text:
Structure of NO2 molecule
Which is the correct structure of $\ce{NO2}$? While searching the internet I found out that

$\ce{NO2}$ have a coordinate and two covalent bonds. $\ce{N}$ will have a positive charge. $\ce{O}$ (coordinate bond) will have a negative charge (this is the  part which I don't understand. As $\ce{O}$'s octet is complete, why will it have a negative charge?)

$\ce{NO2}$ came from $\ce{HNO2}$ that's from $\ce{O}$ have a negative charge. but in this molecule, there is no $\ce{O}$ with coordinate bond.

In one molecule $\ce{N}$ have one electron and positive charge and in other two electrons and no charge?

Are you asking about the Lewis structure together with the formal charges? You can have a non-zero formal charge for atoms that have an octet - the way the electrons are counted is different for those two concepts. It would help if you edited your answer to show the Lewis structures you are talking about. As written, I'm not even sure if you mean $\ce{NO2}$ or $\ce{NO2-}$.

$\ce{NO2}$ is a free radical. Its resonance structures are as follows:

Unsurprisingly, $\ce{NO2}$ reacts readily with a variety of substances, including itself at low temperatures to form $\ce{N2O4}$.

From Wikipedia:

Nitrogen dioxide at −196°C, 0°C, 23°C, 35°C, and 50°C. ($\ce{NO2}$) converts to the colorless dinitrogen tetroxide ($\ce{N2O4}$) at low temperatures, and reverts to $\ce{NO2}$ at higher temperatures.

where did this minus sign came from?

@Bhavay are you referring to the oxygen? It is just the formal charge on it for that resonance structure. There is no true negative charge on the oxygens. It is a neutral molecule. This may help.

n will have a formal charge of +1? right

The molecule is neutral, it can't have an unbalanced negative formal charge. As is the nitrogen should have a positive formal charge (seven valence electrons).

what about hno2 then .is no2- from hno2 and  no2 radicle same ?

@Bhavay No, $\ce{NO_2^-}$ has one additional electron compared to $\ce{^\cdot NO_2}$. They're not the same.

Yes, nitrogen has five Valence electrons (@orthocresol seven total). Please count again, because I only see four electrons around nitrogen in either resonance structure you have drawn. A (neutral) molecule can never have an unbalanced formal charge.

@ orthocresol @Martin-マーチン thank you for pointing that out. I was a little too hasty in making my drawings.

The ONO bond angle is $\approx 134$ degrees and NO bond length $120$ pm. EPR studies show that the unpaired electron is in a $\sigma$ rather than $\pi$ orbital on the N atom, (i.e in an sp$^2$ hybrid orbital), as shown in the middle and left resonance structures in the figure. Walsh diagrams explain this also.


text:
Why is only one of the lone pairs on the oxygen in furan delocalized?

Both the lone pairs on the oxygen are in allylic positions, thus both should be able to participate in resonance. If both lone pairs can participate in resonance, why is only one of the lone pairs delocalized?
Also, for a lone pair to participate resonance, does it have to be in an unhybridized p orbital? If so that would explain why only one of the lone pairs is delocalized as one of the lone pairs would be in an unhybridized p orbital to make it sp2 for oxygen.

Yes, your last statement is exactly why. The p-orbital on the oxygen overlaps with the p-orbitals on the carbons, the sp2 electron pair does not.

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text:
Why is adenine aromatic?

I don't think it follows the $4n + 2$ rule. The double bonds give 8 π electrons. Then the lone pairs on the $\ce{NH}$ and $\ce{NH2}$ groups are delocalized (are they?) so they must contribute 4 π electrons. Thus in total there are 12 π electrons, which doesn't follow Hückel's rule. So, why is it aromatic?

Hückel's rule is for conjugated circuits, so $\ce{NH2}$ group is irrelevant, whereas $\ce{NH}$ is (lone pair on $\mathrm{2p}_z$ orbital), resulting in $n = 10$ and overall aromaticity.

Also, Hückel's rule is for one cycle.

The $4n+2$ rule applies only for a conjugated ring.  If you assume that the amino group contributes a pair of pendant  $\pi$ electrons into the ring system then perforce you do not have a conjugated ring.  And even if you don't assume such a delocalization from the amino group, the cross-link within the ring system is still a stretch for the $4n+2$ rule.

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text:
What will be the major enol form of methyl acetoacetate?

I presume the best enol form would have a double bond between the two $\ce{C=O}$ so they're conjugated, but I'm not sure how the resonance would play out.

The enol forms of beta-diketones and beta -ketoesters are stabilized
by intramolecular hydrogen bonds in its enolic form. Enol structure with greater conjugation is more stable and amajor enol.

update :
In organic chemistry terms, conjugation is used to describe the situation that occurs when π systems (e.g. double bonds) are "linked together".
A conjugated system requires that there is a continuous array of "p" orbitals that can align to produce a π bonding overlap along the whole system.If a position in the chain does not provide a "p" orbital or if geometry prevents the correct alignment, then the conjugation is interrupted (broken) and therefore lost at that point. Due to extra π bonding interactions between the adjacent π systems there is an overall stabilization of the system .This is referred to as the delocalisation energy or the resonance energy or conjugation energy

source:
http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch10/ch10-1-1.html
According to MO theory,1,3-butadiene ,comprised of four overlapping p orbitals are mathematically combined to produce four molecular orbitals.As seen in figure , the four π electrons of a conjugated diene occupy the two lowest energy MOs (the bonding MOs).Specifically, the two electrons that occupy ψ1 are delocalized over four carbon atoms. This delocalization accounts for the observed stabilization energy .

source :
Organic chemistry ,David R. Klein.page no.773
ref :Advanced Organic ,Chemistry FIFTH EDITION ,Part A: Structure and Mechanisms FRANCIS A. CAREY
and RICHARD J. SUNDBERG

Based on abrupt paragraphs with numerous new lines and wrong Unicode characters (e.g. Eszett in place of beta) the answer looks copy-pasted from a poorly OCRed book (the ref you mentioned?). If you are quoting a third-party source, please use corresponding markup (> ...) and do at least minimal corrections to the text to make it more readable/editable; also use proper bibliography referencing with page numbers.

@andselisk i have reformatted .Thankyou.

Ehhm, I see you removed a lot, but it's still not clear 1. what is a quotation and what is your work; 2. what page number(s) were used. Also, the formatting is still lacking proper spacing here and there, plus there is a bunch of links for the images at the end that are not used elsewhere.


text:
Why do we use external pressure when calculating work done by gas?
For irreversible expansion and compression, why is dw= -pextdV but not dw= -pdV ?
Many of the texts and related questions on this site I encountered give a same general idea: because work done= opposing force x displacement. Using this reasoning, I cannot explain my following thought processes:
(a) In the case of compression, the opposing force is obviously caused by p but not pext. Based on Atkins' Physical Chemistry Ninth Edition page 50, we use pext even for compression.
(b) The explanation given by Atkins' Physical Chemistry (same page) is:

'when a gas is compressed, the ability of the surroundings to do work
  is diminished by an amount determined by the weight that is lowered,
  and it is this energy that is transferred into the system.'

Sounds right. Now if I use this reasoning on expansion,

Imagine I have two pistons, two filled with different amount of gases at same volume and so different pressure.
Envisage the external pressure being a load resting on the piston.
When both gases expand by the same amount, the work done by both gases, based on the formula dw= -pextdV and Atkins' reasoning, are the same. This is because the load moved up by same vertical distance and the changes in volume of gases are the same for two piston.
However, logically, that piston with higher pressure should move faster and so does the load on it. This would mean that, that load actually has higher kinetic energy.
With both gases doing the same amount of work, two loads gain the same gravitational potential energy, but one of the loads gains greater kinetic energy, means that one of the loads actually gain greater amount of total energy. How is this possible?

(c)
We always say when p=pext during expansion, the work done by the system is maximum, otherwise 'work wasted'. How can you say that it is wasted when the work actually used to increase the kinetic energy of the load as a whole?

How can one piston have higher pressure on it if the weight is the same? For the two vessels to have the same volume but with different pressure at the start, the weights would have to be different. So, even if they expand by the same amount, the work done will clearly be different as the mass moved is different.

@matt_black The external pressure(=weights) are same for two vessels, whereas the gas pressure are different for two vessels, initially.

@Mithoron I read that post before. Did you actually read the title only or already read it thoroughly? I am looking for answer for part (b) especially.

@The99sLearner What you claim are the initial conditions cannot be true. If the external pressure is the same but the amount of gas is different, then the volumes will be different. The only way to have different amounts of gas at the same volume is to have different pressures and you can only get that by having different weights (the loads resting on the piston).

@matt_black Yes, what you said is true. If the load(=p external) is smaller than pressures of both gases, the gases will expand. I assume you are just saying that the two gases cannot be 'kept still', but I am considering the states during expansion such that, at one particular moment during the expansion, one gas has higher pressure than another.

The ideal gas law (or other equation of state) is valid only when a gas is in thermodynamic equilibrium.  In a rapid irreversible expansion or compression, the pressure depends not only on the volume (and temperature) but also on the rate of change of volume.  This is because a gas has a viscous property which acts to dissipate mechanical energy and convert it to internal energy. In a reversible process, where the rate of change of volume is negligible, the ideal gas law can be used.  But the ideal gas law (or any other equation of state for that matter), cannot be used to determine the pressure (or more precisely, the compressive stress) in an irreversible process.
At the interface between the gas and its surroundings (say the inside face of a piston) where the gas is doing work, by Newton's 3rd law, the pressure of the gas matches the "external pressure," and, if this external pressure is known, the amount of work that the gas does on its surroundings can still be determined. In the "vertical piston" example, if one does a force balance on the piston during the process, one obtains:
$$P_{gas,I}=\frac{Mg}{A}+P_{atm}+\frac{M}{A}\frac{d^2V}{dt^2}=P_{ext}$$ where $P_{gas,I}$ is that gas pressure at the interface with the piston (the subscript I is present because, in an irreversible process, the pressure of the gas within the cylinder may not even be uniform), and the mass times acceleration of the piston is included. The work done by the gas on its surroundings during the process is obtained by integrating this equation between the initial and final equilibrium states to yield$$W=\int{P_{gas,I}dV}=Mg\Delta h+P_{atm}\Delta V=\int{P_{ext}dV}$$This equation acknowledges that, as a result of the viscous property of the gas, the kinetic energy of the piston has been damped out by the time the system reaches its final state.


text:
In a weak acid-strong base titration why is [H+] no longer equal to [A-] after addition of the base? (A2 level chemistry)
I understand that before any base is added $\ce{[H+]}$ is equal to $\ce{[A-]}$ but why is $\ce{[H+]}$ no longer equal to $\ce{[A-]}$ when the strong base is added?
If all the following reactions below are occurring in solution and both $\ce{CH3COO-}$ an $\ce{H+}$ are able to react in either one of two of these different reactions then why should the $\ce{[H+]}$ no loner equal $\ce{[A-]}$ if both have equal opportunities to be consumed? (I am aware that not all the $\ce{CH3COOH}$ will have dissociated but surely this would not affect the concentrations of either $\ce{[H+]}$ or $\ce{[A-]}$?)
$$\ce{Na+ + CH3COO- -> CH3COONa}$$
$$\ce{CH3COO- + H+ -> CH3COOH}$$
$$\ce{H+ + OH- -> H2O}$$
Is it because the salt $\ce{CH3COONa}$ is much more likely to dissociate than $\ce{CH3COOH}$ and $\ce{H2O}$ meaning more $\ce{CH3COO-}$ remains in solution than $\ce{H+}$?

