text:
Why do Rydberg atoms have electrons that can transition above the first ionization energy?
Why do Rydberg atoms have electrons that can transition above the first ionization energy?
This can only happen in a multi-electron system but I don't see how the Rydberg formula would even work for n = infinity + 1. I thought it might have to do with the second ionization energy but the definition of that is just another electron being ionized not anything above the first ionization energy.

Why do Rydberg atoms have electrons that can transition above the first ionization energy? Any refererence to it ? As it is literally asking how $a < b$ can be at the same time $a > b$.

Pretty sure only three-component particles can have bound states above zero energy. The theoretical Rydberg atom certainly cannot. I guess one spin-state of an actual Rydberg-like atom has a slightly larger disscociation limit than the other. Within an electromagnetic field, things might be a bit different, too.

@Poutnik this is something my professor tried to explain. This is translated so it might not be the best working. "there are also internal electronic levels in multi-electron atoms that can be excited. These occur at higher excitation energies than those Rydberg transitions that converge to the first ionization energy (corresponds to the lowest energy to ionize an atom). They are therefore inevitably above the first ionization energy.
Note: This is not possible only in atomic hydrogen and other one-electron atoms, since there is only one electron to be excited."

Remember that too laconic question elaboration is figurally very expensive,
leading to clarifications much longer than the saved words in both characters and time.

A useful concept in Rydberg atoms is the notion of a channel. This concept is borrowed from scattering theory and describes all solutions of the Schrodinger equation for the scattering of the Rydberg electron of specified angular momentum with the ionic core in a certain energetic state. The bound Rydberg states correspond to scattering at negative energy and the scattering at positive energy corresponds to the ion-electron continuum of the channel.
If the ion has multiple electrons, it can have several excited states. Each of these excited states can form a channel. The energy of a Rydberg electron is given by
$E=E_\text{IP} + E' - \frac{hcR}{(n-\delta_\ell)^2}$
Where $E_\text{IP}$ is the first ionization energy, $E'$ is the excitation energy of the ionic core, $R$ is the mass-corrected Rydberg constant, $n$ is the principal quantum number and $\delta_\ell$ is the quantum defect for a Rydberg electron with orbital angular momentum $\ell$.
Each excited state of the ionic core thus gives a new Rydberg series with each own set of principal quantum numbers. So, Rydberg states do exist above the first ionization energy, but their principal quantum number is finite; it just belongs to a different channel.

Image source

if I wanted to know more about this could you reference a book or website.

The classic book on Rydberg atoms is from Tom Gallagher. But this is quite an advanced text. You could also try to google for quantum defect theory.


text:
Gaussian16 Raman calculation from IR checkpoint file
I already did an IR frequency calculation for my molecule. Is there any possibility to just run a Raman frequency calculation with the chk file from my IR calculation?
Or do I have to do the calculation again with the freq=raman keyword? I checked the Gaussian manual and found the raman=polar keyword, but I think this is not exactly what I am looking for.

Related (unanswered) question on researchgate. The paper mentioned suggests it was possible in Gaussian09, so there should be a keyword for this. A combination of freq=SaveNormalModes in the frequency calculation and freq=ReadNormalModes in the Raman calculation may do the trick, but I haven't gotten a chance to test it.

Thanks, I will try it today and will report later how it worked.

@Tyberius @Andrea I tried it with the keywords freq=SaveNormalModesand freq=ReadNormalModes. I used Methanol as an easy example.
First I run an optimization followed by a normal frequency calculation by using the geom=allcheck and the suggested freq=SaveNormalModes keywords. The last calculation a raman with this Routesection
%chk=methanol_raman.chk %oldchk=methanol_freq.chk #p GFINPUT POP(FULL,NBO) B3LYP/6-31G(d,p) freq=(raman,ReadNormalModes) geom=check
The calculation ends with the error FileIO operation on non-existent file..

Raman intensities require polarizabilities to be calculated, IR intensities only dipole moments. Your checkpoint file may therefore not have the properties you require, in which case you'll need to do more calculations. Using the polar=raman keyword would apparently save you a little bit of time, because the program then already knows your force constants and normal modes from the preceding freq calculation, but there will still be computational effort involved in calculating the polarizabilities.
Note that there are some cases where Gaussian also give you the polarizabilities "for free", as the manual states on the polar=raman page:

Note that Polar is done by default when second derivatives are computed analytically.

You can use the chkchk tool to see what's really stored in the checkpoint file.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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text:
Why are α-haloaldehydes unstable?
I'm currently reading through Clayden's Organic Chemistry, and in chapter 30 an assertion is made that

Halo aldehydes are unstable and should be avoided.

I should be clear that the context of this situation is under treatment with an enolate, so basic conditions are being employed. I believe the most likely hypothesis is that the instability has to do with the increased electrophilicity of the α-haloaldehyde's carbonyl group. The increased electrophilicity is supported by the equilibrium constant for the hydrate formation from acetaldehyde and chloroacetaldehyde, which is 1 and 37 respectively. This means that the rate of nucleophilic attack on the carbonyl group will be faster. Is it possible that the rate of nucleophilic attack is fast enough for O-attack on the carbonyl (hard-hard) as opposed to the expected C-C attack at the halogen center (soft-soft)? I have also proposed that α-haloaldehyde could eliminate H-Br to yield the ketene. I am less confident in my second proposal, as I have not seen any literature reporting this, and if so, I would expect equilibrium to lie on the side of the α-haloaldehyde.

I suspect the problem is hydrates and higher oligomers. https://www.osha.gov/dts/sltc/methods/organic/org076/org076.html

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text:
Benchtop NMR spectroscopy parameter explanation; tau and number of points
I have just carried out an exploration of the effect of different solvents and conditions on the 1H caffeine relaxation by benchtop NMR spectroscopy.
When collecting the 1D 1H NMR spectra of the caffeine in CDCl3 solution, the parameter for the number of points was set up, but when the relaxation run for T1 and T2 were done, the parameter was replaced by the value of time delay, tau.
I think I kind of get a clear view from your discussion, as the number of points is needed in determining the right peaks for 1D 1H spectra. But, as for the relaxation time, the intensities behaviour during the time delay is now being the subject of interest, hence the np parameter is neglected.
This is what I understand by far.
#updated
They are the number of points (np) and tau.
I think I kind of understand how np works;

it's the discrete points of analogue FID that are being converted to
digital points along with the FID signal envelope. Therefore it's
associated with the resolution of the spectra which the lesser np used
to define the FID, the lower the resolution.
*Let me know if I interpreted this wrongly.

This question raised when I'm analysing my data though, would really appreciate it if someone can enlighten me.

tau could be anything (usually a name reserved for a time delay, therefore greek "t") for all we know. Provide more info regarding the pulse sequence, maybe a link?

Possibly you apply a technique (equally known as «pulse sequence») already literature known; if this is the case, add this information. Examples of widely used techniques are DEPT, INADEQUATE, APT; keywords like these help to communicate and are used in the literature, too.  They may help you to trace information on-line, too (e.g., https://www.chm.davidson.edu/vce/NMR/InversionRecovery.html, https://www.ucl.ac.uk/nmr/sites/nmr/files/L5_3SH_web_shortened.pdf) and $\tau$ may be the intentionally varied vdlist parameter on page 4 on nmr.chm.jhu.edu/Material/Notes/JHU-NMR-Relaxation.pdf.

Like described here: https://www.nanalysis.com/nmready-blog/part-1-t1-relaxation-definition-measurement-and-practical-implications? https://images.squarespace-cdn.com/content/v1/5707ede0d210b8708e037a1e/1578421129211-76ZUIDJV0VALGIZ5UMSN/ke17ZwdGBToddI8pDm48kB6Pjbjlc1FE7O53-VUEVfwUqsxRUqqbr1mOJYKfIPR7LoDQ9mXPOjoJoqy81S2I8N_N4V1vUb5AoIIIbLZhVYwL8IeDg6_3B-BRuF4nNrNcQkVuAT7tdErd0wQFEGFSnNw0IdTOib5V9FSK4MMSFnRuicwUIHY6BMLw7I1kbnEbFILtcVpNoMqkdXuvWwHDcw/Figure+2.png?format=2500w

@KarstenTheis For my gusto, the example does not show all, because the integrated signal intensities are all for one type of (aromatic) C(sp^2)-bound H.  It would have been much more instructive if it were about H's bound differently to C(sp3), C(sp2) and C(sp) because recording quantitative 1H-NMR often requires to adjust the delay times normally set for recording routine 1H-NMR spectra to check the identity of a product in a synthesis lab. (Of course, recording quantitative 13C-NMR [like https://pubs.acs.org/doi/10.1021/ol403776k ] is a different story.)

@Buttonwood What the question is sorely lacking is the model of the bench top NMR in question. Some are so simple that you can run a couple of canned experiments only, and then it would be a matter of looking those up in the user manual. In the example I found, I'm not sure if you have to run multiple spectra setting the delay time, or if it runs a series of experiments based on a minimum, maximum and increment for the delay time.

Your experiment description in incomplete. There is no single tau in an IR experiment that you would need to care about for evaluation. You need a list of (usually logspaced) recovery delays. If your spectrometer just uses (1...np) x tau for that delay vector, OK, but you need to find out what's going on. In short, RTFM! ;) Until then, I vote to close.

