text:
Is the common way to calculate the pH of a solution with low concentration of acid wrong?
We had a chemistry test and one of the questions was this:

What is the pH of $1,4^{-8} M$ $HCl$ in water?

This is such a small amount of acid that you need to take into concideration the self ionization of water. So the way our chemistry teacher did it on the test was something similar to the first answer on this site: https://www.quora.com/What-is-the-pH-of-10-6-M-HCl-aq

They find the total $[H_3O^+]$ in the solution by adding the $1,0*10^{-7}M$ $H_3O$ from self ionization of water and get:

$1,14*10^{-7}M$.

Then they just find the pH:

$-\log(1,14*10^{-7}M)$

This is $6,943$ which becomes a pH of $6,9$

I did this and got right on the test, but something didn't sit right with me:

If $[H_3O^+]$ was $1,14*10^{-7}M$ and $[OH^-]$ was $1,0*10^{-7}$ (from self ionization of water), then $1,0*10^{-14}M^2≠[H_3O]*[OH^-]$ and the solution is not in equilibrium. So I came up with another way to calculate the pH:

I also first find $[H_3O^+]$ to be $1,14*10^{-7}M$ but then I think that $x$ amounts of $H_3O^+$ and $OH^-$ has to react so the solution is in equilibrium, so I set up an equation:

$1,0*10^{-14}M^2=(1,14*10^{-7}M-x)*(1,0*10^{-7}M-x)$

When I solve for $x$ I find that $6,755*10^{-9}$ mole has reacted per liter. Then I take away that from the original $[H_3O^+]$ and get that the  current $[H_3O^+]$ is $1,07245*10^{-7}M$. Then I just find the pH of the solution:

$-\log(1,07245*10^{-7}M)$

This is $6,9696$ which becomes a pH of $7,0$

A pH of $7,0$! Not $6,9$! And the teacher deducted points for $7,0$! I got $6,9$ on the test and full marks so I'm not that mad anyway, but I'm curious what is the most right answer. I also discussed it with my teacher and I think he somewhat understood my problem, but not my solution.

Edit:

I see that my question was a little unclear. I would like to know which answer is most right and if my thought process is correct.

The simple calculation for strong acids pH =  - log c  neglects water autodissociation and is justified only for c >> 10^-7 mol/L.

I never understood why they keep teaching people 'approximate' methods for pH calculation, which only seems to generate confusion and a lot of playing with sketchy assumptions. Nowadays one can solve exact systems numerically using open-source software, so there is really no need to approximate anything (except the activity vs concentration story). And for a solution of a single strong acid in water, even the exact system is analytical, so... Maybe they should teach students how to build the exact system first, and then show how it can be simplified in special cases.

See for instance https://chemistry.stackexchange.com/a/97840/41751 . The equation that the user derived is valid for any concentration of HCl, you only solve it as a 2nd degree one, without making any assumptions or simplifications. And remember to replace $1$ with the actual $[HCl]_0$.

I believe using approximations and "doing stuff in your head" is the only way to get a proper grasp on the more complex topics and learning about chemistry. There are already too many people using computers without actually knowing what they do. But this is actually not the place to discuss this. However, this question does show that critical thinking is not lost when using approximations.

Knowing what approximations you can use in the given situation is actually a sign you have a good qualitative understanding of the system (what is important, what is not). This is one reason such calculations are still taught as exercize

@user6376297 thx for the link! I'm not totally satisfied with it, but it did help me a little.

The famous example is for weak acids the equation pH = 0.5(pKa - log c), which is derived applying 2  approximations and the user must know what are they and if he/she can afford to use the equation  (which is valid at conditions c >> [H+] >> [OH] )

Expression formatting reference:
Basics/CH+Math expressions/Upright vs italic/Math tutorial
// MathJax is preferred not to be used in CH SE question titles for indexing and question list displaying reasons.

It is a pity this discussion is closed. Because I have redone all the calculation, and I have found  $\ce{[H+] = 1.21·10^{-7}}$ and $\ce{pH = 6.917}$. It is impossible to develop all the calculations under the small space available in the comment : Sorry !

@Maurice Q&A sites are not meant for discussions, this is not a forum. Only questions, answers and temporary comments for clarification. If you still feel chatty, well, there are separate chat rooms, but without decent MathJax support. If your approach is substantially different from the answers in the linked duplicate questions, then do feel free to post your answer to the original question.

@Martin-マーチン : you are absolutely right, there are a lot of people who just apply maths to chemistry and miss the meaning / field knowledge. On the other hand, when I studied chemistry, these approximate equations were taught to me as something to just learn by heart and apply, without understanding them. It was only much later in my career that I finally understood that a chemical system can be mapped to a system of equations. That was my point: tell people how that's done, and then explain how the original system can be simplified. IMO that's better than the other way around.

@Greg : indeed, 'knowing' is the operative word here. If you just learn things by heart, it does not help you understand anything. But I disagree on the 'qualitative' bit of your comment. The topic here is all and only about quantitative matters. In fact, even reducing the exact system to the approximate one requires making assumptions about the relative magnitude of terms in sums, so again quantitative. Purely qualitatively we could only say that adding an acid to water will always lower its pH. But here they are asking the user what the actual pH is, so...

@user6376297 I guess everyones experience is different, but in my education these approximations were explained from high school and up. We can argue day and nigh about how you define “qualitative” in this situation. My point is that knowing that there is a 2-5-10 orders of magnitude difference between two components is part of the understanding of chemistry, and estimating error is a useful and not very difficult to learn skill.

@Greg : I get your point, and there is actually no need to argue. The stubborn fact we cannot ignore is that countless people keep stumbling on the use of these approximate equations, meaning they are not so well understood or obvious to people, in real life / practice. There is probably some stats in SE regarding how often such questions pop up and are immediately closed because they are duplicated. I think the OP here got a nice answer and will eventually get value from it, but that went necessarily via explaining the origin of the exact system and its simplification, again as per my point.

In fact, this is something that happens very often in my line of work, where people are bothered by the 'more complicated' version of a theory, so they learn the 'simplified' version, which only works within a set of well-defined assumptions. Inevitably, they end up forgetting that the assumptions ever existed, and apply the same theory to all cases, including many where the assumptions do not hold (I have a very recent case in mind). So, simplified equations + assumptions = good if used well; simplified equations taken for granted for all cases = recipe for disaster.

If no usable equation is ready, you can always use in acido-basic calculations the scenario below. It is less straight-forward than some ready to use (often simplified) equations, but it is generally valid ( putting aside the Pandora's box of activities).
You start with 3 types of equations

The law of charge conservation - total charge has to be zero
$$[\ce{H+}] = [\ce{OH-}] + [\ce{A-}] \label{1}\tag{1}$$

The law of mass conservation - in a specific form of conservation of total molar amount/concentration of particular forms of given substance:
$$c(\ce{HA}) = [\ce{HA}] + [\ce{A-}] \label{2}\tag{2}$$

The equations for all present acido-basic equilibrii
$$K_\mathrm{w} = [\ce{H+}][\ce{OH-}]=\pu{e-14} \label{3}\tag{3}$$
$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{A^-}]}{[\ce{HA}]} \label{4}\tag{4}$$

Generally, we get a set of N equations for N variable and one often has to apply justified simplifications.

But in our case, things go smoothy to the simple solution:
We have welcome simplification, as we can assume there is not HA present for strong acids as $K_\mathrm{a}$ in \eqref{4} is effectively infinite and therefore
$$c(\ce{HA}) = [\ce{A-}]\label{2a}\tag{2a}$$
After substitution of \eqref{2a} and \eqref{3} to \eqref{1}, we get:
$$[\ce{H+}] = \frac{K_\mathrm{w}}{[\ce{H+}]} + c(\ce{HA}) \label{5}\tag{5}$$
$$[\ce{H+}]^2 - c(\ce{HA}){[\ce{H+}]} - K_\mathrm{w} = 0 \label{6}\tag{6}$$
$$[\ce{H+}] = \frac{c(\ce{HA}) \pm \sqrt{(c(\ce{HA}))^2 + 4K_\mathrm{w}}}{2}  \label{7}\tag{7}$$
We accept only the plus sign variant, the other root is negative.
$$[\ce{H+}] = \frac{c(\ce{HA}) + \sqrt{(c(\ce{HA}))^2 + 4K_\mathrm{w}}}{2}  \label{8}\tag{8}$$
$$\mathrm{pH} = -\log{[\ce{H+}]}\label{9}\tag{9}$$
If it neglect water auto-dissociation and pronounce $K_\ce{w} = 0$ then the equation  \eqref{8} reduces itself to simple:
$$[\ce{H+}] = c(\ce{HA})\label{10}\tag{10}$$


text:
Deriving reduction potential from redox reactions
From a past USNCO test:

40. What is the standard reduction potential of $\ce{Hg^2+(aq)}$ to $\ce{Hg(l)}$?
\begin{align}
\ce{2 Hg^2+(aq) + 2 e- &-> Hg2^2+(aq)} &\quad E^\circ &= \pu{+0.90 V} \\
\ce{Hg2^2+(aq) + 2 e- &-> 2 Hg(l)} &\quad E^\circ &= \pu{+0.80 V}
\end{align}
(A) $\pu{+1.70 V}$
(B) $\pu{+0.85 V}$
(C) $\pu{+0.10 V}$
(D) $\pu{-0.10 V}$

While I understand the method behind it (converting potentials to free energy first, adding, then converting back to $E),$ I don't understand why we have to do that in the first place, since when calculating a typical $E_\text{cell}$, one just adds the potentials.
I've heard people say it's because the number of electrons transferred affects the potential, but it doesn't really make sense to me, as a typical potential is intensive. Why is this case the exception?
The answer is $\pu{0.85 V},$ as calculated from the $\Delta G$ method and $\pu{1.70 V}$ by simply adding the reduction potentials.

It seems to me like the current question wording makes it a different question, compared to the  OP's one.

The electrode potential and the redox potential are closely related but different things.
The redox potential is just an alternative to Gibbs energy of particular redox reaction. It flips signs with flipping reaction direction. It generally needs conversion to Gibbs energy to have additivity to calculate redox potential of other reactions, because of not being additive for different electron counts.
For $$\ce{ox + n/2 H2 -> red + n H+}$$ is
$$\Delta G_\mathrm{r}=-nFE$$
where F is the Faraday constant and E is the redox potential.
as from physics we know that
$$E=qU$$
where E is energy, q is passed charge and U is voltage.
Therefore, if we calculate reaction Gibbs energy as difference of other 2 reaction Gibbs energies:
$$\Delta G_\mathrm{r,3}=\Delta G_\mathrm{r,2} - \Delta G_\mathrm{r,1}$$
then converted to redox potentials it looks this way:
$$-n_3FE_3 = -n_2FE_2 - (-n_1FE_1)$$
and divided by $-F$ :
$$n_3E_3 = n_2E_2 - n_1E_1$$
The electrode potential is the ordinary physical relative electrostatic potential. It's value is in an ideal case value of redox potential of ongoing half reaction in direction $\ce{ox -> red}$, but is affected by various, mostly kinetic factors.
The cell voltage is the difference of electrode potentials without dependency on electron counts of involved redox half reactions. It is not calculation of one redox potential from other two ones. The cell voltage is not a redox potential of a half reaction.

