text:
Can someone explain this answer in detail?
The density of iron is $\pu{7.874 g/cm3}$ and its unit cell volume is $\pu{2.364E-23 cm3}$. These values indicate that there are: two Fe atoms per unit cell and the metal is body-centered cubic.
I am not fully sure why this is correct or how you figure this out by using the data given. I am still very new to the unit cell concept in chem. Any explanation would be great, thanks! Also sorry I was not sure what to tag with this question.

You're given the volume of the unit cell, and the density. So what is the mass of the unit cell?

Work through the answers here, including the mentioned video.  Though on first sight different, it contains the material you need.

Atoms in the center of the cell count fully ... those on the side only half ... those in corners only a eighth, etc., if possible, have a look into the video which showcases the computation on Ne.

Try to invent more meaningful title // A good question title should be quite unique, resembling a magazine article headline, optionally formulated as a question. Avoid  titles saying  everything (like full formulation of the task) or nothing ( like "A question about chemistry")

The density of iron is also the density of the unit cell : $\pu{7.874 g|cm3}$. Multiply this density by the volume of the unit cell. It gives you the mass of this unit cell : $\pu{7.874 g/cm3 · 2.314 · 10^{-23} cm3 = 1.861 · 10^{-22}} g$ . This is exactly twice the mass of an iron atom, as $1$ atom weighs : $\pu{\frac{56 ~g/mol}{6·10^{23} atom/mol}}$ = $\pu{9.33 · 10^{-23} g}$, So there are $2$ $\ce{Fe}$ atoms per unit cell.
This number $2$ can be justified by the structure of the body-centered cubic. This system is made of an infinite succession of identical cubes, piled along the three directions of space Ox, Oy and Oz. The unit cube has 8 vertexes or apexes, all occupied by the center of an iron atom. But around each of these iron atoms, there are 8 cubes starting from it : one above it at left-hand side and before, the second also above and at left-hand side, but behind ; the third is also above but at right-hand side and before ; the fourth is right-hand side, above and behind. The four remaining are obtained by replacing "above" by "under" in the preceding text.
So each vertex of the unit cell is occupied by one atom, but only $1/8$ of this atom is situated inside the first unit cell. In total, There are $8$ times $1/8$ atom in each corner of the cube. Well, in body-centered, there is still one iron atom in the very center of the unit cell. In total, each unit cell is occupied by $8/8 + 1$ iron atoms. So there is on average $2$ iron atoms per unit cell.


text:
Should there be a better explanation on Van Der Waals force?
The explanation on how Van der Waals force forms is this: Van der Waals forces consist of induced, temporary dipoles between neighbouring molecules. These are caused by constantly shifting electron could, relative to the positions of the nuclei.
I am thinking, since the electron clouds of EVERY atom are constantly shifting, and EVERY atom is constantly induced to a temporary dipole. So there are exactly two factors determine the electron cloud. One is the shifting of the electron cloud itself and the other is due to the induction by other shifting electron clouds. So should there be a better explanation/description of the Van der Waals force?

As you think of the electron cloud as shifting, you should imagine it shifting slightly more this way and slightly less that way, according to the induction by other shifting electron clouds. This is the better explanation you are after.

Does this answer your question? How are London Dispersion Forces generated?

This is not just a qualitative explanation, there is actual math/physics behind it. Why do you need “better explanation” if the theory explains well the observations?

Van der Waals force = London force + Debye force + Keesom force.

I don't know much about but at least that is an umbrella term/explanation. Mutual back induction is by the way mentioned by book & teachers. Look @Poutnik comment. What is left out at our level, as well as talking of current ring in aromatics, is the nature of the "movement", as for there is no place for it in the quantum description of molecule that we all know. Perhaps I shall think on how to formulate a question here, about this.

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text:
Decrease in electrical conductivity due to the increase in oxygen content
I am just wondering why there is increase in electrical conductivity of some materials with the decrease in oxygen, or vice versa. When Graphene oxide (which is an insulator) is reduced to rGO (reduced graphene oxide), the rGO becomes electrically conductive. Any suggestions?

It is not limited to oxygen, it is similar with graphite fluoride. They bind to electrons which are not then available for conductivity.

In many metal oxides the oxygen vacancy is a charged and mobile species.

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text:
Relationship of oxygen dissolved in water with bubble time and temperature
I'm using the Winkler method to calculate oxygen dissolved in water. This is at the temperatures 20, 30, 40, and $\pu{50 ^\circ{}C}$ and I bubbled a controlled amount of oxygen in through the reaction of manganese oxide and hydrogen peroxide. At all the different temperatures I bubbled air for 0, 20, 40, 60, or $\pu{80 s}$, respectively. I have found data online for max solubility of oxygen at different temperatures but no data for this sort of relationship. The results I collected for now are only these but they make no sense (I changed the $\ce{O2}$ bubbling after this).

It seems you have a systematic error in titration at 50 C. The solubility of oxygen in water at various temperatures must be known very well.

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I assume the kinetics of oxygen saturation has exponential shape of transition between initial and final value, the latter being T dependent as well as the kinetic constant for that.

@Poutnik do you know where I could find sample data of a similar experiment? I've been looking for it everywhere but I probably just don't know the right keywords to search with. I have found many values for maximum saturation but this shouldn't be what is happening as I am not trying to make it reach maximum saturation. It is to my understanding this has something to do with Henry's Law of solubility of gases.

See engineeringtoolbox for charts and tables of oxygen solibility data.


text:
What is the correct way to write dissociation reactions?
The variants are as follows:
For a salt (NaCl):
$$\ce{NaCl_{(s)}<=>NaCl_{(aq)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{1}$$
$$\ce{NaCl_{(s)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{2}$$
$$\ce{NaCl_{(aq)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{3}$$
$$\ce{NaCl_{(s)} + H_2O_{(l)}<=>NaCl_{(aq)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-} + H_2O_{(l)}}\tag{4}$$
$$\ce{NaCl_{(s)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-} + H_2O_{(l)}}\tag{5}$$
$$\ce{NaCl_{(aq)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-} + H_2O_{(l)}}\tag{6}$$
$$\ce{NaCl_{(s)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{7}$$
$$\ce{NaCl_{(aq)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{8}$$
Which of these are right/wrong and why?

I am afraid you are overthinking about notation. The only problem is that the plus and minus are too far off from the elemental symbol. An chemical equation cannot describe every process. It is just a shorthand notation. You will still need words to explain what is going on.

I am showing the different versions of the equation followed by my comments.

$$\ce{NaCl_{(s)}<=>NaCl_{(aq)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{1}$$

$\ce{NaCl_{(aq)}}$ is the same as $\ce{Na_{(aq)}^{+} + Cl_{(aq)}^{-}}$ so it makes no sense to show them in equilibrium.

$$\ce{NaCl_{(s)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{2}$$

This would be my preference. The physical state is written more conventionally after the charge of the ion, and not as a subscript:
$$\ce{NaCl(s)<=>Na+(aq) + Cl-(aq)}\tag{2a}$$

$$\ce{NaCl_{(aq)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{3}$$

See (1).

$$\ce{NaCl_{(s)} + H_2O_{(l)}<=>NaCl_{(aq)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-} + H_2O_{(l)}}\tag{4}$$

There is no reason to show a single water molecule. This would suggest a 1:1 stoichiometry but there are multiple water molecules in the solvation shells of the two ions.

$$\ce{NaCl_{(s)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-} + H_2O_{(l)}}\tag{5}$$

See (4).

$$\ce{NaCl_{(aq)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-} + H_2O_{(l)}}\tag{6}$$

See (4).

$$\ce{NaCl_{(s)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{7}$$

This one is not balanced.

$$\ce{NaCl_{(aq)} + H_2O_{(l)}<=>Na_{(aq)}^{+} + Cl_{(aq)}^{-}}\tag{8}$$

See (7).

Which of these are right/wrong and why?

Some of them are not balanced, and some, while balanced, are not describing the process as accurately and succinctly as version (2).

I agree completely.

@user545735 either one is fine. (1) shows an H+ transfer from acetic acid to water (Bronsted acid/base). (2) saves some keystrokes and shows the water implicitly.

@user545735 Yes because there are more than one water molecule involved in hydrating or salvation the hydrogen ion.


text:
Haworth projection of Sucrose
I was searching for the Haworth projection of sucrose and I got this image from wikipedia (and much of the internet).
(wikipedia link: https://en.m.wikipedia.org/wiki/Sucrose).
But isn't it wrong!?
The image shows a (#1Carbon to #5C) linkage of glucose to fructose respectively.
What i think the image is showing:
(edited wikipedia image)

But it should show (#1C to #2C) linkage of glucose to fructose respectively.
Should it not be like this image:

Your fructose in the second picture is numbered incorrectly. #2 is #5 and #5 is #2. The correct C2 is a ketal carbon, a masked ketone. Fructose is a 2-ketose.

This may be of help: https://chemistry.stackexchange.com/questions/87178/identifying-the-d-l-form-of-any-aldose-or-ketose-in-cyclic-form/106929#106929

Those are the same, just shown in a different conformation and a different view.

As user55119 points out in the comments, the numbering went awry (in fructose, #2 is the anomeric carbon in the ring form, or the carbonyl carbon in the linear form).
Sucrose has no reducing "ends", which it would have if fructose would link via the #5 carbon.
If you look at one of the crystal structures of sucrose, you will find that the rings are not parallel but rather perpendicular, allowing some nice hydrogen bonds:

Beside the pure chemical perspective of this answer, it is an eye-catching showcase about advantages of vector-based images in formats like .svg.

@Buttonwood The provided figure was pixel-based, so I had to remove the background to make it transparent...

Great animation, much thanks. Which software did you use btw.

@KarstenTheis Oh, but it has so much similarity to the depiction of sucrose on top of the property box in the English edition of Wikipedia, which is available as .svg here ...

@JustCurious The first one may be achieved with inkscape (though, by shape of the symbols, it does not necessarily look like the version in Linux). The second is exporting an animation from Jmol. In both cases, selection of a gif at rate/resolution/frames per second suitable enough to stay within the limit of approx 2 MB. For the former, let's see in the chatroom.

