of which you can draw a series of linear larger or smaller relationships), they are too far apart.

@Jan The  radius given for Xenon is Van Der Waal radius and the radius given for Be is atomic radius. Will it be fair to say Xenon is slightly smaller based on this data?

@Eloise No, it's the calculated atomic radius in both cases. https://en.wikipedia.org/wiki/Atomic_radii_of_the_elements_%28data_page%29

I have been taught that it is not possible to measure atomic radius for noble gases. To measure the atomic radius of Xe, we need a compound in which Xe-Xe bond, which does not exist. Then how is this atomic radius of Xe even calculated?

@Eloise I did make sure to use values in the same column. Other columns make xenon larger. My point is that one cannot predict it going by periodic trends.

If you want to compare the radius of 2 elements, you can usually do it when the elements are either on the same column (group), or the same row (period).
When you move down the column, the element becomes bigger, because it has more energy shells. For example, Oxygen has a smaller radius than Sulfur.
When you move right the row, the element has larger nuclear charge (more protons) for the same number of energy shells. Therefore, each electron is pulled stronger and the radius decreases. For example, Fluorine has a smaller radius than Oxygen.
Now, your elements, Berylium and Xenon, are not in the same column nor row. Nevertheless, Xenon is much lower than Berylium so we might assume that it has a larger radius.

So in trend of atomic radius by group or period, if we comparing with two atom with different colomn and period, we should compare both by how low they are ? For instance, Magnesium and Scandium. Since Scandium is lower than Magnesium by 1 period, hence Scandium has bigger radius. Since Sulfur has lower period than Berylium, hence Sulfur has bigger atom radius. Am i correct ?

No. For example, Strontium has a radius of 200 picometer, and Potassium has a radius of 220 picometer, even though they are in different columns and different rows, and Potassium is in a higher row (values from "Atomic Radius", Wikipedia). However, when there is a very big difference we can say.

So Berylium is bigger than Sulfur because Sulfur is lower than Berylium by 1 period and 4 groups ahead, so nuclear pull dominate. But you said earlier that Xenon is bigger than Berylium. So what should i do if given two element with different group and period. (I'm still learning with trend of group and period, so maybe ignore the numerical value). Also what do you mean by "big difference"? Big difference on period ?

I assumed that Xenon has a larger radius. However, I couldn't find online a measured atomic radius of Xenon. Only a calculated value. And according to the calculated one, Beyllium has a larger radius (112pm) than Xenon (108pm). This shows exactly my point. We cannot compare them. We can only guess. And I was wrong. There is no real value for "big difference", only speculation.


text:
Decomposition speed of ozone

Elementary reactions of decomposition of ozone with chlorine are as follows (substance with $^*$ is reaction intermediate):
  \begin{align}
\ce{Cl2 + O3 &-> ClO^* + ClO2^*}\tag{1}\\
\ce{ClO2^* + O3 &-> ClO3^* + O2}\tag{2}\\
\ce{ClO3^* + O3 &-> ClO2^* + 2O2}\tag{3}\\
\ce{2ClO3^* &-> Cl2 + 3O2}\tag{4}\\
\ce{2ClO^* &-> Cl2 + O2}\tag{5}
\end{align}
  Calculate decomposition rate of ozone using steady-state approximation.

I have this physical chemistry homework, and tried to solve, but it was too complicated for me to do. Could anyone help me?
(I'm sorry, but I'm not good at English, so maybe my sentences are strange.)

Write down the rate laws for all steps. Set the intermediate concentrations to zero. Solve.

Show your calculation in the question. Describe what effort you have done in the question. Otherwise this question will treated as mere homework question and will be closed.

Isn't Cl radical a key factor in the chlorine catalyzed destruction of ozone?


text:
Why atomic number decreases in alpha emission
My teachers taugh me that alpha particle is a helium nucleus carrying net +2 charge, hence if an radioactive subastance emits alpha particle it looses two of its protons and two neutrons but no electrons. Hence its atomic weight will decrease by 4 units. But books says its atomic number also decreases by 2 units and the atom gets transfromed into another atom e.g
Bi(z=213/A=83) >   Tl(Z=209/A=81)
Since electrons are not lost how can the atoms change, only its nucleus must change , and if the atom change then the daughter atom must have -2 charge since the electrons are not lost .
Please help me out.