$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$
$$\ce{NaOH + H3O+ -> Na+ + 2H2O}$$
The base removes the hydrogen ions from solution, and the acid further dissociates to maintain LeChatelier's equilibrium. The $\ce{H3O+}$ ion concentration will decrease as it is continually removed whereas the $\ce{A-}$ concentration will increase until the acid is fully neutralized into a salty solution.
I can't make much sense of your chemical equations, but it looks to me like they are incorrect. There is no reason for formation of the ionic compounds in solution. They should actually be written in reverse. And you have no strong base to react with the hydronium ions.


text:
Are there known compounds of caesium with oxidation number >1?
Given that the left neighbor of caesium, xenon, does have fluorides and oxides, it is not inconceivable that caesium can have oxides or fluorides with an oxidation number higher than 1. Are such compounds known and synthesized? Or, alternatively, are there computations that rule out such compounds as unstable?

Check out this answer and comments, to begin with.

@andselisk: Thanks for the link, unfortunately it is mainly about alkali anions with a short mention of ${\rm Cs}^{3+}$ that is not fleshed out.

As noted in the answer to the other question, alkali metal can exist in higher oxidation state when bonded with polycylic multidendate ligands like cryptands etc. This is an excerpt from an eBook1:

The chemistry of group 1 elements have been dominated by +1 oxidation
states. However, there have been indications that caesium might form
higher oxidation species. Thus, electrochemical oxidation of
$\ce{[CsL]PF6}$ L=18-crown-6 or cryptand-[222] gives evidence for
$\ce{Cs^2+}$ and $\ce{Cs^3+}$. Compounds containing caesium in higher
oxidation state is yet to be isolated.

Caesium is predicted to form polyfluorides but is yet to be needs validated through experiments. This below is from Wikipedia (also citing two references - Ref.2 and Ref.3)

caesium is predicted to behave as a p-block element and capable of forming higher fluorides with higher oxidation states (i.e., $\ce{CsF_n}$ with n > 1) under high pressure. This prediction needs to be validated by further experiments.

Furthermore, it has also been observed by Moock and Seepelt, Angew, Chem. Intl. Ed. Engl. 28, 1676(1989) that $\ce{Cs+}$ can be oxidised in acetonitrile solution to $\ce{Cs^3+}$ at a potential of +3.0 V($\pu{E°}$)4.
References

Introduction to Modern Inorganic Chemistry, 6th edition by R.A. Mackay, W. Henderson, 2017, CRC Press.
Miao, Ms. Caesium in high oxidation states and as a p-block element. Nature Chem 5, 846–852 (2013). DOI: 10.1038/nchem.1754
Forcing Cesium into Higher Oxidation States Using Useful hard x-ray Induced Chemistry under High Pressure by Sneed et. al., J. Phys.: Conf. Ser., 2017. DOI: 10.1088/1742-6596/950/4/042055
Inorganic Chemistry, Gary Wulfsberg, University Science Books, 2000

I think it is worth noting that while all these give hints that Cs+ can, under extreme conditions, be oxidised further, (to my knowledge) no compounds with Cs in a high oxidation state have actually been isolated. I also wish there were a lot more details on how those calculations were done ....

I cannot trace the ebook reference to a primary source because the primary references are on a different page and the book is behind a paywall. So can you please provide the primary source?

@OscarLanzi Commented in the other answer.

Note that IE of Cs+ is near identical to IE of He.

@Poutnik the high second ionization energy of Cs is indeed unfavorable, but the general agreement is that caesium in higher oxidation state will form covalent bonds. This would mitigate the ionization energy effect.

Not known compounds, but Moock and Seppelt[1] have some evidence that at least some short-lived caesium(>I) species exists.
In both Ref. [1] and an earlier work cited by the authors, electrochemical techniques reveal that caesium-bearing electrolytes give an oxidation reaction thar does not appear when the caesium is replaced by potassium or rubidium. The authors note that $\ce{Cs(III) + Cs(I)}$ is more stable than $\ce{2 Cs(II)}$, so they generally speak of caesium being oxidized to the $+3$ oxidation state.
It is possible that caesium ion is actually catalyzing the oxidation of the solvent or the ligand used to complex the caesium (which would account for observations that the oxidation is irreversible in most of the experiments), but such catalysis would likely involve an intermediate caesium(>I) species whose counterpart is absent with lighter alkali metals. Given the fact that no such higher oxidation state species is as yet known as a reaction product, this hypothesis of a short-lived intermediate seems most plausible.
Wikipedia suggests that a francium(V) fluoride complex may he possible, but this is not experimentally verified.
Reference

Moock, K. and Seppelt, K. (1989), "Indications of Cesium in a Higher Oxidation State". Angew. Chem. Int. Ed. Engl., 28: 1676-1678. https://doi.org/10.1002/anie.198916761.


text:
Is calcium chloride an acidic or basic salt?
What I reason:
Calcium chloride is the salt of hydrochloric acid and calcium hydroxide. Calcium hydroxide is usually not considered a strong base, and I believe this is because of it's low solubility. $\ce{HCl}$ is a strong acid, and so the salt should be slightly acidic. Wikipedia states the $\mathrm{p}K_\mathrm{a}$ of $\ce{CaCl2}$ is between 8-9, which is in fact slightly acidic, confirming my theory.
What I've experienced:
I've prepared a number of aqueous calcium chloride solutions from distilled water, all of which are purple in the presence of universal indicator and tested with a calibrated $\mathrm{pH}$  probe to be basic. I've tried multiple sources of calcium chloride and each is basic.
TLDR:
Theoretically and literature values state that $\ce{CaCl2}$ is acidic, but my empirical evidence shows that it is basic.

Calcium hydroxide has solubility about 1.9 g/L. This is enough to create pH above 11, i.e. a strongly basic solution.
$\ce{CaCl2}$ solutions should be very slightly acidic if they were made from pure $\ce{CaCl2}$. This might not be the case. In industry calcium chloride is produces by reaction of calcium hydroxide with ammonium chloride, so industrial-grade calcium chloride is likely to be contaminated with calcium hydroxide.
Another potential reason for contamination is hot drying. Calcium chloride is quite hygroscopic, so it needs to be dried before use. When heated to hight enough temperature, moist calcium chloride hydrolyzes https://link.springer.com/article/10.1007/BF02654424

"Calcium hydroxide has solubility about 1.9 g/L. This is enough to create pH about 10-11" I appreciate to solve and explain it.thanks.

@AdnanAL-Amleh pH+pOH is roughly 14. For Ca(OH)2 solution [OH] is roughly twice molar concentration of calcium hydroxide, which is roughly 0.025 for saturated solution. $pOH = -log [OH]$;  $log_{10} 0.025 = 1.6$;  $14-1.6 > 12$. You don't need much base to create a strongly basic solution.

You mean : $\ce{Ca(OH)2_\mathrm{(aq)} <=>Ca(OH)^+\mathrm{(aq)}  + OH^-\mathrm{(aq)}}$
, so: $$[\ce{OH-}]=[\ce{Ca(OH)2}] = \frac{1.9}{74}=\pu{0.025M}$$

@AdnanAL-Amleh Yes. 0.025 M for $[OH-] $ is a low estimate, it is probably higher because $\ce{Ca(OH)+}$ can dissociate too.


text:
Nucleophilicity and basicity of compounds with or without nitrogen inversion
In nitrogen inversion, the lone pair changes its direction continuously.
Does this in any way delocalize the charge density on the nitrogen making it less nucleophilic and less basic than, say, a bridgehead nitrogen?
Consider the specific case of quinuclidine and triethylamine. Which will be more nucleophilic among them and why?

Related, maybe almost a duplicate, but not quite as it doesn't ask about nucleophilicity: Does nitrogen inversion affect the basicity of amines?

related to https://chemistry.stackexchange.com/questions/38749/does-amine-inversion-affect-basic-strength

@orthocresol yes, I read that. Didn't find the answer too satisfying. It is too vague.

@user226375 I have the same question too. Even in electrostatic maps the negative charges looks like its located above the nitrogen but no density below. Have you find an answer?

Idk why this was overlooked, but please note that quinuclidine does NOT have a bridge head carbon. The bridge head atom is a nitrogen atom.


text:
What is the physical origin of conjugation?
I had a question about the origins of conjugation in conjugated molecules. Take two ethylenes as compared to butadiene. Huckel seemed to argue that conjugation was the result of pi-pi stabilization interactions between the HOMO of the two ethylene molecules. More specifically, he argued that during the creation of Ψ1 and Ψ2, the former is lowered in energy more than the latter is raised in energy, leading to overall stability.

While this explanation satisfies me very much, I have also come across some literature (especially in organic chemistry) citing that π->π* hyperconjugation the main contributor to the stabilization of conjugated molecules. While I do understand this, I "feel" that Huckel's definition is more satisfactory. I am wondering if anyone could give some validity to the latter and if someone could point to the major factor for stabilization in conjugated molecules.

One should keep in mind that most of these texts in organic chemistry contain quite a bit of post-hoc rationalisation. And even Hückel theory is not accurate enough to explain this. Within this, in butadiene the p orbitals will form a conjugated pi system (independent of sigma), and molecular orbitals will be occupied accordingly. Also note that anti-bonding orbitals are always more destabilising than bonding orbitals are stabilising.

@Martin So should we classify $\Psi_2$ as bonding or antibonding?

@KarstenTheis both. It's anti-bonding with respect to the central C-C bond, and bonding with respect to the terminal ones. Vice versa for psi 3.

@Martin That's very Zen.

@KarstenTheis I think it can be considered to be bonding solely because it is energetically lower than the origin carbon p orbitals. What Martin said about being both bonding and antibonding comes about because he was looking at the process by which the MOs were formed, which I suppose is also correct.

I think you have posed an interesting question. Upon looking through some literature, I have chanced upon the answer to your question in Fleming (2010), p. 30-31. The diagram below was also taken from the discussion in the book.
What you have proposed are two ways of constructing the $\pi$ molecular orbitals of butadiene from the $\pi$ molecular orbitals of two ethene molecules. What is normally discussed is the one shown on the left, whereby we consider the interaction between the two filled MOs of exactly the same energy and a similar interaction between the two unfilled MOs of equivalent energy. Considering the bonding and antibonding combinations allows us to derive the four $\pi$ Mos of butadiene.
You also propose considering $\pi$ to $\pi^*$ conjugation as an alternative way of deriving the butadiene $\pi$ MOs. Diagramfmatically, what you are essentially referring to is shown in the MO diagram on the right. Firstly, this interaction cannot be the main interaction between the two ethene $\pi$ systems simply because it would result in the butadiene $\pi$ MOs comprising of two separate levels of two degenerate MOs each, which is quite different from the picture which we are aware of, which has 4 $\pi$ MOs of different energies. Secondly, note that this interaction is considerably weaker due to the large energy difference between the $\pi$ and $\pi^*$ MOs of an ethene molecule. Recall that for a strong interaction between any two orbitals, one of the important conditions is that the energy difference between them must not be too large. In fact, the strongest interaction is obtained when the energy difference between them is zero. This interaction would contribute to the energies of the $\pi$ MOs in butadiene but the contribution would surely be less than the one that is highlighted in the previous paragraph.

