text:
How can an atom have 2 outer shells?
For example, my textbooks says how many electrons in the outer shell in Nickel? and it says the answer is the electrons in both the d shell and s shell. Doesn't shell equal= n quantum number? And n=4 is the outermost shell so shouldn't the answer exclude the electrons in d shell (since they are in the n=3 shell? Also the book says that Iodine has 7 electrons in its outer shell why did they skip over the de shell in this situation?

Electron shells is very obsolete concept, easily misleading to incorrect ideas. See also atomic orbitals

While solving hydrogen atom in quantum mechanics, the total Hamiltonian contains the kinetic energy terms and the nuclear-electron attraction term. There are no terms corresponding to electron-electron repulsion in the Hamiltonian. This gives rise the fact that we can separate the wavefunction into coordinates as $\psi(r,\theta,\phi)=\psi_n(r)Y_l^m(\theta,\phi)$, and the eigenvalue (energy) of the Hamiltonian is dependent only on the nuclear part quantum number $(n)$. This is true for any hydrogenic system ($\ce{H,He+,Li^2+})$, i.e. a system with one electron and one nucleus. The numbering of shells are based on hydrogenic orbitals only. So if we say an electron is in "$3p$" orbital, it means the electron (which is the only electron in the system) has an wave function corresponding to $n=3,l=1$ and $-1\le m \le 1$.
However, as we go to systems with more than one electrons, electron-electron repulsion comes to play and we are unable to separate the wave function into coordinates. Till date, exact solution of Helium atom is an extremely difficult challenge in quantum mechanics. However, even for multi-electron systems, one imagines hydrogenic orbitals while understanding the electronic structure of atoms and that is why we say, "$\ce{C}$ has $1s^22s^22p^4$ electronic configuration". The truth is, however, the orbitals we know of is hydrogenic in nature and we fill up electrons there without even trying to account for electron-electron repulsion and correlation. In facr, the orbitals you imagine may (and most probably will) not look like the orbitals in C atom at all.
So, there is nothing such as outer shell of electrons. There are some available approximate methods like Hartree Fock to estimate the orbitals using discrete basis, however, there are some issues in each of the method we know of today. In other words, we do not know which ones are the "outer shell" yet. However, my best bet in this case will be to take up any one of the standard methods for approximating multiple electron atoms and call the orbital(s) with highest energy level as outer orbitals.


text:
What the product of this reaction?
The reaction is as follows, with (i) being my answer, and (ii) being the correct answer in the book.

My question is why is the product (ii) formed here and elimination not favoured here? I think that the first structure is more stable(has extended conjugation) whereas the latter is relatively unstable. I do understand that the carbocation formed during the formation of (ii) is highly stable (cyclopropyl carbocation and has extended conjugation with phenyl), but is this the reason why (ii) is the major product?

The first structure has a less strained ring???!! Who told you that?

Likely the rearrangement to the more stable cation is the driving force.  In addition to the stability of the rearranged ion, the rearrangement itself is likely to be kinetically facile as it involves a simple electron-pair move and forms the new bond between atoms that are already relatively close together (1,3 in a four-member ring).

@Waylander I get it now, 1st is more strained because of 2 sp2 carbons in a ring. But isn't it more stable because of extended conjugation. Will edit the question now

@OscarLanzi But if a lot of heat is provided can it form (i)? Is (i) more stable than (ii) at all?

@TheQuestioner  The ring strain is a much larger factor than the extended conjugation

I would guess heating would only promote the rearrangement.

Beacause of formation of CPM (Cyclopropylmethyl carbocation) which is extremely stable due to the presence of dancing resonance which is found in this as an exception..

The intermediate is a cyclopropylmethyl cation. It would be reasonable to suggest that you get a mixture of product here.

Wouldn't the ring contracted intermediate be a benzylic substituted cation, which can stabilise to an aromatic tropylium cation?

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text:
Help understanding the synthesis of HOCl
My goal is to add acetic acid to sodium hypochlorite to reach a ph of between 3.8-5.5 and produce hypochlorous acid. I’ll be using bromocresol green as a ph indicator. I was reading about how to synthesize it from an article here by the acidification of hypochlorite, which was noted to be the safest and most controlled way. My initial question was how do I calculate how much acid is needed to reach the specified pH range. However, other users noted that this may dangerously produce chlorine gas, in which case this experiment should be executed under a fume hood with care and proper safety precautions.
Figure 2 from the article notes the production of various byproducts of the reaction at different pH levels. It appears that chlorine gas is produced at both low pH and at high pH, but it is a bit confusing since it is overlapping with another line making it hard to understand.

Ideally, I'd like to reduce any production of Cl$_2$. Is there a way to do that? Anything to help me better understand this reaction is appreciated.
For the moderators,
I don't understand why this question was so poorly received and closed. It clearly lies within the allowed topics listed in the help center. Even though my question was about a potentially dangerous experiment, it is an important one nonetheless. I was able to learn an important detail concerning the safety of the experiment from this post. If anyone else was to follow down the same path as me, they should be able to google about it, see this question, and learn of the safety concerns as well. I think this question should be reopened to educate people instead of being heavily downvoted because OP is an inexperienced chemist trying a dangerous experiment, because that is not the case.

Never add anything to bleach. It is prohibited and clearly written on the bottle.

Ryan, the problem with the speciation diagram is that it is not clear whether it is a real diagram (from real data) or just a simulation from equilibrium euqations. Anyway, you got the decent reply below.

@Ryan Regarding "For the moderators" section: please note that your question has been closed by the community members, not mods. Sometimes the question can be misunderstood or negatively received due to formatting issues. I personally do, however, agree with your reasoning and would like to thank you for taking time and improving your question.

Chlorine is somewhat soluble in water: 14.6 g/100 mL H2O at 0 C, but drops to 0.57 g/100 mL at 30 C. So one way to reduce exposure to Cl2 would be to keep the temperature low and the volume adequate (dilution) to maintain all the chlorine in solution, with a little more volume for safety. There will still be chlorine escape, but this can be minimized by keeping the reaction container fairly well covered or sealed.
A separate issue is safety. It is always good to prepare safety precautions, and wise to imagine a worst-case scenario - like spilling the entire container, or breaking it. Precautions would include more ventilation that you think you will need, rags to wipe up a possible spill; plastic bags to quickly contain the rags. And, of course, smaller quantities to begin with, to make sure that no unexpected reaction occurs.
But a little chlorine is not so evil that it will seep out and destroy you. And the whole point of chemistry is to react things, so being perfectly safe and not mixing things together is sound advice for the non-chemist, but falls on deaf ears for a chemist.

Chlorine bleach is commonly produced via the chloralkali process as a highly alkaline mixture of sodium hypochlorite (NaOCl) and hydroxide (NaOH) along with some unreacted NaCl as well.
It's usually dangerous to lower the pH of chlorine bleach, as you may inadvertently generate chlorine gas
$$ \ce{2H+(aq) + OCl- (aq) + Cl- (aq) \rightleftharpoons Cl2(g) + H2O (\ell)} $$
$$ \ce{4H+(aq) + 4OCl- (aq) \rightleftharpoons  2Cl2(g) + O2(g) + 2H2O (\ell)} $$
And that is why it is kept at a high pH, to help favor the left side of the above equilibria.

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text:
Does Genistein remain in resonance stabilised form in physiological pH or acidic pH?
As a follow up, does the Carbonyl group at 4 position group get deprotonated or Hydroxyl group at 5 position get deprotonated? How does the resonance structure look? I have been looking through literature but have not been able to find any reference to this.
Does the deprotonation start with the carbonyl group or does it always have to start with hydroxyl group?
I may not sound clear as I am an amateur when it comes to organic chemistry.

It would help to have the structure in the question

Added chemical structure

Did help all that much... How go you imagine carbonyl getting deprotonated, for example. Also title doesn't make sense, as there's no form that wouldn't be delocalised.

The first acid dissociation constant for genistein ($\mathrm{p}K_\mathrm{a1}$) is $6.51 \pm 0.20$ (Ref.1), mainly because of hydrogen bonding between $\ce{C_{(4)}}$ carbonyl oxygen and $\ce{C_{(5)}-OH}$, similar to that in acetylacetone (see the diagram):

Therefore, in physiological $\mathrm{pH}$ ($\mathrm{pH} \approx 7.3$) it should be partially ionized. If it is in acedic solution, $\ce{C_{(4)}}$ carbonyl oxygen would be partially or fully protonated based on the acid strength of the solution. Furthermore, the energy minimized three-dimensional structure of genistein shows the hydrogen bonding as depicted in following image (See Figure A; Ref.1):

References:

Yoshiyuki Mizushina, Kazuaki Shiomi, Isoko Kuriyama, Yoshihiro Takahashi, Hiromi Yoshida, “Inhibitory effects of a major soy isoflavone, genistein, on human DNA topoisomerase II activity and cancer cell proliferation,” International Journal of Oncology 2013, 43(6), 1117-1124 (https://doi.org/10.3892/ijo.2013.2032).

Thanks for your answer.
As a follow up questions, since other isoflavones such as Daidzein (https://pubchem.ncbi.nlm.nih.gov/compound/5281708) and Glycitein (https://pubchem.ncbi.nlm.nih.gov/compound/5317750) do not have a hydroxyl group at C(5), how do they ionize?
Another follow up question is that do the hydroxyl group in the C(7) position form phenoxyl radical in all isoflavones?

This should be asked as a separate question.


text:
Finding the sensitivity of zero-order ultra sensitivity and its parameter dependence
I am provided a one-dimensional reaction system showing zero-order ultra sensitivity.
$$\ce{X_1 ->[f_1] X_2}, \quad \quad \quad \ce{X_2 ->[f_2] X_1}$$
The reaction rate function is the reduced enzyme reaction according to Michaelis-Menten's law.
$$f_1(x) = E\left(\frac{x_1}{K_1+x_1}\right)$$
$$f_2(x) = \left(\frac{x_2}{K_2+x_2}\right)$$
$$K_1=K_2=K$$
$$x_1(t) + x_2(t) = x_0$$
where $x_0$ is a constant. Let the steady state of $x_2$ be denoted by $x_2^{\text{st}}$. I need to find the value of $E$ that satisfies
$$x_2^{\text{st}}(E) = \frac{x_0}{2}$$
and the logarithmic sensitivity. I tried to look into some published papers/textbooks but could not find similar problems. Can anyone give some suggestions on directions to look into?

Why not write $\ce{X_1 <=>[f_1][f_2] X_2}$ instead?

I found the answer.
First, we need to find balanced equation for reaction. Then express the concentration of $xˆst$ in the steady state through E.
Then we can arrive to this kind of solution for the value of E.
$\displaystyle{E\frac{x_0 - \frac{x_o}{2}}{K + x_0 - \frac{x_0}{2}}} - {\frac{\frac{x_0}{2}}{K + \frac{x_0}{2}}} = 0.$
$\displaystyle{E\frac{\frac{x_0}{2}}{K + \frac{x_0}{2}}} - {\frac{\frac{x_0}{2}}{K + \frac{x_0}{2}}}= 0. $
E = 1


text:
Does acetoacetic ester give haloform test?
According to my class notes, acetoacetic ester doesn't give haloform test, but I feel that it should.
I know that the methyl group on acetoacetic ester is not responsible for the haloform test. First, the active methylene group is attacked by the base and halogen ion is substituted there. After the attack of $\ce{OH-}$ on the carbonyl carbon we will also get an ethanol molecule (from the ester part), which apparently should give positive haloform test.
I am confused. Which is correct?

