text:
Gaseous reactant in Nernst equation
So I have this equation of a galvanic cell

$\ce{Zn + Cl2 -> Zn^{2+} + 2Cl^-}$

Why does the Nernst equation not consider pressure of $\ce{Cl2}$ here? Only concentrations of zinc and chloride ions are being considered in my textbook.
Am new on SE, so am really sorry for any errors.

A asking B why C has written something wrong or incomplete is a bad concept. // The N. eq. does involve partial pressure on involved redox-active gas, but the writers of textbooks must mention it. You should always refer to the source of questioned info for others to know what is rather to be avoided.

The Nernst equation does involve partial pressure on involved redox-active gas, but the writers of textbooks must mention it. You should always refer to the source of questioned info for others to know what is rather to be avoided.
As exampple, the Nernst equation for the non-standard hydrogen electrode is
$$E = E^° + \frac{RT}{2F} \ln{\frac{a(\ce{H+})^2}{p(\ce{H2})}}$$
Another thing is, that the particular formulation may use implied default gas pressure, not being explicitly expressed in the equation.


text:
Why is scandium not in Group 13 despite being once called "eka-boron"?
When drawing up the Periodic Table, Mendeleyev famously predicted the existence of then-unknown elements such as "eka-boron", which we now know as scandium.
However, in the modern Periodic Table why isn't scandium then part of the same group as boron (Group 13)? For example, gallium (which was eka-aluminium) comes in the same group as aluminium, and likewise for germanium (eka-silicon). But in the case of scandium / eka-boron, aluminium is the element directly below boron. Why is this so?

Historically conditioned but now obsolete thoughts.

Early thoughts before refining the periodic table system.

"Initial versions of the periodic table did not distinguish rare earth elements from transition elements" https://en.wikipedia.org/wiki/Mendeleev%27s_predicted_elements#Ekaboron_and_scandium

At the time, when scandium was eka-boron, there was no group 13.

In older forms of the Periodic Table, boron was not necessarily far away from what we now call scandium. Mendeleev himself used a form where what we now call the fourth and later periods were set up in two rows based on how the stoichiometries of compounds tended to repeat (see here). For instance, copper forms an oxide and a chloride with similar atomic ratios to those of potassium, and Mendeleev's table placed copper in the first column with potassium.

[Source: Källa:Dmitrij Ivanovitj Mendelejev (1834 - 1907). This work is in the public domain in its country of origin and other countries and areas where the copyright term is the author's life plus 70 years or fewer.]
However, the columns are staggered as if to indicate that at least in the later periods, two rows were needed to make a full periodic cycle; rubidium (two rows below potassium) bears a closer resemblance to potassium than copper (one row below), despite all three metals forming (for instance) solid monochlorides.
When we move over to column 3 (elements forming sesquioxides and trichlorides) in Mendeleev's format, where the two elements immediately below aluminum seemed to be unknown, the one immediately below aluminum and two rows below boron was thought to be more like the latter, whereas the element two rows below aluminum would be more like that metal. Mendeleev then labeled these expected elements accordingly.
Today, if we look at melting points, we see a sharp rise from aluminum to eka-boron/scandium (which melts at nearly the same temperature as iron), then back down to low-melting (nearly room-temperature liquid!) for eka-aluminum/gallium.

Note that the periodic table we are seeing today is not the same during Mendeleev's period. If you look at the old periodic table, you will find scandium below boron. So, Mendeleev predicted correctly. It is just that in the new periodic table, it moved to different group.

(image source)

(+1) Very nice old periodic table! Quite elegant, actually.

source: Stack Exchange

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text:
STO-1G in Gaussian
I don't see the STO-1G basis set in the available sets on the Gaussian website and the software (g16) throws an error if I try. I see how it seems possible to use a custom basis but I'm not entirely sure how to use that to run a HF/STO-1G. I was looking on the basissetexchange.org but there is no STO-1G available.
It seems to -e that such a simple set should be available (for technical analysis rather than accuracy, obviously) but I can't figure out how to use it.

I don't believe there is a "STO-1G." Even if there were, it would exhibit poor performance.
The STO-nG basis sets were first published by Hehre, Stewart, and Pople in 1969:
"Self-Consistent Molecular-Orbital Methods. I. Use of Gaussian Expansions of Slater-Type Atomic Orbitals"
Note that this was the first paper to use Gaussian approximations to Slater-type basis sets. There is no mention of an STO-1G.

We have obtained representations of Slater-type orbitals using 2-6
Gaussian functions by least-squares methods...

They conclude that STO-3G is an efficient minimal basis set:

The STO-3G set, in particular, is very economical to use and as a minimal (is, 2s, 2p) basis, can be applied to quite large organic molecules. Any further economy achieved by using an STO-2G set, however, is hardly worthwhile, since the calculation time is then dominated by the solution of the Roothaan equations, rather than integral evaluation.

While it is certainly possible to use one Gaussian to approximate the Slater atomic orbitals, the performance would be poor. Here's an example plot that I use in class.

Note that near the nucleus, the Gaussian-type function does a poor job of describing the wavefunction, and thus the electron density. Additionally, the single GTO is too high at medium range, and shows the wrong asymptotic behavior - it falls off too quickly at long range. (In other words, poor performance everywhere.)
Instead, if we use a best-fit to 3 Gaussian-type functions, we do a much better job at approximating the STO. Pretty much everywhere except very close to the nucleus has a good fit.

Thanks for the answer. I understand that but from today's perspective, this 'sweet spot' argument doesn't really matter. STO-3G is terrible and shouldn't be used for actual calculations, just as much as 2G or 4G shouldn't be used. I need STO-1G for a 'conceptual' calculation rather than an interest in the result. I'd expect that somebody tried this out 'for the science of it' and fitted an STO-1G?

STO-3G can sometimes be useful when you hit convergence issues, but yes I wouldn't recommend it otherwise. I did some searching and cannot find any reference to STO-1G, much less published parameters. It wouldn't be hard to do .. for element of interest, generate the relevant $s$ and $p$ STOs and do a best-fit for the Gaussian.

Oh boy, okay, I never would've thought that I'll be fitting my own basis set. Thanks for the help!

@ste - for something like STO-1G it would be fairly easy. Thus the example I use in class above. Maybe just pick something like water, in which you only need a few basis functions. :-)

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text:
Is there an agreement for a specific temperature referred to in the context of standard Gibbs energy?
I found an exercise in Physical Chemistry by Atkins et al, 8th edition, page 235 that I found confusing. The question is as follows.
"7.2(b) Molecular bromine is 24 per cent dissociated at 1600 K and 1.00 bar in
the equilibrium Br2(g) ⇌ 2 Br(g). Calculate (a) K at 25°C, (b) ΔrG°, (c) K at
2000°C given that ΔrH° = +112 kJ.mol^(−1) over the temperature range."
I initially assumed that to answer (b), the temperature plugged to the equation ΔrG °= -RT.ln(K) is 298.15 K since it's mentioned that the reaction Gibbs energy is standard. However, when I checked my answer with the students' solutions manual, the author of the manual plugged in 1600 K as the temperature for the solution of (b). Who is correct and why? If the solutions manual author is the correct one here, why is there no agreement for a temperature in the context of "standard Gibbs energy?"
Thanks in advance for you who answer or inform if there has been any similar questions to mine in the forum since I haven't found one as I typed this.
Edit (last done on 9.51 PM UTC+7, 8th May 2022): @orthocresol has linked a question different from mine in the sense that what was asked circa 6 years ago was different than what I asked here; however, the answer to one of the linked questions is relevant to my question.

The standard state doesn't define a temperature; so there's no necessity for $\Delta_\mathrm{r}G^\circ$ to be at 298 K. (See also: What is the difference between ∆G and ∆G°?) I must say that the question wording is slightly confusing since that part is sandwiched between two different temperatures...

@orthocresol, I see. Thank you for providing some readings regarding the standard reaction Gibbs energy.

The question to the linked post is somewhat different but the answer is the same.

Note that in specific context of electrochemistry, there is conventionally agreed the standard temperature 25 deg C.


text:
How do mathematical operations involving exact numbers affect significant figures?
I'm asking for assurance that I'm not making any mistakes because I can't find these rules explicitly stated in my textbook or elsewhere, but it seems to me that:

When multiplying/dividing a measured number by an exact number, the product/quotient has the same number of significant figures as the measured factor.

and

When adding an exact number to a measured number, the sum may have a different number of significant figures as compared to the measured addend.

So for instance, if I were to subtract exactly 32 from a measured -7.0 x 10^1 degrees F, the difference would be -102. and contain 3 significant figures, unlike the original measured number which contained 2 significant figures. And if i then divide the answer by exactly 1.8, then the quotient -56.7 degrees C should still contain 3 significant figures.
Are my rules correct?

In practical work, nobody really uses or bothers about these significant number rules the way they are taught in general chemistry courses. Do not worry too much about them. They will have no significance after your course. Focus on better ideas such a statistical treatment of data. These significant figures rules originated when people used mechanical devices to make measurements.

There is a nice book, which is available on Google (hosted by Berkeley Uni) for physics. It is titled Data Reduction and Error Analysis for the Physical Sciences.

As far as rules for sig figs go, the idea is that with mult. & divison, the result has a no. of s.f. equal to the input with the least no. of s.f., and with add.&subtr., the result has a no. of s.f. equal to the input with the least absolute precision (so if one num. goes to the tenths place, and other to the 100ths, the result goes to the tenths). But exact nos. have effectively infinite precision, so don't affect the s.f. of the answer.  Hence when mult. or div. by an exact no., you retain the no. of s.f., and when adding or subtr. you retain the no. of digits before or after the decimal.

So, yes, your calculations are correct.  But your 2nd rule needs to be edited, since it only says what the answer isn't, it doesn't say what it is.  Instead, it should be:  "When adding an exact number to a measured number, the sum will be expressed with the same absolute precision (i.e., be expressed to the same number of digits before or after the decimal place) as that of the measured addend."

The sig figures rules are just  simplified handy applications of more general and more accurate rule of error propagation and additivity of variances: For addition/subtraction of 2 uncorrelated variables,  squares of absolute standard deviations are additive. The same for multiplication/division and squares of relative standard deviations.

Your confusion may lie here:  When adding and subtracting, the answer does not have to do with the number of sig figs in the starting numbers and the answer, it has to do with the "place value" of the last sig fig in each number that is being added and that determines the place value of the last sig fig in the answer:
7.0x10^1 the last sig fig is in the ones place.  32 you said was exact, so its last sig fig is in the "infinite place".  So the number with its last sig fig in the highest place value is 10 and it is in the ones place.  So, you answer must have its last sig fig in the ones place, or -102.
Example:
1.01 + 1.1011 + 1.1 = 3.2111  (3.2 by sig figs)
1.01 last sig fig in the 1/100 place
1.1011 last sig fig in the 1/10000 place
1.1 last sig fig in the 1/10 place
the highest place value of the last sig fig in any number is in the 1/10 place.  so your answer can only be good to the 1/10 place:  3.2
Again, for adding and subtracting, its PLACE VALUE not number of sig figs that determines the answer.


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why is humid air an example of liquid gas solution instead of gas gas solution?
In my textbook, humid air is given as an example to liquid in gas solution. But isn't humidity the amount of water vapour (gas) and therefore should be gas in gas solution?