Have a look at Nyquist-Shannon sampling http://195.134.76.37/applets/AppletNyquist/Appl_Nyquist2.html. which shows how many points are needed to digitise a signal.

AS Karl said read the instrument's manual and add the meaning of the symbols. However there is one important point in discrete Fourier transform (DFT).

I think I kind of understand how np works; it's the discrete points of analogue FID that are being converted to digital points along with the FID signal envelope. Therefore it's associated with the resolution of the spectra. *Let me know if I interpreted this wrongly.

I have been recently discussing this exact point with signal processing engineers and spectroscopist friend with a lot of good constructive discussions. No, the number of points cannot enhance the intrinsic resolution of the signals whether it is NMR or FTIR or any other FT spectroscopic technique. What you increase, when you collect more points, is the frequency resolution on the x-axis, and consequently more points per peak. That is it.
For an N-point DFT, the frequency step is
$$\frac{1}{N\Delta t}$$
For example if your sampling frequency is fixed to 1 s (an arbitrary number), in one experiment you collect 1000 points, the frequency step on the axis is
$$\frac{1}{1000 . 1 s}=0.001 Hz $$
In another experiment, you only collect 250 points with a time step of 1 s
$$\frac{1}{250 . 1 s}= 0.004Hz$$
Note the time to sample the signal in each case is different, because it is given by $N\Delta t$. So by chance, if your original experimental spectrum was "limited" by the number of points per peak by increasing the number of points, you might see an improvement in the peak shape and apparent resolution when you increase the number of points in an FID.


text:
Exergonic vs Endergonic Reactions
Problem
Problem 3, concept check 8.2 from Campbell Biology [1, p. 150]:

Some partygoers wear glow-in-the-dark necklaces that start glowing once they are “activated” by snapping the necklace. This allows two chemicals to react and emit light in the form of chemiluminescence. Is the reaction exergonic or endergonic? Explain.

Answer
Answer key [1, p. A-8]:

The reaction is exergonic because it releases energy—in this case, in the form of light. (This is a nonbiological version of the bioluminescence seen in Figure 8.1.)

From [1, p. 143]:

Figure 8.1 The green glowing spots on the outside of this Brazilian termite mound are larvae of the click beetle, Pyrophorus nyctophanus. These larvae convert the energy stored in organic molecules to light, a process called bioluminescence, which attracts termites that the larvae eat. Bioluminescence and other metabolic activities in a cell are energy transformations that are subject to physical laws.

Question
The answer is an exergonic reaction. What I understand of an exergonic reaction is that not only does it have a negative $ΔG,$ it releases energy. In this case, the energy released is light energy.
My original answer endergonic reaction as for the problem given, to “activate” the necklace, doesn’t it require energy from partygoers to snap the necklace? Is the $ΔG$ value for the light being released much less than the $ΔG$ value of the energy being used to snap the necklace?
Reference

Urry, L. A.; Cain, M. L.; Wasserman, S. A.; Minorsky, P. V.; Orr, R. B.; Campbell, N. A. Campbell Biology, 12th ed.; Pearson: New York, NY, 2020. ISBN 978-0-13-518874-3.

I suspect the "snapping" is just to break a physical barrier that separates the reacting chemicals, rather than the provision of an activation energy for the reaction itself. You wouldn't call a reaction endergonic just because (for example) it required energy for you to mix the two reactants together in the same beaker.

Consider a ball on a table in a bowl. Just because the table keeps the ball from falling to the ground and you have to knock over the bowl or table does not mean that the process is not exergonic.

@orthocresol The «snap» indeed is to break a physical wall in the inner of the tube of a glow stick to allow chemicals to mix which triggers the chemoluminescence.  Demo of a monocolored strick, but there are polychromes around, too.  Used by hikers, too.

Could it be you mean endergonic, not endogenic? (https://www.merriam-webster.com/dictionary/endogenic)

The reaction to consider is the transformation of chemical energy into visible radiation (light) and a little of heat.  A glowing stick consists of at least two compartments which -- on purchase -- a supposed to be kept separate.  The snap only breaks the wall between the two so that chemicals may mix with each other to start chemoluminescence (work principle), a principle similar to a firefly in late summer, too.
To decide if the reaction is exothermic, endothermic; exergonic or endergonic you have to compare the energy of initial state (when the chemicals are separated, no emission of light) with the one after the reaction.  There may be an energetic barrier to trigger the reaction, equally known as activation energy, but in its current form, this detail is not relevant to answer the question.
Take an analogy employing potential energy you might recall from the physics lecture: imagine you ate in a restaurant on a mountain and now walk back home.  It is not relevant that some energy was required to get up from the table, it is only about the comparison between
$$\mbox{state A «being in the restaurant»} \ce{->} \mbox{state B «back in the valley» } + \mbox{energy}$$
Even the pathway, straight downhill, or passing some smaller hills to reach the destination is not important here.  The naming then is, by convention, a formalism.


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Can rate constant depend upon the concentration of reactants (or other species involved in the reaction)?
Transition state theory tells us that the rate constant of an elementary step is $$k_\mathrm{r} = \frac{\kappa k_\mathrm{B}T}{h}\exp\left\{\frac{- \Delta G^\ddagger}{RT}\right\},$$
where $\Delta G^\ddagger$ is the activation free energy of the step, i.e. the difference between the Gibbs free energy of the transition state and the reactant state.
Then the rate law would be $$r = k_\mathrm{r}[A]^a[B]^b[C]^c\dots,$$
where $A$, $B$, $C$, ... are all the reactants that come together in that elementary step. So far so good.
But I have come across reactions where changing the concentration of a reactant (by a large amount) in a multi-step reaction seems to change the rate determining step of the whole reaction.
The explanation offered was that changing concentration changed the Gibbs free energy of the reactant or intermediate states (as Gibbs free energy depends on the concentration in solution). The states which contain the species (whose concentration  is changed) are affected whereas others are not, so the Gibbs free energies of activation are changed, which means the rate constants are changed, which can change the rate determining step.
But if concentrations can affect $k_\mathrm{r}$ then it seems like the rate law cannot be written like that.
Does this sound right, or am I overthinking this?

If the rate constant depends on the concentration of something, it is not a rate constant.

Then how would you define rate constant where the rate determining step can change if the concentration of one component is changed to a large extent?

TST is a crude approximation. It works only for elementary steps, only for the defined potential hyper surface. If you change the concentrations, you change the surface, the approximation breaks. P.S.: Please don't use mhchem for non-chemistry items.

@Martin-マーチン Thank you. So what happens to the rate constant then? It changes, right?

If with "change" you mean it was once that and became that, then: no. If with "change" you mean it is completely different and has absolutely no relationship with anything else, then: yes. That includes the cases, where that rate appears to be equal. I think I need to stress that again: TST only ever works for a single, well defined potential hyper surface.

@Martin-マーチン Sorry, I don't understand what you wrote, could you please explain what you mean? I understand that TST does not work in this case, but I am asking about the rate constant (k) as defined by the rate law (which can be calculated under all conditions). Does the value of k stay the same as it was when the concentrations of the reactants were low?

That is an entirely different question, and I am not equipped to answer it. Sorry.

It seems possible that you mix the terms. There is the reaction rate constant -- in your equation $k_r$, and there is the rate of reaction -- in your equation $r$ and sometimes expressed by $v$.
Depending on the order of a reaction, the overall rate of reaction may, but need not depend on the concentration of the reactants available.  There are reaction types known by experience to proceed typically one kinetic order (e.g., heterogeneous hydrogenation); however, reading the stoichiometric reaction equation often does not provide sufficient information and the rate law has to be determined experimentally.
Examples of kinetic reaction order are zero, one, two, three, but fractional orders are known, too.  It is possible to perform reactions which are of second order ($v = k_r[\ce{A}][\ce{B}]$) in a pseudo-first order regime if the concentration of reagent A is so much much higher, than the one of [B], that you detect after consumption of B still much of A.  This however assumes that the reaction still proceeds by the same pathway, thus requires experimental evidence to be supported.
This contrasts to $k_r$, which is independent from the concentration of the reagents, however depends on the temperature $T$ (keyword Boltzmann, and even more, Arrhenius equation).

If the Gibbs free energy depends on the concentration of the reagents, and $\ce{k_r}$ depends on the Gibbs free energy difference, then surely $\ce{k_r}$ can depend on the concentration?


text:
Negative DBE/Degree of Hydrogen Deficiency?
I tried calculating the Index of Hydrogen Deficiency for C6H15SiCl, and I get negative one. I get this value by subtracting the number of chlorine and hydrogen atoms from that of the carbon atoms, and then adding one. I don't think I am supposed to do anything with the number of silicon atoms.  Is this correct/normal?

SiH4 has the same IHD as CH4, so Si  should be treated as C.
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Index_of_Hydrogen_Deficiency_(IHD)   Always think  thoroughly about possible answers to your question and search for them
before asking.

In case of doubts, analy, how is hydrogen deficiency changed, if H- of  -CH2- are replaced by other atoms or their groups.


text:
Is this diagram correct for difference between 2N and N2

There is 1 $\ce{N2}$ molecule formed with 2 atoms of nitrogen.
2N is 2 separate nitrogen atoms which have not formed any bond.Reasons could be that they didn’t get enough energy to form bond so they are present separately.
I want to confirm is this right ?

So, the question is what's the difference between an atom and a diatomic molecule?