Hello Poutnik. Maybe you forgot to put a minus sign in the last equation? Thereby, you would have $-n_3 E_3 = -n_2 E_2 - (-n_1E_1)$?

Check again. Remember that  redox potential sign as very formal parametr depends on half reaction direction and Gibbs energy sign.

I agree it doesn't change anything. But it seemed to me that Ananta saw it without the - sign and replied. Indeed what is additive is $\Delta_r G$, and you get $E_3^0 =(n_2/n_3)E_2^0 - (n_1/n_3)E_1^0$ for the unknown $E_3^0$. The additivity of standard reduction potentials to get an unknown one, in the other post, is a mistake.

That is why I distinguish redox and electrode potentials.

Comments
This solution is wrong following the discussion in the comments. I have bookmarked the answer and will revisit it soon.
Definition
Let us start with the definition of $E^\circ = -\dfrac{\Delta G^\circ}{nF}$.
Case of the same number of electrons
In the same cell, same number of electrons are being released and captured and the effect of electrons cancels out:
$$E_1^\circ = -\dfrac{\Delta G_1^\circ}{nF};E_2^\circ = -\dfrac{\Delta G_2^\circ}{nF}\text{.}$$
Since the denominators are the same, we can add the two as
$$E^\circ = E_1^\circ + E_2^\circ= -\dfrac{\Delta G_1^\circ}{nF} - \dfrac{\Delta G_2^\circ}{nF} = -\dfrac{(\Delta G_1^\circ + \Delta G_2^\circ)}{nF}\text{,}$$
which makes sense.
Case of different number of electrons
Now consider the general case for two electrochemical preocesses:
$$E_1^\circ = -\dfrac{\Delta G_1^\circ}{n_1F};E_2^\circ = -\dfrac{\Delta G_2^\circ}{n_2F}\text{.}$$
In this case
$$E^\circ = E_1^\circ + E_2^\circ= -\dfrac{\Delta G_1^\circ}{n_1F} - \dfrac{\Delta G_2^\circ}{n_2F} = -\dfrac{n_2\Delta G_1^\circ + n_1\Delta G_2^\circ}{(n_1 \times n_2)F}\text{,}$$

@Poutnik didn't quite understand what you are trying to say here.

@Poutnik is there any problem with my analysis?

@Poutnik I have shown how this calculation is performed in the question. Are you saying this type of calculation is correct or wrong? Please be clear. Also, here, I believe you have to consider $H_2 \rightarrow 2H^+ + 2e^-$ (oxidation reaction) with negative of the reduction potential to obtain the combustion reaction.

Let us continue this discussion in chat.

@ananta Hello Ananta. If you see the OP's question, the answer is $E^0 = 0.85 ; \pu{V}$. If the additivity of standard reduction potential was correct, as you claim, then the answer would be $E^0 = 0.90 ; \pu{V} + 0.80 ; \pu{V} = 1.7 ; \pu{V}$, which is not the answer. How can you explain this?

@MetalStorm as I stated in the answer, I am aware this isn't correct, and I will revisit it soon.

@ananta Oh, that is fine. My bad, call for help if you need it!

The title of your question is incorrect. You are determining a reduction potential from two related reduction potentials.
The fastest way to do this is to take a weighted average. To complete the reaction, electrons participate either in the first or the second reaction, in a 1:1 ratio.
A related question is here, also from UNSCO. In this case, one step requires two electrons and the other just one, so you have to do a 2:1 weighted average.
In some cases, the species undergo comproportionation or disproportionation. Here is the example for copper species:
$$\ce{Cu^2+ (aq) + e– <=> Cu+ (aq)}\ \    E^\circ = +\pu{0.15 V}$$
$$\ce{Cu+ (aq) + e– <=> Cu (s)}\ \  E^\circ = +\pu{0.52 V}$$
The +1 oxidation state is not stable, it falls apart into the zero and +2 oxidation state under these standard conditions.

The point is, it is not the title of his question. The question was somewhat hijacked.

The electrode potential measures the energy change involving the transfer of one mole of electrons. The potential for Hg[II] to Hg[I] is one mole transferred; the potential for Hg[I] to Hg[0] is a second mole transferred.  Concatenating the equations and adding the voltages gives the energy change for two [2] electrons so the composite voltage is divided by 2 to give the standard potential.
Should the electrons differ in each half reaction such as Fe to Fe[II] followed by Fe[II] to Fe[III] the 2 electron potential is multiplied by 2, the one electron potential added and the sum divided by three.

Yes, it is for one coulomb of electrons; allow the poetic license because we all have forgotten physics.

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Energy levels for H atom
Problem Statement:

The number of orbitals (for H-atom) which have greater energy than $3p_y$ orbitals: $3s, 3p_z,3d_{z^{2}},3d_{x^{2}-y^{2}},2p_y, 4d_{z^2}, 4p_{y}$

The solution says that energies of all orbitals with same $n$ (principal quantum number) will be same, only for hydrogen atom.
Can someone explain why is that so? And when not to use the "$n+l$" thing?

In H atom, the energy depends only on the principal quantum number $n$. So 2s and 2p levels have the same energy. 3s, 3p and 3d have also the same energy. It is not the case any more for heavier atoms.

@Maurice but why is that the case, only for hydrogen? Also, will the answer be same for all 1 electron species, like He+ ?

@PurposeHonest It is not evident. You should prove that the bound-state wavefunctions that satisfy $\hat{H}\psi = E\psi$, have energy eigenvalues that only depend on the quantum number $n$. For the hydrogenic atom (1 electron + N protons), the solution still holds, but the energy eigenvalues will change because they are proportional to the squared value of the atomic number $Z$ and the reduced mass $\mu$.

@PurposeHonest. In a poly electronic atom, the interactions between the electrons depend on the shape of each orbital, i.e. on the non-principal quantum numbers.

Since (atomic) hydrogen has only one electron, there is no electron-electron interaction to differentiate internal levels.
Of course, an energy level can be split if there is an external interaction. See the Feynman lectures also. For example, in a magnetic field, the Zeeman effect splits spectral lines. The corresponding Stark effect shows a similar splitting of levels in an electrostatic field.

And due the Lamb shift, ns and np electron energies are not the same even in hydrogen atom.

and there is spin-orbit coupling which is large and does split levels.

There seems to be no simple case. Even positronium has a complex spectrum... with a large peak at 511 keV in its death throes.

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text:
Are the density of states of bonding and the corresponding antibonding states complementary?
As for the bonding and the corresponding antibonding molecular orbitals formed between two types of atomic orbitals, are their density of states (DOS) complementary? This means if the DOS of the bonding states is high, then the DOS of the antibonding states is low, and vice versa.
Intuitively, I think it is a correct statement since the total DOS of the two types of atomic orbitals is a constant, but I can't find a reference which discusses this point.

Since mathematically speaking antibonding orbitals are more antibonding then bonding orbitals are bonding, I want to say no.

@Martin-マーチン I don't fully understand, can you elaborate your point a little bit?

Actually I can't, I don't remember what I wanted to say and I don't understand what I meant. I'm just usually cautious, because there seldom is something ready or some about orbitals.


text:
Striking back of Bunsen burner
I have a question about striking back of Bunsen burner, regarding the reaction between methane and oxygen.
When methane gas reacts with oxygen completely, a clean non-luminous blue flame can be observed.
When methane gas reacts incompletely with oxygen, a sooty luminous flame will be observed instead.
Given the fuel of the Burner burner is methane, when the air hole of a burner burner is fully opened, striking back will occur. I observed that the strike-back flame is green in colour and produces loud noises. Are there any reasons for the green flame colour and noises?

The gas alimented to the Bunsen burner passes a hose, a valve, and a nozzle (where, if permitted, air can mix with the gas).  Often, the nozzle is made of a material containing copper, like brass.  And the green coloration of the flame you report can be due to this - sort of an unintended flame test, but the bright yellow of sodium by salt in the kitchen (e.g., cooking / roasting potatoes over a gas stove) could be familiar with you. If your lab still permits the use of Bunsen burners (borax beads, blow pipe test with char coal, digests from melt; some familiarity with glass blowing/bending, etc), chances are that the burner is a bit dirty within the barrel, too.
The origin of the noise is that the combustion doesn't (only) take place "on top" of this chimney (as it is designed), but right around the air holes, too. It is not the even combustion the burner was designed for; because of potential danger ahead, shut the gas the line.  Once the burner is cold, you can lit it again; gently open the air holes.

Why all this?  Coal gas and natural gas, both which a Bunsen burner can combust, are of different chemical composition.  The former was/is up to half of the volume hydrogen, and 5 to 10 vol% CO; the later mostly methane.  And the pressure these are delivered can differ.  Hence gas stoves for cooking and heating need a different set of burners for one, or the other for safety, and efficiency of combustion.  Depending on the work ahead, it might not matter this much for a Bunsen burner. (I have the impression, some of the tasks mentioned earlier do not work this well with natural gas and new burners, as they used to be in the old days with coal gas and the old ones.)  So maybe you got hands on one of the older ones, where the nozzle still is the one about coal gas.


text:
what are other metal oxide that can be used as electron acceptors?
I am using silver oxide as a cathode because it's common. I was reading that " some solid metal oxides (e.g., silver oxide, lead dioxide, and manganese dioxide) possess high redox potentials. Manganese dioxide and lead dioxide are usually not employed as electron acceptors but as catalysts for oxygen reduction "
Silver oxide/silver (Ag2O/Ag) couple has been used for decades in alkaline batteries. The standard reduction potential of Ag2O/Ag is 0.342 V vs the standard hydrogen electrode (SHE).
What is an example of another electron acceptor metal oxide I can use that might have a higher voltage than silver oxide?

There was used silver(I,III) oxide  // Silver(I,III) oxide or tetrasilver tetroxide is the inorganic compound with the formula Ag4O4. It is a component of silver zinc batteries.