@JustCurious I used commercial software I happen to have, PowerPoint for the first figure, and Camtasia to capture the screen and turn it into an animated gif. The second figure is of SUCROS01 was rendered with the open software Jmol.


text:
Relationship between thermodynamics and steric hindrance in reaction outcomes?
I am investigating reactions with alkyl-substituted amines ($\ce{R} = \ce{Me}, \ce{Et}, \ce{^iPr}, \ce{^tBu}$) and a second reagent, in which product Z is formed. $\Delta G$ for all of these transformations is negative.
When the amine substituent is large ($\ce{R} = \ce{^tBu}$), only product Z is formed. When the amines are small ($\ce{R} = \ce{Me}, \ce{Et}$), product Z does not form at all; rather, a mixture of many components is generated. Clearly, steric hindrance influences the outcomes.
Is it correct to infer that the activation barriers in the reactions leading to the mixture of components (i.e. when $\ce{R} = \ce{Me}, \ce{Et}$) is too high when $\ce{R} = \ce{^tBu}$, which is why that amine reacts differently and forms product Z only?
What is a general statement that can be made about the kinetics and thermodynamics of this reactivity?

To be honest, it's going to be difficult to say anything without knowing what the reaction is. As it stands, I would argue the question is unclear and cannot really be answered. I understand these things are top secret because you don't want me scooping you on the publication, but even just knowing the form of the reactants and products (without specifying reagents or catalysts) might help a lot more already. That said, unless you can provide a full mechanism (or a hypothesised mechanism), whatever answer you get here will still be an incomplete description.


text:
Why is Ion Exchange Chromatography done in a column?
It seems like the idea of Ion Exchange Chromatography is to get a maximum amount of exposure of your gel to your solution. It seems like overhead stirring and a gel bed would be more likely to be successful here.

It seems like you are focussing on the ion exchange part and ignoring the chromatography.

Yes, with a solid ion-exchange bead sitting in solution, you are doing an equilibrium experiment. You do not have a true equilibrium condition in a chromatography column.

Chromatography  is a dynamic separation technique, done on columns or planes. Usage of ion-exchange resin is just circumstance.  There is no separation of analytes or substances in the static gel bed scenario, just different distribution constants.

Do you bath you car's engine with gas, or do you prefer, for practical reasons, to put the gas station's nozzle to this funnel shaped entry to channel the gas into a the tank? (irony!)


text:
Why can two wavefunctions can be interchanged in integrals involving hermitian operators?
In many textbooks, it is said that because operators in QM are hermitian, we can write:
$$\int \psi^*\hat{A}\phi\,\mathrm d\tau = \int\phi^*\hat{A}\psi\,\mathrm d\tau$$
An operator $\hat{A}$ is called hermitian iff $\hat{A}^\dagger = \hat{A}$. Using this definition, I tried to prove the above property but I can't. I started from:
$$\langle\psi|\hat{A}|\phi\rangle = {\langle\phi|\hat{A}|\psi\rangle}^\dagger$$
based on the fact that the operator is hermitian. But then I don't know how to continue. Any ideas?
I have found the first relation in the following textbook: Quantum Chemistry and Molecular Interactions, in page 81.

For any two wavefunctions $\psi_i$ and $\psi_j$ $$\int \psi_i^*\hat{H}\psi_j\,\mathrm d\tau = \int\psi_j^*\hat{H}\psi_i\,\mathrm d\tau$$

You want $\langle\phi_{1}|\hat{A}|\phi_{2}\rangle = \langle\phi_{2}|\hat{A}|\phi_{1}\rangle^{*}$. One simple way to prove this is using real-valued wavefunctions, so there is no complex part. To answer your question: the wavefunctions can be exchanged because, within a factor of $\pm i \rightarrow \mp i$, they are the same and the order of operation doesn't matter.

Are you very sure about this supposed truth? I've tried for ages to prove it, but can't, and indeed the textbooks I have only state the weaker criterion $$\langle \psi | A | \phi \rangle = \langle \phi | A | \psi \rangle^,$$ which is fairly easy to show using the property of the inner product $\langle a | b \rangle = \langle b | a \rangle^$, as well as the Hermiticity of $A$ (I assume that's what you did): $$\langle \psi | A | \phi \rangle = \langle \psi | A\phi \rangle = \langle A\phi | \psi \rangle^* = \langle \phi | A^\dagger | \psi \rangle^* = \langle \phi | A | \psi \rangle^*.$$

@orthocresol In the case of ${\langle \phi | A | \psi \rangle}^*$ in the right integral I must  take its complex conjugate (of the integral). But in the textbooks it swaps them without taking the complex conjugate, which I don't understand. Also for $\langle\psi|\hat{A}|\phi\rangle = {\langle\phi|\hat{A}|\psi\rangle}^\dagger$ I used the property of matrix product $(A \cdot B \cdot C) ^ \dagger = C^\dagger \cdot B^\dagger \cdot A^\dagger$, is this wrong?

Gonna need a citation for "the textbooks". In the integral form it should probably have the complex conjugate too: $$\int \psi A\phi ,\mathrm{d}\tau = \left( \int \phi A\psi ,\mathrm{d}\tau \right)^.$$ For the adjoint bit, no, that's fine. However, note that the resulting integral is just a number (i.e. $1 \times 1$ matrix), so the complex conjugate $\langle \phi | A | \psi \rangle^$ is the same thing as the adjoint / complex transpose $\langle \phi | A | \psi \rangle^\dagger$ (and the former is probably more commonly used).

@orthocresol is correct. In fact the way I learnt it is that the first relation quoted in his/her comment is the definition of a Hermitian operator (see e.g. http://sces.phys.utk.edu/~moreo/mm08/hicks.pdf or Atkins and Friedman "Molecular Quantum Mechanics") If textbooks are quoting the relationship in the question ditch them, they are wrong.

Hmm, there is another way of writing the same relation which I've seen a couple of times: $$\langle \psi | A\phi \rangle = \langle A\psi | \phi\rangle,$$ which may make it look as if the complex conjugate is gone, but if you carefully work through that you'll find that it's exactly the same as what I wrote.

@Anton - See Cohen-Tannoudji (Volume I, Chapter 2, sections B and C specifically, p. 118, equations B-47 and B-49) for some of the steps taken to show that $\langle\psi | A^{\dagger} | \phi\rangle = \langle\phi | A | \psi\rangle^{*}$. Note that this is not what you have written, which might be the source of the error.

Thank you all for the comments. I think the problem is that the textbook isn't clear if the wavefunction is real and complex, as Tyberius points out in his answer.

Yet another possibility is that $\psi_i$ and $\psi_j$ are part of an orthonormal basis set of eigenfunctions such that the integrals are 0 if $i\neq j$ and identical if $i=j$.

I have taken a look at the book and the textbook is indeed giving the wrong definition. I would seriously consider using a different book, if it already fails at such a fundamental and basic concept. Physical chemistry books are often not the best sources for basic quantum mechanics. I highly recommend taking a look at physics books that focus on QM. They typically give much clearer definitions and focus on fewer topics with more detail. For example Quantum Mechanics, Vol. 1  by Claude Cohen-Tannoudji, Bernard Diu, Frank Laloe.

This must be false. As a simple counterexample, if this were true for any Hermitian operator $A$ and any set of basis functions $\phi,\psi$, it must also be true for the identity operator $I$ with $\phi$ a complex function and and $\psi$ a real function. But clearly
$$\int\phi^*\psi d\tau\ne\int\psi^*\phi d\tau$$
since only $\phi$ has an imaginary part. As others have stated in the comments, we can only say that $\langle\phi|A|\psi\rangle=\langle\psi|A|\phi\rangle^*$ for Hermitian operators, with the complex conjugation removed if $\phi$ and $\psi$ are known to be real functions and $A$ is a real symmetric operator.

I dug up this textbook, and I believe that the author is restricting his statement to wavefunctions with no imaginary component, although he does not explicitly state this. He writes on p. 72 that

In our notation, $\Psi$ will often be used to describe the "complete" wavefunction, which may include time-dependent or magnetic terms. Conveniently, we are more often interested only in terms that describe the distribution of the system in space--the spatial wavefunction--for which we will use the lower case $\psi$.

This statement could be interpreted to mean that any use of $\psi$ indicates a real-valued function, and that restriction carries forward to "any two wavefunctions $\psi_i$ and $\psi_j$" on p. 81. Notably, he uses $f(x)$ and $g(x)$ in other cases for generic functions that are not specifically designated as wavefunctions (see for example the definition of linearity of operators which is directly before the quoted definition of Hermitian).

Even if the wavefunctions $\psi$ and $\phi$ are purely real, I don't see why that should mean that the integral $\langle \psi | A | \phi \rangle$ is also real for all Hermitian operators $A$, unless I'm missing something(?). The inner product $\langle \psi | \phi \rangle$ will definitely be real, but I'm not sure why any other integral should be.

@orthocresol - He also mentions that the expectation value $\left< \psi | \hat{A} |\psi \right>$ is real for any observable, so the set of operators being considered may be restricted to those associated with measurable outcomes. I'm trying to give him the benefit of the doubt that he wasn't completely confused when he wrote the quoted text, but rather just wasn't clear about all the assumptions he was making. But it could be that he just doesn't know what he's talking about.


text:
What happens when you mix antimony pentachloride and hydrochloric acid? Will it form chloroantimonic acid?
Learning that fluoroantimonic acid, HSbF6, is produced by the reaction of antimony pentafluoride and hydrofluoric acid, I wondered if it would do the same for the chlorine. I have seen the other question, but it doesn't describe what would happen if the two compounds were mixed together.

No, it doesn't describe what would happen if the two compounds were mixed together.

Antimony pentachloride reacts with water to form antimony pentoxide and hydrochloric acid, which the linked answer shows further converts to meta-chloroantimonic acid.


text:
Precipitation of equimolar amounts of chloride and bromide
I was just wondering about the following situation; if we had a solution containing 0.1M bromide and 0.1M chloride ions, and treated it with excess silver nitrate, what would happen and would quantitative precipitation of both silver salts be possible?
If I understood correctly, as AgBr is less soluble, it precipitates first. I know that the solubility product of AgBr is lower than AgCl, but why does it selectively precipitate first?
Anyway, it precipitates until the point where the remaining bromide (about 0.48% of 0.1M) remains; at this point, AgCl is more insoluble and starts precipitating, until AgBr is more insoluble again, that’s how I understood it.
So basically, am I correct in the assumption that quantitative precipitation of both salts in same solution is not possible because at some point both salts precipitate simultaneously although there is still a significant amount of bromide in solution?