Well, that is pretty muc the difference between two atoms and a molecule. But you don't need energy to form a bond. Bond formation releases energy.

It is also 2N means 2 molecules of nitrogen with 1 atom right @orthocresol

N2 is 1 molecule with 1 atom.

N2 is one molecule with two atoms. There’s no such thing as a molecule with one atom, by definition: molecules are things that have more than one atom.

@user277768 Two hints for future reference. #1) It is fine to use the extended formatting on ChemSE, especially mathjax and mhchem (https://mhchem.github.io/MathJax-mhchem/), in questions, answers, comments.  But because they are something special some programs don't understand well enough for a correct display, don't use them in the title of a question.  #2 If you add a photo, consider a) to crop the important / to trimm off parts which are not relevant to the question with an editor like irfanview or pinta. And b) scale it down.

@orthocresol Jus on question.It is written in my textbook that 2N means two nascent nitrogen.What does that mean then sir?

@orthocresol please help sir

One $\ce{N2}$ molecule formed:
What you have drawn here is (as I have understood) Lewis dot diagram of the $\ce{N2}$ molecule. The correct way of drawing the molecule is as in 1)

Two separate N atoms which have not yet formed any bond:
With the concept of Lewis dot structure formula the Nitrogen atoms are represented as in 2)

Thank you for correcting my diagram but what about the question I wrote you think

Thank you for your appreciation. Your Assumption is
2N is two nitrogen atoms progressing their way to make a bond to become N2.
2N doesn’t mean separate N atoms, because we are in the process of making N2.My Answer:
If you are saying as above, then you mean a quasi or semi bond between two N atoms – which is similar to transition state preceding to the product formation as in reaction mechanisms.

The 2 in 2N as per universal convention means stoichiometric coefficient, and 2 in N2, means atomicity of a molecule.
So when we mention 2N we never mean a molecule or a semi molecule, but two separate atoms instead of writing N N.
As I have understood even if that 2N quasi molecule exists, that has not been trapped yet to get a structure.
Hence in my  opinion 2N is two separate N atoms, which could only be depicted as per the rules of Lewis Structure.

Ok.And N2 is a 1 molecule of two atoms.Can’t we say 2N as 2 molecules ?

More than one atom BONDS together to form a molecule. More than one atom not yet bonded by transfer or sharing of electrons simply means atoms not molecules.This is the case with 2N atoms. There is no such transfer or sharing of electrons when we write 2N. Hence they are atoms.


text:
What are high-energy electrons?
I read that (in cellular respiration) the transported electrons in NADH have a higher energy than those in FADH2.
I can't find a (simple or otherwise) explanation of what a "high-energy" electron is.
I came across a post on Reddit which asks the same question:

I am not a chemist but I am trying to understand what my biochemistry book
means by "high energy electrons" that are obtained from glycolysis and
the citric acid cycle. I am assuming that this "high energy" is not a
property of the electrons themselves, but it is not clear to me in
what way they are "high energy". Do they mean that when these
electrons are transferred to the electron carriers they are part of
molecular structures in which they sit in high energy positions?

(There were no satisfactory answers to the question).
Could anyone come up with a definition understandable by this non-chemist (i.e. me)?

Interesting question. It appears that in the context of biochemistry the definition of "high-energy electrons" is not the same as used for, say, HEED.

I would assume that in analogues reactions the energy gain is higher when NADH in involved instead of FADH2. Or that NADH can undergo a step that FADH2 can't. So what actually has high(er) energy is NADH. Without something more satisfactory consider Biology SE.

@Alchimista Saying that there will be a better answer about biochemistry questions on Biology SE is a trigger for me, compelling me to write an answer right away.

That's funny! I knew there was a chemist/biologist rivalry going on. Great answer, thanks.

The electron has a higher potential energy.

@KarstenTheis meaning was "without something more satisfactory than my comment THEN look at...." but my comment was already in tune. :)

transported electrons in $\ce{NADH}$ have a higher energy than those in $\ce{FADH2}.$

This is jargon describing the redox potential of the electron carrier $\ce{NADH}/\ce{NAD+}$ vs the electron carrier $\ce{FADH2}/\ce{FAD}.$ So if you compare the two reduction half reactions
$$
\begin{align}
\ce{NAD+ + H+ + 2e- &-> NADH} &\quad E^{\circ '}_\mathrm{red} &= \pu{-0.320 V}\tag{1}\\
\ce{FAD + 2H+ + 2e- &-> FADH2} &\quad E^{\circ '}_\mathrm{red} &= \pu{0.031 V}\tag{2}
\end{align}
$$
$\ce{NADH}$ is the stronger reducing agent.

Could anyone come up with a definition understandable by this non-chemist (i.e. me)?

Another way of saying this is that the reaction of $\ce{NADH}$ with dioxygen is more exergonic (the equilibrium lies further on the side of the products, more free energy is available from it) than the reaction of $\ce{FADH2}$ with dioxygen. For completeness, the reduction half reaction for dioxygen/water is:
$$\ce{1/2 O2 + 2 e- + 2H+ -> H2O} \quad E^{\circ '}_\mathrm{red} = \pu{0.816 V}\tag{3}$$

Do they mean that when these electrons are transferred to the electron carriers they are part of molecular structures in which they sit in high energy positions?

It has to do with how tightly the electrons are bound in one and the other electron carrier. The OP is correct to place the term "high-energy" in quotes when referring to electrons; electrons are indistinguishable, so their energy is with respect to the bound state that they are in.

@KarstenTheis So the electrons are less tightly bound in NADH than in FADH2?

@Naj Yes, and most tightly bound in water.

I understand the 'Yes' (electrons are less tightly bound in NADH than in FADH2). I'm not sure I understand 'and most tightly bound in water'. Does this mean 'Electrons are the most tightly bound  when they are within an H2O molecule'?

@Naj Of the three molecules, electrons are mostly tighly bound in water, which is why oxygen is the "terminal electron acceptor" and the final product is water.

@KarstenTheis made it clear by using E0 values of each half reaction. Here I describe what they actually mean by using the phrase "high energy electrons" in biochemistry texts.
Consider Ethyne and Ethane.

Therefore Ethynide has a high energy electron compared to Ethanide
Another Example : H- , OH- , SH- . H- is stronger than OH- and OH- is stronger than HS- in the terms of basicity. We can say that H- has more active electrons compared to those 2. Therefore in the opinion of biology systems , its electron has more energy.(discussed more in the upcoming paragraphs)
The more thermodynamically unfavorable reaction gives out the more instable product.
In turn, the instable product participates in other reactions , in a fast rate.
High Energy Electrons do mean the unstable electrons present in the biochemicals. These biochemicals are unstable because they've missed aromaticity ,their previous octet configuration or became radicals etc. therefore they have high Gibbs energy.
Look at NAD+ , FAD and their reactions
NAD+ ---> NADH

FAD ---> FADH2

NAD+ loses its entire aromaticity by the reduction. This makes the reduction reaction ,to some extents, thermodynamically unfavorable for NAD+ (compared to FAD). This is why E0 of FAD is bigger than E0 of NAD+.
As I said Reduction of NAD+ is less favorable than reduction of FAD. Therefore NADH is more unstable/has high energy electrons than FADH2 and you can say that electrons in NADH are more accessible because they want to gain their aromaticity back.(not because of the radius).
So NADH is more reactant than FADH2.
EDIT: This just came to my mind. In biochemistry texts , usually they say that "covalent bond between carbon and carbon in sugars, has a lot of energy that eventually gives the energy to NADH,FADH2 and later for ATP."
This highness of energy comes from the oxidation number of carbons in sugars(e.g. glucose) . See this

Each of them eventually participate in a CO2 by-product. The oxidation number in CO2 is +4. This results in huge movement of electrons from glucose to 6 CO2 molecules, hence the term "high energy covalent bonds in glucose". In one instance i.e. reduction of NAD+ to NADH in Step 6 of glycolysis , there is a nucleophilic substitution , HPO4(-2) acting as nucleophile and interestingly H- as leaving group.
Here, the attack of Pi and the exit of H-(carrying electrons) , oxidizes the aldehyde carbon located at the end of the molecule.
Step 6
this unstable product containing high energy electrons (H-) , reacts with NAD+ at an extremely high rate.(the more unstable the reagent , the faster the reaction). {H- = H+ + 2e-)
Hydrid  acting as a leaving group , is an interesting concept. look at Cannizzaro reaction . In the Step 6 of glycolysis, Aldehyde becomes oxidized into phosphonic acid and later carboxylic acid (reducing another particle- NAD+) ,While in Cannizzaro reaction , an aldehyde becomes oxidized into a carboxylic acid ( reducing another particle- another aldehyde).


text:
What could these letters "S" in red circles mean in a biochemical diagram?
What could be the meaning of the red circles with letters S in them in the diagram below? I searched in the text but could not find.
From "Role of TREK-1 in Health and Disease, Focus on the Central Nervous System", Figure 2:

See https://pubmed.ncbi.nlm.nih.gov/16006563/

It almost certainly refers to the amino acid (residue) serine. The numbers (300, 333, 351) refer to the residue numbers. The COOH at the end signifies the C-terminus of the protein (subunit).
The black P's next to it are phosphate groups; both protein kinase A and C ("PKA" and "PKC") are involved in control of protein activity precisely via phosphorylation of serine residues in proteins (amongst others).

https://en.wikipedia.org/wiki/Protein_kinase_A
https://en.wikipedia.org/wiki/Protein_kinase_C

There are some other hints. The letters S and E are on a line that begins with NH2 and ends in COOH. There is no element E, so S is probably no element either. There is a phosphate near the S, and serines (just like threonines and tyrosines) are targets for phosphorylation.