A solution is a single phase by definition. So as soon as there is liquid water (droplets) in air, it is not a solution.

A asking B why C has written something wrong will not provide the answer unless B knows the history or context of such writing.

Would be nice if you post the identity of the source textbook.

Humid air is just a mixture of gases. It is not a solution (neither of a gas nor of a liquid) in the sense that a solvent (air) is needed to dissolve a solute (water).  In this sense, water vapour is not dissolved in air; it is just another gas that is mixed with the other gases in the air.
The idea of air "holding" water or being "saturated" with water is misleading. 100 % relative humidity does not correspond to a solubility limit of water in air. It just corresponds to the equilibrium vapour pressure of water at the given temperature.
The equilibrium concentration of water in the atmosphere depends on temperature. It is almost independent of the amount of air (or other gases) that is present and therefore it is almost independent of the total ambient pressure. A vacuum has approximately the same equilibrium concentration of water vapour as the same volume filled with air. At $10\ \mathrm{^\circ C}$, it is about $9.4\ \mathrm{g/m^3}$, at $20\ \mathrm{^\circ C}$, it is about $17.3\ \mathrm{g/m^3}$, and at $30\ \mathrm{^\circ C}$, it is about $30.4\ \mathrm{g/m^3}$.

Note that not always the water will be in gaseous state (in form of water vapor) in humid air. The moisture we speak of in humid air may be in gaseous state (as water vapor) or liquid state (miniscule water droplets). The former will be considered a homogenous mixture and latter a heterogenous mixture. In case of extremely humid air, the moisture will mostly be in liquid form (as water droplets) and that's when you will consider humid air as liquid-in-gas colloidal system. In less humid air, the moisture will most likely be water vapor rather than water droplets.
You can see my previous answer or this physics.SE question for more information.

"In case of extremely humid air, the moisture will mostly be in liquid form (as water droplets) and that's when you will consider humid air as liquid-in-gas solution" As you say it yourself, this is a heterogenous mixture and therefore not a solution.

@Snijderfrey it is a colloidal solution and not true solution. I have edited my answer just to be clear.


text:
How are membranes for Polymer electrolyte membrane electrolysis produced?
Polymer electrolyte membrane electrolysis is a process used for creating hydrogen with a thin metal membrane. This membrane is often includes a thin metal sheet of Platinum. My question is, how these ultra-thin membranes are produced?

Please be more clear. Are you wondering how thin platinum sheets are fabricated, specifically? Otherwise this appears to be a rather broad question. The technology undoubtedly has evolved and production methods might vary. References in the Wikipedia article you linked to provide some information on the subject, for instance https://www.sciencedirect.com/science/article/pii/S037877531630012X?via%3Dihub

If it's Pt thin films you are interested in, there is this for instance: https://www.sciencedirect.com/science/article/pii/S0360319920312532


text:
Etching polished marble?
My wife's cousin owns a marble company, so when they told us we could have free cutoff pieces, we thought it could make a cool patio.
Problem is, they're SUPER slippery.
Someone suggested pouring acid on them to 'etch' them so they'll roughen up.
I've bought and tried muriatic acid from a hardware store, and 1M sulfuric (available on Amazon, amazingly), but neither seems to do anything.
Any ideas what I can do here?

Mechanical abrasion. Sand paper and rent one of those big ones from Home Depot. Or go easy and use a belt sander, will give it a more random/natural-ish look.

Do you know anything about the polishing? There’s probably a sealant on it that prevents the acid from contacting the actual marble surface. You may be able to remove the polish with a solvent if you know what it is.

Good question... I don't know, but I will ask. Thanks!

sounds funny but is your marble "marbled" or uniform in color and material. If it has streaks of other material types in it (which is usually what makes it so attractive) then if you can find out what those other materials are then some thing that differentially dissolves one material over the other might be helpful. You can add a snapshot to your question here if you like - I think that would be quite helpful.

Any acid will dissolve some calcium carbonate. But you will still have a slippery surface. Blasting with very coarse grit may roughen it enough. It is impractical to make a non-slippery floor tile with the marble. I have some non-slip floor tile , it is very rough. It is so rough that it tore up any type sponge I tried to use to smooth grout seams. When the Amoco (Aon)  building in Chicago replaced a thousand of tons of  marble , some was made into floor tiles. It was done by crushing and making terrazzo. If there was some easy way to pour some acid on it to make floor tile, I think they would have done that. PS; the marble makes an excellent board for rolling pastry.


text:
How to theoretically determine if a hydrate of a salt precipitates out of solution
In a solid-liquid extraction simulation process, there are multiple salts that precipitate in the solution. But how do I determine if these salts are hydrates or not? To give an example:
$$\ce{Ca3(PO4)2(aq) + 3 H2SO4(aq) -> 3 CaSO4(s) + 2 H3PO4(aq)}$$
$$\ce{Ca3(PO4)2(aq) + 3 H2SO4(aq) + 6 H2O -> 3 CaSO4.2H2O(s) + 2 H3PO4(aq)}$$
To give another example, $\ce{MgSO4(aq)}$ also forms but is made solid when it is dried afterwards. It has more than 10 hydrate forms besides the anhydrous form.

When produced in water at room temperature, precipitates are always under a hydrated form. For example $\ce{CaSO4}$ is always precipitated as $\ce{CaSO4·2H2O}$


text:
What is the inherent mechanism in carbocatalysis?
When checking non-metallic catalysis materials, one mentioned example is that of carbocatalysis; using heterogeneous carbon structure to catalyze a reaction (Not to be confused as a support surface for other metallic catalysts, like palladium).
I imagine that the aromatic conjugation of pi electron's has a part to play; yet the mechanism is clearly different from established metal catalysts.

@Maurice https://en.wikipedia.org/wiki/Carbocatalysis

I am aware of the wikipedia link. Yet, there is no mention of the mechanism, just the reaction itself

The mechanism is not clearly understood. Yet, you may find some insight in this review.


text:
Descriptive explanation of different Carbon Nanotubes Conductivity
I'm soon having a talk about Carbon Nanotube FETs and wanted to give a short introduction into CNTs in general. Of course for device operation it's important that some types of single-walled CNTs are conductive while others behave more like a semiconductor. Since the talk should not be very long I don't want to dive into the tight binding calculations. So I want to know if there is a more picturesque (even if not fully correct or detailed) way of explaining the different band structures of zigzag, armchair and chiral Carbon Nanotubes. Maybe via orbital overlap or something?
Thanks in advance!


text:
Why is NMP developing a yellow color after distillation?
I got an aqueous mother liquor of N-methyl-2-pyrrolidone (NMP) from some reaction I was running and when I redistilled it under vacuum, it came out colorless but turned yellow the next day. I am checking the composition and the GC is normal at $>99.6\,\%,$ but the color is so hideously yellow.
Also, since my reaction product also sometimes gets discolored, I am having problems with it being yellow when I run the reaction with yellow NMP. I am using molecular sieves to extra dry it and that seems to just worsen the problem.
What to do to reduce/remove the color? I've tried activated carbon, Hyflo® Super Cel® filter aid etc. I just need to get rid of the color.

Search this topic in Google Patents: https://patents.google.com/patent/JPH1171346A/en

The method in the patent didn't work, but I complexed my NMP in CaCl2 solution prepared in situ via HCl and CaO and extracted NMP from the complex. It's clear now.

We have the same problem with Recycled MMA but we are not able to solve this issue? Can anyone guide uu

Okay so I have solved my problem. I did various things and here's a rundown.

Instead of Short Path Distillation Setup, I installed a 4 equivalent tray column on the recovery head. The color seemed to clear a bit and the initial distillate was colorless but over a period of 9 days, it again became hideously yellow. Also, this was the heart of the distillate and I rejected the heads and tails of 99.6% NMP (I was re-distilling my already recovered NMP)

Just before my water recovery was complete, I checked the pH of my solution and it was 4.8 so I adjusted it to 11 using CaO. I recovered almost all water and filtered the solution (It was sticky. Would not recommend) and then distilled NMP from the resulting highly viscous cake like substance. Even without a column, my water by KF was about 0.08% and NMP was colorless and pure (99.8%). It has retained its colorless appearance for over 35 days now

I dissolved my distilled colored NMP in CaCl2 in about 5 NMP to 1 CaCl2 molar ratio. The resulting solution was difficult to dissolve but it dissolved at 80C under some vacuum (100 torr absolute pressure). I then distilled NMP from this solution and it was again colorless (been colorless for about 34 days now)

So yeah. Its a strange chemical.

Why is 1-methyl-2-pyrrolidone (N-methyl-2-pyrrolidone; NMP) developing a yellow color after distillation?

NMP is a biodegradable liquid with  high power for solubilizing chemicals, and thus can be used as a safe aprotic solvent in many reactions. Yet, it is very hygroscopic and is slowly oxidized by air (Ref.1). Even though it is thermally stable, steel  parts of apparatuses used in the presence of NMP as the solvent have been reported to undergo a corrosive reaction at temperatures above $\pu{300 ^\circ C}$, which restricts its applications at higher temperatures in a metal-containing setups (Ref.1). Also, NMP has been shown to oxidize in the presence of transition metals under nontropospheric conditions by well-known catalytic mechanisms (e.g., Ref.2), as well as in the presence of UV-light and hydroxyl radical initiators. Thus, it is possible that NMP tends to oxidize in the presence of air giving the yellow color by the oxidized product(s).

What to do to reduce/remove the color?

The answer is within the question: "when I redistilled it under vacuum, it came out colorless but turned yellow the next day." I believe this color change is due to its exposure to air. According to the literature, it is better off if you use NMP freshly distilled. If it is difficult to do so, keep freshly distilled solvent in nitrogen atmosphere for the longer use (Ref.3). The following image shows the NMP resistance to oxidize even in the presence of better metal catalyst under $\ce{N2}$ (Ref.3):

Keep in mind that the work in Ref.3 has demonstrated that oxidation of NMP was shown to correlate with decreasing pH, and this image demonstrate that this oxidation can be quenched by using an inert nitrogen atmosphere.
References:

Abolghasem Jouyban, Mohammad A. A. Fakhree, and Ali Shayanfar, "Review of Pharmaceutical Applications of N-Methyl-2-Pyrrolidone," Journal of Pharmacy & Pharmaceutical Sciences 2010, 13(4), 524-535 (DOI: https://doi.org/10.18433/J3P306).
Russell S. Drago, "Homogeneous metal-catalyzed oxidations by $\ce{O2}$," Coordination Chemistry Reviews 1992, 117, 185-213 (DOI: https://doi.org/10.1016/0010-8545(92)80024-L).
Gavin Lennon, Shannon Willox, Ragini Ramdas, Scott J. Funston, Matthew Klun, Robert Pieh, Stewart Fairlie, Sara Dobbin, and Diego F. Cobice, "Assessing the Oxidative Degradation of N-Methylpyrrolidone (NMP) in Microelectronic Fabrication Processes by Using a Multiplatform Analytical Approach," Journal of Analytical Methods in Chemistry 2020, Article ID 8265054, 12 pages (DOI: https://doi.org/10.1155/2020/8265054).

