text:
Why does the colour of gold sol change with particle size?
Regarding the colour of gold sol, my book says

Finest gold sol is red in colour; as the size of particles increases, it appears purple, then blue and finally golden.

This is the image I found on Wikipedia which portrays the same phenomenon.

The same page also discusses the effect of size etc. on the colour, but I don't understand exactly what's happening.
Why shouldn't the colour be the same for all sizes? Isn't it characteristic of the molecules which make the sol as the light absorbed and reflected depends on them?
Does this mean that when we break down any other thing to very small particles, it's sol will change in colour?

Generally, the electronic structure of a small particle is a function of size - the high fraction of atoms  that are on the surface alters the overall bonding. So, yes, changing size can change the color you see since the electronic transitions in the small volume are different.

Yes, the color of colloids is dependent on size. Besides absorption, there is scattering as well. @Natasha, read about the term plasmon, which is collective oscillation of charge density in an electrical conductor. This is the beauty of nanochemistry although Faraday the great knew about this property long time ago.

The color of conductive nanoparticles depends on plasmonics. In brief, quantum behavior predominates in the few electrons of a nanoparticle, rather than as a continuous conduction band. Surface plasmons strongly influence the color and polarization of light in sols and colloids. For example, for hundreds of years, stained glass has been made with gold or selenium to get pinks and reds.

Does it mean that if we can create nanoparticles of something else, it'll behave in the same way? Or is it characteristic to gold?

@Natasha Not for, eg, a covalent crystal. Take sugar or carotene, in this case the predominant transition are those of the molecules. Those effects are evident in solid with extended bonding such as metals. Be careful to isolate various mechanisms. The final colour of the sample can rise from abs and various scattering mechanisms. The point is Au nanoparticles can be loosely    seen as "molecules". Look what happens at the gold levels when you combine an increasing number of gold atoms. One gets a reduction of the en. gap and a shift in absorption.

@Natasha Yes. There are many nanoparticle systems where the behaviour is dominated by the size of the particle. Semiconductor nanoparticles can be tuned to particular colours by size, for example.


text:
Hardened polyurethane remover
I've looked at the MSDS of a couple of products for removing hardened polyurethane (PU) foam. Both contained only ethanolamine as a dangerous substance, but only between 1-5%. So I'm guessing this might not be the active substance. The Wikipedia page on polyurethane doesn't indicate any degraders (and the page on polymer degradation doesn't mention polyurethane), so what (active) substances are typically used to degrade hardened polyurethane in commercial cleaners?
I also found that some metal acetate-ethanolamine complexes can be used as catalysts for preparing PU foam. So possibly ethanolamine is the active substance or just a catalyst for removal as well. Can someone shed some light on the purpose of ethanolamine?

Example for one of the products I was looking at "Soudal PU remover SDS"; there's also a consumer leaflet for that one which might have additional hints. It's "very active", "odourless", "free of aromatic hydrocarbons", pH 10.5.

The one solvent I see referenced the most for polyurethane foams is N-methyl-2-pyrrolidone or as the the cool kids call it: NMP. You don't have to just take my word for it a certain supplier that I have no affiliation with sells it for this purpose: Spray Foam Systems NMP.
As far as the purpose of ethanolamine, it must either be reacting to reduce chain length.  If you look at the urethane group below the Carbon bonded to two oxygens and a nitrogen is susceptible to attack by the basic ethanolamine. This reaction would break the molecular bonds making the polymer chains shorter and thus more mobile and more easily dissolved, thus saving time. This explination is a little hand-wavy but I didn't want to go deep into organic reaction mechanisms.

NMP has been banned in the EU in consumer products. (To be more precise they set a 0.3% limit.) Which is probably why the stuff I looked at doesn't have it. OECD notes that the limit is equivalent to a "total ban" http://www.oecd.org/officialdocuments/publicdisplaydocumentpdf/?cote=ENV/WKP(2018)5&docLanguage=En

@Fizz Well that is interesting. This product lists acetone but I believe it is for uncured foam. Is there a smell?

I'm adding a MSDS link to the question. It took a while to find it again (online)... And it's not for uncured. No acetone.

@Fizz This one has even more ethanolamine

There's also one which adds carbitol. Datasheet only in German. This isn't one of those two I had found initially. I can't remember what the other one was called.

I rememebered now it's this one; datasheet also only in German. Top rated on Amazon.de so it can't be complete nonsense.

Since carbitol is apparently non-toxic (for the purpose of SDS), it's possible that the other products have it as well, they might just not mention it. It seems likely given that the last two products I mentioned (those with datasheets in German) are both brands of the same Dutch company. Carbitol concentration was "<25%" in the one that did mention it.

Ether based solvents seem to be common.

source: Stack Exchange

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sort: votes


text:
Electronic configuration of period 3 element from its successive ionization energies

Given below are the first five successive ionization energies $(\mathrm{IE})$ in $\pu{kJ mol-1}$ of an element in period 3.
$$
\begin{array}{ccccc}
\hline
\mathrm{IE_1} & \mathrm{IE_2} & \mathrm{IE_3} & \mathrm{IE_4} & \mathrm{IE_5} \\
\hline
801 & 2427 & 3659 & 25022 & 32822 \\
\hline
\end{array}
$$
Deduce the electronic configuration of the valence energy level of the element.
  Explain how you came to this answer.

I need clarification on this question. I have explained into using my own knowledge, therefore correct me if am wrong.
I believe that the element in period three would be Al, since there is a big jump from $\pu{3659 kJ/mol}$ to $\pu{25022 kJ/mol}$. For the first three electron $\mathrm{(3s)^2(3p)^1}$, they are all in the same energy level which is the third shell and that takes $\pu{801 kJ/mol} + \pu{2427 kJ/mol} + \pu{3659 kJ/mol}$ to remove them, after that we notice a big jump after the third ionization energy indicating the its entering a new level of energy which is the second energy level.
Please advice me if its correct or not.

Yes, your reasoning is logical and makes sense, but the question must have a typo because those IE values are the exact match for boron B, an element from period 2 right above aluminium Al (data from [1, p. 10-204], converted from $\pu{eV}$ to $\pu{kJ mol-1}$):
\begin{array}{lccccc}
\hline
\text{Element} & \mathrm{IE_1} & \mathrm{IE_2} & \mathrm{IE_3} & \mathrm{IE_4} & \mathrm{IE_5} \\
\hline
\ce{B} & 801 & 2427 & 3659 & 25022 & 32822 \\
\ce{Al} & 578 & 1816 & 2745 & 11580 & 14842 \\
\ce{Ga} & 580 & 1980 & 2965 & 6102 & 8299 \\
\ce{In} & 559 & 1821 & 2704 & 5210 & \\
\ce{Tl} & 589 & 1971 & 2878 & & \\
\hline
\end{array}

Another hint that it's not aluminium is the absolute value of the jump in energy between $\mathrm{IE_3}$ and $\mathrm{IE_4}$, which is too high for latter group 13 elements.
Probably the author meant group 13, not period 3, but that's just a speculation at this point.
References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97. ISBN 978-1-4987-5429-3.

Appendix
Source code for the plot, pgfplots:
\documentclass{standalone}
\usepackage{pgfplots}
\usepackage{siunitx}

\begin{document}

\begin{tikzpicture}
    \begin{axis}[
        title={Successive ionization energies of group 13 elements},
        xmin=1, xmax=5,
        ymin=0, ymax=35000,
        xlabel = $i$,
        ylabel = {$\mathrm{IE}_i$/\si{\kilo\joule\per\mole}},
        minor tick num = 1,
        mark = *,
        grid,
        legend cell align = {left},
        legend pos = north west,
    ]

    \addplot coordinates {
        (1,801)
        (2,2427)
        (3,3659)
        (4,25022)
        (5,32822)
    };
    \addlegendentry{Boron}

    \addplot coordinates {
        (1,578)
        (2,1816)
        (3,2745)
        (4,11580)
        (5,14842)
    };
    \addlegendentry{Aluminium}

    \addplot coordinates {
        (1,580)
        (2,1980)
        (3,2965)
        (4,6102)
        (5,8299)
    };
    \addlegendentry{Gallium}

    \addplot coordinates {
        (1,559)
        (2,1821)
        (3,2704)
        (4,5210)
    };
    \addlegendentry{Indium}

    \addplot coordinates {
        (1,589)
        (2,1971)
        (3,2878)
    };
    \addlegendentry{Thallium}

    \end{axis}
\end{tikzpicture}

\end{document}

Converted to PNG with the following command:
magick convert -density 120 G13_ionization_energies.pdf -quality 100 G13_ionization_energies.png

thank you, so much i will have to confront my teacher about this. Again thank you

@Georgy No prob, just don't be harsh with your teacher: it shouldn't be a confrontation, rather a discussion supplied by your argumentation. We are people and make mistakes sometimes.

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text:
Can I lock up CO2 into silicon carbide?
Can it be possible to use atmospheric $\ce{CO2}$ in a reaction (or series) with silicon dioxide, both of which are plentiful, to sequester the carbon in a solid and more useful material, silicon carbide?

You seem to be missing the very point of the problem. The question is not "what compounds of carbon are more useful or less harmful than CO2" (there are thousands of possible answers to that, and SiC is in no way exceptional or interesting). The question is "where are the money".

The issues with carbon sequestration are usually economic not chemical. One of the better ideas (because it is easy and cheap to do) is to grow biomass and then turn it into charcoal. This can be buried to permanently sequester the carbon. No complicated chemistry required.

@matt_black only slightly better idea is to not burn coal in the first place.

@A.K. True, but it is a little late for that.

Yes, but not practically.
The reduction of silica and carbon dioxide to silicon carbide is a conducted at high temperatures (commonly higher than 1500 °C, reaching 2500 °C). The energy required to reach these temperatures is immense. Furthermore, this is done in a nearly pure carbon dioxide atmosphere. I remind you that our atmosphere has about 400 ppm of carbon dioxide.
Silica is not abundant as you think. The main source for silica, sand, is actually quite rare and many countries are experiencing environmental damage due to illegal or legal mining of sand. Sand is commonly host to fragile and unique ecosystems. You can mine sandstone, but you will need to crush it to sand. Pulverising is the most energy intensive phase of any mining operation. For silicon carbide to form efficiently, it has to be crushed to a fine powder. Sand size is not enough. This is extremely energy intensive.
This is going to be an extremely expensive undertaking, and since the source of most of our energy is still fossil fuels, then this would be counter productive.

Silica is common as are other silicates. High quality, pure silica for glassmaking or electronics is what we are short of.

@matt_black doesn't really matter. You want a carbon sink, not to make a new Intel chip. Even if it's 90% pure, good enough to put the carbon in. However, this is so energy intensive you will probably release more carbon to the atmosphere than sequester in the carbide.

I wish we could do it in desert. There are plenty of sand and sun. If we could build a robot to manufacture silicon carbide with solar power, print it into a building structure or building module. I don't know if it could replace concrete but it would be great if it can


text:
Helmholtz double layer
I don't quite understand what is meant by the Helmholtz double layer.
Textbooks states something along the lines of:

the colloidal particles attain charge by selective adsorption, due to
  this it attracts a layer of opposite charge,this layer is called the
  Helmholtz double layer and the potential between them is called Zeta Potential

What stops the oppositely charged layers from combining (neutralizing) with each other?