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How come Cyclopropenylhydroxide exists and not cyclopropenol?

I'm currently studying aromatic compounds via online recorded lectures, and my teacher brought up about cyclopropenol and cyclopropenylhydroxide while teaching carbocylic non benzenoid compounds. He said the latter exists while the former doesn't exist; and due to time restrictions they cut off the part where he explains the reason, out of the video.
So why doesn't it exist?

What is "cyclopropenylhydroxy"? That's not a valid chemical name. "Cyclopropenol" could refer to one of two isomers, which is it? Do you have structures which you could draw?

@orthocresol Thank you for suggesting the changes. I'm a high school student and am pretty much new to all this stuff, so forgive me if I didn't provide all details required in order to help people answer the question properly. I've made the changes and attached an ss of my video lecture, so I hope that helps ;)

OK, no problem. Now, I'm really not sure whether what you're being taught is legit. The former seems to exist just fine at low temperatures, and was apparently reported in 1984, although I can't find any references to it since then. I would be extremely skeptical of the latter actually existing. I highly doubt that a carbocation can form a stable salt with something as basic as hydroxide ion.

@orthocresol the teacher said that the former dissassociates into the latter cause of the aromatic character [ Unusual stability ] since the latter follows all the 4 rules for aromaticity [ cyclic, huckel's rule, planar and conjugated pi electrons ]. He added that the former doesn't follow the huckel's rule which I'm finding trouble understanding since I couldn't get the difference in no. of pi electrons in both the compounds since there is only double bond [ away from the OH atom/cation ].

Aromaticity and all that is fine, but it's still a carbocation. Aromaticity doesn't suddenly make a carbocation more stable than a neutral compound. To use a different example, this is almost like suggesting that cyclopentadiene should spontaneously dissociate into cyclopentadienyl plus a proton, because the anion is aromatic. But, you probably know enough chemistry to look up the pKa of cyclopentadiene and figure out where that equilibrium lies. (Hint: the pKa is 16.) As for Huckel's rule in the alcohol, it is not relevant at all, because the pi system does not extend over the whole ring.

@orthocresol So ig the question and the basis behind it is irrelevant. Thanks ;) was pondering whether I understood the topic properly.


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Can a crystal structure of a substance appear or disappear?
I read somewhere before that some substances no longer crystallize naturally in some specific structures, although it do that readily before. Instead, new structures that never happen before become widespread.
Edit: @Buttonwood get the correct name for this phenomenon: disappearing polymorphs.
So:

Is it true?
If it is true, what is this phenomenon called?
If it is true, what is some high profile event of this phenomenon.

"I read somewhere before" gives others very little to go on.  Please give a specific references. There is nothing like data and a good references.

@Hal I read it in an answer in worldbuilding stack. I cannot find the answer again, so I think I should ask here.

I read somewhere about an army of spear-wielding tribesmen mounted on bears. Wait, I think it was on worldbuilding.SE too. Seriously, no.

@IvanNeretin It may sound far-fetched.  But in my interpretation of this vague question lacking a lot of detail, it is about disappearing polymorphs; which is in some fields a considerable bump in the road to success.

Does not the question imply behaviour of substance is time dependent, what would be breach of time translation symmetry and therefore breach of mass and energy conservation law ?

Well, life didn't exist before and now it's widespread. And if life ever arose again body plans might be different. Is this roughly what you mean?

@Buttonwood Thank you. I believe you get the right keyword for this.

The phenomenon is well known. But there are some weird explanations for it including the work of Rupert Sheldrake who considered it part of a more general idea called morphic resonance which few scientists take seriously. But his work popularised some of the evidence on disappearing polymorphs.

@Poutnik If a substance crystallising in a different polymorph today than it did last year broke the laws of physics then there is going to be a hell of a fight with chemists. It is clearly a real, repeatedly observed, and sometime economically significant phenomenon.  Why it happens is something of a mystery.

@matt_black The question is, if substances behaves differently at the same or at different conditions ? The different conditions may not be obvious as different for the observers.

@Poutnik The problem has occured in drug manufacture where conditions are usually tightly controlled. What is different is clearly not at all obvious and isn't just some easily observable temperature or solvent change (where getting a different structure is unsurprising). Hence why it is a mystery.

@matt_black OR, the substance behaves randomly in the same way now as before, due quantum effects, so the time translation symmety is saved. :-D Also the system settings would have random deviations. Or different vendor and quality of substances can cause such an effect.

In short: Yes.
But it is not possible to offer an exhaustive answer to all aspects intervening here. But scratching barely the surface of the large topics of crystallography / materials science / pharmacokinetics let's try this one:

Substances may solidify in a regular pattern, i.e., from a given point within the bulk, you may then predict both local structure in proximity (a few Angstroms, $ \pu{1 Å} = \pu{10^{-10}m}$) as well as in far distance (e.g., for a macroscopic crystal).  This regular pattern may be described by a model, the crystal structure.

When solidifying, atoms, ions, molecules intend to do so minimizing the energy.  Especially for flexible molecules (think e.g., about alkyl chains, phenyl rings), this may be tricky.  Maybe you recall the Newman projections of simple, flexible alkanes

(credit)
where the energy about an individual molecule is plot: often, there are multiple  energetic minima, but only one global energetic minimum.  For the solidification of a material, the situation is even more complex, because minimization of the energetic minimum comprises both the molecule in question, as well as its interactions with nearest molecules around.  If the rate of crystallization is too high (e.g., just crushing out a compound as a precipitate), it may happen that the molecules in the solid state again found an energetic minimum, but not not the global one.  Like maybe an alkyl group still is in gauch orientation to an other.

The possibility that the same molecules arrange in space to yield different pattern is called polymorphism.  It is not something limited to organic molecules, though; it is equally observed for minerals ($\ce{TiO2}$ as rutile or anatase) and elements (e.g., sulfur) and obtaining one, or an other polymorph (to be read as «particular arrangement of molecules in the solid state») may depend on parameters like temperature, pressure, concentration of other substances during the crystallization.

There are attempts to predict how molecules crystallize (e.g., IUCr's contests about small molecules example, a new one is in preparation; which is not the same as CASP14 about protein folding like AlphaFold2), but the reliability of these computational predictions depend on multiple parameters.

There are multiple properties which may alter greatly by a different arrangement of the same molecules in the solid state, in different polymorphs.  For solid pharmaceuticals, for example, the rate these active ingredients dissolve by ingestion, and thus enter the anabolism of the patient may vary a lot.  Companies are very interested to obtain not only the same, chemically pure compounds for the pills, but equally to obtain again and again the same polymorph with the tested properties. Equally because in some jurisdictions both a compound, as well as the polymorph may be protected by a patent (an entry into the topic).

Finally, yes it already happened to large businesses that they were unable to crystallize the molecules in the intended polymorph, a.k.a. disappearing polymorphs.  As you may imagine, consequences for for the company and for customers / patients may be significant.  PubMed allows you to read a survey publication about the topic without a paywall.

This is really just a reflection of our inability to completely control and accurately reproduce experimental conditions.


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Conditions for spontaneity of redox reaction among redox couples with equal standard reduction potentials
A question from the university course:

Could a redox reaction spontaneously occur among the redox couples $\ce{A_{ox}}/\ce{A_{red}}$ and $\ce{B_{ox}}/\ce{B_{red}}$ within non-standard circumstances if they have the exact same standard reduction potentials? So then, under which conditions?

The lecturer used the Nernst equation as follows for this question:
$$\Delta E = \Delta E^{\circ\prime} + \frac{RT}{zF} \ln{\frac{[\ce{A_{ox}}]}{[\ce{A_{red}}]}} - \frac{RT}{zF} \ln{\frac{[\ce{B_{ox}}]}{\ce{[B_{red}}]}}$$
From my understanding he is calculating the energy that is needed to run the reaction spontaneously, but I don't understand how he came to this equation.
If the reduction potentials of two half-reactions are the same, is it still possible for them to run spontaneously under certain conditions?

Yes, it is possible to construct two half cells with the same standard reduction potentials. For example, we can half cells made of copper dipping in copper sulfate solution. The caveat is that if both the cells are under standard conditions, the resulting electrode potential would be zero.
The trick is to use different concentration of copper ions in each half cell. With the help of Nernst equation you can determine which side will be the anode and which side will be the cathode. Read more about "concentration cells" as they are called.


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stereochem of epoxide opening
Here I have an epoxide opening reaction, but I was wondering that when a nucleophile attacks, How would the stereochemistry work? I have 2 products I predicted and I'm not sure which one would be correct. It's either the first one predicted in the reaction of the second picture I have attached. Or are they actually the same thing?

Where did you get the second one from?

@orthocresol It's just another prediction I had, so I wasn't sure which one is right

Yes, and why did you predict that? For example in your first one, you have a (very good!) mechanism which explains why you’re coming up with that particular stereochemistry.

@orthocresol I predicted that because I thought that with inversion, the bond would get flipped to the other side. So when you have a SN2 would only the dash and the wedge exchange if the leaving group is on the dash/wedge?

