text:
The test of metal hydroxides and metal oxides
Is testing metal oxide different from testing metal hydroxide using a litmus paper?

Litmus paper does not test for either metal oxides or metal hydroxides; indeed, it will not change color at all. Try it: put dry $\ce{CaO}$ on dry litmus paper. Litmus does change color with pH+, the negative log10 ratio of $\ce{H+}$ ions to $\ce{OH-}$ ions (the ions may actually be more complex, such as $\ce{H3O+}$, but the idea is the same). This means the substance is in aqueous solution, i.e. in water.
Litmus also changes color with non-metals in solution, such as $\ce{NH3}$, which forms $\ce{NH4OH}$ in water, or with acetic acid, having an excess of $\ce{H+}$ ions.


text:
Stability between benzyl free radical and tertiary free radical
My teacher told me that the hyperconjugation of 9 alpha hydrogen domintes over resonance (for a carbocation )and so the stability of a tertiary carbocation is more than benzyl carbocations due to exceptional hyperconjugation .
I was convinced to the fact and started applying to the questions but then i got stuck and was confused when i saw the stability order of the following groups of free radical .
According to me the order told by my teacher for carbocation should still be valid for free radical .but in the following picture benzyl free radical is ahead of tertiary free radical in terms of stability.

Not entirely sure but from my experience, resonance provides a far better delocalisation of electrons (or absence of them). So I would agree with the image you've presented.

this issue has been addressed on ChemSE: https://chemistry.stackexchange.com/questions/26570/how-is-a-free-radical-on-allylic-position-more-stable-than-at-benzylic-position/87653#87653

Yes correct....in case of t-butyl cation it is more stable than benzylic cation due to 9 alpha hydrogen hyperconjugation, 3 methyl gropu inductive effect...it is proved by hydride ion affinity that t-butyl cation have ∆H -231 kcal/mol while benzylic cation have ∆H -234 kcal/mol...if talk about any other tertiary cation excpt t-butyl cation then benzylic cation have more stability....and it is the same case in free redical

source: Stack Exchange

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

authors: Αντώνιος Κελεσίδης

sort: votes


text:
Effect of H-bonding on o-nitrobenzoic acid

My argument here is regarding the acidity of o-nitrobenzoic acid being greater than p,m-nitrobenzoic acid. In o-nitrobenzoic acid, there is an intramolecular hydrogen bond which makes the bond stronger and resists the release of H+ ion. Also going by that along with ortho effect in check shouldn't acidity of o-nitrobenzoic acid be relatively lesser? To what extent does hydrogen bonding affect the acidity?

The  intramolecular Hydrogen bonding here is not possible to occur. If you don't see it obviously, I will suggest you to make a $3D$ model to visualise. The hydrogen is far away from the oxygen on the nitro group, as both the group doesn't remain in the same plane. Also, by ortho effect, the acidity doesn't at all become lesser, it is increased for sure.

@Soumik Das. But H can form Hydrogen bond with nitrogen right

@dhaneshvijay Not really.. Because for forming Hydrogen bonds, the other atom should be highly electronegative and most importantly rich in electron density. Here the Nitrogen in Nitro group is positively charged, which makes it electron deficienct and not suitable for hydrogen bonding in this case.

Steric inhibition due to resonance is more important here rather than the H-bonding. Also won't ortho effect make ortho most acidic?

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


text:
How does base size affect the ratio of kinetic/thermodynamic enolate?
One of the factors that favor kinetic products is bulky bases. So, I'd expect that, for a certain substrate, the use of a more bulky base to favor even more the kinetic product.
However, in the table below we see that as the base becomes even more sterically hindered, the thermodynamic product is formed. It's never the main product but it does get produced a little bit more (1% to 10%).
Why is that? Why in the case of Ph3CLi we get more B than in the case of LDA?

Consider an even larger base where the steric interactions in the transition state to form A or B would be close to identical. I think you're getting closer to that limit.

I thought that since either can be formed, the more the reaction is delayed the more the thermodynamic (stabler) product would form. So since bigger bases means more difficult and slower displacement, then more time is given to the reaction to form B.

The temperature is low enough that you're likely not getting any equilibration. What you really want to see is the relative rates between the first 3 rows.

As I've mentioned, I don't expect the rate to become vice versa for the two products, since there is not much energy for it. Regarding my comment it's the only reasonable approach I thought.

It's frustratingly difficult to find a source for the information (it's in Carey/Sundberg Vol A, which in turn cites Augustine's Carbon-Carbon Bond Formation, which I can't access)... In the absence of further information, I think both arguments make sense: either the reaction is purely under kinetic control and increasing the steric bulk decreases $\Delta \Delta G^\ddagger$, or there is a thermodynamic component to it arising from equilibration between kinetic enolate + unreacted ketone.

@orthocresol Could you please elaborate on the decrease of ΔΔG‡ in relation to the bulkiness of the base? As I'm not sure I fully understood the first comment as well. Even in the case of a bulkier base, wouldn't the sterical interactions be larger for B. It is meant that as the base becomes more hindered the energy difference will become smaller (as both TS will experience large steric) so both products would be "able" to form?

Yup, that's precisely what is meant. $\Delta G^\ddagger$ is the Gibbs energy of activation, so $\Delta\Delta G^\ddagger$ is the difference in the Gibbs energy of activation, which is what determines the product ratio under kinetic conditions.

@orthocresol I've checked Carey's book as well and apart from what you've mentioned it is mentioned several times that as the base becomes bulkier, kinetic product is favoured (e.g. page 287, 595). It's perplexing.

One problem with comparing product ratios using $\ce{LiN($i$Pr)2}$ vs $\ce{Ph3CLi}$ is that it's not really an apples-to-apples comparison - one is a lithium amide base, whereas the other is a alkyllithium that is highly stabilised by conjugation.
I think this illustrates two important points:

Things are more complicated than they first appear, ie. are the species we draw really representative of the species undergoing reaction?
Analysis of substrate and reagent structures will only get you so far, eg. for kinetically controlled processes, analysis of competing transition states is necessary.

Regarding the first point, there have been dozens of papers looking into the structure and reactivity of lithium amide and alkyllithium bases, particularly by the research group of David Collum. LDA is probably the most heavily studied. It turns out that these species are much more complex than what is implied by the structural formula. For instance, LDA in a coordinating solvent, such as THF, exists as a disolvated dimer:

How does the steric bulk of this species compare to the LDA structure we normally draw? Likewise, what is the actual structure of $\ce{Ph3CLi}$ in solution?
Even with a particular solvent and conditions, transition states have been put forward for the deprotonation event that vary markedly in structural features between substrates. For example, in a review of LDA (Angew. Chem. Int. Ed. 2007, 46, 3002-3017), Collum and coworkers show the following two different transition state structures for deprotonation of an ester using LDA in $\ce{tBuOMe}$ vs THF:

So, if we can get a good idea of what the transition state for LDA vs $\ce{Ph3CLi}$ deprotonation looks like, the complexity is such that high level calculations, eg. DFT, would be needed to explore the difference in selectivity. Even if we could qualitatively say which transition state involves the more bulky species, there may be features of the transition state, eg. conformational, stereoelectronic, that only become apparent through the modelling. Sometimes, it's not just which reagent is bigger, but details of how the two reactants come together at the transition state that determines the favoured outcome.

I wouldn’t consider (monomeric) Ph3CLi particularly resonance-stabilised, because there’s no way that the lone pair can be conjugated with all three phenyl groups at once. But the inductive withdrawal of the phenyl groups definitely pulls the pKa down quite a bit. Unrelated: Collum’s papers have an amazing sense of humour sometimes.

Yes, he can be quite conversational in some of them. He's also quite forthright on certain issues on social media. Point taken re: degree of resonance stabilisation of trityllithium, although there will be partial overlap of each phenyl ring pi system, so I wouldn't say that the lone pair can't be conjugated with all three at once. Stability of the trityl cation is evidence that the conjugation afforded by partial overlap due to lack of co-planarity of the phenyl rings is still significant.

Fair point re trityl cation!

Goodness me. 2018 me did NOT know about Collum's tweets at all. This is certainly offtopic for SE but I will leave a warning for people to not go near it, and I 100% don't endorse any of it...

authors: Αντώνιος Κελεσίδης


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Why is hexane called skellysolve-B
https://pubchem.ncbi.nlm.nih.gov/compound/hexane
Another name for hexane is skellysolve-B, is that just to be silly?

Presumably that was/is a trade name.

It's probably considerably notorious historical trade name. In many older publications, e.g. "Skellysolve B (petroleum ether)"[1,2] appears for solvent used.
It was a product or trade name of Skelly Oil Company[3]
 founded in 1919.
It's not a pure hexane, rather narrow-boiling hexane oil fraction.[3]
I could find analysis of similar product Skellysolve C, which is C7 heptane fraction:[3]

24.9% heptane
22.9%

2,3-dimethylpentane
2-methylhexane
3-methylhexane

19.3%

trans-1,2-dimethylcyclopentane
trans-1,3-dimethylcyclopentane

15.0% methylcyclohexane
5.5% cyclohexane
5.2% heavy nonaromatics
2.7% toluene
2.5% hexane
…

Other ones:[4]

Skellysolve A is essentially n-pentane, b 28-30°/atm,
Skellysolve B is essentially n-hexane, b 60-68°/atm,
Skellysolve C is essentially n-heptane, b 90-100°/atm,
Skellysolve D is mixed heptanes, b 75-115°/atm,
Skellysolve E is mixed octanes, b 100-140°/atm,
Skellysolve F is petroleum ether, b 30-60°/atm,
Skellysolve G is petroleum ether, b 40-75°/atm,
Skellysolve H is hexanes and heptanes, b 69-96°/atm,
Skellysolve L is essentially octanes, b 95-127°/atm.