I trust my Calcium Complex recovered NMP a lot since I have kept it in open air for about 30 days now. I'll do one better and place it in an open plastic and open iron container for long. I am trying to simulate industry application here so asking for nitrogen blankets and metal elimination is not a chance for me. I'll report back with the results

Mate about 5 days in, open glass 50 ml Test Tube and its retaining its transparent color. Checked and APHA 44 color. This boy holding good

It has degraded when kept in open. I have freshly distilled a clear batch today and added some additives to it. I will update on the situation.


text:
Convention for writing concentration units
1st part of my question:
Q) The concentration of an NaCl solution is 1M. Express this in percent wt. by vol. unit.
Ans:
$$\text{In %(w/v) unit}=\frac{58.5\pu g}{1000\pu{mL}}\times \frac{100}{100}=5.85\ \pu{\%(w/v)}\tag{1}$$
$$\text{In (w/v)% unit}=\frac{58.5\pu g}{1000\pu{mL}}\times \frac{100}{100}=5.85\ \pu{\%(w/v)}\tag{2}$$
2nd part of my question:
Q) The concentration of an NaCl solution is 1M. Express this in ppm unit.
Ans:
$$\text{In ppm(w/v) unit}=\frac{58500\pu{mg}}{1\pu{L}}=58500\ \pu{ppm}\tag{3}$$
$$\text{In (w/v)ppm unit}=\frac{58500\pu{mg}}{1\pu{L}}=58500\ \pu{ppm}\tag{4}$$
My questions:

Which of $(1)$ and $(2)$ is correct?
Also, the unit of molarity is $\pu M$, or the unit of mass is $\pu {kg}$. Similarly, what is the unit in the first part of my question? Is it $\text{%(w/v)}$? Is $\text{%(w/v)}$ a unit just like $\pu{kg}$ and $\pu{M}$?
Which of $(3)$ and $(4)$ is correct?
Also, is it okay if I append/suffix (w/v) at the end of ppm? For example, if I say the concentration of NaCl is $58500\ \pu{ppm(w/v)}$ instead of saying that the concentration of NaCl is $58500\ \pu{ppm}$, will it be okay?

By not explicitly presenting own attempts to answer it, you put the question in closing danger.

None of the notations, ppm (w/v) or (w/v) ppm, is a standard notation. I don't see this used much in analytical chemistry publications. Very rarely, you might see ppm (w/v) in some papers. If you are doing your own calculations, or writing a report, one should clearly explain how you are expressing concentrations. The extra label (w/v) is not needed because it is understood that for aqueous solutions it will be w/v (for convenience). For water, ppm by w/v and ppm by w/w are almost identical.
The mg/ L approximation for ppm does not apply very well to solutions whose density is significantly different from 1. One should be careful about it.
The International Union of Pure and Applied Chemistry book '"Quantities, Units and Symbols in Physical Chemistry" says

"Deprecated usage Adding extra labels to ppm and similar symbols,
such as ppmv (meaning ppm by volume) should be avoided. Qualifying
labels may be added to symbols for physical quantities, but never to
units.

Thank you so much for your answer! Also, could you please tell me which of the statements $(1)$ and $(2)$ seem more correct to you?

@tryingtobeastoic, 5.85 w/v % looks fine without brackets.

Understood. Also, should I write
$$\text{In % w/v unit}=\frac{58.5\pu g}{1000\pu{mL}}\times \frac{100}{100}=5.85\ \pu{% w/v}\tag{1}$$
or
$$\text{In w/v % unit}=\frac{58.5\pu g}{1000\pu{mL}}\times \frac{100}{100}=5.85\ \pu{% w/v}\tag{2}$$

2 is better because you can read it as weight by volume percent. Don't worry about these minor things...there are no hard and fast rules as long as you are clear.


text:
Two atoms exist in the same coordinate position in the lattice?
I am trying to simulate the properties of FeF3(H2O)3, so I download its crystal structure file from Crystallography Open Database, but it seems that in the lattice, the O atom and F atom exist at the same coordinate position.
It also shows the F atom and H2O are at the same coordinate in the original paper, which measures the structure of FeF3(H2O)3,
Here is the chemical formula information of FeF3(H2O)3. it seems the H atoms would be lost when forming a solid crystal.
Chemical name   Fe F3 (H2 O)3
Formula F3 Fe H6 O3
Calculated formula  F3 Fe O3

My question is:
Is it possible that two atoms exist at the same coordinate in the lattice?

That's what we call "disorder". In some unit cells, there is an O, and in some an F.

The paper states that four ligands of the iron are occupied by a 50:50 statistical mixture of water and fluoride (the double circles). They could have added the partial occupancy to the coordinates if the file format had allowed it:

Because fluoride and water have distinct net charges, the stoichiometry is fixed, and there is probably some local order to maintain local charge neutrality as well, but not at the atom-by-atom level.
The hydrogen atoms might not be resolved using the method employed here, but they certainly are part of the crystal.

Ivan Neretin sums it up well in his comment. There can be either one type of atom or molecule or another at the relevant position in the unit cell.
It's actually fairly commonplace. Consider an ordinary mild steel, which consists of the metallic ferrite phase reinforced by a hard, semi-ceramic cementite (carbide) phase. Typically the steel will contain manganese alloyed with the iron. Then in the body-centered ferrite phase each lattice point could have either an iron atom or a manganese atom (or an atom of any other metal added to the alloy), and the same holds true for the metal-atom positions in the orthorhombic cementite phase. We could render this by labeling each metal position as $\ce{(Fe,Mn)}$ (with additional elements as necessary for the alloy).


text:
pH for optimal kinetic for enzyme reaction
If I am given the following enzyme reaction:

where K$_{EH,1}$=K$_1$ = $10^{-4}$ and K$_{EH,2}$=K$_2$ = $10^{-8.2}$. Iwant to calculate at which pH I have the optimal kinetics by calculating $\frac{v'_{max}}{v_{max}}$ at different pH levels (this is done in Matlab with the code pH=0:0.1:14; K1=10^-4; K2=10^-8.2; v=1./(1+K2./10.^-pH+10.^-pH./K1); [m,n]=max(v); pH(n))
But if the rate equation is:
$v= \frac{k_2 [E]_0 [S]}{K_M(1+ [H+]/K_1 + K_2/[H+])  + [S]}$   (1)
Then how do you get v'$_{max}$?
From a similar problem, the rate equation was:
$v= \frac{k_2 [E]_0 [S]}{[S](1+ [H+]/K_1 + K_2/[H+])  + K_M}$ (2)
So then knowing that v$_{max}$ = $k_2 [E]_0$ when [S]>>K$_M$, you get from eq. (2) that v'$_{max}$ is:
$v'_{max}= \frac{v_{max}}{1+ K_2/[H+] + [H+]/K_1}$
However in my case, by doing this to eq. (1) then I will only be left with $v= k_2[E]_0$ which I can't use in the same way as is done for eq. (2).
Is there another way to calculate the pH or how should I get v'$_{max}$? All help is appreciated!

I don't think the rate equation is

$v= \frac{k_2 [E]_0 [S]}{K_M(1+ [H+]/K_1 + K_2/[H+])  + [S]}$

I suspect it is rather:
$$v= \frac{k_2 [E]_0 [S]}{(K_M  + [S]) (1+ [\ce{H+}]/K_1 + K_2/[\ce{H+}])}$$
The term $(1+ [\ce{H+}]/K_1 + K_2/[\ce{H+}])^{-1}$ describes the fraction of the enzyme in the productive protonation state.
You don't have to plot this to figure out the optimal pH. The optimal pH is always at the midpoint between the two $\mathrm{p}K_\mathrm{a}$ values. That is because that is the pH value with the highest fraction of the once-protonated enzyme.


text:
Reactivity vs Stability: Chlorine Trifluoride
Chlorine trifluoride shows up on many pop science content, which is how I learned about it just today, as an incredibly reactive compound capable of oxidizing seemingly incombustible substances (glass, concrete, ashes, asbestos). One source explicitly mentions that it is more reactive than fluorine, the strongest oxidizer of all elements.
I got confused when I read that chlorine trifluoride can be produced by heating chlorine and fluoride together. From what I could gather, the reaction is also exothermic. So if I hadn't known that chlorine trifluoride was a nightmare fire chemical, I would've assumed it was more stable than chlorine or fluorine.
I think the confusion might be down to the terms reactivity and stability. Some sources will treat them like antonyms, but some treat them as a kinetic and thermodynamic property, respectively. Is the latter the proper explanation, that chlorine trifluoride is thermodynamically more stable but it takes less energy to break its bonds and react with other chemicals? It sounds reasonable, but I'd like some clarification.

Well, fluorine is perfectly stable ( no tendency for spontaneous changing to something else ) and extremely reactive ( tendency to react with other stuff). Generally, both stability and reactivity can be considered from both thermodynamic and kinetic point of view.


text:
What is the mechanism of sodium reacting with water vapor?
We all know that
$$\ce{2Na +2H2O(g)->2NaOH +H2}$$
The question is:
What is the mechanism (not the reason that sodium is more 'electropositive' or more reactive than hydrogen, as in this question; and not in liquid phase involving $\ce{H+}$, which is mentioned also in that question) of this reaction?
The question of why sodium reacts with water is a thermodynamics question, but the question of how is a kinetics / mechanistic question. It is known that solvated electrons are very stable in liquid ammonia, but in water they are more shortived. From experiments such as pulse radiolysis it is known that solvated electrons in water react with solavted protons to form hydrogen atoms which then can combine to form hydrogen gas. If at the surface of a lump of sodium in water we have electrons being cast off into the water then we will have a source of solvated electrons and also sodium cations.
But if a sodium surface is exposed to water vapour then the reaction will be different as we can have water moelcules on the surface rather than bulk water. What will the mechanism be under these conditions.
See A Non-Exploding Alkali Metal Drop on Water: From Blue Solvated Electrons to Bursting Molten Hydroxide, Dr. Philip E. Mason,Tillmann Buttersack,Prof. Sigurd Bauerecker,Prof. Pavel Jungwirth, 2016 https://doi.org/10.1002/anie.201605986
Is it the same as the following two reactions? If not, what's the difference?
$$\ce{3Fe + 4H2O(g) ->[\Delta]Fe3O4 + 4H2}$$
$$\ce{Zr + 2 H2O(g) ->[\Delta]ZrO2 + 2 H2}$$

Note that element symbols in ch. reactions should be upright, using mhchem package ( \ce{} )//  Useful links for text and formula formatting:
Notation basics ,  Formatting of math/chem expressions  and
upright vs italic

Try something like this:
$$\ce{2 Fe + 3 H2O(g) ->[\Delta]Fe2O3 + 3 H2}$$
$$\ce{2 Fe + 3 H2O(g) ->[\Delta]Fe2O3 + 3 H2}$$

Related: https://chemistry.stackexchange.com/questions/132500/how-to-predict-what-metals-form-when-they-react-with-water

Fe reacts with water to give Fe++ not Fe+++; Steam gives Fe3O4.  It seems to get to Fe2O3 O2 is needed.

@jimchmst thanks for the correcting.