Imagine a NaCl crystal, why don't Na+ and Cl- neutralize each other? The answer is that the bulk crystal is electrically neutral yet the ions retain their charge at a microscopic level. In the same way bulk solution containing colloidal particles is always neutral. Helmholtz layer is a microscopic view of the interface of a solid surface and the solution containing ions. The ions adsorbed on the surface are rather big and cannot fully neutralize the surface charge, thus there is a net charge on a particle. (Note zeta potential  is not the surface charge). Check Youtube for excellent lectures.

@M.Farooq Another follow up question,which is not quite related probably,but still... If the layers adsorb on the top of one other,then the distance b/w the charges would be tending to zero. This amounts to a very large potential gradient being set up,and hence a large amount of potential energy would be needed to separate the two layers. Yet we say the 2nd layer is "diffused" and is easily movable,and can be removed easily.. Why?

@M.Farooq Thank you for the reply and suggestion of watching videos,actually I did try to watch some lectures but they were all college level and I am still in High School. I have the same doubt as YASUF HASAN, if a potential develops shouldn't there be a force opposing this potential in order to stop both of them from interacting?

@Yusuf and Vaishakh, people have studied electrical double for almost hundred years which can give you an idea that it took a really long time to understand electrical double layer. Perhaps it is not fully understood yet. They used hardcore mathematics and electrostatics and developed expressions for potential changes near the surface of the electrode or colloid surface. There is a theory behind it called DLVO. See this link for a non mathematical explanation https://www.sciencedirect.com/topics/chemistry/electric-double-layer

Thanks for the answer M. Farooq. While many analogies are imperfect I'm still confused by the NaCl analogy. Solid NaCl is not a conductor so electrons can not move but the electrode is a conductor. Why can't charges move from the electrode to neutralize the adsorbed ions?

@GeorgeAlbercook Please open a new question and refrain from asking in the answer section.


text:
Hydrate decomposition equation steps
When a hydrate decomposes it loses water molecules in steps.
Does each step have a chemical equation?
If so is the following equations for decomposition of cupric sulfate hydrate correct?
$$
\begin{align}
\ce{CuSO4 * 5 H2O &→ CuSO4 * 4 H2O + H2O}\\
\ce{CuSO4 * 4 H2O &→ CuSO4 * 3 H2O + H2O}\\
\ce{CuSO4 * 3 H2O &→ CuSO4 * 2 H2O + H2O}\\
\ce{CuSO4 * 2 H2O &→ CuSO4 * H2O + H2O}\\
\ce{CuSO4 * H2O &→ CuSO4 + H2O}
\end{align}
$$

Possible duplicate of How much heat is required to dehydrate a hydrate?

One can rarely tell a priori what the steps are, and the results obtained experimentally may seem not intuitive at all if one tries to deduce the dehydration steps solely from the crystal structure (e.g. from the arrangement of the aqua ligands and water molecules trapped within the crystal lattice).¹
TG-DTG-DTA techniques, often coupled with powder XRD, are commonly used to determine the exact process behind each step interpreting mass change over time.
Alternatively, you can deduce the possible steps from the initial structure, but this approach is less accurate.
So, for the classical textbook example of thermal decomposition of copper(II) sulfate pentahydrate the following steps of dehydration (1–2) and calcination (3–4) can be distinguished [1]:
$$
\begin{align}
\ce{CuSO4 * 5 H2O &→ CuSO4 * H2O + 4 H2O}\tag{1}\\
\ce{CuSO4 * H2O &→ CuSO4 + H2O}\tag{2}\\
\ce{2 CuSO4 &→ CuO * CuSO4 + SO3}\tag{3}\\
\ce{CuO * CuSO4 &→ 2 CuO + SO3}\tag{4}
\end{align}
$$

You could also assume the loss of four water molecules as a single step since in the crystal structure copper(II) atom has octahedral environment with four $\ce{H2O}$ and two $\ce{SO4^2-}$ ligands.
Writing chemical equation for each act of the water molecule loss is not a sin, but it makes little practical sense as the intermediate crystallohydrates are unstable and rapidly undergo further dehydration.
Notes
¹ From the crystal structure one may assume that during the first step only one water molecule is eliminated from the lattice (Figure 2a) since copper(II) is coordinated octahedrally by four water molecules in equatorial area and has two sulfate ligands in apical positions (Figure 2b).

Figure 2. Crystal structure of $\ce{CuSO4 * 5 H2O}$, data from CIF for Chalcanthite. Color scheme: $\color{#EEEEEE}{\Large\bullet}~\ce{H}$; $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#FFFF30}{\Large\bullet}~\ce{S}$; $\color{#C88033}{\Large\bullet}~\ce{Cu}$. a: Unit cell content. b: Coordination environment and nearest neighboring water molecules of the copper(II) center.
However, this is not the case even though lattice water molecules are kept in place via H-bonds and should posses seemingly weaker interaction with the cationic part.
In reality many assumptions suitable for estimation of melting points of molecular crystals, e.g. regarding the number of bonds and their types don't work well for ionic crystals and so the dehydration patterns can only be established exactly experimentally.
References

Gadalla, A. M. Kinetics of Thermal Decomposition of $\ce{CuSO4 · 5H2O}$ to $\ce{CuO}$. International Journal of Chemical Kinetics 1984, 16 (6), 655–668. https://doi.org/10.1002/kin.550160604.

Copper sulphate pentahydrate dehydrates in steps of 2, 2, 1 according to Wikipedia and the reference mentioned in this question: https://chemistry.stackexchange.com/questions/171016/how-heating-of-copper-sulphate-pentahydrate-blue-vitriol-can-be-explained?

@Apurvium I don't see any thermal analysis data in the linked post. Can you please provide the primary source (neither Wikipedia nor a textbook is one) for what you call "steps of 2, 2, 1"? I suppose someone might have done TGA later with even slower heating rates and managed to observe an extra peak on DTA. If so, I'd edit my answer to include better data.

I can only find this resource: https://books.google.co.in/books?id=i9nyvTYBQtAC&pg=PA229&redir_esc=y#v=onepage&q&f=false      in which the existence of crystals of trihydrate depends on the water vapour pressure.

@Apurvium I get "You have either reached a page that is unavailable for viewing or reached your viewing limit for this book." Either way, a primary source with outlined experiment is needed to draw any further conclusions.


text:
Shortcut method to find atomic term symbols ignoring the spin orbit coupling
Is there any shortcut method by which we can find the atomic term symbols ignoring the spin-orbit coupling, i.e. ignoring the values of J.
For instance, when writing the d-d transitions for any complex, we write the atomic term symbols and them split them further into other states. I know the lengthy procedure to find the term symbols by considering the values of Ml, and finding the exact states including the spin-orbit coupling but is there a shortcut way by which we may find the atomic term symbols?

Like, in the above picture, how did we find the terms for d2 to be 1S, 1G, 3P, 1D, and 3F

As far as I know, no.

After searching on the internet, and going through textbooks, I understand there is no "shortcut-method" to get the atomic term symbols in atoms with a large number of electrons.


text:
Why does sp hybridised orbital have more directional character than p orbital?
It states so in my textbook. But since s orbital is non directional, shouldn't sp hybridized orbital be less directional in character than p orbital?

Yeah.. Seems your book is wrong here

It a "free" use of the term "direction", but it conveys the message especially if bond formed by it is foretoken. Just to say it a not precise term. I liked the answer by @amanasci

Ignore everything else in image , this image is just for shape comparison.
As you can see in image that so orbital is a lobe with unsymmetrical shape this leads to its more directionality.
P orbital possess equal Electron Density on both sides of the orbital , but in case of sp orbital electron density isn't equal and that's the reason why sp is more directional than p .
That's also the reason for their different Electronegativity.

The image appears to be plagiarized from this presentation, slide 8. You also might want to edit your illustrations instead of asking the reader to ignore a part of it. There is no white space before punctuation marks, and please use proper capitalization and consistent notations (e.g. don't mix up "P" and "p" for a p orbital).


text:
Is common ion effect applicable on two strong electrolytes

Consider the nitration of benzene using mixed conc. $\ce{H2SO4}$
and $\ce{HNO3}$.If a large amount of $\ce{KHSO4}$ is added to the
mixture the rate of nitration will be :

Faster
Slower
Unchanged
doubled

The answer given in my textbook is option 2 , i.e. the reaction will become slow. my teacher told me that it is due to common ion effect.
But in common ion effect, a strong electrolyte decrease the dissociation of weak electrolyte. So,how can common ion effect possibly be the reason here($\ce{KHSO4}$ and $\ce{H2SO4}$ both are strong electrolytes). In my view the answer should be no effect and I'd love to have my concepts about Ionic Equilibrium be corrected if I'm wrong.

I just want to commend you for self-removing the homework tag.

I removed it because then people won't even read my question and just move forward even if I would have mentioned my thoughts in the question. I can add it again though but before I do that I'll ask you for a bit of help. It's been quite a while using stack exchange and I'm still not able to figure out that how do you guys subscript and superscript and how do you make sentence appear in yellow box to highlight it. Please help me with it and I'll add the Homework tag again.

Don't add the Homework tag. WE are working to perminately remove it from the site. Here are some Meta posts that may help you get started with  equations:  https://chemistry.meta.stackexchange.com/q/86/23561 and a more advanced markup post https://chemistry.meta.stackexchange.com/q/3044/23561

Certainly not.
Suppression of dissociation of a strong electrolyte is not a very rational thing to discuss, owing to the unmeasurably large value of $K$.
To see what I mean, try looking up the $K_{a1}$ of $\ce{H2SO4}$.
(At least I couldn't find it!)
Your textbook's given answer is correct.
The equilibrium suppressed by the introduction of $\ce{KHSO4}$ is not the dissociation of $\ce{H2SO4}$, but rather this one:
$$
\ce{HNO3} \ + \  \ce{H2SO4} \rightleftharpoons \ce{NO2+} \ + \ \ce{HSO4-} \ + \ \ce{H2O}
$$
Nothing too auspicious: all that remains is the run-of-the-mill common ion effect we've known from Kindergarten days!
PS.
Conclusion: $\ce{KHSO4}$ sucessfully decreases the concentration of $\ce{NO2+}$, which happens to be the electrophile and a key participant in the rds(rate determining step) of Nitration of benzene.
$$
Rate=k[\ce{NO2+}][\ce{C6H6}]
$$

Thanks a lot.I looked for Ka1 of H2SO4 in my textbook and it's value was given as Very Large (LOL). Then I looked on internet and a website said 10^3. Link: https://depts.washington.edu/eooptic/links/acidstrength.html


text:
What will be unit of volume in V=nRT/p if p is in Torr instead of atm?
$V=nRT/p$, where $n$ is amount of gas in $\mathrm{mol}$, $R$ is gas constant, $V$ is volume in $\pu{dm3}$, and $p$ is pressure in $\pu{atm}$.
What happens to $R$ if $p$ is in $\pu{Torr}$ instead of $\pu{atm}$?
Does this question also have some relation with my question above in title (What will be unit of volume in $V=nRT/p$ if $p$ is in $\pu{Torr}$ instead of $\pu{atm}$?)? If yes, please answer this too.