In practice, the very first drop of AgNO3 solution will create a huge oversaturation in both AgBr and AgCl, so both will start precipitating chaotically.

It can be said that precipitation of chlorides may be preferred kinetically even before it is preferred thermodynamically.

@KarstenTheis It is even her own question......

All depends on the particular scenario.
Let assume the concentrations of both halogenides and silver nitrate are low enough, so only the solubility product of silver bromide is reached. Then silver bromide is precipitated selectively ( if we neglect coprecipitation effects ).
Similar effect is achieved, if nitrate is being added so slowly the reprecipitation is faster than primary precipitation.
Otherwise, both salts are precipitated simultaneously. Saying this, be aware there will be ongoing fast or slow recrystallization/reprecipitation.
More soluble chloride will gradually dissolve in favour of bromide precipitation until chloride/bromide concentration ratio is equal to ratio of the respective solubility products.

And if we add excess we are basically precipitating both of them at the same time, regardless of relative solubility?

If you think otherwise I am curious about your reasons. // Who asks too much, thinks too little.

Well, in my lab experiment (measuring iodide and chloride) I conducted potentiometry and we measured the potential change of the solution. There were two equivalence points after 1.82 and 3ml of AgNO3 0.1M. That would mean that first the iodide, then the chloride precipitates, we also only gave amounts of 0.1ml of AgNO3 0.1M into it.
In my handbook it says that measuring AgBr with this method wouldn’t be possible because the solubility of AgBr and AgCl are closer together. But I didn’t really understand if they meant giving in AgNO3 as excess or how you put it, as infinitely small additions

Yes, potentiometry is preferred way here. 2 equivalence points do not mean there is no AgCl(s) before the first one. All depends on the scenario details. The point just means I- are done, Generally both nitrate and hologenides are too concentrated.

If some $\ce{AgCl(s)}$ is introduced into a bromide solution, it reacts with $\ce{Br-}$ ions according to : $$\ce{AgCl(s) + Br- -> AgBr(s) + Cl-}$$ so that $\ce{AgCl}$ is quickly transformed into $\ce{AgBr}$ which is nearly insoluble in, for example,  $\ce{NH3}$ solution. So if some $\ce{AgNO3}$ is added to a solution containing both $\ce{Br-}$ and $\ce{Cl-}$ ions, both may react in the beginning with $\ce{Ag+}$ ions, but after some time the precipitate is only made of pure $\ce{AgBr}$, because the $\ce{AgCl}$ precipitate (formed in the beginning) has all reacted with bromide ions. $\ce{AgCl}$ will be obtained only when the $\ce{Br-}$ ions have all been consumed.


text:
How to find the temperature for when Kc = 1
The formulas I can use are $\Delta$G = H - T$\Delta$S, and $\Delta$G = $\Delta$G(s.t.d) + RTlnK
I understand that when Kc is 1, the reaction is barely spontaneous and thus I can use the first formula for when $\Delta$G = 0 to find T. However, I have trouble understanding why the second formula can't be used here, or if it can, I don't know how exactly.
Overall, with my current understanding of the formulas, I am running into contradictions whenever trying to calculate something.
If someone can also explain to me as to how the first equation and second equation relate to each other, and provide a comprehensive explanation of gibbs free energy at equilibrium, it would be very helpful.

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What do you call $\pu{\Delta G(s.t.d)}$ ?

Standard Gibbs free energy change.

It is better to write $\Delta G°$ or $\Delta G^{\circ}$   to be displayed as $\Delta G^{\circ}$  // $\Delta G^{\circ}= -RT\ln{K}$


text:
Testing for lead in alcohol
I want to test for lead in alcohol rather than water. The reason for this is that I have many antique crystal decanters and flasks. Crystal is made with lead oxide (PbO) added to the glass.
Normally when crystal is tested, the procedure is to leave pure alcohol (ethanol) in the vessel for a fixed amount of time (like 3 months) and then test the alcohol to see if it has any lead in it.
Can I just use regular retail lead strips for this (they test water)? Or will the alcohol invalidate the results?
Is there a chemical test I can do that will be equally or more sensitive to the commercial strips? I have a fairly decently equipped small laboratory, so I can prepare a test if it is straightforward.

I do not think stripes would be usable. The best bet, but expensive, would be to get it analyzed by AAS, e.g. in water treatment lab.

I'd (1) test the mixture; (2) distill off the ethanol; (3) retest the water and ethanol fractions; and finally (4) adjust for volume and determine if the lead showed up in either. Seems like some folks have done similar with bourbon to some measure of success.

Lead compounds are insoluble in alcohol. It means that the amount of lead dissolved in alcohol is extremely weak. It should even be impossible to detect by usual tests, like colored strips. The only way of detecting it in alcohol is probably atomic absorption spectroscopy. And it needs a highly specialized apparatus managed by specialists : is not feasible for amateurs.

@Maurice If lead does not dissolve into alcohol why are there are all kinds of scare stories about how crystal is deadly or dangerous, like this one: https://www.latimes.com/archives/la-xpm-1991-02-08-vw-829-story.html. Is this just environmental bullshit with no chemical reality?

In the article, they say that lead dissolves in water at a level difficult to measure : 1 to 5 microgram per liter. It is 1 milligram in a cubic meter (1000 kg water). Can you imagine how small it is ? One milligram is the smallest amount to be seen.  One micrograms is simply to tiny to be visible, And in alcohol, lead is even less soluble. In my toxicology documents, lead is said to be toxic when more than 1 mg lead is absorbed per day. You will have to drink 1000 kg water to be intoxicated by the afore mentioned lead solution. Try to draw your own conclusion !

In ancient times, “sugar of lead” (lead (II) acetate) was deliberately used to sweeten wine. ;-(

I think this might be useful http://www.currentseparations.com/issues/16-3/cs16-3a.pdf introduction at least gives references.


text:
How to name radical
The name of biradical —CO—CH2–O—CH2–CO— is Oxybis (1-oxo-2,1-ethanediyl). Why is it wrong to write 1,2-ethanediyl instead of 2,1?
Thanks in advance!

The relevant rule for the numbering in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

P-15.3.1.2.2.4 Numbering of the components of a multipart di- or trivalent
multiplicative substituent group, when necessary, is achieved by attributing lowest locants to the atoms that are at the end of the component nearest to the multiplied parent structure, except where the component has a fixed numbering. The locants attached to the multiplied parent structure are cited last. (…)


text:
Loss of entropy and solvation energy in proteins
I am reading a chapter on protein stability. One section outlines the role of salt bridges or ion pairs in the stability of a protein. The excerpt goes as follows (italization added for emphasis):

The association of two ionic protein groups of opposite charge (e.g., Lys and Asp) is known as an ion pair or a salt bridge. About 75% of the charged residues in proteins are members of ion pairs that are located mostly on the protein surface (Fig. 6-36). Despite the strong electrostatic attraction between the oppositely charged members of an ion pair, these interactions contribute little to the stability of a native protein. This is because the free energy of an ion pair's charge—charge interactions usually fails to compensate for the loss of entropy of the side chains and the loss of solvation free energy when the charged groups form an ion pair. This accounts for the observation that ion pairs are poorly conserved among homologous proteins.

My question is, how does this look visually? I can read it but I am not able to completely understand the physical representation of the ion pair vs loss of entropy and solvation free energy.

Would be good if you could provide a ref to the text. Also, consider that generalizations, such as in this case, are to be interpreted with care. What is the particular function of any given ionic pairing interaction? Also, when you have many such interactions the effects might be synergistic.

[OP] I am not able to completely understand the physical representation of the ion pair vs loss of entropy and solvation free energy

In an ion pair, two amino acid side chains are in proximity. This means they have limited range of motion compared to side chains on the surface of the protein that interact exclusivly with solvent. It also means they are less solvated. If a salt bridge is observed, it means that the Gibbs energy of formation is negative. This could be some combination of entropy and enthalpy contributions.

[textbook] This accounts for the observation that ion pairs are poorly conserved among homologous proteins.

When a specific interaction is crucial for the native structure or the function of the protein, you would expect the responsible amino acids to be found in a set of homologous proteins, i.e. they would be conserved. Here, these structures form but are not crucial, so there is no evolutionary pressure to conserve the ion pairs in the same locations.

[textbook] these interactions contribute little to the stability of a native protein. This is because the free energy of an ion pair's charge—charge interactions usually fails to compensate for the loss of entropy of the side chains and the loss of solvation free energy when the charged groups form an ion pair.

The second sentence is a bit inaccurate. Instead of saying "fails to compensate", it would be better to say "barely compensates". A broader view is offered e.g. in this publication.

My question is, how does this look visually?

Protein structural data shows a static view, so it does not directly visualize conformational freedom. It also does not show all solvation (just water molecules that are ordered or somewhat ordered). In fact, some of the side chains lacking salt bridges will not be visible in a crystal structure because they themselves are disordered; lysine is a frequent example. To see a static but 3D view of the salt bridges in myoglobin mentioned in the textbook, you can view the interactive 3D figures of this short article in Proteopedia. Some of the amino acid side chains are shown in two orientations because they are disordered.

So they are making a comparison between favorable interactions in the bridged and unbridged form, claiming that these interactions are roughly equal in strength.
For the bridged form, the favorable interaction is just the attraction between a positive ion and a negative ion on different parts of the protein.
For the unbridged form, we no longer have this ion-ion interaction. However, in it's place, these ions can now interact with surrounding solvent molecules, which is also stabilizing. In addition, since the two groups with the ions are no longer linked together, they can move more freely, increasing entropy. This again is favorable energetically.
According to the authors, these effects are roughly equal in magnitude (with the bridged tending to be less favorable, but not impossible), so there isn't a strong preference for the bridged form for proteins with these kinds of charged sites. I would interpret, "poorly conserved" to mean something like 10:90 bridged: unbridged, rather than say 1:999.

Is it though? The authors state that the salt bridge itself is not a major stabilizing force because entropy and solvation fight against it. Therefore, "ion pairs are poorly conserved among homologous proteins". In other words, the magnitudes of both opposing forces are different. My question is more of why is that the case. Although correct me if I'm wrong.