Solvated electrons do not exist in gas phase.
In the water vapor, the reaction is simply an electron transfer from the sodium atom $\ce{Na}$ to one of the $\ce{OH}$ bonds of the $\ce{H2O}$ molecule. Then it is then broken into an $\ce{H}$ atom and an $\ce{OH-}$ ion. Later on, two $\ce{H}$ atoms are joined to yield a $\ce{H2}$ molecule.


text:
Converting benzaldehyde to benzophenone in no more than two steps
This is a question from my book, as stated in the title:

How will you convert benzaldehyde to benzophenone in no more than two steps?

I can do it in three steps - first using  $\ce{PhMgBr}$ on benzaldehyde to get $\ce{Ph_2CH(OMgBr)}$ (sorry for this symbolic mess, I don't know how to format organic synthesis) followed by hydrolysis to $\ce{Ph_2CHOH}$ which is oxidised to $\ce{Ph_2CO}$.
I tried thinking of other ways, but really couldn't find anything. Hints would be appreciated. Thanks!

Hydrolysis is generally done one pot as a follow-up to the Grignard reaction. It's not usually considered a separate step.

I agree with @Zhe

@Zhe: I thought of that, but like still does there exist some synthesis without this "follow-up" thing?

Pretty much all reactions need some kind of follow-up to remove excess reagents/side product and isolate the desired product

In practice, no, because reactions are never as clean in glassware as on paper. Generally, for best results, you must use an excess of at least one reagent and reactions don't go to completion, so you'll need to follow-up the reaction with isolation and purification of the desired product.

I could give you a satisfactory answer, but I don't say that this is the best method.

Conversion of benzaldehyde to benzoyl chloride:You can achieve this by adding $\ce{Cl2}$ or $\ce{SO2Cl2}$ (or other suitable chlorinating agents) to benzaldehyde. For more details refer this.
By Friedel-Crafts acylation: Reaction between benzene and the above obtained benzoyl chloride in presence of Lewis acids like $\ce{AlCl3}$ gives you benzophenone.

Related to step 1: https://chemistry.stackexchange.com/questions/42471/synthesize-acyl-chloride-from-aldehyde

Thank you! This "aldehyde to acyl chloride" I thought of it but couldn't find a way to do this.


text:
Does carboxyl group interfere with reaction of 2-carboxybenzaldehyde with NaCN/HCl?
The aldehyde group of 2-carboxybenzaldehyde reacts with $\ce{NaCN/HCl}$ forming cyanohydrin. Does the carboxyl group remain intact? Does it react separately or with the cyanohydrin?
I suspect possible intervention of the lone pair from the $\ce{-CN}$ group to form an aromatic ring. Does it (or something else) happen or not?

formation of the 3-cyanophthalide looks like a possibility.

@Waylander: If it happens, it should proceed by protonation of the $\ce{-OH}$ and then attack of the carboxylic acid, removing $\ce{H_2O}$ right? (That's what I was thinking in the first place)

That seems a reasonable mechanism. I do not have database access to try to find an example, maybe someone else who has can give a definitive answer.

@Waylander: The reaction is known to give what you predict but with a twist. The crude, presumable cyanohydrin, is isolated. The residue is treated with DCC to effect lactonization. Chem. Absts. is unaware of the cyanohydrin. [Y. Li, A. Linden, M. Hesse, Helv. Chim. Acta, 2003, 86, 579.]

@Waylander's suggestion that 3-cyanophthalide [3-oxo-1,3-dihydroisobenzofuran-1-carbonitrile] 6 is expected as the product of the reaction is well-reasoned. This reaction has been reported by Hesse, et al.[1] The reaction of aldehyde 1 in the presence of sufficient aq. HCl to generate HCN and enough to act as a catalyst for the reaction does not provide 6 directly. The cyanophthalide 6 is not formed under the conditions employed but, seemingly, the cyanohydrin 2, which was isolated by extraction with ethyl acetate. The crude material was lactonized with dicyclohexylcarbodiimide [DCC] 3 as shown in the steps 2 $\rightarrow$ 6.

Y. Li, A. Linden, M. Hesse, Helv. Chim. Acta, 2003, 86, 579.

I understand this, but in absence of 3, would it not be possible for the hydroxy group in cyanohydrin to get protonated, followed by intramolecular attack of the carboxylic acid?

Perhaps under some set of conditions but not likely in my opinion. The facts are what they are!


text:
Substitution v/s elimination
For SN2 pathway we use an strong nucleophile (OH-,CN-) so they can easily substitute X(halogen) group from rear side.
While for elimination bimolecular reactions we use a strong conjugated base (RCH2O- or alc.kOH) so that it takes proton from carbon  near by to halogen,thus forming an alkene .
While in my text book I had found

I was agree with first diagram stating that bulky group will more tend to show elimination rather that substitution but in second diagram they show ethoxide ion giving substitution reaction .
To which I disagree, According to me it be should also show elimination reaction.

Welcome to CH SE site!
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@Poutnik I have issue with the diagram  part of it . Text is only for giving context of what I'm talking about.

I see. It has looked for me like the essential part of the question.

If would like to know what you take on this problem.

I would think you get a mixture of both, but I would be more biased towards the elimination than substitution.

So ,as per concept I should sum up that in either of the situation elimination should be major product .

The right scenario probably depends a lot on solvent. My guess is that someone probably carried out this study in the 30s and there's likely a reference.

Check out the Williamson ether synthesis, conditions [and reactants] can be adjusted to some extent to affect the elimination substitution rates.

I think both elimination and substitution reaction will occur any of them can dominate cos ethoxide ion is a strong nucleophile and in presence of a alcohol as a solvent. So the reaction which will dominate depends upon solvent which is not mentioned. Moreover it is also a 2°alkyl halide so solvent is much more important factor.


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Resource estimate for ab initio calculation
I have recently come across this very intriguing paper “The Fundamental Vibration of Molecular Hydrogen”.
I have asked the authors the following naïve questions directly via email?

What was the amount of computer memory used by the full ab initio calculation?
What was the time-to-solution of the full ab initio calculation?

While I wait for their response, could anyone help me to make a ball park estimate of those two answers?

The folks over at Matter Modeling might have estimates.

I’m voting to close this question because it's been cross posted and answered on matter modeling


text:
Calculating zero-point energy for the Varshini potential
As per my understanding, for a harmonic oscillator, we can find the zero-point energy by the relation below:
$$ E_0 = \left(0 + \frac{1}{2}\right)\hbar \omega$$
The equation for the potential in question is:
$$ V(x) = D_e \left(1 - \frac{r_e}{x} e^{-\alpha(x^2-r_e^2)}\right)^2 + T_e$$
How would one go about finding the value of $E_0$ for this potential surface?
Edit: All values aside from $x$ are known.

Can you explain what these symbols mean...?

You should look at the paper by Varshini in Reviews of Modern Physics, vol 29, 664 1957 where this function is described for diatomic molecules along with many others.


text:
Is potassium permanganate an oxidizing agent or is the permanganate ion the oxidizing agent?
I have seen both potassium permanganate and the permanganate ion being called the oxidizing agent. Isn't it more appropriate, however, to call the permanganate ion the oxidizing agent only as the potassium ion doesn't play any role in oxidizing?
Similarly, is potassium dichromate an oxidizing agent or is the dichromate ion the oxidizing agent?

It may be more precise to specify that it's the permanganate ion, but it's not wrong to say that KMnO4 is an oxidising agent. At the end of the day, I don't think anybody cares enough about it to make this distinction. You can't have permanganate without a counterion anyway.

Permanganate with any counterion is an oxidising agent as it is the manganese that is the active agent. Similarly with dichromate.

In context of substances, KMnO4 and not MnO4^- is the oxidizing agent. In context of ions, MnO4^- and not KMnO4 is the oxidizing agent. So, choose the context. The same for chromium.

@orthocresol I just realized that one might argue further that not MnO4-, but rather Mn is the oxidizing agent.

P.S.: ... and there is the 3rd context of elements changing oxidation number, where the respective element is the oxidizing agent.

In the context of substances, $\ce{KMnO4}$ and not $\ce{MnO4-}$ is the oxidizing agent.
In the context of ions, $\ce{MnO4-}$ and not $\ce{KMnO4}$ is the oxidizing agent.
In the context of elements changing their oxidation number, $\ce{Mn^{VII}}$ is the oxidizing agent.

So, choose the context.
The same respective contexts are there for the dichromate case.


text:
Why did my book write S2- if hydrogen sulfide isn't ionic?
Problem:
Balance the following reaction using the ion-electron method:
$$\ce{KMnO4 + H2SO4 + H2S -> MnSO4 + S + K2SO4 +H2O}$$
My book's solution:
Oxidation half-reaction:
$$\ce{S^{2-}->S + 2e^{-}}\tag{1}$$
$$...$$
My question:

Is writing $(1)$ correct? I'm asking this because $\ce{H2S}$ is not an ionic compound. How can my book write $\ce{S^2-}$ then?

Your book is wrong. And you are right. In acidic solution $\ce{H2S}$ is not dissociated, and not deprotonized. It reacts as $\ce{H2S}$. Anyway the ion $\ce{S^{2-}}$ is nearly inexistant even in rather basic solution. The main ion produced by $\ce{H2S}$ in basic solution is $\ce{HS-}$

It is indeed better to render (1) as
$\ce{H2S -> S +2H^+ + 2e^-}.$
Since you did not ask, I assume you know the permanganate reduction equation. When you combine this reduction with the correctly written reaction for hydrogen sulfide oxidation in acid  you should get
$\ce{5 H2S + 2 MnO4^- + 6H^+ -> 5 S + 2 Mn^{2+} + 8 H2O,}$
where the net hydrogen ions on the left side are interpreted as coming from the sulfuric acid (which is not oxidized or reduced, so does not appear directly in the half-equations). You should then infer that the balanced ratio of reactants is $\ce{5H2S:2KMnO4:3H2SO4}$.


text:
How to find pH for acetate buffer?

Calculate the pH of the solution that results when $\pu{40.0 mL}$ of $\pu{0.100 M}$ $\ce{CH3COOH}$ is mixed with $\pu{30.0 mL}$ of $\pu{0.200 M}$ $\ce{NaCH3COO}.$ $K_\mathrm{a}(\ce{CH3COOH}) = \pu{1.8E-5}.$

Can I use Henderson–Hasselbalch equation to solve this?
It would be
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{CH3COOH}]}{[\ce{NaCH3COO}]} \quad\implies\quad \mathrm{pH} = 4.44,$$
but if I set up the ICE table
\begin{array}{rrrrr}
\ce{&CH3COOH &-> &H+ &+ &CH3COO-}\\
\text{I} & 0.100      && 0  && 0.200 \\
\text{C} & -x         && +x && -x \\
\text{E} & 0.100 - x  && x  && 0.200 + x
\end{array}
and solve for $x,$ it gives me a different $\mathrm{pH}.$
The volumes threw me off and make me think that I have to find the total volume and new molarity before solving for $\mathrm{pH}.$

Yes, you should take into consideration the effect of dilution when you mix the two. Note however that your could also consider the molar ratio of acid and conjugate base. This should become obvious if your write the HH equation out explicitly.

Calculate the amount of acid and of acetate in moles. And then divide by the total volume. The use the HH equation.

Thank you, Maurice. That actually makes a lot of sense. Does my equation look right?