You make up your own unit or you use the gas constant in the appropriate units. Either way, it is way easier to convert from torr to Pa or bar.

Regarding the universal gas constant $R$, you may find it expressed in multiple units -- and consequentially, sometimes numerically quite different -- in textbooks about Physical Chemistry, or even right on top right of the corresponding entry of wikipedia.

(source)
Regarding your second point, of course you may use cubic metres as unit of volume, or non-SI but still SI accepted units like the litre (cf. here).  What you shall do then (a check useful even if using SI units only, too) is to perform a dimensional analysis.  Simply speaking, your equation is true if not only the numbers, but equally the units on the left, and on the right side of the equation sign match each other.  For the units, though, some of them may cancel out each other if related by multiplication and division, as a later example of the second reference (accidentally about the Universal Gas Law), too:

(source)

What happens to $\mathrm{R}$ if $\mathrm{P}$ is in $\pu{Torr}$ instead of $\pu{atm}$?

Short answer: you can use the value of $\mathrm{R}= \pu{62.32~Torr\cdot{L}\cdot{K^{-1}}\cdot{mol^{-1}}}$ and the unit of $\mathrm{V}$ in  litere.
These are gas constants $\mathrm{R}$ and each is used with the appropriate units.
\begin{align}
\mathrm{R}&=\frac{\mathrm{PV}}{\mathrm{T\cdot{n}}}\\
&=\frac{\pu{atm}\times{22.4L}}{\pu{273K\cdot{mol}}}\\
&=\pu{0.082atm\cdot{L}\cdot{K^{-1}}\cdot{mol^{-1}}}\\
&=\pu{0.082\times{\pu{760Torr}}\cdot{L}\cdot{K^{-1}}\cdot{m^{-1}}}\\
&=\pu{62.32~Torr\cdot{L}\cdot{K^{-1}}\cdot{mol^{-1}}}\\
&=\pu{62.32~mlHg\cdot{L}\cdot{K^{-1}}\cdot{mol^{-1}}}\\
&=(\pu{62.32\times{10^{-3}~m})(\pu{1.35951\times{10^4Kg/m^3}})(\pu{9.81m/s^2}})\pu{K^{-1}\cdot{mol^{-1}}}\\
&=\pu{8.31\times{10^3}\times{\frac{Kg\cdot{m}}{m^2\cdot{s^2}}}}\times{\pu{L\cdot{K^{-1}}\cdot{mol^{-1}}}}\\
&=\pu{8.31\times{10^3\times{\frac{N}{m^2}}}}\times{\pu{L\cdot{K^{-1}}mol^{-1}}}\\
&=8.31\times{10^3}~\pu{Pa}\cdot{L}\cdot{K^{-1}}\pu{mol^{-1}}\\
&=\pu{8.31KPa}~.\pu{L\cdot{K^{-1}}mol^{-1}}\\
&=\pu{8.31J}~.\pu{K^{-1}mol^{-1}}\\
&=\pu{1.98Cal.}~\pu{K^{-1}mol^{-1}}\\
\end{align}


text:
The oxidation of aldehydes and alpha-diketones with peroxy compounds
Recently, I have been reading up on the Baeyer-Villiger (BV) oxidation. The oxidation is most commonly discussed for ketones. However, the oxidation also works for aldehydes and $\alpha$-diketones (Rojas, 2015).

The above image is Scheme 2.2 from Rojas (2015).
I have mainly two queries:

Does the oxidation of aldehydes to carboxylic acids by peroxy compounds, such as peracids, proceed via a BV oxidation where the hydride is the migrating group? This would be seemingly different from the oxidation mechanism for the oxidation by $\ce {MnO4^-}$ or $\ce {CrO3}$ where the aldehyde is oxidised when it is in the gem-diol form.
Why does the oxidation of $\alpha$-diketones yield the corresponding anhydride and not an $\alpha$-ketoester? This is extremely puzzling because the acyl group is terribly bad at stabilising any partial positive charges on the acyl carbon. How is it possible that such a BV oxidation can occur?

Reference
Rojas, C. M. (2015). Molecular Rearrangements in Organic Synthesis (1st ed.). Hoboken, New Jersey: John Wiley & Sons, Inc.

Hydride migrates to leave the more stable cation. 2) Acyl migrates for the same reason.

The Baeyer-Villiger oxidation is conducted under acidic conditions such as an alkyl peroxide in the presence of a mineral acid (e.g.; H2SO4) or a peracid (e.g.; peracetic acid or m-chloroperbenzoic acid). In the reactions described in the diagram, the reactions of aldehyde 1 or α-diketone 7 are initiated by proton of the carbonyl oxygen followed by addition of peroxide (an alkylperoxide is illustrated). Ultimately, the oxygen of the RO-moiety is protonated and is the effective leaving group (ROH) in the rearrangement. In both sequences, the red route is favored; the blue route disfavored. Cation 3 is a more stable species than cation 5 because of hyperconjugation with the alkyl group or resonance stabilization if R'=aryl. Cation 3 affords a carboxylic acid while cation 4 would lead to formate ester 6.
In the case of peroxyhemiketal 8, acyl migration leads to cation 9 which is more stable than species 11, which has a positive charge adjacent to a carbonyl group, which is destabilizing. Thus, anhydride 10 is formed in preference to α-ketoester 12.

Thank you for your answer. I suppose migration of the acyl cation is favourable due to stabilising interactions from the lone pair on the carbonyl oxygen?

Because you haven't quite answered my query with regards to the favourability of acyl migration

I appreciate your hesitancy about accepting the existence of an acyl anion. I agree. But in the rearrangement there is neither an R' nor acyl anion per se. As either of these electron pairs become partial bonded to oxygen, a positive charge begins to develop in the transition state leading to 9 and 11. The positive charge developing in 11 is not good. Clearly, the activation energy to 9 is lower than to 11. It is a competitive, irreversible process. PS: An acyl cation is not involved!


text:
Pauli exclusion principle
Pauli exclusion principle states that 'No two electron can exist in same state' or 'No two electron can have same set of quantum numbers'.
But in reality there is no state of electron. The state is of whole system which is composed of many electrons. It is our approximation that each electron is described by a single wave-function which we called spinorbital. So what does that statement even mean in reality?
Quantum numbers is only exact for hydrogen atom. What does quantum number even mean for a electron in a many electron system.

Wikipedia has the correct definition:

A more rigorous statement is that with respect to exchange of two identical particles the total wave function is antisymmetric for fermions, and symmetric for bosons. This means that if the space and spin co-ordinates of two identical particles are interchanged, then the wave function changes its sign for fermions and does not change for bosons.

The textbook shorthand you quote, 'No two electron can have same set of quantum numbers', only makes sense when you are using the one-electron approximation. As you move beyond that approximation, you also have to move beyond this shorthand definition.


text:
Does hyperconjugation into sigma-star explain methylamine's increased basicity relative to ammonia?
The inductive effect with the polarisation of the C-N bond obviously plays a role in stabilising the positive charge of the conjugate acid. But given the similar electronegativities for C and H, CH3 group should donate as much electron density to N as H does (roughly), if only the bonding atoms are considered. And if we're considering atoms other than the two bonding atoms in their contribution to stability, is it not a question then of hyperconjugation?
This question seems to deny hyperconjugation plays a role in stabilising ammonium derivatives. However, it takes a very narrow view of hyperconjugation and focuses only on electron density donation into bonding orbitals. But if more substituted double bonds are more stable because of the donation of alkyl groups into pi-star orbitals, antibonding orbitals must be viable recipients. And similarly in methylammonium, we have sigma-star orbitals on the positive nitrogen which the C-H bonds of the methyl should be able to donate into, destabilising the N-H bond in methylammonium but overall stabilising the system.
So, my question is does hyperconjugation from sigma C-H bonds into sigma-star orbitals of sp3 centres occur, and then from that - does hyperconjugation explain methylammonia's strength (or alkylammonias' strength, more generally) relative to ammonia?

Actually hyperconjugation explain the inductive effect of methyl.

I'm not sure I understand what you mean. Are you saying it explains the inductive effect of methyl when bound to an sp3 centre like in methylammonium. Because in that case, you've answered my question but haven't shown why the answer I linked, which contradicts that answer, is wrong.Could you expand on what you mean?

I was actually saying that. But indeed that is the standard for interpreting the donating effect to a (partially) empty orbital. I realised that the electron donating effect of alkyl gr. might be more obscure than commonly assumed :) I have to think with calm :)

AFAIR differences in basicity of alkylamines are rather attributed to solvation effects.

@Mithoron right, but if I remember it only counts between tert and sec .. and we shall reason in gas phase. I am realising that thinking of charge dispersion without ascribing it to "electron donating" should be more natural.  In vacum, bigger a ion lower the charge density, whatever extent charge can distribute from the inner to perifery, the bigger the better, in this context.  But what you say is also correct. Indeed I have to think, and so do also for the cited Q. I even found that +I of alkyl are indeed questionable, and I feel better :)  Fortunately the facts remain, too.

@Mithoron But wouldn't ammonium be a more charge dense conjugate acid than methylammonium/alkylammoniums, and thus be better solvated in water - so how does a solvation argument lead to saying methylammonium is a more stable conjugate acid than ammonium, and thus methylamine is a stronger base than ammonia?

Cheap Will indeed that is true between di and tri methylamine. The straightforward order is seen in gas phases.  Matter of compromise. ...

Yes, I once thought about this interesting phenomenon also. There is no good reason why hyperconjugative interactions cannot take place between the $\ce {N}$ lone pair orbital and the $\ce {C-H}$ $\sigma$. However, the interaction may be less favoured if you consider the geometry, compared to the the usual hyperconjugation to alkene $\pi$ systems or to carbocation $\ce {p}$ orbitals. Although geometry does not help to favour this interaction as much, the change of atom from $\ce {C}$ in the usual case to $\ce {N}$ does help to decrease the energy gap for the interaction since $\ce {N}$ is more electronegative than $\ce {C}$ and thus, its atomic orbitals would all be of a lower energy compared to $\ce {C}$.
Raising the energy of the HOMO of a general nucleophile would increase its nucleophilicity by decreasing the energy difference between its HOMO and the LUMO of the electrophile. In the same way, by this filled orbital-filled orbital interaction, the HOMO of the methylamine is now raised, allowing it to more easily interact with acids. Thus, $\ce {MO}$ theory can be used to explain the increasing basicity of amines with increasing alkyl substitution.


text:
Possible erratum in solutions to Orbital Interactions in Chemistry 2ed
I was attempting the problems in Chapter 1 of Orbital Interactions in Chemistry, and found a slight problem.
ftp://ftp.wiley.com/public/sci_tech_med/orbital_interactions_2e/Chapter%201%20-%20Answers.pdf
For Q1 (d), I obtained the same secular determinant as the question. However, my answer was slightly different as solving the determinant yielded completely different answers for me.
For the question, H12 (resonance integral) was given as -14.18 eV, S12 (overlap) was given as 0.596, and the energy of the 1s hydrogen atomic orbital was given as -13.60eV. However, plugging these numbers into the secular determinant yielded me orbital energies of -18.26eV, -13.6eV and 41.07eV (the answer given was -33.657eV, -13.6eV, 6.457eV).
Here's the question, for reference. Is my answer incorrect, and if so, how is it incorrect? Thanks!
Here's the question.