To construct the MOs of the butadiene, it is incomplete to just consider the stronger interactions shown on the left. For completeness, we would have to then mix the resultant MOs obtained from these two interactions (i.e. both left and right diagrams), factoring in the extent of contribution of each interaction. Naturally, we would be curious as to what is the effect of the weaker interaction on the MO energies derived from the stronger interaction. Fleming (2010) describes the effects rather detailedly in the text:

Mixing these two sets together, and allowing for the greater contribution from the stronger interactions, we get the set of orbitals shown in the figure below. Thus, to take just the filled orbitals, we see that $\psi_1$ is derived by the interaction of $\pi$ with $\pi$ in a bonding sense ($\psi_a$), lowering the energy of $\psi_1$ below that of the $\pi$ orbital, and by the interaction of $\pi$ with $\pi^*$ in a bonding sense ($\psi_w$), also lowering the energy below that of the $\pi$ orbital. Since the former is a strong interaction and the latter weak, the net effect is to lower the energy of $\psi_1$ below the $\pi$ level, but by a little more than the amount ($\beta$ in simple Huckel theory) that a $\pi$ orbital is lowered below the p level in making the $\pi$ bond of ethylene. However, $\psi_2$ is derived from the interaction of $\pi$ with $\pi$ in an antibonding sense ($\psi_b$), raising the energy above that of the $\pi$ orbital, and by the interaction of $\pi^*$ with $\pi$ in a bonding sense ($\psi_x$), lowering it again. Since the former is a strong interaction and the latter weak, the net effect is to raise the energy of $\psi_2$ above the $\pi$ level, but not by as much as a $\pi^*$ orbital is raised above the p level in making the p bond of ethylene. Yet another way of looking at this system is to say that the orbitals $\psi_1$ and $\psi_2$ and the orbitals $\psi_3^*$ and $\psi_4^*$ mutually repel each other.

Does it really matter if we factor in the weaker interaction into our considerations? Yes, it does. If we were to not do that, we would observe that the stabilisation predicted would not be as great as the one which we would predict from just looking at the MO diagram shown on the left. In fact, as Martin writes in the comments, the interaction between orbitals would produce an antibonding MO that is more antibonding than the bonding MO produced is bonding. If we were to not consider the weaker interaction, then we would predict that $\psi_2$ is raised more in energy from the original $\pi$ level than $\psi_1$ is lowered in energy from the original $\pi$ level (i.e. $\ce {E_2}$ is larger than $\ce {E_1}$), and this is clearly not the case.
Of course, as you mentioned, we can also build the $\pi$ MOs of butadiene from first principles via the Huckel method. I am not very well-versed with the mathematics behind the method but I do believe that the method discussed above and the original Huckel method would converge in terms of the results obtained eventually. Hence, they are equally valid approaches to the problem.
Reference
Fleming, I. Molecular Orbitals and Organic Chemical Reactions (Reference ed.). United Kingdom : Wiley, 2010.

This summarized the book by Fleming (2010) beautifully. I happened to read the book prior to posting this question, but I must have skipped over the section detailing this. So it looks like pi->pi* hyperconjugation is responsible for the "extra" overall stabilization in butadiene over two ethylenes. Also, this is a organic chemistry approach, and as Martin said in the comments, an ad hoc approach to summarizing first-principle results.


text:
How addition of a soluble salt affects equilibrium involving precipitation with a base sharing the same cation?
I have the following reaction at equilibrium:
$$\ce{Ba(NO3)2(aq) + KOH(aq) <=> Ba(OH)2(s) + 2 KNO3(aq)}$$
Will there be a change in the equilibrium if the salt $\ce{KCl}$ is added? Will the reaction shift in a certain direction to restore equilibrium?
I know that with adding $\ce{KCl}$, $\ce{K+}$ ions are added, but I am not sure if that is related to the answer.

Start by writing the net ionic equation.

And also balance your equation


text:
Why do we use LiCl as a source of chloride ions as nucleophile but not other alkali metal chlorides?
$\ce{LiCl}$ has higher covalent character than other alkali-metal chlorides, so in polar solutions like ethanol, its ioniztion will be less, and so less chloride ions will be available for participating in substitution reaction.
So why don't we use sodium chloride or other alkali metal halides which are more ionic?

Could you edit your question to give an example of a reaction that would work with LiCl but not other chlorides, and add why you think it might work with those? Right now, it's unclear what you are asking.

One reason is lithium chloride is more soluble in the organic solvents often used for such reactions, and getting the chloride source into the reaction medium is more important than having it be perfectly ionic.
Let's roll the tape.  Wikipedia identifies lithium chloride as soluble in a variety of polar solvents and gives numerical values for several of them.  The solubility reaches hundreds of grams per liter in some of them, including over 400 grams per liter in methanol.
Much less data are listed here for sodium chloride, but the solubility of sodium chloride in methanol is listed as 14.9 grams per liter under ambient conditions, about 30 times less on a mass basis than lithium chloride.  And the chloridecomponent shows an even bigger discrepancy because of the heavier mass per mole of chlorine in sodium chloride versus lithium chloride.
Clearly lithium chloride's solubility carries over better from water to polar organic solvents.  The reason for this is lithium chloride is more adaptable.  The small size and higher polarizing power of the lithium ion imparts greater covalent character in lithium chloride than in its heavier alkali metal congenres.  Alternatively, in a purely ionic model, it promotes the formation of ion pairs which are uncharged overall because of the equal but opposite charges of the constituent ions.  Either way, the lithium chloride adopts a conformation that resembles a polar covalent molecule, which fits better with the polar but weakly ion-solvating solvents organic chemists generally use than the dissociated ions favored by chlorides of heavier alkali metals.

There could be also stronger solvation of Li+ ions by alkohol molecules because of the small ion and strong field.

In aqueous solutions ions are both solvated and paired, so this seems reasonable.  We would still get alcohol solvation of an ion-pair dipole, as ion pairing does not hog up the majority of the coordination sphere.

You probably mean some small minor equilibrium concentration of hydrated such pairs, or concentrated solutions with water deficit.

400 g/L in the methanol case is 5M.  There are going to be lots of ion pairs.

As I said. For 0.1 M, probably less than lots, especially in water. I suppose there will be some equilibrium M+(solv) + X-(solv) <=> MX(solv).


text:
Removing ozone from a gas flow
I have a potential problem in that I need to reliably remove ozone from a stream of ionized gases, mostly air. The flow rate is quite low, less than 400 sccm. I was thinking of something catalytic, maybe platinum mesh? There may well be a related need to remove any $\ce{NO_x}$ created. I am open to suggestions — it's mainly a health and safety issue if it vents into the air.

I was thinking about cheap antioxidants.
You might be interested in this paper:
https://www.sciencedirect.com/science/article/abs/pii/S0003986106001652

400 ccm is not a flow rate. Per hour? Per microsecond?

@Karl Should be sccm

You should look for catalytic decomposition of ozone into oxygen rather than searching for chemical reactions. See this article Gas phase ozone decomposition catalysts, https://www.sciencedirect.com/science/article/pii/S0926337396000446

Ozonolysis is a synthetic procedure deploying diluted ozone as a reagent.  This reaction may be relevant in the context of your question, because i) during the reaction, the gas typically is bubbled across a solution (and you have some exhaust to deal with), and ii) even more importantly, after the reaction, the organic products are to be isolated safely, too.
Here, I would like to suggest to use what is otherwise described as reductive workup.  Ozone, which may dissolve in solvents, hereby is quenched (deactivated) by a material which in turn is oxidized by ozone.  Because ozone is a powerful oxidizer, the deactivation is performed while keeping the concentration of ozone entering the quench solution low, and simultaneously keeping the temperature of this solution low, too.
The provision of the flux of 400 mL/min only partly describes your problem.  It neither states the concentration of ozone in this input, nor the regime of this input -- it may be in batches, it could be a contiously working line.  With these uncertainties in hand, bubbling the ozone-containing gas input into an aqueous solution of sodium hydrogensulfite (with "i", $\ce{NaHSO3}$) seems both simple to setup, manage, and scale.  The reaction basically will yield sodium sulfate ($\ce{Na2SO4}$).
As an example, Ragan et al. published about an Ozonolysis at large industrial scale I hereby refer to.  For about 19 h, their recation was running at low temperature with a constant flow of 15-20 L/min, 4-5 wt% $\ce{O3}$ in air bubbled across their reaction vessel.  After completion of reaction, their workup reported was pouring the reaction solution slowly (over 30 min) to a prechilled quench solution of 3.0 kg $\ce{NaHSO3}$ in 6 L water at 0 degC.  Only then  they allowed the mixture to warm slowly to room temperature.
Depending on the conditions in your example, maybe you do not need to cool your sulfite solution as much, yet monitoring the temperature of the stirred quench solution is recommended.  With the parameters missing in your question, the minimal quantity of bisulfite you need might be estimated.  Equally, there are KI/starch indicator stripes, which may be used to check if the concentration of oxidizing ozone still is significant; if present, they will change from colourless to dark blue (ref, example).
Literature reference: Ragan et al., Safe Execution of a Large-Scale Ozonolysis: Preparation of the Bisulfite Adduct of 2-Hydroxyindan-2-carboxaldehyde and Its Utility in a Reductive
Amination, Organic Process Research & Development 2003, 7, 155−160 (DOI: 10.1021/op0202235).


text:
Is it possible to arrest the post-actinic effects of photodegradation in ABS polymers?
Polymers like acrylonitrile butadiene styrene (ABS) are adversely effected by UV exposure, both visually (by yellowing) and with regard to other material properties [1]. What's worse is that the degradation of the material continues for a long time after the initial UV exposure (see [2] for quantitative observations of this effect in other polymers). Can these effects be stopped after the fact?
Context: The unsightly degradation of ABS is a problem for collectors of old computer hardware. "Retrobrighting" techniques [3] (using hydrogen peroxide and UV/heat) can de-yellow the old plastic very effectively (albeit with unknown side-effects on the material) but the improvement has turned out to be temporary. The re-yellowing occurs in the same pattern as the original yellowing from photodegradation and it does so even if the material is protected from further UV exposure! [4]
Discussion: There are many dubious claims about the causes of ABS yellowing and the effect of "retrobrighting". Most often the yellowing is claimed to be caused by bromine and the hydrogen peroxide is claimed to "bond with bromine free radicals and restore the color" [eg. 3, 5]. I did not find any evidence to support these claims so far.
While radicals play a role in the photodegradation, I think the yellowing could rather be due to chromophores like polyenes being built up [6]. The hydrogen peroxide might convert the carbon-carbon double bonds of the polyenes to epoxy links, thereby destroying the chromophores [7] and restoring the original color.
(Intense exposure to direct sunlight also seems to destroy the chromophores temporarily(?) [8]). The problem is that this does not remove the causes of the post-exposure yellowing and therefore over time the yellowing reoccurs [4]. (If the hydrogen peroxide would truly capture the radicals as often claimed, I'd assume that the yellowing would stop permanently.)
Question: Is there any known and practical method to arrest the degradation of polymers after UV exposure? In particular, if this continuing degradation is caused by the presence of radicals in the material, can these radicals be neutralized or removed somehow?
Update: I'm reading about hindered amine light stabilizers[9] (HALS) which are thought to scavenge the radicals involved in photodegradation. I wonder if HALS have any effect if applied to already aged and exposed plastic. Does anyone know?

References

Yousif E, Haddad R. Photodegradation and photostabilization of polymers, especially polystyrene: review. Springerplus. 2013;2:398. Published 2013 Aug 23. https://doi.org/10.1186/2193-1801-2-398
James E. Picket, Reversible post-exposure yellowing of weathered polymers, Polymer Degradation and Stability, Volume 85, Issue 1, July 2004, Pages 681-687, https://doi.org/10.1016/j.polymdegradstab.2004.03.008, PDF (linked under YouTube video [8]): https://drive.google.com/file/d/1vE41ASE14kZVGjEFz_tsiLRLK3c928Eh/view
"HOW TO CLEAN & WHITEN YELLOWED PLASTICS" http://www.retrofixes.com/2013/10/how-to-clean-whiten-yellowed-plastics.html
"Retr0Bright (or RetroBright) treated plastics re-yellowing even with minimal light exposure?" https://www.classic-computers.org.nz/blog/2013-01-15-retr0bright-only-temporary.htm
"What makes ABS plastic turn yellow?" https://chemistry.stackexchange.com/a/74620
https://www.u-helmich.de/che/Q2/farbe/02/farbigkeit-03.html
"Peroxide", https://mysite.du.edu/~jcalvert/phys/perox.htm
Perifractic's Retro Recipes, "Incredible Retrobrighting Discovery! Deyellow Plastic with just Sun?! No bleach/dismantling!", May 4, 2019, YouTube, https://youtu.be/8P1OVj0IcqY
https://en.wikipedia.org/wiki/Hindered_amine_light_stabilizers

My chemistry education shouts "ozonolysis" of the double bonds in the polybutadiene block. And that bromine stuff sounds crazy ... unless there are brominated flame retardants involved. But still.