A few oxides do beat silver oxide (or at least the silver(I) oxide reported in the question) in terms of reduction potential. Here, examples are given for basic solutions and for acid solutions. Reduction potential data are taken from https://wisc.pb.unizin.org/chem103and104/chapter/appendix-k-standard-electrode-potentials-in-acidic-or-basic-solutions/.
Basic solution
Nickel dioxide, $\ce{NiO2}$, is reduced to nickel(II) hydroxide at +0.49 V versus the standard hydrogen electrode with the latter at pH 0. For basic cells it would ge more appropriate to correct the hydrogen electrode voltage for a pH of 14, which the Nernst Equation renders as 0.83 V. Thus $\ce{NiO2}$ comes in at +1.32 V over the pH-14 hydrogen reduction, versus +1.17 V against tge same reference for $\ce{Ag2O}$ reduction. Note that in actual silver-zinc batteries may the silver oxide there is silver(I,III) oxide, $\ce{Ag^+(AgO2^-)}$, so the silver oxide reduction potential may be different from that given in the question.
Acidic solution
Lead dioxide should nit ge so easily dismissed. It is the electron acceptor in lead-acid batteries commonly seen in cars. Helped along by precipitating lead sulfate from the sulfuric acid solution electrolyte, $\ce{PbO2}$ reduction in that acid weighs in at +1.69 V over the hydrogen electrode in acid. $\ce{NiO2}$ reduction to nickel(II) ion in acid solution is close behind at +1.68 V, although $\ce{NiO2}$ is usually used with basic electrolytes.


text:
Optical activity of (1r,3R,5S)-1-methyl-3,5-dipropylcyclohexane (or other long alkyl groups)
Consider a cyclohexane with a methyl group at C1 position and a propyl group at C3 and C5 position and each of the first bond bonded to the cyclohexane in alkyl groups has wedge bond. I do not think this can be called a meso compound as the second and third bond of the propyls rotate freely and thus there is no plane of symmetry and it is chiral. Despite of that, some sources tend to regard those two propyls as a single unit that is called as Pr and hence the compound is meso. Giving those second and third bond of the propyls with mixed wedge-wedge, wedge-dash, dash-wedge, and dash-dash also give same chemical name as (1r,3R,5S)-1-methyl-3,5-dipropylcyclohexane.
I am wondering if those variants will be chiral or not. Whether those variants are chiral or not, I am also interested in how much is the optical activity of each variant.

The compound is meso even though there are many conformations of the propyl groups. What ever conformation one propyl group adopts, the other one can adopt. On the other hand, imagine each propyl group having a different conformations making that molecule chiral, frozen in time. But there is another molecule that is its mirror image. These two chiral conformations form a racemic pair....no optical activity. Similarly, the gauche conformations of n-butane form a racemate while the anti conformation has a plane of symmetry.

Unless rotation is blacked, you have no optical activity.

Regard less methyl is wedge or dash, if both isopropyl groups are in wedge positions, this compound is meso. The compound is optically active when one isopropyl group is a wedge and the other is a dash.


text:
How long can a silicone and white spirit solution be stored without hardening?
I’ve heard that some people use a mixture of silicone and white spirit to waterproof their gear.
How long this solution can be stored without hardening if it is stored in a closed hermetic bottle?
If the solution does harden over time, what would be the cause of this? Does anyone have any information or experience with this?
(Additionally, since it’s not really possible to preserve an opened tube of caulk ―which I’m only using for this purpose―, I wonder if I can increase my chances of not spoiling it by storing it in this state).

In my test, it did not last 24h

"White Spirit" or naphtha will dissolve silicone oils. Water will not.
Therein lies the secret of using silicone as a water proofing agent.
It is applied, much like paint, and the solvent evaporates, leaving a film.
Best chances of preserving it would be to store in a cool place, away from sunlight or heat, in a tightly sealed container, again much like an oil based paint.

Thanks but you don't answer the question: "How long can it be stored without hardening? [in this condition]". (As long as sealed container of oil based paint?)

@JinSnow obviously shelf life is highly dependent on a number of factors.  I suggest it may be possible that a silicone/spirit mixture may have a shelf life in the range of a sealed container of oil based paint.  As always, common sense indicates that more definitive answers may be had with actual testing, as compared with internet thread sources (it may not harden at all as long as solvent is present and the silicone does not degrade).


text:
Reaction of Sodium Silicate (water glass) with Zinc Oxide?
I was reading about different additives for Sodium Silicate (water glass) and noticed this line in particular:

The water-resistance of sodium silicate solution adhesives may also be
improved by the addition of fine powders such as zinc oxide or
amorphous calcium carbonate. These additives will react slowly with
the silicate to form an insoluble mass

I would like to understand what kind of insoluble mass is formed by the addition of Zinc Oxide.
What I've found so far points towards "disodium zinc silicate" (Na2(ZnSiO4)), which according to this paper forms as a precipitate from the combination of Na2SiO3 + ZnO + NaOH. Could this also occur without the NaOH present as implied by the additives website? I don't currently have access to the full paper, so I'm just going off of the info in the abstract.
I also see very little info online about "disodium zinc silicate". Is it electrically or thermally conductive? Is it stable at high temperatures?
Does it really do anything different for the properties of the Sodium Silicate than adding an inert filler?

As a side note, watch a crystal garden grow in sodium silicate: https://www.youtube.com/watch?v=sAr7mvBjWq8 . Consider that a mass of interlocked crystals can be tougher (less brittle) than single crystals.


text:
The overall Gibbs free energy in Gluconeogenesis
Gluconeogenesis is an exergonic process but I wonder what is the overall Gibbs free energy in step 3 and step 10?
I know there’s about $\pu{2\times ( -31.4 ) kJ/mol}$ for 2 pyruvates convert to 2 phosphoenolpyruvates (PEP). The reverse reaction path would vary in gluconeogenesis, but there’s no information on the web on this reverse energy.
Gluconeogenesis step 10:
$\ce{2 Pyruvate + 4 ATP + 2 GTP ->[Acetyl CoA、Biotin Pyruvate carboxylase][Mg^{2+}] 2PEP + 4ADP + 2GDP + 6Pi}$
Step 3:
$\ce{\mathrm{F1,6BP} + H2O ->[Fructo-1,6-bisphosphatase] \mathrm{F6P} + Pi}$

Would be nice if you elaborated what those steps are.

What is the meaning of the commas (,), after CoA and between carboxylase and Mg2+ ? Why do you write arrows from time to time with and sometimes without the final arrow (>) ?

Please see https://chemistry.meta.stackexchange.com/questions/86/faq-how-can-i-format-math-chemistry-expressions-on-chemistry-stack-exchange on how to format equations and chemical expressions.

I tried to improve your post with some edits but as Maurice points out, there are some odd formatting features and nonstandard chemical notation. Please edit this for clarity, those features might be the result of copy-pasting.

Also, the question in the title and body don't match. Please edit that.


text:
Need for AINR-based simulations (vs. laboratory experimentation)
The Results section of this paper related to ab initio molecular dynamics states that "16 $\ce{H_2O}$ and 15 $\ce{HCN}$ molecules were taken together, and the system was allowed to evolve for 750 ps" in order to find pathways for RNA precursors. The question might be naive but, what is the need of such a complex and expensive simulation if water and hydrogen cyanide can be mixed in the laboratory and study what results from this homogeneous mixture? Is it environmental conditions, or the temporal resolution of the study what matters here?

Your question might be better suited for matter modeling SE, as they might provide input into what makes simulations grand as opposed to experiments. Simulations, if accurate (always arguable), provide information about events at the molecular level that would be hard and even more expensive to obtain by experimental means. They might also stimulate experiments (to verify what is predicted by the sims). Theory sometimes leaps ahead of experiments.

RNA precursors need catalyzers to let HCN and H2O react.


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Improve the leaving capability of a leaving group (F-) with electron density donating groups (EDP) in close proximity (neighboring group effect)?
I have a situation in which I want to hydrolyze a phosphonic acid halide under basic conditions. In my specific case, the halide is a fluorine substituent. Other halides like chlorine and bromine are known to hydrolyze quite quickly under aqueous conditions. Until now fluorine abstraction was not possible, even under strongly basic conditions. An additional substituent on the phosphorous is an oxido moiety which carries a negative charge. I am wondering about two things: Is there the possibility to increase the leaving capacity of fluorine by electron-donating groups in close proximity (bond to the same atom)? When the oxido moiety is replaced by an alkyl group R- or Alkoxygroup RO-, the molecule loses a negative charge, which might stabilize the fluorine and inhibits the hydrolyzation. What are the underlying principles that might explain the problems of hydrolyzation when an oxido moiety is present? Are there other strategies to make hydrolyzation an energetically more favored process?

You are trying to get a -ve charged (or at least electron-rich) nucleophile to approach an already -ve charged molecule. Can you put a protecting group on the O- species, maybe silyl?

I see that 2 negative charges by two oxido substituents would create repulsion and is therefore energetically disfavoured. Oxiodo can be protonated to form hydroxide OH. Mechanistically I would expect that OH is attached to the molecule when F- leaves and is therefore present for a specific amount of time. Silylprotecting groups are usually cleaved by fluoride, which is a side product of the reaction. But for the next reactions I indeed plan to mask the oxido unit.

If you have an OH group the proton is quite likely to transfer to the incoming nucleophile and inactivate it.


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Vacuum distillation of azeotropic mixtures
I was wondering if vacuum distillation can seperate azeotropic mixtures.
Consider the ethanol/water system. At 95.6% ethanol content, the system forms an azeotrope because the vapor composition matches the liquid composition:

As one can see, no further seperation occurs in the condensed ethanol phase (going to the right of the diagram).
One of my questions is: If we start with an 50/50 (v/v) ethanol/water mixture, will eventually all the ethanol boil off, i.e will the starting flask consist completely of water after distillation finished? I get that the received ethanol will only contain 95.6% ethanol, but will the starting mixture contain 100% water after ethanol is boiled off (going to the left of the diagram).
And another question is: At lower pressures, do the vapor composition curve and liquid composition curve intersect or is fractional distillation of ethanol/water possible? At pressures of down to 10 mbar?

Sorry I meant 10 mbar, that’s what membrane pumps in our lab use

Vapor pressure at 20 Deg C is about 20 mbar.// Ethanol content would approach 0% asymptotically. // Position of azeotrop  composition is pressure dependent, cannot tell in which way.

You will receive 95.6% ethanol only using reactivation column, otherwise the distillate composition will reflect the partial vapor pressure ratio over the liquid, according the phase diagram.

One of my questions is: If we start with an 50/50 (v/v) ethanol/water mixture, will eventually all the ethanol boil off, i.e will the starting flask consist completely of water after distillation finished? I get that the received ethanol will only contain 95.6% ethanol, but will the starting mixture contain 100% water after ethanol is boiled off (going to the left of the diagram).

Practically yes. Imagine you register the temperature along the distillation. After almost all the ethanol has gone to the  other flask, you will register that the tempereature will reach a plateau around $ T \approx 100 \; ^\circ\pu{C} $. This means that the initial flask has now pure water, if the experiment is performed at ambient pressure.

And another question is: At lower pressures, do the vapor composition curve and liquid composition curve intersect or is azeotropic distillation of ethanol/water possible? At pressures of down to 10 mbar?

I will try your claim by the two following elements:

The gas phase is modelled by the Soave-Redlick-Kwong EOS.
The liquid phase is modelled by the NRTL model.
There are the results in Txy diagrams. I will zoom on purpose the values of $ x_\pu{ethanol} \in (0.8,1.0) $, to see the evolution of the minimum azeotrope more clearly:

Thus, the azeotrope disappears as the pressure decreases. However, working at this values of pressure, is industrially discourarged. As a first rule, you will always avoid running distillation columns in vacuum, because of the energetic costs that it will demand. However, from a thermodynamic point of view, you are correct.

Wow thanks a lot for this answer :)

This is this first time I've seen a liquid-vapor equilibrium diagram for a mixture at different pressures, and it's presented really well. Thank you for this contribution!