@BrianBlumberg if these factors were equal in magnitude, we would expect the ratio of bridged to unbridged to be 50:50. I'm assuming "poorly conserved" would be roughly within this order of magnitude, so even something like 10:90 would mean they have roughly equal energy. If "poorly conserved" is more like 1:999, then I might be interpreting the passage wrong, but I believe it saying the energies in these cases are not too different (so no preference when comparing similar proteins), rather than the bridged being very unstable in comparison.

I think that the assumption the author makes is that if we only consider ion pairs, then the opposing force is greater. In other words, the ion pair is like putting on a seatbelt the wrong way. It might work preventing very slight forces but it'll just unclip if something major occurs (which is often the case in biological systems). So, the bridged form isn't unstable, just that the ion pair does little for the stability. Hopefully, Ph.D. Karsten Theis can explain this (he's edited my post and seems to know quite a bit about biochem).


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Does fluoroantimonic acid dissolve noble metals?
Hello,Does fluoroantimonic acid dissolve noble metals? Nernst equation suggests that platinum would require pH ≈ −20.1 to be dissolved, using Hammett acidity function as an extension of pH scale it seems that it should dissolve in fluoroantimonic acid. On the other hand possiblility of dissolving gold depends on the exact value of acidity function as it would require pH ≈ −25.7 to be dissolved (fluoroantimonic is sometimes stated to have H0 = −31.3). However wikipedia says that Hammett acidity function of fluoroantimonic acid is in range: −21 > H0 > −23.
Calculations:

Reaction
Standard electrode potential
"pH" required to dissolve

$Pt^{2+}+2e^-\rightarrow Pt$
$E^0=1.188V$
$pH=-\frac{1.188}{0.05916}\approx-20.1$

$Au^{3+}+3e^-\rightarrow Au$
$E^0=1.52V$
$pH=-\frac{1.52}{0.05916}\approx-25.7$

Sources:
https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)
https://en.wikipedia.org/wiki/Nernst_equation
https://en.wikipedia.org/wiki/Hammett_acidity_function

I, for one, would not use potentials in aqueous solutions to predict redox thermodynamics in what is clearly a nonaqueous system (presumably HF solvent).

One of the components of fluoroantimonic acid, antimony pentafluoride ($\ce{SbF5
}$) is itself quite a strong oxidiser, and may end up being what causes the dissolution to occur, if any.

(Partial answer. Maybe someone can post a complete answer)
Strength of an acid does not necessarily indicate its corrosiveness. Very strong acids like carborane acid will barely react with many materials, while weak acids like hydrofluoric acid can easily eat through glass. Theoretically, fluoroantimonic acid should react with metals but they are rather slow and dissolve much lesser amount of metal when compared to other acids (here is a comparison of how much steel will 1kg of fluoroantimonic acid dissolve vs 1kg sulfuric acid). For a metal to dissolve in acid, the acid needs to have some oxidizing power, and fluoroantimonic acid doesn't really have any, aside from the dissociated hydrogen ions themselves. The hydrogen ions in fluoroantimonic acid are not potent enough oxidizers to corrode metals. There can be another reason of not reacting violently because of formation of an oxide layer which will prevent the metal from reaction. Some may even form insoluble fluoride salts which will settle in acids and does not react further.
For comparison, see this video of the reaction of manganese in triflic acid
Also see this video: THE STRONGEST ACID IN THE WORLD Fluoroantimonic acid (shows the reaction of $\ce{HSbF6}$ with magnesium, sodium and potassium)
More links: Quora, Reddit

I don't see really any base for your claims, just speculation. In particular this link to some random forum is lame.

While passivation with fluoride may be possible here, same can't be told about oxide. You also undermine your point mentioning aggressiveness of HF, which is indeed the main component of  fluoroantimonic acid.


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A question on Lewis dot structures

In the image shown is a carboxylate ion, I don't understand why we draw 6 electrons on the oxygen and say it has a formal negative charge on it instead of putting 5 dots to represent it used one of its valence electrons? Like is it wrong to put an odd number of electrons on the atom in a Lewis dot structure, or is there some other reason?

It is the other way around: we are told that it has a negative charge on it, hence we draw 6 electrons instead of 5.

For future reference: consider to familiarize yourself with programs to draw chemical formulae (a comparison of some).  If your school does not offer to participate in a campus license, there are some versions with reduced functionality (compared to the full program), e.g., ChemDraw JS, or Marvin JS I used for the answer below running from any web browser allowing JavaScript.

Starting from acetic acid, there is single bond between oxygen and hydrogen because each of them contribute a single electron:

When the acid donates this proton, both electrons remain on the more electronegative atom (in the Pauling scale, oxygen: 3.44, hydrogen 2.20, reference).  Thus, this totals to six.
Note, when drawing structure formulae with electrons and electron pairs, do not confuse the dash indicating the charge, with the dash to represent two electrons.


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The oxidation number of N and Cl in NOClO4
My doubt is not about the solution itself, actually I was able to find the solution myself.
We could split NOClO4 as (NO+) + (ClO4-) and then proceed, but here I "knew" this split. Is there another approach to this question? If I'm unaware of this split, Is it actually possible to solve this stuff?

You are correct in the formulation

If I was unaware of this "split", is there a method to solve this?

No. The formula itself does not tell you whether it is (NO+)(ClO4-) or the other way around. You are supposed to deduce that using your knowledge of chemistry.

You have to remember things like ClO4- is a stable non-nucleophile anion that is often used as a counterion for reactive cations

The rule is simply to give the electrons to the negative end of any polar bond. In the case of the central nitrogen-oxygen bond in the formulation
$\ce{O=N\color{blue}{-}O-ClO3},$
the electrons would be accorded to the oxygen end and thus the split $\ce{NO^+, ClO4^-}$ follows. You also split the bonds within each component "ion" with the same rule until you find that each oxgen atom ends with $-2$, the nitrogen with $+3$ and chlorine with $+7$ as you seem to already know. (The nitrogen is $+3$ instead of $+5$ because it retains a nonbonding pair.)
Nitrosyl bonded with perchlorate is easy. How the electrons are precisely arranged in a metal nitrosyl complex is nontrivial. For instance, my understanding is that the nitrosyl moiety used to be deemed $\ce{NO^+}$ in the iron-nitrosyl "brown ring" complex, but now it's rendered $\ce{NO^-}$, with iron in a correspondingly higher oxidation state, instead.

First, ClO4 takes minus 1. The rest of ion is NO. It is plus 1. You should equal ClO4 to -1. Cl is +7. Also, you should equal NO to +1. N is +3. Thus, you are right. There isn't any way that I know to solve this issue-Fr0zenF0x


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Gas chromatography of samples containg water
I have read a few articles where it is stated that samples containing water are detrimental for a GC experimental results and equipment.
For example, in the paper Identification of essential oil components by gas chromatography/mass spectrometry, ed. 4.1 by Robert P. Adams, it is written: "Oxygen and water can severely affect the mass spectrometer both in the quality of spectra obtained and in permanent damage to components such as the filament."
But why is it so damaging with water in a GC machine? What happens with the filament, and how is the column damaged?

I assume searching for water damage GC MS column brings a lot of valuable hits.

yes, that is true, "The myth about water GC incompatibility is historical and comes from the old column technology, which were coated and not chemically bonded. " , from https://phenomenex.blog/2018/01/30/inject-water-gc/. So if I check that I have a compatible column, there should be no problems? Still in papers from 2017 and forward, they claim water is a serious problem...

In depends what stationary phase is used. Polar phases would not be happy about water. MS part is another topic. Water may affect MS sample input preparation and ioniation chamber, but I am not expert here.

It depends on the temperature and the polarity of the column, Apolar capillary columns contains usually polyethylene or similar polymer. They are not sensitive to water, Polar columns may be made of heavy esters. When working around 100°C, water has no effect. But at about 300°C, the risk of hydrolysis is not negligible. It depends of course of the nature of the stationary phase.

Obviously, I'm a novice... The sample is a water-based solution containing volatile components, such as aldehydes, esters, and an unknown pesticide. The column is of the apolar capillary type and the detector is built around a TCD design, the carrier gas is nitrogen, and the temperature is set to 120 C. Will I see a water peak? Or will the water just be diluted in the mobile phase and look like a drifting baseline?

This is a very specific case that you quote about GC-MS. Many mass spectrometers are used daily with water (called electrospray ionization MS), the issue is that of a particular gas chromatography-electron impact mass spectrometer.
First of all many GC columns have been invented that are meant to analyze water in solvents. We are talking about parts per million levels of water. However water is a low molecular weight compound and MS detector with electron impact ionization is not a good choice. So, the GC instrument itself with detectors such as barrier discharge ionization detectors are compatabile with water (of course very tiny amounts).
The issue is that GC temperature is pretty high (> 150-250 Celcius), and the filaments in mass spectrometers are made of tungsten/rhenium and other fancy metals/alloys. Heated water vapor is corrosive to metals. These metals are not totally inert and they are sitting at high temperature. They too react with the analytes slowly including hydrocarbons. Halogenated compounds are quite reactive too. Even organic solvents can damage them. Presence of water may accelerate corrosion.


text:
Are Wacker oxidations tolerant of carbonyl groups?
I was wondering if there was anything in the mechanism of a Wacker oxidation that would interfere with ketones or aldehydes already present in the molecule. Ketones and aldehydes can't be oxidized any further under normal circumstances, so I think it should be tolerant of these groups, but I am unsure.

Aldehydes can be further oxidised rather readily


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How to Combine an Emission spectrum into a colour?
I am trying to wrap my head around the flame test, and for the most part, it is making sense. My question is their a way to figure out the colour of a flame of an unknown element given the emission spectrum. In other words, how does the eye combine the mess of an emission spectrum into a single colour? I have been researching for a while now, so I am prepared for some pretty heavy maths. Some links would also be helpful for further reading.

Your brain receives information from your eye’s rods and cones and you see colors. What you see with an element’s flame atomic emission is just whatever your brain makes it. The element thallium gets its name from a Greek word that means “green twig” because it has a dominant strong emission at 535 nm. The story goes that it was identified spectroscopically, but the discoverer was color blind, so they called over their assistant to see it. Then the assistant apparently tried to lay claim to the discovery, being first to see the green emission. They lost!

I am afraid this is not really a question about chemistry, but about the physiology of vision (probably better suited for Biology StackExchange). Anyhow, take a look at various colour spaces that connect spectrum to the perceived colour, such as CIE 1931.