@Karl, there are reports of ozone gas + UV/heat having been used with some success for de-yellowing the plastic, which hints at C=C bonds being responsible for the color, doesn't it? According to Nintendo, there are brominated flame retardants in their plastics at least but I don't know which role they play, if any. LOL, I learned the word "actinic" from reference [2].

And I learned it from you. ;-) No idea what exactly the colour is. Many organic compounds are somewhat yellow-y, even without obvious chromophores. I'd suspect detoriation products of the polystyrene in ABS.


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Difference in Periodic Tables in Block Formation (split in d-block)
I had a question. When I was learning Chemistry and blocks in the periodic table, my professors always broke the table down like the image below

Recently, when I looked up the periodic table (in block form), I noticed that the first group of the d-block was separated off from the rest of the d-block. So, rather than it going very neatly s, f, d, p (as it does in the picture above), it now goes s, one group of d, f, the other 9 groups of d, then p (like the image below).

Can someone please explain this to me? I am not a chemistry major or even good at chemistry, just curious. I thought it may have something to do with electron configurations (1s2 2s2 2p6... etc) but when I tried it out, I couldn't understand. If you could explain it in layman terms, I would greatly appreciate it.
Thank you

Just different formatting. You can see from your own examples that the lower table is almost readable, the upper table isn't even close.

Thank you. I understand that aspect (as in why block F is placed below the periodic table in order to fit on an 8.5x11).

What I do not understand is why, in the extended form, does the F-block sometimes come after the s-block but in more modern periodic tables it falls inbetween the d-block (as I mentioned in the description, it goes s-block, one group in the d-block, THEN the f-block, followed by the rest of the d-block....)


text:
Why is it bad to charge a phone to 100%?
I see everywhere that a phone should ideally be between 20 and 80% to work efficiently, but I don't understand why, in an electrochemistry point of view. (I'm not talking about the fact that overcharging a phone is bad).

It depends largely on what the cathode and anode materials are. In general, when you charge the phone you remove lithium ions from the cathode. This produces a change in the unit-cell of the material, producing internal strain. This becomes more severe at high states of charge. This strain can cause fracturing in the particles, which causes regions of the particle to become disconnect from the current pathway. The same is true at the graphite anode.
One example I work with is the layered NCA material. Beyond 60-80% there's a dramatic drop in the c lattice parameter[1], which can be described as the layers "collapsing". This produces internal fractures, which become dramatic at high states of charge[2] and builds up over multiple cycles causing loss of capacity as portions of the electrode become disconnected[3].
[1] https://dx.doi.org/10.1021/cm503833b
[2] https://pubs.rsc.org/en/content/articlehtml/2018/ee/c8ee00001h
[3] https://pubs.acs.org/doi/abs/10.1021/acs.nanolett.7b00379


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Corrosion of iron: no electron pushing mechanism?
Why does the oxidation-reduction reaction between iron, water, and $\ce{O2}$ gas to form rusts lack an electron pushing mechanism and a specific sequence of steps?
$$\ce{4Fe^0(s) + 3 O2(g) + 2n H2O(l) -> 2 Fe2O3·nH2O(s)}$$
I've noticed that most of the reactions described in undergraduate organic chemistry textbooks have detailed and widely accepted electron pushing mechanisms whereas most inorganic reactions (such as the corrosion of iron) lack electron pushing mechanisms. Why?
Also, which type of $\ce{O2}$ is involved in the corrosion of iron? The singlet excited state or the triplet ground state or doesn't matter?
$$
\underset{\text{triplet oxygen}~\ce{^3O2}}{
\ce{
·\overset{\Large .\!\!.}{\underset{\Large .\!\!.}{O}}-\overset{\Large .\!\!.}{\underset{\Large .\!\!.}{O}}·
}
}
\qquad
\underset{\text{singlet oxygen}~\ce{^1O2}}{
\ce{
\overset{\Large .\!\!.}{\underset{\Large .\!\!.}{O}}-\overset{\Large .\!\!.}{\underset{\Large .\!\!.}{O}}
}
}
$$

My tip to the hat (i.e., a compliment) for typing this nicely the two Lewis formulae with plain LaTeX' \underset instead of using a mol sketcher.

Arrow pushing mechanism is just a formal way of thinking how the reaction might be occurring. Unfortunately, organic chemistry textbooks present it as if all of these reactions were occurring in real time, step by step, and we are able to watch it like a movie. The reality must be far more complex. If you were truly investigating a reaction mechanism, we have to design very clever experiments (there are books on this topic). It is experimental data that determines the true outcome of an organic reaction not the arrow pushing formalism. Usually, when chemists find out the experimental products, then they postulate arrow pushing mechanisms.
Coming to the rust question. Rust has puzzled chemists and engineers for decades because it causes millions of dollars of losses every year.  Your equation is a very simplistic way of thinking that rust forms in a single step.
$$\ce{4Fe^0(s) + 3 O2(g) + 2n H2O(l) -> 2 Fe2O3·nH2O(s)}$$
It does not proceed in a single step. Our atmospheric chemistry is very complex, you have carbon dioxide, you have sulfur dioxide, nitrogen oxides, ozone, water vapors, sunlight, plenty of free radicals and plenty of minor trace components. Then there are different phases of rust too. You can have a quick look at Google Scholar and one such representative abstract [1].
In short, people still do PhD in this field and one can only imagine how complex corrosion science is.
References

Misawa, T.; Hashimoto, K.; Shimodaira, S. On the Mechanism of Atmospheric Rusting of Iron and Protective Rust Layer on Low Alloy Steels. Corrosion Engineering 1974, 23 (1), 17–27. https://doi.org/10.3323/jcorr1974.23.1_17.


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Flame test color for non-ionic compounds
What color would a non-ionic compound, such as $\ce{NH4OH}$ or $\ce{HCl}$, produce in a flame test? Is no color produced, or is it different for each compound?

I'm confused. I would consider both $\ce{NH4OH}$ and $\ce{HCl}$ to be ionic compounds.

Flame emission (or better atomic emission) does not care whether your starting compound was ionic or not. Everything decomposes into flame either into atoms or very simple molecules. The blue color you see from the flame is also from very small molecules which exist in the flame.
https://en.wikipedia.org/wiki/File:Spectrum_of_blue_flame.png
Ordinary flame like the Bunsen burner is way too cold for non-metals. However if we were to heat HCl or ammonium hydroxide in a confined "container" at very high temperatures say, 5000 - 10,000 $^oC$, eventually you would start to see atomic emission from hydrogen, oxygen, nitrogen and chlorine. Most of the non-metal emission (p-block) lies in the UV or deep UV. You won't be able to see anything with bare eyes.


text:
Do alkenes have priority over alkynes in R/S naming?
I am studying for an upcoming exam and I came upon a confusing question for the drawing of the molecule (3R)-hex-1-en-5-yn-3-ol.

Why is the stereocenter (3-carbon) R? Shouldn't it be S? After looking at it, I see that the priority is OH > alkyne > alkene. I'm really confused.

The carbon 3 is not directly connected to alkyne. It I'd connected to CH2 group which is connected to alkyne. Alkene(CH=CH2) has higher priority than CH2 group

The priority goes for this molecule as: i) $\ce{OH}$; ii) $\ce{C,C,H}$; iii) $\ce{C,H,H}$ and therefore, (R)- is the correct assignment.

Please don’t write answers in comments! It bypasses our quality measures by not having voting (both up and down) available on comments. Comments are for clarifying and improving the question; please don’t use them for other purposes.


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Organic reaction rate equations needed
Where to find a list of common organic reaction rate laws?? Anybody has this?? I could not find one on internet

There is a plenty of information you can find by serching for "rate equations"! One such website is: https://courses.lumenlearning.com/boundless-chemistry/chapter/the-rate-law-concentration-and-time/

The Mayr group at LMU Munich (Germany) spent a great deal determining rate constants for organic reactions between nucleophiles and electophiles.  They established a public, searchable database (explained here), and grouped their findings on posters (C-, N-, P-, O-nucleophiles; entries relevant to organometallic chemistry), too.

(picture source)

Even if it wasn't me who asked the question, thank you, great resource!


text:
Using a gas stove instead of Bunsen burner for a flame test?
In all chemistry labs we use the Bunsen burner. What's the difference between it and a good old gas stove used in home? Can the stove also be used to get positive flame tests for some cations (like $\ce{Ca^2+}$)?

There's nothing special, actually newer versions of burners (eg. Mecker)  made it kinda obsolete.

@mithoron Then, what I intend to ask is whether the currently used Burners in Chem labs are any different from a normal household stove.

I can see that, but you should edit your post, so that title wouldn't be ambiguous.

@evamPUNdit
When you set it to yellow flame (by closing the air intake) it's performing incomplete combustion. The glow is small, hot particles of unburnt carbon. Soot, basically. With the blue flame there is sufficient oxygen that gaseous carbon oxides are formed.

The key thing that Bunsen burner is that it allows you to adjust the amount of air mixing with the gas so you get a hot flame with complete combustion. A gas cooker doesn't allow any adjustment of the air/fuel mix: but it is set to always use the right mix to give complete combustion given the intended fuel. The biggest problem when using a stove for a flame test is pervasive contamination of your cooker with other cooking ingredients (especially salt) that will contaminate your flame colour.

Yes, you can use a common stove to test for cations.
But a stove is designed to minimize the risk of incomplete combustion (which could lead to production of carbon monoxide), hence its flame always appears as an intense blue flame.
Such color contamination could be problematic when testing for cations.
In contrast, the combustion (and the color of the flame) can be regulated in a Bunsen burner.
If you want to experiment at home, an old (but gold) method is that of placing a wick through the metal lid of a closed glass vial, thus creating a burner. You can put alcohol in it, and light the wick.
You can adjust the flame by moving the wick.
Plus, as reported in
[1],
a solution of the salt in alcohol produces a long lasting colored flame in that setting.
References

Dragojlovic, V.; Richard F. Jones. Flame Tests Using Improvised Alcohol Burners. J. Chem. Educ. 1999, 76 (7), 929. https://doi.org/10.1021/ed076p929.

Gas boilers have a pilot flame which, contrasting to the burners of a cooking stove, is vertically aligned.  Neither large, nor very powerful, but steadly and sometimes sufficent for a flame test using a stick of magnesium oxide previously dipped into the salt solution to analyze.

But does that mean that a bunsen burner would allow incomplete combustion?(at some specific position of the knob)

@evamPUNdit Better than the portable, wick-based burner described by The_Vinz, the Bunsen burner allows you to regulate the ratio of natural gas to air.  Next to the valve regulating the flux of the gaz, the Faraday form of the Bunsen burner as an adjustable window (typically in a barrel-like form around the foot of the exhaust) well below the foot of the flame.  Per default, this opening is closed, yielding a lightening flame -- like a candle (incomplete combustion.  Opening this window allows air to mixed with the gas, offering better combustion, yielding a hotter, yet less visible flame.