With just dil. $\ce {NaOH}$ at r.t.p., the hydrolysis may be rather slow? Only with heating may sufficient ethanol be produced to give a positive iodoform test. I would agree with you that under heating, with $\ce {NaOH}$ (aq), $\ce {I2}$ (aq), iodoform test would give a positive result.

Does this answer your question? Does acetic acid give a positive result with the iodoform test?

This has an accepted answer.

@MathewMahindaratne granthium is not asking about acetic acid. granthium is asking if iodoform test reagents would be able to hydrolyse the acetoacetic ester and produce suffiicient ethanol to give a positive iodoform test result.

Sorry, my mistake.:-)

I think this question will help you. It doesn't have an answer yet, but the discussion there seems relevant.

Look at this: https://chemistry.stackexchange.com/questions/135626/do-cyclohexane-1-3-diones-give-positive-haloform-reactions ---- Search "acetoacetic" in reference i PDF. NaOCl gives dichloroacetic acid. It is unlikely that the diiodide, under iodoform conditions, is ever formed although the diiodide of AAE is known. The methyl group, which is less acidic than the methylene group, is not involved.

As @user55119 says it is unlikely the di-iodide is formed under iodoform conditions.
According to this paper1, acetoacetate gives only mono-iodination at the methylene centre under more powerful iodinating conditions than the standard iodoform $\ce{I2/OH-}$. The mono-iodo product in the presence of excess base will exist as the iodo-enolate effectively stopping iodination at the methyl ketone centre. Thus no iodoform product will be formed.
References

Goswami, Papori & Ali, Shahzad & Khan, Dr. M. Musawwer & Khan, Abu. (2007). Selective and effective oxone-catalysed α-iodination of ketones and 1, 3-dicarbonyl compounds in the solid state. General Papers ARKIVOC. 2007. 82-89. (Link)

what about any other haloform test?

i wanted to know for any halogen ( i have now edited the question)

See the comment from @user55119 above. The reaction works with NaOCl and Brominating agents.


text:
Which will be major and minor alkene formed if E1 reaction occurs on 2-chloro-2,4,4-trimethylpentane?
More substituted alkene should be major product while the less substituted as minor product as it is E1 mechanism.
But in my answer sheet it is given reverse, i.e. major as less substituted and minor as more substituted by giving the reason that it is happening because of steric hindrance of tert-butyl group. But I am confused now that it should not be the case because there is no role of steric hindrance in E1 mechanism, and answer should be as usual.
Please help me in this question with other reason.

Perhaps, you could provide the name of the base that is used for the elimination reaction. The size of the base is also crucial in allowing us to determine whether the Zaitsev product or Hofmann product is formed. For example, a large, bulky base (e.g. DBU) would offer the Zaitsez product.

Actually base was not present in reaction..rather AcOH/∆ was written above arrow and below arrow it was written E1..can some reaction happen with this?

@TanYongBoon i added a screenshot of ques..please check it

@AniruddhaDeb is it totally experimental or any reason for this..? Other than steric effect as it does not affects E1...

@AniruddhaDeb ohk buddy..but plz let me know if you find anything...as doubt still persists..

@TanYongBoon This is basically solvolysis, so the solvent is the base. Furthermore, as an E1 reaction, the size of the base is significantly less important.

@Zhe What would the base be here? Would it be the chloride ion that popped off the substrate? Certainly, that is more basic than acetic acid.

Or acetic acid, since it's probably the solvent. Remember you're also heating.

The major product in this case is favoured due to Steric hindrance in the product. Drawing the full structure of the molecule in the reaction makes this clearer

We can compare the degree of steric hindrance in the major product and the minor product by using A-values. The Wikipedia article gives the following A-values for common substituents:
$$\begin{array}{|c|c|}
\hline
\text{Substituent} & \text{A-value} \\
\hline
\ce{-CH3} & 1.7 \\
\ce{-CH2C(CH3)3} & 2 \\
\ce{-C(CH3)3} & >4 \\
\hline
\end{array}$$
Clearly, the major product has a lower set of A-values for it's substituents (2 & 1.7) compared to the minor product (4 & 1.7). Also, the minor product does not have any scope for forming stereoisomers, as both the groups on one end of the double bond are methyl. If one of them was a hydrogen group, then the E isomer would be the major product.

Even clearer is if one were to draw out the hydrogen atoms on all of the methyl groups. Not suggesting that you do that in the answer...

Even clearer if you show that there can be rotation about the C-C single bond such that the bulky t-butyl group can be positioned further away from the methyl group.


text:
What is the number of electrons being transferred in electrolysis of aluminum oxide?
$$\ce{ 2/3 Al2O3 -> 4/3 Al +O2}$$
As we know, to find the Gibbs free energy, we need to use this formula:
$$ G= -nFE^\circ$$
What would $n$ be in this case? After breaking it down into the half reactions $n = 12,$ but this doesn't seem right.

12 is wrong indeed. But I think I understand how you got there. How do you break this into half-reactions, to begin with?

I guess the misunderstanding arises from the fact that you treat $n$ as the same entity as the lowest common multiple (LCM) used to balance the net equation using half-reactions for the electrolysis of molten aluminium(III) oxide:
$$
\begin{align}
&\text{cathode:} &\ce{Al^3+(melt) + 3 e- &-> Al^0(l)} &|\cdot4\tag{red}\\
&\text{anode:} &\ce{2 O^2-(melt) &-> O2^0(g) + 4 e-} &|\cdot3\tag{ox}\\
\hline
& &\ce{4 Al^3+ + 6 O^2- &-> 4 Al^0 + 3 O2^0}\tag{redox}
\end{align}
$$
or, finally, with the smallest whole-number coefficients:
$$\ce{2 Al2O3(l) -> 4 Al(l) + 3 O2(g)}\label{rxn:R1}\tag{R1}$$
as well as normalized to a single formula unit:
$$\ce{Al2O3(l) -> 2 Al(l) + 3/2 O2(g)}\label{rxn:R2}\tag{R2}$$
In the equation
$$Δ_\mathrm rG = -nFE$$
$n$ is the number that matches the amount of electrons (in mol — since we are using molar convention for Gibbs free energy) transferred in the balanced equation from a reducing agent to an oxidizing agent.
From the balanced reaction \eqref{rxn:R2} it is evident that the number of electrons $n$ transferred from $\ce{O^2-}$ to $\ce{Al^3+}$ is $3 \cdot 2 = 6,$ where $3$ is the number of oxygens per formula unit, and $2$ is the number of electrons each oxygen donates.

But my textbooks says n=4?

@Leah I suspect your textbook is not correct, but could you please provide a reference? What textbook and what page number? It's important to see the context.


text:
Average energy concept
Problem

When the number of electrons in bonding molecular orbitals is equal to number of electrons in anti-bonding molecular orbitals bond cannot be formed between two atoms because
A) Bonding effect by the electrons in bonding M.O.s is cancelled by the antibonding effect of electrons in antibonding M.O.s
B) Antibonding effect by the electrons in antibonding orbitals is more than the bonding effect by the electrons in bonding M.O.s
C) Average energy of bonding and anti-bonding M.O.s is more than the average energy of atomic orbitals from which they are formed
D) Average energy of bonding and antibonding M.O.s is equal to the average energy of atomic orbitals from which they are formed

Answer

 B and C

My thoughts
I get the reason for option B as anti-bonding orbitals have higher energy comparing to the bonding orbitals, so naturally their destabilizing effect must be greater than the stabilizing effect offered by the bonding orbitals.
But I really can't understand the average energy concept of orbitals as expressed in the option C . could you help clarify what exactly is the energy concept regarding the molecular orbital and atomic orbital ? and how they can be compared?
Correct me wherever I'm wrong.

I took a liberty to rewrite the text from the screenshot, correct layout and title. However, I fail to see the question here. "I don't understand X" isn't really a question or a constructive way to help others help you. could you probably elaborate a bit more as to what exactly you are experiencing difficulty understanding?

I took your advise and i hope i am somewhere near the point now? (thanks for the input)

Choices B and C are two different ways of saying the same thing. Say $E_{MO}(A)$ and $E_{MO}(AB)$ are the energies of electrons in bonding and anti-bonding orbitals, respectively, relative to the electrons in the original atomic orbitals from which the molecular orbitals are formed, or, to make matters simple, set the energy of electrons in the original atomic orbitals to zero, that is, set as a reference $E_{AO}(1)$ and $E_{AO}(2)=0$ (assuming for simplicity that there are two MOs formed from two AOs).
To say bonding occurs is equivalent to saying that the energy of electrons in the separate atomic orbitals is higher than the energy of the electrons in the bonding molecular orbital. Using that reference system, positive energies (>0) disrupt bonding, and negative lead to bonding. Then saying that the effect of the antibonding electrons is greater than that of the bonding ones is equivalent to the statement
$$0\lt E_{MO}(B) + E_{MO}(AB)$$
or equivalently
$$0\lt  \frac{E_{MO}(B) + E_{MO}(AB)}{2} = E_{MO,\text{avg}} $$
but this is just what choice C says, since the average energy of the atomic orbitals is just zero, that is
$$  \frac{E_{MO}(B) + E_{MO}(AB)}{2} = E_{MO,\text{avg}} \gt 0 = E_{AO,\text{avg}}  $$

When a test question is so vague that you can't figure out what answer is best, it's difficult to formulate a question about the answers to the question. The test question specifies bonding electrons equal to antibonding electrons. No energy statements. There needs to be one more answer:
E. All of the above, although some answers are a little better than others.
In order to develop a bond, the decrease in energy, i.e., the stabilization, of bonding electrons, needs to be enough to prevent disruption of the bond. If the "bond" is stabilized and destabilized by the same amount, there is nothing holding it together. If the destabilization is greater than the stabilization, the "bond" never formed.
Even if, under extreme conditions, like high pressure, or very cold temperatures, molecules are formed that are metastable for short periods, bonding forces hold them together only until some vibration puts them into a state which is  less stabilized; then the atoms move apart more and more until they are separated - no bond any more.


text:
Assigning R/S configurations to Fischer projections of 3‐chloro‐2‐hydroxybutanoic acid
A question requires you to assign the R/S configuration for 3‐chloro‐2‐hydroxybutanoic acid:

In the answer, while assigning the configuration to carbon #1, apparently the group below is given the third priority, lower than the COOH group above.
Using the CIP rules, doesn't the Cl atom in the bulky group give it a higher priority than the COOH group?