Unfortunately, Chemical Abstracts Service, mixes names for different substances, including this tradename: “n-hexane”, “hexane (all isomers)”, “Skellysolve B” under single CAS RN 110-54-3.[5]

References:

Newman, H.; Zilversmit, D. Quantitative Aspects of Cholesterol Flux in Rabbit Atheromatous Lesions. Journal of Biological Chemistry 1962, 237 (7), 2078–2084.
Holick, M. F.; DeLUCA, H. F. A New Chromatographic System for Vitamin D3, and Its Metabolites: Resolution of a New Vitamin D3, Metabolite. J. Lipid Res. 1971, 12 (4), 460–465.
Griswold, J.; Andres, D.; Berg, C. F. V.; Kasch, J. E. Pure Hydrocarbons from Petroleum - Separation of Straight-Run Fractions by Distex Process. Ind. Eng. Chem. 1946, 38 (1), 65–70.
Armarego, W. L. Purification of Laboratory Chemicals; Butterworth-Heinemann, 2017.
CommonChemistry.org – CAS Registry Number: 110-54-3


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Why is fluorine one of the most electronegative atoms even though it doesn't have the largest effective nuclear charge?
Why doesn't electronegativity increase as the effective nuclear charge increases? Most atoms with a higher atomic number than fluorine have a higher effective nuclear charge. However, flourine is a more electronegative atom.

Fluorine is the most electronegative element because the definition of electronegativity makes it so. The electronengativity scales are defined based on experimentally determined properties of the elements. Fluorine has appropriate values for all of the common scales to ensure it has the highest electronegativity.
The Pauling scale, which is the first electronegativity scale proposed, relates the difference in electronegativity $\Delta\chi_{A,B}$ between two atoms $(A\ \& \ B)$ is related to the bond dissociation energies $(E_d)$ of the diatomic species $AA,\ BB,\ \& \ AB$:
$$\Delta \chi_{A,B}=|\chi_A - \chi_B| = (\text{eV})^{-1/2}\sqrt{E_d(AB)-[E_d(AA)+E_d(BB)]/2}$$
Since only differences can be determined, one element (hydrogen) was chosen to have an arbitrarily set value. On the Pauling Scale Hydrogen has a value of $2.20$. Since the $\ce{H-F}$ bond is much stronger $(E_d=586\ \text{kJ/mol})$ than the average of the  $\ce{H-H}$ bond $(E_d=436\ \text{kJ/mol})$ and the $\ce{F-F}$ bond $(E_d=157\ \text{kJ/mol})$. Data from here. In general, since fluorine forms stronger bonds to all other elements than it does to itself, fluorine will have the highest electronegativity.
The Mulliken scale relates electronegativity to two other properties: electron affinity $E_{ea}$ and ionization energy $E_i$:
$$\chi_A = \frac{E_{ea}(A)+E_i(A)}{2}$$
Fluorine has a high electron affinity (easy to gain an electron) and a high ionization energy (hard to lose an electron).


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Chemistry: What's the difference between "binding energy" and "ionization energy"?
In respect to PhotoElectron Spectroscopy (PES), we are analyzing graphs with "binding energy (MegaJoules)" on the x-axis, and with "relative number of electrons" on the y-axis.
I have also seen the x-axis be labeled as "ionization energy." I am really familiar with what ionization energy is (it's the energy required to strip off the highest-energy electron from an atom), but I'm not really sure what binding energy  means. My chemistry instructor says that ionization energy and binding energy are different things, but we were not sure why; since 2 different PES graphs used ionization energy and binding energy interchangeably on the x-axis, are they the same? I've also noticed the trends in ionization and binding energy follow each other-- for instance, lower-energy orbitals (like 1s) have higher binding and ionization energies. Thus, what's the difference between these two terms?

Binding energy is usually used in the context of nuclear physics (fission/fusion calculations) and as a measure of the stability of nuclei. Whereas ionisation energy deals with the energy needed to remove electrons from an atom. So the terms deal with different particles - nucleons VS electrons.


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Regioselectivity in Diels–Alder reaction between 2-nitrobuta-1,3-diene and acrylonitrile

Since $\ce{-CN}$ shows strong −I effect, there will be δ+ charge on the carbon adjacent to it. Similarly, $\ce{-NO2}$ shows strong −I and –M effects. So, I proposed that π-electrons are being donated to the electron-deficient parts:

But the answer given in my book has $\ce{-CN}$ group at the third position to $\ce{-NO2}$ group. Can someone please tell me where I have got my concepts wrong?

Regardless of any simple electronic models, since DA part of what we call frontier orbital chemistry it should be dependend on orbital coefficients. There are tables for dienes and dienophiles and if I simply connect the sides with the biggest coefficients of the same symmetry I end up with your solution.
Source: 10.1021/ja00793a069

Not convinced by this DA reaction; both dienophile and diene are very electron-deficient...

You pointed out correctly that $-CN$ has a great $-I$ effect but it won't bring a $+\delta$ on adjacent carbon atom, reason being it's instability. Apply electromeric effects on propanenitrile.
If you need help then just comment below :-)


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Why does the alkaline hydrolysis of benzamide yield ammonia?
Nucleophillic addition to the carbonyl group is governed by the rule that a less basic group is displaced by a more basic group since $\ce{NH2^-}$ is more basic than $\ce{OH-}$ the reaction should not proceed forward but it does why so?

If the reaction were simply having the form
$\ce{RC(O)NH2 + OH^- <=> RC(O)OH + NH2^-}$
then your reasoning might well apply.  But, in the real world, the carboxylic acid $\ce{RC(O)OH}$ deprotonates (more than the amide $\ce{RC(O)NH2}$) and the proposed, strongly basic amide ion of course protonates to ammonia.  These proton transfer reactions push the displacement forward giving the true reaction
$\ce{RC(O)NH2 + OH^- -> RC(O)O- + NH3}$

This reaction only proceeds under forcing conditions - concentrated OH- and high temperature. Every reaction is an equilibrium so a small concentration of NH2- will be formed. This is immediately protonated to NH3, and because of the elevated temperature this is boiled out of the reaction mixture. Thus the reaction proceeds (if only slowly) to completion as there is never an equilibrium concentration of NH2-.  Le Chatelier's Principle applies.

And as @oscar-lanzi says the RCO2H is also very quickly removed from the equilirium as RCO2-


text:
HOMO / LUMO energies in a qualitative view
I have a question regarding the HOMO and LUMO energies depending on various effects like substituents and electronegativity. It seems to be a somewhat known topic but I cannot really find a clear
answer anywhere that would just summarize those effects. So I started to wrap my head around various qualitative effects that you can attribute to either a large or a small HOMO-LUMO gap.
The easiest approach would therefore be considering Koopman's theorem, where the negative HOMO energy approximates the ionization potential and the negative LUMO energy approximates the electron affinity.
If we then move to Mulliken's definition of electronegativity we can see that the sum of both IP and EA (divided by two) serve as a good approximation to the electronegativity. Which means that upon many other effects the electronegativity
seems to be a key feature to describe the HOMO and LUMO energy. And this is also what I remember from lectures.
In my memory a more electronegative element will lower the energy of the LUMO. That is why in for example photocatalysis many conjugated heterocyclic (N-substituted) ligands are used. The lower π*-orbitals give rise to MLCT transitions.
On the other hand the HOMO energy should be increased as well.
If we move to the Klopman-Salem equation we can derive that according to HSAB concept a large HOMO-LUMO gap will be a pair of a hard acid and a hard base and a small HOMO-LUMO gap will be a soft-soft pair.
One simplification to this is the case of a Lewis-acid and a Lewis-base reacting, or an electrophile and a nucleophile. Here the HOMO of the nucleophile will interact with the LUMO of the electrophile.
This means that a high nucleophilicity requires a high HOMO energy. Unfortunately there is no good description besides lone-pairs on real nucleophilic strenght.
So I thought I could perhaps find other examples and therefore moved to carbenes and their analogues. We destinguish singulett and triplett carbenes. A singulett carbene has a larger HOMO-LUMO gap.
Moving to heavier elements we get more singulett character. But this can't be used here since this effect is due to more complicated effects. As Si-Si-bonds and double bonds tend to be be much more reactive and are easily photo-activated as well the HOMO-LUMO gap should be smaller than in corresponding C-C-bonds.
But what about substituents on the carbene? Here we find that electronegative ligands will stabilize the a1-orbital in the carbene and in for example aminocarbenes, the π-donor-function of the $\ce{NH2}$-group increases the energy of the b1-orbital.
So from these examples I would summarize that in general the inductive effect of an electronegative substituent will stabilize and therefore descrease the energy of the HOMO.
And a π-donor will increase the energy of the LUMO by destabilizing it.
Comparing this to ligand field theory I would conclude that having an alkyl group will destabilize and therefore increse the HOMO and a π-acceptor will descrease the energy of the LUMO.
Are my findings and ideas somewhat applicable and true? And are there other very simple effects that I can take into consideration if I want to describe the relative positions of the HOMO and LUMO?


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If temperature and volume are same, the pressure of gas obeying van der Waal's equation is smaller than that of an ideal gas"
Can you state the reasons for it. I think " It is smaller because mutual forces of attraction would resist the collision of particle and container".

Your statement is false. There are cases where the pressure from the Van der Waals pressure is higher than the pressure from the Ideal Gas Law.
The Van der Waals equation introduces two corrections to the Ideal Gas Law:

It accounts for attractive forces between gas particles. This means that the Van der Waals pressure can be lower than the Ideal Gas pressure.
It accounts for the small but nonzero volume of the particles. This means that the Van der Waals pressure can be higher than the Ideal Gas pressure.

Depending on the amount and nature of your gas, the volume and the temperature, either one of these effects can dominate.
Qualitatively:
In a situation where gas molecules are squished together very closely, the Van der Waals equation yields:

Higher pressures for molecules that are big but not sticky, like isobutane, and
Lower pressures for molecules that are small but sticky, like hydrogen chloride.

Quantitatively:
I'm attaching a spreadsheet with a bunch of examples; pardon the German. For a number of situations described by:

substance, column A
volume (liters), column B
amount (moles), column C
temperature (°C), column D

the pressure is given according to the Ideal Gas Law (column I) and the Van der Waals equation (column J).
There are examples that follow your 'rule', and the Van der Waals pressure is lower than the Ideal Gas Law pressure; e.g., row 6.
There are examples that do not follow your 'rule', and the Van der Waals pressure is lower than the Ideal Gas Law pressure; e.g., row 11 or 22.
Keep in mind that both equations only apply to gases; look up phase diagrams and make sure the gases in question don't condense under the conditions given. Rows 12, 13, 23 are supercritical fluids.
You'll find that the Ideal Gas Law is quite precise in most cases, and the cases where it's not, are those where the gas is close to not being a gas anymore.

Is there a way to attach a spreadsheet to an answer?


text:
Can the inverse of a micelle be formed with non-polar solvents?
A micelle is formed due to the clustering of soap molecules wherein the interior of the cluster is formed by the hydrophobic end and the ionic ends are on the surface of the cluster.
So, if a non-polar solvent is used (Assuming non-polar solvents dissolve non-polar solvents), can the reverse of micelle be formed? That is, wherein the ionic end forms the interior and the hydrocarbon end the exterior?

yes, a micelle can go either way.