Any help would be appreciated.

I don't see why it's too broad (even before the newer edit). 1d and 1e are very closely related questions, and the answer to them is practically the same, as I've already illustrated in my answer. The only thing that might be useful is if the question were typed out (instead of attached as a screenshot), but that has nothing to do with it being too broad or not.

I agree with orthocresol. The question itself is somewhat hard to read and would benefit from an entire reshape. Wiley is (unfortunately) also not very well known for stable links, so including only the relevant information from that pdf file would be advisable. All in all, I am leaning towards reopening, but I would at the moment not want to overrule a community decision.

I ran the calculations through the MATLAB command window and obtained the same answers as you:
>> H = [-13.60 -14.18 0; -14.18 -13.60 -14.18; 0 -14.18 -13.60]
>> S = [1 0.596 0; 0.596 1 0.596; 0 0.596 1]
>> energies = eig(H,S)

energies =

  -18.2615
  -13.6000
   41.0717

On the other hand, if you assume that $S_{12} = 0$ i.e. $S = I_3$ (3x3 identity matrix), then you get the same result as the authors:
>> energies = eig(H)       % or eig(H, eye(3))

energies =

  -33.6535
  -13.6000
    6.4535

I find it slightly telling that the authors explicitly comment that the bonding orbital is stabilised by exactly as much as the antibonding orbital is destabilised. If you solve the secular equations without explicitly substituting in numbers, you should find that this can only be true if the overlap matrix is set to the identity matrix.
The same is true of part (e); you only get the authors' answer by setting $S = I_3$.
I would consider this an erratum in the provided answers.

To find the eigenvalues, use instead [energies, MOs] = eig(H,S). You can try this block of code online. Note that this website has Octave instead of MATLAB, so energies and MOs need to be reversed:
H = [-13.60 -14.18 0; -14.18 -13.60 -14.18; 0 -14.18 -13.60];
S = [1 0.596 0; 0.596 1 0.596; 0 0.596 1];
[MOs, energies] = eig(H,S);
disp(energies);
disp(MOs);

energies is a 3x3 matrix with the eigenvalues as the diagonal elements. MOs is a 3x3 matrix, where each column vector is the MO corresponding to the relevant eigenvalue (i.e. the left-most column is the MO corresponding to the first eigenvalue in energies).

could I inquire what the correct wave functions for (d) would then be? I got the Bonding MO as 0.36841 + 0.52072 + 0.36843.

@ANZGCFlyingFalcon, if you learn a little bit of coding, it's easy to verify it for yourself instead of having to rely on me or anybody else. This is around five lines of code. For now, though, please see the addendum to the original answer.


text:
What is the logical way to understand Iso, Sec, Tert and Neo prefixes?
I'm only familiar with Alkanes at a basic level currently.
What seems logical to me is that an Alkyl molecule with the:
"iso" prefix means its an isomer of that alkyl, the usual way i'd see it would be the alkyl group with a substituent (the way an hexane could be isopentane if  I got it right)
as for the "sec" prefix, the alkyl group is connected to the main molecule through a secondary carbon (meaning the carbon on the alkyl is bonded with two other carbons in the alkyl, no consideration given to the main chain here)
I find the "tert" prefix similar to the latter, it is connected to the maine chain through a tertiary carbon (bonded with three carbons on the alkyl)
I couldn't find a basic logical way to address the definition of the "neo" prefix.
Are the ways I described the rest of the prefixes correct?

Sec- and tert- are exactly what you say they are. There's no good way to understand the others. Consider that leucine has a isobutyl group, but isoleucine has a sec-butyl group.

https://chemistry.stackexchange.com/questions/98362/when-to-use-iso-and-sec-while-naming-compounds-in-alkyl-halides

The prefixes, iso-, sec-, tert-, neo- are all traditional names when structural organic chemistry was still developing. You have already rationalized all except neo-. Neo- is Greek term for new. I looked up the OED etymological dictionary for older meanings, and the earliest use is traced to 1876. Now you can see in no 2, that neo- was a good option.

Used in the names of chemical compounds, minerals, etc., which are closely related to, or new forms of, known substances.
neocatechu  n. (Obsolete) a kind of catechu.
Used in the names of compounds and radicals in which one carbon atom is linked to four others.
Not part of standard chemical nomenclature although well established in neohexane and neopentane (and related terms).
(a)
Categories »

† neo-acid  n. Obsolete
1876   W. Odling in London, Edinb., & Dublin Philos. Mag. 5th Ser. 1 216   The names which call to mind that the normal, iso-, and neo-acids are mono-, di-, and trimethyl & etc. derivatives would seem to be most advantageous.

Neo- even it is used today represents a terminal tert-butyl group, just like the prefix iso- represents a iso-propyl group at the end.
IUPAC names have superseded all.

I only regularly see "neo" as a prefix to indicate neopentane derivatives such as neopentylamine


text:
Techniques to extract lipids
I am planning an experiment on how to isolate casein and part of the process involves removing any lipids that may still be in it after filtering the precipitated casein out of the milk. I have read that one can use 95% ethanol to dissolve the lipids into it, since like-dissolves-like. This would not remove all lipids though and thus a 25 mL 1:1 solution of diethyl ether would be necessary to remove the remaining lipids for a certain amount of the collected precipitate.
What exactly occurs in the reaction when lipids are dissolved into the alcohol? What other alcohols could be used? How does one figure out how much alcohol they should use and at what concentration?
These questions may seem silly to some, yet I have conducted a considerable amount of research online trying to figure it out myself — to no avail though. I appreciate all well-intentioned answers.

You may use rubbing alcohol (2-propanol) instead.

Casein is a mixture of phosphoproteins found in milk that accounts for approximately 80% of the total protein content of cow milk (cow milk contains 3.3% total protein). Hence, it is easy to prepare casein as a pure protein from cow milk.
Casein exists in milk as the water soluble calcium salt of a phosphoprotein (UC, Davis). Thus, it is usually prepared by acid precipitation. Acid treatment removes the calcium cation, leaving a water insoluble phosphoprotein. Note that casein is amphoteric, i.e., it forms salts with both acids and alkalines.
Generally, acetic acid is used in the casein precipitation, because its mildness and less harsh property compared to hydrochloric acid. Although, most people extract casein precipitation in the cold acid solutions, such practice would lead to the formation of protein with poor physical properties. Softness and rather incomplete separation of the casein are problems encountered.
If the precipitation is performed in warm solutions, the casein separates as a single large colloid, with complete separation, leaving a yellow solution of whey behind (which primarily consists of milk-sugar, lactose).
Keep in mind that usual casein precipitation is comparatively easy but delicate. If too much acid is added or if the acid is added too quickly, or if the acid is too strong, part of casein mass will redissolve in whey. Also, casein may be extracted from whole milk. Disadvantages are that whole milk contains a relatively large concentration of fat. This fat will be physically included in the casein curd as it separates. Thus, I'd recommend to use skim milk (if possible), which would be one solution for your co-precipitation of fat problem.
A procedure for casein precipitation in UC, Davis is as follows:
First, prepare a solution of $\pu{7.0 mL}$ glacial acetic acid in $\pu{50.0 mL}$ deionized water. This is your acid solution to be used for the precipitation.
Place $\pu{600.0 mL}$ of skim milk in a $1$-$\mathrm{L}$ beaker. Warm the milk to $\pu{110 ^{\circ}F}$ (approximately $\pu{43 ^{\circ}C}$; Note that temperature should not be $\gt \pu{55 ^{\circ}C}$). At this temperature, add the acetic acid solution (prepared earlier) drop-wise and with stirring. At a certain point, casein will separate as an amorphous mass leaving clear supernatant (light yellowish). Gather the solid from the beaker with a glass stirring rod or separate it using suction filtration.
Caution: Do not add any more acid than necessary to obtain a good curd. A few minutes must be given between acid additions to allow for stabilization.
Press the removed mass of casein between paper tissues to remove as much water as possible. Place the mass of casein in a food blender with $\pu{500 mL}$ of water. Suspend the casein in the water by blending as much as possible.
Separate the casein from the water with a filter and wash the cake with 2-propanol. If 2-propanol is not available use ethanol or some low molecular weight alcohol.
Let the casein powder air dry. You may wash the dried cake with two portion of ether if necessary. You may suspend the dry powder in ether and let it soak well before filtering.
Yield of casein in $\pu{600.0 mL}$ of skim milk is reported as $\pu{13.8 g}$.


text:
Identifying Lewis acid
What is the Lewis acid on the reactants side?
$$\ce{HNO2(aq) + HPO4^{2-}(aq) <=> NO2-(aq) + H2PO4-(aq)}$$
The book says it is the $\ce{H+}$ only, not the whole molecule of the nitrous acid, how so?

That's a Bronsted not Lewis equilibrium.

@Mithoron the reaction can be interpreted in terms of the Lewis theory.  But yes, when that theory identifies $\ce{H^+}$ as the acid, the Bronsted interpretation is usually simpler.

Can you plzz tell which book this question is from...?

the book is called Chemical Principals, The Quest for insight, 7th ed.

What we have, in the Lewis theory, is an acid-base displacement reaction.  The Lewis acid, $\ce{H^+}$, is initially combined with one Lewis base, $\ce{NO2^-}$.  Then another Lewis base, $\ce{HPO4^{2-}}$, takes the $\ce{H^+}$ away displacing the $\ce{NO2^-}$.
Similarly when $\ce{HCl}$ is dissolved in water and dissociates, the water is displacing the chloride ion.  The Lewis acid involved in this displacement is again $\ce{H^+}$.


text:
Pyrophoric nature of Raney Nickel
I am about to do a hydrogenation lab, and a, struggling with how to prepare my Raney Nickel so it does not spontaneously ignite during my lab. Most websites say to just submerge it in water– is this correct?

IT won't catch fire when wet, but it might not work either. If you need it to be dry, use a vacuum line and then fill the vessel with nitrogen. The dry product will catch fire very rapidly if you allow any air into the mixture.