Shouldn’t it be $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{NaCH3COO}]}{[\ce{CH3COOH}]}$$?


text:
Purifying ferrous sulfate
For my wood staining experiments I have purchased 1 kilo of ferrous sulphate as a source of iron(II). It turned to be quite oxidised (yellowish) and the its solution is muddy yellow instead of green. Does it make sense to try to recrystallise it in a hope that oxidation products are less soluble or insoluble in water?

No, it probably doesn't. You could try to reduce it, though.

Ferrous sulfate solutions are unstable with respect to oxidation in water at neutral pH. No cure for the brown stuff. The oxidized products are already insoluble in water.

BTW, podzol, the soil of pine forests, is often rich in ferrous iron, and when first exposed, beneath a layer of forest duff, it's first a pale green. Within fifteen minutes, it darkens to a reddish-tan as the iron is oxidized by the air. as @AChem states, most ferrous salts are not air-stable.

How about filtering it? Ferric ions tend to aggregate; it might be easy, even if the rate is slow, to get what you need as you need it.

If you were to dissolve it in dilute sulfuric acid then shake it with a solution of DEHPA (bis-2-ethylhexyl hydrogen phosphate) in an aliphatic kerosene then I am sure that the Fe(III) will extract while the Fe(II) will stay in the aqueous solution. You would have to work out a method of keeping the oxygen away from the purified product. You might want to read about the Fricke dosemeter. It so happens that radiation increases the rate at which ferrous sulfate reacts with oxygen

You better make your turbid solution boil with some iron wires $\ce{Fe}$ and a bit of sulfuric acid. The turbidity observed in your solution is due to a mixture of two precipitates containing $\ce{Fe^{3+}}$ ions, produced by the oxidation reaction $$\ce{4 Fe^{2+} +  O2 + H2O ->  4 [Fe(OH)]^{2+}}$$ then $$\ce{[Fe(OH)]^{2+} + 2 H2O ->  Fe(OH)3(s) + 2H+}$$  or $$\ce{[Fe(OH)]^{2+} + SO4^{2-} ->  Fe(OH)SO4}(s)$$ Whatever the chosen reaction, both are producing a turbid solution. And this turbid solution may be destroyed to produce $\ce{Fe^{2+}}$ ions if boiled with some iron wires or nails dipped into the acidified solution, which will reduce ferric substances into ferrous ions, according to : $$\ce{2 Fe(OH)3 +  Fe + 6 H^+ ->  3 Fe^{2+} + 6 H2O}$$ or $$\ce{2 Fe(OH)SO4 +  Fe + 2 H^+ ->  3 Fe^{2+}  + 2 SO4^{2-} + 2 H2O}$$ In both cases, the precipitate has disappeared and all the iron in solution is under the form of ferrous ions $\ce{Fe^{2+}}$ and only these ions


text:
Should I write formula for oxalate or oxalic acid while balancing a redox reaction using the ion-electron method?
Problem:
Balance the following reaction using the ion-electron method:
$$\ce{H2C2O4 + KMnO4 + H2SO4 -> CO2 + MnSO4 + K2SO4 + H2O}$$
My book's solution:
Oxidation half-reaction:
$$\ce{C2O4^2- -> 2CO2 +2e^-}\tag{1}$$
$$...$$
My question:

Should oxalate ion $(\ce{C2O4^2-})$ be written in $(1)$? I'm asking this because oxalic acid $(\ce{H2C2O4})$ is a weak acid and doesn't dissociate/ionize much into $\ce{H+}$ and $\ce{C2O4^2-}$ ions. So, should I write $\ce{H2C2O4}$ instead of writing $\ce{C2O4^2-}$? To elucidate $$\ce{H2C2O4->2CO2 + 2H+ +2e^-}\tag{2}$$ Should I write $(2)$ instead of $(1)$?

Related

Both 1 and 2 are fine. Since you are dealing with "ion" electron mehod, follow your book.

It is nearly identical case as your recent question is-potassium-permanganate-an-oxidizing-agent-or-is-the-permanganate-ion-the-oxid // Note that oxalic acid is very strong weak acid with pKa1=1.25.

One should report the species which is the most important one in the given system. As oxalic acid has two $\mathrm{p}K_\mathrm{a}$ values ($1.23$, and $4.19$), it means that you should report the molecular formula $\ce{H2C2O4}$ if the pH is lower than $1.23$. You should write the ionic formula $\ce{HC2O4^{-}}$ is the pH is between $1.23$ and $4.19$. It is improbable that you should work at a pH greater than $4.19$, because the reaction requires $\ce{H2SO4}$ to proceed, and this acid is not supposed to be present at such a low concentration (about $\pu{10^{-5} M}$).
Please note that, whatever your choice, the final equation would be the same, as your equation has no ions.


text:
How can I increase dissolved nitrogen in flowing water?
Will injecting pure nitrogen gas using a diffusion stone increase the dissolved nitrogen content in water while the water is flowing through a pipe?

Yes it should. Look up Henry's law.

If it's already saturated, no.


text:
Can eating soap protect you if you ingested radioisotopes?
In “Pass the Vegetables, Please”, the 71st episode of Gilligan's Island, Gilligan and the gang all ate seeds of radioactive vegetables, and the professor suggested eating soap. He says the hydrocarbons absorb radiation:

Professor:
I've applied the principles of theoretic chemistry to the problem of radioactivity. Now the best preventative for the internal bombardment of radioactive particles is through their containment by hydrocarbons.
Gilligan:
Would you mind repeating that?
Professor:
Not at all. I've applied the principals of theoretic chemistry to the problem of radioactivity.
Skipper:
Oh come on Professor, never mind all the scientific lingo.  What are you trying to tell us?
Professor:
That our best protection against the vegetables we've eaten is through hydrocar…  well that is to say through plant fats. And the best source of plant fat is the soap we've been making here on the island.
Gilligan:
Do you mean we have to take a bath with that soap?
Professor:
No we shall have to consume it internally.
Gilligan:
Oh thank goodness. For a minute there I thought we were going to have to eat it!

Does that have any basics in reality?

It would cause diarrhea and help purging the digestive system. OTOH, it would left it in desolate state biochemically.

If only it were that easy we should not have to worry about radioactivity at all.

Gilligan and the gang did end up getting rescued, none the worse for wear. Notably, they were all healthy 14 years later when the sequels were produced. I thereby infer that the Professor was correct.

Some of these sentences are non-sense. For example : Now the best preventative for the internal bombardment of radioactive particles is through their containment by hydrocarbons. First remark : there exist three types of particles coming from the radioactive substances : 1) the alpha particules, made of 2 protons and two neutrons, which can cross about 0.1 millimeter matter (and will have an effect only over this distance. 2) the beta particules, made of usual electrons, which can travel and cross water and human tissues about one centimeter thick, and this value is changing a lot with the energy of the beta ray. 3) the gamma rays, which are similar to the light, although with a much higher energy ; they can cross water and human tissues many meters thick, although this exact value changes with the energy of the ray. Some radioactive isotopes are emitting alpha, beta or gamma rays with high energy, and some with a low energy. So their effect depends mainly of their energy. It does not depend so much on the target (hydrocarbons or soap or living tissues). So the sentence printed in italics does not have any meaning.
Edit : As a consequence, the question was : Can eating soap protect you if you ingested radioisotopes ? The answer is : NO !
Why are some of the readers criticizing my explanations ? They are all scientifically correct. Maybe they are oversimplified. I could of course develop them, but it would take time. I am a radioprotection expert in my country.

Well, I like your explanation. The sentences are appropriate for a comedy show for scientific illiterates. However... if they had cancer, maybe the radioisotopes could cure them, and maybe that's why they lived so long.

I think the issue is that you aren't answering the question directly, you've just picked on that one sentence. Now, it does kind of indirectly answer the question, but maybe being more explicit would help.


text:
Corruption of TLC plates or solvent
I have been conducting TLC experiments, and sometimes I observed some lines when I use mixture of eluents.

And I found that some lines are observed after I develop the TLC plates with some eluents. And these lines are detectable via both 254 nm and 365 nm UV.

The plates might be corrupted when I cut it, so I test a small piece of unused plate. In this time I didn't develop the plate, rather immersed it quickly. Then the borderline was detected by UV. Even, the borderline turned yellow after it is immersed in methanol or ethyl acetate.

Are the plates corrupted due to long-term storage? These plates existed before I joined the lab, probably stored more than 2 years. I test the unopened solvents although they are stored for some periods after purchase.

I have noticed it in past regularly, if the eluent was mix of very polar minor solvent and much less polar major solvent. It seemed to me like imporities of solvent, plate sorbent binder or adsorbed  stuff over the time.


text:
Hund's Rule for Determining Term Symbol Energy Order
In order to determine the energy order based on J-value for term symbols we employ Hund's rule:

In a less than $\frac12$-filled subshell Lowest J-value is Lowest energy
In a more than $\frac12$-filled subshell Highest J-value is Lowest energy

For configurations where only one orbital is involved such as $p^5,d^3$ etc. this is easy to apply. But what about configurations where two orbitals are involved such as $p^5d^1$ or $d^9f^1$? In these cases to which orbital do we apply the less/more than $\frac 12$-filled condition?


text:
Tmax of parallel reactions/Steady state reactions

Below are four reactions, to this experiment inhibitor is added so that k$_3$ = $0$ (but the other reactions are unaffected) calculate [D] when the reaction is over. The initial concentration of A is $0.016$M.
A → B              k$_1$= $0.110$ s$^{-1}$
A → C              k$_2$= $0.025$ s$^{-1}$
B → C               k$_3$= $75.2$ s$^{-1}$
C → D              k$_4$= $17.2$ s$^{-1}$

I am having trouble understanding how to solve this problem. I know the first step is to calculate the time $t$ however I do not know how to apply what I know from other problems to this one.
I know that t$_{max}$ will be when dD/dt = $0$, and if I remove reaction $3$, then I get that dD/dt = k$_2$[A]. This is where I get confused.
Should I now replace [A] with the integrated equation of A from the reactions, meaning:
-dA/dt = [A](k$_1$ + k$_2$) => ln[A]-ln[A]$_0$ = -(k$_1$ + k$_2$)t => [A]=[A]$_0$e$^{-(k_1 + k_2)t}$
But then if I substitute that into dD/dt = k$_2$[A] =$0$ then I can't solve for $t$
How would one solve this?
Thank you!
EDIT
Have I derived the wrong expression for dD/dt?
I get that: dB/dt = $k_1$[A], dC/dt=k$_2$[A]-k$_4$[C], dD/dt= k$_4$[C]
However, here I have to set both dB/dt and dC/dt equal to zero. So that means that [A] has to be zero and that would in turn lead to [C] also having to be zero. So how should the expression be derived?

Is this the same problem as before? Or did I misread the question the first time? If you write the full reaction, and the the reaction goes to completion, what is the concentration of D? Isn't that just stoichiometry?

@Zhe this is not the same question as before, then a time was given and I think I managed to solve it. Also I don't quite understand what you mean, here I have to calculate a time t in order to determine the concentration of D when then reaction goes to completion.

Why? Is not completion determined fully by the amount of A? Why are we even bothering to treat this as a kinetics problem?

Oh, I see because k3 = 0 now...

You won't be able to solve for a time when dD/dt = 0. It should be $t = \infty$. To first order, A is partitioned between B and C. There is no backward reaction, so the relative rates determine what becomes B and what becomes D.