Hmm, yes, you could put it that way. If you're not fully convinced, my suggestion would be to make a model of it. That's the ultimate source of truth.

A few points to ponder. The stereochemistry of the epoxide implies a single enantiomer. If the reaction is indeed SN2, then there is inversion of stereochemistry. If you know how to assign R/S configurations, compare your two bromohydrin structures. Are they the same? What if the reaction is SN1? Lastly, what is the significance of the bold wedge in the bromohydrins?

In these three-membered ring reactions, it's an SN2 substitution reaction that happens. Hence, there is an inversion of the configuration associated.
Edit: Perhaps the mechanism is more complicated than SN2, contrary to Klein's book which states that it's SN2, this online source says that the actual mechanism is between sn1 and SN2 (leaning to SN2 side) to explain the preference of nucleophile to attack tertiary carbon though more sterically hindered.

Reference: See page-431 of Organic Chemistry David-R-Klein

Well, it could have some SN1, given the relative stability of a benzylic cation. I think that is what user55119's comment hints at. I'm not sure about the relative extents.

@orthocresol: I was not pushing for an SN1 mechanism but rather for the OP to ponder the consequences of an SN1 mechanism vs. an SN2 route.

Why the down vote? If there is any mistake in my answer I'd be more than happy to correct it.


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Confusion over the rate law
I am really confused about the rate law. From my understanding, the rate law gives an equation that tells us the rate of the disappearance of the reactants. So, for example, the rate law of the reaction $\ce{2 NO + O2 -> 2 NO2}$ is as follows
$$\text{rate of disappearance of reactants} = k \, [\ce{NO}]^2 [\ce{O2}]$$
My biggest confusion stems from the fact that the rate law gives one uniform rate for the disappearance of all the reactants, which makes no sense to me. For every $\ce{O2}$ used, two $\ce{NO}$ are used, so how can they be disappearing at the same rate?
Hopefully, it makes sense where I am coming from, and I would appreciate any help trying to clarify this for me.

Rodrigo de Azevedo's answer can be stated with words by saying that the rate of the reaction is the rate of disappearance of $1$ mole $\ce{O2},$ and also the rate of disappearance of $2$ moles $\ce{NO}$. That is why the coefficient $2$ appears in the law written for $\ce{NO}:$
$$\frac{d}{dt}[\ce{NO}] = - 2k[\ce{NO}]^2[\ce{O2}]$$

The general case for a reaction $aA+bB=cC$ is
$$r=-\frac{1}{a}\frac{d[A]}{dt}=\cdots=+\frac{1}{c}\frac{d[C]}{dt}=k[A]^\alpha[B]^\beta[C]^\gamma$$
where $n=\alpha+\beta+\gamma$ is the total order and $\alpha,\;\beta $ etc are constants that can be positive, negative or fractional and are the orders of A, B etc. The rate constant has units (concentration)$^{1-n}$/time. This type of rate equation tells us nothing about the actual mechanism of the reaction.

$$\begin{aligned} \frac{{\rm d}}{{\rm d}t} [\ce{NO}] &= - \color{magenta}{2} k \, [\ce{NO}]^2 [\ce{O2}]\\\\ \frac{{\rm d}}{{\rm d}t} [\ce{O2}] &= - k \, [\ce{NO}]^2 [\ce{O2}] \end{aligned}$$

The rate is determined by the rate limiting reagent, in this case, the oxygen species will dictate the rate. The consumption of the nitrous oxide is mediated by the reaction of the oxygen, therefore, both reagents will dissipate at the rate that oxygen is consumed.

The statement as by Rodrigo de Azevedo, demonstrated the proportion of this consumption as it pertains to the rate constant for some K.

This answer should have a short description of the equations. Maurice's answer did that very well, so why not this one.

@Antimon Because I wrote before Maurice and assumed that the OP knows enough calculus to arrive at a conclusion.

Here is a better definition: The rate law of a reaction is the quantity which when scaled by the stoichiometric coefficient of reactant/ product (with sign) gives the rate at which that particular quantity varies with time.
Or, it's the rate of a chemical substance's concentration changing normalized by the number of its moles in the chemical reaction in which it is participating.


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SMARTS string matching an aldehyde or a ketone where the target alpha carbon is less substituted than the other carbon
I need a SMARTS string that matches the following critereon:

It makes an aldehyde or a ketone only
The target carbon is an alpha carbon
The other alpha carbon must be more substituted (or be an aldehyde)
The target carbon must not be more substited than a secondary carbon

I came up with the following string:
[C;X4;$([H3,H2;$(CC(=O)[H])]),$([H3;$(CC(=O)C)]),$([H2;$(CC(=O)[C;!H3])])]

Translated, this is saying, the target carbon:

Has four bonds
Lives in any one of the following environments:

Is has two or three hydrogens and is on an aldehyde
Has three hydrogens and is on a ketone
Has two hydrogens and is on a ketone where the other alpha carbon does not have three hydrogens

But that seems very long for what seems like a simple critereon.
Any suggestions? Are there any cool SMARTS tips I'm missing?
Thanks,

You don't say whether the target carbon must be aliphatic, but your SMARTS string assumes that it does.  (You are using C instead of [#6] to specify carbon atoms.).  I suppose "not more substituted than a secondary carbon" implies that it does?

Also, I don't think it seems long.  It is not a simple criterion.  It's fairly complicated, and if I came up with that SMARTS pattern, I'd feel pretty proud of myself.

You might want to check out SMARTS viewer which can help with visualizing and debugging SMARTS. I also usually try the patterns on a few data sets and make sure it looks like a good match. (BTW, it's not that long for what you describe.)


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PLP-Dependent Cyclization and Decarboxylation
I am looking at Compound A (see below) and am trying to figure out a decarboxylation mechanism that results in Compound B (see below). I have come up with an internal mechanism that might work that I have attached to this question. However, I am trying to connect it to enzyme catalysis, and I am thinking that a PLP-dependent enzyme is involved I just cannot quite figure out the mechanism.

A PLP-dependent enzyme could certainly accomplish a reaction like this. An example of something similar is the PLP-dependent enzyme KAPA synthase in the biotin biosynthesis pathway. In that reaction, alanine is decarboxylated and a new bond is formed between the alpha carbon and an electrophilic thioester. Interestingly, the proposed mechanism involves deprotonation of the alpha carbon rather than decarboxylation as the initial step. Only after the new C-C bond is formed is the carboxylate lost:

1: Source: Webster, S.P, et al. (1998) Characterisation of S-amino-7-oxononanoate synthase: A bacterial PLP-dependent, acyl CoA condensing enzyme. Biochemical Society Transactions 26(3):S268. DOI: 10.1042/bst026s268
Although this is an intermolecular reaction, your proposed intramolecular reaction should be even easier to accomplish. An example of a PLP-dependent enzyme catalyzed reaction that results in intramolecular cyclization (but not decarboxylation) is that of aminocyclopropane carboxylate synthase.

One thing you do not want to propose is an unstabilized carbanion, either enzymatically or non-enzymatically,  as the nucleophile in this $\ce{S_N2^'}$ cyclization. @Andrew has demonstrated how pyridoxal phosphate (PLP) can provide stabilized anionic character at the desired carbon site followed by acylation or, in this instance, cyclization with subsequent decarboxylation. Upon examining your structures, I suspected that this reaction may have been part of a biosynthetic study on the formation of ergot alkaloids.
Kozikowski, et. al.[1] synthesized racemic amino acids 1a and 1b to determine if one of the enantiomers was capable of incorporation into the ergot skeleton via the agency of Claviceps sp. strain SD58. A doped sample of 1a/1b produced elymoclavine 5 that displayed molecular ions for M and M+3. The presumed pathway involves dehydration of 1 to diene 2, which is ostensibly oxidized to vinyl epoxide 3 with cytochrome P-450. Decarboxylative cyclization proceeds to give (E)-allylic alcohol 4, chanoclavine-I. The (Z)-allylic alcohol of 4, isochanoclavine-I, has been proposed[2] to form by isomerization of (E)-4 or may form directly from vinyl epoxide depending on the conformation of the epoxide during the cyclization. These latter two points are significant because it is known[2] that the carbon of the hydroxymethyl group of 4 becomes $\ce{C7}$ in elymoclavine 5.

A. P. Kozikowski and J-P. Wu, M. Shibuya and H. G. Floss, J. Am. Chem. Soc., 1988, 110, 1970-1971. https://doi.org/10.1021/ja00214a054
H. G. Floss, Tetrahedron, 1976, 32, 873-912. https://doi.org/10.1016/0040-4020(76)85047-8


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Why hydrogen bonds are stronger than Van der Waals forces?
Why the hydrogen bonds are more intense than Van der Waals forces in the case where the molecules with the Van der Waals forces have a stronger dipolar moment than the molecules with the hydrogen bond?
For example, ethanol (hydrogen bond) has a boiling point of 78 ºC and ethanal 20 ºC (Van der Waals forces) but their dipole moments are respectively 1,69 D and 2,69 D.
I thought the intensity of the hydrogen bond was due to the high polarization of the bond between a small and very electronegative atom like N, O, F, but this is not the case in the previous example, as I understand.

So, you're asking how dipole moment works, or about boiling point? I think title doesn't reflect the main point.