Yes, they are aptly enough called inverse micelles, eg such as formed by the detergent AOT. While you retain the hydrophobic/philic balance typical of a water-soluble detergent, wherein you have a long hydrophobic tail and hydrophilic headgroup, you generally have to alter the headgroup area (making it smaller relative to the tails) to favor inverted phases of this type. In addition if charged the headgroups will strongly repel, requiring a cosolvent or some counterions to reduce such interactions.


text:
Is there a single word antonym for dissociate?
Teaching ionic and covalent compounds, and one of the differences is that some ionic compounds will dissociate when dissolved but covalent compounds usually do not (salt water vs. sugar water, just basic stuff).
Although I'm taking a very basic approach, I would still like to use proper scientific language with my students. Is there a single-word synonym for "stay bonded" or a single-word antonym for "dissociate"? My own searches either come up with only 2-word terms or with synonyms for "come together" which aren't appropriate since the atoms are already together before dissolution.
Example sentence: Some ionic compounds dissociate when dissolved, but covalent compounds __ (stay together/don't dissociate) __.

I'm voting to close this question as off-topic because it is related to semantics and not an actual chemistry question

A proper antonym for dissociate is associate, however, that is not what you are looking for. I do not think that there is a word for what you are describing; I also don't really understand in what way it would be used.

@NilayGhosh Chemistry vocabulary is off-topic?

The usual term is dissociate, not disassociate. And I agree with @Martin-マーチン, I cannot think of another term for what you want. I think the clearest and most concise way of putting it would just be "covalent compounds do not dissociate in water". That is a huge generalisation, of course, but may be acceptable for new students.


text:
Non-covalent interactions
According to Keesom, Debye and London effect, the non-covalent reactions should be stronger, the bigger are electric charge differences between two interacting particles. E.g. Ion-induced dipole are stronger than dipole-induced dipole interactions.
So shouldn't be Van der Waals forces stronger than hydrophobic interactions? It is said that if we count with greater amounts of a substance, the hydrophobic interactions can be even stronger than hydrogen bonds. But aren't hydrophobic powers actually composed of the above mentioned effects? If two particles were absolutely neutral, they would not react since only opposite characteristics are attracted towards each other. These weak opposite charges can be caused also in non-polar substances by London effect. But that effect should be weaker than Keesom effect between permanent dipoles which contribute to VDW forces, for example in water.
Is it true that Van der Waals forces are stronger than hydrophobic interactions?


text:
Calculating molar solubility of calcium hydroxide from solubility product
On the Wikipedia page for calcium hydroxide, the stated solubility product for $\ce{Ca(OH)2}$ is $K_\text{sp} = \pu{5.5 * 10^{-6} M^3}$. It doesn't stipulate the temperature, but I have a chemistry book that says the exact same value for $\pu{25 ^\circ C}$.
However, when I calculate the solubility of $\ce{Ca(OH)2}$ in water using this value ($\pu{5.5 * 10^-6 M^3}$), I get a result of $\ce{0.082 g/100ml}$.
However, according to solubility tables on Wikipedia, calcium hydroxide is only this soluble at around $\pu{85 ^\circ C}$. How can this apparent contradiction be explained?

I think this is very curious, but it seems you answered the question yourself already in https://chemistry.stackexchange.com/questions/104055/what-is-the-solubility-product-for-calcium-hydroxide … . Personally I'd trust the National Lime Association before the wikipedia, after all their money is getting lime right.

But according to the National Lime Association, it still doesn't make any sense.

The only way to sort discrepancies like this is to ignore any data that doesn't reference an original source. In others words you looking for data from a scientific report, not just a data table. Now with the sources you have to look at the experimental errors and techniques used to decide which data is more reliable. You then somehow have to merge all the data into the "accepted values."

PS - Kudos for noting the discrepancy!!

A cursory search suggests the solubilities in g/100 mL on the wikipedia are closer to a consensus, for what it's worth:

solubility 0.2% ($0^oC$):
https://www.cdc.gov/niosh/npg/npgd0092.html
solubility 0.159 g/100 mL
(0.16% by weight) ($25^oC$):
https://www.mdpi.com/1996-1944/10/10/1219/pdf
the value you reported from the National Lime association.

These are reasonably consistent with the values in
https://en.wikipedia.org/wiki/Solubility_table#C
but not with the $K_{sp}=5.5 \times 10^{-6} M^3$ on the wiki page.
Skoog, West and Holler's Analytical Chemistry textbook (7th Edition) reports $K_{sp} = 6.5 \times 10^{-6} M^3$ at 25$^oC$. They provide a ref:

most of the data were taken from A.E. Martell and R. M. Smith, Critical Stability Constants, Vol. 3-6.

Note the apparent inconsistency recurs (I called this discrepancy an "error" before but am no longer so sure):
http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf
http://www5.csudh.edu/oliver/chemdata/data-ksp.htm
It reminds you that you should generally reference your data sources if you want to be helpful to fellow researchers.
The value for $K_{sp}$ in the wikipedia (which is not referenced as far as I could tell) and in a highly respected textbook which references its source is smaller by roughly an order of magnitude compared to values derived using standard equations from reported solubilities in a number of other sources noted above. This inconsistency may be because some of the reported data is for hydrates with higher Mw. Not sure how you could explain such a discrepancy otherwise. If the $K_{sp}$ were much smaller you could argue that the dissociation constant of water affected the solubility.
My current guess is that there is a reasonable explanation for the discrepancy (ie it has a chemical or physical explanation and is not due to human error). It is unfortunate that the solution is not more obvious given that $\ce{Ca(OH)2}$ is a recurrent example in problems (textbook calculations or experiments) teaching solubility to undergraduates.
See also:
Solubility of calcium hydroxide determined by titration

Even if it was off by an order of magnitude it still wouldn't make any sense if you try to calculate it.

Yea there has to be an explanation. I just wish someone knew it:/ Thanks for the help anyway, I appreciate it.

I think you search a little further you may stumble on the explanation. I'd try your teacher.

I’ll try to search a little more, and my teacher doesn’t know the explanation


text:
What is correct cyclopenta[a]phenanthrene numbering?
IUPAC BlueBook 2013 states:

P-25.3.3 Numbering fused ring systems
Fused ring systems with retained names, systematic names, or fused names are systematically numbered in the same manner. Anthracene, phenanthrene, acridine, carbazole, xanthene and its chalcogen analogues, purine, and cyclopenta[a]phenanthrene are exceptions; traditional numberings are retained.
(…)

All except the last one, cyclopenta[a]phenanthrene are trivial retained names, this one is systematically created name (P-25).
All except cyclopenta[a]phenanthrene have the numbering explicitly depicted (in P-25.1.1 or P-25.2.1; anthracene and phenanthrene in errata).
It's natural and expected that there can be different numberings for the same compound for different naming system chosen (e.g. decahydronaphthalene vs bicyclo[4.4.0]decane – fusion carbons C-4a and C-8a vs C-1 and C-6).
So, what is (your qualified opinion on) the correct numbering of cyclopenta[a]phenanthrene? In other words, for example, what is the correct name for this compound:

1H-cyclopenta[a]phenanthrene
3H-cyclopenta[a]phenanthrene
17H-cyclopenta[a]phenanthrene

or other one?
(1.) would be correct systematic numbering (if I'm not mistaken; used by current MarvinSketch), (2.) a historic alternative one, (3.) would be for steroid (gonane, alternative name sterane in pre-IUPAC 1951 steroid nomenclature) numbering.
If it's (3.) that would be, in my knowledge, only exception from systematic numbering for systematically created name, however extraordinary claims require extraordinary evidence, which is not seen in the book. And, the book can have yet uncorrected mistakes.
Curiously, as gonane numbering is well defined (Appendix 3), most unambigous name seems to be 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16-hexadecadehydrogonane (or gona-1,3,5(10),6,8,11,13,15-octaene).


text:
Wittig reaction with an alkyl dibromide
Recently, I have come across a problem, in which a ketone needs to be converted into an alkene. However, in the alkene product, there is also another bromide functional group, at the end of the alkyl chain. I was thinking of using a dibromide (an alkyl chain with bromine atoms at both ends) and performing the Wittig reaction. However, I foresee that both bromide functional groups would participate in the reaction. Would one solution to this be to use a protecting group? Or would the problem be solved if I manipulate the stoichiometry of the reactants (e.g. ensure that the dibromide is in excess etc.)?

Use a bromo alcohol with the alcohol protected as THP ether. After the Witting, deprotect and form the required bromide.

@Waylander Would manipulation of the stoichiometry of the reactants work as well? Just wondering.

There is a danger that when you deprotonate the Phosphonium bromide to do the Wittig that the anion will react with the remote bromide, either to cyclise or intermolecularly.

A structure would have been helpful. You might not need a protecting group. Possible that the alkoxy ylid is fine.

It seems that the method to circumvent this problem is indeed to use a protecting group to protect one end of the alkyl chain, which is the end to be converted to a bromide after the Wittig reaction.
As Waylander aptly mentioned, another problem which comes with using an alkyl dibromide is that there may be an intramolecular substitution reaction, in which the anionic carbon in the phosphonium ylide reacts with the other end of the alkyl dibromide.
To circumvent this problem, we could use an alkyl bromide with an alcohol functional group at the other end of the chain. This alcohol group can then be protected using various protecting groups, such as a tetrahydropyran ether or trimethylsilyl ether. After the Wittig reaction, the compound can then be deprotected using acidic hydrolysis. After which, the alcohol functional group can then be converted to a bromide using $\ce {PBr3}$. The desired compound is thus obtained.


text:
What does it mean to equilibrate a column in size exclusion chromatography and why is a buffer used?
I'm reading a procedure on size exclusion chromatography and it says that the first step is to equilibrate the column with buffer. What does this mean?

To equilibrate the column with buffer just means to run the column with the buffer only for a while.

Why is this done?