@NicolauSakerNeto My pleasure, thanks for the compliment.


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The reaction of HNO3 with H2S?
I am totally confused about the gas that would be released in this reaction. It is given NO in one book and NO2 in other. Even the internet has mixed opinions. Help me out guys.

There are mixed opinions because the reality is mixed, and not just mixed in a simple way, but mixed in a mixed way, too. Do you think you know the product for S? Wrong.

Nitric acid reacts with plenty of substances. This reaction may produce both $\ce{NO}$ and $\ce{NO2}$. It is a question of concentration, because $\ce{NO}$ may react with $\ce{HNO3}$ according to :$$\ce{NO + 2 HNO3 -> 3 NO2 + H2O}$$ and $\ce{NO2}$ may react with water according to the same reaction, written backwards : $$\ce{3 NO2 + H2O -> NO + 2 HNO3}$$ So both gas are always produced simultaneously. And if there is a majority of $\ce{NO}$, this colorless gaz does react with oxygen from air, producing brown fumes of $\ce{NO2}$ according to $$\ce{2NO + O2 -> 2 NO2}$$

According to Ref.1, the reaction of $\ce{HNO3}$ with $\ce{H2S}$ has been processed in multi steps, which agrees with Maurice's suggestion in the first answer. The authors found that:

Solutions containing 5-30% nitric acid show no signs of reaction for
some time but; if nitrogen peroxide $(\ce{NO2})$ is added, a violent reaction with rise of temperature follows.
When a 40% solution is used the reaction takes place after an interval of apparent passivity called in this paper the “induction period."

The "induction period" became longer when, before passing hydrogen sulphide, carbon dioxide was bubbled through the solution to remove any $\ce{NO2}$ resulting from the decomposition of nitric acid (keep in mind that before passing $\ce{H2S}$,
the air in the vessels was always displaced by carbon dioxide throughout this experiment).

Milkiness due to sulphur appeared on the surface and spread rapidly through the liquid. When, after stopping the flow of hydrogen sulphide, the vessel was shaken, the atmosphere above the solution turned brown owing to the presence of nitrogen peroxide $(\ce{NO2})$:
$$\ce{HNO3 + H2S -> HNO2 + S} \tag1$$
$$\ce{HNO3 + HNO2 <=> H2O + 2NO2} \tag2$$

The $\ce{HNO2}$ may also break down according to the equation:
$$\ce{3HNO2  <=> 2NO + H2O + HNO3} \tag3$$
On passing a further supply of $\ce{H2S}$, the brown color disappeared as $\ce{NO2}$ reacts with $\ce{H2S}$ giving sulphur, water, and nitric oxide $(\ce{NO})$:
$$\ce{NO2 + H2S -> NO + S + H2O} \tag4$$
Thus, either $\ce{NO2}$ or $\ce{NO}$  or both could be present in the reaction, basd on the strength of the nitric acid and how long $\ce{H2S}$ has been passed through. Also, keep in mind that solutions containing less than 5% nitric acid (dilute conditions) are not reactive with $\ce{H2S}$ even if traces of $\ce{NO2}$ are added to the reaction (Ref.1)·
References:

H. B. Dunnicliff and Sardar Mohammad, "The Action of Hydrogen Sulphide on Solutions of Nitric Acid," J. Phys. Chem. 1929, 33(9), 1343–1362 (DOI: https://doi.org/10.1021/j150303a005).


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How does industry separate glucose from fructose using liquid chromatography when producing high-fructose corn syrup?
Can someone explain how the separation of glucose and fructose is achieved using liquid chromatography in the industrial production of high-fructose corn syrup?
I've seen references to ion-exchange chromatography, but both glucose and fructose are primarily neutral molecules, so I am not sure how that would work. Other have stated size-exclusion, which makes even less sense.
I've also come across mentions of simulated moving bed chromatography being used. I'm interested in understanding what the stationary phase is and the exact separation method employed in this process.

If you add a quote please include the source.

Are you citing https://www.sciencedirect.com/science/article/abs/pii/S138358661100164X ?

As an example, see the abstract to https://pubs.acs.org/doi/10.1021/ja01645a051

Chromatography is not used to produce high fructose corn syrup (HFCS): it is only used on a small scale to analyse the product. Bulk production uses an enzyme to convert some glucose to fructose to get the final product.

source for quote ChatGPT4, I tried to find where and why it would say, but did not.
Problem is that industrial processes used on farms would have been published by scientists in the 1950s. All of the science publications I could find were too new and too narrow focused for industrial scale.

I am glad that you realized that ion-chromatography is not an effective chromatographic mode for separating neutral molecules. However, we should not forget the backbone of cation or anion exchange resins. It is an aromatic group backbone of ethylene divinyl benzene polymers (among other possibilities). In general these resins can interact strongly with organic compounds. Coming back to sugars. Sugars can be ionized in a special way- if you use an strongly alkaline solution like dilute NaOH. Once ionized by losing the protons of the hydroxyl groups, these sugar anions are amenable to ion chromatography, so the organic backbone, plus their ionized state contributes to retention and infact pretty good peak shapes in alkaline medium.
Search Thermo Fisher's website for separating sugars via ion chromatography. They have dedicated columns This is a general answer which should be good enough to let you search deeper.

I appreciate the attempt, but I doubt it would be cost effective to purify sugar by jacking up the pH to over 12 using NaOH. Wikipedia says that glucose has a pKa of 12.16, I would think fructose has a similar pKa, but could not find a source.
Nonetheless, I think I found my answer, see below.

@vossman77, You are right, this is how sugars are analyzed at low concentrations, as we say under "analytical conditions". Your answer is good and this is called preparative chromatography.

I was surprised, I got feedback so quickly, I thought I was asking into the void.
The production of HFCS-55 goes through this process:

Corn kernels are ground into corn starch
α-Amylase enzyme hydrolyzes α-bonds in starch, producing oligosaccharides
Glucoamylase breaks down oligosaccharides into glucose
Xylose isomerase converts glucose to ~42% fructose and ~50% glucose
Liquid chromatography purifies fructose to ~90% concentration
90% fructose solution mixed with 42% fructose to create 55% fructose "High Fructose Corn Syrup"

I was really curious about the "liquid chromatography"
I think I finally found the answer when I ran across this PDF document, entitled: AmberLiteTM Uniform Particle Size Ion Exchange Resins - Chromatographic Separation of Fructose and Glucose with AmberLiteTM CR99 Ion Exchange Resins -
Technical Manual and the section on "3.1 The Chromatographic Mechanism"

[...] Inside the bead, the dissolved sugars interact with the calcium ions held by the resin. Fructose, glucose and water form weak ligand complexes with the calcium ion. A stronger interaction in the fructose/calcium ion complex than in the glucose/calcium ion complex is the basis of the mechanism of separation of fructose from glucose1. This mechanism of separation is called ligand exchange chromatography.

While technically it is a cation exchange column using Ca2+, the interaction between fructose/glucose and the Ca ions, is an affinity relationship. The footnote from the DuPont technical manual clears this up:

1 Chromatographic separation resins do not “exchange” ions as do the resins used in deashing and mixed bed polishing. They function by adsorbing and “slowing” fructose as it moves down the column. The syrup doesn’t exchange ions in the process.

I also found a nice webpage from PuroLite, that goes through the process.

This sweetener is produced by passing a 42% fructose solution through a calcium form strong acid cation resin to effect a separation and create a 75-90% fructose solution which can be blended back with additional 42% fructose to produce a 55% fructose purity.

and

Owing to a greater number of sites available for hydrogen bonding, the fructose molecule will form a coordination complex with calcium ions fixed onto a strong acid cation resin. This results in a preferential affinity of the resin for the fructose molecule over the glucose molecule and hence a chromatographic separation of the two sweeteners as they pass through the resin bed. From a feed solution containing a purity of 42% fructose by weight, the fructose in the product fraction can reach in excess of 99% purity.

When producing 55% fructose, the optimum productivity and efficiency of the system is achieved by enriching to an 85-90% fructose concentration.

some other highlights:

In the 1970s, the technology was developed for continuous chromatographic separation of fructose utilizing a simulated moving bed (SMB) separation system.

let me know if I missed anything.


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How are gravimetric analyses of particulate matter performed, in order to determine particulate matter concentration in air?
I've been surfing Google Scholar a bit and learned that a common and comparatively less expensive method of measuring particulate matter concentration in air was analysis through gravimetric methods. I was wondering, what sort of materials are used to filter the particulate matter from the air, in an indoor setting, for particles of diameter 2.5 μm? And does the technique consist of measuring the weight difference between a clean vs. used filter? Any reading recommendations would be highly appreciated!

The American Society of Testing and Materials ASTM has standard methods on such analyses in great detail. Not suprisingly, those documents are pretty expensive to purchase and none of them are free. You will have to access via a decent univeristy library.

@AChem Appreciate the recommendation, thank you!


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Mechanism of Darzen's Reaction
I was dealing with Darzen's Reaction for the preparation of alkyl chlorides.
$$\ce{R-OH ->[SOCl2][Pyridine] R-Cl}$$
Now I encountered this as the mechanism of the reaction
$$\ce{R-OH + SOCl2 -> RO+HSO^-Cl2 ->[-Cl^-] R-O+HSOCl ->[NC5H5]R-O-SOCl}$$
$$\ce{R-O-SOCl ->[NC5H5][-Cl^-] ROSON+C5H5 ->[Cl^-] R-Cl}$$
My question is that why does the reaction in the second line occur? In other words, why does $\ce{R-O-SOCl -> ROSON+C5H5}$? I am unable to find any reasons to explain why this reaction would be favourable. What major factor allows pyridine to displace the chlorine atom?

https://chemistry.stackexchange.com/questions/69063/if-socl2-reacts-with-alcohols-via-sni-why-doesnt-pocl3

@Mithoron Thanks for the post! But could you please elaborate on the stability of $\ce{ R-O-SOCl}$ in comparison to $\ce{ROSON+C5H5}$?


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Does the word "restraint" have any special meaning in the case of protein or biochemistry?
The following screenshot is taken from the book The Encyclopedia of Physical Science & Technology, volume: Biochemistry, Edition: 3rd, Page-197.

The text says:

FIGURE 3 Schematic representation of the peptide bond and the observed restraints on the conformational.

On the other hand, the research article

Discrete restraint‐based protein modeling and the Cα‐trace problem. (2003)

also mentions and repeatedly uses the word restraint.