@evamPUNdit Even better: watch a welder or a glas blower: The freshly ignited torch is clearly visible, because of the incomplete combustion.  But then to work with, a second valve is opened, allowing oxygen to mix to the gas.  Again, then forming a sharp, very hot (oxy-fuel cutting up to 3500 C), sometimes barely visible flame because of an improved combustion process.


text:
Reaction of borax with NaOH
How does borax  react with NaOH?
My textbook says that sodium peroxoborate is formed. I couldn't find the reaction anywhere else.
Can anyone verify as well as explain it?

This is nonsense.

The following reaction schemes are adapted from [1, pp. 76–77].
Amorphous borax $\ce{Na2B4O7}$ forms tetrahydroxoborate in concentrated sodium hydroxide solution:
$$\ce{Na2B4O7 + 7 H2O + 2 NaOH → 4 Na[B(OH)4]}$$
Upon fusion sodium metaborate is formed:
$$\ce{Na2B4O7 + 2 NaOH →[\pu{700-750 °C}] 4 NaBO2 + H2O}$$
In order to form peroxoborate, a peroxide should be introduced in the first place, e.g. by using hydrogen peroxide (see also Preparation of Sodium Peroxoborate):
$$\ce{Na2B4O7 + 2 NaOH + 4 H2O2 + 11 H2O → 2 Na2[B2(O2)2(OH)4] * 6H2O}$$
References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)

Andselisk has given a more satisfactory answer for this question, which has been accepted. Yet, I'd like to give some interesting facts and view about borax, if some readers are interested.
The name borax is derived from the Persian word borak, meaning white. IUPAC name of borax is sodium tetraborate decahydrate ($\ce{Na2B4O7.10H2O}$). Also note that, after analyzing its crystal structure, borax is more properly formulated as $\ce{Na2[B4O5(OH)4].8 H2O}$ and given its systematic name 2,4,6,8,9-pentaoxa-1,3,5,7-tetraborabicyclo[3.3.1]nonane-3,7-diolate, sodium salt, hydrate (1:2:10), accordingly.
When dissolved in water, borax is subject to completely dissociate, and then,  hydrolyzes to orthoboric acid ($\ce{B(OH)3}$) and $\ce{ OH-}$ anions, yielding a $\mathrm{pH}$ of about 9.13 according to the equation:
$$\ce{ Na2[B4O5(OH)4].8 H2O (s) -> 2Na+ (aq) + B4O5(OH)4^{2-} (aq) + 8H2O (l)}$$
$$\ce{ 2Na+ (aq) + B4O5(OH)4^{2-} (aq) + 5H2O (l) <=>  2Na+ (aq) + 4B(OH)3 (aq) + 2 OH- (aq)}$$
Therefore, borax reacts with strong acid solutions. For example:
$$\ce{ 2Na+ (aq) + B4O5(OH)4^{2-} (aq) + 2 HCl (aq) + 5H2O (l) -> 4B(OH)3 (aq)  + 2Na+ (aq) + Cl- (aq)}$$
The liberated $\ce{B(OH)3 (aq)}$ acts as a Lewis acid on $\ce{ OH-}$ ions from autoionization of water, and hence, in equilibrium with water:
$$\ce{ B(OH)3 (aq) + 2H2O (l) <=> B(OH)4- (aq) + H3O+ (aq)}$$
Thus, it can also be back-titrated with alkali solutions such as $\ce{NaOH}$. However, note that the boric acid is a weak monoprotic acid ($K_\mathrm{a} \approx 6.4 \times 10^{-10}$). Therefore, it cannot be directly titrated with standard alkali for analytical purposes. Yet, by the addition of an organic polyhydroxy compound such as 1,2- or 1,3-diol, it is converted to a much stronger acid, which can be titrated using phenolphthalein. This is due to the complex formation between hydrated borate ion and either 1,2- or 1,3-diol (see illustration in the diagram):

It is noteworthy that the combination of these two reactions is commonly used by some analytical chemists to standardize their acidic/basic solutions.
Now, consider the following two reactions:
$$\ce{ 2Na+ (aq) + B4O5(OH)4^{2-} (aq) + 5H2O (l) -> 2Na+ (aq) + 4B(OH)3 (aq) + 2 OH- (aq)}$$
$$\ce{ 4B(OH)3 (aq) + 8H2O (l) <=> 4 B(OH)4- (aq) + 4 H3O+ (aq)}$$
If you add these together, you’d get:
$$\ce{ B4O5(OH)4^{2-} (aq) + 9 H2O (l) -> 4B(OH)4- (aq) + 2 H3O+ (aq)}$$
To cancel $\ce{2H3O+ (aq)}$, add $\ce{2NaOH (aq)}$ to both side. Now, you’d get:
$$\ce{ 2Na+ (aq) + 2OH- (aq) + B4O5(OH)4^{2-} (aq) + 5 H2O (l) -> 2Na+ (aq) + 4B(OH)4- (aq) }$$
Now, compare the equation Andselisk wrote in his answer for the borax reaction with strong alkali solutions. For example, $\ce{2NaOH (aq)}$:
$$\ce{ Na2B4O7 + 7H2O + 2 NaOH -> 4NaB(OH)4 }$$
Both are acceptable ionic and condensed formations, respectively.
Sodium peroxoborate ($\ce{ Na2[(OH)2B(O-O)2B(OH)2]}$), on the other hand, can be prepare by the oxidation of sodium metaborate ($\ce{NaBO2}$) using hydrogen peroxide ($\ce{H2O2}$):
$$\ce{2NaBO2 + 2H2O2 -> Na2[(OH)2B(O-O)2B(OH)2]}$$
Sodium peroxoborate can be precipitated out of water as hexahydrate: $\ce{ Na2[(OH)2B(O-O)2B(OH)2].6H2O}$. Sodium metaborate ($\ce{NaBO2}$) can be prepared by heating borax and $\ce{NaOH}$ together:
$$\ce{ 2NaOH + Na2B4O5(OH)4.8H2O  -> 4 NaB(OH)4 + 3H2O -> 4 NaBO2 + 11 H2O}$$

Did you write thus answer based in memory or from a source? I would like to read the source if possible, to study more(if possible).

It was not one source. There are few of them. I made a Google search under few key words including the title of the question. Trick is you need to be in a campus computer to be able to see all the references. Sorry on that part.

What does a campus computer change? I have never used one.

They have campus' subscribed journal access, of which you can't have with your personal computer at home. That's all.


text:
Why is the HOMO always lower in energy than the LUMO?
We say that increasing the energy of the HOMO of molecules increases their reactivity towards electrophiles while decreasing the energy of the LUMO increases their reactivity towards nucleophiles. However, this seems to be built on the implicit assumption that all molecules will always have a HOMO that is lower in energy relative to all LUMOs in every single other molecule. How can that be justified?

Have a look at https://en.wikipedia.org/wiki/Molecular_orbital_diagram, especially the bit mentioning the Aufbau principle, and then have a think ...

@IanBush Please bear in mind that I am not referring to the HOMO and LUMO of a single molecule. An example to consider specifically: Why is the C-Pb $\sigma^$ MO necessarily lower in energy than the C-F $\sigma^$?

Haven't thought about this in many years, but in the combined system (when they are interacting) if the HOMO were higher in energy than the LUMO it wouldn't remain as the HOMO for very long (symmetry and very small hopping integral exclusions apply) - in which case your argument above still applies, and you need not make the assumption that you think is required. Hopefully somebody who has looked at this in the last 3 decades can point out where I am wrong.

While HOMO and LUMO have an absolute meaning for each (neutral) molecule, "electronegative" and "nucleophilic" are relative concepts, they are categories used to classify molecules. So if I understand your question correctly, you are right. It is a built in assumption (based on a comparison of different molecules) that a nucleophile will be nucleophilic in the presence of just the right kind of partner, an electrophile.

Not every molecule, but every one that you would reasonably regard as a possible reaction partner.

Sorry I was referring to the energy difference between C-Pb $\sigma$ MO and the C-F $\sigma^*$.

That is by definition.
HOMO means highest occupied MO.
MO are calculated and accommodated by energy, from lowest to highest.
According to the multiplicity of the state of your molecule, it must have N paired electrons and M unpaired electrons.
In the ground state, you need to obtain the electronic state of the state with lowest energy. Then you start filling in MO from those with low energy, using the N paired electrons, two at a time, until you fill the N/2 MO with lowest energy.
Then you place your M electrons (one at a time) in the next MO. The last orbital that still gets electrons is your HOMO (by definition), the MO above it (in energy) will not get electrons in the ground state, that is the Lowest Unoccupied MO (LUMO).
Only in excited states, electrons form lower MO can be promoted to MO higher in energy, the most likely transition is from HOMO to LUMO, because their are next in energy, and therefore the energy gap is the smallest possible gap. But that does not mean you cannot promote an electron to a MO above the LUMO, or even make it to leave the molecule realm (ionization). Of course, there are methods to calculate all of these cases.
There is no mystery there.

There is no such assumption whatsoever. Nucleophile and electrophile are names for the compounds that have higher and lower LUMO/HOMO orbitals in this exact relationship between those two compounds. It is just like oxidizers and reductants - there is no absolutes, it's the relative potentials that matters.

For a reaction to take place, molecules must:
• overcome their electronic repulsion by charge attraction and/or orbital overlap
• have orbitals of appropriate energy to interact—a filled orbital on the nucleophile and an empty orbital on the electrophile
• approach each other such that these orbitals can overlap to form a bonding interaction.
What does this mean for nucleophiles and electrophiles? Well, in general, filled orbitals tend to be low in energy—that is after all why they are filled! Conversely, empty orbitals tend to be high in energy. So the best interaction (the one that gains the new molecule the most energy) is likely to be between the highest in energy of all the filled orbitals—an orbital we can term the ‘highest occupied molecular orbital’ or HOMO for short—and the lowest in energy of all of the unfilled orbitals—the ‘lowest unoccupied molecular orbital’ or LUMO for short. Nucleophiles are either negatively charged or neutral species with a pair of electrons in a high-energy orbital (the HOMO)Few organic compounds have vacant atomic orbitals and in most organic electrophiles the LUMOs are instead low-energy anti-bonding orbitals associated with electronegative atoms. These anti-bonding orbitals can be either π* orbitals or σ* orbitals—in other words, molecules which make good electrophiles might have a double or a single bond to an electronegative atom such as O, N, Cl, or Br.
To conclude a basic albeit arbitrary assumption would be that since HOMO is filled, it has the stabilisation energy as opposed to the LUMO.


text:
How To Calculate "İntermolecular Energy" as QM and MM?
In this article: http://ambermd.org/antechamber/gaff.pdf, in Table-7, authors have calculated "intermolecular energy" of a molecule set. I wonder how can I calculate this energy by using gaussian09 and amber-17? Is there any tutorial ? or what is the way for this? I need to calculate this energy to prove my "new force field parameters are correct" but I really don't know this.
Thank you.

Did you read the materials and methods section of the pdf you linked to?

Yes, and I tested them too but I really couldn't understand how to do it.Because when I calculate relative energy I found positive delta G result but in the article somehow all delta G values are negative. I did pes scan and restrain scan with sander.

I'm voting to close this question as off-topic because it is a question about software and not chemistry.