While chemical properties are determined by the electronic structure of an element, that structure is, in turn, caused by the number of protons in the nucleus of the element (which is the atomic number). And it isn't the number of electrons that characterises and element (strip electrons from an iron atom to give Fe2+ or Fe3+ and we still recognise the element as iron.)
What matters in radioactive decay is what changes in the nucleus. If a nucleus loses two electrons it becomes a different element and the electronic structure will be different. Usually the orbital electrons are ignored as what happens during the decay involves the nucleus only. After the nuclear decay the orbital electrons may rearrange in chemical processes.
We don't normally account for the orbital electrons in nuclear reactions as the orbital electrons are engaging in chemical not nuclear processes and. Besides, there are nuclear processes that involve electrons (neutrons in some nuclei can emit electrons and protons in others can capture them: these processes do involve electrons but it would be confusing to mix accounting for them with accounting for the chemical changes that result from things happening to orbital electrons).
Nuclear chemists only count what happens to the nucleus and ignore the ensuing chemical changes.

In nuclear decay we are normally interested in the behavior of atomic nuclei.
Electrons from the atom's electron shell are considered to be spectators only and are ignored.
Since the number of protons in the atomic nucleus decreases by two in alpha emission, the atomic number also decreases by two. However, the principle of electroneutrality tells us that the atom will also lose two electrons from its electron shell.

So electrons will also be emiited correct. Hence we should get spots for negative and positve charge on electric plates placed to trace particles.


text:
Why does benzyne dimerise and not trimerise?
Why does benzyne prefer to undergo dimerisation to stabilise itself, even though the product formed (biphenylene) is anti-aromatic?

Why does it not undergo trimerisation through the following pathway, resulting in an aromatic and presumably more stable compound (triphenylene)?

Central rings of biphenylene and triphenylene are approximately non-aromatic, I can see why you could think otherwise though

Actually, trimerisation does happen. See e.g. Chem. Rev. 1962, 62, 81. Biphenylene as a whole is considered to be aromatic. See e.g. Z. Naturforsch. 1973, 28B, 478 or J. Phys. Org. Chem. 2010, 24, 263 or J. Am. Chem. Soc. 1996, 118, 2903. It is a pretty stable compound in all respects.

Also, 3-body collisions are unlikely to occur, even more so when a reactive intermediate such as o-benzyne is involved.

related https://chemistry.stackexchange.com/questions/57678/is-compound-aromatic-if-it-has-also-anti-aromatic-ring

So, is bipheylene the kinetic product and triphenylene the thermodynamic one?

When reactive intermediates such as o-benzyne are involved, there is no equilibrium, hence thermodynamics are not involved.  All products produced are kinetically derived

English does not capitalise many words (but therefore capitalises some other languages don’t). Most notably, chemical terms such as dimerisation or anti-aromatic are practically never capitalised. I therefore rolled back your breaking (not fixing) error.

As per Wikipedia,

Biphenylene is quite stable both chemically and thermally, and behaves in many ways like a traditional polycyclic aromatic hydrocarbon. However, both the spectral and chemical properties show the influence of the central [4n] ring, leading to considerable interest in the system in terms of its degree of lessened aromaticity.

This implies that biphenylene is rather stable, despite its anti-aromaticity.
As for triphenylene, I'm not sure, and it doesn't say so anywhere, but perhaps the free energy of formation of triphenylene is much more, as it requires more molecules to be put together, leading to the reaction being non-spontaneous. However, benzyne does trimerize as well, forming triphenylene.
So benzyne can actually undergo both dimerization and trimerization, forming two different, but stable products.
As requested, a more primary source is available here. This site cites other sources, whose links I have provided in the quotation.
These lines say more or less the same thing:

Several recent publications (2 3 4 5 6) concluded that biphenylene possesses signiﬁcant strain energy of the order of 50–60 kcal/mol. These numbers are at least confusing since molecules that possess such tremendous strain can hardly exist. In the meantime, biphenylene is a stable molecule that has been synthesized a long time ago while its chemical properties show typical aromatic behaviour.

Whether or not what's written in Wikipedia is accurate, please provide references to more reliable primary literature if possible.

Many times when I quote from WP, it includes its own references, which I then look up in the References section of the article and compile in the answer as Cited References.