I have reformulated the question. Hope in this way is clearer.

"Dipole moment" not "momentum". It's a fair question.

You are mixing different interactions, or the other way around things are mixed and coexisting. 1) VdW forces, or whatever goes under the umbrella of dispersive forces, do not require any permanent electrical dipole over the whole molecule nor over part of it; 2) H bond involves a dipole along the X-H and another electronegative atom, dipoles along the bonds. In principle you can imagine a molecule with two such occurrences and no el dipole mom. 3) shape and size of a molecule are involved in the overall strength of local dispersive forces. This point influences also the B.P.,togheter with Mw

See London dispersion, Debye and Keesom forces, that together are covered by the broader term van der Waal's force, sometimes with Casimir effect. OTOH, the hydrogen bond is the true chemical bond.

Distance
In vacuum, the calculated (classical) interaction between to aligned dipoles decreases with the square of the distance. If the dipoles can't come very close to each other, the interaction will be weak. If you treat hydrogen bonds as a special case of dipole-dipole interaction, you will find that a N-H ... O=C interaction is much stronger than a C-H ... O=C interaction. For one, the N-H group has a larger dipole moment than the C-H group. For another, the distance between the donor and acceptor are larger for the weak C-H hydrogen bond.
When the dipoles of two ethanol molecules approach, they can come fairly close because the hydrogen on the hydroxyl group lacks inner electrons. For ethanal, in contrast, the carbonyl carbon of one molecule has to come close to the oxygen atom of the other. Both atoms have filled inner shells (and more electrons in the outer shell), so that approach is limited by electronic repulsion.
Covalent character
The hydrogen bond has covalent character, explaining why it is a directional interaction. This covalent character (like all covalent bonds) can not be fully explained using classical (Coulomb interaction) arguments. So in a hydrogen bond, we have an additional attractive term that is lacking in pure dipole-dipole interactions. In extreme cases, you might think of the hydrogen bond as a 4 electron-3 center bond (the 4 electrons are the lone pair of the acceptor and the bonding electrons of the covalent bond of hydrogen with the rest of the molecule; the three atoms are the hydrogen, the electronegative atom it is covalently bound to, and the acceptor atom).
Nomenclature
Some use Van der Waals forces as an umbrella term for (permanent) dipole-dipole, dipole-induced dipole, and temporary dipole-induced dipole interactions. Others use Van der Waals forces (or dispersion forces or London interactions) as a separate and weaker kind of interaction and contrast it with the stronger dipole-dipole interaction.

When a $\ce{H}$ atom is bound to an electronegative atom like oxygen, its only electron is mostly placed between the two nuclei. Outside of this $\ce{H}$ atom, the $\ce{H}$ nucleus is nearly "naked". Speaking naively, there is no other electron to occupy the space around the nucleus towards outside. It looks as if the nucleus $\ce{H}$ is not protected any more against approaching negative charges. On the contrary it can attract external negative charges, like the outer doublets of the oxygen atom from another molecule. This attraction is purely electrostatic, and it is strong, much stronger than Van der Wals forces, which are the result of attracting and repulsing forces.
This happens in water and ethanol.
In ethanal $\ce{CH3CHO}$, the $\ce{H}$ atom is bound to an atom (carbon) which has nearly the same electronegativity. The electron of this $\ce{H}$ atom is not moved so strongly between the two atoms $\ce{C}$ and $\ce{H}$. So this electron protects the nucleus from approaching negative charges. The H-bond in ethanal is quite weak.

But @Maurice ethanal has a high dipole moment due to the C-O bond that is highly polariced (C and O atoms have very different electronegativity). In both molecules there are C-H weak bonds that do not do not contribute to the total moment of the molecule (almost).

@fich Yes the dipole moment C=O is important. It has no effect on the possibility of H from CHO to form a H-Bond. The H atom is protected by its electron which is not "only" in the region between C and H.

Yes, I understand that in ethanal there is no highly polarice -H bond. But how do you explain then that the intermolecular forces in ethanal are weaker than in ethanol being ethanal a more polarized molecule than ethanol (as we can see by the dipolar moments)?

@fich Dipolar forces are not important with respect to pure Coulomb-type electrostatic forces. In dipole interactions, there is always an attraction AND a repulsion between charges. In electrostatic forces like in H-bonds there is only a plus-minus attraction between H nucleus and electron from next Oxygen atoms. This is much stronger than dipole - dipole interactions.

Thanks for the responses and effort. I have selected Karsten response because I did not understand why do you treated H-bonds as if they were not polarized (they are strong dipoles in fact). The key point is distance between dipoles, as Kasrten stated. Maybe you wanted to explain this.

Dipole moments are considered only in the absence of intermolecular hydrogen bonding.
Hydrogen bonding is much stronger than other dipole-dipole interactions (Van der Waals forces of attraction). Then energy liberated by dipole-dipole interactions pales in comparison to that liberated by hydrogen bonding. The stronger dipole moment argument is used where neither of the molecules that we are comparing has any form of hydrogen bonding (inter or intra).
Hope it helps!

Thanks for the response. But I want to know why happens what you explain. The energy of the molecules with hydrogen bond is not due to the electromagnetic interaction between their highly polar molecules?

Dipole-dipole interactions occur between any two atoms having a partially positive and a partially negative charge. On the other hydrogen bonding is a specific subset of this where the atom having a partial positive charge is a hydrogen atom bound to a highly electronegative atom and the atom having a partial negative charge is another highly electronegative atom, which greatly increases the polarity of the bond. Greater the polarity of the bond, stronger is the bond.

This is the main reason why hydrogen bonding is mostly seen in molecules that contain hydrogens bonded to N, O or F and N, O or F. The high polarity means that hydrogen bonds are very strong (at more than 10% strength of a typical covalent bond length) and hence we use the terms 'bonds' and not 'interactions' for them.

@Aldiboy But as you say, the high polarity implies a very strong bond in the hydrogen bond, but in the example I wrote, the polarity is greater in ethanal than ethanol and only the last has hydrogen bond and therfore greater boiling point. This is the point I do not get.


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Oxidation number approximately
I argued with my lecturer when he is saying that during the calculation of oxidation number, and your answer give you like 3.46 you can approximate it to whole number, I know tell him NO that if the answer is in decimal that means your calculations is wrong, he now tell me to go and research on my thought.
So please can anyone help me out

related: https://chemistry.stackexchange.com/questions/5357/are-fractional-oxidation-states-possible

Thank you I ever find that question before

For complex or non stoichiometric ( many metal oxides ) compounds, it makes sense to calculate the average oxidation number that can be a rational number.
This can be useful if we are to enumerate chemical reactions with non-stoichiometric metal oxides, or if just a summary elemental composition is known, like hydrocarbon mixtures.
Oxidation numbers of particular atoms are integer numbers, with exceptions when   extra electrons ( or theirs deficit ) are shared among different number of atoms. The particular example of this are superoxides $\ce{M^{I}O2}$.


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Does using same calibration curve for chemical analysis introduce correlation in sample error/uncertainty values?
I am using a hypothetical case to explain my question
Let us consider a chemical analysis method A (using a certain instrument), which is used to quantify the value of a particular species b. According the instrument manufacturer's notes, the instrument used needs to be calibrated once a month. I am analyzing several samples of b in the duration of a month, and using the same calibration curve to obtain the quantity of b along with the uncertainty bound for the value.
I want to understand the following,

If number of samples (over the period of a month) of b are using the same calibration curve to quantify it value, are their errors/uncertainties correlated (by any means, the calibration curve introduces it, however small it may be) ?
Additionally, method A has the ability to quantify several species (b, c, d,$\dots $), does errors/uncertainty of species from the same sample (containing different species like b, c,$\dots$) correlated in any way?
If there exists even a very small correlation, does it make sense to consider it for any further analysis?

I am new to the realm of chemistry (analytical chemistry to be precise), I request your help.
Thank you.
PS: Apologies if the tags are inaccurate

1.) Of course, 2.) very likely, but you need to be a lot more specific about A, 3.) how would you do that? You need to have a superior analytical method to only find that correlation.

You should distinguish between errors and uncertainties. If your calibration is somehow systematically wrong (an error has been introduced while determining the relation between observable and expected values) then all values subsequently derived from that calibration will suffer from the same systematic error. This does not mean that the uncertainties due to noise are otherwise correlated.

Please note that calibrating an instrument once a month is another story as compared to constructing the calibration curve from that instrument. The former is meant to check the accuracy of the instrument's output and the latter serves to determine the concentration of a substance by noting the response of the instrument to 6-7 standards.
Also note that all instrumental analysis techniques are relative with very very few exceptions (gravimetry, microwave spectroscopy, coulometry which are absolute etc.). The rest all need standards for comparison.
In analytical chemistry, it will be considered a bad practice to make a calibration curve one day and use it for the entire month. The good practice is to make a fresh calibration curve for every batch of samples and there should be a "reference" check which keeps on validating the calibration curve after a certain number of samples have been analyzed.
The main reason is that most instruments will show what is called as the drift i.e., a low frequency and hence long term noise in the baseline. The baseline response can rise and decline slowly.
Also the word correlation is perhaps not suitable in your situation. Correlation has a very specific meaning in signal processing/instrumental analysis. What you imply is the influence of interferences and something called as the matric effect. Suppose, you have b, along with c,d,e,...,f. Yes, the presence of the additional components can indeed influence the signal of b. In most cases, the signal of b is decreased causing a lower result than expected.
Beyond your hypothetical situation, one can say nothing. Which instrument you had mind?