@Zhe But won't D reach its maximum value at the completion of the reaction? In other problems I have seen that dD/dt=0 is used to derive an expression for tmax.. it seems this cannot be used here but how do I use t = infinity?

You can't use steady-state approx on B because it is a product if k3=0 not an intermediate.

But won't D reach maximum when A is consumed? You could also just set $t=\infty$ and evaluate. That should just work.

@Andrew Oh yeah of course, so then only C is an intermediate. In that case my expression for dD/dt seems to be correct..

@Zhe I'm sorry but I don't understand how to "evaluate" it. What equation should I use if I want to evaluate it when t = ∞? I apologize if this is an obvious solution but I am really struggling to understand it all

Based on what you wrote, you can evaluate [A] by taking the limit $t \rightarrow \infty$. If it's not clear, I don't think I understand what you're asking.

The amount of D produced eventually is the fraction that goes to C initially, i.e. $k_2/(k_1+k_2) $ and as C only goes to D the amount of D is that initially in A, i.e $A_0$ multiplied by the fraction $k_2/(k_1+k_2) $. You should also realise that C appears with the rate constant $k_1+k_2$  although you do not need this.

@porphyrin Thank you it became much clearer now!


text:
Is hydrolysis of acid chlorides still favored when the resulting acids cannot donate their protons?
For example, take the reaction:
MeCOCl + H₂O -> MeCOOH + HCl
Bonds broken: C-Cl + H-O -> 330 + 460 = 790 kJ/mol
Bonds formed: C-O + H-Cl -> 350 + 432 = 782 kJ/mol
I realize that BDE is only an estimate and does not take into account the fact that HCl (and MeCOOH to some extent) will undergo heterolysis and donate a proton to water (or whatever other species is available to accept it), forming a stable anion. I assume it is this process that makes the reaction favorable, correct?
If so, then would this reaction in the gas phase favor the MeCOCl + H₂O side?
I will admit I don't immediately recognize the products as having more favorable solvent interactions when considered as neutral species:

H-Cl makes weaker hydrogen bonds than H-O (bad)
O is more electron-withdrawing and leaves C more electron-deficient (bad)
but creates a stronger dipole (good)
and less diffuse electron density because it is smaller than Cl (good)
but HCl has a weaker dipole than H2O (bad)
and less concentrated electron density (bad)

This is why I ask how important the ability of MeCOOH and HCl to donate their proton is for hydrolysis to be favorable.
How would using an aprotic solvent affect the equilibrium?

BDE's are for homolysis, not heterolysis.

Yes, but it should still be a valid indicator for the relative stability of each species, shouldn't it?

"it should still be a valid indicator for the relative stability of each species, shouldn't it?" Absolutely not. You're completely ignoring all of the components that depend on charge, and that's particularly important, particularly in a protic or polar solvent.

I'm aware that these effects exist, that's why I was asking about the reaction in the gas-phase, where they become irrelevant.

(additional info moved to question)

https://chemistry.stackexchange.com/questions/164729/is-hydrolysis-of-acid-chlorides-still-favored-when-the-resulting-acids-cannot-donate-their-protons

@user123795 ...?

The data more or less explain why acid chlorides are made by reaction of acids with PCl5, SOCl2 and similar compounds rather than anhydrous HCl. In water or similar solvents [see Schotten-Baumann method] removal of the H by a base definitely promotes the reaction. I have often wondered Why what seem to be very reactive compounds when one is exposed to them are not so active in anhydrous reactions and why the need for the base. The final driving force is the reaction with the strong acid HCl or, in the case of sulfonyl chlorides the sulfonic acid. The same thoughts explain the lower reactivity of acid anhydrides.


text:
Is RCOOH + H2SO4 + heat a viable alternative to RCOCl + AlCl3 in Friedel-Crafts acylation?
Can concentrated strong acid like H₂SO₄ be used to protonate a carboxylic acid to RC(OH)₂⁺, forming RCO⁺ by removal of H₂O, analogous to the reaction with nitric acid to produce nitronium?
Would this be a viable way of acylating an aromatic compound?

Sure it is possible.
It is best known for intramolecular cyclisation using Polyphosphoric Acid (PPA) or Methanesulfonic acid more here.
Examples are known of intermolecular Friedel-Craft acylation using strong acid on a carboxylic acid substrate - PPA is used in this example


text:
H+ concentration decreasing over time reasoning in weak acid strong base titration
We know that, when $\ce{CH3COOH}$ is just present in water, it is in equilibirum with water and having some $\ce{H+}$ dissociated to give some $\mathrm{pH}$ to it. Now when we start adding $\ce{NaOH}$ to it, we say as $\ce{OH-}$ is having tendency to react with $\ce{H+}$ formed intially, so now onwards $\ce{H+}$ starts decreasing, but by Le Chatlier this means weak acid should increase its dissociation. The $\mathrm{pH}$ is given by $\mathrm pK_\mathrm a + \log\frac{\ce{[salt]}}{\ce{[acid]}} $. But how to show that net $\ce{H+}$ concentration will decrease  at any time? So that $\mathrm{pH}$ will keep on increasing, as its not easy to see that $\ce{H+}$ conc will decrease over time from just the two chemical equations given by -
$$ \ce{CH3COOH <<=>  CH3COO- +  H+}$$
$$ \ce{H+ + OH- <<=> H2O}$$
Because forming rate of $\ce{H+}$ from $\ce{CH3COOH}$ is different from reacting rate with $\ce{OH-}$ isn't?

See https://chemistry.stackexchange.com/a/27802/79678.

There i am not getting how there are two more equlibirum setup being considered , also there also its being said that as H+ is consuming , so it makes H+ decreasing , which is not answering my question :( @EdV specially my reason of rate thing which is not very obvious

Useful links for text and formula formatting (not to be applied to titles):
Notation basics ,  Formatting of math/chem expressions  and
upright vs italic // For more, see Math SE MathJax tutorial.

By providing H+, that was consumed by OH-, there cannot be reached the original concentration , as ratio of acid and it's salt is changing. Therefore pH increases. // Thoughts about reaction rates are misleading. Acidobasic reactions are very fast and when solution is properly mixed, it can be considered to be already at equilibrium with negligible error.

@Poutnik surely i will use those stuff for fomatting thanks , The ratio of acid and salt, that was used in the pH formula isnt? , you are saying that with respect to formula isnt ?  How do you prove that lets say intially there were x moles of H+ , after a bit titration occured it will never reach greater than x moles of H+ ? Volume remains same , so concentration directly depends on moles of H+ . In my doubt above i didnt mention any equation of acid base isnt? Or you mean there are three reactions going on inside , two are equlibrium ones and one is acid base ?or there are more than that ?

And i need to consider the effect of  all? Which i didnt did in above ?

rhetorical question: Cant there be the same pH with different ratio of conjugate acid/base?  //At any point of titration, the following conditions have to be met: 1/ charge balance equation - net charge is zero 2/ molar amount balance equations like total acid concentration = sum of c. of conj. acid and base. 3/ acid dissociation equilibrium equation 4/ water autodissociation equilibrium equation.

I think i will get the answer if you may provide the above four equations which you are talking off Sir ? As i didnt derive at all the formula with the way you are saying , i will definitely want to see how its shows it , so may you for once please give the equations(mathematical equations the four conditions you said above just wrt to the above problem ?) @Poutnik

Let us continue this discussion in chat.

If starting generally with Le Chatelier's principle, the reaction of the system against the external change compensates it just partially. If we increase pressure of a gaseous reaction mixture in equilibrium two times, it will react to partially decrease the pressure, but will not return to the original pressure.
Back to the reaction: Note that the net reaction during the titration is:
$$\ce{CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l)} \tag{1}$$
As you have written yourself, the equation for $\mathrm{pH}$ is:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]}}\tag{2}$$
which can be directly derived from the dissociation equilibrium reaction:
$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}\tag{3}$$
belonging to the equilibrium reaction:
$$\ce{CH3COOH(aq) <=> CH3COO-(aq) + H+(aq)} \tag{4}$$
which partially resupplies H+(aq) ions consumed by the neutralization reaction
$$\ce{H+(aq) + OH-(aq) -> H2O}\tag{5}$$
which has its own equilibrium equation
$$K_\mathrm{w} = [\ce{H+}][\ce{OH-}]\tag{6}$$
The reaction (4) resupplies H+ just partially, because as the titration is progressing, the value of the logarithm in (2) increases and therefore $\mathrm{pH}$ increases.

What about the equlibirum reaction of OH- reacting with CH3COOH ? I mean you have written that in top but that is by combining the below two reactions in the post of yours Sir . i mean as a separate reaction it will happens too isnt ? We already have CH3COOH and H+ inside the container when OH- is coming it will react with both isnt ? So it will have some other equlibirum constant isnt ? Why we are combining it , as its actually happening as a reaction eith the other two going on

That equilibrium is not independent and it's eq. constant is the reciprocal value of the acetate basicity constant, i.e. 1/Kb=Ka/Kw.

I see so actually there are three reactions going on two are eqb ones ch3cooh disscoation and ch3cooh reaction with oh- and one is water one , but here the thing is that one eqb is depedent on another hence we write just two and its sufficient for calculations ? Am i right ? Thats what books do by not writing some equations because its dependent on others which we already wrote? @Poutnik Sir

Also @Poutnik Sir The answer given here for a question which was told to me to be seen regarding my query ,  do you think its fully correct Sir ? https://chemistry.stackexchange.com/a/112360/123521

Sir may you once check this too ?

Checking it and telling you it is right or wrong will not do for you any good. You go to check it and realize why it is right or wrong.


text:
Spectroscopic methods for quantifying peptides/proteins with or without Tryptophan or Tyrosine content
I have several peptides (20-50 amino acids long) which I want to quantify the solubility/concentration in a solvent at certain temperature and pH.
These peptides may or may not contain Tryptophan  or Tyrosine (aromatic residues) in them.
What's the preferred quantification method for it?
I came to UV-Vis method (e.g. NanoDrop) and Fluorescence method (e.g. Qubit).
Do those methods strictly require the presence of Tryptophan or Tyrosine?

Do you have a spectrophotometer or fluorimeter or both? If you have only one or the other  than you have no choice if both measuring fluorescence is generally more sensitive than measuring absorbance.

@porphyrin I have both. Does fluorescence require aromatic residues?
Nanodrop-UV seems to require aromatic.

What about phenylalanine? Do you have it in your peptides? And do you have an approximate idea of the concentrations to be expected? Of course there are the typical assays for protein quantification (Bradford, BCA...) which you could use.

@Snijderfrey some have some not. Yes I have idea concentration to be expected.
I'm looking a generic method without amino-acid content restriction. I wonder if fluorescence is the one.

Expensive methods: Mass Spec or FT-IR.

Generally yes, fluorescence requires aromatic residues, you should excite from 300 to 310 nm approx to excite tryptophan and not phenylalanine or tyrosine.  Detection should be at about 380 nm to see just trp  but you should look up details of spectra. It will be hard to tell if your protein contains several trp or just one as the environment changes the fluorescence yield. You may need to do absorption as well.


text:
The cleansing action of soap
My textbook says that:

The formation of micelles takes place only above a particular
temperature called Kraft temperature (Tk ) and above a particular
concentration called critical micelle concentration (CMC). On
dilution, these colloids revert back to individual ions. Surface active
agents such as soaps and synthetic detergents belong to this class.
For soaps, the CMC is $10^{–4}$ to $\pu{10^{–3} mol L^–1}$.