You also need to be careful with your residues. Do not pull air through them on a filter, they will catch fire if they dry out. Keep them under water in a sealed jar labelled for disposal

Please speak to a supervisor or a demonstrator who has been trained in the use of RaNi. The amount of help you can get over the Internet is very limited.


text:
Overall enthalpy balance on a mixing process
Let's say I have a mixing process involving $\ce{A}$ and $\ce{B}$ at temperatures $T_\ce{A}$ and $T_\ce{B}$, respectively. The $\ce{AB}$ mix is at temperature $T_p$. Our $ΔH_\mathrm{mix}$ is defined at $\pu{25 °C}$ (i.e., our reactants/products are not at the reference temperature) and we know its value. What would be the overall balance?
It is my understanding that we would have 3 steps:

Use the amounts and heat capacities of $\ce{A}$ and $\ce{B}$ and find $ΔH$ (open system) from their specified temperatures to the reference temperature $\pu{25 °C}$. This is $ΔH_1.$
Use the amount and heat capacity of $\ce{AB}$ mix to find $ΔH$ from the reference temperature to $T_p$. This is $ΔH_2.$
Finally, $ΔH = ΔH_2 + ΔH_\mathrm{mix} + ΔH_1.$

That is to say, the overall $ΔH$ is the sum of all the steps required to get the initial state(s) to the final state.
I have been told by my lecturer, however, that it should be
$$ΔH = ΔH_2 - ΔH_\mathrm{mix} - ΔH_1$$
Why are $ΔH_\mathrm{mix}$ and $ΔH_1$ subtracted?

I think this is just a definition issue. Look closely at how your lecturer defined the $\Delta H$ for each step. Is it for the reaction in the same direction that you used? Using the opposite sign for the last step is one way to emphasize that the $T_p$ is dependent on how much heat was released or absorbed in the first two steps, because that is the amount of heat that must be absorbed or released in the last step, ie $\Delta H_2$ is equal to $-(\Delta H_1 + \Delta H_{mix})$ using your sign convention. (I'm assuming that the system is insulated.)

Let’s use a trivial case to test the two possible results. If we set $T_\ce{A}$, $T_\ce{B}$, and $T_p$ all to room temperature, the first and third term would be zero (sign does not matter), and you should simply get the enthalpy of mixing as a result. The result you arrived at does that, but the answer you were given changes the sign.
Unless there are some unexpected conventions about the signs of the terms, your result makes more sense.


text:
Calculating the Presence of Anthocyanin in a Solution
I've produced a red cabbage solution (boiling red cabbage in water basically), and would like to calculate the concentration of the red cabbage/anthocyanin in the solution. How is this accomplished?
Thanks

It is not a trivial exercise because it is not a single compound in red cabbage which causes this color (anthocyanin is a class not a single compound). There is a whole master's thesis dedicated for isolating red cabbage pigments from the University of Ohio. https://etd.ohiolink.edu/!etd.send_file?accession=osu1437655932&disposition=inline
One can determine monomeric anthocyanin using UV-Vis spectrophotometry by measuring absorbances at two wavelengths. It is always good to mention what apparatus your school has.
Giusti MM, Wrolstad RE. 1996. Characterization of red radish anthocyanins. J Food Sci 61(2):322-6.
Giusti MM and Wrolstad RE. 2005. Characterization and measurement of anthocyanins by UVvisible spectroscopy. In: Handbook of Food Analytical Chemistry. RE Wrolstad, SJ Schwartz (eds). John Wiley & Sons Inc. New York .p: F1.2.1-F1.2.13


text:
Comparison of acidities
My book says

To compare the acidities of two compounds, compare the stability of the anions formed by loss of a proton with respect to the original compound.

Suppose we have acetic acid and ethyl alcohol. Now on losing a proton, they form the respective anions:

My book again says

The acetate anion is resonance stabilized whereas in the ethoxide ion the charge is localized on the oxygen atom. Hence, acetate ion is more stable as compared to the ethoxide ion and hence, acetic acid is the stronger acid.

The problem is we have only compared the stabilities of the anions with respect to each other and not with respect to the original compounds. So how do we know that it is correct?

Comparing basicities does give you information about related acidities.

Once you understand that the acetate ion is more stable the the ethoxide ion, we understand the fact that after releasing The acidic hydrogen the final product formed is more stable for acetic acid as compared to ethanol.
This would lead to the conclusion that acetic acid is a better acid as compared to ethanol.
While comparing the acidity of two species, only the comparative study of conjugate base is enough, since its simply the comparison between compounds.
Comparing the stability with the original compounds will lead you to the conclusion that whether the compound itself has a tendency to act as an acid or base.
Alternately one could compare the stability of the original compound and its conjugate(Which is the acid dissociation constant) and compare its value with the acid dissociation constant of other compunds to compare its acidity.
In any case comparing the stability of the conjugate ions will tell you which one of them is comparatively more acidic/basic than others. It will not let you conclude whether its the nature of the compound to naturally act as an acid/base or not. For that one should compare the stability with original compund or just check its dissociation constant.

Just sthg that I would like to quip here: The answerer has not explicitly mentioned why comparing stabilities of an acid and it's conjugate base gives it's acidic/basic nature. This is because we consider acidity-basicity when a dynamic equilibrium exists b/w an acid and it's Brönsted base... And here,the reaction is driven towards that side where lower Gibbs free energy is available, or which side is more thermodynamically "stable". The arguments stated above would not work at any other reaction coordinate where kinetics would be the governing factor. Otherwise, +1


text:
What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to ground state?
When an electron in a higher energy state jumps to a lower energy state in Bohr’s model of hydrogen atom, it emits a photon which has energy equal to the difference in the energy between the final and initial energy states.
$$h\nu = E_f - E_i$$
where $\nu$ = frequency of photon emitted
Now what I know is each photon emitted corresponds to an emission line in the spectrum.
So, an electron can make the transition from a higher energy state to any lower energy state provided it emits a photon given by the above equation. In the question given, if an electron is in higher energy state; the maximum number of spectral lines is when the electron jumps to the next lowest energy state and so on as then photons will be emitted during each transition.
If an electron moves from $n = 6$ to $5$ we get a spectral line. Then from $n = 5$ to $4$ we get one and so on giving us a maximum of $5$ spectral lines. Now if the electron made the transition directly from $n = 6$ to $n = 4$ or any other lower energy state we would get less than $5$ spectral lines.
However, on looking up answer to the question it was given $15$. How can it produce $15$ spectral lines? When an electron from $n = 6$ jumps it has $5$ options and example it makes the jump to $n = 3$ then it has only two options $n = 2$ or $n = 1$. It can’t go back to $n = 6$ unless energy is supplied and it absorbs energy. So how can more than 5 lines be observed in the emission spectrum?
I am missing a key concept in such problems. Can someone tell me what I am missing?

The question asks about all possible transitions. Whenever you look into such deexcitation of electrons, obviously we are not concerned about one single electron,  but an ensemble of electrons which are present in Hydrogen gas. Thus the total transition from $i^{th}$ state is $i-1$. So, total no. of transitions is $\sum_{i=1}^{6} (i-1) = 15$.

Also see Number of spectral lines. If you expand the sum, you end up with the formula $$N = \frac{n(n-1)}{2} = \frac{6(6-1)}{2} = 15$$

@SoumikDas The question asks for an excited electron in a H atom as I have mentioned in the title. Then shouldn’t the answer be 5?

Please note that these answers are only true in the Bohr model, a better quantum model shows that there will be $n^2$ transitions. The  orbitals are now labelled with increasing orbital angular momentum, $\ell=0,1,2,3 ...$ as s, p, d, f, g, h.. (e.g 1s, 2s,2p, 3s,3p,3d ...with restriction $\ell\lt n$) and selection rules to govern what transitions are allowed. No doubt you will learn about this soon.

Since comments caused certain level of confusion, I guess I'll try to provide a further illustration.
You should consider all possibilities for an electron "jumping" down the excited energy state $n$ to the ground state $n = 1$.
Electron doesn't get stuck forever on any of the levels with $n > 1$.
Besides that, spectra is not a characteristic of a single excited atom, but an ensemble of many and many excited hydrogen atoms.
In some atoms electrons jump directly from $n = 6$ to $n = 1$, whereas in some others electrons undergo a cascade of quantized steps of energy loss, say, $6 → 5 → 1$ or $6 → 4 → 2 → 1$.
The goal is to achieve the low energy state, but there is a finite number of ways $N$ of doing this.
I put together a rough drawing in Inkscape to illustrate all possible transitions*:

I suppose it's clear now that each energy level $E_i$ is responsible for $n_i - 1$ transitions (try counting the colored dots).
To determine $N$, you need to sum the states, as Soumik Das rightfully commented:
$$N = \sum_{i = 1}^{n}(n_i - 1) = n - 1 + n - 2 + \ldots + 1 + 0 = \frac{n(n-1)}{2}$$
For $n = 6$:
$$N = \frac{6(6-1)}{2} = 15$$
Obviously the same result is obtained by taking the sum directly.

* Not to scale; colors don't correspond to either emission spectra wavelenghts or spectral series and solely used for distinction between electron cascades used for the derivation of the formula for $N$.

I’m still not able to understand for a single H atom containing 1 electron which is at n=6 as per question, maximum emission lines corresponds to 6 to 5 to 4 so on.. 1, right? Obviously the electron won’t get stuck there hence it jumps to lower energy state. For eg: the electron in n=6 makes jump to n=5, now it can’t go back up to n=6 right (unless you provide energy)?So it has to make jump from n=5 to lower state and hence maximum spectral lines will correspond to 5->4->3->2->1. Can you clarify please?

If it made jump from n=6 to n=3, now it can go to n=2 or to n=1 hence now the number of spectral lines will be less. It can’t go back to n=4 or n=5 or to n=6 from n=4, so why are we including all such lines? Once it makes a jump, it can’t go back it has to move to a lower energy state and so only way to maximise number of lines is maximise number of jumps and hence 6> 5-> 4-> 3> 2-> 1 giving 5 jumps and a line per jump giving 5 emission lines. Please note I am speaking about a Hydrogen atom containing 1 electron as given in the question.

If you artificially limit yourself to a single atom, then yes, maximum number of transitions is 5, but I'm not aware of such experiment for the emission from a single excited hydrogen atom. A resonance fluorescence spectrum from a trapped heavier atom/ion — maybe; but it isn't what the question from your textbook is about. Again, classically you are supposed to review a large number of excited atoms and find the number of the spectral lines by judging all possible transitions en masse.

Yes I suppose you’re right. The question asks for lines emitted when considering many such H atoms yielding maximum of 15 lines, however upon reading I assumed it asks for one atom and it made me wonder how 15 such lines can be emitted.

Yep, this is pretty common among textbooks' levels of abstraction. Author first talks about a single instance for simplicity, but forgets to switch back to the real world with the superposition of such instances.

According to me in the above question there are two things that should  be noted
1 the question  is saying  A Hydrogen atom that means a single hydrogen atomm
2 maximum  number of spectral lines
Which is clearly telling that electron will jump from n=6 to 5 and 4 and so on till it reaches ground  state
Hence correct  answer should be 5
Now If it isn't  mentioned  in question  about single  atom (or told that a sample of hydrogen) than consider it as a bunch in this case consider all possibilities hence in this situation answer will be 15

If you want to consider just one atom measured only once, an odd sort of experiment, nevertheless the number of possible transitions is a number between 1 and 15. The atom may transition from $n=6 \to 1$ or cascade down this way and that way using every other possibility.  Of course we would not know much about the atom from such a single experiment, in fact doing it twice might be very confusing! Luckily its hard to work with single atoms.


text:
What are the total amount of resonance structures for these two cations and which is the most stable cation?
Please see the following two cations.