You want to make sure that your protein does not get damaged in the separation process. The protein travels faster than the small solutes in its solution, and will encounter the small molecules that are on the column, i.e. it will encounter different ionic strength, pH etc. So before you apply the protein solution to the column, you make sure the column is treated with an appropriate solution.
The solution is often called the buffer because it contains pH-controlling buffer substance. You can use size exclusion chromatography to separate by size, but also to switch from one buffer (or more general: one set of small solutes) to another. This is called buffer exchange, and there are short thick columns for that purpose (for separation, you would use a long thin column).


text:
Significance of this unusual matrix element
Recently, I was solving some questions on Rigid Rotor. I found this question very amusing.
Evaluate the integral
i) <$Y_{l,m+2}$ | $L_{x}^{2}$ | $Y_{l,m}$ >
I evaluated it, found that the value is
$$\hbar^2/4 \sqrt{[(l(l+1)-(m)(m+1)][(l(l+1)-(m+2)(m+1)]}$$
But I didn't get the point of evaluating it. Is there any application of such kind of integrals?
Is there any physical significance of it?
I know similar kind of integral are evaluated in case of transition dipole integral  where we take two different wave function.
What is actual significance of such operations ?

Strictly speaking this is not an expectation value. I would just call it a matrix element.

@Feodoran Sorry for my primitive knowledge.

I would have expected $\hbar^2$, did you expect the result to be complex for some $\ell$ and $m$ ?

@porphyrin Thanks for pointing out, I really messed it. Now I have corrected it. I was really sleepy when I solved it.

These kind of elements appear when constructing the Hamiltonian of for the rotation of an asymmetric top molecule. This term appears Because the moments of inertia along the spatial axes are different you cannot express the rotational energy in terms of $L^2$ and $L_z^2$ alone as you would do for a linear or symmetric top molecule.


text:
Why does the temperature remain constant during freezing?
I know it's the most silly question one could ever ask but why is that so?
I am aware of what latent heat means and I also know it has something to do with freezing too.
Say I have with me, a mug of hot water and I put it to freeze. And so the temperature reduces first to 273 K and then remains constant to convert into ice and decreases once again.
 But why does the temperature remain constant though? Does this has something to do in molecular level?

In a broader sense, everything that happens to us and around us has something to do in molecular level, but I'd rather not put it this way. The temperature stays constant because it just can't go down. How would it go down? You can't have liquid water at below-freezing temperature (except in some special cases, but those are out of scope for now).

Creating an ordered solid from a disordered liquid releases energy
At the molecular level this is what is happening when you cool something like water.
At first the cooling reduces the average kinetic energy of the water molecules so they slow down. Basically, temperature is the average kinetic energy so removing energy lowers the temperature.
At some point, though, water starts to form ice where the molecules are not wandering about randomly and get rigidly stuck together in ice crystals. But this process also releases energy: the ordered ice crystal has a significantly lower energy than water at the same temperature. Converting liquid water to ice releases that energy.
What this means is that when extracting energy from warm water the result is simply a lower temperature. But one you get to 0°C the extra energy extracted is used to convert the liquid to the solid rather than reducing the temperature. Once all the water has solidified, further removing energy reduces the temperature again.
The key is that the process of freezing itself involves energy changes so instead of reducing the temperature, the energy comes from the conversion of liquid to solid.

When you write "But this process releases energy" when talking about melting, it seems as if you are using this to distinguish it from simple cooling.  But note that both cooling and melting release energy. Thus it's not the release of energy that makes melting qualitatively different from cooling.

@theorist Assume that you are removing energy at a steady rate. The energy comes from cooling the water. At the crystallisation point you are still removing energy at that rate. But a new process (crystallisation) starts in the liquid which, from the previous point of view, releases extra energy that stops the temperature falling steadily. Alternatively, you switch between two different energy releasing processes (cooling and crystallisation) at that point.

I agree with what you wrote in your comment.  I don't doubt you understand the process.  My point is that the way you've written the portion of your answer that discusses melting ("But this process releases energy") makes it sound like what's distinct about the melting (vs. the cooling) is that the melting releases energy. (because you start with "but" and emphasize "releases").   So I'd recommend you instead write something like "As with the cooling, this process also releases energy. But what's different is...."

To give an analogy, suppose you were talking about the difference between two different versions of some car (let's call them the LX and SX models) and, in talking about the SX, wrote "But the SX uses gasoline..."  That would make perfect sense if the LX didn't use gasoline (say, it was a diesel).  But if both used gasoline, you would not want to write  "But the SX uses gasoline...", because it implies the LX somehow does not.

@theorist OK good point. I've made a minor edit.


text:
Why do nonmetal oxides react with water to form acidic compounds?
I am confused about the molecular basis by which nonmetal oxides react with water to form acidic compounds. For example, I understand that sulfur trioxide reacts with water to form sulfuric acid as shown in the following reaction:
SO3 (g) + H2O (l) --> H2SO4 (aq)
I recognize that the polarized nature of the O-H bond in sulfuric acid makes it thermodynamically favorable for H2O to abstract a proton and subsequently form an acidic solution with the formation of H3O+. However, I fail to recognize how reactions like this with a nonmetal oxide proceeds to form an acidic oxide. Namely, I fail to understand the role of the covalent bond in the nonmetal oxide for the mechanism of this reaction. I recognize that metal oxides dissolve in water to form the highly polarizing, high charge density ion O2-, which polarizes the O-H bond in H2O to form hydroxide. Can someone give a similar explanation for nonmetal oxides and their dominant acidic nature in water?

Well, in terms of this reaction, sulfur in $\ce{SO3}$ acts as a Lewis acid for an incoming $\ce{OH-}$ nucleophile. The reaction mechanism seems to be any old nucleophilic addition reaction:

$\ce{H2O <=> H+ + OH-}$ : (Water can disassociate into its ions)
$\ce{SO3 + OH- -> SO2(OH)O-}$ : (I couldn't decide on whether this is the right
way to write this intermediate, but I'll be happy to add a diagram for clarity)
$\ce{SO2(OH)O- + H+ -> SO2(OH)2}$ : (This is the final step of the mechanism to yield $\ce{H2SO4}$)

So, is it that the polar covalent nature of bonds in nonmetal oxides (such as CO2 and N2O5) allow for the molecule to act as a lewis acid and subsequently form an acid (H2CO3 and HNO3)? Then why do metal oxides which are even more polarized not act as lewis acids and form acids as well? One more question--why do some nonmetal oxide molecules such as carbon monoxide not form acids?

See it like this - non-metal oxides are acid anhydrides, and so adding non-metal oxides to water will form the subsequent acid. Metal-oxides don't act as Lewis acids because of the fact that the metals don't have any 'empty' orbitals for electron pairs to be donated to. Technically speaking, adding carbon monoxide to water will produce carbon dioxide - an acidic gas - which can then proceed to form carbonic acid as mentioned.

@KlanStevens The dissociation constant for water is very low, would the $\ce{H+}$ and $\ce{OH-}$ from water be enough to turn $\ce{SO3}$ into sulfuric acid with appreciable concentration? Or would the amount of sulfuric acid produced be very low?


text:
Linear algebra in Lewis structures?
In general, I have come across a phenomenon in chemistry where solutions to certain types of problems can often seem frustrating when trying to work out by hand, as they seem to depend on each other far too much, but are actually easily representable formally using a linear relation.
In general, I have noticed that when I am trying combinations of things to see if it works that these relations appear.
For example, balancing an equation:
$$
\ce{C3H8 + O2 -> CO2 + H2O}
$$
$$
\left[
\begin{array}{c|ccc|c}
 & \ce{C3H8} & \ce{O2} & \ce{CO2} & \ce{H2O}\\
\hline
\ce{C} & 3 & 0 & -1 & 0  \\
\ce{H} & 8 & 0 &  0 & -2 \\
\ce{O} & 0 & 2 & -2 & -1 \\
\end{array}
\right]
$$
Row reducing the matrix gives a solution vector of
$$
\left[
\begin{array}{c|c}
\ce{C3H8} & -1/4\\
\ce{O2} & -5/4\\
\ce{CO2} & -3/4 \\
\end{array}
\right]
$$
$\ce{H2O}$ has an implied coefficient of $-1$, so in order to make the coefficients integers, we will multiply everything by $-4$, and get
$$
\ce{1C3H8 + 5O2 -> 3CO2 + 4H2O}
$$
which is correctly balanced. There is no guesswork in doing this.
Similarly, problems involving Hess's law can be solved using linear combinations:
$$
\begin{array}{cccc}
\ce{CH4_{(g)} + 2O2_{(g)} -> CO2_{(g)} + 2H2O_{(l)}} & {\Delta}\ce{H} = & \pu{-890 kJ mol^{-1}} & (1)\\
\ce{2CO_{(g)} + O2_{(g)} -> 2CO_{(g)}} & {\Delta}\ce{H} = & \pu{-566 kJ mol^{-1}} & (2) \\
\ce{2CH4_{(g)} + 3O2_{(g)} -> 2CO_{(g)} + 4H2O_{(l)}} & {\Delta}\ce{H} = & x\; \pu{kJ mol^{-1}} & (3)
\end{array}
$$
$$
\left[
\begin{array}{c|cc|c}
& \pu{1} & \pu{2} & \pu{3} \\
\hline
\ce{CH4_{(g)}} & -1 & 0 & -2\\
\ce{O2_{(g)}} & -2 & -1 & -3\\
\ce{CO2_{(g)}} & +1 & +2 & 0\\
\ce{H2O_{(l)}} & +2 & 0 & +4\\
\ce{CO_{(g)}} & 0 & -2 & +2
\end{array}
\right]
$$
Row reducing the matrix gives a solution vector of
$$
\left[
\begin{array}{c|c}
\pu{1} & +2\\
\pu{2} & -1\\
\end{array}
\right]
$$
The dot product with the original enthalpy vector then becomes the desired enthalpy of change, $x$:
$$
x =
\left[
\begin{array}{c}
+2 \\ -1
\end{array}
\right]
\left[
\begin{array}{cc}
-890 & -566
\end{array}
\right]
= \pu{-1214 kJ mol^{-1}}.
$$
Again, there is no guesswork. A computer could do this.
Is there a similar linear relation possible for expressing the possible number(s) of bonding electrons for each atom in a Lewis structure? (By possible number, I mean that it follows the octet rule if applicable, or in the case of hydrogen only forms a single bond.)
If so, how could it be set up?

This is a good question, but I think it would be better suited to post this on math.stackexchange.com, and explain the rules of lewis structures in your question.

@AbhigyanC Will do, though Math.SE doesn't have chem support :(

To give a concrete example is your question about: from the structure N-N-O how many ways can I make bonds (here 2: N-triple-N-single-O vs N=N=O) or: I have C3H6 and how many octet rule obeying isomers (a ring vs propene)?