I don't think this should have been closed as a homework, or as a software question (I question the validity of this close reason, but that is a topic for another time). Creation & validation of force fields is a task performed in chemistry and should be equally on topic as Qs about distillation. However, this Q lacks a lot of context and is either too broad, or too unclear. Please add what you have done, what your results of this approach are and why you think they are wrong, and what you would have expected. And I don't consider 'theirs are negative and mine are positive' as enough context.


text:
1,4-addition vs 1,2-addition aldol/aldol type reactions
In the aldol/cross aldol reaction, of $α,β$-unsaturated carbonyl compound, under what circumstances does a 1,2-addition i.e. aldol condensation (carbanion formed directly attacking carbon of carbonyl group) and 1,4-addition i.e. Michael condensation (carbanion formed attacking $β$ carbon)?
These are the following points I considered:

Stability of resulting product.
The amount of conjugation in the compound. What I feel, extended conjugation ($γ,δ$-unsaturated double bond as well) would always lead to 1,4-addition.
The group attached to the $β$ carbon. For example, an electronic withdrawing group would increase the probability of 1,4-addition and an electronic donating group would reduce the probability, favoring a 1,2-addition.
The steric crowding around the carbonyl or the unsaturated carbon system.

Are these arguments valid?

I have added a few examples (I have just written the final product and not the mechanism as my question lies in the very step after generating carbanion). In examples 1, 2, and 4, aldol condensation (1,2) has taken place but in example 3, Michael condensation (1,4) has taken place.
Is there a way we can account for this?

Technical note: reactions 1, 2, 4 can be written with MathJax; reaction scheme 3 can be drawn with MarvinSketch (free, GUI, see licensing) or chemfig (free, $\mathrm\LaTeX$ package).

As for using a photo on ChemSE: if you do this as a resort (vide infra), the upload function here allows you to see how it will look like in your post -- have an eye on the dimension (as you see, it need not be huge) and orientation (now easier to read...).  Better, though, to use a sketcher to export reaction schemes in formats accepted here.  Invest some time to learn one sufficiently well, which will be helpful in your studies anyway (not only ChemSE).  In addition to andselisk's comment, have a look at https://en.wikipedia.org/wiki/Molecule_editor and http://www.gunda.hu/dprogs/.


text:
Uncertainties Calculation
Recently I did an experiment on isolating casein and now that I am processing the results, I feel my method of uncertainty calculation is not feasible. The casein was isolated 3 times and the results are as follows: 2.386, 2.628 and 2.164 grams. The scale had an uncertainty of 3 decimal places. Furthermore, the casein was isolated from 100 mL of milk, measured using a measuring cylinder with an uncertainty of 1 mL. The mean uncertainty was found using the following formula:
$$Δx_\mathrm{avg} = \frac{Δx}{\sqrt{N}} = \frac{R}{2\sqrt{N}}$$
Hence, my calculations were:
$$\mathrm{Range} = 2.628 - 2.164 = 0.464$$
Thus the uncertainty for the average would be:
$$\frac{0.464}{2\sqrt{3}} = 0.133945262 = 0.2$$
This would result in a final value for the average amount of casein in the isolated milk of $\pu{2.4 ± 0.2 g}$.
However, this error seems unreasonable considering that my measurements were to 3 decimal places. Furthermore, how would I add the uncertainty of the 100 mL of milk to this value? Would I simply add the uncertainties?

You need to do two different things here: Error propagation and statistical comparison. If your statistical error is larger than expected from the error propagation, then your assumptions for possible errors are incomplete or wrong.

Resolution of scales has minimal influence on the statistical error introduced by the whole process. You can see the results differ at the first decimal place.

If we have 2 variables
with absolute uncertainties(+)
$A \pm \sigma_A$ and $B \pm \sigma_B$
(+)  expressed as standard deviations, but applicable to confidence intervals as well.
Then $C = A \pm B$ has uncertainty
$$\sigma_C = \sqrt{{\sigma_A}^2 + {\sigma_B}^2}$$
For $C=A/B$ or $C=A\cdot B$,
$$\sigma_C = C \cdot \sqrt{\left(\frac {\sigma_A}{A}\right)^2+ \left(\frac {\sigma_B}{B}\right)^2 }$$
Let $A$ is content of cassein in 100 ml of milk.
Let $B$ is the actual volume of used milk.
Let $C=A\cdot B$ is the mass of  cassein in the volume of milk used in analysis.
Then
$\sigma_C$
is the estimation of the error of the amount of cassein (known as the result error)
$\sigma_A$
is the estimation of the  error of the cassein mass in 100 ml of  milk (unknown)
$\sigma_B$
is the estimation of the  error of the milk volume  (1 ml )
The error introduced by scales can be neglected, or introduced as additive variable $D=0$ with absolute error estimation $\sigma_D=0.001~ \text{g}$.
Another, the biggest  factor is the used analytical method with its bias $E$ and error $\sigma_E$.
$$\begin{align}
C &= D + E + A\cdot B \\
\sigma_{AB} &= A\cdot B \cdot \sqrt{\left(\frac {\sigma_A}{A}\right)^2+ \left(\frac {\sigma_B}{B}\right)^2 }\\
\sigma_C &= \sqrt{{\sigma_D}^2 + {\sigma_E}^2 + {\sigma_{AB}}^2}\\
\sigma_{AB} &= 2.4 \cdot \sqrt{\left(\frac {\sigma_A}{2.4}\right)^2+ \left(\frac {1}{100}\right)^2 }\\
0.13 &= \sqrt{{0.001}^2 +{\sigma_E}^2 + {\sigma_{AB}}^2}\\
\end{align}$$


text:
Why does Azo coupling of β-naphthol takes place at alpha position and not at gamma position

I have searched for coupling reaction of $\beta$-naphthol with benzene diazonium salt at quite a few places. But everywhere the coupling has been shown at alpha position. Why doesn't it take place at gamma position to minimise steric hindrance.
Chemistry LibreTexts

Chemguide

My book

It is already addressed here https://chemistry.stackexchange.com/questions/97067/mechanism-of-formation-of-2-naphthol-red-dye-aka-sudan-1

Coupling reaction of β-naphthol with benzene diazonium is an example of electrophilic aromatic substitution.
If the electrophile attacks at alpha position ,then two resonance structures 1 and 2 , with aromatic rings are possible.

If the electrophile attacks at gamma position ,only one resonance structures 3 , with aromatic ring is possible and 4 is not aromatic.

Therefore attack at alpha position is the major product.
References:
https://pubs.rsc.org/en/content/articlehtml/2015/gc/c4gc02381a
https://en.wikipedia.org/wiki/Azo_coupling
Mechanism of formation of 2-naphthol red dye (aka Sudan 1)


text:
Finding molarity of hyposulfite solution used in titration

$\pu{0.1 g}$ of $\ce{KIO3}$ is treated with excess $\ce{KI}$ solution. The liberated iodine required $\pu{44 ml}$ of $x$ molar hypo solution. Calculate the value of $x$.

I have used the equivalent concept and got the answer as $\pu{0.026 M}$.
But the correct answer to this question is $\pu{0.063 M}$.
Can you tell me where I have gone wrong in my solution?
$$
\begin{array}{cc}
\ce{&KIO3 &+ KI -> &I2}\\
&\frac{0.5}{214}~\mathrm{meq} & &\frac{0.5}{214}~\mathrm{meq}
\end{array}
$$
$$
\begin{array}{cc}
\ce{&I2 &+ &2 Na2S2O3 &-> 2 NaI + Na2S4O6}\\
&\frac{0.5}{214}~\mathrm{meq} & &\frac{0.5}{214}~\mathrm{meq}
\end{array}
$$
$$∴\frac{0.5}{214} = 44\times10^{-3}\times x\times 2 \to n\text{-factor of hypo}$$
$$∴x = \pu{0.026 M}$$

Is it that meq(KIO3) + meq(kI) = meq(I2)?

I have no idea what "meq" is, but a hint is that your first reaction looks off.

meq is milli equivalents...

Your method in using the equivalent concept is correct, but the miscorrect in your solution is in determining the n-factor.

Refer to the above equations and the calculation N-factorPoutnik
N-factor Is the change of atom oxidation state multiplied by the number of atoms changing this state.
Because the iodine is just an intermediate product to be titrated by $\ce{Na2S2O3}$, the oxidation number change is by  $6$.Therefore, n-factor of  $\ce{KIO3}$ is $6$.So, the equivalent mass is M/6.And , n-factor of  $\ce{Na2S2O3}$ is $1$.So, the equivalent mass of $\ce{Na2S2O3}$ is equal to its molar mass.
$$
\begin{array}{cc}
\ce{&KIO3 &+ 5KI + 3H2SO4 -> &3I2 +&3K2SO4 + 3H2O}\\
&\frac{0.6}{214}~\mathrm{meq} & &\frac{0.6}{214}~\mathrm{meq}
\end{array}
$$
$$
\begin{array}{cc}
\ce{&I2 &+ &2 Na2S2O3 &-> 2 NaI + Na2S4O6}\\
&\frac{0.6}{214}~\mathrm{meq} & &\frac{0.6}{214}~\mathrm{meq}
\end{array}
$$
$$∴\frac{0.6}{214} = 44\times10^{-3}\times x\times 1 \to n\text{-factor of hypo}$$
$$∴x \approx{\pu{0.064 M}}$$

To find the molarity $c$, you need to know the amount $n$ required for a titration:
$$c(\ce{Na2S2O3}) = \frac{n(\ce{Na2S2O3})}{V(\ce{Na2S2O3})}$$
Unknown amount $n(\ce{Na2S2O3})$ can be found from the second balanced reaction:
$$\ce{I2 + 2 Na2S2O3 -> 2 NaI + Na2S4O6}$$
$$n(\ce{Na2S2O3}) = 2\cdot n(\ce{I2})$$
Finding $n(\ce{I2})$ is trivial if you would've written the first redox equation correctly, and that's where the problem in your solution arises.
The incomplete first equation resulting in wrong stoichiometry and false ratio between $\ce{KIO3}$ and $\ce{I2}$.
The reaction takes place in acidic media (I added [diprotic] sulfuric acid, but basicity of acid doesn't matter), and the ratio between $\ce{KIO3}$ and $\ce{I2}$ won't be $1:1$:1
$$\ce{KIO3 + 5 KI + 3 H2SO4 -> 3 I2 + 3 K2SO4 + 3 H2O}$$
$$n(\ce{I2}) = 3\cdot n(\ce{KIO3}) = \frac{3\cdot m(\ce{KIO3})}{M(\ce{KIO3})}$$
With all the steps sorted out, the final expression for the molarity of hyposulfite can be rewritten algebraically and solved by plugging in the values:
$$
\begin{align}
c(\ce{Na2S2O3}) &= \frac{2\cdot 3\cdot m(\ce{KIO3})}{M(\ce{KIO3})\cdot V({\ce{Na2S2O3})}}\\
  &= \frac{2\times 3\times \pu{0.1 g}}{\pu{214.0 g mol-1}\times\pu{4.4E-2 L}} \\
  &\approx \pu{6.4E-2 mol L-1}
\end{align}
$$

1 All you need to do to prove that is to write half-reactions for the redox process between iodate and iodide:
$$
\begin{align}
\ce{2\overset{+5}{I}O3- + 12 H+ + 10 e- &-> \overset{0}{I}_2 + 6 H2O} & \tag{red}\\
\ce{2\overset{-1}{I}^- &-> \overset{0}{I}_2 + 2 e-} &|\cdot 5 \tag{ox}\\
\hline
\ce{2 IO3- + 10I- + 12 H+ &-> 6 I2 + 6 H2O} \tag{redox}
\end{align}
$$
or, after dividing the total redox reaction by 2, finally
$$\ce{IO3- + 5I- + 6 H+ -> 3 I2 + 3 H2O}$$

Thanks, but is there a way to solve this using gram equivalent concept?

Or could you find the mistake in my solution?