I have proposed the following resonance structures.
Please let me know if I have missed something.
Thank you.


text:
Why does the degree of dissociation change when we dilute a weak acid even though the equilibrium constant is constant?
$K$ represents the ratio of concentrations of molecules in a solution at equilibrium, which means that $Q_\mathrm{r}$ (that ratio at any given point) looks to be identical to $K$. In other words, the molecules in that solution react accordingly so that they reach equilibrium and the ratio of their concentrations is equal to $K$.
If $K$ is large enough (bigger than $10^4$ in my curriculum), this means that the the concentration of the reactants is almost zero. In other words, the  equilibrium position of that solution looks very much like a reaction that went to completion.
The more we dilute an acidic/basic solution, the higher the degree of dissociation, even though $K$ stays the same. So, does that mean that the more we dilute a solution the harder it is for it to reach the point of equilibrium for that specific molecule/solution or what?
For instance, say you found $K$ of solution to be $10^{-5}$. This means that when the reaction happens there are lots of reactants left, and not much products produced, which means that the degree of dissociation is low. But the more we dilute a solution, the closer it gets to a "complete reaction" (if you pour a small amount of weak-acid molecules into a large tank of water, it's certain that all of the weak-acid molecules are going to react with the water, i.e. the degree of dissociation approaches $100\%$).
So, how come $K$ can be independent of the initial reactants concentrations, and tell if a reaction was complete or not, when the "completion" of a reaction (the degree of dissociation) depends on the initial concentrations of reactants?

These two videos might help : https://www.khanacademy.org/science/chemistry/thermodynamics-chemistry/gibbs-free-energy/v/changes-in-free-energy-and-the-reaction-quotient  ; https://www.khanacademy.org/science/chemistry/thermodynamics-chemistry/gibbs-free-energy/v/standard-change-in-free-energy-and-the-equilibrium-constant . It's down to the chemical potential. Two systems with different concentrations of reactants and products may both be at equilibrium, there is nothing contradictory about that. The equilibrium constant tells you how to identify such systems.

If K is large enough (bigger than $10^4$ in my curriculum), this means that the the concentration of the reactants is almost zero.

This statement is not always true - it depends on the stoichiometry of the reaction.
For the reaction $$\ce{A(aq) <=> B(aq)}$$
the concentration of A is much smaller (ten thousand times) than that of B if $K = 10^4$, and this is independent of diluting the solution. It's fair to say that there is almost no A compared to B.
For the reaction $$\ce{4A(aq) <=> 4B(aq)}$$
the concentration of A is just ten times smaller than that of B if $K = 10^4$, and this is also independent of diluting the solution. However, there is quite a bit of A compared to B, and it is a bit misleading to say the the concentration of A is almost zero.
Both of these reactions had the same number of reactant species as product species (all in the same solution, so all equally affected by dilution).

The more we dilute an acidic/basic solution, the more the reaction is "complete", the more the reactants disappear, but K stays the same.

Here, "complete" would mean the absence of reactants, i.e. a reaction that goes to completion. Weak acids are defined by not completely dissociating, in contrast to strong acids. The general reaction for a weak acid is:
$$\ce{AH(aq) <=> H+(aq) + A-(aq)}$$
Notice that there are more product particles than reactant particles. For those type of reactions, dilution favors the products. In fact, if $K = 10^-5$ for this reaction and all concentrations are $\pu{10^-5 M}$, the reaction is at equilibrium. This means even though the equilibrium constant is much smaller than one, you still can have reactants and products at the same concentrations.
On the other hand, if there are more reactant particles than product particles, dilution has the opposite effect. For a complex formation reaction between a metal M and a ligand L
$$\ce{M(aq) + 4 L(aq) <=> ML4(aq)}$$
dilution will cause the complex to fall apart. If $K = 10^4$ and the free ligand concentration is one molar, there will be a lot of complex and very little free metal. On the other hand, if the free ligand concentration is one millimolar (0.001 M), there is hardly any complex and most of the metal will be in the form of the free metal.

Why does the degree of dissociation change when we dilute a weak acid even though the equilibrium constant K is constant?

As the examples above illustrated, this is because for these specific reactions, the sum of exponents for the products is higher than that of the reactants in the equilibrium constant expression. As you dilute, Q will decrease, and the reaction will go forward to reach K again. At infinite dilution, the product will be favored no matter what the value of the equilibrium constant is.
If you want to have a general statement about what the value of K means, it would be something like "as K increases, the equilibrium concentration of products will increase and those of reactants will decrease".

See my comment above.
As a numerical example, take acetic acid ($\ce{AcOH}$), which has $K_\mathrm{a} = 1.8 \cdot 10^{-5}$.
This means that:
$$K_\mathrm{a} = \frac{[\ce{AcO-}][\ce{H+}]}{[\ce{AcOH}]}$$
And the total nominal concentration of acid is:
$$C_\mathrm{a} = [\ce{AcOH}] + [\ce{AcO-}]$$
Combining these two equations, you can see that the % of dissociated acid is:
$$\alpha = \frac{[\ce{AcO-}]}{C_\mathrm{a}} = \frac{K_\mathrm{a}}{K_\mathrm{a} + [\ce{H+}]}$$
i.e. not a constant, but a value that depends on the $\mathrm{pH}$. So no, there is no need for the 'degree of completion' of the reaction to be a constant.
You can even calculate the explicit concentrations of all the species.
Given that:
$$C_\mathrm{a} = \left(\frac{1}{[\ce{H+}]} + \frac{1}{K_\mathrm{a}}\right) \cdot ( [\ce{H+}]^2 - K_\mathrm{w}^2)$$
you can see that, for $\mathrm{pH} = 5$ you need:
$$C_\mathrm{a} = \left(\frac{1}{10^{-5}} + \frac{1}{1.8 \cdot 10^{-5}}\right) \cdot \left((10^{-5})^2 - (10^{-14})^2\right) = 1.5554 \cdot 10^{-5}$$
Then:
$$[\ce{AcOH}] = \frac{C_\mathrm{a} \cdot [\ce{H+}]}{K_\mathrm{a} + [\ce{H+}]} = 5.555 \cdot 10^{-6}$$
$$[\ce{AcO-}] = C_\mathrm{a} - [\ce{AcOH}] = 9.999 \cdot 10^{-6}$$
So the % of dissociated $\ce{AcOH}$ is:
$$\alpha = \frac{[\ce{AcO-}]}{C_\mathrm{a}} \approx 64 \%$$
Now repeat for $\mathrm{pH} = 3$:
$$C_\mathrm{a} = \left(\frac{1}{10^{-3}} + \frac{1}{1.8 \cdot 10^{-5}}\right) \cdot \left((10^{-3})^2 - (10^{-14})^2\right) \approx 0.05656$$
$$[\ce{AcOH}] = \frac{C_\mathrm{a} \cdot [\ce{H+}]}{K_\mathrm{a} + [\ce{H+}]} \approx 0.05556$$
$$[\ce{AcO-}] = C_\mathrm{a} - [\ce{AcOH}] \approx 0.001$$
$$\alpha = \frac{[\ce{AcO-}]}{C_\mathrm{a}} \approx 1.8 \%$$
So you see, in a more dilute mixture the $\mathrm{pH}$ is higher, and the dissociation reaction of acetic acid is 'more complete', whereas at higher concentration the pH is lower, and there is proportionally less acetate anion.
All of these systems are at equilibrium.

$K$, the equilibrium constant, is independent of the composition of the system. It simply describes the preference of the system for reactants or products.
On the other hand, there is the reaction quotient, $Q$, which describes the position of the system relative to equilibrium. $Q$ is the value that is variable and that you adjust to be closer to $K$. Technically, $K$ can also be computed from the standard free energy change of the reaction, $\Delta G ^{\circ}$.
What's confusing is the $Q$ and $K$ are computed in the exact same way, with the sole difference that for $Q$, you plug in the values you have right now, and for $K$, you plug in the values at any equilibrium.
If $Q=K$, you are at equilibrium.
If $Q < K$, you need to push the reaction towards more products.
If $Q > K$, you need to push the reaction towards more reactants.
When you dilute a solution (or make other changes), you may change the value of $Q$, which may put you farther from equilibrium, but the opposite may also be true. It depends on the specific reaction and the form of the equilibrium expression.
EDIT:
Your confusion is because you're confusing $Q$ measured at different points in time. At the end of the reaction $t = \infty$, $Q = K$. At the beginning of the reaction, $Q_{0} < K$. Over time, as the reaction goes towards equilibrium, the value of $Q$ will get closer to $K$. If at the beginning of the reaction, you dilute the solution, you may get a different value of $Q$. If it's smaller, then you have farther to go to get to equilibrium. But again, this depends on the molecularity of the reaction.
$$\ce{A + B -> C}$$
$$\ce{A -> C}$$
$$\ce{A -> C + D}$$
Here are three different reactions.
In the first reaction, at the beginning of the reaction, with $Q < K$, dilution pushes you towards equilibrium because reactant concentrations enter into the equilibrium expression twice.
$$Q = \frac{\ce{[C]}}{\ce{[A][B]}}$$
If you halve all three concentrations, the value of $Q$ will get bigger.
In a similar vein, the second equation is unaffected by dilution. In fact, at equilibrium, dilution preserves equilibrium.
And for reaction 3, dilution pushes you farther away from equilibrium.
For all three reactions, at the end of the reaction, you will have $Q = K$. But for reactions 1 and 3, dilution at equilibrium changes the value of $Q$ and the reaction will need to adjust again to reach equilibrium.

"You may change the value of Q" the Q at the end of the diluted-solution's reaction? If so, then does that mean that if you dilute a solution to a certain point, Q stops going towards K? Also, If K is small (say smaller than 10⁴), can you dilute a solution to a certain concentration so that it's reaction is complete (α=1)?
If not, then how can this saying be explained: "K is only affected by temperature and not the initial concentration of the reactants, while the 'advancement' of a reaction (α) is affected by both temperature and the initial concentration of the reactants"?

No, immediately after dilution. $Q$ will be $K$ at the end. You cannot dilute the reaction to completion. But it might get closer or farther depending on coefficients of the balanced reaction. I think you are confusing $K$ and $Q$. The first is constant. The latter is not.

Hmm, I think I'm having trouble wrapping my head around how can a diluted-reaction be closer to completion with Q at the end of that reaction still being the same as before

Let me add an edit. I think that will help.

The question is inactive, I doubt anyone will see it, but go ahead

A really handwavy explanation would be that the system, attempting to maximize its entropy, will prefer dissociated states as the system dilutes since there will be an increasing number of microscopic configurations with atoms $A$ and $B$ far apart (i.e. molecule $AB$ dissociated) as opposed to close together (i.e. atoms $A$ and $B$ bound together into molecule $AB$).


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Energy of photons of given frequency and number
I’m having trouble with calculating the energy of $\pu{5.00E10}$ photons of frequency $\pu{1.00E9 s-1}$.
I know that to calculate the energy $E = hf$ is to be used. However, I am unaware as to what to do with the $\pu{5.00E10}$ value and its meaning.