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Can molecule have just h-bonds between its parts
I'm just starting to study chemistry so sorry for the probably obvious question to you, guys.
According to the definition of molecule in Wikipedia:

A molecule is an electrically neutral group of two or more atoms held
together by chemical bonds.

According to that definition, I assume that parts of the molecule could be theoretically connected just with the h-bonds. As far as I understand the example of that case is DNA two strands of which are connected by only h-bonds (though many and many ones).
According to Khan Academy:

When atoms combine by forming covalent bonds, the resulting collection of atoms is called a molecule.

And now, as a newbie, I'm a little bit confused with all of that. So, could you help me with the following questions:

Is that true that two strands of DNA do not have covalent bonds to each other and connected only with h-bonds?
If the p. 1 is true could parts of molecule be connected only by h-bonds by definition?
If the p. 3 is true then why DNA is a molecule but two water compounds (H2-O***H-O-H) are not?

In short: (1) Yes. (2) Depends on the definition. (3) Because there are pretty many of them in one DNA strand.

@IvanNeretin, then what definition is correct and/or official?
Are there any formal parameters to split between the case of two water compounds and DNA strands? Maybe, the number of h-bonds or some kind of h-bonds total strength or something like this?

The definition you quoted is official enough to me. But it is not complete. You may look for the complete one; you may even found some, but they won't be complete either. There are always borderline cases. You are looking for strict formal criteria where there are none.

@ademchenko 1)True   2)NO. DNA is a polymeric macro molecule, It is consisted of 2 strands connected by Hydrogen bonds. Each Strand is also a polymeric macro molecule. therefore the term "molecule" results from covalent bonds of phosphodiester in each strand. I suppose you're not familiar with proteins 4 types of structure    see : https://en.wikipedia.org/wiki/Protein_structure. The hydrogen bonds are created in 2nd, 3rd , 4th structure of a protein but 1st structure ALWAYS contains ONLY covalent bonds.Almost the same thing goes with DNA. 3) Because H2O**H2O is not polymer as DNA is.

@Sam, I'm sorry about that, I'm really confused because of my lack of knowledge. But do you mean that polymeric molecule is not a molecule by definition? I know what the polymeric molecule is but I thought it is a type of molecule. So, I thought the term "molecule" suits "polymeric molecule" as well.
So, according to what you have said could we say that: the only type of molecule which parts can be connected by exclusively h-bonds (no covalent bonds are presented) is the polymeric one?

@ademchenko yes. Still I've heard in inorganic chemistry there are some exceptions . In any of these exceptions at least one covalent bond is present. HF usually appears as multimers with hydrogen bonds. BH3 usually exists as BH3 - BH3 . BH3 molecule does not exist by itself . (this is not hydrogen bond though). PLUS the definition of hydrogen bond  always requires 2 polar covalent bonds having H. therefore you can not have hydrogen bonds in the absence of covalent bonds in any way.

WIKIPEDIA : A hydrogen bond (often informally abbreviated H-bond) is a primarily electrostatic force of attraction between a hydrogen (H) atom which is COVALENTLY BOUND to a more electronegative atom or group, particularly the second-row elements nitrogen (N), oxygen (O), or fluorine (F)—the hydrogen bond donor (Dn)—and another electronegative atom bearing a lone pair of electrons—the hydrogen bond acceptor (Ac)

Sam, IvanNeretin, thank you, guys, for your answers!

Just to smooth thingd, as you likely aren't puzzled anymore. If you don't consider H-bond as a truly chemical bond, then no molecule can be held by only H-bonds. Conversely, there can be so many of them, that their breaking becomes difficult That would be a special case. For most of the case and depending on the function under discussion you can refer to DNA as one molecule of it. But this won't pose any problem to those seriously discussing molecules, DNA, proteins or whatever ensemble having a chemical physical identity.

There are some grey areas in the normal usage of the word 'molecule'
The normal usage of the term molecule is accurately described by your quote from the Khan Academy. The majority of molecules chemists talk about are groups of atoms joined with covalent bonds. But, like many things in chemistry, there are some grey areas and the normal usage gets bent a little.
One of the major areas where the terminology gets fuzzy is in polymers. While on a strict interpretation of the normal definition a single polymer strand (be it DNA or a single strand of polyethylene) is a molecule the terminology is not particularly helpful as the key things that matter in bulk polymer properties are not easily reduced to the properties of individual 'molecules' so the very idea of molecules is less relevant. In polyethylene (a very simple polymer) there will not be a mix of identical molecules but, rather, a wide range of very similar molecules (even for linear polyethylene there will be a wide range of chain lengths (say from 1,000 to 10,000 carbons). So polymer chemists tend not to talk about molecules much.
DNA is a polymer but, by its nature, one where the structure of a given strand is constant (and can be precisely replicated). We do tend to say DNA molecule at least in casual conversation despite the fact that the two strands are connected only by hydrogen bonds. But there are good reasons why this is not an abuse of the normal definition of molecule. One is that many hydrogen bonds along a polymeric chain add up to a strong overall bond even though the individual bonds are merely hydrogen bonds. But the relative weakness of each bond is a key part of the process of replicating DNA as the two strands can be easily unzipped by the machinery in the cell nucleus. You could argue that a DNA strand consists of two molecules using the normal definition but this doesn't help much and, in normal discussion, we talk about both a DNA molecule with two hydrogen-bonded strands and a DNA strand as a molecule. This mild abuse of a strict definition of "molecule" doesn't matter much as it doesn't interfere with any understanding of how DNA works.
Mostly, though, chemists don't consider hydrogen-bonded units as "molecules" even when the bonds have significant effects on the bulk properties of the substance. The physical properties of, for example, acetic acid are heavily influenced by hydrogen bonding. In solution molecules often form pairs (joined by two hydrogen bonds) and this leads to a big elevation in their expected boiling point compared to a similar hypothetical molecule of the same size that couldn't do hydrogen bonds. But there is no benefit in describing this a "molecule" and nobody does this.
But the main point is that many properties in chemistry have an almost continuous range with no clear point where a black and white distinction can be made. The definition of what counts as a molecule is like this. Describing hydrogen bonded DNA as a molecule isn't a terrible use of the terminology; describing an acetic acid hydrogen-bonded pair as a molecule probably is. The term molecule can be useful in some circumstances but not others. The utility of the term is what matters not a simple black or white definition.


text:
Red Compound from Glycine and Glycerol?
I got an unexpected reaction which I'm trying to explain to myself:
I mixed 10 parts glycerol with 1 part glycine at ~90C and to my great surprise the mixture turned from transparent to a tint of red. I had maybe 1-2% water in there too. There might have been a very slight evolution of gas but that most probably was just the water boiling out.
I googled to try to find an answer to no avail. Does anyone know what could have happened?

No idea nor I want to start a search but this could be a starting point https://www.semanticscholar.org/paper/Maillard-reaction-of-sugars-with-glycine-in-system-Wijayanti/6f1dfa5424284c569abc94c5e8e723405cf2ed53?p2df

@Alchimista: I'll read into it. Thank you!!

Your temperature might be too low, but a little colour can go a long way and acrolein can form from glycerol, but usually catalysts are involved. An imine or enamine between glycine and an acrolein intermediate might not be impossible


text:
Molar Mass & Pressure-Volume Relationship
How do differences in molar mass affect the pressure-volume relationship between four different gases?

Always think  thoroughly about possible answers to your question and search for them in textbooks/ this site and generally on internet before asking. This would prevent redundant information provided in eventual answers.

Under ideal conditions smaller molecules will move faster and have more collisions which will increase pressure and volume. If pressure increases, the molecules have less free space to move and volumes may increase slightly beyond what the ideal gas law predicts. If volume decreases and molecules have less space to move, intermolecular forces can come into play and decrease the force with which collisions occur which will cause the pressure to be less than ideal gas law predictions. Larger masses, larger force of collision.
A lot of circumstances to consider.


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Can Steel be Created from Germanium?
I've tried searching with Google as well as on this website, but I haven't been able to find even a question asking about it. It was something that I was just curious about, but my limited knowledge in Chemistry has prevented me from testing anything or realizing how to phrase my question in a way that an answer can be found.
All in all, can steel be made also from the other elements in the carbon group, such as silicon or germanium?
And would the steel be stronger or weaker if it can be made?

Germanium is not in the same group as Iron. Neither ruthenium nor osmium has the same phase diagram with carbon as iron does.

No ! Steel is made of iron, plus some other elements

Well, hypothetically, if we would break heavier germanium nuclei to nuclei of light elements e.g. by high energy proton bombardment, and then if we  fused these light nuclei in particle accelerators to iron nuclei, we could make steel from germanium by some carbon addition. The problem is, current technology would be able to provide just few iron atoms this way.

Since this question has been revived, a simple answer is that even carbon is slightly too large for the interstitial voids in iron which it occupies in steels. Germanium is larger even than iron and would require much greater distortion of the iron lattice in order to fit between the iron atoms.