In second carbon(C2), Cl is first priority group due to its higher atomic number than C1 attached to C2, secomd priority group is C1 and third is CH3. This giving anticlockwise rotation, but H is in front of the observer(on horizontal line), so the overall rotation is clockwise. Thus it is R configuration.

You can also think the other way, instead of seeing the Fischer projection from above the paper, you can see it from below the paper, then horizontal line groups are away from you and vertical line groups are toward you. Then the anticlockwise rotation seems to be clockwise from below the plane of paper.

Sorry, the picture's not clear; I was talking about the other carbon. I'll edit it.

I undid andselisk's edit. I meant carbon #1.

In carbon#1,  OH is first priority group , C#2 is second priority group due to Cl attached to it, COOH is third priority group. So the rotation (move finger from first to third priority groups) is clockwise but as H is towards the observer so the rotation is anticlockwise so the configuration is S.

Thanks! That was what I was looking for; can you put that down as an answer so I can mark it as one?

The compound mentioned is (2S,3R)-3-chloro-2-hydroxybutanoic acid. For Carbon atom #2, the CIP priority order is

$\ce{Cl}$
Carbon #1
$\ce{CH3}$
$\ce{H}$

The methyl group is lower priority because Carbon #1 has a hydroxyl group attached to it, which is higher priority than Hydrogen. This implies that the orientation at carbon atom #2 is R.
For more in-depth information, you might want to check out Wikipedia

Sorry, the diagram was unclear when you posted; I meant the other carbon.

I meant carbon #1, and no one edit that, please!

My first answer was for Carbon #1, after which you edited to carbon #2. Anyways, I think you have enough hints now to give it a try yourself.


text:
Decay scheme of sodium-24
Sodium-24 decays into an excited nucleus of magnesium-24. Magnesium reaches nuclear ground level via two consecutive gamma emissions. What are the half-lives (time scales) for these two processes? I think that those gamma photons are emitted very rapidly after Mg-24 is formed. I'm curious to know but I cannot find data. I'm a Chem undergrad with interests in nuclear chemistry.

https://en.wikipedia.org/wiki/Nuclear_isomer suggests that if it's usual de-excitation, then it's on order of picoseconds.

For energy levels in nuclei (and their gamma emission) the first place to look is at the Evaluated Nuclear Structure Data Files. I usually use the Brookhaven site, but there may be a closer mirror to you. Enter 24 into the Nuclide or Mass box and hit Enter. Check the 'Adopted Levels, Gamma' box by $^{24}$Mg, and request the pdf version (back up at the top).
On the last page you will find the lower levels like this (although rotated by 90 degrees):

So, the 4122.889 keV level has a decay time of 22 fs - this is the level that the first gamma decays from. The 1368.672 keV level has a decay time of 1.33 ps - this is the second gamma in your sequence.

You can estimate the lifetime from the classical equation for radiative decay rate constant , $\displaystyle \frac{1}{\tau}= \frac{(e\omega)^2 }{6\pi\epsilon_0m_ec^3}$ where $\omega$ is the frequency (Hz), $m_e$ electron mass and $\epsilon_0$ the permittivity of free space.
This value gives a lower limit (i.e. fastest possible value)  on the lifetime $\tau$, and this works out in the region of $\approx 4\cdot 10^{-14}$ s or $40$ fs for the 2.76 Mev transition. This formula is somewhat simpler than calculating the dipole vector integral and Einstein coefficients, and produces lifetimes that are equal to or larger than the classical value.


text:
Why does nital need alcohol when etching metals?
I was reading about HNO3 and I saw it can be mixed with alcohol to form "nital", which is useful when etching metals, especially carbon steel.
I wonder why alcohol is needed? HNO3 itself is corrosive for metals. So I did a little search and some college students claimed that alcohol is for holding back the reaction rate to get a better result, but it raised more questions for me... My thought now is alcohol is not as polar as water, so HNO3 would probably be less readily dissociated. But I can't figure out the rest.

Speculation, it might not be the answer: The alcohol may remove grease and other lipophilic matter sitting on the surface, thus easing the access of the aqueous solution of $\ce{HNO3}$ to the metal.

I thought it would form alkyl nitrates too. The term nital seems to be the name of the solution, not some functional group. It's said to be used a lot in etching industry, so I guess it works somehow...some also mentioned it's dangerous and can be explosive when made. I'm more interested in what role does alcohol play.

OK, I found it, but it'd be good if you put some reference in future posts.

So-called “nital” is commonly used to etch iron-nickel meteorites to display the characteristic etch patterns. But @Mithoron is indeed correct. I have always assumed the ethanol or methanol served to prevent passivation of the iron or steel or Fe-Ni meteorite. No clue how, but this should not be done without all the safety precautions and PPE.


text:
How to write the formula of a mixed metal salt?
I suppose the formula of the Sulphate salt of Sodium & Potassium is $\ce{KNaSO4}$ as the Pubchem site says so.
So, is there any rule that in the formulae such mixed metal salts we need to write the metals in the order of electropositivity?

"$\ce{KNaSO4}$" is correct, but not because of electronegativity order, but rather the alphabetical one.
From IUPAC "Red Book", section IR-2.15 ORDERING PRINCIPLES [1, p. 40]:

IR-2.15.2 Alphabetical order
Alphabetical order is used in formulae as follows.
(a) Within the group of cations and within the group of anions, respectively, in formulae of salts and double salts.
Deviations from this rule may be acceptable if it is desired to convey specific structural information, as in Example 5 below.
Examples:

$\ce{BiClO}$ (anions $\ce{Cl-}$ and $\ce{O^2-})$
$\ce{NaOCl}$ (the anion is $\ce{OCl-},$ cf. Section IR-4.4.3.1)
$\ce{KNa4Cl(SO4)2}$
$\ce{CaTiO3}$ (perovskite type)
$\ce{SrFeO3}$ (perovskite type)

Reference

IUPAC “Red Book” Nomenclature of Inorganic Chemistry, 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; IUPAC Recommendations; Royal Society of Chemistry: Cambridge, UK, 2005. (PDF)

Related: Is it NaOCl or NaClO?


text:
Who was C.H. Kao?
I am editing a Wikipedia article about the Chen-Kao reaction that is used to identify ephedrine, but I am unable to find biographical info on C. H. Kao:
K.K. Chen and C.H. Kao - Ephedrine and Pseudoephedrine, their Isolation, Constitution, Isomerism, Properties, Derivatives and Synthesis - Journal of the American Pharmaceutical Association volume=15 issue=8 year=1926 pages=625–639
K.K. Chen and C.H. Kao, Pharm. Zentralhalle, Vol.70, p.27 (1929)

The basic info such as secondary name (Gao Chongxi), years of life (1901–1952), as well as occupation (professor of chemistry at Tsinghua University) are easily retrievable. Would you like to find something on the personal life, or anything else in particular? Probably someone with a solid knowledge of Chinese could help as well.

I would really like to reference an autobiography of some sort, so I or anyone else can write an article later on, possibly. Although full name will suffice

You will have to write to the library archivist at the University of Wisconsin at Madison. Write a formal email with all your proper affiliation and the purpose. Provide reference to several papers. Usually you will get a positive response.

@andselisk This is C.H. Kao, you mention Gao Chongxi. What is meant by secondary name in Asian (Chinese) culture?

@M.Farooq Hope it does not get to that. I mean that person has a reaction named after him, he should be mentioned somewhere...

According to the wikipedia article I'm editing it is either  Chung-Hsi Kao or Cheng-Heng Kao

Bear in mind that English transliterations of Chinese names have changed since the early 20th Century. The name "Kao" in the former Wade-Giles transliteration is spelled "Gao" in the modern pinyin transliteration. Same for Xi/Hsi and Chong/Chung. So "Gao Chongxi" and "Kao Chung-Hsi" are the same name transliterated two different ways.

@Andrew, interesting. But this gentleman C.H. Kao consistently uses this spelling in his papers so it should not be hard to locate him by contacting the UW library archivist.

@NikolaiFrolov, I have asked a couple of times, even from MIT archivists to get photos for a paper I was planning to write. They might charge a little bit but if you are serious, university library is the best way.

You will have to consult "The Study of Change Chemistry in China, 1840-1949
By James Reardon-Anderson, 2003." Andselisk found the right information about his professorship in Tsinghua University but his spellings and my search spellings are different. Our Chinese fellows can elaborate more on it.
As I mentioned in the comments, he studied from the University of Wisconsin, Madison. If you wish to seriously pursue his biography, you will have to contact the library archivist.
From Google Books one can see the relevant information:
On page 224 and 235

As a Mandarin speaker, I can assure you that Andrew's comment on the main post is entirely accurate. Maybe you could provide a link to this Google book... (or even better, type out/copy the text)?

Yeah, probably what I'll do. Even if I dig up his biography, one reference won't satisfy the notoriety requirements of Wikipedia

@NikolaiFrolov, For Wiki, you have to link the new entry on Dr. Kao with another existing entry. For example, notable U of Wisconsin aluminus or may be generate a name reaction page first.


text:
How would you know if the light emitted is of one wavelength or if it is a mixture of two or more wavelengths?
I am just learning some chemistry after a long time without studying it and I have thought about something.
Some copper compounds emit green light when heated to flame. So, I would like to know if it is known if an emitted light is of one wavelength or if it is a mixture of two or more wavelengths.
Thanks in advance

It is quite easy to measure the wavelength of light with a spectrophotometer. You might want to review how they work if you're interested.

Way back around 1666 to 1671, Sir Isaac Newton figured out that a prism resolved sunlight into its constituent colors. Same thing here and in the spectroscopic discovery of elements, e.g., Tl, Cs, Rb, etc.

Or try a cheap diffraction grating, e.g. https://www.amazon.com/Diffraction-Grating-Slide-Excellent-demonstrating/dp/B006ZBDRN0/ref=sr_1_38

Newton’s prisms: https://royalsocietypublishing.org/doi/pdf/10.1098/rsnr.1981.0002.

@M.Farooq I haven't downvoted but it is clear the question has flaws. For instance, what is meant by "two or more wavelengths"? Clearly "blue+yellow" versus just "green" visible wavelength(s). Your answer doesn't really dig into that point, rather touches on how you might determine the spectrum, which is not the question.