@AbhigyanC Crossposting  is discouraged - if post is closed here it may be migrated. I think it has nothing to do with matrices and whatnots. Thing is, there are molecules that don't follow any rules.

@user213305 the former

In agreement with @AbhigyanC I'd say this in essence a graph theoretical problem that will have a known mathematical solution. However I disagree with AbhigyanC that it will be feasible to explain the "rules of Lewis structures" on MSE. I expect that this will be a medium sized project of months rather weeks and there will be continuous requests to go back to CSE. At the same time I suggest to try to check the literature, there is a huge body of graph theoretical works in chemistry. Ivan Gutman might be an expert to name at this stage.

@Rudi_Birnbaum There are many mathematicians that are also chemists, and there are many chemists that are also mathematicians. It wouldn't be so hard to explain it to them... There should be a stack exchange for inter-disciplinary stuff

@Mithoron is right. There is no way to predict how many electrons will be used in bonding unless we already know the structure. There are indeed many molecules that don't follow any rules.


text:
Xylene leaves a stain after evaporating?
I'm using xylene to clean a sacrificial polymer layer off of a PMGI surface in a ventilated hood.
For some reason, when xylene dries naturally from the surface there is a liquid stain that can be seen visibly on the PMGI. The stain can be removed by spraying with more xylene and then blowing it off with N2.
This stain will happen even if I apply xylene to a clean PMGI surface that has never been coated in the sacrificial layer so it isn't just incomplete cleaning. It seems like a contaminant is coming from the xylene.
My xylene is specified to have only 5 ppm of residue after evaporation. Is this enough to leave a visible stain after a thin layer evaporates?

Have you tried a drop of xylene on clean glass? That would show whether it's the solvent or something in the substrate leaving the stain.


text:
Dilute FeCl3 colour is darker at 5e-4M than at 5e-3M. Why?
I did two different dilutions of $\ce{FeCl3}$ (within a $\pu{1h}$ timespan) from $\pu{0.05 M}$, and the one with $\pu{5E-4 M}$ concentration looked a darker orange than the one with $\pu{5E-3 M}$.
The picture shows on the left: $\pu{5E-3 M}$ (I erased the label on the flask); on the right: $\pu{5E-4 M}$

I did a test with litmus paper, which showed that the more concentrated one ($\pu{E-3}$) was more acidic, which makes sense. I also did a rates test ($\ce{FeCl3 + H2O2}$), and the more concentrated one made the reaction happen faster, which again makes sense. Why, then, does the more concentrated $\ce{FeCl3}$ look lighter coloured?
Since I had made the more dilute one ($\pu{E-4}$) before, I thought there might have been more time for the ferric hydroxide to precipitate and form a darker colour. But I had only left it for $\pu{1h}$. I also thought that since the more concentrated one looked greenish, $\ce{FeCl2}$ might have been made, yet this does not seem plausible as $\ce{FeCl3}$ is more stable.

FeCl3 hydrolyses if pH is too low, so if you non acidified solution dilute it too much, it precipitates. I.e. it's darker because of precipitate of Fe2O3 (aq).


text:
How can I find the equilibrium constant Kp without temperature?
The question I'm looking at is
I found that that Kc = 139M^-1, but how can I determine Kp? There is no temperature given, but I was told that it is still possible to calculate.

Why do you think you need the temperature in this case to calculate $K_p$?

HINT - Assume ideal gas behavior.

$K_p=K_c(RT)^{\Delta n_g}$(try to prove this yourself) where $\Delta n_g$ is number of gaseous products -Number of gaseous reactants.
In your question, $\Delta n_g=0$ so $K_p=K_c=2.43$

Thank you! If the number of gaseous products was not 0, would this still be solvable? For example, in this question: https://i.gyazo.com/0cf8ffd4c95e126583a973e3b768b30a.png

Yes. Kc can directly be found.Hint: No of moles directly proportional to concentration. Kp = Kx * (P_total/(total number of moles))^∆ng


text:
Rearrangement of anions

My book says a. But i think it is wrong.
Rearrangement is not  possible in anion me, ans also
 there is group migration but here the group has been  divided .

why its yellow?

The most fascinating thing about aldol reaction is that it is reversible. So, when the aldol product is kept in a basic medium, it can actually undergo reverse mechanistic pathway to regenerate the two carbonyl compounds between which condensation occured to form the product . Then those two carbonyl compounds can literally undergo condensation in a different way to produce some different aldol product. This phenomenon is called as Retro-aldol reaction.
In this case, also if you observe carefully, the compound (A) is an $\alpha-\beta $ unsaturated ketone (basically a dehydrated aldol product), which by hydrolysis can convert itself into an aldol and thus undergo retro-aldol reaction. Here the aldol condensation was done intramolecularly, so the two ketones produced undergoes intramolecular aldol condensation in a different way to produce another isomer (B).
The mechanism of formation of (B) is as follows,

Thus, option (a) should be the correct answer.


text:
GVP(PP) orbital energies?
Hartree-Fock orbital energies can be interpreted physically, e.g. in Koopmans theorem, while one cannot assign active MCSCF orbitals to energies in general. What about GVB, say for example GVB(PP) orbitals, does it make sense to assign energies to those?


text:
How does hyperconjugation explain the stability of alkenes?
The more the number of alpha hydrogens, the higher is the stability of alkenes, because the more the number of hyperconjugative structures. Why is this so? Has it got something to do with the electrons being easily delocalised?

H+ leaves giving you a negative charge on Carbon which participates in resonance

https://chemistry.stackexchange.com/questions/20362/the-effect-of-hyperconjugation-on-the-stability-of-alkenes-with-mo-theory

Does the above link provides you with your answer ?

As always in chemistry one phenomenon (experimental observation) can be explained with different models, that live in different "physical model worlds", and more often than not these models are not consistent among themselves and might be even bare of a physical reality, especially if they are analyzed in rigorous way, nevertheless might be very useful tools, that take the justification of existence out of the fact that they work in many cases.
That said,  especially the concept of hyperconjugation itself can be employed at different levels. A very primitive model, but one that works astoundingly well, is to assign a positive inductive (electron releasing effect) to alkyl chains that is increasing with the degree of substitution of the ipso atom. That leads to the idea that this (however small) charge that is pushed to the sp$^2$ carbon atom stabilizes its slight positive polarization (since the majority of electrons in the double bond is above and below the plane of the atoms engaged attached to the doubly bound atoms. The physical reality of this model if often drawn into doubt, which is good to know, but still its simple and powerful and thus there is no reason to abandon it.
At a higher level of model, using a MO picture the idea is that the (empty, anti-bonding) $\sigma^*$ orbitals that come inevitably with the C-C and C-H $\sigma$ bonds can act as kind of "acceptors" for the $\pi$-electrons of the double bonds if in the right orientation. It is important to note that in this idea to actual overlap that mediates the interaction is thought to be established with the part of $\sigma^*$ orbital that is opposing the bound atom. That donor-acceptor interaction ultimately leads to a stabilization of the system. Again the physical reality is also here at least disputed, but however less than in the former case. But still both have in common that the Substituent acts as the donor and that this causes stabilization.
So yes in this higher level of description of the hyperconjugation the reason for the stability is the increasing number of possibilities to form stabilizing overlap.

From your explanation of the first model, I understand that more substituted carbons, or lesser number of alpha hydrogens, mean greater +I effect, and hence greater stability for the adjacent slightly positively polarised sp2 hybridised carbon. But that seems to imply that more substitution or lesser number of alpha hydrogens equals more stability, which is not true. Have I misunderstood you? And as for the antibonding model, the sigma start orbital wants the C-H bond to break in order to accept pi electrons and achieve stability. Is that right?


text:
Are we using an assumption in calculating the ratio of first state excited to unexcited chemical species using the Boltmann distribution?
In calculating the ratio of first state excited to unexcited chemical species using the Boltzmann distribution, are we assuming that we only have species in the unexcited and first state excited states? It seems to me that this must be the case, because there's no part of the equation that accounts for the population of e.g. the second excited state.

we don't need to know the population of any other states to determine the ratio between the population of two states.

The expression for the Boltzmann equation is:
$\eta_{i} = \dfrac{g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{\sum_{i} g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}$,
wherein $\eta_{i}$ the probability is to encounter the ensemble in state $i$ is in the ensemble, $g_{i}$ the degeneracy of state $i$, and $\epsilon_{i}$ its energy (I reckon the other variables are familiar to you).
To calculate the ratio between two states, you can simply divide the above expression for two different states by which you obtain:
$\dfrac{\eta_{i}}{\eta_{j}} = \dfrac{g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{\sum_{i} g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)} \cdot \dfrac{\sum_{i} g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{g_{j} \exp \left( \dfrac{-\epsilon_{j}}{k_{b}T} \right)} = \dfrac{g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{g_{j} \exp \left( \dfrac{-\epsilon_{j}}{k_{b}T} \right)}$
The expression above shows that because states $i$ and $j$ are of the same ensemble, you do not have to use (or even know) the energy levels of the other states in that ensemble if you are only interested in the probability ratio between the two states.


text:
Is a time domain spectrum obtainable from a frequency domain spectrum?
I know for a fact that a frequency domain spectrum can be obtained from a time domain spectrum using a Fourier transform  - but can you do the reverse?
Also what are the advantages and disadvantages of the frequency and time domain spectra?

Can you do the reverse?

Yes, mathematically the Fourier transform can be reversed. If we define the spectrum $S(\omega)$ as the Fourier transform of a time-domain signal $f(t)$,
$$S(\omega) = \mathcal{F}[f(t)] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)\mathrm{e}^{-\mathrm{i}\omega t}\,\mathrm{d}t$$
then the inverse Fourier transform is simply given by
$$f(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}S(\omega)\mathrm{e}^{\mathrm{i}\omega t}\,\mathrm{d}\omega$$
Wikipedia has a page on this which is rather technical, but any decent book on Fourier transforms should cover this, so a mathematics book targeted at physicists/chemists may be easier to digest.

What are the advantages and disadvantages of the frequency and time domain spectra?

The time domain signal (in NMR, "free induction decay") is for the most part, not particularly useful. It is the Fourier transform which makes it valuable for chemists, because the frequencies that appear in the spectrum are related to transitions between different energy states:
$$\Delta E = \hbar\omega$$
In general, probing these energy levels is the main point of spectroscopy, regardless of what range of energies you are using.