Just to add to the above answer with a possible theory of why this might be happening-:
The central ring in biphenylene is majorly non aromatic as mithoron pointed out in his comment. My teacher says that the central ring has two bond lengths as a result of Jahn-Teller effect. Because of the longer bonds in between the rings, conjugation is broken between the p orbitals of the atoms in the central ring making it non-aromatic. Hence it does not share its properties with other anti-aromatic compounds.
And biphenylene being the major product during benzyne polymerisation reaction is due to the less probability of 3-body collisions as pointed by ron. Benzyne being a di-radical is highly reactive hence kinetics plays a major role as compared to thermodynamics.


text:
If a gas always occupies the volume of its container, will its volume always be 22.4L at STP?
According to the Avogadro's law, an equal volume of gases have a equal number  of moles (with constant pressure and temperature), therefore STP molar volume is $\pu{22.414 dm3 mol-1}$.
Think this: There is a container of volume $\pu{40 dm3}$ without any gases. I fill it with 1 mole of $\ce{O2}$ gas ($\pu{32 g}$), therefore according to the STP molar volume, the volume should be $\pu{22.414 dm3}$.
But we know that a gas always take the volume of its container. In above problem, the container volume is $\pu{40 dm3}$. So the volume of the gas should be $\pu{40 dm3}$. But why $\pu{22.414 dm3}$? I am confused with it.

The pressure in STP is no longer one atmosphere = 101.3 kPa.  In 1982 it was changed to exactly 100 kPa.  That raises thd standard volume to 22.71 L.

According to Avogadro's law, equal volumes of gases at the same temperature and pressure have an equal number of particles. If you had one mole of particles at STP it would occupy $\pu{22.4dm3}$
However according to Boyle's law, pressure is inversely proportional to volume. Hence, by increasing the volume like you did, the vessel's pressure will decrease, thereby not making the measurement to be at STP. Hence, it will not be $\pu{22.4dm3}$.


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Query in Chemical Kinetics
I have been recently studying chemical kinetics. I've understood what the rate law and the rate constant mean, and also that, kinetics depends completely on experimental observations, due to which we have zero order reactions as well as fractional order reactions.
Take for example, this: $$\text{Rate of a certain chemical reaction} = k [A]^p [B]^q [C]^r $$
Here, $[A]$, $[B]$ and $[C]$ are the molar concentrations of the three reactants involved, while $p$, $q$ and $r$ are the number of molecules of each respectively taking part in the reaction.
So, $\text{the order of the reaction} =p+q+r $.
Now, this $p$, $q$ and $r$ are completely determined experimentally, and cannot be determined theoretically.
How are these values determined experimentally? How does one find out how many molecules of each reactant are taking part in the reaction? Is there any special instrument to do this? If not, then how do scientists find the order of a reaction?
As an attempt to shorten out the answering area, consider this reaction: $$\ce {NO2 +CO -> NO +CO2} $$ Now, experimentally it has been determined that $\text {the rate of this reaction} = k[\ce{NO2}]^2$.  I want to know how have scientists determined that the rate depends on the square of the concentration of $\ce {NO2} $ and not on the other reactant.
N.B.: My question is how it can be experimentally determined, and not why it has to be experimentally determined.

One important thing to note is that p, q, and r are only the number of molecules of each involved if this the rate law for an elementary reaction (something which takes place in one step). Otherwise, they are as you said just empirical parameters that can take on negative and fractional values and don't directly relate to how many molecules are involved in the reaction.

@Tyberius OK, I get that. I just called them number of molecules to express myself. You can edit to express them better.

Chemguide has a decent intro: http://www.chemguide.co.uk/physical/basicrates/experimental.html

If you are looking more in terms of the experimental setup, I would look at orthocresol's link. If you are more interested in the generic methodology, the Wikipedia article on order of reaction gives a good description.

The comment to this question answers some of your question https://chemistry.stackexchange.com/questions/117236/measurement-of-concentration-during-a-given-reaction-to-determine-rate-laws#comment221973_117236

Are you asking about the logic of the experiment (independent variable, dependent variable) or about the details of conducting a kinetic measurement? If it is the former, @user80623 answers your question if I understand what it says. If it is the latter, your question is too broad for this format. Please edit your question so that it is possible to answer.

Order of a reaction can be found by the following methods:
1. Plotting a graph of concentration of reactant versus time
For example, let us take decomposition of $\ce{N2O}$ to $\ce{N2}$ and $\ce{O2}$ on $\ce{Pt}$ surface.
$$\ce{2N2O ->[Pt] 2N2 + O2}$$
The graph will look similar to the following:

The slope of the graph is constant. That is, it does not vary with time.
This is possible when the reaction is zeroth order. Hence, we can conclude that the reaction is zero-order.
For a $\mathrm{1^{st}}$ order reaction, the slope of the graph drawn between $\ln[\mathrm{Reactant}]$ with time is constant, for a $\mathrm{2^{nd}}$ order reaction, the slope of the graph drawn between $\mathrm{\frac{1}{[Reactant]}}$ with time is constant.
In general, for an $n^{\text{th}}$ order reaction, the slope of the graph drawn between $\mathrm{\frac{1}{[Reactant]^{n-1}}}$ with time is constant (where $n \neq 1$).
2. Half life method
This is useful only when a single reactant is present.
In this method, half life of a reaction is measured, varying initial concentration of the reactant.
$$t_{\frac{1}{2}} \propto \frac{1}{\mathrm{[A]^{n-1}}}$$ where $n$ is the order of the reaction and $\mathrm{[A]}$ is the initial concentration of the reactant.
3. Initial rate method
In this method, the initial rate of the reaction is measured at different initial concentrations of the reactant.
Let the reaction be $n^{th}$ order w.r.t. a reactant $\mathrm{A}$, then
$$\mathrm{r_0 = k[A]^n}$$ at different initial concentration of $\mathrm{A}$, say $a_1$ and $a_2$, the observed initial rates be $r_1$ and $r_2$, then
$$n = \frac{\log(r_1)-\log(r_2)}{\log(a_1)-\log(a_2)}$$
In case of multiple reactants, say for a hypothetical reaction
$$\ce{11A + 20 B -> 5 C + 9 D },$$ you want to find the order of it.
Let the rate equation be $$ r = k[\ce{A}]^n[\ce{B}]^m$$
First take a reactant in excess and find the order of reaction with respect to other reactant by comparing initial rates or half lives.

You know that rate of a reaction is proportional to molar concentrations of each of the reactant with each term raised to the number of molecules participating in the simultaneous collision i.e reaction. This law is called Law of Mass Action. This is true only for elementary reactions, where all reactant molecules simultaneously collide to form products.
For example:
$$\ce{H2(g) + I2(g) -> 2HI(g)}$$
is a elementary reaction and its
$$\text{rate}=k[\ce{H2}]^1[\ce{I2}]^1$$
which matches with experimentally determined rate.
But the complex reaction:
$$\ce{NO2 + CO-> NO + CO2}$$
is a two step reaction with the formation of an intermediate; $\ce{NO3}$. The mechanism of the reaction is as follows:
Step 1(Rate determining step):
$$\ce{2NO2\xrightarrow[ ]{\text{slow}}NO3 + NO}\label{eqn:1}\tag{1}$$
This step is slower compared to $(\ref{eqn:2})$. Hence the rate of the overall reaction depends on this step.
Therefore overall,
$$\text{rate} =k[\ce{NO2}]^2$$
Step 2:
$$\ce{NO3 + CO\xrightarrow[ ]{\text{fast}} NO + CO2}\label{eqn:2}\tag{2}$$
These are the steps through which the reactants have to go in order to form products. The rate determining step(RDS) will be always the slowest because it has the highest activation energy (forms an unstable product).
Note that each step is an elementary step and law of mass action is separately applicable to each step.
In the case  of elementary reactions reactant molecules collide simultaneously giving product molecules with no formation of an intermediate. Hence, law of mass action can predict rates of only such reactions.
Mechanisms can be verified experimentally, with the help of intermediates. There are various method of determining rate of a reaction:

Initial rate method: Maintaining concentration of one reactant
constant and varying the other.
Optical rotation method (if reactants are optically active):
Measuring the net optical activity of reaction mixture.
Titration: Titrating the reaction mixture.
Barometry: Measuring pressure of container of gaseous reactants.

Thus the exponents in the rate law can be determined.


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How is group theory (and matrices) used in symmetry operations?
I am a high school IB student. For my 2-year math course, there is an internal assessment called "exploration" which should be max. 12 pages long. The goal of this assignment is for students to explore areas or uses of mathematics beyond the course.
Now onto the important stuff. After doing some research on topics to investigate in linear algebra, number theory, graph theory or group theory, I came across a potential subject: "Investigating the use of group theory (and of matrices) in symmetry operations in chemistry."
I know my question very broad but I am here looking for material to learn from. Please post any useful links or names of topics to look at please!

related https://chemistry.stackexchange.com/questions/19477/symmetry-representations

This is primarily opinion-based/too broad, there are too many equally valid answers. It would suit a forum very well, but probably not Stack Exchange. On topic, you're better served by looking up a good textbook on the matter. Cotton's Chemical Applications of Group Theory is the classic, but it is rather tough to read, I don't expect an IB student to be able to read it. The group theory chapter in Atkins' Molecular Quantum Mechanics might be more accessible. As for websites, a simple Google search for "molecular symmetry and group theory" should turn up plenty of resources.