@BuckThorn, Interesting point. I don't believe these simple color mixing schemes apply in flame spectroscopy. That is why I did not invoke this point. "Green" can mean a spectrum corresponding to green wavelengths only without having blue or yellow in it or implying the absence of green color in the flame. Check this flame spectrum of copper, there is no yellow, but a heavy presence of red. It is tricky! https://www.sciencesource.com/archive/Flame-emission-spectrum-of-copper-SS2529806.html#/SearchResult&ITEMID=SS2529806

@M.Farooq Unfortunately your link is behind a paywall but I see what you mean: https://www.sciencephoto.com/media/1750/view

The green light from copper flame is sadly not a single wavelength. It is very good question though because not many students even think about it. In order to "see" what wavelengths are present in a light source, you either need a spectroscope (old school way) or a monochromator with a detector (modern way). Of course you can search about Bunsen-Kirchoff spectroscope. It uses a prism to disperse the light falling on the prism.
Depending on how much interest and money you have you can easily construct a spectroscope in a shoebox (several published articles out there). Ten years ago or so I used to make CD diffraction spectroscopes in shoeboxes or cardboard boxes for seeking answers to queries like yours.
If you want spent some money ($400-500) you can buy a high quality pocket spectroscope with a wavelength scale . Then you can enjoy all light sources (except the Sun). It is fascinating to see and explore so many phenomena. You will find out by looking at the moon that it has some dark lines (never look at the Sun from a spectroscope). I am not suggesting to buy third rate products from Amazon because twice I purchased from them and the wavelengths were all wrong!
The monochromator is a fancier product. I never had the courage to buy one. They are very expensive. The most common design is a Czerny-Turner design. I am sure Prof. EdV had even fancier products which we have never fathomed :-)
This picture shows white light entering a slit, it falls on a mirror which makes all the rays parallel. The mirror sends the light to a high quality diffraction grating, the reflection diffraction grating separates wavelengths in space. The light from reflection grating falls on a focusing mirror, and the mirror focuses lies on the exit slit. The exit slit has a detector. You would slowly rotate the grating. If there are multiple wavelengths in the light source (say a flame) the detector will record a spike every time a different wavelength reaches the detector.

(+1) Glad you did an answer! Tom O'Havers' Animated difraction grating is also useful. I used to use the Wingz version in teaching.

The awe of demonstrating that with a real open monochromator with xenon arc lamp is at another level.


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Interaction of trifluoroacetates with acetylcholinesterase
There exists a substance called TMTFA, or 3-(N,N,N-Trimethylammonio)-2,2,2-trifluoroacetophenone. It is known for being able to inhibit acetylcholinesterase at femtomolar concentrations. The TMTFA-AChE adduct resembles the tetrahedral transition state through which the hydrolysis of acetylcholine by the enzyme normally proceeds, and this adduct is stabilized by the electron-withdrawing trifluoromethyl group present in the TMTFA molecule. This inhibition is reversible.
While there have been studies regarding trifluoroacetophenones and their interactions with AChE, I haven't been able to find studies exploring the interactions between AChE and trifluoroacetates, such as 3-(N,N,N-trimethylammonio)phenyl trifluoroacetate or 2,2,2-trifluoroacetylcholine. Would these, and other, esters of trifluoroacetic acid inhibit AChE or be hydrolyzed normally? If it's the former, then what are the typical inhibitory concentrations of such esters? What is the shape of the adduct of the ester with AChE? How long would they remain bound to the enzyme? Have any studies been done regarding the bioactivity of such esters?
References

Nair, H. K.; Seravalli, J.; Arbuckle, T.; Quinn, D. M. Molecular Recognition in Acetylcholinesterase Catalysis: Free-Energy Correlations for Substrate Turnover and Inhibition by Trifluoro Ketone Transition-State Analogs. Biochemistry 1994, 33(28), 8566–8576. DOI: 10.1021/bi00194a023.
Harel, M.; Quinn, D. M.; Nair, H. K.; Silman, I.; Sussman, J. L. The X-Ray Structure of a Transition State Analog Complex Reveals the Molecular Origins of the Catalytic Power and Substrate Specificity of Acetylcholinesterase. J. Am. Chem. Soc. 1996, 118(10), 2340–2346. DOI: 10.1021/ja952232h.

According to Ref.2, TMTFA binds titly to AcE from a Pacific electric ray, Torpedo californica. It might not do the same to Human AcE. I believe your intention is for inhibitors of Human AcE.


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Stability of F- Cl- Br- I-? In aqueous state
If we talk about hydrolysis F- will be more heavily hydrolysed so it should be more stable in aqueous state.
But I know that HI is a stronger acid than HF. So the stability of conjugate acid ie I- should be more than F-. As dissociation takes place in water.

If a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected from the author to elaborate the topic of the question by doing at least basic own topic review, writing what he/she has found and understood, and what is the stumble stone. The quick questions without explicitly expressed particular effort are not very welcome, and may be closed.

Your logic is all upside down. A thing that is easier to destroy is less stable.

If we consider $\ce{HX}$ as the common formula for hydrogen halogenides, then there is an equilibrium
$$\ce{HX(aq) + H2O <=> X-(aq) + H3O+(aq)}$$
that can be expressed also as:
$$\ce{acid1 + base2 <=> base1 + acid2}$$
The equilibrium is competition of 2 acids, $\ce{HX}$ ansd $\ce{H3O+}$. $\ce{H3O+}$ is the strongest acid stable in aquaeous environment. Acids we consider as strong acids ($\ce{HCl}$, $\ce{H2SO4}$, $\ce{HNO3}$ ) are much stronger acids than $\ce{H3O+}$. Therefore for $\ce{HCl}$, $\ce{HBr}$, $\ce{HI}$, the equilibrium is practicaly fully pushed toward right:
$$\ce{HX(aq) + H2O -> X-(aq) + H3O+(aq)}$$
On the other hand, $\ce{HF(aq)}$ is a weak acid, weaker than $\ce{H3O+}$, with dissociation constant $\mathrm{p}K_\ce{a}=3.17$, for which the equilibrium is rather on the left side:
$$\ce{HF(aq) + H2O <<=> F-(aq) + H3O+(aq)}$$
Highly concentrated $\ce{HF(aq)}$ is rather a strong acid:
$$\ce{2 HF(aq) + H2O <=>> HF2-(aq) + H3O+(aq)}$$
and liquid HF is very strong acid ( H0 = −11 ) :
$$\ce{3 HF <=>> HF2- + H2F+}$$

Fluorine ion and chlorine ion are very corrosive in nature, so In water they easily form the hydrohaloacid as order of acidic strength in water is  $hi>hbr>hcl>hf$ hence the stability order will be $F^{-} >Cl^{-} >Br^{-} >I^{-}$ which also shows stability of their acids.

The only halogenides forming acid in water are fluorides: $\ce{F- + H2O <=> HF + OH-}$, with the resulting pH alkaline, not acidic.

@poutnik sorry I did mistake in hurry . is it fine now?

How does NaCl easily form HCl ? Or rather H3O+, as H3O+ is the strongest acid stable in water.

From where the NaCl arrives?@poutnik

It is not relevant, @Yuvraj. NaCl does not undergo hydrolysis. Its corrosiveness is not due acidic properties, but because of electrochemical impact.

@poutnik indeed so I was asking is my answer is satisfying or you find it still incomplete?


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Reaction Of Carbonic Acid With NaCl
I was just wondering if it was possible for anyone to specify what reaction conditions are required for the reaction between $\ce{H2CO3}$ (Carbonic Acid) and $\ce{NaCl}$ (Sodium chloride)?
Does this reaction happen naturally, or is an input of energy or catalyst required?
Will this reaction still proceed if the $\ce{NaCl}$ is in an aqueous solution i.e. dissolved in water? As both substances exist in the sea, does this reaction occur in the ocean as well?

For the record, $\ce{H2CO3}$ does not exist as a separate entity. It is actually a solution of carbon dioxide in water maintaining equilibrium.

Also, refer to these papers: 1, 2.

Keeping aside this question is very close to the one posted by you earlier (https://chemistry.stackexchange.com/questions/135748/collecting-the-products-of-a-reaction-before-they-themselves-react) -- and actually, the answer as well -- you could compute the Gibbs / free reaction enthalpy of this reaction postulated, too.  This would give you a sense if energetically speaking the reaction were favourable, or not.  (There are some reactions, like diamond to graphite, which are favourable at RT but face a kinetic barrier this high that they don't occur at significant rate, though, too.)

$\ce{H2CO3}$ and $\ce{NaCl}$ do not react.
The sodium chloride provides a surface with high friction which allows the $\ce{CO2}$ to transform into a gas. The same thing happens with mentos and fizzy drinks.
One might say that the sodium chloride is serving as a catalyst for the transformation of $\ce{H2CO3}$ into $\ce{H+(aq)}$ and $\ce{CO2(g)}$.

The answers on the remote link are s...uboptimal, but your content here looks good. I remove the link and format your answer. It would help a lot, if you would then provide a better link.

No, the question was not correctly answered now or previously! The answer is an outline of the Solway Process centering on the introduction of NH3 to the mix of aqueous NaCl and CO2. Wikipedia link: https://en.wikipedia.org/wiki/Solvay_process .


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Multiday lab courses in classic chemistry
This time of the year I usually get some extra income I can use on interesting courses. Usually rescue or sport courses. However, I also have a master degree in chemistry, and would like to spend some days or a week just to do some old fashioned chemistry with a great teacher. But I have never found a "summer school" or something like that.
Does anyone has a any suggestion? I can gladly expand "classical chemistry" in any direction, as metallurgy or microbiology.

Beside asking upfront the research groups, there are summer schools in normal (non-covid 19) circumstances.  For example Oxford's Chemistry Summer School, or ACS' Green Chemistry Summer School.  Given your account name, I speculate you reside in Europe; if so, a suitable directories page not only about core-chemistry may be  www.summerschoolsineurope.eu.  Sadly, because of the current pandemic, many scientific events are cancelled, though (example).  Still, references of courses in the past are retained (example), so you could find help by the former organizers as well.
Maybe you are an outlier by age in comparison to the «typical attendee».  However, the experience you gained after your master thesis, both in terms of chemistry under the hood as well as e.g., project and data management, may be of great benefit to the class providing additional perspectives.  In particular, I think about planning the experiments in the smaller groups, or discussing the results.  Equally, you may become a role model for younger participants in the way you choose to stay continuously engaged in learning.

Thank You, This was a useful answer. I think I will just WoW to a couple of the links you gave :-)


text:
Column Dispersive Mass Balance analytical / numerical solution
Is there any analytical solution for the most common expression of column mass balance:
$\dfrac{\partial C}{\partial t} + \left(\dfrac{1+\varepsilon}{\varepsilon}\right)\dfrac{\partial q} {\partial t} + u \dfrac{\partial C}{\partial t} \:\:=\:\: D_{\mathrm{axial}}\:\dfrac{\partial^2 C}{\partial t^2} $
This is the most common isothermal form for just one gaseous solution (two expressions for concentration change between the phases, convective flow and flow dispersion). If you consider adsorption from the "side" of adsorbent (solid phase), the density of fixed bed $\rho_{{\mathrm{bed}}}$ should be considered. $\partial q\big/\partial t$ is expressed by LDF (linear driving force) model which simplifies PDE to ODE.
What I would like to know:

I know that these equation can be always solved numerically, but is
it possible to solve this balance analytically for certain boudary /
initial conditions? If so, how would you express $\partial
   C\big/\partial t$ ? (I cannot find a direct question anywhere, some suggest it may be solved analytically with certain simplifications without giving any details)
There are several mathematical models for numerical solution. Which one would be suitable for adsorption? Unfortunately, literatures I found does not go into details.

JoshE, You should consult the book of Guiochon, Golshan Shirazi and Atilla Felinger (postdocs of Guiochon). The book is titled "Fundamentals of Preparative and Nonlinear Chromatography". He spent his in solving these type of problems.