I just like to add to @orthocresol's excellent answer that there might be situations in which you can extract time-domain information from a frequency spectrum. The lifetime of excited states is for instance inversely proportional to the linewidth of a spectral line (this follows from the FT). This principle can, for instance, be used in studies on chemical kinetics to probe very fast processes.


text:
Elimination reactions of vicinal dibromide
When a vicinal dibromide undergoes elimination reactions in vigorous basic conditions, say in ammonia with $\ce {NaNH2}$, is the product an alkyne or is it a conjugated diene? I would think that the answer depends on thermodynamic and kinetic factors. Considering bond energies of the $\ce {C-C}$ $\pi$ bonds, perhaps the conjugated diene would be more favourably formed?

That reaction isn't reversible, so it's not going to dependent on thermodynamic factors.

After the first elimination, you have a vinylic bromide, so there are two options for the second elimination. The two acidic hydrogens that the base can abstract are one attached to sp2 carbon and one to sp3. Since sp2 hybridized carbon is more electronegative, the base abstracts the hydrogen attached to it, and then bromine leaves being a good leaving group, which gives alkyne as the major product.


text:
Does resonance destabilize a molecule?

According to this table, more heat is released in breaking a single double bond in 1,3-butadiene than in trans-2-butene. Does this mean that the resonance of 1,3 butadiene leads to a destabilization of the compound?

Your comparison is not correct. You are comparing two molecules with difference in internal bonding. You will not get trans-2-butene by removing 1 double bond from 1,3-butadiene. You should have compared 1-butene and 1,3-butadiene. Theoretically, addition of one double bond to 1-butene should give 1,3-butadiene, i.e. $\Delta H^0_{theo} = -60.6 $ kcal/mol. But practically it is lesser ($-57 $ kcal/mol) due to its higher stability due to conjugation.

None of the comparisons you might make suggests any sort of deatabilization by conjugation.
Let's look at the four-carbon compounds.  Depending on which source you believe for 1,3-butadiene, you could release up to 27.8 kcal/mol by hydrogenating one double bond (the figures given for this compound refer to hydrogenating both double bonds).  But then you release 30.3 kcal/mol hydrogenating the remaining double bond in 1-butene.  The combination of two double bonds in 1,3-butadiene has less energy to be released by hydrogenation than two independent bonds like the bond in 1-butene has.
Similarly for the five-carbon compounds.  When you hydrogenated both double bonds in the pentadiene isomers, a pair of conjugated double bonds in 1,3-cyclopentadiene has less energy to release than the non-conjugated double bonds in 1,4-cyclopentadiene.
Finally, something a little more subtle is going on between 1-butene and 2-butene.  There can no conjugation like that in 1,3-butadiene when only one double bond is present, yet 2-butene has clearly less energy to release by hydrogenation than 1-butene.  Thinking of carbocation nomenclature, the former is a like a "secondary carbication by having an alkyl group on only one end of the double bonds, while the latter is "tertiary" with an alkyl group on both ends.  Hyperconjugation, like that in carbocations, is at work here.


text:
Can Le Chatelier's principle be derived?
Le Chatelier's principle says :

If a constraint (such as a change in pressure, temperature, or concentration of a reactant) is applied to a system in equilibrium, the equilibrium will shift so as to tend to counteract the effect of the constraint.

This seems as an analogue of Newton's first law of motion where we talk of inertia resisting a change in the state of rest or motion of a body. Now in physics we have General Relativity whence we can derive Newton's laws. Is there a corresponding general theory in the domain of chemical kinetics too, based on sounder mathematical principles rather than just empirical observations, whence we can derive the Le Chatelier's principle as a special case?
Any suggestions are thankfully invited.

What kind of derivation are you looking for, exactly?

One based on mathematical abstractions, like we do in the quantum theory.

The equilibrium constant is derived from the chemical potential and with expressions such as the Van't Hoff isochore  (temperature dependence of equilibrium constant) explains Le Chatelier.

Yes. To prove Le Chatelier's principle doesn't require a general theory of chemical kinetics, just an understanding of (thermodynamic!) fluctuations and response. As usual, Callen's Thermodynamics and an Introduction to Thermostatistics is a good reference. Ch. 8.5 is relevant, and I shall simply recapitulate Callen's development below.
(Incidentally, I think this principle is closer to Newton's third law of action and reaction, but in general thermodynamic laws have a very different flavor from mechanical laws because of their underlying structure. Thermodynamics deals with nebulous relationships; mechanics with definite equations of motion.)

In fact, we will prove an extra statement, the Le Chatelier-Braun principle. This principle states that the secondary effects induced by a perturbation also serve to curtail it. This augments the Le Chatelier principle, that the primary effect induced by a perturbation serves to curtail it.
I will steal Callen's example of a system immersed within a pressure and temperature reservoir, with diathermal walls and a movable domain wall with which to control its volume. The wall is moved slightly outward, causing a positive volume change $\text{d}V$. The primary effect is the decrease in pressure of the system, which leads to a driving force to decrease the system's volume. A secondary effect is the change in temperature $\text{d}T$ resulting from this change in volume, $$\text{d}T = \left(\frac{\partial T}{\partial V}\right)_S\text{d}V = -\frac{T\alpha}{Nc_v\kappa_T}\text{d}V.$$ The prefactor is unimportant; we care only about the sign of the result. All variables are positive except for $\alpha$, which is of variable sign. For now, we can assume it positive. The reduction of temperature in the system then drives heat flow into it, which will itself affect the pressure of the system: $$\text{d}P = \left(\frac{\partial P}{\partial S}\right)_V\frac{\text{d}Q}{T} = \frac{\alpha}{NT^2c_v\kappa_T}\text{d}Q.$$ This change in pressure is positive and diminishes the effect of the original perturbation. The same result is obtained if one takes $\alpha$ negative.

Let us consider a thermodynamic system with first law $$\text{d}U = f_1\,\text{d}X_1 + f_2\,\text{d}X_2.$$ It is coupled to a reservoir, itself with first law $$\text{d}U' = f_1'\,\text{d}X_1' + f_2'\,\text{d}X_2' = -f_1'\,\text{d}X_1 - f_2'\,\text{d}X_2,$$ noting that $\text{d}X_i' = -\text{d}X_i$, because $X_i+X_i'$ is assumed fixed. A perturbation $\text{d}X_1^\text{pert}$ drives fluctuations $$\text{d}f_1^\text{fluc} = \frac{\partial f_1}{\partial X_1}\text{d}X_1^\text{pert} \quad \text{and} \quad \text{d}f_2^\text{fluc} = \frac{\partial f_2}{\partial X_1}\text{d}X_1^\text{pert}.$$ These fluctuations in intensive quantities themselves lead to responses $\text{d}X_1^\text{resp}$ and $\text{d}X_2^\text{resp}$. The signs of these responses can be determined by minimizing the total energy of the system and reservoir given the initial perturbation, $$\text{d}(U+U') = (f_1-f_1')\,\text{d}X_1^\text{resp} + (f_2-f_2')\,\text{d}X_2^\text{resp} = \text{d}f_1^\text{fluc}\text{d}X_1^\text{resp} + \text{d}f_2^\text{fluc}\text{d}X_2^\text{resp} \leq 0.$$ Because $X_1$ and $X_2$ are independent variables, each term in the sum must be negative, and we have $$\text{d}f_1^\text{fluc}\text{d}X_1^\text{resp} \leq 0 \quad \text{and} \quad \text{d}f_2^\text{fluc}\text{d}X_2^\text{resp} \leq 0.$$ The first equation yields $$0 \geq \text{d}f_1^\text{fluc}\text{d}X_1^\text{resp} \quad \Longleftrightarrow \quad 0 \geq \text{d}f_1^\text{fluc}\frac{\partial f_1}{\partial X_1}\text{d}X_1^\text{resp} = \text{d}f_1^\text{fluc}\text{d}f_1^{\text{resp}(1)},$$ where we have multiplied both sides of the inequality by $\partial f_1/\partial X_1$, which must be positive by stability. $\text{d}f_1^{\text{resp}(1)}$ is the response of $f_1$ to the fluctuation due to $X_1$ only, and so we have the Le Chatelier principle. The second equation yields $$0 \geq \text{d}f_2^\text{fluc}\text{d}X_2^\text{resp} = \frac{\partial f_2}{\partial X_1}\text{d}X_1^\text{pert}\text{d}X_2^\text{resp},$$ and hence that $$0 \geq \frac{\partial f_1}{\partial X_1}\text{d}X_1^\text{pert}\frac{\partial f_1}{\partial X_2}\text{d}X_2^\text{resp} = \text{d}f_1^\text{fluc}\text{d}f_1^{\text{resp}(2)},$$ where we have both used a Maxwell relation and multiplied both sides of the inequality by $\partial f_1/\partial X_1$. This is the Le Chatelier-Braun principle.

Note that both dT and dP as written cannot be correct since $(∂T∂V)_S = -(∂P∂S)_V$


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How to align povidone-iodine molecules in order to look for optical polarization
In this answer I mention how early polarizing filters (e.g. sunglasses) were made by doping a polymer with iodine atoms and then stretching the film to orient the polymer molecules.
Electrons donated by the iodine could then respond to the electric field of the incident light preferentially in the direction parallel to the molecules.
This and this answer mention that one common house-hold form of antiseptic iodine is a povidone iodine solution.
I don't have a feeling for how long those povidone chains are in over-the-copunter povidone iodine solutions are, but I would like to ask how I might try to remove the solvent, which is in this case just water, and see if I could make a film of the povidone-iodine complex, align or stretch it somehow, and see if the result has any polarizing activity.
If I tried to evaporate some, would the iodine remain attached to the povidone chains, or evaporate as well?
An alternative that would align the molecules in a solution somehow would also be interesting.

Source "Skeletal formula of povidone-iodine (PVP-I) — a widely used iodine-containing antiseptic and disinfectant."

The wikipedia article says the PVP-I complex can be loaded into hydrogels. I have some hydrogel tape that stretches readily and would likely easily take up the PVP-I just by dipping the tape (or soaking it) in the PVP-I solution. I can do a quick proxy test with the tape and food coloring. As for removing the solvent from the PVP-I solution, that would likely destroy the complex because the HI is needed to link the polymer to the iodine molecules. If I see the PVP-I in my local pharmacy, I will pick up a small bottle and try my luck at making a hydrogel polarizer!

@EdV wonderful! I haven't a clue what "hydrogel tape" is much less have some lying around, looking forward to news of what happens.