Lots of information at physics stack exchange. Example: https://physics.stackexchange.com/a/488695/313612. Notice at the linked answer that the answerer says the purest yellow is 571 nm and that is shown in the figure. But sodium’s famous yellow D lines are at 589.0 and 589.6 nm and and no other sodium lines are nearly as intense. They look pure yellow to me.

Sorry, I just want to turn an atomic emission spectrum into a wavelength. What spectral line does a flame choose, and why?

It does not pick a wavelength and is not a single color! What you perceive depends on your eyes and your brain. Look at some questions and answers such as this one: https://physics.stackexchange.com/q/564362/313612.

I am still learning, so sorry if this is a weird question, but how could you describe this sort of colour. I want a connection between atomic emission spectra and the colour,  It doesn't need to be too specific, just this atomic emission spectra produces the colour yellow because... Also, as Domen answered previously, could we plot this on a colour space like CIE1931. Sorry if this is repetitive, I'm just struggling to wrap my head around it.

There is no way to do what you want and trying to wrap your head around it can only result in a headache. It is like asking someone to summarize their favorite song by stating the violin (or piano or guitar or …) note to which it is equivalent. Look at the screen you have before you: the white background is from tiny red, green and blue LEDS. Use a magnifying glass to see this. Each emits a range of colors. By carefully adjusting those ranges and intensities, we perceive that mixture as white.

Thanks a lot for your help, Although you where a bit late for the headache!

Not a bad question, but off-topic.

It is not simple -- for example, to many people, red + blue spectral lines appear purple/violet, i.e., shorter wavelength than pure blue! Though you could assign a color name to a pure spectral line, as you state, the appearance of mixed spectral lines becomes much more difficult to describe.
In this color wheel, names are assigned to mixed colors.
An interesting case is didymium glass, used in photography and by glass-workers to block the bright sodium yellow lines. Under incandescent illumination, the glass has a distinctly pink appearance, and under skylight, it's decidedly mauve.
So take description of spectral appearance with a grain of salt... oops, that's definitely yellow!

(+1) Nice answer! Glad you see the sodium D lines emission as yellow and my Nd:glass laser rod behaves exactly as you describe in terms of colors. Also, if I remember correctly, Isaac Newton saw only 5 colors in the rainbow, so even our terms and names for colors are not so intrinsic as might be hoped.

I would have thought we could predict the colour of a flame by the emission spectrum, is there any way to get an approximate colour. The colour between people with normal vision doesn't vary dramatically for example I don't think anyone sees a sodium flame as green, so can we pinpoint the colour with RGB or something.  I am probably misunderstanding this so if you could clarify is this is impossible that would be benificial.

@EdV Hmm, I have never perceived sodium lines as pure yellow, always seeing there some warmer orangish/pinkish/salmonish tints.

@Yak I think the "forget" part was for you operatively. The Cie space in principle serves to the purpose for the average colour seeing people. Also, while the lines structure makes the prediction difficult for elemental emission, it is generally easier to predict the colour appearance by looking at the abs/em of molecules. Obviously roughly, is very hard to tell the exact nuance.

@Poutnik Maybe you are one of the very rare people with Tetrachromacy. ;-) Or we just perceive colors a bit different.

Light mixed from 2 simple light color can be usually represented by a single light somewhere between. But mixed red and blue light triggers perception of magenta/purple color which is not present  in the spectrum line range, but is the evaluation of the brain. It kind of joins the blue/violet and red ends of the spectrum into a closed circle.

@EdV Hmm, I am aware about the TCH, but I would rather go here:  "Yellow: 570–590 nm, Orange: 590–620 nm." . The sodium doublet 589.0 + 589.6 nm is at the orange border.

There are also impossible colors and so on. But getting back to chemistry, a digital photograph of a green-colored flame can have its intensity of green quantified by intensities of RGB pixels, yet this says nothing about what actually caused the green flame color, i.e., boron, barium, copper, thallium, etc.

@Poutnik Orange border for you, but pure yellow for me!

@EdV, thanks for pointing out "impossible colors." The crossed-eye color illusions are interesting! BTW, seven is considered a "magical" number, so colors such as indigo were added to make the spectrum more "complete". There was no color orange in English until the fruit (naranj, نارنج, and then misnamed) was imported,

As I understand your issue, you need to convert spectra into RGB triplets. Quantitative colorimetry theory addressing that issue is quite thoroughly stated in this wikipedia article on CIE 1931 color space

In other words, how does the eye combine the mess of an emission
spectrum into a single colour?

Although your query is related to vision and others have addressed it, let me clarify a couple of things that might be causing confusion. First, you are talking about simple flame tests for alkali metals, alkaline earth metals, and a few elements like boron and copper. Bunsen flame spectra are not a forest of emission lines but a very simple spectrum (indeed not a mess at all). You may have been misled by Google images search. Most of them are just cartoons and fake. Bunsen flame is a low-temperature flame. Not many lines are expected.
If you have a pocket spectroscope or even a homemade CD spectroscope, you will see that sodium flame is just one orange-yellow line around 589 nm. Similarly, potassium flame spectrum is just one or two lines. If you are lucky with sensitive eyes, you might the deep red and violet lines.
See this excellent video of the true flame colors Flame emission spectra. The video shows the simple spectra. You will also note that the yellow sodium line is usually present in most spectra...Na is everywhere.
You can try making a CD spectroscope to try it yourself. Internet is full of such easy constructions, but the spectra on Google Images is disappointly cartoonish (or rather software generated) from at a first glance.
Another misconception to be noticed is that many people think that the Bunsen flame colors are always due to atomic emission. This point is not valid. Ca, Ba, Sr, Cu, B, flame colors are due to small molecular compounds of those elements, not their atoms. Only the alkali metal emission is atomic emission, and the rest are molecular emission in a Bunsen burner.
For real analytical work, Bunsen flame color is useless. We have to use a high-temperature from a spark or electric discharge or a plasma around 5000 K to see true atomic emission of most of the elements in the periodic table.  Then, indeed, the emission spectrum is a mess, mainly in the ultraviolet, and you certainly need a device to separate wavelengths. The concept of color then vanishes because we cannot see ultraviolet.

Sorry, I cannot see sodium emission photos. Which link are you referring to?

This link is useful: https://www.rapidtables.com/web/color/RGB_Color.html. I got it from an answer over at electrical engineering stack exchange. The question there is https://electronics.stackexchange.com/q/590739/223146.

One last very informative link: http://hyperphysics.phy-astr.gsu.edu/hbase/vision/cie.html. As stated there, RGB space is smaller than human perception of colors. Pretty neat!


text:
Virial expansion of real gas equation and volume work
Hi I am currently working on a thermodynamics problem, and it asks me to first find the volume work done on a gas that behaves like the simplified Dieterici equation (see below) predicts on isothermal compression.

I already started working on it, but am not sure if it is correct; I started with the idea that the volume work done on a gas during compression is the negative integral of p over dV (Where Vi is the volume the gas started with and Vf the volume of the gas after the compression):

First of all; would that be the correct expression for the volume work? I am honestly not sure.
In the second part of the problem, it asks me to find the virial expansion of the first given equation (Dieterici equation) of p to quantify the difference of the volume work between an ideal gas vs a Dieterici gas. I am unsure how to do that, and why virial expansions even matter in this context. How would I even start with that?
Thank you all in advance.

That is not the Dieterici equation, since the Deiterici eqn. contains an exponential.   Rather, it is the first two terms of a Taylor expansion of the Dieterici equation about $T=\infty$.  I.e., it is an approximation of a Dieterici gas.  2) Even with that, it's still not quite correct to say your expression gives the work for that approximation of a Dieterici gas.    Rather, what you've derived is the correct expression for the work done on a two-term series approximation of a Dieterici gas about $T=\infty$, when it undergoes a reversible and isothermal compression or expansion.

Screenshots are discouraged, since they are not searchable.  It is encouraged that you instead use MathJax to show equations.

Hello thanks for your help. Yeah it’s true that it’s about isothermal compression. You’re right, and how would I formulate the virial expansion for the simplified dieterici equation

I'll leave that to someone else, but I will say that, if you want an answer, you need to do three things:

Search this site to see if your question has already been at least partially answered, and include references to those answers in your post.
Show your own work on formulating a virial expansion as a starting point.
See my 2nd comment about screenshots.

You may find useful these links:
notation ,  formulas/expressions ( not to be applied to titles ) and
upright vs italic

What is your understanding of the general form of the virial expansion?


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Why does the reaction of Brady's reagent and cyclohexanone not also form an enamine-type product?
In this reaction in a Khan academy video, cyclohexanone reacts with Brady's reagent to form hydrazone product.

My question is why doea Brady's reagent + cyclohexanone not also form an enamine product where the central, secondary amine group attacks the carbonyl?
My two guesses would be sterics and/or electron withdrawing effects of the ring, but is there another reason? Would this product exist, but minimally?

Perhaps the nitrogen next to the aromatic ring could attack the carbonyl group. However:

The bulky phenyl group is closer to the reaction site, and thus causes more steric interference, than would be the case with the primary nitrogen attack.

The nitrogen next to the phenyl ring conjugates its electron pair to the ring just as in a typical aromatic amine, so it is electronically deactivated as a nucleophile. This is especially the case when the ring has strongly electron-withdrawing ligands at other position.

Arylhydrazines are thus much more reactive nucleophiles at the nitrogen that is farther from the aromatic ring.


text:
Why doesn't 2,3-pentadiene show geometrical/cis-trans isomerism?
There are two different groups bonded to each terminal C in the C=C=C system, and there is restricted rotation about the C=C=C. So why doesn't it show cis-trans isomerism?
The most common answer I get for this is that the set of bonds C-CH3 and C-H on C1 and C3 in the C=C=C are not in the same plane. Why do they have to be in the same plane for the exhibition of cis-trans isomerism?

Because of the way cis-trans isomerism about a double bond is (sensibly) defined, you need a plane. Note that in all cases, there are two enantiomers.

The cis-trans isomerism makes sense when the final substituents are in the same plane. In 2,3-penradiene, the final substituants $\ce{H}$ and $\ce{CH3}$ are not in the same plane. They are in planes which are perpendicular to one another.