@M.Farooq Thank you for your suggestion. I think I have seen this book in the past. I have gone through the chapters concerning mass balance and description of these equations and each expressions is very detailed. In Chapter 10, there are different solution methods (mostly mathematically based). However, despite defining initial and boundary conditions, I still miss the middle-step where someone would show step-by-step solution (even if it is dimensionless form - many other literatures suggest to rewrite mass balance) leading to obtain component gas conc. as a function of time, $C(t)$.

If I understand your sentence about LDF correctly, it considers dq/dt a constant?  If so, the equation becomes a homogeneous second-order differential equation in C, or a first-order homogeneous differential equation w.r.t. dC/dt, and would thus have analytical solutions as long as epsilon, u, and D_axial are fixed.  Or does LDF assume that dq/dt can be approximated as a function of C?

@CurtF. The term $\dfrac{\partial q_i}{\partial t}=k_i\left(q_i^* - q_i\right)$ is not constant. What I learned, the transport coefficient $k_i$ can be expressed as $k_i=f\left(D_{\rm eff}\right)=f\left(\rm Re, Sc\right)$ such that the column geometry and velocity of gas matters. The maximal adsorbed conc. is expressed by the adsorption isotherm, i.e. $q_i^*=f\left(P,T\right)=f\left(p_i,y_i,T\right)$. According to, e.g. Ruthven(1984), the average concentration through particle is defined as integral over microparticle radius $q_i = \dfrac{3}{r_c^3},\displaystyle\int_0^{r_c}qr^2,\mathrm{d}r$.


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Which software to use to output molecular orbital diagrams?
I want to output a molecular orbital diagram. Can anyone recommend a software to do this? Also I searched for a python module, but didn't found a pure solution. Maybe I am using the wrong keywords.
I want to connect two atoms with a specific electron configuration, to see at which energy levels they have pi- and sigma-bindings.

That depends, do you know what it's supposed to look like or are you asking about construction one where you don't know how it looks like? In the first case I've always been creative, for example using excel to draw the positions of the energy levels and then connect the rest by hand. If it's the latter case I fear you'd have to rely on computational chemistry first to determine the orbital energies. I'm using Orca to calculate that and Chemissian to draw the MO diagram afterwards.

@Justanotherchemist Thank you for questioning my motivation! I want to connect two atoms with specific electron configurations, to see at which energy levels they have pi- and sigma bindings.

I guess I still don't understand. Do you want to do a quantitative MO Diagram? Like one where you calculate the energy levels of the free fragments, then the energy levels of the corresponding MO, then use the linear combinations of the MO to draw the correlation lines? (Please don't use edit statements, simply extend your question. If anyone cares, every post has a history.)

I guess he's up to that. I'd say for a very simple view, so not taking the absolut energies for granted you could go with a quick calculation from any of the free softwares like Orca. And then use something like Avogadro to visualize the orbitals, check which ones are atom centered and which ones are molecular ones and take their energies.

@Justanotherchemist is right. I want just a simple view, to see if my solution for some exercises is correct. There is no need for a quantitative MO Diagram. I will take a look into Orca and Avogadro. Thank you!

Note, we have also a Matter Modelling site.

@Seminom I'd encourage you to edit that into your question in a way which you like; right now people are reading your question and thinking that it's not specific enough. Clarification in comments is good, but comments are temporary, so clarification in the question itself is even better.

If your interest is a qualitative reprepresentation, you may reach out for an editor like inkscape and use lines, squares and Unicode to create a picture like the following one

In text fields you may enter Ctrl + U followed by an unicode code like
2191 (for $\uparrow$), 2193 (for $\downarrow$), 21C5 (for $\uparrow\downarrow$).*)
If you have access to LaTeX, you may be interested in modiagram, especially if you are interested in an output as .[e]ps or .pdf.  Only a few parameters are needed for an illustration like

(credit)
(It is possible to construct more advanced orbital diagrams like this and this, too.)

For the quantitative, and more programmatic approach you seek, especially with Python, there are solutions with the matplotlib library which might require some adaptation by you (example).  One of them is PyEnergyDiagrams once shown here on ChemSE:

the other EnergyLeveller.py:

Addition:  With the special chars used here, maybe a couple like Ipe + LaTeX is more handy to use, than inkscape.

+1, and thank you for the open-source solutions, and especially for not suggesting to use ChemDraw for drawing MOs.

Thank you! That looks great. Do I have to calculate the levels before plotting or is PyEnergyDiagrams / EnergyLeveller able to calculate this just with information about the elements and their electron configuration?

@andselisk Beside open-source solutions, given the profile description by @Seminom, tools more likely accessible to a physicist than to a chemist appeared the better choice.  The idea with ChemDraw did not occur to me, as the arrows / harpoons are available elsewhere with less effort, too.  After all, the aim wasn't to check Powerpoint for Turing-completeness (short version: https://www.youtube.com/watch?v=uNjxe8ShM-8, long version: https://www.youtube.com/watch?v=_3loq22TxSc).

@Buttonwood I saw the one about WordTeX before, but I missed these videos; they are brilliant:)

@Seminom No, each of the program expects you to provide the electron levels and splits.  This is something quantum chemical programs like the free NWChem (http://www.nwchem-sw.org/index.php/Main_Page), Psi4 (http://www.psicode.org/); or the commercial Gaussian may compute, and to some degree other physicist are capable to record experimentally.

@andselisk The longer video shows some of the hidden features not presented in the shorter one.  As I recall, even using the transition of one irregular polygon into an other.  On the other hand Tom Wildenhain went further, checks about compiling C with Powerpoint (https://www.youtube.com/watch?v=LArkm4v5mWA, https://github.com/TomWildenhain/pptcc) or «just» computing fractals (https://www.youtube.com/watch?v=O8l_awjgoMI, https://github.com/TomWildenhain/PowerPoint-Fractals).


text:
How does sodium rosinate enhances lathering property?
Why don't we use sodium carbonate or sodium stearate? What does sodium rosinate contain that the other two don't?

It is hard to understand how you might think of sodium carbonate versus sodium stereate as alternatives. Rosinate and stereate are detergents, carbonate is not.

Rosin acids, or more commonly, resin acids, are more complicated structures that are able to protect an air bubble mechanically. Abietic acid, a common resin acid, looks like a platey structure with more strength than a linear chain like stearate or oleate, two other common soap anions.

In fact, resin acids were, and still are, used very commonly to maintain a stable air content in concrete, to minimize freeze-thaw damage. Ordinary surfactants would give unpredictable air content: either too low because the fine particles of cement broke the air bubbles, or too high because there was so much surfactant  that it overwhelmed the fine particle disruptions.
Some synthetic foam stabilizers are polymeric anions or have other strong intermolecular attractions.


text:
Why isn't carbon-14 included in CIAAW list of isotopic abundances?
According to Wikipedia, standard atomic weight is based on the published isotopic abundances of CIAAW:

The standard atomic weight $(A_\mathrm{r},$ standard) of a chemical element is the weighted arithmetic mean of the relative atomic masses $(A_\mathrm{r})$ of all isotopes of that element weighted by each isotope's abundance on Earth. The standard atomic weight of each chemical element is determined and published by the Commission on Isotopic Abundances and Atomic Weights (CIAAW) of the International Union of Pure and Applied Chemistry (IUPAC) based on natural, stable, terrestrial sources of the element.

However, this table of isotopic abundances appears to be incomplete — for example, it's missing carbon-14.
Are they only listing primordial isotopes, or maybe those above some threshold of abundance? Is there some other criteria?
Why is carbon-14 not listed?

I guess this is because we have very few of it.

According to this, the abundance is about a part per trillion, so the imprecision in the abundances of the two main isotopes dwarfs that.

Maybe C-14 isn't that abundant.

Also, what was quoted also says stable ...

The CIAAW source specifically says under carbon: "Yet, this isotope never occurs in normal carbon sources in concentrations high enough to affect significantly the Ar(C) value."

@JonCuster Thanks! I hadn't seen that, but searching for your text I find it on https://ciaaw.org/carbon.htm

Also, 14C abundance varies with location: almost none in geological sources, highest in green, growing plants or other carbon sinks.

It's what Ed says: C-14 is not stable.

@BuckThorn I'm confused by that statement. The table includes three isotopes of uranium, all of which are radioactive. Is there some other meaning for "stable" besides "not radioactive"? And if not, and only stable isotopes factor into the standard atomic weight, how can we even have a standard atomic weight for elements like uranium which have no stable isotopes?

No, I was thinking specifically radioactive. I suppose with uranium one doesn't really have a choice since all isotopes are radioactive, but they are always found together in the same proportions? But I retract my previous comment, since you are right. The alternative explanations (extremely low and variable natural abundance) must account for the absence of the isotope in the table. Actually, the comments in the table explain that natural sources of U can show variability outside of the tabulated limits.

As per wikipedia, Re, In and Te only have minority abundance stable isotopes, so stable has to yield to pragmatic reality: theory guides, experiment decides, as they say. For uranium, all the isotopes are radioactive, so stable is not applicable. For carbon, the relative abundances make it a moot point.

From the wikipedia articles on Carbon and Isotopes of carbon, the atomic weight of carbon, in Daltons, is just the weighted average. Thus, AW = (0.9893 x 12) + (0.0107 x 13.00335483521) + (1 ppt x 14.003241988) = 11.872 + 0.139 + 10 ppt = 12.011, as in the periodic table.

On a final note, the OP’s linked table of isotopic abundances shows that no listed isotope has an abundance given to more than 6 decimal places, except for La and Ta, which both have 7 decimal place listed abundances. So carbon 14’s abundance is in the 12-th decimal place and it is entirely negligible.


text:
How can I measure the scale in water?
I have hard water in my house. All the taps have tartar around. It generates some problems in some of my home appliances so I have bought polyphosphate salt but when I boil water I don´t get the expected results.
I´m looking for some product or filter that would make the water in my house better for drink and use. That´s is why I´m evaluating to install a filter with polyphosphate salt or a floating dispenser (like a buoy).
I´m now trying to make a test before installing the floating dispenser I have recently bought in my water tank so, this is what I have done:
First I have filled a pan with water and boil it until it got empty and this was the result:

Then, I have use the floating dispenser in a pan with polyphosphate salt for 30 mins. After that I took out the dispenser and boil it until it got empty. This was the result:

It seems like there is no difference.
How can I make a better measurement or treatment? Is this the right product for my problem? 30 minutes is enough to the polyphosphate salt? Do you think that will be better a floating dispenser or the filter with that product? Maybe I have to buy another kind of filter?