Would someone please tell me:

How this apply practically to real life situations when we use
soaps or detergents to wash laundry?

Is the Kraft temperature for soaps so less that it is attained by
friction and heat generated while scrubbing the clothes during
washing?

And since the critical micelle concentration seems to be low,
does lathering of soap two or three times satisfy the requirements
for attaining micelle formation?

First, let's get the spelling correct: it's the Krafft temperature.
Second, let's imagine that there are two kinds of solubility curves of surfactants in water: one type of behavior is that solubility increases as temperature rises (this type of surfactant has a Krafft temperature), but there is another type of surfactant whose solubility decreases as the temperature rises, especially at certain temperatures.
The second type of surfactant is generally composed of a hydrophobic portion like an alkylated phenol connected to a long chain of ethylene oxide units. Such surfactants have cloud points: temperatures above which they become more insoluble in water, leading to formation of larger and larger micelles, and eventually a liquid layer. Micelles are tiny agglomerations of surfactant molecules, not individual molecules in solution. These surfactants are useful in dishwashing machines, where hot water is beneficial to removing fats by melting them, but foaming is bad because it clogs the spraying action, so the insolubilization of the surfactant provides a liquid layer with defoaming action in addition to providing micelles to emulsify the fats.
My experience has been mostly with this second type of surfactant, but it suggests that the opposite type of behavior might also occur.
And when it does, a given solution of surfactant might be cooled enough to become cloudy. A warmer solution can dissolve more of this kind of surfactant, so the temperature at which the surfactant precipitates out will vary.
Now there is an important concentration which differs for each surfactant: the critical micelle concentration (CMC). Each surfactant equilibrates between solubility (complete dispersion on a molecular level) and micelle formation (aggregation into micelles, which are composed of several/many molecules, forming tiny aggregates which disperse well in water, and do not totally agglomerate into one phase).
The Krafft temperature is the temperature at which the solubility of the surfactant does not exceed the CMC, so lowering the temperature of this solution will not enable micelles to form. Only above the Krafft temperature will there be enough surfactant in solution so that if the temperature is reduced, surfactant will come out of solution into micelles.
The three questions are good ones:
Laundering with cold water suggests that enough surfactant should be added to generate micelles, but the manufacturer should be sure to use a surfactant that has enough solubility to be an effective cleanser.
Friction and heat would raise the temperature, so would minimize micelle formation. Whether this would reduce washing ability is debatable, because emulsification can occur even from soluble molecules.
Lathering - generating foam or bubbles - suggests that air bubbles are coated with a continuous film, not really individual molecules and not micelles. Foam is helpful in removing dirt, but it is not clear whether this friction/rubbing is dependent on micellization more than just plain solubility. If you have a concentrated solution, even without micelles, friction and rubbing could solubilize a lot of fats and oils.
Also see https://en.wikipedia.org/wiki/Krafft_temperature.

Thank you so much! Ah, and one more thing please... both my teacher and my textbook spell it as "Kraft temperature" so may I please know if there is an alternative spelling or something like that?

Well, Wikipedia used 2 f's because that temperature was named after the German chemist Friedrich Krafft (1852-1923). He also did significant other work: https://en.wikipedia.org/wiki/Friedrich_Krafft           Friedrich Kraft (one f) brings up an entirely different set of people and companies! And you are welcome - it was a pleasure to think about this effect.


text:
What polyatomic substances have the lowest known freezing points at standard pressure?
At standard pressure, helium never freezes, so that's clearly the coldest liquid—but it's also a monatomic substance.
Diatomic hydrogen freezes at 13 K.
Neon freezes at 24 K.
And then the next coldest liquid I can find documented is $\ce{OF2},$ which is liquid between 49 K and 128 K (which I only know because Robert Forward references it in the science fiction novel Camelot 30K).
Are there any other known compounds which have freezing/melting points between 0 K and 50 K?

What non-monoatomic gases do you know? Hint: you are looking for nonpolar, small, and light molecules. I can think of just four in all that make sense to take into account.

I don't like the wording of the question. It sounds like HNQ bait. Also liquids can be supercooled well below melting point.

@Karl Well, O2, N2, CO2, CH4, CF4, and CCl4 all freeze at higher temperatures. OCl2 freezes at a higher temperature and is not non-polar. The analogous nitrogen compounds are all polar, and freeze at higher temperatures. What are you thinking of?

@Mithoron How would you improve it?

Lowest freezing point room temperature liquid?

@andselisk That's a very different question, for which I would expect little or no overlap in answers. Neither hydrogen, nor neon, nor OF2 are liquid at room temperature, for example.

The title and the text ask two very different questions. One is about a substance in the known set of substances, and the other is absolute.

@Zhe The title was edited without my consent in a way that contradicts the body. Having not received any suggestions on how to improve it, I guess I'll take a while guess and edit it again...


text:
Measurement of actual carbon dioxide mass emitted from a diesel engine
I was wondering if there is a method to measure the actual total mass of carbon dioxide $(\ce{CO2})$ emitted from a diesel engine used for power generation. Infrared $\ce{CO2}$ sensors generally measure the gas concentration, describing the amount of gas by volume in the air, which is not useful in my application. Flow sensors are already installed in the diesel supply and return lines to determine the amount of diesel consumed, and hence the amount of $\ce{CO2}$ created, but I need to correlate these measurements with actual measurements of $\ce{CO2}$ mass produced at the output of the exhaust system.
Any idea on how to approach this problem would be greatly appreciated.
The question can be reformulated as to measure the $\ce{CO2}$ emission rate from the engine exhaust system. A typical solution in practice is to find a dedicated sensor or a monitoring system that does that.

RGA measurements of the exhaust would give you the relative concentrations, combined with a mass flow measurements gives you mass.

Would you please let me know how to conduct RGA and mass flow measurements in practice?

There are only a few possible products of burning diesel: CO2, CO, unburned fuel and particulates. Knowing how much fuel is burned is the most accurate way of assessing CO2 output as the others are not major products. But CO and CO2 are measurable in the exhaust (but less accurately) so you can at least triangulate the estimate made from fuel burned.

@matt_black: Would you please let me know how "CO and CO2 are measured in the exhaust"?
As I mentioned in the question, the consumed fuel is already being measured, and the CO2 being estimated knowing the amount of carbon per gallon of diesel. However, I am looking for a redundant accurate method to measure the gas mass emitted from the exhaust stack.

@AEW CO is routinely measured in car exhausts using standard meters during routine inspections mandated in many countries (in the UK cars older than 3 years have to have MOT tests and the measurement is part of those). So commercial devices specifically measuring CO are widely available. Commercial CO2 meters are also common and have been used to tune engine performance. Neither, though, are as accurate as measuring the amount of fuel burned.


text:
Why is octasulfur present in equation for standard enthalpy of formation of sulfuric acid?

Write a balanced chemical equation corresponding to the standard enthalpy of formation of $\ce{H2SO4}.$

\eqref{rxn:r1} is the given answer, \eqref{rxn:r2} is my answer:
\begin{align}
\ce{1/8 S8 (s) + H2(g) + 2 O2(g) &-> H2SO4(l)} \label{rxn:r1}\tag{R1}\\
\ce{S(s) + H2(g) + 2 O2(g) &-> H2SO4(l)} \label{rxn:r2}\tag{R2}
\end{align}
How can I know I have to use $\ce{S8}$ rather than simply $\ce{S}$?

Your answer would have been correct if $\ce{S}$ was available as a pure substance. The trouble is that the free sulfur atom tends to join to other sulfur atoms at room temperature to make $\ce{S8}$ octogones. Nobody knows why. So $\ce{S(s)}$ does not exist.

It is important to know what is and what isn't important. Writing S as S8, for one thing, isn't.

You know it the same way you know to write  $\ce{O2}$ instead of  $\ce{O}$: from knowledge of the chemistry.  And it is important to know this, since the free energy for that reaction includes breaking the covalent bonds within $\ce{S8}$, which wouldn't be the case if it were instead $\ce{S}$.

Well, IMHO, if we cannot use S(s) with implied whatever structure the solid sulfur has, we cannot use NaCl(s) nor NaHCO3(s) either, as there is no NaCl nor NaHCO3 in these solid salts.

See: https://chemistry.stackexchange.com/questions/8438/reaction-between-zinc-and-sulfur

Here is a definition of standard enthalpy of formation from Wikipedia.

The standard enthalpy of a compound is the change of enthalpy in formation of one mole of the substance from its constituent elements, with all substances in their standard states.

So according to it, the constituent elements should be in their standard states. Now coming to question, the standard state of sulphur is rhombic sulphur and it exists as $\ce{S8}$ molecule in nature but not simply as individual atoms.

Note that by IUPAC molecule definition,  there are no monoatomic molecules. I do not personally like it, but I am not an authority here.

So, which terminology is used here?

Atoms. Or possibly the more general term particles.


text:
Why does a big protein move during gel electrophoresis but to a lesses extent through a size exclusion column?
A big protein going through gel electrophoresis will be forced through the gel, it will drift a small way and will show on the top of the gel, while in size exclusion, the big protein is not forced to go through the pores and will elute first. Why is the protein forced through one medium and not through the other?

Please pay attention to grammar and punctuation in the future. Use a spell checker if necessary.

In the polyacrylamide gel, there is no way to avoid the narrow passages. The gel is made in situ by a chemical crosslinking step. In size exclusion, the solid phase is made of beads with diameter ranging from 50 to 200 µm diameter. The voids between these beads are connected and sufficient larger for the protein to go through.
Another difference between the two methods is the for SDS gel electrophoresis, the protein is denatured while for size exclusion, it is usually in the native state, so the shape and the flexibility is different, leading to different mobility.


text:
Is there an acid (or other solvent) that dissolves nickel but not brass?
Preferrably one that is common or easy to make using common chemicals.
My application is to remove nickel plating from brass parts. Right now the best solution I have is to scrape and chip it off, but that sometimes takes up to 4 hours per part, and it often scratches the brass (sometimes badly).
An ideal solution would be to simply dunk the part in a solvent for 20 minutes and take it out clean!
Does such a chemical exist?

I don't think it would work, or at least be that simple.

Why did you ask without doing research? Or at least telling what kind of brass is this, exactly.

I am not an expert in this field, which is why I am asking on a website that is supposedly tailored to the field.
Who are you to assume I have not researched this?  And what degree of "research" is acceptable before I am allowed to ask a question on StackExchange?
I don't know the alloy: all I know is it is "brass" and 99.9% "pure nickel".
I do not know how to identify exactly what kind of brass it is.
If you can't or won't answer, I'll go to a different website. This website seems awfully pedantic.

Removing nickel plating from brass - "I have often stripped nickel from copper and brass. A thin layer will come off quickly in dilute hydrochloric acid. ... A common method of stripping thicker layers is a solution of 3 parts sulfuric acid to 2 parts water by volume. Apply a positive DC voltage to the part and use a lead cathode. About 6 volts should work."