The (cis)/(trans) nomenclature (or, more universally applicable, (E)/(Z) nomenclature) works fine for isolated and conjugated double bonds.  In the case of 2,3-pentadiene, however, where double bonds literally are adjacent to each other, you have a cumulene.  As a result, the mutual orientation of two terminal methyl groups is close to perpendicular.
To put things in perspective (literally), build a model with a model kit:*

and apply the nomenclature for axial chirality instead.  You assign the methyl groups a higher priority than the hydrogen atoms (CIP rules), then, while looking along atoms (2) and (3), the «rotation» along the sequence 1 -> 2 -> 3 -> 4 is clockwise, thus it is (2P)-penta-2,3-diene (where P (as in plus, to the right, clockwise), and M (as in minus)  is used to describe the  axial chirality and helical chirality.
The relevant rules in the Blue Book (see resources page) are P-92.1.2.1, P-92.1.2.1.2, P-92.1.2.2.1, and P-92.1.2.2.3.  If you want to continue to use (R) and (S) to assign axial chirality, add a subscript to the label (here, they correspond inversely and hence (2P)-penta-2,3-diene is ($2S_\text{a}$)-penta-2,3-diene).
*In case you do not have access to a model kit, you may paste the following SMILES string (a structure description a [dedicated] computer program understands) CC=[C@@]=CC as input for molcal.org.  Don't mind that the angle enclosed by the methyl groups differs from the one you get with a physical model kit (example), focus on the relative orientation.
Reference:
Favre, H. A., Powel, W. H. Nomenclature of Organic Chemistry. IUPAC Recommendations and Preferred Names 2013. Excerpt prepared by Moss G. P. at https://iupac.qmul.ac.uk/BlueBook/

Thanks. Actually what exactly is cis-trans isomerism? Is being on the same plane a pre-requisite?

I get that it's chiral, but don't get why it cannot show cis-trans.

To show cis/trans, you need two substituents at each end of the double bond.

Eventually, it is convention to use (R) and (S) here to provide an unambiguous description about the atoms' arrangement.  cis is about from the same side of a reference plane, and trans about from the other side. For but-2-ene, this reference plane a would be orthogonal to a plane b passing the double bond and one of the two methyl groups, while a would pass the double bond. If you consider wiggling the 2,3-pentadiene of the methyl groups (conformational changes, not constitutional ones), this however could yield a fuzzy description despite no bonds were detached/rebuilt.

@Buttonwood: Did you mean M/P? See the first two entries here: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/MP14221.html

@user55119 For cumulenes and axial chirality, I'm more used to (R) and (S); on the other hand, (P) and ($\Delta$) / (M) and ($\Lambda$) more helical chirality like in the case of helicenes (original publication). This being said, I perceive using M/P on cumulenes as equally suitable tools.

Are you sure this is (R)? It looks the groups on the far end of the chiral axis are mislabeled, that methyl group should be 3. Then the 1-2-3 rotation is counterclockwise and so the configuration looks like (S) to me.

@OscarLanzi Thank you for the reporting the error. A turn to the right to yield a helical P, which here indeed is (in CIP) an $aS$ (or alternatively, $S_a$).

Let me speak of geometrical isomerism as such the discussion is more general.
Geometrical isomerism of the cis-trans / E-Z type can happen about a plane containing two atoms each with two different substituents (though in cyclic compounds the latter requisite is merely formal). In fact the two identical substituents or the pair with high priority can be at the same or at the opposite sides of that plane, leading to different compounds.
If such a plane does not exist, or better if there are not such two atoms, there are not geometrical isomers (as properly intended*).
*Other isomerisms lead to different geometrical arrangements of atoms and moieties in molecules, of course. This often confuses newbies. The term geometrical in geometrical isomerism solely refers to the arrangement discussed above.


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Are there examples of molecular formulas with nested groups?
Sometimes a molecular formula is written using groups, for instance $\ce{Al2(SO4)3}$.
Are there examples of formulas with nested groups, such as $\ce{ABC[MNO2(XYZ)4]3}$ or is this avoided?

Very good question. My first thought was organometallic complexes or products of incongruent weathering of igneous and metamorphic minerals but I've not been able to find an example. Maybe the organic realm?

I'm not quite 100% sure if this is what the OP want but there's  Na[HFe(CO)4] on https://en.wikipedia.org/wiki/Metal_carbonyl_hydride - Pull off the second proton from the carbonyl hydride and you are there, but it might qualify as is

@IanBush On that same page there is HCo(CO)3(P(OPh)3) that could qualify? Or are nested groups written with square brackets for the outer group and parentheses for the inner one?

@xenoid You wrote the question, you judge what qualifies ...

@IanBush I'm not a chemist, just trying to figure out how formulas could look like for a program that will process them.

@xenoid then you could have asked a more straight question, ie how connectivity is input, or how chemicals are listed in databases

@xenoid If you are looking for a way to programmatically build molecular structures from chemical formulae, SMILES is one commonly used option

Too long for a comment:  Chemical formulas for polymers can utilize nested enclosing glyphs but, from what I've seen, the accepted format is to use parentheses for the inner glyphs, and square brackets for the outer glyphs, rather than nested parentheses.
An example would be polystyrene, whose monomers look like this (source: Wikipedia: Polystyrene):

Given this, the formula for polystyrene could be given as:
$$\ce{CH3–(C8H8)_{n}\! -CH3}$$
But some might prefer to instead give it, more informatively, as the following, in which case the nested enclosing glphys make an appearance:
$$\ce{CH3–[CH2-CH(C6H5)]_n –CH3}$$
[See William Reusch's polymers website (MSU), Section 2. Writing Formulas for Polymeric Macromolecules , which provides many such examples.]
Using parentheses inside square brackets is also the convention given here for complex ions:
Wikipedia: Chemical formula
Since you're trying to write a program, you might find something here to be of use:
Stack Overflow: Can nested parentheticals be parsed in chemical formulae?

The closest I could find is mixed cation-anion salt:

$\ce{NaM2[PuO4(OH)2]·4H2O}$ where $\ce{M = Rb,Cs}$ and
$\ce{NaRb5[PuO4(OH)2]2·6H2O}$.
$\ce{(NO)_{0.25}(NO2)_{0.75}Cu(SbF6)3}$, $\ce{(H3O)_{0.125}(NO)_{0.125}(NO2)_{0.75}Zn(SbF6)3}$, $\ce{(H3O)_{0.25}(NO2)_{0.75}M(SbF6)3}$ where $\ce{(M=Cu, Zn)}$

Also make sure to check double salts and this question.
References

Grigor’ev, M.S., Krot, N.N. & Charushnikova, I.A. Synthesis and Structure of Mixed-Cation Salts with [PuO4(OH)2]3– Anions. Radiochemistry 60, 233–240 (2018). DOI: 10.1134/S1066362218030025
Z. Mazej, E. Goreshnik, Eur. J. Inorg. Chem. 2021, 1776. DOI: 10.1002/ejic.202100139


text:
Where can i find accurate bond angles?
I am trying to find places where I can have accurate bond angles of certain molecules such as $\ce{F2Se}$, $\ce{Cl2S}$, or even $\ce{S2Se}. I have looked everywhere and have found only 3 sources that give me reliable and accurate bond angles but I need a couple more.
So far I have the databases I have consulted are:

Molview simulation
The Cccbdb database
Understanding trends in molecular bond angles paper
TEACHING THE VSEPR MODEL AND ELECTRON DENSITIES.

(I am aware that Molview is a simulation.) I need this as I am writing a chemistry paper in high school where I am testing the accuracy of the VSEPR theory. According to the people correcting my paper, I need five sources in total.

Do you need five databases or just 5 sources for your paper? You could certainly use other sources that describe the theoretical background.

Hi, welcome to Chemistry SE! Now I understand your school policies, but I would recommend using sources/examples that the students are aware or something that you have discussed on the class, of it's for an exam. After all they are only students. The ones that come to top of my head is Hydrides of oxygen family, where bond angles become towards 90 deg down the group.

As noted in your previous post of this same question, search the primary literature. You can use Google Scholar for example to do this. In just one search, for example, I find this paper which has the bond angles of F2Se and Cl2S along with many others (see Table 2 in the pdf which you can download for free): https://www.ingentaconnect.com/content/apcs/apcs/2016/00000032/00000007/art00011#

Why did you delete and repost your question? Note that single line breaks between text aren't reflected in the post. You can see for yourself that your list of sources has been condensed all into one single paragraph, which makes it difficult to read. On top of just reading the text that you type in, it's important to also check how it looks after being posted (there is even a preview screen to help you with that). Also, you should have added links to these sources, instead of making people Google them.

There's a great open access paper from 2020 on the subject:
"Understanding Trends in Molecular Bond Angles"
J. Phys. Chem. A 2020, 124, 7, 1306–1311
The authors calculated a wide variety of species using accurate quantum chemical methods (CASSCF/CASPT2 method, ANO-RCC basis set).
They rationalize the trends as a function of the polarizability of the atoms involved.
Not sure about $\ce{S2Se}$ but it's a nice article explaining trends in VSEPR bond angles. Maybe a bit technical in places, but the figures, tables, and supporting information are all quite detailed.

That's apparently one of OP's sources listed in their question, although I'll certainly agree the lack of formatting makes it difficult to spot. A bit annoying, because I edited their previous version of the question to make it easier to read, only for them to delete it and repost the unedited text.


text:
What are the resulting conjugate acid and base of phenol and 4-nitrophenol
I have to complete the equation
4-nitrophenol + phenol --->
and then label the acid, the base, the conjugate acid and the conjugate base.
I think the answer might be
4-nitrophenol + phenol ---> 4-nitrophenoxide + [C6H5OH2]+
I don't know what the name of the last compound would be and it looks unlikely, but it's the only option I can think of.
If this is right, what would the name of the resulting conjugate acid be? If it's wrong, why? and what would the right answer be?
Thank you in advance

I would not expect neutral phenol molecules to be protonated to a large extent by an acid as weak as 4-nitrophenol. Second reactant should be phenolate ion maybe?

@OscarLanzi would you know of any alternate solutions?