Possible duplicate of Resources for learning Chemistry

also have a look at the answers/comments  to this question https://chemistry.stackexchange.com/questions/58609/understanding-group-theory-easily-and-quickly/58613#58613


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What happens when aluminum reacts with tert butyl alcohol?
I wanted to know that what happens when aluminum reacts with tert butyl alcohol..?

Organic Syntheses 1941, 21, 8
Aluminium and tBuOH heated in benzene in the presence of catalytic Mercuric Chloride forms Aluminium tris t-butoxide


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How would we know if an interhalogen compound would be a gas/liquid at a particular temperature?
Ex- how can we understand that ClF is a gas @ 298K? Is there some logic behind it? Please explain.

If you know that chlorine is a gas then you can infer that ClF is also a gas.  The most important intermolecular force in interhalogens is dispersion.  Chlorine has stronger dispersion forces than ClF.  The dipole-dipole forces can be neglected in almost all cases.


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Which is more likely to form: silver chloride or silver phosphate?
If silver nitrate is added to an aqueous solution (such as tap water) that contains both chloride ions and phosphate ions, which precipitate is more likely to form: silver chloride, silver phosphate or both should form?
And how do you know?
This is NOT a homework question. I'm 32 and learning chemistry on my own as a hobby and doing experiments using a Chemistry Kit for kids.
The reason I'm asking the question is because I did the experiment on the tap water at home and I didn't get any silver phosphate precipitate, but I did get a milky white appearance which I'm assuming is silver chloride (or another halide). I'm wondering if I got no silver phosphate precipitates because all the silver ions were "used up" by the chloride ions or simply because the concentration of phosphate ions is either zero or too small.
I tried to figure out if silver chloride formation takes precedence (so to speak) over forming silver phosphates using enthalpies of formation, but I got lost with that a bit.

Depends on the concentrations of the ions and the solubility products of the precipitates.

The key concept here is solubility products which are a form of chemical equilibria. Once you have understood that, you will realise that it depends on the ions’ concentrations and relative solubilities which one will precipitate first.
In general, you have to consider the following principal equilibria:
$$\ce{Ag3PO4 v + 3 Cl- (aq) <=>[$k_1$] 3 Ag+ (aq) + 3 Cl- (aq) + PO4^3- <=>[$k_2$] 3 AgCl v + PO4^3-}\tag{1}$$
and futher ones that derive from the acidity equilibrium of the phosphate anions as well as the formation of silver hydroxide at higher $\mathrm{pH}$ values. I will choose to ignore these.
Equilibrium $k_1$ is actually only the precipitation of silver phosphate; we might instead write it as:
$$\ce{3 Ag+ (aq) + PO4^3- (aq) <=>[$k_1$] Ag3PO4 v}\tag{2}$$
The equilibrium constant is thus:
$$K_1 = \frac{a(\ce{Ag3PO4})}{a(\ce{Ag+})^3 \times a(\ce{PO4^3-})}\tag{3}$$
We are dealing with activities here; the activity of a solid is, by definition, always $1$.
This equilibrium is usually given in the inverse form in which the dissolved ions are the product and the solid salt is the reactant; in that case $K_1{}^{-1}$ is what you might know as the solubility product.
The same discussion can be made for the other equilibrium:
$$\begin{gather}\ce{Ag+ (aq) + Cl- (aq) <=> AgCl v}\tag{4}\\[1em]
K_2 = \frac{a(\ce{AgCl})}{a(\ce{Ag+}) \times a(\ce{Cl-})}\tag{5}\end{gather}$$
Again, the more common form is $K_2{}^{-1}$ which is also known as the solubility constant.
Knowing the values of both $K$ as well as the initial concentrations of all ions will allow you to calculate in which direction the reaction goes although the calculation may be tedious.


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Why can't oxalate ion donate two pairs of electrons from the two double-bonded oxygen atoms
So, I read that an oxalate ion is a bidentate ligand but since I saw that it has 4 oxygen atoms I was confused as to why it isn't a tetradentate ligand instead.

I searched it on Google and found this:
Oxalate ion dentiticity - why is it bidentate and not tetradentate?
Now, in the link it's implied that the oxalate ion can donate the 2 pairs of electrons either from the 2 oxygen atoms having negative formal charge or from 1 "double-bonded" O atom and 1 negatively charged O atom.
But why can't the two pairs of electrons be donated from Both the "double-bonded" O atoms?
 Why isn't this "third" bonding mode possible?
EDIT:  From the comment below, I infer that because of resonance, there is no double-bonded or negatively charged O atom in the Oxalate ion; rather all of the O atoms have a partial negative formal charge and double-bond character, is that correct?
So, can electrons be donated from any O atom in the ion?
(And so did the answer in the provided link just ignore this for the sake of simplicity?)