Abstract
We present a novel de novo method to generate protein models from sparse, discretized restraints on the conformation of the main chain and side chain atoms. We focus on Cα-trace generation, the problem of constructing an accurate and complete model from approximate knowledge of the positions of the Cα atoms and, in some cases, the side chain centroids. Spatial restraints on the Cα atoms and side chain centroids are supplemented by constraints on main chain geometry, φ/ξ angles, rotameric side chain conformations, and inter-atomic separations derived from analyses of known protein structures. A novel conformational search algorithm, combining features of tree-search and genetic algorithms, generates models consistent with these restraints by propensity-weighted dihedral angle sampling. Models with ideal geometry, good φ/ξ angles, and no inter-atomic overlaps are produced with 0.8 Å main chain and, with side chain centroid restraints, 1.0 Å all-atom root-mean-square deviation (RMSD) from the crystal structure over a diverse set of target proteins. The mean model derived from 50 independently generated models is closer to the crystal structure than any individual model, with 0.5 Å main chain RMSD under only Cα restraints and 0.7 Å all-atom RMSD under both Cα and centroid restraints. The method is insensitive to randomly distributed errors of up to 4 Å in the Cα restraints. The conformational search algorithm is efficient, with computational cost increasing linearly with protein size. Issues relating to decoy set generation, experimental structure determination, efficiency of conformational sampling, and homology modeling are discussed.

I have two questions:

Does the word restraint mean the same things in the above two sources?
Was the word restraint used as a general term that means a measure or condition that keeps someone or something under control,  or does it mean any specific property of protein or a biochemical element?

I’m voting to close this question because it's cross-posted at https://mattermodeling.stackexchange.com/questions/10756/what-does-it-mean-by-a-restraint-in-the-case-of-a-protein-chain. The policy of disallowing cross-posts is, ironically, a restraint pertaining to rules, regulations, and commonly-accepted behaviors on SE.

This is an interesting question and slightly different from the version on MM SE (imo particularly interesting the difference in definition of constraint and restraint)

3 occurrences of the word (root) 'constraint' vs. 26 for 'constraint' on the linked cross-post. And the OP has bolded and italicized the word 'restraint' in their summary of their questions. I really don't see how this is materially different than the cross-post.

@ToddMinehardt I'm content to disagree with you on this one. I agree the post here differs marginally from that over at MM SE. Question is what to do about it (where does it fit better?) Note that using constraints/restraints is standard practice to derive structures from experimental data, so arguably a better fit here.

[OP] Does the word restraint mean the same things in the above two sources?

Restraint is used in two contexts in chemistry in general, and biochemistry specifically. Your first example shows the first context, and your second example shows the second.
The first context is in the analysis of observed conformations. Some conformations are not observed because they would result in clashes of non-bonded atoms or would require to break a partial double bond, as is the case for the peptide bond. We rationalize this observation by saying that there is a restraint. There may or may not be a theoretical model explaining the restraint.
The second context is related and refers to restraints applied in simulating the dynamics of molecules, e.g. in molecular dynamics. The restraints could be based on generally known properties (like trying to keep the bond distances in a reasonable range with a quadratic penalty function), or on specific experimental data (like turning an observed NOE from NMR data into a distance restraint of non-bonded atoms, or having a penalty function based on X-ray diffraction data).
So you can distinguish the two contexts of "restraint" based on whether nature imposes the restraints on a real molecule, or researchers impose it on a model of a molecule.

[OP] Was the word restraint used as a general term that means a measure or condition that keeps someone or something under control, or does it mean any specific property of protein or a biochemical element?

The general definition explains pretty well what is meant (in either context). For a specific type of molecule (like a protein), there are specific restraints that you would expect from the combined knowledge on that type of molecule.

In molecular simulations, a restraint is a potential energy term introduced to limit the range of possible values that a geometrical feature of a molecular structure is allowed to have. In practice it is a penalty function that makes extending the value of the geometrical feature (typically a distance, angle or dihedral) beyond a specified ideal range incur an increasing or very large value of potential energy, thereby discouraging such values. The ideal range can be described mathematically as $x = x_{\textrm{ideal}}$, $x \leq x_{\textrm{max}}$ or  $x \geq x_{\textrm{min}}$ , sometimes in combination. The restraints$^*$ do not imply that the value of $x$ cannot deviate from the ideal point or range, same as a pendulum might swing away from an energy minimum. However by imposing a sufficiently large penalty such a restraint may become a strict constraint $^*$.
The potential itself can take many forms, commonly a quadratic function that increases away from a central value, sometimes with a cutoff penalizing extreme values, and sometimes with a constant intermediate potential.
The meaning of restraint can be understood as being roughly the same as in ordinary contexts, a device that keeps something from straying from a desired region, for instance a leash for a dog on a leash. It allows the dog to move around but not just anywhere. The word has the same meaning in both contexts in the question (as explained by Karsten) but in one case it refers to the actual potential function that limits observed values of a geometry (in the figure, backbone dihedral values $\phi$ and $\psi$) whereas in the second it refers to potential functions used during a simulation to reflect what is already known about geometrical features (typical bond lengths, angles, dihedrals, minimum vdW contact distances) or other new experimental information (typically distances between atoms on the same or different residues), but both ultimately obtained via experiments, same as in the first example.
$^*$ [edited] In the nomenclature of molecular simulations, constraint may have a more specific meaning and refer to geometric features that are enforced to have a fixed value. For instance, from the Biosym Discovery manual:

The seminal difference between a constraint and a restraint is that a constraint is an absolute restriction imposed on the calculation, while a restraint is an energetic bias that tends to force the calculation toward a certain restriction.

Could you add some references in support of your answer?

@user366312 the meaning of "restraint" is defined mathematically in any textbook on molecular simulations, see for some examples https://chemistry.meta.stackexchange.com/questions/2889/resources-for-learning-chemistry


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In a voltaic cell with standard hydrogen electrode is the measure half cell always the cathode?
I am a bit confused. My openstax textbook states:

"When the half-cell X is under standard-state conditions, its potential is the standard electrode potential, E°X. Since the definition of cell potential requires the half-cells function as cathodes, these potentials are sometimes called standard reduction potentials."

I would greatly appreciate if someone can explain to me why the half
cell has to be a cathode if the measured reduction table has both
positive and negative values.

Formal negative reduction potential (Positive reaction Gibbs energy as $\Delta G^{\circ}_r = -nFE^{\circ}$) for being cathode means positive oxidation potential (negative reaction Gibbs energy) for being actually anode, as the latter is the direction of spontaneity. Reaction H2 + 2 Na+ -> 2 H+ + 2 Na has highly positive reaction Gibbs energy and highly negative formal reduction potential.

Sorry I am not quite understanding. So its the cathode even though it had less "electron wanting desire" because hydrogen is the lowest?

It is formal cathode, but actually anode, as the Gibbs energy change and related potential are formally evaluated in this case in the direction opposite to spontaneity. If we used absolute values of reduction potentials instead, it would bring huge mess.

So whether if an electrode is a cathode or anode it will always be a "formal" cathode. However, when you calculate the E° the value would either be neg or positive. If its an actual "anode" the electrode would be having electrons flowing out from that end and the cell will have a negative E°?

The half cell reduction potential can be seen as the relative external potential (wrt SHE), that has to be set on the considered half cell to keep the half-cell at the edge between being a cathode and being an anode. The halfcell is an anode if Eext > E and vice versa. This addresses just TD, but not kinetic aspects. Many potentials are due frozen kinetics purely formal.

See results of the search site:chemistry.stackexchange.com cathode potential cell negative galvanic OR voltaic and generally also the Related section on the right side of the CH SE content page.

In aqueous solution the potential of the standard hydrogen electrode is defined as 0. volts. The standard conditions are: all activities, fugacities, molarities equal to one [1.]. By DEFINITION in a cell Oxidation is at the Anode and Reduction is at the Cathode. Now comes the dilemma How do we write a standard redox equation?
Na +1/2H3O+ = Na+ +  1/2H2 + H2O;  E = ~+2.8V  oxidation at the anode. Or
Na+ +1/2H2 + H2O  = Na  +  H3O+    E = ~-2.8V  reduction at the cathode!
Way back when, the Europeans liked the first and the Americans liked the second. As usual they fought it out and decided on the second [It was preferred that Sodium ions did not react with water than that Sodium did react explosively with water.] but you can still reverse everything when necessary.
Physicists are still confused because they consider electric current as the flow of +charge not as [we know it] the flow of electrons.
Sodium reacts violently with water. It is oxidized. the oxidation potential is positive [+]. Sodium ion is reduced with difficulty in water. The reduction potential is negative [-].
Finally, it is the nature of aqueous solution that the potentials for H2 are "central". Some species [Na+, Ca++ etc.]  are reduced with difficulty and have negative reduction potentials. Others such as O2, F2, Cl2, MnO4- etc. are reduced easily and have positive reduction potentials. There is a large range in between. Of course, if one prefers to work with oxidation potentials everything is reversed.

You have it the other way round: Europeans preferred the reduction potential (Ostwald school of thought), not the Americans (Gibbs and Latimer school) and it was not simply not a matter of reversing the sign. It is the actual electrostatic sign of the electrode with respect to the hydrogen electrode and this is why all the reduction potentials have been adopted. So the European convention won not the American :-)

Thanks! I was working from memory of my Quant analysis course, the potential table was oxidation, and a long ago discussion. The way the equation is written has no effect on what is happening at the electrode.


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Why so many intermolecular forces?
I was recently learning about intramolecular and intermolecular forces. It seems there are a variety of interactions ( Dipole-Dipole. Ion-Dipole etc ). Why must this be the case? Couldn't all chemical species just interact in the same manner. I feel like this would be more intuitive.

"Why must this be the case?" — I mean, that's just how it is. A rainbow has 7 colours, we don't go around asking why can't it just have 1.

They all do interact in the same manner. Ion-dipole and other smart names are just labels that we humans put on things.

all the forces involved are the same electromagnetic forces. It is the details that are complex. Nature and quantum mechanics don't have to be simple and the normal classifications used in chemistry are a way to simplify the exceedingly complex nature of quantum scale electromagnetic interactions.

Bound displaced charges are more conveniently evaluated as a complex object with internal charge structure than as a set of independent charges.

There aren't different intermolecular forces. But there are many, many levels of approximations and in chemistry it often happens that the approximations are taken at face value due to the way its taught and presented in lectures. All these interactions that you list are part of the multipole expansion of the electromagnetic interaction, which is truncated. The terms retained in your approximation depend on the situation that you want to model. For intermolecular interaction these are typically dipole and point charge terms, since these are the dominant terms.

Everything we have today is up to the limit of knowledge and discoveries that are made in the world. Won't be surprised in the future, if there's much more than all that is being uncovered. Look at the evolution of these findings, interactions and forces. They have taken various shapes to solve a problem with minimum energy. Those may have evolved to support something essential for a state of equilibrium etc. Many QM theorists don't agree with the simple force fields based simulations, because FF systems are as far as it goes and doesn't necessarily serve certain aspects of the actual system

@HansWurst says it the best.
In chemistry, there are essentially only one interaction, which is Electromagnetic interaction. However, the complexity of the quantum system (probability, charge distribution, etc.) necessitates multiple description of the same interaction. Both dipole-dipole and ionic interaction are basically Coulomb interaction. However, in dipole-dipole interaction, the charge is not in separate atoms.