Most drinking water has residue like what you are showing.  Boiling water shows there are ions in the water, but it does not show what kind of ions are present.  Boiling water is practically useless when determining scaling potential and it is not safe to do using cooking pots and pans.  Part of the reason laboratory equipment is more expensive than cooking pots and pans is because lab equipment has much better fire prevention measures.  Boiling down water in this manner has the potential to start a fire.
pH is a better measure for scaling potential.  If you want to get technical, water pH is used to determine the Langelier Saturation Index or the Ryzner Index which are indicators of scaling potential.
Your water is probably too alkaline. Alkalinity and acidity need to be balanced because if water is too alkaline then too much scale forms; however, if the water is too acid pipes corrode.
For a household, filtering water with an activated carbon filter first, then an ion exchange filter might work.  Ion exchange filters contain a resin which exchange scale forming ions like calcium with ions like sodium. These filters can be easily fouled by organic material, so the organic material will need to be removed by something like an activated carbon filter.

Thank you @Agriculturist! I have updated my question to be more clear. As I understand from what you said I have to look for another type of filter and not just Siliphos product.

@Ignacio  This question should probably be edited to remove mention of brand names.  On this exchange calling a compound "bleach" is appropriate, but calling it "clorox" is not.

Thanks again @Agriculturist! I wasn´t sure how to call that product and that is why I put the name. If you suggest me another name I will change it (polyphosphate salt perhaps?).

That name is better.  Polyphosphates bind to metal ions and metal surfaces so as to inhibit scale formation, but they do not remove those ions from the water.  There are a number of factors which influence the effectiveness of polyphosphates and there are multiple types of polyphosphates.  Figuring all these factors out would require testing the water in a lab set up to do that type of testing and someone familiar with that particular technology.


text:
Name for H2O- radical? (H2O+ radical is Oxoniumyl)
Radical are atoms with unpaired electrons, and they have short lifetimes because the unpaired electrons are highly chemically reactive. The $\ce{H2O^{.+}}$ radical is named oxoniumyl. Is there a name for a hypothetical $\ce{H2O^{.-}}$ radical?

$\ce{H2O^{.-}}$ does not exist.

That is has not been proven to exist could be one thing (I have not found anything about it being proven to exist, did a quick and very shallow Google search. ) But that it cannot exist, is there theoretical proof for that?

It might be a very unstable intermediate in some reduction of water reaction.

There is gas-phase hydrated radical anion $\ce{(H2O)_n^{.-}}$. I don't know if the anion exist for n=1.

This question was apparently banned. Nothing about it is out of place. It is a very concise question. I would guess moderation error was the case there.

@Haverse, what do you mean by that? Your quesion hasn't been closed, nobody has voted to close it, and nobody has even downvoted it.


text:
why lawrencium is placed in f-block elements although its last electron enters in 6d-subshell?

The electronic configuration of lawrencium ($\ce{Lr}$) is $\mathrm{[Rn] 7s^2 5f^{14} 6d^1}$. As its last electron enters the $\mathrm{6d}$ sub shell, it should be a part of $\mathrm{d}$-block elements, but in the periodic table it is given as a part of $\mathrm{f}$-block elements. Why?

I noticed that if we place lawrencium in $\mathrm{d}$-block elements the symmetry and similarity in properties of many elements will be broken. Is that the reason to place it like this? Or something else? Is lawrencium an exception? How? Please explain.

As noted in https://en.wikipedia.org/wiki/Block_(periodic_table), "Not everyone agrees on the exact membership of each set of elements." and "Sometimes the f-block is considered to be between groups 2 and 3 instead of 3 and 4, which means that lanthanum and actinium are f-block elements, and lutetium and lawrencium are d-block elements instead located under yttrium." Ultimately, what you call it doesn't really matter much at all.

Actually, the last electron isn't even in the 6d shell but in the 7p shell according to this https://www.rsc.org/periodic-table/element/103/Lawrencium


text:
When we cannot use Friedel-Crafts acylation to insert alkyl group?

Questions:

Show a probable synthethic pathway to obtain the first molecule, I.
Will the same strategy be used to insert the alkyl group for J. Explain.

My answers (please tell me if there is any incorrect answer):

Friedel-Crafts acylation -> Clemmenson reduction -> Nitration -> Reduction of Nitro Benzene

No, to insert alkyl group for molecule J, it must undergo Friedel-Crafts alkylation. In Friedel-Crafts alkylation, there is carbocation rearrangement to ensure the carbocation is stable and tertiary. In J, the carbon that will directly attach to benzene is already stable, so the C(CH3)3 substituent will attach directly to benzene. (I'm not sure what to explain more)

Is this correct for alkylation step in J?

My questions:

For no. 2, is it true if we say that, in this case, no rearrangement will occur in Friedel-Crafts Alkylation step because the carbocation is tertiary and stable so it can proceed with this step (Friedel-Crafts Alkylation). Is it possible that rearrangement will not occur in Friedel-Crafts Alkylation?
Why we cannot use Friedel-Crafts acylation to insert alkyl group in J? I am confused, if we use (CH3)3Cl, there will be no carbocation rearrangement, but why the question asked about acylation, I think it is not possible to undergo acylation step because there is no carbocation rearrangement, so the substituent attached to J will be CH3CH2CH2???

tBu-benzene is prepared exactly as you describe, see this preparation here. Follow this by nitration and reduction.
There is no obviously comparable strategy of acylation that can compete with this. Making a tBu group from acetophenone is not straightforward and is illogical when there is a perfectly good direct route.

If a Friedel-Crafts reaction is made with $\ce{CH3-CH2-CH2-Cl}$ and $\ce{AlCl3}$, isopropyl benzene will be obtained, and not propylbenzene, because n-propyl cation may isomerize into isopropyl cation. That is why  $\ce{CH3-CH2-CO-Cl}$ should be used instead. And -CO- is reduced later on.


text:
Reagent choice in the formation of Wittig reagent

During the formation of Wittig reagent, BuLi is used as it is a VERY strong base (pKa = 50). According to Wikipedia, however, sulfur ylide is 'especially conveniently' made from dimsyl anion (pKa = 35), which is a relatively weaker base although it is also used for preparation of Wittig reagent.
If there is a stronger base is allowed (BuLi), why is dimsyl anion used 'conveniently'? To prevent side reactions or just by what is available at hand?

Typically phosphorus ylids are made from PPh3. What is the purpose of having 3 phenyl groups linked to the phosphorus atom in Wittig reagent? Can't be as small as sulfonium ylide molecule? I have no idea what is the use of adding 3 bulky and sterically-hindered groups in the ylide.

I am fairly sure that for both, there are practicality issues at hand.
This is almost certainly so for the choice of phosphine. Even though PPh3 is absolutely terrible from an atom economy point of view, and even though PPh3O can be a real pain to remove after the reaction is done, at least it's a nice white solid which doesn't smell. On the other hand, the simpler alternative PMe3 is a volatile, pyrophoric, and extremely stinky liquid. PPh3 is also substantially cheaper than other alternative phosphines. (There are similar considerations for the Mitsunobu reaction which I have some experience with.)
As for the base, $\ce{BuLi}$ is just one of the most accessible strong bases to use. I don't think you can buy dimsyl sodium from the average chemical supplier. My suspicion is that the main reason why dimsyl anion is somewhat associated with the sulfur ylids is because E. J. Corey and his postdoc M. Chaykovsky popularised them together (e.g., Ref.1). Also note that $\ce{NaH/DMSO}$ mixtures are associated with significant explosion hazards (Ref.2).
Others with more experience will be able to give fuller answers.
References:

E. J. Corey, Michael Chaykovsky, “Methylsulfinyl Carbanion ($\ce{CH3-SO-CH2-}$). Formation and Applications to Organic Synthesis,” J. Am. Chem. Soc. 1965, 87(6), 1345–1353 (https://doi.org/10.1021/ja01084a033).
Qiang Yang, Min Sheng, James J. Henkelis, Siyu Tu, Eric Wiensch, Honglu Zhang, Yiqun Zhang, Craig Tucker, David E. Ejeh, “Explosion Hazards of Sodium Hydride in Dimethyl Sulfoxide, N,N-Dimethylformamide, and N,N-Dimethylacetamide,” Org. Process Res. Dev. 2019, 23(10), 2210–2217 (https://doi.org/10.1021/acs.oprd.9b00276).

I presume you never collected that disgustingly smelling triphenylphosphane oxide? Or was what I smelled residual triphenylphosphane?

@Jan I collected plenty of it (not quite kg's, but we did multigram scale PPh3 reactions)... but I can't say I've ever noticed a smell.

Now I wonder if it’s genetic like HCN …

Indeed ;-) TsCl is also one that seems to affect people to different extents...

Oh god, yes, that’s also terrible D=

Wittig reactions can be performed with all sorts of phosphanes. The resulting phosphonium salts can be deprotonated by all sorts of bases, as long as the base is basic enough.
Triphenylphosphane has the advantage that its phosphonium salt can only be deprotonated on one carbon atom: the three ipso-phenyl carbon atoms are quarternary and thus cannot be deprotonated. In addition, it is extremely cheap and easy to handle. However, the three phenyl groups also make it somewhat bulky and the Wittig reaction may not be as effective as using other phosphanes.
Trimethylphosphane is another candidate that eliminates all problems of bulkiness. However, it comes with its own set of problems like being a volatile, pyrophoric and toxic liquid that reacts with ambient oxygen to form $\ce{OPMe3}$, meaning that it must be kept under inert gas.
An in-between choice is tributylphosphane. While it is still unpleasant, requires inert gas handling and is flammable, it is less toxic and less susceptible to oxidation than trimethylphosphane. For an example of tributylphosphane being used in total synthesis, check out Gieseler and Kalesse’s synthesis of angiolam.[1]
As for the choice of base, it depends a lot on the compound in question. As you can tell in the same paper by Gieseler and Kalesse (experimental procedures in the supplementary information should be accessible free of charge), they use $\ce{KOtBu}$ as a base to convert the tributylphosphanyl salt into the corresponding ylide – but they have an activated Wittig reaction thanks to the neighbouring carbonyl group. For unactivated Wittigs, BuLi may be your only choice but if it is not, a milder base will typically cause less side-reactions in a sensitive substrate.
Which base is most convenient to use often depends on the laboratory in question. If there is a constant influx of students that have to titrate BuLi to learn the method, and you have large stocks prepared because you need it often, BuLi might be your method of choice. If you just established 234ff-base as a new named compound in your laboratory, you will probably have ample amounts of that and a strong desire to try that one first. And so on.
[1]: M. T. Gieseler, M. Kalesse, Org. Lett. 2014, 16, 548–551. DOI: 10.1021/ol403423r.


text:
Which is the major product of bromination of 3-(dimethylamino)benzamide, ortho or para?

I know that A will most probably not happen because of high steric hindrance. However between B and C, will B become the major product? Because $\ce{-N(CH3)2}$ is less bulky compared to $\ce{C=O-NH2}$, so there will be less steric hindrance.

With E already twice substituted, the use of ortho, meta, para to designate the entry of the new (then third) substitutent requires to be a bit more specific.  You have luck with A, Br were introduced in a position that is ortho to both $\ce{NMe2}$ and the amide.  This contrasts to B, for example, were Br is ortho to $\ce{NMe2}$, and para to the amide.