@Gallus Construction in ChemDraw JS of the protonated phenol (for reference, SMILES string this version assigns is [H][O+]([H])C1=CC=CC=C1) is auto-named phenyloxonium.  For future reference, among ACS' organicchemistrydata.org are Reich's compilations of pKa, e.g., here including the two phenols in question.


text:
Does changes in pH affects cyclic form of glucose in water solution?
Glucose in water solution is mainly in cyclic form. Both base and acid can catalize formation of hemiacetal, but in distinct mechanisms. I found information that monosacharides eg. glucose exist mainly in cyclic form under neutral pH eg. human organism pH~7,40, but in alkaline and acidic environment chain becomes open and open chain is the form that is most abundant under this conditions. Is that true and if it is how it can be explained?

I would be curious to know if you still have that source claiming the glucose becomes uncyclized. In any case, I asked my biochem prof your question and he said nothing really changes especially for cyclized glucose. Here's why nothing would change: the pKa for a primary alcohol is around 15. This means you're going to have to increase the pH to at least 15 if you want the alcohol to be deprotonated. Also, that pKa value only increases as you switch from primary to secondary, to tertiary alcohols (here's a site with a lot more pKa values: https://www.engineeringtoolbox.com/paraffinic-benzoic-hydroxy-dioic-acids-structure-pka-carboxylic-dissociation-constant-alcohol-phenol-d_1948.html). Recall that the majority of hydroxyl groups on glucose are secondary alcohols. In other words, you're going to be needing a solution that is so powerfully alkaline if you want to see any changes to a glucose compound. Further, it seems unlikely that increasing pH (even extreme increases) would uncyclize glucose and even more unlikely that it would uncyclize polymerized glucose. If you wanted to open a glucose ring, you're looking at something along the lines of an SN2 reaction, not merely an increase in pH.


text:
Is my understanding of conjugate acid-base pairs correct?
$$\ce{HCl_{(aq)} + H_2O_{(l)}<=>H_3O^{+}_{(aq)} +Cl^{-}_{(aq)}}$$
Cl- is the conjugate base of HCl. Now, can I say that HCl is the conjugate acid of Cl-?

If you have performed a simple search, the Wikipedia page Conjugate acid would tell you you can.

In the Arrhenius sense, exactly yes.
Replace Cl- with some arbitrary species A-, and HCl with HA, and you get the exact Arrheniusian definition.

Rather by the Broensted-Lawry model. By the Arrhenius model, Cl- nor NH3 are not bases, only fictitious NH4OH is.


text:
Why are most organic molecules diamagnetic, when most of them have non-paired nucleus spins?
I was studying some coordination chemistry and somewhere around the text it was stated that one could know the magnetic moment of a complex from its metal because since all ligands were often diamagnetic, closed shell molecules, the unpaired electrons were centered on the metal anyway. It is implied that the spin of all of those unpaired electrons is the source of the paramagnetism.
So my question is this: From the common isotopes of C, H, N, S, O and P only oxygen-16 and carbon-12 have zero spin. Then why don't the spin of the other nuclei do not contribute to the paramagnetism of the sample the way electrons do? Particularly hydrogen, which is often quite abundant in organic ligands and has a spin of 1/2, same as the electron.

The paragraph was obvisouly talking about electron spin, not nuclear spin. Nuclear spin has very little contribution to magnetic properties, as nuclear magneton is much smaller than the corresponding electronic magnetic moment (Bohr magneton), due to the much higher charge/mass ratio of the electron

Do you refer to compounds like TEMPO, subject to electron paramagnetic resonance (EPR)/electron spin resonance (ESR)?

See also Nuclear_magneton

@Greg I believe you have nailed the question. Would you mind to write an answer covering those bases so I can close the question?

As requested, this is my answer:
The question is essentially about why nuclear contributions to magnetic behavior is generally neglected.
It is true that both nuclei and the electrons can carry unpaired spin in a material. The magnetic moment associated with these spin is proportional to two constants: the nuclear magneton in the case of nucleus and the Bohr magneton in case of electrons. These are the constants essentially telling you how strong the given spin interact with the magnetic field, and they are inversely proportionate with the mass of the particle. In other words, they depend on how big is the charge of the particle compared to its mass. Since electrons are about 1800 times lighter than a nucleon, the magnetic properties of a material will primarily dependent on the electronic spin and the contribution of nuclear spin is negligible in most cases.

So if I understand it well, nonzero nucleus spin leads to very slight paramagnetism contribution, which in total effect just very slightly decreases the diamagnetism of the material, but cannot cause paramagnetism without electron contribution.

@Poutnik yes, essentially that is what happens

@Poutnik and about the electrons, there is the spin and also the orbital magnetic moments to consider. It's just that, for light elements, the orbital component is often negligible in comparison to the spin component.

@urquiza Yes, I am aware about both electron angular momenta.

@urquiza Orbital momentum generally breaks down due to low symmetry, its contribution is low even in lighter transition metal complexes (except cobalt, but that is it).


text:
why can water heated to a certain temperature in microwave explode when removing cup , bowl etc to room temperature air?
Studied a lot about $\mathrm{pH}$ , molecule dipoles properties $\ce{H2O}$. never saw anything along lines of energy storage, except ice contains heat energy , ionization of water always happens. boiling is vapor pressure of water overcoming that atmospheric pressure , why can not do it (boil) at high altitudes. Can you get the water hot though at a  high altitude?  Question about microwave exploding was main question I wish answered ,tks. Water buoy Doug.

Try to clean and organize your thoughts more while writing a question. This is rather a mess. If it was like four time longer, it would not be readable.

"Opposite" and related https://en.m.wikipedia.org/wiki/Glaze_(ice)

Water in a microwave oven can get easily overheated above boiling point and then by a random trigger starts boiling violently.
With increasing altitude and decreasing pressure, boiling point of water gets lower. If low enough, cooking of various food becomes impossible, unless pressurized cookware is used.
If a vacuum pump decreases air pressure enough, water boils even at room temperature. See also Wikipedia: Armstrong limit for atmospheric biological limitations:

The Armstrong limit or Armstrong's line is a measure of altitude above which atmospheric pressure is sufficiently low that water boils at the normal temperature of the human body. Exposure to pressure below this limit results in a rapid loss of consciousness, followed by a series of changes to cardiovascular and neurological functions, and eventually death, unless pressure is restored within 60–90 seconds.On Earth, the limit is around 18–19 km (11–12 mi; 59,000–62,000 ft) above sea level, above which atmospheric air pressure drops below 0.0618 atm (6.3 kPa, 47 mmHg, or about 1 psi). The U.S. Standard Atmospheric model sets the Armstrong pressure at an altitude of 63,000 feet (19,202 m).


text:
Oxidation number of methylene blue
I came across methylene blue's structure when it reacts with oxygen.

The left is the reduced state, and the right is the oxidized state.
However, when I looked at them carefully, the oxidation number of O2 changes normally, but
I couldn't find the change of oxidation number in methylene blue.
( the middle N is -3, S is -2, the right N is -3, in both structures. The H is always +1.)
What am I missing here?

Count the number of atoms. It has changed.

2H missing from methylene blue, and added to O2 to make H2O2.
But their oxidation numbers are +1 on both molecules. Does that explain the oxidation of methylene blue?
( I know losing H is one kind of definition of oxidation, but I want to know how to justify it when it contradicts the oxidation number definition.)

The concept of oxidation numbers is rarely used in organic chemistry. That being said, you surely can assign them, and the result won't contradict the overall picture that we already know. If you insist on doing that, then look again at those carbons, especially the one which is doubly bonded to N in the right structure, but not in the left one.

Thank you! I missed the change of carbon atoms!

You missed the carbon atoms, and specifically the carbon atoms in the central ring which are bonded to nitrogen and sulfur. When the methylene blue is in its reduced leuco state, each of these has one bond to the more electronegative atoms and none to hydrogen (the only other element in the molecule), thus you count out the oxidation state as +1 for all four of these carbons.
Now you shake the bottle and it turns blue. Note that in this form two of the central carbons now have two bonds to the nitrogen or sulfur, so they have been oxidized from +1 to +2. Which two carbons get oxidized depends on which of multiple contributing structures you draw, but the net result is always you oxidize two carbon atoms by one unit.
Half the available contributing structures have the two carbon atoms on the left side of the central ring oxidized, and half have the carbon atoms on the right half oxidized. Since the two halves are equivalent, if you are into fractional oxidation numbers you can say that "on average" the four central carbon atoms each go from +1 to +1.5.


text:
Given type and number of carbon atoms and degree of unsaturation of a hydrocarbon, write structural formula
Write a structural formula of the hydrocarbon that contains 4 primary carbon atoms, 2 ternary carbon atoms, 3 quaternary carbon atoms, and has degree of unsaturation 3.
This has been given at the chemistry test today. Only one person succeeded the problem.
Correct solutions are:

In my attempts I made a triangle and a square, but none of them seemed to work to satisfy the type of carbon atoms and I stopped.
My question is - is there any algorithm to solve this type of problems, or are there any other information that we can extract from the problem to help us draw the structural formula, aside from the number of hydrogen atoms (in this case, 14)?
As an unexperienced programmer who worked with graphs, I can only imagine a brute force for the bonds. For example, a ternary carbon atom could have a triple bond, or a single and a double. The primary carbon atoms are easy to solve, we must link them to a single bond, but how about the other ones? Is there an official software for this type of computations?
We have just begun organic chemistry, and the teacher missed 4 classes, that is 8 lost hours of chemistry that we do not recover. Please understand that I am an organic chemistry noob.
Thanks!

This solution is a blatant, outrageous, malignant nonsense. As for the general approach, brute force will do.

@Ivan Neretin. Why do you call it nonsense ? I agree the bent C=C=C bonds are non realistic. But why call it a non-sense ?

@Maurice Because profane swearing is forbidden here, or so I was told.

@IvanNeretin Sorry, I did not know it is impossible to bend C=C=C, no one told me. I have edited the post providing another formula.

Now that's another story.

If you have a sum formula then you indeed can generate/write down quite a number more or less plausible molecules. The numbers of entries may be huge (e.g., GDB13), and imposing any structural features may reduce them greatly (e.g., molgen as one of the generators) to sets easier to manage, e.g., in mass spectroscopy.

This answer will give you little insight about chemistry, and maybe a little insight into problem solving strategies.

The primary carbon atoms are easy to solve, we must link them to a single bond, but how about the other ones?