Intermolecular forces are of a diverse variety:

ion-ion interaction
ion-dipole interaction
dipole-dipole interaction
ion-induced dipole interaction
dipole-induced dipole interaction
induced dipole-induced dipole interaction (dispersion/London forces)

Encyclopedia Britannica explains some of them.
But all these forces essentially stem from the same quantum mechanical effect. What makes them distinct from each other is the level of complexity involved in the species that the forces act on. The effect we are seeking is the displacement/distortion of electron cloud of a species.
Take, for example, Argon which is an inert gas but it liquifies at $-185.8$°C which states there is some level of inter-atomic forces acting among atoms to liquify the gas. These forces are the London forces that are caused by the temporary and random distortion of electron clouds within the atoms.
In single atom, the electron cloud is swirling around but this distribution of electron cloud is not perfectly symmetric due to wave-particle duality of electrons and the uncertainty principle. When this symmetry breaks, the electron cloud redistributed in terms of the electron density in the atom. Where the density is higher, there develops a partial negative charge (δ-) in that region whereas the region where the cloud density decreased gets an equally partial positive charge (δ+). Then a nearby atom that was perfectly symmetric would break it's symmetry in response to the temporary polarity generated in the first atom. Where there is (δ-) in the first atom, it will develop a (δ+) in it's vicinity and vice-versa. In this way more and more subsequent atoms are weakly held together over larger distances.
Now, talking about the aforementioned forces these are just variants of the same effect with some complexity of the species involved. But the same principle is acting. Either temporary or permanent redistribution of electron clouds (or electrons in case of ions) within the same species or from one species to the other.

I think this is a good answer, but quadrupole-quadrupole (e.g., benzene dimer) and other kinds of electrostatics also show up, not to mention hydrogen bonds, chalcogen interactions, sigma holes, … also not all intermolecular interactions require displacement / distortion of the electrostatic distribution - some are just undistorted electrostatics.

I agree that my answer covers only a few basic interactions because I wanted to give only a basic idea of how it works. There are many left out such as metallic bonding that are bit complex and my knowledge does not well extend to them. I request if anyone could elaborate on those forces, they may add their answer here.

Why must this be the case? Couldn't all chemical species just interact in the same manner.

It would be fine to say that intermolecular interactions are all based on how electrons and nuclei (specifically, protons) interact. However, this would be of little use when you are interested in the specific interactions between two species.

Why so many intermolecular forces?

By classifying intermolecular forces into different types, you can get a sense of the strength of interactions (e.g. ionic interactions are stronger than dispersion interactions when comparing a single pairwise interaction). You can also get a sense of how the interactions get weaker with distance (e.g. some have a $\frac{1}{r}$ law, some have a $\frac{1}{r^2}$ law).
You can also classify molecules or moieties by the kind of intermolecular forces they support, and arrive at rules of thumb such as "like dissolves in like".


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Enthalpy change of neutralisation?
Consider this problem.
The enthalpy of neutralisation is defined as the enthalpy change when one mole of water is formed through the react of an acid and base at standard conditions. As the coefficient of water in the products is 2. I feel like the answer is $\frac{x}{2}$. However, I feel confused. How do I know how many moles is formed. The reactions only give me the ratio. So even if this reaction produced one mole of water, it would still have the same ratio. Hence $x$ should be the answer. What's really going on?

The enthalpy change of neutralisation for the reaction 2NaOH + H2SO4 $->$ Na2SO4 + 2H2O is $x$. But how do we know the reaction above is not producing a mole of water? The reaction only gives the ratio between reactants and products. Why are we assuming 2NaOH means 2 moles?

Denoting $z$ the enthalpy value of the products, then, the enthalpy of neutralization can be calculated by the heights of the diagram
\begin{align}
  \Delta_\pu{neut} H &= \text{Enthalpy of the products} - \text{Enthalpy of the reactants} \\
  \Delta_\pu{neut} H &= z - (-x + z) \\
  \Delta_\pu{neut} H &= x
\end{align}
By the information of the energy diagram, this corresponds to the following reaction
$$2\ce{NaOH} + \ce{H2SO4} \rightarrow \ce{Na2SO4} + 2\ce{H2O} \hspace{1 cm}
   \Delta_\pu{neut} H $$
However, this reaction is not in agreement with your definition, so we have to "divide" the chemical equation by two
$$ \ce{NaOH} + \frac{1}{2} \ce{H2SO4} \rightarrow \frac{1}{2}\ce{Na2SO4} + \ce{H2O} \hspace{1 cm} \frac{\Delta_\pu{neut} H}{2} $$
Therefore, the answer is
$$ \boxed{\frac{\Delta_\pu{neut} H}{2} = \frac{x}{2}} $$


text:
How many lone pairs does NO have?
In an exam, I have been asked about the number of lone pairs in the molecule NO, the answer has been given as 2. According to Wikipedia, there is a structure relating a three electron bond as a third bond, but doesn't it violate the octet of oxygen?
I kind of understand that the octet rule is old and has been disproven, I haven't found any explanation for this phenomenon, the only other fact that supports it (which I have found) is the bond order from MOT, but I don't understand it.

Are you primarily asking about the bond order or about the number of lone pairs? You can correlate bond order (a concept we made up) to something real, the bond length. If you compare the bond lengths of $\ce{NO+, NO*, NO-}$, and maybe also look at isoelectronic species (e.g. $\ce{O2, N2, CO}$), you can get a sense of what it going on without having to rely on bonding theories you don't yet understand.

I have assumed both the bond order and number of lone pairs to be correlated, but I am not sure about it, then what do we say about the number of lone pairs in NO molecule, and I have edited the question

Here is the most common way of writing the Lewis structure of NO, a molecule with an uneven number of valence electrons:

Source: http://www.chem.ucla.edu/~harding/IGOC/N/nitric_oxide01.png
From that image, you would say that there are two lone pairs on the oxygen atom (that is the typical binding pattern) and one lone pair and an unpaired electron on the nitrogen.
However, the Lewis structure does not reflect that the bond is a bit shorter than the N=O double bond found in other molecules and ions (and a bit longer than the triple bond).
If you take a look at the MO-diagram, you will see three pairs of electrons designated as bonding, and a single electron as antibonding (if we call the electrons in the 2s orbitals non-bonding for now).

Source: https://chemistry.stackexchange.com/a/15317
There is no good way to draw this as a Lewis structure. It seems like Wikipedia tried something without explaining it:

Source: https://commons.wikimedia.org/wiki/File:Nitric_oxide.svg
This depiction, however, is confusing. There should never be three electrons in an orbital.
You could invent a new way of drawing Lewis structures, showing a triple bond plus an electron that is designated as antibonding. It would look like this (but it should be emphasized that this is a non-standard depiction):

However, if you take this as the Lewis structure, you would count two lone pairs. Hopefully, this discussion shows you that lone pairs are not well-defined if you can't draw a Lewis structure that captures the molecule well.


text:
How can I find vapour properties of a compound molten salt?
I'm carrying out thermal-hydraulic simulations of the Aircraft Reactor Experiment (a nuclear molten salt reactor) based on circulating molten fuel salt: NaF 53 mol%, ZrF4 41 mol%, UF4 6 mol%.
While the liquid salt thermodynamic properties have been measured and are hence available, the properties at saturation and the properties of the vapour phase are unknown. Due to numerical needs, I now need to switch from a pure liquid simulation to a "fake" liquid + vapour simulation ("fake" as the operating conditions should be well below the saturation temperature).
Hence, I need to introduce sensible (but not necessarily precise) thermodynamic properties for the vapour phase, as well, so that the simulation effectively remains in pure liquid.
I have all the liquid laws needed, for the following properties:

specific enthalpy $h_l(p,T) = (34.5 + 0.235 \times (T-273.15) )\times4184.$
density $\rho_l(p,T) = ((3.98 - 0.00093(T-273.15))\times1000.)\times\exp(10^{-9}p)$
specific isobaric heat capacity $c_{p,l}(p,h)$ (simply $=\dfrac{\partial h_l}{\partial T}$)
thermal conductivity, dynamic viscosity.

Now, similarly, for the vapour phase I need:

specific enthalpy $h_v(p,T)$
density $\rho_v(p,T)$
specific isobaric heat capacity $c_{p,v}(P)$ (which however could also be taken as the derivative of the enthalpy?)
[thermal conductivity and dynamic viscosity, but these are less relevant for the simulation]

Whereas at saturation I need the following:

pressure at saturation temperature $P(T_{sat})$
temperature at saturation pressure $T(P_{sat})$
liquid and vapour specific enthalpies at saturation pressure $h_{l,sat}(P_{sat})$, $h_{v,sat}(P_{sat})$
similarly, liquid and vapour densities and isobaric heat capacities $\rho_l(P_{sat})$, $\rho_v(P_{sat})$, $c_{p,l}(P_{sat})$, $c_{p,v}(P_{sat})$

Since reasonably, we should be able to deduce the properties at saturation from the properties of the two phases, this makes the vapour properties the real goal.
I've started looking here for the vapour laws. For instance, I've found properties for a single component of the salt like NaF here, but I am unsure whether the properties of single molecules suit the compound salt.
So, finally: what should I do?

Sadly, having glanced through a number of papers modeling molten salt binary/ternary/higher order phase diagrams (often mentioning various experiments as motivation), I find no gas phase data available in them. And it would appear that there are gas phase reactions, making modeling of the gas phase trickier than one might normally expect.


text:
Specifying which compound is oxidized/reduced in a redox reaction
I wrote the oxidation numbers in a redox reaction:
$\rm{2 \overset{+7}{Mn} \overset{-2}{O_4^{-}}~+~5\overset{+4}{S} \overset{-2}{O_2}~+~6\overset{+1}{H}_2 \overset{-2}{O}~~\rightarrow~~ 5\overset{+6}{S}\overset{-2}{O_4^{2-}}~+~2\overset{+2}{Mn^{2+}}~+~4\overset{+1}{H}_3 \overset{-2}{O^{+}}}$
It is clear that, S is oxidized and Mn is reduced.
But the text says that:
$\rm{SO_2}$ is oxidized and $\rm{MnO_4^{-}}$ is reduced.
It is easy to see that S suffers oxidation. But writing "$\rm{SO_2}$" as a whole seems to take our concentration away from S.
Is there any rule by which we can write the "compound as a whole" which is oxidized/reduced?

It is $\ce{Mn^7+}$ within  $\ce{MnO4^1-}$ that is oxidized. I'd accept both answers.

There is the mighty rule of context if there are  considered oxidized/reduced elements OR particles like molecules or ions  OR the whole ionic/dissociated compounds like solutions of salts/acids/bases.

It is useful to recall that a reducing agent is an electron donor and is itself oxidised.

Both concepts can be accepted. Some chemists state that Sulfur atom is oxidized and the Manganese atom is reduced. Some other chemists may state that $\ce{SO2}$ is oxidized, and permanganate ion is reduced. These two concepts are not contradictory : both can be accepted and considered as valid.

If you can separate the redox reaction into a reduction and an oxidation half reaction, you can say that the entire molecular entity undergoes reduction or oxidation.