Analogy: you have 5 bottles of coke 1 L each; what is the total volume?
The formula you shown $(E = hf)$ is derived for exactly one photon.
Now, since you have 50 billion of photons $(N = \pu{5.00e10}),$ guess what the total energy would be?

 $$E_\mathrm{tot} = Nhf = (\pu{5.00E10})(\pu{6.63E-34 J s})(\pu{1.00E9 s-1}) = \pu{3.32E-14 J}$$

How about if I am given instead of a frequency value a wavelength value along with the photon value, how do I answer solve for the energy.

$$f = \frac{c}{λ},$$ where $c$ – speed of light; $λ$ – wavelength.

For the speed of light do I use c= 3.00E8 and if I do, do I use the frequency value gained to solve for the energy?

Yes, you do, but mind the units. If you take $c = \pu{3.00e8 m s-1}$, then make sure you plug in the wavelength in meters.


text:
Reduction of triphenylmethane dyes in ink
Right now I'm trying to explain why the ink from an ink pen turns colourless in water.
I know that the pH is responsible as it breaks apart the conjugated system of double bonds. I'm just not sure on which part, to $\ce{OH-}$ comes into play (example: Water blue).
My suggestion would be that $\ce{OH-}$ can be reversible added to double bond in the center forming two phenyl rings and one ring with 2 double bonds (which will rearrange into a phenyl ring with the double bond from the following amide-group) with a $\ce{C-OH}$ in between.
Under acetic conditions, the $\ce{OH-}$ is eliminated.
Is that correct and does anyone know a mechanism, explanation etc. for that reaction somewhere on the Internet?


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Reactions between Ca(OH)2 and CO2, and Sr(OH)2 and CO2
So I'm doing an investigation, and it involves using supersaturated solutions of Ca(OH)2 and of Sr(OH)2. I notice that there's always a flaky precipitate formed on top. I need to know about the reactions with carbon dioxide. Is it carbonate that's formed? Or is it hydrogen carbonate? Why?

In my opinion nothing prevents both carbonates and bicarbonates from forming those flakes on the surface; out of curiosity, why is it so crucial to determine which one is it?

Although knowing the identity of the precipitate is not really significant to my results in any ways, it's very important that I discuss it since the investigation is for my IB Chemistry Extended Essay. I am required to discuss basically everything going on in the solution in as much detail as possible. I just thought I'd probably be marked down if it I had no idea what's going on in my solutions..

Interesting, I didn't think of the possibility that it's both. Could be. Thanks!

it is calcium carbonate that is forming.

Your solid is the respective carbonate. Group II bicarbonates can form aqueous solutions, but do not form stable solids, ergo the solid flakes must be carbonate and formed along the lines of:
$$\ce{Ca(OH)2 (aq)+ 2CO2 (aq/g) <=> Ca^2+ + 2(HCO3)- (aq) <=> CaCO3 (s) + CO2 (aq/g)}$$


text:
Is there an inexpensive solvent I can use to dissolve thermal paste without scrubbing?
I'm removing large voltage regulators from circuit boards which are often coated with a generous amount of thermal grease. I've found that 70% isopropyl alcohol works great for softening this and removing it with some light scrubbing.
The problem I have is that, without a lot of additional effort and scrubbing, the surfaces will often retain a film from the grease. There are also areas which I may have missed or were unable to reach when scrubbing.
This type of grease is often electrically conductive and can cause a short circuit if accidentally brushed over the pins of a chip. It's also just plain annoying when it gets on your hands.
Is there an inexpensive solvent that I can use to dissolve the thermal paste without having to scrub the components? I felt that it would be more appropriate to ask this here on the chemistry stack exchange as I'm also interested in understanding how other solvents would work better for this.
I tried leaving the components in a bag of 70% isopropyl alcohol but this confirms that it only softens it. The thermal paste don't actually free itself from the surface without some mechanical force.
See here for a photo where the surface looks clean but you can see where I scraped my finger across it through the bag and it removed more of the grease.
The color of the component becomes a dull grey due to the grease becoming embedded in the pores of the plastic which is what it's supposed to do. I would like to know if there is a way to simply drop these into a chemical and have the grease dissolve away on its own without damaging the package.

I think you are already using one of the best solvents for the job. Isopropyl alcohol is affordable, non-corrosive, easy to remove and dissolves grease fairly easy. To improve the efficiency you may want to try to use an ultrasonic cleaner, and add some cleaner concentrate (surfactant) to isopropyl alcohol.

You also have to consider that you don't want to dissolve the plastic of the electronic case.

For cheap solvent for grease, gasoline or kerosene works wonders for auto mechanics.

First the sad part - there is no magic compound that will dissolve thermal paste like a champ and leave your electronics alone. If that was no "sine non qua" as they say, we could have suggested a varied assortement of highly aggressive halogenated solvents which we would mix with something else for the extra bang, some hydrogen peroxide perhaps, or the good old fuming nitric acid. Things would happen, rapidly.
As for all-round no-nonsense solvents, isopropyl alcohol is a big favourite, it is fairly environmentally friendly especially if you consider your hands and your tabletop as part of the environment. And of course it solvates a wide range of icky substances that needs removal. As a small step up the agressive solvent ladder would be to consider acetone, "citrus solvent" (basically oils and acid of lemons, it kicks ass) or acetic acid based solvents. But none of them will magically clean the surface, you'll have to use some elbow grease. Or thumb grease, I guess.


text:
Struggling to understand wavevector K for deriving energy levels in reciprocal space
I don't really understand why, when deriving wavefunctions for arrays of atoms, the wavevector k is introduced and the general idea surrounding momentum space. please help!!

I think at the moment your question is a too broad. Could you give a little more detail on what is giving you trouble? @hphillips997

Because regular arrays have additional requirements on the wave functions that describe them. This translational symmetry induced momentum constraints.

The Hamiltonian for a particle in a periodic array of atoms is invariant (doesn't change) under translation by any vector of the form $n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3\vec{a}_3$, where the vectors $\vec{a}_1,\vec{a}_2,\vec{a}_3$ for the unit cell and $n_1,n_2,n_3$ are any integers. The set of all such translations form a group.
When a Hamiltonian is invariant under all elements of some group, it follows that its eigenfunctions can be written as functions of definite symmetry with respect to that group. A function of definite symmetry with respect to the group described above turns out to be characterized by a "wavevector" $\vec{k}$. This is completely analogous to situation for the hydrogen atom, where spherical symmetry implies $[L^2,H]=0$ which in turn implies that total angular momentum is conserved and that our eigenfunctions possess a quantum number $l$ (along with $n$ and $m$) that quantifies the wavefunction's total angular momentum. To make the analogy clear:

Hamiltonian of hydrogen atom commutes with arbitrary rotations $\to$ eigenfunctions of Hamiltonian have quantum number $l$
Hamiltonian of particle in periodic potential commutes with arbitrary translations $n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3\vec{a}_3$ $\to$ eigenfunctions of Hamiltonian have wavevector $\vec{k}$.

A function $\phi_{\vec{k}}$ of wavevector $\vec{k}$ happens to have a special behavior under group operations:
$$
\phi_{\vec{k}}(\vec{x} + n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3\vec{a}_3) = e^{i\vec{k}\cdot(n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3\vec{a}_3)} \phi_{\vec{k}}(\vec{x})
$$
At this point we usually note that if two wavevectors $\vec{k}$ and $\vec{k}'$ differ by some $m_1 \vec{b_1} + m_2 \vec{b_2} + m_1 \vec{b_2}$, where
$$
\vec{b}_i = 2 \pi \frac{ \sum_{j,k=1}^3 \epsilon_{ijk} \vec{a}_j \times \vec{a}_k}{\sum_{m,n=1}^3 \epsilon_{imn} \vec{a}_i \cdot \left( \vec{a}_m \times \vec{a}_n \right)}
$$
and $m_1,m_2,m_3$ are integers, then
$$
\left( \vec{k'} - \vec{k} \right) \cdot(n_1 \vec{a}_1 + n_2 \vec{a}_2 + n_3\vec{a}_3) = 2 \pi \left( n_1 m_1 + n_2 m_2 + n_3 m_3 \right)
$$
so that the wave vectors have equivalent symmetry properties. This allows us to restrict our consideration of wavevectors to those of the form $c_1 \vec{b}_1 + c_2 \vec{b}_2 + c_3 \vec{b}_3$, where $ 0 \leq c_j \lt 1 $. More commonly we equivalently restrict the possible wavevectors to lie in the Brillouin zone.
As to why the concept is introduced, the answer is that recognizing that our energy eigenfunctions can be classified by some symmetry often enormously simplifies problem solving. Using again the hydrogen atom as an example, recognizing the Hamiltonian's spherical symmetry allowed us to break up a 3-dimensional Schrodinger equation into three 1-dimensional differential equations, which is an enormous simplification. For a periodic pontential, knowledge that our eigenfunctions have a wavevector $\vec{k}$ allows us to find the eigenfunctions by solving a differential equation over a single unit cell (as opposed to all space).


text:
How to convert soluble colorless [HgI₄]²⁻ back to insoluble red HgI₂?
When a solution contains $\ce{Hg^2+}$ and $\ce{I-}$, depend on the concentration ratio of $\ce{Hg^2+}$ and $\ce{I-}$, the following reactions can happen:
$$\ce{Hg^2+ + 2 I- ⟶ \underset{insoluble red color solid}{HgI2}}$$
If there is excess $\ce{I-}$, then this red solid will continue to react with excess $\ce{I-}$ to form an colorless soluble complex:
$$\ce{HgI2 + I- ⟶ [HgI3]^-}$$
$$\ce{[HgI3]^- +  I- ⟶ [HgI4]^2-}$$
Is there a way to revert $\ce{[HgI4]^2-}$ back to $\ce{HgI2}$?

Have you any ideas? There is rather trivial solution...

What I can think of is adding the product [hgi4]2- in the solution so that the equilibrium will be pushed to the left and hgI2 will be formed again.

No, you should add something else to eliminate the excess, think about proportion of reagents here.

I can add more Hg2+ in solution to react with the excess I-.

Indeed, it should work. Technically they react with these complex anions, polymerising until solid particles precipitate AFAIK.

As noted in the comments, to attain mercuric iodide, you need to maintain a balance between the concentration of $\ce{Hg^2+}$ and iodide ion. If there is slight excess of iodide ion, it will form a complex (much like $\ce{KI3}$ from $\ce{KI}$). The complex is popularly known as Nessler's reagent and is prepared pretty much in this way.
Another solution is to decompose the complex to revert back into original products. But I won't recommend this solution. Though, it is expected to give the original reactants, occasionally, decomposition reactions give rise to undesired side products and may release toxic fumes of mercury, iodine, iodide, potassium oxide($\ce{K2O)}$ etc. Do not try to heat the complex. Instead opt for the 1st solution.

Thank you very much for your comment. I will stick with maintaining the balance between the concentration of mercury and iodide ion.


text:
Is there another name for compound "delta-7 sterine", does it exist alone?
Is there a more common name for "delta-7 sterine"? Is this spelled correct?
I read all over the internet and this compound exists in many articles, but I cannot find any information on it at all, also it doesn't show any information it. Can you please direct me to more information on this compound on what it is, it's more common names, chemical makeup, etc. All I could find in articles is that it's a steroid found in pumpkin seeds.
I would like to buy this compound for experimenting but cannot find a supplier that recognizes this name of the compound, does it exist alone in this form?