This is a good insight. The entire structure has 9 carbons and 14 hydrogens. To simplify it, we can replace the 5 methyl groups by hydrogen, giving us the formula $\ce{C5H4}$. This core structure will harbor the rings and double/triple bonds, and the methyl groups can be added later.

For example, a ternary carbon atom could have a triple bond, or a single and a double.

Rather than going for an by-atom strategy, it might make sense to think about the total number of double and triple bonds as well as rings. The degree of unsaturation is three. Each double bond and ring contributes one while a triple bond contributes two. For a total of three, you could have:

a triple and a double bond
a triple bond and a ring
three double bonds
two double bonds and a ring
one double bond and two rings
three rings

With five carbon atoms to work with, three double bonds implies an allene. Here are some examples of the structures you would start with. With the small numbers, it might not be necessary to use an algorithm. Many of these are ridiculously strained, so there would have to be a filter to remove the impossible structures.

Adding the methyl groups
To each of the 5-atom structures you would have to add the methyl groups (replace a hydrogen with a methyl group) in a way that no CH2 groups remain. Sometimes, there are multiple ways and sometimes there is only one. In the end, you get more than the two example structures, some of which are plausible (for example, the structures with a triple and a double bond lacking any rings).


text:
Could supercritical CO2 or N2 support biogenesis?
At the surface of Venus, at a pressure of 93 bar and a temperature of 737 K, both carbon dioxide and nitrogen are supercritical fluids.
Now, supercritical $\ce{CO2}$ is used as a solvent in processes such as decaffeination of coffee beans. But is it possible that the supercritical $\ce{CO2}$ on the surface of Venus could support more complex chemical reactions by dissolving substances, mixing them, and promoting reactions between them? Is it possible for supercritical $\ce{CO2}$ to play a similar role to what water does on Earth?

Comments are not for extended discussion; this conversation has been moved to chat.

What you are missing is that supercritical fluid can be "gas-like" or "liquid-like". Liquid-like CO2 is used for extraction, but in such high temperature it would be gaslike, similar to the one you're breathing. You'd need way more pressure to make CO2 that hot, to be liquid-like.

If you are interested on the "bio" aspect, I would rather focus on the T.

...  and acidic fumes of sulphuric acid.

Mithoron, I admit I've never been sure how to conceptualize a supercritical fluid, but that helps.

@Greg Most of "permanent gases " are at room temperature supercritical fluids. At low pressure, their properties converge to properties of ideal gas. At high pressure, their (at least some) properties converge to properties of liquids. All without any sudden phase change. Imagine if you had a vapor, being progressively pressurized and  progressively becoming a liquid, without any condensation.

@Poutnik Some phase diagrams have dashed or even solid lines extending horizontally and vertically from the critical point. I'm guessing those lines don't really have a physical meaning. A gas doesn't suddenly gain or lose some property when it crosses one. So it doesn't really make sense to point to the part of the diagram labeled "gas" and say "it is not supercritical here"?

Sure those lines have no physical meaning just nomenclature one. Supercritical CO2 used for some extractions does not principally differ from supercritical nitrogen, if pressurized to comparable densities. The truth is, if T is much higher than Tc, we can practically forget it is still in supercritical region and consider is as permenent, near ideal gas.

It is better to denote the respective regions as liquid phase, gaseous phase and supercritical phase, with the latter being neither of the former two and having continuous convergence to both.

By other words, a supercritical fluid can be gas-like without being a gaseous phase, or it can be a liquid-like without being a liquid phase.

If your key question is just the solvation ability of supercritical carbon dioxide and nitrogen for biogenic reactions, keep in mind that pure supercritical carbon dioxide is a very poor solvent. It is close to a solvent like pentane. I am sure supercritical nitrogen would be even worse solvent for polar compounds. The beauty of supercritical fluids is that their solvation ability depends on the density as well. Higher the pressure, better is their solvation ability in general.
Those who do supercritical fluid chromatography with carbon dioxide know that most bio-molecules / amino acids/ nucleic acids/sugars etc. are not soluble in such a system. Secondly, supercritical carbon dioxide is not miscible with water, you have to add small amount of alcohols to allow some water with pressurized carbon dioxide, in that case it is no longer a supercritical fluid.

Thank you. I think you understood the question perfectly, and it looks like the answer is not what I hoped it would be, and also more complicated than I thought it was. And, whoever edited the title, it does look a little classier now.


text:
Is there a canonical variable for period and group?
For example, "Z" is the standard symbol for atomic number. I'm writing a manuscript that uses the group and period of elements within some equations, and so far I'm just denoting them as $G_{table}$ and $P_{table}$, defining them in the paragraph before the equation is presented. But there are a couple of concerns with that, namely:

If there is a canonical symbol that I forgot after undergrad, I'll look rather silly to a reviewer
A reader skimming the paper might confuse those symbols for Gibbs free energy or pressure

Is there a standard variable that's commonly used to denote the group or period of an element within an equation?

Don't worry too much about a reviewer. They are after human beings like us. No there is no symbol for periods or groups.

That's what I was thinking, but I just wanted to make sure I wasn't totally failing at google!

There isn't a standard symbol. However, if you choose these current symbols, please note that "table" should be placed in upright font, not italics:

correct: G_{\text{table}} $G_{\text{table}}$
wrong: G_{table} $G_{table}$

The reason is because "table" is a plain old English word. You can see Which symbols are written in roman (upright) font and which are italicized? for more information; even though that post was written mainly for Chem.SE, the information there is applicable to all of chemistry. The IUPAC Green Book writes (Section 1.3.1)

Subscripts and superscripts that are themselves symbols for physical quantities or for numbers should be printed in italic type; other subscripts and superscripts should be printed in Roman (upright) type.

and other sources such as the ACS Style Guide have similar recommendations.
Personally, if I were you I'd just cut the subscript altogether, unless it's likely to lead to confusion. No need for clunky symbols. For example, Euler's polyhedron formula usually just reads
$$V - E + F = 2$$
instead of something like
$$n_\text{vertex} - n_\text{edge} + n_\text{face} = 2.$$


text:
Inductive Effect of -CH3 group
3° carbocations are the most stable because of the +I effect of -CH3 group. This group gives a slight negative charge to the central carbon atom, which makes it less positively charged and thus, more stable.
But, in the case of -N(CH3)3 and -NH3, the central N atom has a positive charge.
-CH3 group has +I effect on the central N atom, making N atom slightly less positively charged. So, -N(CH3)3 group should have lesser -I effect. Isn't this statement correct?
But, -N(CH3)3 group has greater -I effect than -NH3. Why does this happen?
Why does -N(CH3)3 have greater -I effect than -NH3?

The R3N- group has a full positive charge that you omitted from the structure.

Yeah, but still the +I effect of R will make the N atom less positively charged, similar to the case of carbocations. So, in R3N- group, the central N atom, being slightly less positively charged than that of H3N- group, should give lesser -I effect than H3N- , right?

Sorry if I have made any mistake. I am in high school now. I cannot understand this phenomenon. Why does -N(CH3)3 gives more -I effect than -NH3?

Are you speaking of the positive $\ce{NH3^+}$ ion, or of the negative  (-)$\ce{NH3^-}$ ion, or of the neutral molecule $\ce{NH3}$ ? Or even of the $\ce{R-NH3^+}$ ion ?

NH3+ ion and NR3+ ion. "-NH3" and "-NR3" are common ways of representation of those groups.

The methyl group carbon has a partial negative charge due to the polarity present in the C-H bonds. This partial negative charge is preferentially donated, leading the methyl group to have the +I effect.
An important way to consider the stability of a tertiary carbocation is through the process of hyperconjugation. Neighboring C-H or C-C bonds, parallel to the vacant p-orbital of a carbocation, can donate some of their electron density to the empty p-orbital, thus stabilizing the positive charge. Tertiary carbocations have a greater capability of providing this electron density, and thus are more stable.
The positive charge present on the nitrogen in either the -NR3+ or -NH3+ groups induces the extent of the -I effect exhibited by both groups. Although it may seem that the presence of alkyl groups would lend a greater +I effect to the -NR3+ group, there are other factors that result in this "anomaly" of sorts.
The factor that causes this is the issue of steric hindrance. The alkyl groups surrounding the nitrogen are relatively large and do not want to be in the vicinity of each other. Although the structure of the -NR3+ group would appear to be tetrahedral, as the alkyl groups push away from each other, they move towards what seems to be more of a planar structure. This is an increase in bond angle from the tetrahedral structure.
It is important to note that the greater the bond angle, the greater %s character. S orbitals are typically closer to the nucleus as compared to the p orbitals, and thus, the electrons are held closer -- the electronegativity of the nitrogen atom will be greater than if it was in the -NH3+ group. As such, it exhibits a greater -I effect.


text:
Electronegativity, polarity, pi bonds, and nitriles
While googling some numbers, I found that the dipole moment of acetonitrile is, according to wikipedia 3.92D. Compare this to the dipole moment of ethylamine, according to NIST of 1.220D. Now, I've read elsewhere (and perhaps this explanation is wrong) that polar molecules with pi bonds tend to be more polar than counterpart molecules with only sigma bonds, because there is an overall greater displacement of electron density towards the more electronegative element. In other words, more electrons = more unequal sharing = more polarity.
While this is a quick explanation for this difference, it leaves me scratching my head a little. The difference in electronegativities of carbon and nitrogen is ~0.49 (weakly polar), which in my imagination isn't nearly enough to justify the huge differences between dipole moments of ethylamine and acetonitrile. Do two pi bonds really result in a nearly 3 fold increase in dipole moment?
Compare this with the behavior of carbon with a much more electronegative element, oxygen. The dipole moment of methanol is 1.69D while the dipole moment of carbon monoxide is 0.122D. The more electronegative element produces a smaller dipole in a molecule with greater electron density between the carbon and oxygen, relative to a similar molecule with less electron density. In other words, our previous explanation of more electrons = more unequal sharing = more polarity breaks down in this case.
This is the crux of my dilemma. How can a more electronegative element produce a smaller  dipole than a less electronegative element? To ask this question another way, how can a less electronegative element produce such a large dipole? Is carbon monoxide a strange case, or have I confused myself (which isn't so uncommon)?
Thanks!

Related discussion about the dipole moment of $\ce{CO}$: https://chemistry.stackexchange.com/q/30797/41556

CO is quite different, one could say carbon is withdrawing mesomerically.