Well, you seem to don't understand that in oxalate all 4 atoms are identical. Write mesomeric structures if you need.

@Mithoron, so in an oxalate ion there is no two "double-bonded" oxygen atoms as such, rather all of the oxygen atoms have partial negative charge and also partial double-bond character? Then did the answer in the provided link just ignore this in order to give a simple explanation and in fact the two pairs of electrons can get donated from Any two adjacent O atoms?

Yeah, it's pretty much as you say.

I am very tempted to find a cool crystal structure with tetradentate oxalate (but it’s past midnight and I need to sleep). However, for coordination to a single metal cation, the four-membered ring that would form if you were to use both oxygens of a carboxy group is too small. Hence bidentate ligands always form at least a five-membered ring with the metal.

Oxalate is a planar molecule. The four oxygen atoms share the pi electron density through resonance, as noted in comments, and thus are all identical. Imagine those four oxygen as four corners of a table. Now you are a Lewis acid, perhaps a transition metal cation, and you walk up to the table. How many of the four corners ligate you? Perhaps you imagine the planar ligand dropping on top of you, so that the four corners are around you. For effective overlap with all four corners of the table, you'd need to be a pretty big cation. In almost all cases you'd achieve better orbital overlap, and a more energetically favorable coordination complex, if you just cozy up to two of the donor oxygens.

I don't understand this answer. Where do the tables come from?

It's a metaphor, the four oxygen atoms can be likened to the four corners of a table


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Improving Elephant Toothpaste experiment
we are trying to perform the Elephant Toothpaste experiment involving the reaction of yeast and Hydrogen Peroxide for the students at school.
However, the results have been very underwhelming, likely because we have to use the 3% version of Hydrogen Peroxide.
My question is: What are some other ways to help create a stronger propulsion or thicker foam?
Some things I have been experimenting with are reducing the water mixed with the yeast, and swirling the bottle to mix the yeast and Hydrogen Peroxide. Any advice is much appreciated.

You might try horseradish as a catalyst, rather than yeast. Use the fresh root, not the prepared sauce, and test a few different plants - some are more effective than others.
Cosmetic supply stores stock 6% (20 volume) $\ce{H2O2}$, but that concentration is more likely to cause skin irritation or bleach fabrics.
The most effective catalyst for decomposing  $\ce{H2O2}$ that I've used is potassium permanganate, $\ce{KMnO4}$, but it has a few liabilities:

It's a strong oxidant and can start a fire when mixed with organic chemicals.
It stains skin and fabric anything from pinkish-purple to black. The stain can be removed with sodium bisulfite, $\ce{NaHSO3}$, sold as an iron-stain remover. That can be a fun demonstration but is not suitable for young students to use.

I spent a considerable amount of effort improving this demonstration and ended up with this (10 sec. video).
The best result I found was from the following:

Concentrate store-bought 3% $\ce{H2O2}$ to ~30% $\ce{H2O2}$ (or buy the food-grade 30-35% online). Meticulous method is here (6m video).
Make a solution of 1:1 $\ce{KI:H2O}$ by mass, e.g. 50g of each.
Add some dish soap and food coloring to a large Erlenmeyer flask, then add 100 mL of the 30% $\ce{H2O2}$ solution.
Add the 50g/50g solution of $\ce{KI}$ in water.

$\ce{KI}$ is easily purchased online. If you want more of a "toothpaste" effect than the "snowy" effect in the above video, mix the soap/food-coloring/$\ce{H2O2}$ thoroughly before adding $\ce{KI}$.
Safety notes:
KI is mostly harmless -- but as in all cases, completely avoid contact with eyes/skin. The iodide will dye your skin brown-yellow (like the hospital stuff used for disinfecting the skin). Concentrated $\ce{H2O2}$ can be very nasty and oxidize your skin, turning it white for a few hours. So use gloves and goggles in all cases. The clean-up for the above video wasn't bad at all but we tossed the towels used to do it.