Here's the 3D structure of the substrate:

As you can see, the $\ce{-N(Me)2}$ group is clearly larger than the $\ce{-CONH2}$ group. Also notice that the $\ce{-N(Me)2}$ group is planar with the benzene ring. This is because this structure is the most favourable for $\mathrm{p-\pi}$ conjugation 1.
Your guess that A would not form is correct. The large size of the $\ce{-N(Me)2}$ group would also point to the fact that C is the major product. For a better explanation of the major product, we can take a look at the 3D structures of B and C:

You can clearly see that in B, the bromine atom pushes the dimethylamine group out of it's stable configuration, where it is coplanar with the benzene ring. This requires more energy and also produces an unstable product. Therefore, C is the major product in this reaction.
References:

Novikov, V. P., et al. “Molecular Structure and Conformation of N,N-Dimethylaniline by Gas-Phase Electron Diffraction and Quantum-Chemical Calculations.” Russian Journal of General Chemistry, vol. 74, no. 8,2004, pp. 1247–53. doi:10.1007/s11176-005-0146-9.

This answer relies on steric hinderance to figure out which product may be formed.  My anticipation was to see an answer equally considering that electronic effects by the amine (electron push) and the amide (electron withdrawing) leading 1) to a preferred direction substitution pattern by either one if the other were absent.  And then 2) «because both amine and amide already are present ...» the preferred entering direction of Br as third substitutent. Vaguely I recall seeing a table where these effects of an electrophilic aromatic substitution were quantified, too.


text:
Crossing at critical temperature of a gas
This is written in the book 'Treatise on Thermodynamics' by Max Plank-

the crossing of the critical temperature at any pressure other than the critical pressure differs in no way from the crossing of any other temperature.

But from the phase diagram of the states, it seems that if we change the temperature at a pressure greater than the critical pressure, then on crossing the critical temperature the gas will liquify where as crossing at other temperatures the gaseous state won't change.
So is the statement in the book wrong?

No, the book's statement is correct. If the temperature is below the critical temperature then the critical pressure won't be needed to liquefy the gas. If the temperature is above the critical temperature then no pressure will liquefy the gas. Only at the critical temperature is the critical pressure needed.

At the critical temperature/pressure, all properties of liquid and gaseous phase  mutually converge to the same values. There is just one phase at these temperatures, neither liquid, neither gas. Or, both at the same time.
Changing  pressure at temperature above the critical temperature causes continuous changes of the supercritical phase properties without the edge of the gas/liquid phase transition.
Changing  temperature at pressure above the critical pressure causes continuous changes of the supercritical phase properties without the edge of the gas/liquid phase transition.

What they refer to is how properties such as density and viscosity change when you pass through the critical point.
Say you have water initially at 25°C under one atmosphere pressure.  You heat it up and, of course, it boils into steam at 100°C.  During the boiling process the density and viscosity drop discontinuously from those values for the liquid to the values for the gas, but once you have formed steam any further changes become continuous and at a finite rate.  That includes reaching the critical temperature of 374°C, which really isn't critical at all unless you are also at the critical pressure of 218 atmospheres.
So, now try the same experiment but with 218 atmospheres (ctitical) pressure so that the water "boils" at the critical temperature of 374°C.  Now you find the density and viscosity decreasing continuously, with no apparent break due to any phase change.  But, at precisely the critical point, those properties are momentarily dropping at an infinite rate, much like the cube root of a number dropping at an infinite rate when the number passes through zero.  The critical point is the one point where you get this infinite rate of change within a single phase.


text:
Positive work in the Carnot cycle
In the Carnot cycle the overall work done on the piston is positive. I am having trouble with the real meaning of that. Does positive work mean the piston is moved up and has, over the course of the cycle, gained potential energy?

The Carnot cycle uses an idealized heat engine to illustrate basic thermodynamic principles. Net work is done either by or on the system, depending on the direction in which it is run. As part of this idealization, the piston is massless and frictionless. The piston merely serves as a moveable impermeable boundary between the system and the surroundings—i.e., as an interface that allows the system to do work on the surroundings, and visa versa. The piston exerts no force of its own, so no work is done by or on the piston; the piston merely serves as a conduit for force exerted by the system or surroundings, and is not part of either.
To more specifically address your question, since the piston is massless, its gravitational potential energy is zero, independent of position.
Also, as this is a cyclic process, the system is returned to its original state at the end of each cycle, and thus the piston is returned to its original height. [Hence, as Chet Miller points out in his comment, even if the piston had mass, its PE would be unchanged at the end of each cycle.]

so basically it just means that work is done on the surrounding and is caused by the system? Or rather the hot and cold reservoirs and the conversion of heat into work which then leaves the system

Even if the piston has mass, over a cycle of the entire, its potential energy doesn't change.

@muhammadhaider Yes, when run in the "forward" direction, net work is done by the system on the surrondings, at the cost of a transfer of thermal energy from the high T to the low T reservoir (both of which are in the surroundings).  So you are converting a flow of heat into a flow of work or, more colloquially, heat into work.  That's why this is called a heat engine.

It is not the piston that the working fluid is doing work on.  Over an entire Carnot Cycle, the potential energy of the piston does not change.  But, if you were adding and removing tiny pebbles to and from the piston at its various elevations during the cycle, there would be net work done in increasing the potential energy of the pebbles.


text:
How much does oxygen enrichment improve some combustion engines?
This question is about how much the combustion efficiency increases with oxygen enrichment in modern combustion engine designs. A desirable way to answer is to give the fuel efficiency dependency on oxygen percentage of air volume intake.
Some clues can maybe be had from

Engineering Science and Technology,
Volume 19, Issue 1, March 2016, Pages 438-443
Effects of oxygen enriched combustion on pollution and performance characteristics of a diesel engine  P.Baskar, A.Senthilkumar

and

THE EFFECT OF OXYGEN-ENRICHED AIR ON THE PERFORMANCE AND
EXHAUST EMISSIONS OF INTERNAL COMBUSTION ENGINES
,VARADARAJA SETTY, B.E.
MASTER THESIS
IN
MECHANICAL ENGINEERING
, Texas Tech University

Reports like the following are less interesting, since spark timing is the regulating mechanism with increased oxygen levels. Spark timing techniques are considered outmoded.

J.X. Zhou, B. Moreau, Christine Mounaïm-Rousselle, Fabrice Foucher. Combustion, Performance
and Emission Analysis of an Oxygen-Controlling Downsized SI Engine. Oil & Gas Science and
Technology - Revue d’IFP Energies nouvelles, Institut Français du Pétrole, 2016, 71 (4), pp.49.
ff10.2516/ogst/2015035ff. ffhal-01597165f

Let me get this straight: you want to know the relative efficiency of 1) combustion with air as oxidizer and 2) combustion with oxygen-enriched air, with the energy cost of the oxygen enrichment process taken into account, i.e., not just ignored? If my understanding is correct, then it comes down to doing some detailed energy cost accounting. I have no expertise in that, but maybe someone here does. I hope you get a helpful answer.

If you like cars, check out turbo charger

I am splitting this question into two questions.

What does "energy efficienct" mean?
It depends on what you mean by "efficient".
For combustion reactions that involve engines, if "efficiency" means how much work (the "good" type of energy) that comes out of the engine per unit input of fuel, then yes oxygen-enriched mixtures would be more efficient than regular old air.
If instead "efficiency" means how much of the fuel is combusted all the way to $\ce{CO2}$, the answer is also probably yes, but for different reasons.
The efficiency of engines
To understand why, keep in mind that no engine can ever be more efficient than an ideal Carnot engine.  The efficiency of a Carnot engine is $\frac{T_H - T_C}{T_H} = 1 - \frac{T_C}{T_H}$, where $T_C$ is the temperature of a cold reservoir (e.g. our everyday environment)   in Kelvins and $T_H$ is the temperature in Kelvins of a hot reservoir.
Thus, the hotter you can get the "hot" side of Carnot engine, the better.  In a car engine or other Otto engines, this would correspond to the maximum temperature of the combusted air (or oxygen)-fuel mixture just before the expansion stroke.
Adiabatic flame temperature
With air instead of pure oxygen as the oxidizer, some of the energy of combustion goes into heating up inert components of the air, mostly nitrogen.  This lowers the maximum temperature of the air-fuel mixture and decreases theoretically attainable efficiency.
The adiabatic flame temperature is a quantitative number behind this idea.  Wikipedia lists the adiabatic flame temperature, the maximum temperature attainable in the combustion of certain fuels, and the values for fuels burning in air are a bit less than the same fuels burning in oxygen.
Limits of this analysis
Of course, there are many factors that make real engine performance different than ideal engine performance.  In fact, designing engines that can operate with at the extremely high temperatures that would maximize efficiency is tough because of materials problems.  (The adiabatic flame temperature of gasoline in pure oxygen is somewhere around 3800 K, far higher than the melting temperature of steel, for example.). So whether a given real engine designed to work under particular sets of conditions will necessarily do better with pure oxygen instead of air is less clear.  Probably yes?  And for sure an engine designed appropriately for use with pure oxygen would be more efficient than an equivalent one designed for us with air.

With respect to your question "if a combustion process with air as oxidant can benefit by oxygen enriching the air intake?", the answer is possibly yes, as to quote from Wikipedia on gas mixes associated with the fuel gas hydrogen, as an example:

The flame temperature is high, about 2,000 °C for hydrogen gas in air at atmospheric pressure, and up to 2800 °C when pre-mixed in a 2:1 ratio with pure oxygen (oxyhydrogen). Hydrogen is not used for welding steels and other ferrous materials, because it causes hydrogen embrittlement.

where one could claim that obtaining a higher combustion temperature is likely consistent with achieving an increase in energy efficiency. Note, with respect to 'net' accounting of total energy employed to achieve the elevated temperature, that is likely a technology-specific based answer (see, for example, here) on how, for example, was oxygen concentration achieved.
Interestingly, one can expand beyond the conceptional use of pure oxygen also as noted in this application employing the fuel gas acetylene:

Air/acetylene and nitrous-oxide/acetylene are the gas mixtures used as a fuel source in Flame Atomic Absorption (FLAA).

where NH3/air/catalyst is a possible path to NO (as derived from air). With respect to FLAA, it has been described, per a reference, as:

Flame Atomic Absorption is a very common technique for detecting metals present in samples. The technique is based on the principle that ground state metals absorb light at a specific wavelength. Metal ions in a solution are converted to atomic state by means of a flame.

Note that pure oxygen is used in iron production, as net energy gain is positive, More energy is saved  than spent by oxygen separation.

Poutnik: Thanks for your cited example, as my 'net' energy accounting qualification may imply that any positive net energy gain cases may not actually exist in practice.


text:
Birch reduction of 1,2-Cyclopentanedione and Diketones
This answer given to this problem was that only one of the double bonds is reduced into an alcohol. I
think i understand how the double bond is reduced into an alcohol. An electron is transferred to the carbonyl carbon and the double bond cleaves homolytically, forming a carbanion (stabilised by resonance ) and the oxygen with an extra electron which then will gain another electron and both get hydrogens to form the alcohol.
First of all is my method right? I don't think so as both the double bonds would have been reduced into alcohols. Also i was unable to see any sources for the birch reduction of di-ketones, I was only able to find it for unsaturated ketones.