Well, a 24 hour soak of the hydrogel tape in blue food coloring (in water) resulted in no uptake of the coloring into the tape. It happens that the PVP-I is available, so I might try it directly. I wonder if the world building SE ever considered a truly bizarre world where most experiments worked, even some on the first try? Nah, far beyond ken! ;-)


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How is fluorine gas contained and transported, why aren't the metal cylinder and tube in this video attacked?
I included the GIF below as part of an answer to Can paper burn without oxygen or air?. It was made from the video Fluorine - Periodic Table of Videos.
I noticed that the cylinder containing the fluorine gas "filled up on the roof" in the video looks like it is made out of metal, as is the tube feeding the gas to the combustion sample.
The liquification experiment also shown in the video uses a flexible tube that appears to be plastic.
I know hydrofluoric acid will dissolve gas, but can dry fluorine gas be contained in glass? Since metals, even precious metals are generally attacked by fluorine, how is it stored, and how are the metal cylinder and tube in the video not also consumed?

Incidentally you can look up individual elements on Periodic Table of Videos at http://www.periodicvideos.com/

Related: https://chemistry.stackexchange.com/questions/19517/is-there-a-way-to-contain-fluorine-gas-for-long-term-so-that-it-can-be-visually

I think it is mentioned somewhat around 4:35 or so in the link you added. The metal cylinder that has to store elemental fluorine will have to be made from a special metal like nickel or monel. Those form metal fluoride layers on the inside but they remain on the metal and allow no further fluorine to attack the rest of the metal which is called passivation.
About the plastic tube, it could be Teflon but I don't know if that works or not. Basically, the problem with fluorine especially about demonstrations was always the container. You could have easily done experiments in metal containers but then no one would be able to see through. Then there were, of course, fluorinated plastics but they tend to be quite rough and it all becomes blurred. Which is probably also the reason why there is no real fluorine ampoule for element collectors. That was at least how I understood it until quite recently when I was listening to a talk by Florian Kraus, who is a fluorine chemist. And they are using a special polymer that can hold fluorine and other highly fluorinating interhalogens but is much more transparent. So I guess by now there is quite a variety of polymers available that will hold fluorine.
Elemental fluorine could be stored in glass at least theoretically. It is not the fluorine that attacks the glass but hydrogen fluoride or I think even more precise something like $\ce{H2F^+}$. You could avoid that to some degree by drying the fluorine but the problem is the glass itself. As glass forms a polymer structure of $\ce{[SiO4]^4-}$-tetrahedra that join all vertices there has to be a sudden end once there is no more repeating unit added. But that would mean that the outside of a glass surface is all negatively charged (ignoring metal cation additives). Therefore there are many $\ce{Si-OH}$ groups on the outside which tend to be acidic. When fluorine reacts with them it will form hydrogen fluoride that will attack the glass.

I see what you mean about the glass as well, thank you for this thorough answer! References in, and answers to pH probe bulb - what is happening within the glass? are reminders that hydrogen is not easily off-put by glass, and if thin enough, has its way eventually.


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On the stability of carbocations
I have a question regarding the stability of carbocations. It is shown below.

I am confused between options II and III. The structure shown in option II has more resonance structures. Thus, I believe it should be more stable than that in option III. However, the given answer is option III.

Related: https://chemistry.stackexchange.com/questions/24549/why-are-sp-hybridized-carbocations-high-energy?noredirect=1&lq=1 https://chemistry.stackexchange.com/questions/10509/why-are-vinylic-and-arylic-carbocations-highly-unstable

Electronegativity of the atom which holds the positive charge is a dominating factor over resonance, in option 2, the positive charge is in an sp hybridized carbon atom which is more electronegative than th sp2 hybridized carbon atom on which the positive charge resides in option 3, and positive charge on more electronegative atom is very unstable. Hence 3 is way more stable than 2. Vinylic carbocations are very unstable and rarely encountered.


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A problem on stability of carbanion

Determine the most stable of the two carbanions.
I found second one as more stable because of inductive effect. But the given answee is first one.

Do you mean carbocation or carbanion?

Resonance is always a dominating factor over inductive effect. In the first option there is a d-orbital resonance of chlorine atom. The d-orbital of chlorine atom aligns itself parallel to the overlapped p-orbitals of the pi-bond and the p-orbital of the negative charge. Clearly, they are conjugated. Though resonance involving d-orbitals and p- orbitals are weak as compared to resonance involving all p-orbitals, this effect will still dominate over the inductive effect of fluorine. As fluorine does not have d-orbitals, it won't show any d-orbital resonance.


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Rate constant converion
I  need to calculate the half life for a first order reaction for which I need the rate constant $k$. From literature I observed that for a very closely simulated experiment I obtained the numerical value of $k$ but the unit of $k$ was given in $\frac1{\mathrm s}\cdot\frac1{\mathrm{kPa}}$.
I am only familiar with $k$ in $\frac1{\mathrm s}$. from which I can easily estimate the $t_{1/2}$.
Can somebody please explain how I convert the $\frac1{\mathrm s}\cdot\frac1{\mathrm{kPa}}$ unit to $\frac1{\mathrm s}$.
Additional informaton: The laboratory experiment was conducted at approx $85\ \mathrm{kpa}$ and approx $100\ \mathrm{kpa}$ is $1\ \mathrm{atm}$.

First order rate constants are only in  1/time, i.e. the inclusion of pressure does not make any sense if it is first order.
If $X$ is a quantity proportional to   the concentration then for a first order reaction $X=X_0e^{-kt}$ where $X_0$ is the amount at time zero. Hence $\ln(X/X_0) = -kt$ from which you can see that the units of $k$ are 1/time.


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Can someone tell the me the data bank which has all the STO-3G basis sets for many electron atoms?
I am trying to express the slater orbitals of Helium atom in terms of 3 gaussian functions (STO-3G basis set). Is there any data bank or reference table in which I can find the exponents and coefficients in it ?
Also in primitive cartesian gaussian , it seems like both $2s$ and $2p$ has the same functional form. For $2p$ it is $x\exp(-\alpha x^{2})$ , $y\exp(-\alpha y^{2})$ and $z\exp(-\alpha z^{2})$. What is for  $2s$ orbital in primitive gaussian function ?

A wide range of Gaussian basis sets for all elements can be found at the Basis Set Exchange. Alternatively, you can try to extract that data from the basis set library of a quantum chemistry software package.
The functional form of a primitive cartesian gaussian with $l=0$ is $\exp(-\alpha r^2)$, independent of the $n$ quantum number.

Extremely sorry


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Roasting pyrite - product magnetite
When roasting pyrite, the products are usually Fe$_{2}$O$_{3}$ and SO$_{2}$, but is it possible to get magnetite and SO$_{2}$ from roasting pyrite? Thanks for answers.

No. The roasting must go all the way to Fe2O3, ie Fe+3, to get rid of all the S in the pyrite.


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Depression of freezing point
Recently, I learnt that by adding a non-volatile solute to a solvent, the  vapour pressure of the solution is decreased and consequently its boiling point.
Facts and definitions
freezing point: "The temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase."
So as the solution cools down, the vapour pressure decreases and eventually equals the vapour pressure of the solid.
My argument as to why depression in freezing point shouldn’t take place
consider this argument:the atmospheric pressure is constant regardless of the temperature of the solution.
As we cool down a solution, we are basically removing the kinetic energy of the solution's particles  so due to less kinetic energy and due to atmospheric pressure the Van der Waal's forces come to play, hence turning the solution into a solid state.
the pressure exerted on the liquid by the atmosphere is the same whether you add a solute or not, so no depression of freezing point should take place.
Reality check
I know that this argument is not true, as experiments proves the colligative properties are true.
Where and why my "intuitive" argument is failing?

Kindly reopen the question , I have tried my best to make it understandable and aesthetic.

I still don't get it, but there is one obvious error in your reasoning: The atmospheric pressure is totally irrelevant, only the partial pressure of the substance in question plays a role.

You argument is largely based on enthalpy, whereas freezing point depression and boiling point elevation is (in the situation when the solute only weakly interacts with the solvent) mainly an entropic phenomenon.
Because of the increased entropy by adding a solute (i.e. there are more configurations by having two types of molecules instead of one), the solution will liquify at a lower temperature as compared to the pure solvent and with a similar reasoning, the boiling point will increase with respect to the pure solvent.

but the reason given by my teacher and and textbooks is that , add solute -vapour pressure decreases-bp and mp decreases-as they are dependent on the pressure above the solution

Perhaps I am overlooking something, but to the best of my knowledge the boiling point increases when you add a weak-interacting solute to a solvent. For instance, have a look at this page: https://en.wikipedia.org/wiki/Boiling-point_elevation Did your teacher (or textbook) by any chance mentioned some additional details?

yeah boiling of liquids takes place when vapour pressure is equal or greater than atmospheric pressure

but adding of solute decreases the vapor pressure. now the elevation of boiling point completely makes sense to me whereas depression in freezing point does not

this statement "weak-interacting solute to a solvent" is true for raoult's law-ideal solutions

For strongly interacting solutes, you would indeed expect the boiling point to decrease as more energy is required to bring the solvent into the gas phase. That does not readily explain the freezing point depression, as by the same reasoning you would expect the freezing point to increase if the interaction is stronger. If on the other hand the interaction between solute and solvent is similar in both liquid as well as the solid phase, then the entropy argument comes back into play.

my bad sorry it goes like this, the reason why the vapor pressure (total) decreases because the non-volatile solute takes up some space on the the surface hence not much solvent particles are permitted to escape. also i read somewhere that even solvent and solute interaction can decreases vapor pressure. im not sure of second one but first one is true

No need to apologise. I am happy you found the reason. The devil lies in the details. :-)

also the statement "weak-interacting solute to a solvent" doesn't apply for colligative  property, the property only depends on concentration of solute in solvent . the interactions between them doesn't matter. also bp and mp increases and decreases resp. only when you add non volatile solute

Let us continue this discussion in chat.


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nickel-copper fabric: shielding viability after being subjected to weathering
I know the solution can be as simple as a well-sealed garbage can, but I am contemplating making a more malleable fabric based faraday-cage-esque contraption. The plan is to use nickle-copper fabric to achieve the shielding, I'll experiment with how many layers/folds create the best seal. It is to be placed outside, exposed to the elements, potentially incurring moderate to heavy rain. Before I get too heavily invested in this, I would like to really think about the shielding viability of the fabric in the long-term.
For reference, here is a generic picture of the nickel-copper fabric. Amazon has its fair share too, for those interested.