@PeterMortensen But you don't mention the part that says that such methods also attack brass. That's the thing - whatever you'd use can dissolve both.

@Mithoron A fair point, but not a proof it can't work. If a reagent attacks the coating faster than the brass, careful use of it might be effective.

@Mithoron: It was primarily to demonstrate the "research" part (after all the drama on MSE (now limited visibility (10,000 reputation points required to see it)))—finding this required very little effort.

I think the most practical approach is to improve the mechanical technique The nickel plate will be thin so a buffing wheel with about 180 mesh silicon carbide or other abrasive would be a good starting point. Use coarser or finer abrasive as necessary.  Random production of nickel monoxide does not sound good for amateur use.

This is a bit more complicated than dip and rinse, but if you have a lot to strip, maybe you can make it work.

Nickel is mined, going from the oxide to the metal, by reacting the ore with water gas, a mixture of H2 and CO (made by adding H2O to hot and/or burning coal - at the relatively low temperature of 50ºC and atmospheric pressure). Now the OP already has metallic nickel, so this step is redundant. The next step, formation of Ni(CO)4 obviously takes place at temperatures below 230ºC, because it decomposes there to nickel metal. Ref 1
The proposed operation is similar to mining, except that the nickel is discarded and the depleted ore (the brass item) is retained. First step, get some coal; start it burning (with a propane torch). Shut off its air supply and feed steam or water mist onto the coal (which will cool it, but also generate H2O and CO) and pass the water gas over the nickel-coated item. The burning coal will cool, so you have to run some air over it to heat it up again; then more water mist or steam to get more water gas to pass over the nickel-coated item. Repeat until all the nickel is gone from the item and deposited somewhere else, far, far away, because the Ni(CO)4 is terribly toxic and you don't want to have any contact with it - at all.
As I ponder the set-up, this method should work, but it does seem more like a theoretical exercise or possibly an industrial project, rather than a do-it-yourself type of chemistry. It could be improved, or simplified, perhaps, by getting a cylinder of carbon monoxide and passing it slowly over the nickel-coated brass item at a temperature high enough to form Ni(CO)4, but not decompose it, then lead the gas (excess CO plus the Ni(CO)4) thru a very long pipe to a hot zone where the Ni(CO)4 would be decomposed to metallic nickel and the relatively less toxic carbon monoxide.
If you try this, do appreciate the toxicity of nickel tetra carbonyl (Ref 2) and CO (Ref 3).
Ref 1. http://wwwchem.uwimona.edu.jm/courses/nickel.html
Ref 2. https://en.wikipedia.org/wiki/Nickel_tetracarbonyl
Ref 3. https://en.wikipedia.org/wiki/Carbon_monoxide

I'm not an experimentalist, so somebody who knows more here please correct, but recommending Ni(CO)4 as a suitable substance for use in a home lab really makes me worry.

I'm just commenting here on the technical stuff and not on the content. You've chosen to make an image of a text that you later refer to by a link which proves that you just as easily (or more so) could have copied it from and pasted it into an appropriate environment. Why? Also, we have MathJax with mhchem enabled, they make setting reaction equations and formulae rather easy.

@Martin: Sorry, I'm not so well versed in all the options.

OP would be better off dumping all his brassware into an acid bath and performing a good ol' separation than work with metal carbonyls. Should they not want both their health and starting material in shambles, my two cents would be using acidification to dissolve minute amounts of nickel, then complexation (e. g. with oxalate) to recover more of the nickel.

@TheRelentlessNucleophile: I agree. I was thinking that adding a chelating agent in a known amount might complex off the Ni without affecting the Cu or Zn, but I think it would take a long time, and might need some acid - could get too complicated .

Find a plating shop that does nickel plating and ask them how they strip their rejects. Brass is difficult because anything acidic attacks the zinc faster than the nickel and complexing agents attack the copper faster than the nickel. The answer might be tumbling with a fine abrasive with a complexing agent such as citric acid.


text:
Why would tellurium + sodium hydroxide have worked as a good anti-knock gasoline additive (if it wasn't so smelly)?
The April 22, 2022 Veritasium video The Man Who Accidentally Killed The Most People In History mentions several aspects of the historical use of tetra-ethyl lead in gasoline, the resulting widespread distribution of lead in the environment and within people, and the consequences of that.
As part of the story, it mentions that during Thomas Midgley Jr.'s search1 for a gasoline additive that would reduce premature detonation in a high-compression gasoline engine, a tellurium-based additive was tried. The graphic in the video indicates "tellurium + sodium hydroxide".

Kettering wanted to find an additive which would increase the octane rating of ordinary fuel and eliminate knocking in high-compression engines. So he hired 27-year-old engineer Thomas Midgley Jr. Midgley experimented with all sorts of compounds, from melted butter and camphor, to ethyl acetate and aluminum chloride. He later wrote "Most of them had no more effect than spitting in The Great Lakes."
Ethanol was an interesting exception, it did stop the knocking, but you needed a lot of it, about 10% of the fuel  mixture to be effective. That much ethanol would be expensive and hard to turn a profit on. And Midgley was really after an additive that was cheap, easy to produce and effective even at low concentrations. So he kept trying.
Then he hit on tellurium. It worked wonderfully as an anti-knocking agent, but it had a terrible smell. You couldn't get rid of it by changing clothes or bathing. His wide was so offended by the stench that he had to sleep in the basement for seven months. Midgley wrote: "I don't think that, although this doubled fuel economy, humanity would suffer this smell."

Question: Why would tellurium + sodium hydroxide have worked as a good anti-knock gasoline additive (if it wasn't so smelly)?
According to the second quote from Midgley, in addition to reducing/stopping engine knocking it also improved mileage substantially. By what mechanism is a tellurium-based additive able to do this?

1at the behest of Charles Kettering after the higher compression engine of the Model 30 Cadillac exhibited problems with knocking (premature detonation of fuel-air mixture before the electrical spark was triggered)

The only reference to sodium hydroxide in the video is this visual cue in the animation:

Additional supporting screenshots:1, 2

related in meta: Would a {metalloid} and/or {tellurium} tag be appropriate for the site, and /or for my question about tellurium?

Not an answer, but I am glad Te never got used as an anti-knock ingredient! Besides Te being rather rare, its compounds are notoriously nasty smelling: hydrogen telluride supposedly smells like rotten horseradish. Worse still, exposure results in the human body slowly, over months, eliminating the tellurium compounds in the breath and sweat. We lucked out on this one!

See/listen also Episodictable.com: 52. Tellurium: To Hell You Ride

Tellurium was an effective additive which Midgley knew because he tested engine performance with it as an additive. It probably works in ways similar to lead additives. but the side-effect of small was unacceptable.

See this for the actual mechanism: Callear, A. B., and R. G. W. Norrish. “The Behaviour of Additives in Explosions and the Mechanism of Antiknock. I.” Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences, vol. 259, no. 1298, 1960, pp. 304–24, http://www.jstor.org/stable/2413734. Accessed 28 Apr. 2022.

companion question: How did Sinclair Oil put "liquid nickel" into gasoline, and what were the chemical mechanism(s) by which it was supposed to reduce engine wear?

Midgley tried various compounds, especially
$\ce{(C2H5)2Se}$,
$\ce{(C2H5)2Te}$,
$\ce{(C2H5)4Sn}$, and
$\ce{(C2H5)4Pb}$,
which all work as anti-knock agents with similar mechanisms.
The oldest method for the synthesis of dialkyl tellurium is the reaction of disodium telluride with alkyl halides. The required solution of disodium telluride can be prepared from tellurium and a reducing agent in aqueous sodium hydroxide.
So I guess they simply put a part of the first step (tellurium and sodium hydroxide) on the label for the video.
Midgley, T.; Boyd, T. A. The Chemical Control of Gaseous Detonation with Particular Reference to the Internal-Combustion Engine. J. Ind. Eng. Chem. 1922, 14 (10), 894–898.

See this for the actual mechanism: Callear, A. B., and R. G. W. Norrish. “The Behaviour of Additives in Explosions and the Mechanism of Antiknock. I.” Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences, vol. 259, no. 1298, 1960, pp. 304–24, http://www.jstor.org/stable/2413734. Accessed 28 Apr. 2022.


text:
Finding relation between Gibb's free energy and equilibrium constant for a homogeneous gaseous phase reaction
I was trying to find relation between Gibb's free energy change and equilibrium constant for the following homogeneous gaseous phase reaction at constant temperature
$$\ce{aA +bB }\rightleftharpoons \ce{cC + dD}$$
Let the initial partial pressures are $P_{1,i}$ and final partial pressures are $P_{2,i}$ for $i=A,B,C,D$
Now Gibb's free energy change of reaction is given by
$$\Delta_rG=\Delta G_{reac} + \Delta G_{prod}$$
However when I tried to find Gibb's free energy change for $\ce{A}$, I got stuck at first step
Using the relation $dG=VdP-SdT$
$$\Delta G_{A}=\int_1^2VdP=\int_1^2\frac{n_ART}{P_A}dP$$
However I cannot perform this integration as $n_A$ is not constant.
I saw other answers relating to same question but cannot gather much from them
Is my method correct ? If yes, then how can I continue
?

Are you familiar with the effect of concentration of a species on the free energy of a mixture of gases?

The chemical potential (in this case free energy of the gas divided by amount of gas in moles) remains constant when you evaluate the Gibbs free energy of reaction. It is the amount of each gas that changes.

@BuckThorn  $\mu_i=\mu^o_i+RT\ln P_i$ , as the reaction proceeds in forward direction  the partial pressure of $i$ decreases and hence chemical potential should decrease , shouldn't?

Please see my answer. By the way, it's Gibbs' (although the apostrophe is usually dropped), not Gibb's.

It's sometimes not obvious because of the way the subject may be taught, but $\Delta_rG$ is what's called a differential form, so it actually describes how the free energy changes as the reaction progresses by an infinitely small change in the amount of reactants and products at a given composition, $T$ and $p$.
The equation presented for the chemical potential $\mu_i=\mu^o_i+RT\ln p_i$ is correct, but the partial pressures are constant when evaluating $\Delta_rG$. In fact, that equation can be used to derive the relation between the reaction quotient and the Gibbs free energy of reaction:
$$\Delta_rG = \sum_i c_i \mu_i= \sum_i c_i \left( \mu^o_i+RT\ln p_i \right)\\ = \sum_i c_i \mu^o_i + RT \sum_i c_i\ln p_i \\ = \Delta_rG ^o  + RT \ln Q $$
where $Q = \sum_i c_i\ln p_i$ is the reaction quotient and the $c_i$ are the coefficients of the reaction (not to be confused with the amount $n_i$ of each reactant or product, which remains constant).
This idea can be conceptually difficult at first. Another way to think of it is to imagine a very (=infinitely) large reaction vessel for which you measure the change in free energy upon changing the amount of reactants and products by $c_i$  (see for instance the textbook by Klotz and Rosenberg). Since the $c_i$ are finite and the vessel infinite, the composition remains constant. It's a mathematical sleight of hand equivalent to adding an infinitely small amount of substance to a finite system.

so basically $dG=\Delta_rG d \xi$ and change in $\Delta G \neq \Delta_rG$ ($\xi$ is extent of reaction and $\Delta G$ is overall change in Gibbs' energy) , am I right?

Yes, that's right.