Gram-equivalent is an obsolete concept and shouldn't be used; however it wasn't the problem, as I already pointed out, the wrong reaction between $\ce{KIO3}$ and $\ce{KI}$ was. You can adapt your solution taking the correct stoichiometry into account.


text:
What kind of chemical reaction happens when combining Tahini and Date syrup?
While both Tahini and Date syrup are quite thin, when mixed together they form a paste-like substance that is very delicious, and quite thick. But what kind of chemical reaction leads to such sudden change in the viscosity of the mixture?

A reference found by googling: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.485.6346&rep=rep1&type=pdf

I'm voting to close this question as off-topic because you can easily Google it as demonstrated by Apoorv Potnis.

After a quick inspection of the article linked to by Apoorv all I could see was a report of physical characterization indicating that the mixture forms a thixotropic paste. It only confirms but does not explain the observation made by the OP. While the OP seems to be potentially misguided (looking for an explanation in the form of a "chemical reaction") there is potentially interesting and complex physical chemistry associated with the answer, which I'd like to see. Plus, I love tahini, I think I'll have it for lunch tomorrow now.

I’m glad I could spark some interest in Tahini!  On a more serious note, I did see the mentioned paper, but I found no answer to this question in it. Forgive me if the question is a bit vague. Chemistry is not really my field of study. I’m just a very curious fat guy not unlike Winnie the pooh.

This is a halvah question to ask!

Effects replicated with grape syrup instead of date syrup.


text:
Nuclear Fusion and Fission
When a heavy nucleus  undergoes nuclear fission , it splits to two smaller, more stable nuclei and produces heat, similarly, when two light nuclei fuse, to make a heavier, more stable nucleus and produces energy.
The produced energy, is due to the mass defect between the reactants and the products, which is transformed to energy, according the mass-energy equivalence.

But if the products of each process(fission, and fusion) are more stable, that means that they have  more binding energy, thus, shouldn't the mass defect be transformed to nucleus binding energy, rather than emitted as heat or another sort of energy ?

The question is rather off topic belonging to the physics SE site.

Isn’t it nuclear chemistry?

Nuclear chemistry covers chemical aspects of processes where nuclear reactions take part.

I was taught this in a chemistry course at high school

A scope of lectures does not always match scope of scientific domains. Part of taught nuclear chemistry is not chemistry, but physics. Similar taught interdomain part of science is physical chemistry.

Note that binding energy of nucleons does not mean contained energy, but the energy needed to break it.
Therefore kernels with lower binding energy per nucleon contain more nuclear energy per nucleon and have higher mass per nucleon.
It means forming kernels with higher binding energy per nucleon releases energy and vice versa.
It is the same as for chemical bonds, just some 6-7 orders higher.
E.g. nitroglycerine has lower binding energy per atom than reaction products formed by its explosion.

How does a lower binding energy mean a stronger nuclear force? it doesn’t make sense

It does not. It means higher energy, but weaker nuclear force binding them together. Similarly, nitroglycerine had higher energy and weaker chemical bonds

How does this answer my question? From where does the energy produced come from?

the energy come the same way as the energy of chemical reactions. Just the force is not electromagnetic force, but nuclear force.

Then why its calculated by the mass defect of the reaction, where did this lost mass go, if to the produced energy, then how did the products have a larger binding energy? I was taught that the binding energy is a result of a mass defect in the nucleus.

Again, it is the same as for chemical reactions. The mass and energy of free electron and proton are bigger than the mass and energy of proton-electron pair, bound in a hydrogen atom. The same for hydrogen atoms versus hydrogen molecules. The same for hydrogen/oxygen molecules versus water molecules.

So the energy released by the reaction is the energy that is released when forming the stable nuclei (binding energy) which is also equivalent to the mass defect?

Yes, with objection that the term mass defect is misleading. There is no physical defect. Mass is determined by the internal energy of the object.

When a heavy nucleus undergoes nuclear fission , it splits to two smaller, more stable nuclei and produces heat, similarly, when two light nuclei fuse, to make a heavier, more stable nucleus and produces energy.

That is correct. But note that there is a distinction between "light nuclei" and "heavy nuclei."
In other words "light nuclei" are light because when they fuse they release energy.  But this doesn't say that  you can keep fusing the "light products" ad infinitum and that you will keep getting another "light" nucleus. (Think about two mass 4 atoms fusing to make a mass 8 nucleus. Then two mass 8 atoms fusing to make a mass 16 nucleus, and so on....)
And "heavy nuclei" are heavy because when they fuse they release energy. But this doesn't say that  you can keep splitting the "less heavy products" ad infinitum and that you will keep getting another "heavy" nucleus. (Think about a mass 256 atoms splitting to make two mass 128 nucleus. Then a mass 128 atoms splitting to make two mass 64 nucleus, and so on....)
The nuclear binding energy curve is shown below.

So in a star fusing light elements to form heavier isotopes releases energy until $\ce{^{56}Fe}$. To make heavier elements requires energy. Thus elements heavier than $\ce{^{56}Fe}$ are not made by stellar fusion, but from super nova explosions and the collision of neutron starts.

But this, doesn’t answer my question, I ask about the origin of the energy that is produced by the reaction.

The origin of the energy is Einstein's famous equation $E = mc^2$. So $\ce{^{56}Fe}$ has 26 proton and 30 neutrons. However its mass is less than the rest mass of those nucleons. If you fuse two atoms of $\ce{^{56}Fe}$ you get $\ce{^{112}Te}$. However one atom of $\ce{^{112}Te}$ is more massive than two atoms of $\ce{^{56}Fe}$. Conversely you can split one atom of $\ce{^{56}Fe}$ into two atoms of $\ce{^{28}P}$ which weigh more than the $\ce{^{56}Fe}$ atom. So fusing the two $\ce{^{28}P}$ atoms releases energy.


text:
How to reduce surface tension of oil?
I would like to create tiny colored drops slowly falling down in a transparent liquid.
The diameter of the droplets must be around 2mm, and their speed must be approximately 0.5 mm / sec.
For that, I need the colored liquid to be slightly more dense than the static liquid.
The easiest setup is to drop water into oil with a syringe.
However, when I drop water from air, the oil surface tension prevents water drops to fall slowly into the oil.
And when I put the needle in the oil, the water accumulates around it and creates huge drops.
I would also like to pump back the water at the bottom to re-inject it later on; so the two liquids must be immiscible.
Is there a simple way to solve this problem?
A better setup would be to increase the oil density so that I can drop it in water (which is cheaper, so much better if I need to scale up my system). I asked another question for that purpose but I could not find any cheap heavy compound miscible with oil.

You may ease the droplets of water to "fall further down" across the oil by adding some of dishwasher into the water.
As just tested with a glass filled with some of sunflower oil, the water-dishwater droplets reaching the surface air./.oil will deform and flatten when they fall from about 30 cm above on said surface. Eventually however, cohesion and surface tension of water contracts these and they pass across the oil in droplike shape to reach the bottom of the glass where multiple water-dishwater droplets may coalesce.
Limitation:  Since dishwasher is a surfactant, however, it renders the two phases miscible in each other.  Stirring the content of this glas will distroy the bounderies of the two condensed phases.  (Which might be not your concern, though, reading "I could not find any cheap heavy compound miscible with oil [fit for this purpose]".)  It won't create you a thing similar like a "sand clock" which may be used again and again.
Addition since the original post was edited:  James Guides indeed added a good thought to the table.  Perhaps you do not need a liquid A imiscible to an other liquid B to study the fall of droplets, and you may use spheres instead.  Depending on perspective, this then is a problem of fluid mechanics (mécanique des fluides) as in applied biochemistry (e.g., here) and rheology (e.g., sphere viscosimeter).  In a steady state, the settling velocity $V_s$ may be described as
\begin{equation}
V_s = \frac{2}{9} \frac{r^2 g (\varrho_p - \varrho_l)}{\mu}
\end{equation}
with
$r$ the Stokes radius of the particle
$g$ earth's gravitational accleration
$\varrho$ the densities of of the particle ($\varrho_p$), or of the liquid ($\varrho_l$), and
$\mu$, the dynamic fluid viscosity (which depends even more on temperature, than $\varrho$).
Other than the two fields mentioned above, this relation is of importance havesting small crystals for X-ray diffraction analysis.  From there I know about documenting examples as in the screen photo below

taken for a large, a medium, or a small sphere of Teflon falling in a column of glycerol.  It equally allows to understand why you can't harvest small objects in a viscous medium from a bottom of a container with a spoon and have to resort to a fork.  (And with 2 mm of diameter, Brownian motion wont inhibit yet their settlment.)

Thanks for your answer! I just tried and it does not work well enough to create the slow and tiny drops that I want. Maybe I should be more precise: I need to create 2 mm wide droplets, falling at 0.5 mm / sec. Moreover, I would like to be able to pump the water at the bottom to re-inject it later on.

@arthur.sw As I see in the diffview of the question now, and the intial post, now, additional parameters to solve the question come into play.  Please clarify #1: Is the assumption correct your interest primary focus is how these fall?  #2: Falling droplets equally reminds me to Mullikan's oildrop experiments; but you do not intend to apply an electric field to (de)acclerate the droplets, do you? (https://en.wikipedia.org/wiki/Oil_drop_experiment)

Thanks again for your help! Since I would like to pump back the dropped liquid, I cannot use spheres. And I would like to keep things as simple as possible, so I will avoid electric fields. I hoped there is a simple way to reduce oil surface tension...

Maybe you know a way to increase oil density to make it heavier than water? This would be even better!

@arthur.sw As for aqueous solutions used to determine the density of crystals: it is finding something both soluble in the solvent and denser than it to increase the density.  Here, I would give chloroform a try (already a colorless liquid at room temperature, about 1.5 g/mL), sparingly solube in water, and dissolving well with a wide range of organic substances, and not flammable.  It would however require a closed system, both to limit its evaporation (boiling point 61 C) as exposition (depends on local jurisdiction [I am not a lawyer]: listed as toxic by inhalation, potential cancerogene).

2 mm drops of any liquid in any other liquid will be very unstable, and will grow to be an entire layer. Perhaps you could buy some plastic spheres of the size you desire, made from plastic slightly heavier than water. Then you can adjust the rate of fall by thickening the water. If you use an organic thickener, like xanthan gum, you should probably add a biocide.
If the plastic spheres are lighter than water, you could add ethanol or isopropanol to decrease density (but viscosity will also decrease). Glycerine has a specific gravity of 1.24, and can be mixed with water to increase density and viscosity.

+1, Your perspective recommending spheres en lieu of droplets (with their possibly wider size / geometry dispersion than commercial) is a very good one, especially since the original question now states the expected sinking rate and size of the objects of study.  It was considered for an addition to my answer.

Thanks! This would be a possibility if I would not need to pump back the dropped liquid.


text:
First order reaction
I am little bit confuse about relating the half life time period of a first order reaction something like
$$\ce{2A -> B + C}$$
Here the rate of reaction will be half of the rate at which $\ce{A}$ is used up.
$$\text{Rate of reaction} = k[\ce{A}]$$
$$\text{Rate of disapperance of}~\ce{A} = 2k[\ce{A}]$$
Integrating it half life will be $\displaystyle\frac{\ln 2}{2k}$.
Am I wrong somewhere?

It is absolutely correct.

Is there something else you are confused about?  Your math may be correct, but you seem to be implying that there is a problem with your answer.

Am I being obtuse here but $\ce{A+A\to \cdots}$ is a second order reaction, not a first order one which would be $\ce{A\to\cdots} $. You can also calculate the half life of a second order reaction.


text:
What is the difference between radii of Gyration and hydrodynamic radii of a polymer?
Why do you consider that the relationship between RH and RG should depend on the quality of the solvent?