I think this needs a diagram

@Waylander I have added it now.

The course of the reaction depends on whether or not a proton source other than ammonia is present. In the absence of an alcohol (ethanol, tert-butanol) the α-dione 1 adds an electron from sodium to give a resonance stabilized radical anion 2. A second electron is added to afford enediolate 3. [This is the same species formed with sodium in the acyloin condensation on diethyl glutarate.]  Water workup provides the α-hydroxyketone 4. In the presence of alcohol, enolate 3 is protonated to form 5. [The reason I am reluctant to protonate the alkoxide of 5 is to avoid reductive elimination of hydroxide from 4 to leave the enolate of cyclopentanone. Further protonation/reduction may lead to the alkoxide of cyclopentanol.]  One-electron reduction of 5 yields ketyl 6, which via protonation and a second reduction leads to dialkoxide 8. Workup affords the diol 9.
The bottom line is if you want to form 4 by metal/ammonia reduction, avoid alcohol.

Just to clarify. If the alkoxide of 5 is protonated, the OR- group will act as a base and eliminate the -OH group as water which leads to the formation of an alpha-beta unsaturated ketone that leads to an enolate which on further reaction forms the alkoxide of cyclopentanol. But the same can also happpen with h2o right?(3 to 4). Hydroxide acts as a base forms the enolate again?

could u pls clear the above doubt? Thanks!

If 5 is protonated on oxygen by an alcohol to form 4, then adding an electron to the double bond forms a ketyl (like 6 but protonated on oxygen). This species could eliminate hydroxide and after the addition of a second electron one has the enolate of cyclopentanone, which can protonate with ROH and cyclopentanone can reduce to the alkoxide of cyclopentanol. Aqueous workup gives cyclopentanol. Enediolate 3 is stable to Na/NH3 in the absence of ROH. Aqueous workup gives 4. I'm just showing possible outcomes.

Your second structure with the electron added to oxygen is my enediolate 3.

thx a lot. got it now


text:
Heat Transfer Coefficient Problem
The Problem

An unusual reference gives the formula below for the heat transfer coefficient of an emulsion.
$$H = 0.0061\left(X^{0.6} + 0.038T\right)\cdot V^{0.35}$$
In the formula $T$ is the temperature in $\pu{^\circ F}$, $V$ is the flow rate of the emulsion in $\pu{\frac{ft}{min}}$, $X$ is a characteristic dimension in inches and $H$ is the heat transfer coefficient in $\pu{BTU h^-1 ft^2 ^\circ F}$.
Calculate the heat transfer coefficient in $\pu{W m-2 K-1}$ when emulsion at $\pu{30 ^\circ C}$ flows at $\pu{1.25 m s-1}$ and the characteristic dimension is $\pu{0.8 mm}$.

My Attempt

I converted the characteristic dimension ($X$) from $\pu{mm}$ to $\pu{in}$.

$$\pu{0.8 mm} ⋅ \frac{\pu{39.37 in}}{\pu{1000 mm}} = \pu{0.03 in}$$

I converted the temperature ($T$) from $\pu{^\circ C}$ to $\pu{^\circ F}$.

$$\frac{(n \ \pu{^\circ F} - 32)}{1.8} = 30 \ \pu{^\circ C} $$
$$n \ \pu{^\circ F} = \pu{86 ^\circ F}$$

I converted the flow rate of the emulsion ($V$) from $\pu{m/s}$ to $\pu{ft/min}$.

$$\frac{\pu{1.25 m}}{\pu{s}} ⋅ \frac{\pu{3.28 ft}}{\pu{m}} ⋅ \frac{\pu{60 s}}{\pu{min}} = \frac{\pu{246 ft}}{\pu{min}}$$

I plugged the values of $X$, $T$, and $V$ into the equation to solve for $H$.

$$H = 0.0061\left(0.03^{0.6}\ \mathrm{in} + 0.038(86\ \mathrm{^\circ F})\right)\left(\frac{246\ \mathrm{ft}}{\mathrm{min}}\right)^{0.35}$$
Is my work correct so far? If yes, I'm a little confused as to what I should do next. How do I simplify the units in my last step in order to then convert to a value with the desired units $\pu{W m-2 K^-1}$?

there is a little typo in the last equation , the mm should be ft. Numerically then you have H in BTU /hr ft$^2$F. Now convert this into SI unit by substituting with 3600 s /hour,  0.0254 *12m /ft etc and convert temperature in F into Kelvin.

Shouldn't the mm actually be inches not ft (the value I calculated in step 1 is in inches because the problem defines X as the 'characteristic dimension in inches')?

however, you want values in ft as used in the other part of the equation and to get consistent values to get BTU/hr/ft$^2$ then you change each of these to SI units.

Is my work correct so far?

Yes, your conversions to the units for the inputs to the equation (#1 to #3) look correct to me.

How do I simplify the units in my last step in order to then convert to a value with the desired units...?

You don't.
This sort of expression embodies a purely empirical relationship -- it has no particular basis of connection to the underlying physical relationships between these various quantities.  The reason the units of each of the input quantities are specified ($^\circ\mathrm{F}$, $\mathrm{\frac{ft}{min}}$, etc.) is because you're supposed to use just the numerical values of those quantities, after conversion to the indicated units, and substitute those into the equation. So, instead of the equation you have in #4, it would just be:
$$
H = 0.0061 \cdot \left(0.03^{0.6} + 0.038 \cdot 86\right) \cdot 246^{0.35}
$$
Then, you take the purely numerical result you obtain for $H$ from this equation and tack on the indicated units; in this case, $\mathrm{BTU \cdot {hr}^{-1} \cdot {ft}^{-2} \cdot \ {^\circ F}^{-1}}$. From there, you can carry out a straightforward conversion to the metric units asked for in the final answer.
These sorts of empirical correlations are extremely common in chemical engineering practice, where one doesn't especially care if an equation has solid theoretical underpinnings, as long as it performs reasonably well in predicting quantities of interest.
To note, while you can construct internally consistent units for the various constants in the formula, they usually don't end up having any particular useful physical interpretation. For example, since it's only possible to add two quantities together if they have the same units, the $0.6$ exponent on $X$ means that the $0.038$ coefficient on $T$ would have to have a dimensional value of $0.038\ \mathrm{{in}^{0.6} \over {^\circ F}}$, which has no particular physical significance that I can determine. I'm not even sure what sense there is to be made of units such as "$\mathrm{{in}^{0.6}}$", "$\left(\mathrm{ft\over min}\right)^{0.35}$", etc.

That clears it up a lot since this problem is for an assignment in a chemical engineering course I'm taking. =) But does that mean that if that was done for more theoretical  purposes in chemistry, I should simplify the units in the last step?

@air_nomad See edit. tldr; No. :-)


text:
Friedel-Crafts acylation of furan
Can furan undergo Friedel–Crafts acylation with benzoyl chloride?
Some people say it will have no reaction. Is it true?
If so, why cannot it undergo Friedel–Crafts acylation like other ketones?

Classical Friedel-Crafts acylation conditions fail completely in the case of the sensitive and polymerization-prone furan, or give only rather unsatisfying yields only.

Quoted from the abstract of this recent paper$\ce{^{[1]}}$ which describes a mild method for the acylation of furans with free carboxylic acids. Furan is not very stable to classical F-C conditions with $\ce{AlCl3}$. $\ce{BF3}$ is a better catalyst$\ce{^{[2]}}$.
References

Martina Opietnik, Alois Jungbauer, Kurt Mereiter and Thomas Rosenau, “Mild Friedel-Crafts Acylation of Furan with Carboxylic Acids and the Heterogeneous Catalyst Couple AlPW12O40 / Mg(OH)2”, Current Organic Chemistry (2012) 16: 2739. DOI: https://doi.org/10.2174/138527212804004490
THE ACYLATION OF FURAN AND THIOPHENE WITH ALIPHATIC ANHYDRIDES IN THE PRESENCE OF BORON TRIFLUORIDE-ETHERATE by JOHN V. HEID and ROBERT LEVINE
The Journal of Organic Chemistry, 1948 13 (3), 409-412
DOI: 10.1021/jo01161a015

Sir, I was wondering will chlorobenzene undergo fridel crafts acylation reaction as chlorine exerts both (-I) effect and (+M) effect?

Yes it will. Here is an example https://www.sciencedirect.com/science/article/abs/pii/S0040403998008818


text:
How are half lifes calculated for radioactive substances that decay over many years
How are the half lives (or whole lives) calculated for slowly decaying radioactive substances such as tellurium? I must be having a fundamental misunderstanding of how we are able to predict the amount of time it will take for an isotope or radioactive material to decay fully including for ones that can remain unstable for practically eons.

Same way as any other half-life: by counting the decay events. Did you think that the half-life of 1600 years for radium was obtained by counting atoms in a flask left by Marie Curie that long ago? Well, no.

@IvanNeretin How in the world does somebody practically measure these decay events?

"How in the world does somebody practically measure these decay events?"...Well by using instruments.

Well, 226Ra is fun, but 128Te with T1/2=2.2e24 years less so. In 1 mol of 128Te, the decay rate is 1 kernel in several years. It is worse than catching neutrinos.

Even 209Bi is fun, compared to 128Te.

Suppose you have a sample of $N$ atoms of a radioactive material. As it is radioactive, you may state that a small fraction $k$ is decomposed by second. This fraction does not depend on the total number of atoms. If this sample is placed in a Geiger counter, each decomposition of its atoms produces a ray that crosses the inner volume of the Geiger counter, and it will be detected and counted. In a given time, the number of rays is equal to the number of disintegrations. If $\mathrm dN/\mathrm dt$ is the number of atoms disintegrated per second (or per minute), the factor $k$ can be calculated by dividing the number of rays detected every second in the counter, by the total number of atoms $N$ of the radioactive material.
$$k=\frac{\mathrm dN}{\mathrm dt}\cdot\frac1N$$
Once $k$ is known, the half-life $T$ can be derived from the integrated law, if $N_0$ is the original number to atoms at $t=0$
$$N/N_0=\mathrm e^{-kt}$$
The half-life is the time $T$ after which $N=N_0/2$, which gives $$1/2=\mathrm e^{-kT}$$
with the following result:
$$T=\frac{\ln2}k=0.693/k$$
So the period may be obtained from the knowledge of $k$. It is not necessary to wait up to the radioactive material has decreased to one half.
For example, if $0.0238\ \mathrm g$, or $6.02·10^{19}$ atoms $\ce{U-238}$ is inserted in a Geiger counter, $292$ rays will be detected per second. From this measurements, the constant $k$ can be calculated:
$$k=\frac{292\ \mathrm{s^{-1}}}{6.02\times10^{19}}=4.87\times10^{-18}\ \mathrm{s^{-1}}$$
and the half-life is $$T=\frac{0.693}{4.87\times10^{-18}\ \mathrm{s^{-1}}}=1.423\times10^{17}\ \mathrm s=4.51\times10^{9}\ \mathrm a$$