H2O2 can cause eye damage. When used with reasonable care, it is unlikely to cause anything but temporary skin bleaching - if kept away from the eyes. That and maybe some ruined clothing (or other surfaces). I'd not use 30% around children, although as a controlled demonstration -physically separated from the audience, it can be safely used. First thing you could do, depending on your resources, is to determine if your 3% is actually near 3%. There are tables which list its density, if you can weigh some out into a volumetric flask, say. If you have a chemical refrigerator, it should be stored cold (and of course, away from sunlight). I guess if it were me, I'd buy some fresh 30% and dilute it for the kids to use (prior to the session). I'd use potassium iodide rather than yeast (what is it that you're trying to teach?) and possibly consider FeCl3 (if the lesson is about catalysts) - although staining would be more of an issue with the iron salt. Make sure that the peroxide isn't phosphate stabilized (if possible). If you really want to use yeast, make sure it's fresh, and I'd give it at least a minute or two in 35°C (95°F, which is just warm, not hot) water. Try different brands. Also consider your liquid dishwashing soap - they're not all the same. Some sources suggest adding glycerol to stabilize the bubbles - I've not tried that, but it seems reasonable (my old bubble recipe calls for 0.5 cups DW soap and 2 Tablespoons glycerine...). No reason you shouldn't add the KI to the soap (technically, it's a detergent) - you'd have to futz around to see if any water was needed and if so how much. At 3%, that means (do the math) that 100 ml will generate ~2 liters of O2 (at room temperature) - so a 20X total expansion maximum (ignoring volume increase due to steam formation from exothermicity). Figure the reaction is only partially complete and that figure drops. Seems to me the important thing is mixing - how the two solutions are added to one another. If it were me, I'd add the soap to the peroxide...although if I was leading a bunch of kids maybe I'd use them to try out different techniques. Proper eye protection is, imho, eye goggles - not safety glasses. This experiment is energetic and uncontrolled, use caution (with anything higher than 3%-6% peroxide).


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Does aqueous component contribute to the equilibrium constant?
Consider an equilibrium reaction with an aqueous solution $\ce{A(aq)}$ in equilibrium phase with gas:
$$\ce{A(aq) + B(g) <=>[$K_\mathrm{eq}$] C(g)}$$
Would the equilibrium constant have aqueous solution $\ce{A(aq)}$ contributing to it?
In my opinion I think it would. Can anyone please correct me? Also, please help me  find literature which explains equilibrium constant in details.

When a substance is distributed between two phases chemists define a partition coefficient.

If you are talking about a species dissolved in a aqueous solution then yes it would feature in the constant .
However if you are referring to the solvent itself then no it wouldn't feature in the constant this is because the concentration of the solvent is constant , as well as the fact that it is present on both sides of the equation since both reactants and products are dissolved in the same solvent
If you are referring to a partial pressure equillibrium constant , then a hybrid one will have to be created that has the partial pressure of the gases and the concentration of species that are not gases
Using a Kc value would mean simply taking the concentration of both the gases and the solutes , to end up with a constant , however solvents are excluded because the concentration of solvents is constant and is thereby thought as being absorbed in the eq constant itself
Derivation of the equilibrium constant requires probability and is quite lengthy so maybe you can check the video on khan academy that discusses the derivation

I am referring to the solute and it is the concentration equilibrium constant,so It is factored in.If it is partial pressure equillibrium constant ,How would you state the pressure of the solute ?

As I stated before it would be a hybrid constant , one that takes in the partial pressure of gases and the concentration of the solute , so you don't take the pressure or the solute but the concentration of it


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Dangers of corroded mercury battery
Disclaimer: Sorry if this is not a pure chemistry question. If it's better suited to another SE, I'm happy to have it migrated there.
I'm in the process of cleaning my father's workshop, and recently discovered an old Duracell mercury battery (picture from Wikipedia):
The battery is corroded, and I would say that the color of the corrosion is brown or dark orange.
The battery was lying in a pile of other things on a workbench. Presumably, the battery has been lying there for many years. As things have been moved around on the workbench over the years, one can expect the corrosion to have been scraped off and been mixed with sawdust, smeared on tools etc.
My concern is that I should basically treat the whole workshop as being contaminated, but still I feel it might be a bit over the top.
I would like some information on just how toxic the corrosion is, and some suggestions for what I should do. Could I swab some sample surfaces to have them tested? How expensive would this be?