E-commerce and related questions about buying/selling are considered off-topic on Chemistry.SE (I edited your question accordingly to meet the requirements).

"delta-7 sterine" seems to be an adaptation of German term "Sterine" (plural for "Sterin") used by German scientists who elucidated the structures of those steroids using NMR [1].
In English, it seems, the name should be written as Δ⁷–sterol and is not a single compound but a group of compounds (for more info I'd suggest searching Google Scholar).
References

Sucrow, W.; Reimerdes, A. Δ⁷-Sterine Aus Cucurbitaceen. Zeitschrift für Naturforschung B 1967, 23 (1), 42–45. https://doi.org/10.1515/znb-1968-0109. (Open Access)


text:
Is the iodide and persulfate reaction bimolecular?
Given the following reaction in an aqueous state:
$$ \mathrm{2I^- + S_2O_8^{-2}\rightarrow  I_2 + 2SO_4^{2-}}$$
Is it bimolecular o trimolecular? Because two molecules of iodide react with one molecule of persulfate, so I would say trimolecular.
Furthermore, I know it is a second order reaction because of the Iodide Clock experiment I performed, so would that mean that the reaction is complex and not an elementary reaction?

You already know the answer. Just say it.

So are my assumptions correct?

Which one? You have experimental evidence, trust it.

Yes, but a second order reaction does not imply is bimolecular if the reaction is not elementary

Not exactly. Bimolecular or trimolecular is not  relevant for nonelementar reactions. If the whole complex reaction is of second order, then it's usually a safe bet that there is no trimolecular elementar reaction in it.


text:
Determining ratio of competing fragmentation processes in mass spectrometry

In a photoionization experiment, three photons of wavelength 266 nm were absorbed by a neutral molecule of 6-methyl-2-heptanone. If the kinetic energy of the electron removed during ionization is 1.4 eV, what is the ratio of the simple cleavage to the rearrangement produced fragment ions? The following plot describes the fragmentation of the molecular ion of 6-methyl-2-heptanone. The IE of this compound is 9.10 eV.

My intuition would just be to use the value of 9.10 eV provided in order to look at the rates of cleavage to rearrangement in order to determine ratios. I'm not sure how the 1.4 eV and the 266 nm photons come into play.

Your three photons represent the total energy added to the system:
$$E_\mathrm{photon} = 3 \cdot \frac{hc}{\lambda} = \pu{13.98 eV}$$
That, less the sum of ionization energy and kinetic energy of the removed electron, leaves the change of the internal energy of the cation:
$$E_\mathrm{int} = \pu{13.98 eV} - \pu{9.10 eV} - \pu{1.4 eV} = \pu{3.48 eV}$$
Finding the value of $\log k$ for both curves at this energy will give the relative rates of simple cleavage vs rearrangement.


text:
At what position in the distillation apparatus does the temperature match the boiling point of the distilled substance?

A student is separating $\ce{CHCl3}$ (bp = $\pu{61 ^{\circ}C}$) from $\ce{CHCl2CHCl2}$ (bp = $\pu{146 ^{\circ}C}$) by distillation. She has just begun to collect the first distillate in the receiving flask. At what position in the apparatus will the temperature be $\pu{61 ^{\circ}C}$?

The answer to this questions is B.
I narrowed down the options to either A or B, because I know that the temperature at A would be slightly higher than $\pu{61 ^{\circ}C}$, but lower than $\pu{146 ^{\circ}C}$, and the temperature at B would be slightly cooler than the temperature at A. C and D would be much lower in comparison.
What makes the answer to this questions B rather than A? I think that the two temperatures should be somewhat close to each other, but is there some intricacy to this that I am missing?

You want to find the place where chloroform is present both as pure liquid and as pure gas at normal pressure. If you boil pure chloroform, this would be at the liquid gas interface of the bubbles rising in the liquid.

The temperature in area A can well exceed B all right, especially if the distillation adapter is long or if there is a fractioning column or another condenser before the three-way adapter.
However, no target component is going to make it through the condenser unless its vapors reach the side arm (the one with the joint to condenser).
Apparently, it's only possible if the temperature of that area coincides with the boiling point of the target, that's why the thermocouple or thermometer probe is installed exactly at that area (B).
As you rightfully suggested the temperatures in areas C and D are arbitrary: it only should be above the m.p. to prevent clogging and less than b.p. so that the target is not evaporating away.
P. S. there is a vacuum adapter shown on the picture (area C), so there also might be distillation at lower pressure, which would also lower the b.p. of the mix constituents.
Since the question doesn't mention its usage, atmospheric pressure is assumed.

A vacuum adapter shown on the picture (in area C) is for pressure equilibrium in this case due to the closed circuit otherwise.

What makes the answer to this questions B rather than A? I think that the two temperatures should be somewhat close to each other, but is there some intricacy to this that I am missing?
The factor that you're missing is heat loss. Because the apparatus is in the air it is losing heat to the atmosphere. As the vapor rises through the three way adapter it cools and drips back into the pot. So there is a temperature gradient from the pot temperature to the condenser. So Point B will be colder than the liquid or the vapor just above it.
The other thing here is that you are not going to get perfect separation since 1,1,2,2 tetracholoroethane has an appreciable vapor pressure at $\pu{61 ^{\circ}C}$. The separation that can be achieved is measured by the number of plates of the distillation apparatus. So the greater number of plates, the better the separation.
Since the condensate will be a mixture, you want the thermometer where the gas phase will start traveling down the condenser. This will give you the best information about the composition of condensate.
What you are trying to do is to adjust the applied heat to keep Point B at or just above $\pu{61 ^{\circ}C}$ so that the condensate is as rich as possible in trichloromethane.


text:
What gas was produced in this orbiting space-balloon from 33 pounds of benzoic acid and anthraquinone? Why these two chemicals?
As pointed out in this answer, on page 4 of NASA Technical Note TND-1366 The Orbital Behavior of the Echo I Satellite and its Rocket Casing During the First 500 Days (June, 1962) says:

PHYSICAL CHARACTERISTICS

Echo I has an effective cross-sectional area of 7,854 square feet. The launch weight of 157 pounds decreased to 124 pounds with the loss of 33 pounds of benzoic acid and anthraquinone which were used to maintain inflation for the first few  weeks in orbit. Thus, the initial ratio of the weight to mean drag area for Echo I was 0.020 pound per square foot and reduced to 0.016 pound per square foot after several weeks in orbit.

Echo-1 was about 100 feet in diameter, so that change in weight of 0.004 pound per square foot of drag area corresponds to the 35 pounds mentioned.
Questions:

What kind gas would have been produced by this mixture, and what fraction of the initial mass would have been converted to gas? Could it have been all of it as suggested? (the idea is that after the gas leaked out the balloon would have become rigid enough to retain its shape while in orbit.
Why both benzoic acid and anthraquinone? Did they react? Did they sublimate only? Is there any chemistry-related reason why both substances were used?

below x2: cropped and full size from Echo, NASA's First Communications Satellite

From the answer to the queistion you have linked, it would seem that there was no reaction; that the two substances merely sublimated. I cannot think of a reaction that would provide a gaseous product per se. The choice for these two substances may simply be down to: they evaporate at the right conditions and do not cost too much.

@TAR86 I've just stumbled upon this question, edited and undeleted it. I think it can be answered but there may be some interesting reasons why two compounds were used.

There are three materials to investigate: the two solids in question have been described as having fast and slow rates of sublimation. The balloon was sent up with 4.5 kg of benzoic acid (bp 250C) and 10.6 kg of anthraquinone (bp 380C). https://en.wikipedia.org/wiki/Project_Echo
The third material is the film: 12.7 micron biaxially oriented polyethylene terephthalate metallized with 0.2 micron aluminum. The two sublimatible solids were used to rapidly inflate the balloon and then to keep it inflated in case of micrometeorite strikes.
Echo-2 was made with a thinner PET film (9 microns) sandwiched between two thicker aluminum films (4.5 microns each), and the balloon was inflated to a pressure that slightly deformed the aluminum sandwich and kept the whole structure more rigid, so continual sublimation was not required.
The PET films are so thin! By comparison, a human hair is typically 100 microns in diameter. Perhaps the choice of sublimatible materials was made for non-reactivity (with the film) rather than with any reaction to form a gas. But the rationale is still not clear - it certainly couldn't have anything to do with cost.

Interesting! If the fast and slow rates of sublimation were very different, one might be for initial inflation so it could start performing as a signal reflector quickly, and the other might be to maintain inflation over time. But If I understand correctly the metalization was thought to maintain the shape even if the pressure went to zero. Did Echo 2 remain spherical without requiring gas pressure? If so, how is this known to be true? The answer there is extensive!

https://i.stack.imgur.com/CZ9t6.png talks about the rationale of having a fast and a slow sublimating substance. Source: https://digitalcommons.odu.edu/cgi/viewcontent.cgi?article=1028&context=mae_etds


text:
CO2 IR extinction coefficients near the 15 micrometer band
I have searched multiple times for a spectrum or a table of the gaseous carbon dioxide Lambert-Beer extinction coefficients in the IR Earth thermal region.
All I found was absorption spectra.
Is this available anywhere with free access? I want it for illustrative/education purposes to make some Excel calculations.

Related: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6174548/


text:
Does the mole fraction of a solution change with Vant Hoff factor?
I reason that as mole fraction is a measure of colligative properties, increase in the Vant Hoff factor ( and hence increase in the value of colligative properties) should also leave a mark on the mole fraction of the solute components.
However, my instructor tells differently.

Equal doesn't always means causation.  Vant Hoff factor does not change for very dilute solutions. It does change when mole fraction is high enough to cause non ideal behaviour, but it can decrease, for instance when x is high enough that the solute does not fully dissociate.

That is not what I asked for. My instructor told me that if a solute component dissociates in solution to yield ions, the mole fraction of the solute increases, and this is so because the no. of solute particles increases,but the total no. of moles remain constant (why?) If the no of moles of solute increases, shouldn't the total no of moles also increase? I know that will still lead to a greater mole fraction, but doesn't the total no of moles (solute + solvent) also increase?

It depends on how you define the mole fraction, in NaCl:water or sodium ion : water chloride ion . I don't see much sense on the latter here because is account for dissociation and solvent molecules are in excess. What do you call the solute? Is NaCl, and you account for dissociation. I am afraid you mean in case you add practically more solute. Then of course you change the overall number of moles, not only x.

I will explain with an example. Suppose 'x' moles of glucose are dissolved in 'y' moles of a solvent (assume water, also assume x<y). The mole fraction of glucose (solute) is x/(x+y). Now if in a similar situation, 'x' moles of NaCl are dissolved in 'y' moles of water and 100% dissociation is assumed, then is the mole fraction of solute components 2x/(2x+y) or 2x/y ?(In either case, the mole fraction increases)

In your example, for each ion it will be x/(2x+y). So the latter. Consider that in dilute solutions x + y is about y, too.