Please post your thoughts regarding the answer- the site has a policy regarding "homework"-type questions which requires you show your effort to solve the problem.


text:
Why does the reduction of Dichromate to Chromium 2 need an acid?
The Ecell shows that Zn is enough. Then why, are both of them required?
Also, can I use any acid ? Or only HCl + H2SO4

A big factor is kinetics.  A lump of zinc or even zinc powder sits there undissolved in a neutral solution and does not provide good contact with the chromium species with which you want the zinc to react.  Adding acid creates nascent hydrogen in the solution, and this is what really does the job of reducing the chromium species.  This source, drawn from a related answer by Nilay Ghosh, explores the matter more fully, and I recommend going there.  In addition to the nascent hydrogen, the acid enables keeping the zinc ions in solution as a salt, where otherwise they would form a hydroxide precipitate that interferes with the reaction.
Not every acid is equally good.  To get the most nascent hydrogen you need a strong acid and, even more important, one that does not eat up its hydrogen with other oxidizing functions (nitric acid, you're out!).  Add the fact that cheap is also good, and hydrochloric and sulfuric acids are the usual choices.


text:
Is there a light cured resin (UV or another) that I could make in my house with easy to find chemicals?
Is there a light cured resin that I could make in my house with easy to find chemicals? It does not matter if the finished product have practical or commercial applications as long that is not too toxic.
I want to use it for SLA 3D printing of prototypes.

Unlikely. Perhaps you can be a bit more specific about what you want to use it for.

I want to use it for SLA 3D printing of prototypes

I don't know what that is. Please elaborate.

Here is a good explanation https://www.youtube.com/watch?v=fUxz_1UfdaM and here https://www.youtube.com/watch?v=yW4EbCWaJHE

There are simple light-curing chemicals, such as the gelatin-dichromate system, that you could make at home. However, the specifications for stereolithography are rather rigid, and it is doubtful that you could synthesize a monomer that doesn't corrode the system, that has the correct viscosity, has long-term stability and is not toxic.
You might do better taking up gum dichromate photography, which is 2D printing.


text:
Solubility in different pressure conditions
I am working on my scholarship exam practice. I believe this exam assumes high school + first year university knowledge. And I'm not quite sure what I did wrong. Could you please have a look?

The solubility of oxygen in 1.0 L water is 28 mL at 25 °C and 1.0 atm.
  How much oxygen can be dissolved in 1.0 L of water at 25 °C and 4.0 atm?

The answer remains 28 mL but IMO I thought when pressure goes up 4 times while other factors are constant, the solubility would do that as well. Please help on where I missed and what topics I should focus on specifically.

This is a confusing question because, while solubilities can be reported in mL/L, there can be ambiguity when choosing a pressure during conversion to this unit, for instance using the following equation to convert from molarity $c$ to volume/volume units:
$$ \rho =  \frac{cRT}{p}$$
In this online data page, for instance , in some columns the solubility is reported in mL/L, converted to this unit using throughout a pressure of $\pu{1 atm}$ (the source of the data cannot be verified), even as the partial pressure of oxygen $p_{O_2}$ is increased above $\pu{1 atm}$.
In the OP the volume refers presumably to an equivalent volume of oxygen gas at the (partial) pressure of the gas above the liquid.
Using the numbers from the OP, assuming the gas is ideal, then
$ c=\frac{101325\times28\times 10^{-6}}{8.3145\times298.15} \pu{M} =\pu{ 1.1 \times 10^{-3}M}$
when $p = \pu{1 atm}$.
On the other hand, if $p = \pu{4 atm}$
$ c=\frac{4\times101325\times28\times 10^{-6}}{8.3145\times298.15} \pu{M} =\pu{ 4.6\times 10^{-3} M}$
so the solubility is the same ($\pu{28 mL/L}$) if described in terms of volume at the given pressure, but $\times 4$ greater when regarded as a molar concentration.
Note by the way that according to a number of sources the solubility at $\pu{25 ^\circ C}$ is $\pu{258 \mu M}$ (~$\pu{8.2 mm Hg}$) at $p_{O_2}=\pu{1 atm}$, and $\pu{1.0 mM}$ at $p_{O_2}=\pu{4 atm}$.

As noted by Wikipedia,

"Henry's Law is a gas law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid."

So you are absolutely right. 4 times the pressure of oxygen should result in 4 times as much oxygen being dissolved.
The answer key for the problem must be wrong.

User Delta_G made an interesting observation/correction. The problem is asking about the volume of gas that dissolves, not the concentration of oxygen in the water. So 28 ml at 4 atm of oxygen gas contains four times as much oxygen as 28 ml at 1 atm of the gas. So in either case 28 ml of the gas phase dissolves into the water.

Do we need to think about Boyle's law too?  In the problem as written it gives the amount of oxygen in mL.  At 4 times the pressure, the same volume of oxygen contains a lot more oxygen.

@Delta_G - ah ha! I half-read the problem and was just thinking about the fact that the water would contain four times as much oxygen. But you are absolutely right. If at 1 atm of oxygen 28 ml of oxygen gas dissolves, then when you increase the pressure to 4 atm, then 28 ml of the 4 atm oxygen gas will dissolve too. So 4 times as much oxygen in the water, but only 28 ml of the gas phase dissolves in either case. // Thank you for the correction!

Let me reiterate your point again. If we increase the pressure, the solubility as in grams or moles would increase but not for volume. Please correct me if I am not on the right track. How does this link to Boyle's law though? Since Boyle's law state that pressure is inversely proportional to the volume, providing that other variables are constant. This means volume would decrease if we increase the pressure.

@TreyAnupong - That's the point. If 18 ml oxygen gas at 1 atm contains X grams of gas, then 18 ml of gas at 4 atm contains 4X grams of gas. So 4 times as much pressure, 4 times as much mass per volume in the gas.

Sorry if my question is amateur, why would we only set the volume constant though? If we keep the mass of gas constant, the volume would now decrease. For example, if 28 mL oxygen gas at 1 atm contains X grams of gas, then for the same amount (X g) 7 mL of gas at 4 atm could be dissolved in water.

@TreyAnupong - Yes, but using Henry's Law the water itself should contain 4 times as much oxygen when the pressure is 4 atm as it does when the oxygen has a pressure of 1 atm. So 28 ml of 1 atm gas dissolve into the water, and 28 ml of 4 atm gas dissolve into the water.


text:
How are lowercase m and uppercase M used in general chemistry courses?
Where do lowercase m and uppercase M make appearances in general chemistry, and what do they signify?

Lowercase m

$m$ is the official symbol for the quantity mass.
m is an SI-prefix that stands for milli ($\pu{e-3}$), for example in mg (milligram).
$m$ is sometimes uses as the symbol for molality. This is not IUPAC recommended (they recommend $b$) and leads to the confusing definition $$m_\text{solute} = n_\text{solute} / m_\text{solvent}$$ where the first $m$ is for molality and the second one for mass.

Uppercase M

$M$ is used as symbol for the quantity molar mass (dimensions mass per amount of substance).
M is an SI-prefix that stands for mega ($\pu{e6}$), for example in MW (megawatts).
M is often used as an abbreviation for mol/L, and mM for mmol/L.
$M$ is sometimes used as a symbol for the quantity molarity (dimensions are amount of substance per volume, official IUPAC name is amount of substance concentration, and official IUPAC symbol is $c$). Textbooks that follow this convention sometimes use MM or ℳ as an alternate symbol for molar mass to avoid name conflicts. Combining the quantity $M$ with the unit M gives silly statements such as: $$M = 5~\text{M}$$

Other useful information

Make sure your lowercase m and uppercase M look different when you write or type, otherwise you are confusing your readers. If you type, quantities should be italicized and units should not. That also helps to distinguish the different meanings of m.
You can't cancel m with m if they don't have the same meaning. It is fine to cancel the milli in mg/mL to obtain g/L, but not cancel $m$/mg to obtain 1/g.

In addition to the original answer:
Lowercase

m is a symbol for for meta- prefix (locant).
m is a symbol for metastable [isotope].
m is an abbreviation for "medium" or "multiplet" in spectroscopy.
$\pu{m}$ sometimes used in place of $\pu{mi}$ for mile, e.g. mph — miles-per-hour.
$m$ is a magnetic quantum number in NMR spectroscopy.

Uppercase

$\ce{M}$ signifies [central] metal (not $\ce{Me}$ — this one is methyl).
M is used to denote mesomeric effect.
M is a symbol for aminoacid methionine.
M (minus, or Λ) is a left-handed helix in axial chirality.

Small caps

$\mathrm{\small M}$ is used for concentration (amount-of-substance concentration) of substance.
It's rarely mentioned that when "M" is used in place of  mol/L, it's supposed to be small caps "M" ($\mathrm{\small M}$), not just capital "M", but nobody seems to care.
Also note that both the term and the symbol are deprecated [1, p. 27].

The term molarity and the symbol $\mathrm{\small M}$ should no longer be used because they, too, are obsolete. One should use instead amount-of-substance concentration of $\ce{B}$ and such units as $\rm mol/dm^3$, $\rm kmol/m^3$,
  or $\rm mol/L$.

Related questions

What does the M in μM-Sux stand for?
What does the unit M stand for?

References

Thompson, A.; Taylor, B. N. Guide for the use of the international system of units (SI). NIST Special Publication 811 2008. (NIST)


text:
How are lowercase k and uppercase K used in general chemistry courses?
Where do lowercase k and uppercase K make appearances in general chemistry, and what do they signify?

In addition to the original answer I also allowed myself to invite lowercase Greek letter kappa κ to the table as typographically it has a similar-looking appearance and add some more definitions:
Lowercase k

$k$ is one of the Miller indices.
$k$ is a symbol for thermal conductivity.

Uppercase K

$\ce{K}$ is a symbol for potassium.
$\pu{K}$ is a symbol for Kaiser, unit of energy $(\pu{1 K} = \pu{1 cm-1})$.
K is a symbol for aminoacid lysine.
K is used in spectrometry to denote spectral lines.
$K$ is sometimes used as a symbol for decadic absorption coefficient.

Lowercase Greek letter kappa κ

κ is used to denote denticity.
$κ$ is a symbol for [specific] conductivity.
$κ$ is a symbol for molar Napierian absorption coefficient.
$κ$ is a symbol for asymmetry parameter.
$κ$  is sometimes used as a symbol for bulk viscosity.
$κ$ is sometimes used as a symbol for magnetic susceptibility.

Uppercase Greek letter kappa Κ

Κ is a symbol for a kappa number characterizing bleachability.

Both lowercase k and lowercase Greek letter kappa κ

$k$ and $κ$ is a symbol for imaginary refractive index.

Both lowercase k and uppercase K

$k$ and $K$ is in general may refer to a coefficient (German "Koeffizient"); indices are often used to denote which one, e.g. mass transfer coefficient $k_\mathrm{d}$, octanol/water partition coefficient $K_\mathrm{ow}$.

Lowercase k

$k$ is used as the symbol for rate constants (dimensions vary, concentration to some power per time).
k is an SI-prefix that stands for kilo ($\pu{e3}$), for example in kJ (kilojoule).

Uppercase K

$K$ is used as symbol for the thermodynamic equilibrium constant (dimensionless quantity). Other versions of the equilibrium constant sometimes have units (e.g. dissociation constant $K_\mathrm{d}$ in biochemistry).
K is an SI-unit symbol for kelvin (unit for absolute temperature).

Other useful information

Make sure your lowercase k and uppercase K look different when you write or type, otherwise you are confusing your readers. If you type, quantities should be italicized and units should not. That also helps to distinguish the different meanings of k.
You can't cancel K with K if they don't have the same meaning. It is fine to cancel the kilo in kJ/kV to obtain J/V, but do not cancel kJ/(mol K) to obtain J/mol.