$\rm{2 \overset{+7}{Mn} \overset{-2}{O_4^{-}}~+~5\overset{+4}{S} \overset{-2}{O_2}~+~6\overset{+1}{H}_2 \overset{-2}{O}~~\rightarrow~~ 5\overset{+6}{S}\overset{-2}{O_4^{2-}}~+~2\overset{+2}{Mn^{2+}}~+~4\overset{+1}{H}_3 \overset{-2}{O^{+}}}$

can be obtained from the two half-reactions
$$\ce{MnO4- + 5 e- + 8H3O+ -> Mn^2+ + 12 H2O}$$
and
$$\ce{SO2 + 6 H2O -> SO4^2- + 4H3O+ + 2e-}$$
We know that hydronium and water are acid/base pairs, so they don't show changes in oxidation state. By elimination, the electrons are coming from the sulfur dioxide and are going to the permanganate.
It is not always trivial to separate a redox reaction into oxidation and reduction half reaction, for example if one part of a molecule gets oxidized and another gets reduced. In those cases, it is best to assign oxidation states (atom by atom) and specify the atom that gets reduced or oxidized.

Chemical reactions are reactions of MOLECULES. A useful definition being the identity of a chemical species in the system being studied. Oxidation numbers are an artifact invented to simplify certain types of reactions.  The oxidation number of an atom is dependent on the molecular structure it is in. In the case of SO2 or MnO4- the molecule or molecular ion is reduced or oxidized and our current understanding of atomic and molecular structure allows the simplification that most of the change involves the sulfur or manganese atoms. This is manifested in assigning constant oxidation numbers to oxygen[-2] and hydrogen[+1] atoms in most compounds.
This idea carries over to Organic chemistry where the reactions of certain functional groups are to a large extent independent of the bulk of the molecule. As one studies and practices chemistry One learns that there are nuances and variations.

Ions are not molecules.....

@Poutnik Strangely enough, they are molecular entities, though. https://goldbook.iupac.org/terms/view/M03986

@Karsten Yes, they are. But so are atoms.

For certain properties such as colligative properties ions act as molecules Also ions such as sulfate, permanganate are molecular ions.  Atoms can be monatomic molecules in the gas phase [noble gases] or in amalgams.  A more general idea is what fits into PV =nRT or into colligative properties or the entity that affects the chemical potential; not the textbook definition.

IUPAC Gold Book authoritative definition of molecule  says An electrically neutral entity consisting of more than one atom (n>1). // ions like sulfate are not molecules, but molecular entities (as Karsten said) what is a superset containing atoms, molecules, ions, ion pairs and similar structures.

Perhaps the gold book needs amending.

Thank you for the comments and replies.
Based on the comments and replies, I thought about five simple steps which may be applicable to each species in any redox reaction.
Applying the steps to $\rm{SO_2}$:

The partners are S and O.
During the reaction, S loses electrons.
During the reaction, the other partner O neither loses nor gains electrons.
So the net effect is that, $\rm{SO_2}$ loses electrons.
Consequently, we write: $\rm{SO_2}$ is oxidized.

Applying the steps to $\rm{MnO_4^{-}}$:

The partners are Mn and O.
During the reaction, Mn gains electrons.
During the reaction, the other partner O neither loses nor gains electrons.
So the net effect is that, $\rm{MnO_4^{-}}$ gains electrons.
Consequently, we write: $\rm{MnO_4^{-}}$ is reduced.

Are these five steps valid? Can we apply these steps to any redox reaction?

You are definitely over thinking this. SO2 to sulfate has not lost any electrons add them up.  It is how the electrons are distributed in the molecule or ion.  In sulfate more of the valence electrons of sulfur are shared with more O atoms. The different electronegativities of the atoms cause bond polarities and charge separation in the molecule.


text:
Why does the conjugate base of peracetic acid not have any resonance structures?
Peroxyacetic acid, $\ce{CH3C(O)OOH}$ has its anion form when a proton is detached, like $\ce{CH3C(O)OO-}$.
I think it can have two resonance forms like I drew (even though they have several charge separations, they're at least possible), but several websites address that it cannot have any resonance structures.

1st one would be OK if not for lack of positive formal charge. In 2nd the same issue gets more pronounced...

I don't think first structure ever exists.

@mat I was thinking about first "alternate" one. You mean the anion itself? While maybe not terribly important in solution, afaict there are salts available.

There is resonance in CO, so there should be resonance in RC(O)R. If that description is to be applied. It's just a mathematical model after all.

When you are considering resonance structures, you have to maintain certain aspects of the molecule; you have to maintain molecular formula/connectivity, number of electrons, and net charge. You maintain the molecular formula and number of electrons throughout your proposed structures, but what you fail to account for is net charge. In the true structure, there is a net charge of -1 resulting from the rightmost oxygen. In your structures, one example has a net charge of -2 while the other has -3. Here is a good source with the rules for resonance structures.
With that information, you can dismiss the resonance structures as invalid from the outset. This is because a resonance structure is the result of molecules "spreading" their electrons to achieve a more stable state; net charge will not get bigger. When you make the net charge bigger, as you have done, you make the molecule less stable, as there are now more lone pairs and charge is less evenly distributed. Furthermore, carbon will not pull an electron away from oxygen as you have done in your bottom drawing, at least in normal conditions; oxygen has a much higher electronegativity than carbon and will hold onto its electrons too tightly for carbon to remove them like that.
For these reasons, paracetic acid's conjugate does not show resonance because there is no resonance structure that follows the rules of resonance; the potential resonance structures you have considered are destabilizing, which is the opposite of the effect of resonance (resonance will always make a molecule's structure more stable).

When you are applying the concept of resonance, you have to maintain every aspect of the molecular. Resonance is a mathematical description resulting from an incomplete model. As such the stabilising effect is purely hypothetical. Unfortunately I have to disagree with this answer. To me there is no doubt that there is pi-delocalisation in this molecule.

What do you mean when you say "maintain every aspect of the molecular?" As well, are you saying that these proposed structures would exist in the molecule? Or are you only saying that to apply the concept of resonance as I have done is incorrect, and that the molecule will have delocalized electrons even if resonance cannot explain that delocalization?

Oh sorry, there is a typo, it should be: When you are applying the concept of resonance, you have to maintain every aspect of the molecule. This is due to you writing "certain aspects", which gives the impression that there are some aspects that differ between resonance structures. Resonance is a concept stemming from Valence Bond Theory. It captures the delocalisation of electrons within this theory; something which is inherent in Molecular Orbital Theory. I recommend the following: https://chemistry.stackexchange.com/q/51632/4945

Ah yes I see haha. That wasn't my intention when I wrote "certain" but I see what you mean. Cheers.

Before we suggested any resonance structures for the conjugate base of peracetic acid (PAA), we have to prove it physically exists. The $\mathrm{p}K_\mathrm{a}$ of peracetic acid is found to be 8.3 (Ref.1). According to Ref.1, PAA is spontaneously decomposed to products given in equation $(1)$ in the $\mathrm{pH}$ range of 5 to 10:
$$\ce{2CH3C(O)O-O-H -> 2CH3C(O)OH + O2} \tag1$$
The authors have shown that the maximum rate have occurred at $\mathrm{p}K_\mathrm{a}$ of PAA, and the mechanism of decomposition is gone according to following figure using labelled $\ce{CH3C(O)O^{18}-O^{18}-H}$.

Accordingly, nucleophilic peroxy $\ce{O-}$ would attack the electrophilic carbonyl carbon of acetic part of another molecule to give the product as shown. The mechanism was confirmed by finding 87% of decomposed oxygen has been in $\ce{O^{18}-O^{18}}$ formula. However, a different group (Ref.2) has proved that there is no peroxy $\ce{O-}$ exists during the decomposition during the $\mathrm{pH} \le 7.5$:

[...] in the preparation of PAA, the $\ce{H+}$ concentration in the system is usually more than $\pu{0.1 mol L−1}$. It indicates that PAA is present as molecular form and no peraetic anion is found in the reaction solution.

Thus, the reaction mechanism of PAA spontaneous decomposition in an acid system can be predicted as two-step reaction. Step 1 is as follows:

The protonated form $\mathrm{F_1}$ can be attacked by another unprotonated molecule in step 2 to give the active intermediate as follows:

Thus, it is very unlikely that significant amount of peracetyl anion exist in these conditions before it self-decompose below its $\mathrm{p}K_\mathrm{a}$. That means it doesn't have a chance to show or form any resonance activity!
References:

E. Koubek, M. L. Haggett, C. J. Battaglia, Khairat M. Ibne-Rasa, H. Y. Pyun, and J. O. Edwards, "Kinetics and Mechanism of the Spontaneous Decompositions of Some Peroxoacids, Hydrogen Peroxide and t-Butyl Hydroperoxide," J. Am. Chem. Soc. 1963, 85(15), 2263–2268 (DOI: https://doi.org/10.1021/ja00898a016).
Xuebing Zhao, Keke Cheng, Junbin Hao, and Dehua Liu, "Preparation of peracetic acid from hydrogen peroxide, part II: Kinetics for spontaneous decomposition of peracetic acid in the liquid phase," Journal of Molecular Catalysis A: Chemical 2008, 284(1-2), 58–68 (DOI: https://doi.org/10.1016/j.molcata.2008.01.003).

This is all great information, but I disagree with the final takeaway message. Just because something is under certain conditions reactive, it doesn't mean that this something isn't showing a stabilizing effect in the first place. I would be very surprised if there wasn't any pi-delocalisation in paa.


text:
What are the products of 1 mol of vitamin C (C6H8O6) and 1 mol of hydrogen peroxide (H2O2)?
Basic stoichiometry suggests that  $\pu{1 mol}$ of vitamin C($\ce{C6H8O6}$) react with $\ce{1 mol}$ hydrogen peroxide ($\ce{H2O2}$) to produce one $\ce{1 mol}$ dehydroascorbic acid DHA($\ce{C6H6O6}$)?

Other sources suggest that the product of the above chemical reaction DHA ($\ce{C6H6O6}$), is more susceptible to further oxidation than vitamin C  ($\ce{C6H8O6}$), leading to further reduction.
("Both dehydroascorbic acid and hydrolyzed dehydroascorbic acid (diketogulonic acid) are more susceptible to hydrogen peroxide oxidation than ascorbic acid.")
Would the reaction change if it took place in 1 liter of $\ce{H2O}$?
How could I control the reaction to produce DHA?
 See here

Avoid confusion of O and 0 in chemical formulas. Note that mole is the unit like kilogram, mol is the unit symbol like kg.

Thank you. I appreciate your feedback. Why am I getting down voted?

I corrected capitalisation, but title doesn't exactly match the rest... Hiding bare urls would also be good idea.

Thank you for your feedback.

If you quote material, you should include a proper citation, too.


text:
pH of a mixture of a weak acid and a different (non-conjugate) weak base
Let's say I have 1000 ml 1 M acetic acid ($\mathrm{p}K_\mathrm{a}$ = 4.76) and add 1000 ml 1M aniline ($\mathrm{p}K_\mathrm{aH}$ = 4.65). To what proportion is aniline protonated?
1M acetic acid has a pH of:
$$\mathrm{pH} \approx 0.5*(4.76-\log 1)=2.38$$
Using Henderson Hasselbalch to determine fraction of protonated aniline:
$$2.38=4.65-\log x$$
$$ x=\mathrm{[HA]/[A]}=186$$
so ~ 0.5 % would be aniline and 99.5 % would be protonated?
This can not be true. As aniline judging by its $\mathrm{p}K_\mathrm{a}$ does protonate less than the acetate anion. Does anilin get protonated after all?
Where is the mistake?