Based on the color alone, the corrosion could be mercury (II) oxide, which is nasty. As with other mercury salts, it can end up in your body in a variety of new and exciting ways, such as by inhalation (that is not to say that elemental mercury cannot do so, it is simply more difficult), which make it more dangerous than elemental mercury. For this reason, do not touch it, even if it ends up being another salt.
What I (and I am no hazmat professional) would do is search around on the Internet for my local hazardous materials disposal facility (it may go under a different name) and would call them. I would ask them about the battery whilst giving as much information as possible about it, even if it seems unrelated. If (and only if) they ask you to bring it in, do the following:

Wear at least two pairs of gloves while handling it, and only do so briefly.
Put it in several layers of baggage, something that you will be okay with losing; and make sure it will not break.
Bring it in carefully, being sure not to breach the membrane. Also, be sure not to touch anything with the gloves used.

In addition, be careful with any of the tools that it touched, they could also be highly contaminated. If you want to learn more about mercury (II) oxide, the Wikipedia article seems relatively accurate (the same could be said about mercury batteries): https://en.wikipedia.org/wiki/Mercury(II)_oxide
Disclaimer: As stated, I am no hazmat professional, and my advice does not take authority over the advice of a professional.

Thanks! I decided not to accept your answer yet in order to see if someone else might have anything to add. I was wearing a pair of work gloves when I accidentally picked up the battery, so hopefully the oxide never came in contact with my skin. It currently sits in a glass jar.


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Does carbon monoxide form or carbon dioxide forms when carbon reduces metal oxide?
I would like to discuss this in context of ellingham diagrams.
When $\ce{C -> CO}$ line is below metal oxide line, CO is produced.
Similarly, in other case $\ce{CO2}$ is produced.
But when both lines are below metal oxide, is carbon monoxide produced or carbon dioxide?

If both carbon monoxide and carbon dioxide are thermodynamically favored, then carbon monoxide comes "first" in the sense that the metal oxide first oxidizes carbon to $\ce{CO}$.  This is because the reaction is carried out at relatively high temperature where any potential carbon dioxide would itself react with carbon to form the monoxide.  If more oxide is available after all the carbon is converted to $\ce{CO}$ and further oxidation to $\ce{CO2}$ is favorable, the latter reaction then occurs.

So is it like carbon dioxide forms first but due to high temperature it becomes monoxide on reacting with carbon? Or does monoxide form directly?

No, I am saying that if any carbon dioxide were to form it would not last in the presence of carbon.


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What is the difference between a glassy, amorphous and a vitreous phase?
I have been reading about some inorganic materials lately, and these three terms confuse me. If they all mean the same thing, would still one of them have preference over others?

After some reading I have decided to answer this question myself for anyone else with the same confusion. The info is from the downvoted answer of alphonse and the wiki-link, but with a better overview.

Both vitreous and glassy are amorphous. Amorhous is the broadest term to rule them all.

Some amorphous materials have a Glass Transition Temperature (GTT). Above this temperature the material becomes a viscous liquid. note that both below and above the GTT the material is amorphous.

An amorphous solid that has a GTT is called a glass.
The transition (heating) over the GTT results in a glassy phase.
The reverse transition (cooling) corresponding to the GTT is called vitrifcation. The result is a vitreous phase.

If I made mistakes please comment and I will update the answer.

Mostly a matter of preference. BUT. Consider the Glass Transition temperature. (see https://en.wikipedia.org/wiki/Glass_transition). It is important that you understand that while all glasses are, by definition, amorphous, that not all amorphous solids are glasses. Vitrification is the transition from liquid to a glass, such a glass could equally be called vitreous. I wouldn't call a solid, for example a rubber solid, below its Tg to be vitreous, no matter that it is amorphous. (Of course, if a solid is crystalline, then it's outside the context of this discussion.) There are contextual complexities, of course. Silica glass in it's molten state is still glass, while at the same time is not in the glassy state. So, there are different meanings for the term "glass". For instance, I'd not use it for polymers, even though polymer glasses is a legitimate area of molecular dynamics  study. Glass in the strictest sense is an solid, amorphous, brittle, rigid, inorganic, transparent material. As you remove those qualifiers the term tends to be used less often. (Some would even argue that glass has to be SiO containing, but I wouldn't go quite that far.) I wouldn't describe a solid as vitreous unless its appearance was "shiny" (and it was brittle), but that may be a personal preference rather than being representative of my technical background. My preference is that amorphous materials be called amorphous. If something can exist in both the liquid and the amorphous solid state, then I'd be comfortable with calling the solid state a glass. For polymers, the question is quite a bit trickier. Many polymers are semi-crystalline. And some (quite a few, actually) amorphous polymers do not exhibit a true liquid state (for example, due to physical cross-links). So, I'd be very cautious about the use of the term glass for them, although in certain contexts its appropriate. I'd avoid the use of "vitreous solid" completely, unless speaking about something after having vitrified.