Question:
Unlike its ferrous counterparts, nickel/copper do not rust but rather have a patina or tarnish that develops after weathering. Will nickel-copper fabric tarnish like we might expect a penny to tarnish? If so, what compromises to shielding would be sustained? Also, if so, would that compromise the structural integrity of the fabric (more prone to tears, ect)?
Further Clarifications:
 - exact chemical makeup of the fabric is uncertain, the one I bought was simply marketed as "nickel-copper fabric" so I assume its mostly nickel and copper, but it seems to have a fabric-like feel, so perhaps its mixed with some kind of substrate or something. If the consensus is that this information is crucial I'll try to track down the manufacturer and get specifics, but at the moment I don't have a practical means of going about that(manufacturer is unknown). You are free to speculate about common implementations of nickel-copper fabric and state your assumptions accordingly

shielding: I'm concerned that if the fabric weathers poorly, the shielding will be affected
tear resistance: I'm concerned that if the fabric weathers poorly there may be more points of failure, leading to easier tears in the fabric
health: the fabric will occasionally come in to close contact with humans (me) and I'm not sure if tarnished nickel-copper has any adverse health risks
set up: my design is too hypothetical to put into words at the moment, but you can imagine a tarp-like thing, outside layer is cordura (water-resistant, not water-proof) and inside layer is nickel-copper fabric
placement: outside, exposed to the elements, incurring rain

Apparently the fabric you are referencing is made from metal-coated polymer (PET) fibers. Unless covered with protection layer, it is likely to degrade over time. || Also, you might want to look for aluminium wire mesh.

Your alloy should be sufficient. For one, cupronickel alloys are very corrosion resistant and is used to make many different coins across the world. occasionally yes, a nickel or dime will tarnish but this is usually from exposure to soda, salt or some other corrosive liquid, other wise they can sit in a fountain for weeks without tarnishing. AS long as the acid rain isn't too severe, you should be okay.
As for the product you are considering, the description on amazon describes it as:

Copper+Nickel+Polyester Plated Fabric

Now plated metals do not necessarily have the same properties as the pure casted alloy since to plate on polymer requires using electroless plating and does not guarantee the same microstructure as a casted alloy, but corrosion should be unaffected.
See also: Monel

Weathering probably won't be a problem as long as it isn't exposed to acids too much. This alloy should be resistant to tarnishing especially well. On your health concern, it really should be fine; I wouldn't worry about that. Overall, sounds like a decent setup.


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UV Absorption of iodine with some solvent
$I_2$ in solution is typically absorbing around $525 \; nm$ this is why it appear to be magenta, I recently learn that with some solvent :

Benzene
Toluene
$o$-Xylene
$p$-Xylene

The coordination complex between the solvent and $I_2$ is also absorbing around $300 \; nm$ I guess there is some explanation with the MO diagram ?

Iodine is highly polarisable and will form a (Mulliken) charge-transfer complex with many aromatics. Very many pairs of donor - acceptor pairs do this and have been studied for approx $100$ years, see for example tetracyanoethylene–pyrene (J. Chem. Phys. 105,1996 p2287) for an ultrafast time resolved study.  In  absorbing a photon the ground state complex DA transfers  an electron directly to a charge transfer state $DA \overset{h\nu} \to \mathrm D^+A^-$. The interaction is thus that of all the orbitals in one molecule with all those in another.
Books
R. S. Mulliken and W. B. Person, 'Molecular Complexes' Wiley, New
York, 1969. and G. Briegleb, 'Elektronen Donator Acceptor Komplexe' Springer, Berlin, 1961.


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Order Of Reaction
A+B ———> C+D
If A and B are taken in excess then what will be the order of reaction?
I thought since it is taken in excess, their change wont really be significant and so answer would be 0.

Having both A and B in excess makes no sense. There are only three possible cases. (1) A > B, (2) A = B, or (3) A<B. If (1) Then A is in excess. If (3) then B is in excess.

Reaction orders are not related to excesses at all. The question is wrong in its premises.

No like pseudo kind of  rate order@IvanNeretin

If the  reaction is a kind of heterogeneous catalytic reaction or photochemical reaction your answer correct

3 doesn't make any sense for the reaction order.
Given the elementary reaction
$$\ce{aA + bB -> \text{products}}$$
then the rate equation would be
$$r = k\ce{[A]^a[B]^b}$$
since by definition an elementary reaction is a reaction for which the stoichometric coefficients are the same as exponents for the concentrations in the rate equation. Thus the order of the reaction is $a+b$.
So for the case of the elementary reaction
$$\ce{A + B -> \text{products}}$$
$a=1$ and $b=1$ so the reaction order is 2.

Reaction orders are defined per component. If both A and B are in excess (but available in roughly the same concentration), the reaction order would be one in either of them.
If A or B is in excess, then the reaction order is one for the compound in excess and zero for the other compound.


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Reaction between HCl and NaHCO3
I want to know why in the reaction between HCl and NaHCO3, we have CO2 and H2O instead of H2CO3 ?

see https://en.wikipedia.org/wiki/Carbonic_acid#Chemical_equilibrium

H2CO3 is stable at high pressure or excess of water only


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How can iodized oil be made?
I want to combine solid elemental iodine with olive oil, to form an iodized oil - what is called an iodostearic acid or iodooctadecanoic acid,  here the double bond in the oil becomes saturated by iodine. I have not done this before and I am looking for guidance on how to do this with basic kitchen tools. Would this be as simple as adding iodine and oil in the ratio of 8 parts iodine to 10 parts oil (as the iodine number for olive oil is around 80), and then waiting several hours or days for the reaction to take place? Is heat or some catalyst needed? Please advise! Thank you.

Why would you waste olive for that?

See also W: Iodine value

Essentialy yes, this will work for an unsaturated acid. It will work better if you dissolve your acid substrate in a volatile organic solvent (ether or dichloromethane would be good) and stir it efficiently, wash with sodium thiosulfate solution to remove any unreacted iodine, then allow the solvent to evaporate.

Yes, but the OP was talking about starting with olive oil which is unsaturated but which will end up saturated if he adds enough iodine.

My bad. Edited my answer

doi:10.1016/S0140-6736(97)02427-6 might also be interesting, in that it (1) describes the consequence of switching oils (2) mentions the actual procedure, i.e. reacting hydroiodic acid with alcoholic esters of oil FAs


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Feasibility of Electrophilic attack on the Nitrogen over the Phenyl ring of Aniline
I have learnt that the '-NH2' substituent attached to the phenyl ring, owing to the substitutent's highly positive mesomeric effect, activates the phenyl ring quite a lot for Electrophilic Aromatic Substitution, so much so that it easily forms a tri-substituted product when Br+ is used as an electrophile with Aniline, and that a protecting group such as Acetyl (introduced using Acetic Anhydride) has to be introduced on the Nitrogen to deactivate the ring for just monobromination to occur.
This made me wonder that if such is the influence of the '-NH2' on the benzene ring then why would the Nitrosonium ion generated in diazotization prefer attacking the lone pair of Nitrogen instead of the electronically enriched benzene ring on the ortho or para position?
Initially I lazily (and wrongly) attributed it to perhaps some stronger bonds forming between similar atoms, i.e Nitrogen-Nitrogen bonds being stronger than the Carbon-Nitrogen bonds.
But, more interestingly I realized that this line of reasoning wouldn't apply to my above stated acetylation of the Nitrogen in Aniline, for in that case too there isn't an Aromatic Substitution taking place and there are no Nitrogen-Nitrogen bonds.
In short, what is the reason for the feasibility of Diazotization and Acetylation in the way they occur with Aniline? Why aren't electrophilic aromatic substitution reactions taking place? I may have erred in my basics, please help me out.

Your capitalisation seems particularly bad, also a waste of time as no purely chemical name is capitalised (this of course doesn't count other reasons like  start of sentence, or reactions named after people.

Electrophilic Aromatic substitution in Aniline will occur when the $\pi$-electron cloud of Benzene ring attacks the electrophile generated (mostly in-situ)  instead of the lone pair on the Nitrogen atom. Thus for occurance of Electrophilic aromatic substitution, there always remains a competition between the $\ce{N}$ atom and the $\pi$-electron rich benzene ring. You may think that the lone pair of nitogen remains fully delocalised through the Benzene ring, but there always remains a high enough electronic charge density on the Nitrogen atom to behave as a nucleophile and in fact in greater extent the electronic density of the ring (not always for Oxygen though, being highly electronegative, its less nucleophilic, but Nitrogen is certainly more nucleophilic).
Thus,you will observe that it is difficult to perform electrophilic aromatic substitution reaction on  unprotected Aniline. Thus, most of the reactions on Aniline are initiated by attacking of the elctrophile by Nitrogen atom instead of the ring. It is its general trend. Thus to do substitution in the ring, we first need to decrease electronic charge density on the Nitrogen atom and that's why we first treat the Aniline with $\ce{CH3COCl}$ in Pyridine medium to form acetanilide, and then perform any other reaction on the ring like Bromination, Chlorination, Alkylation etc.
You will also observe that Aniline doesn't undergo Friedel Crafts alkylation/acylation to a great extent. As at the moment you add Lewis acid, Nitogen atom behave as Lewis Base and forms the adduct, which make the ring highly electron deficient due to induced positive charge on Nitrogen.
Thus, the Diazotization  and Acetylation are the normal/expected reaction of Aniline as a effect of higher nucleophilicity of $\ce{N}$(due to lesser elctronegativity and weaker basicity). Instead if you have to perform Electrophilic aromatic substiution in the ring, you have to think about how to perform it.

Same thing as in earlier comment ^

Adding to Soumik Das' great answer,
The ring would have to be activated to a large extent to overcome the nucleophilicity of aniline's nitrogen if you would like to add the nitrosonium ion at the benzene, as the aromaticity of the ring would have to broken in the rate determining step, which would have significant activation energy demands.And indeed, it has been observed that for tertiary aryl amines, electrophilic aromatic substitution of the NO+ ion takes place majorly.
As a sidenote, if you consider diazotization in the case of tertiary amines, then according to the mechanism, we need to have atleast one hydrogen present on the nitrogen(of the amine) for the tautomeric step to take place. But due to the lack of H in tertiary amines, the addition of electrophile and it's removal from the amine will probably exist as a parasitic equilibrium