Looks like only thing you got right is pKa of acetic acid.

Most pH related ready to use formulas are intended for specific scenarios AND specific simplifying conditions. When using them, you must know you can use them and why you can use them. Generally, you have to solve a set of nonlinear equations, based of equilibrium equations, substance amount inventory equations and charge balance equation.

e.g. Famous pH=0.5(pKa - log c) is pH of weak acid solution IF there is no other pH active component AND IF c >> [H+] >> [OH-].(ignoring Pandora's box of activity coefficients.)

Before you use any formula to calculate a pH, you first have to figure out what the major species are, and whether more than a small fraction of the species will react.
You calculated the pH of the acetic acid solution as 2.38. Try calculating the pH of the aniline solution. As you said, aniline is the deprotonated form, so the $\mathrm{p}K_\mathrm{a}$ given is that of the protonated form, and aniline is a weak base. So the pH is about 14 - 2.32 = 11.68.
This (and the fact that acetic acid and aniline are at the same concentration) tells you that there will be a reaction, and the pH will be somewhere in between.
As a first approximation, imagine that instead of acetic acid and aniline, you had acetic acid and acetate. In that case, you would have a 1:1 buffer with a pH of 4.76. If it were aniline and protonated aniline, you would have a 1:1 buffer with a pH of 4.65. The pH will be in that region, so the major species will be two weak acids and their two conjugate weak bases. This does not allow for using a weak acid formula or a buffer formula directly.
If acetic acid were to react completely with aniline, you would get an amphoteric salt of protonated aniline with acetate (at a concentration of 0.5 mol/L). It turns out there is a formula to estimate the pH of amphoteric salts and the pH of salts of weak acids and weak bases that you can try.
Once you have the pH, you can check if all equilibrium expressions (that for auto-dissociation of water, for the acetic acid and for the aniline) are satisfied simultaneously. Or you can follow the classical engineering approach with mass balance, charge balance and multiple equilibria (solve 3 equations simultaneously).
The $\mathrm{p}K_\mathrm{a}$ values are probably extrapolated to infinitely dilute solutions (ionic strength 0). At a concentration of 1 mol/L, the ionic strength is far from zero, so you can estimate the pH to one or two significant figures, say pH = 5. To get a better value, you could just make the solution and measure the pH, if the aniline is soluble at this concentration at a slightly acidic pH.

thanks. It may seem like a trivial question but non of my graduate chemistry colleagues could tell me

mass balance and multiple equilibria  Charge balance is additional equation which is "free of charge". (Sum of charges of anions and cations is zero. [H+]+[BH+]=[OH-]+[A-])

One more thing that can be added maybe:
You can predict the position of equilibrium between two moderately weak acid and base by comparing the pKa's and assume that oxonium and hydroxide is not the major species as explained here:
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/02%3A_Polar_Covalent_Bonds_Acids_and_Bases/2.09%3A_Predicting_Acid-Base_Reactions_from_pKa_Values

And then determine the pH from Henderson Hasselbalch for one or the other acid/base pair.

It is similar as pH of Na2HPO4 solution. One can approximate concentrations of H2PO4- and PO4^3- are the same. In our case, acetate and anilinium.


text:
G09 how to I prevent Na-O bond formation doing DFT on a sodium carboxylate?
I am trying to determine Gibbs free energy of solvation for a sodium salt of an alkylcarboxylate. B3LYP/6-31g. The minimized structure makes covalent bonds between sodium and carboxylate oxygens. How do I prevent this?
Thanks

How did you come to the conclusion that the bonds are covalent? Even an inaccurate level of theory like B3LYP/6-31G doesn't produce these kinds of errors.

I think it's important to remember that DFT calculations (e.g., Gaussian) produce a set of atoms and XYZ coordinates. We like to see "bonds" in visualization tools, but they are inferred based on interatomic distances.

see my answer below about covalent

How do I prevent this?

You don't.
Ionic bonding is an idealization we do not expect to be realized in pure form between individual atoms, even in cases where the electronegativity differential would seem to be quite large. As described here, purely ionic bonding is more closely approached between multiatomic entities for which the molecular-orbital structure favors such bonding.
Moreover, the boundary between "ionic" and "covalent" bonding is really indistinct; what appears to be ionic bonding may in many cases have a covalent interpretation. The lattice structure of calcium carbide offers one such case.

Reminds me about https://chemistry.stackexchange.com/q/40729/4945 However, I think OP is not interpreting their calculation correctly, and to me the question isn't really clear.

how do I know it is covalent? when I look at fchk file, visualization draws two bonds between oxygens and sodium and the atoms no longer bear any charge.  i have no problem modeling ammonium bromides, just getting stuck with sodium carboxylates.

@NonclassicalCarbocation a visualisation tool will create an image based on interatomic distances. At least with the information usually included in a fsck file. You need to analyse the calculation further to judge the bonding situation. You must not equate a line in a program between two balls as a covalent bond. Study your system, one way of doing that is described in the question I've linked, but there are others. However, be aware that there will always be room for interpretation. As with your original goal, you should follow the literature to find the best approach.


text:
What do the units of enthalpy kJ/mol represent?
Enthalpy is making me slightly confused.

Here $\Delta H$ refers to the change in enthalpy. Does this refer to the enthalpy change of reaction? Are they the same? It seems to be change in enthalpy, standard enthalpy change and standard enthalpy change of reaction are all interchangeable. However, standard enthalpy changes of combustion, neutralisation, formation etc.. have different definitions. Is this correct?
Now the diagram above, shows the reaction between $\ce{Cl2 + H2 -> 2HCl}$. With my understanding, $\Delta H$ refers to the change in enthalpy when when one mole of $\ce{Cl2}$ reacts with one mole of $\ce{H2}$ to form two moles of $\ce{HCl}$.
$$\ce{Cl2 + H2 -> 2HCl}\quad\Delta H=x \ \mathrm{kJ/mol}$$
Where $x$ has some numerical value. I think $\Delta H$ is measured in $\mathrm{kJ/mol}$ as the exact quantities of each element are unknown. However, If we were to have exactly one mole of $\ce{Cl2}$ and one mole of $\ce{H2}$ then the change in enthalpy would be $x\ \mathrm{kJ}$. Without the $\mathrm{/mol}$. So enthalpy can be measured in either $\mathrm{kJ}$ or $\mathrm{kJ/mol}$ right?
Finally what if we have the reaction: $\ce{CH4 + 2O2 -> CO2 + 2H2O}$ where $\Delta H=-890\ \mathrm{kJ/mol}$. What does the per mole now refer to? Because now the reaction is between one mole of $\ce{CH4}$ and two moles of $\ce{O2}$ so it doesn't make sense to me to talk about enthalpy change per mole anymore. Could someone explain intuitively what the per mole now represents?

Expression formatting reference:
Basics/CH+Math expressions/Upright vs italic/Math tutorial
// MathJax is preferred not to be used in CH SE question titles for indexing and question list displaying reasons.

$\ce{Cl2 + H2 -> 2HCl}$ and $\Delta H=\pu{-890 kJ/mol}$ makes $\ce{Cl2 + H2 -> 2HCl}$ and $\Delta H=\pu{-890 kJ/mol}$

Imagine the reaction:
$$\ce{CH4 +2O2 -> CO2 + 2H2O}$$
Now one can interpret that one molecule of $\ce{CH4}$ reacts with 2 molecules of $\ce{O2}$ to produce one molecular of $\ce{CO2}$ and two molecules of $\ce{H2O}$. It will be equally valid if another person says that,
$$\ce{1 mol CH4 +2 mol O2 -> 1 mol CO2 + 2 mol H2O}$$,
say we can measure the heat released during the reaction, and it can be expressed as a number with a unit Joule. Say this reaction releases 890,000 Joules. Since the heat is released, let us insert a negative sign, i.e., - 890 Joules. Now we can write
$\ce{1 mol CH4 +2 mol O2 -> 1 mol CO2 + 2 mol H2O}$ with -890 kJ.
Can I divide the whole statement with mol? What do we get, now?
$\ce{1 CH4 +2  O2 -> 1 CO2 + 2 H2O}$ with -890 kJ/mol.
So, the per mol in the enthalpy, is per mol of the entire reaction. It confirms that original equation's coefficients were understood to be mol rather than the number of molecules.


text:
How to read 1H-NMR- and IR-spectre (for ibuprofen)
I have read the answers to related/similar questions. They didn't help.
I have to read a proton-NMR-spectrum and an IR-spectrum of ibuprofen. The structure formula is given to:

I have not been formally taught how to do this, nor have I solved any problems beforehand. This is because it's required for my project, but we haven't had the training like the former classes have for reasons I won't waste time explaining.
I know the very basics of the two spectroscopy methods as I am a high school student.
My questions at the moment have become the following:

(proton-NMR) How does di-substituted aromatic rings affect the splitting of the signals from A and B?
(IR) Should I only focus on confirming the aromatic ring and carboxylic acid? And do I only have to figure out the fingerprint area or do I have to analyze the whole IR-spectrum?

Before you take a look at the spectra, I recommend looking at the last two pages of this link
The spectras are as follows:
proton-NMR-spectrum: (Source)

IR-spectrum: (Source)

Is there a question and describe your project. The link gives an excellent nmr explanation, Your spectrum does not show splitting and integration so is useless. The infrared in Nujol obscures important information in the C-H stretch and C-C stretch; you need an atr or KBr pellet spectrum.

To a very crude first approximation, A and B appear as doublets in the NMR. That's because A couples to B and vice versa, and to a first approximation, the AB pair on one side doesn't interact with the AB pair on the other side.
This is not entirely true, of course. The system is not really two separate AB pairs, but is properly described as AA'BB' (or AA'XX', depending on the exact frequencies involved). This website: https://organicchemistrydata.org/hansreich/resources/nmr/?page=05-hmr-15-aabb/ and this question: How can multiplets in para-disubstituted benzene rings be described? discuss it in more detail, but at your current level, you almost certainly don't need to worry about that.

For the IR, I'd focus on picking out whatever you can pick out; what you mentioned is probably enough. It's rare to use the messy region of the IR for structural elucidation. IR still has its uses, but there's a reason why NMR is nowadays considered the 'primary' characterisation technique for typical organic molecules.

So do you suggest that I just call A and B doublets, since that's what our knowledge of the topic can explain? Also I don't understand how C is quartet either. I have read the "answer" but I need more explanation.

If you don't understand why C is a quartet, I'd suggest reading some websites on basic NMR, google 'n+1 rule nmr' for some examples.

I figured it out myself, but thanks for the tip anyway :) I think I'm just tunnel visioning heavily.

@Daniel: Only structural feature you can abstract from this IR is the characteristic peak at around $\pu{1720 cm-1}$ for the presence of carbonyl peak.