No.
Steel is a generic term referring to an alloy containing iron and carbon (with iron as the main component) and various types/qualities of steel contain many other additional elements, e.g. nickel, chrome, manganese, etc. to give or enhance different properties in the alloy.
If it is not iron-based, it is not steel.
As for alloys using metals from the same group as iron, firstly, germanium is not part of that group.
All of the elements that do form part of that group (Fe, Co, Ni, Ru, Rh, Pd, Os, Ir, Pt) can be and are used in alloys, (although carbon addition does not have the same influence on their properties), osmium not so much as it is a pain to store or process.  Some of those alloys are indeed "stronger than steel", for some definitions of stronger, but most of these elements are a lot less abundant and a lot more expensive than iron.
No one is going to use platinum, palladium, rhodium, etc. to make rebar or railway lines.
Edit:
Germanium, as pointed out by Martin below, is however part of the same group as carbon.  It and the other elements in that group can also be used in alloys, like carbon, but their properties are different (carbon is non-metallic, silicon and germanium are mettaloids and the heavier elements are metals) and they do not have the same effect on the properties of iron in an alloy.  Germanium, when added to tin alloys, however, has been shown to significantly increase the hardness of the alloy, and as discussed below, it helps tarnish and firescale resistance in silver. One also needs to remember that while silicon and carbon are abundant, germanium is not, and as it is critical in many high tech applications, what is available is not likely to be squandered on experiments for improving steel.
Now, I am going to go away and think about simple, not too technical ways of explaining why, it works this way or finding references that will do it better and will further edit this answer appropriately later.

And if you thought cable theft is bad, imagine how quickly thieves would make off with your railway lines if they did use platinum in the alloy ...

Platinum-osmium alloys are a thing (very hard), but certainly aren't steel.

@Jon Custer Yup. But seriously. Storing and processing osmium is a b ... bear. Plus you have to practically replace the walls every few months and you don't wanna know what the filters for the respirators cost for the personnel working with it. Ask anyone who had the joy of working in a precious metals refinery.

for sure, the whole OsO4 sublimating at room temperature and being teratogenic is pretty off-putting...

Indeed. Most refineries tend to just store it under lower temperature, because there really isn't that much of a market and safely disposing of it is a whole other can of worms. And the walls still slowly turn green regardles of the refridgeration ...

@Ed V But it isn't steel and isn't stronger than steel (and germanium still isn't part of the same group as iron) so I don't think it's really relevant. The OP wasn't asking if germanium could be used in alloys with other metals, they specifically wanted to know if "it and other elements in that group" could be used in place of iron to make steel.

The question (admittedly unclearly phrased) makes sense if you realise that germanium is in the same group as carbon, not in the same group as iron.

@Martin Kochanski ugh true. Yeesh. I suppose because the concept steel is linked to iron so firmly in our minds no one even considered that might have been what the OP was after. I'd better go edit that answer. Although the why in this case will be a pain to try explaining when the OP claims to not know much chemistry ...

Silicon, the element in Grouo 14 directly below carbon, is widely used in steels, especially in electrical steels and more recently in high-strength automotive steels.
Silicon does not replace carbon; it's a separate alloying element just as chromium or manganese would be. Although silicon is in the same group as carbon, the difference in atomic size means it does not fit interstially into the fcc iron lattice (where carbon has significant solubility when hot) or form the hard phases that carbon does upon cooling for strength. Instead, like most metals used in steel, it substitutes for iron atoms in the lattice. But it can impact the surface texture grain structure that impacts electrical steel applications, as well as the phase structure created in the presence of the carbon for the high-strength applications.
Heavier Group 14 elements are less commonly alloyed into steel. Germanuum is expensive, and tin and lead are often undesirable because the formation of low-melting liquids and impact hot mill processing or welding (hot shortness). However, they may be used in specialty applications such as lwad in free-machining steels.


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Why do we not using concentrated HCl and zinc reaction to prepare hydrogen?
First apologize for this kind silly question.One of my student ask me this question but I was unable to get a perfect answer from internet.Generally dilute HCl and impure zinc are using for hydrogen production in laboratory.My question is why  do reaction stoped after some little time reaction if we use pure zinc and concentrated HCl?
Is it for coverup zinc by zinc sulphate?
It will be thankful for any kind help.

Your title has little to do with your question. Make up your mind. We don't use concentrated HCl because the reaction would be too vigorous. As for why the reaction stops: because we make it stop.

Sorry, what? zinc sulphide ?!? Where is that coming from?

Sorry zinc sulphate

Thank you Ivan can you explain vigorous means?

As in, heats up and ignites the hydrogen - causing big exposions

And tries to get out of the reaction vessel

Hold on, why would zinc sulfate build up in a system of "pure zinc and concentrated HCl"?

Up vote: This is actually a very interesting question, NOT well addressed per my research, that involves, I suspect, some good understanding in various areas including electrochemistry and I even have a reference from a 2005 article in Physical Chemistry Chemical Physics Journal. Or, one can be content with Maurice's 'answer', which appears to be popular.

If metallic zinc $\ce{Zn}$ is in contact with concentrated $\ce{HCl}$ solution, $\ce{H2}$ is produced, but the reaction is exothermic  : the solution gets hot. As $\ce{HCl}$ is not so highly soluble in hot water, a fraction of the dissolved  $\ce{HCl}$ will be vaporized. So the gas produced will be a mixture of $\ce{H2}$ and $\ce{HCl}$, which is not wanted.
Anyhow, the reaction between pure $\ce{Zn}$ and pure concentrated $\ce{HCl}$ is not really rapid. And, after some time, when the solution contains a certain amount of $\ce{Zn^{2+}}$, the reaction rate decreases in such a manner that the reaction looks finished. It is rather surprising, and even deceiving.
However, if some metallic impurities are present in solution, like $\ce{Co^{2+}}$ or $\ce{Cu^{2+}}$, the zinc metal reduces these ions according to $$\ce{Zn + Co^{2+} -> Co  +  Zn^{2+}}$$or $$\ce{Zn + Cu^{2+} -> Cu  +  Zn^{2+}}$$ As a result, some small amounts of metallic cobalt or copper are deposited on the zinc metal. The contact of the two metals $\ce{Zn + Co}$ or $\ce{Cu}$ produces a galvanic cell, and the reaction rate increases a lot. $\ce{Zn}$ gets dissolved much quicker, producing electrons that go onto the copper or cobalt spots, where they reduce $\ce{H+}$ to produce a lot of $\ce{H2}$.
So the best way of producing H2 by the reaction $\ce{Zn + HCl}$ is to use not too concentrated solutions of $\ce{HCl}$ solution, with small amounts of copper or cobalt salts added into the acidic solution.
The same phenomena of galvanic cell happens if  $\ce{HCl}$ is replaced by $\ce{H2SO4}$

HCl is still highly soluble in hot water, as about 20% HCl forms azeotrop with maximum boiling point approx 120 Deg C. So gas above heated diluted HCl has decreased the relative content of HCl, but it is relatively enriched for concentrated HCl.

Good anecdotal recount, however, adding a source reference is my recommendation (and then, you may be able to actually correctly account for some of the chemistry, or just read my answer).

As to why it is better not to employ concentrated Hydrochloric acid is likely due to the exothermic nature of its reaction with Zinc metal and the volatility of $\ce{HCl}$ itself fostering a possible loss of Hydrogen Chloride.
Further, the very presence of water may be beneficial (see Hydrogen formation in the reaction of Zn+(H2O)n with HCl) in allowing a Zinc ion to be moved into a $\ce{[Zn(H2O)6](2+)}$ hydration sphere which can even further interact in a ligand exchange reaction. Relatedly, to expound per Libre Text on the reaction:

If you add concentrated hydrochloric acid to a solution containing hexaaquacobalt(II) ions (for example, cobalt(II) chloride solution), the solution turns from its original pink color to a dark rich blue. The six aqua molecules are replaced by four chloro ions. The reaction taking place is reversible.

$\ce{[Co(H2O)6](2+) + 4 Cl- <=> [Co(H2O)_6Cl_4](2-) + 6 H2O }$

And further on the above reaction, to continue quoting:

Concentrated hydrochloric acid is used as the source of chloride ions because it provides a very high concentration compared to what is possible with, say, sodium chloride solution. Concentrated hydrochloric acid has a chloride ion concentration of approximately 10 mol dm-3. The high chloride ion concentration pushes the position of the equilibrium to the right according to Le Chatelier's Principle.

where I would expect a similar reaction sequence (as reported in the cited 2005 article from the Journal of Physical Chemistry Chemical Physics) with copper in place of cobalt.
Also, a cited improvement in reaction rate, relating to the use of impure zinc, is perhaps best illustrated in an related experiment where the acid is, albeit, $\ce{H2SO4}$ acting on Zinc (as a solid piece of $\ce{Zn}$ metal) in the presence of a source of a copper impurity. Note, the experiment employs three test tubes containing $\ce{Zn}$ + Acid, but one without any Copper presence, one with low surface area Copper turnings, and the 3rd with aqueous $\ce{CuSO4}$.
Surprisingly to some, it is claimed that the 3rd test tube may actually be the largest source of Hydrogen gas! Interestingly, in the 3rd test tube, there is no starting copper metal. However, there is cited a displacement formation of new Cu (black in color) metal by Zinc interacting with the cupric ions of $\ce{CuSO4}$. This newly formed black Copper, however, has a decidedly high surface area compared to both the piece of Zinc metal and the Copper turnings.
As such, my cited explanation of the accelerated reaction is based on the electrochemical cell formed with an anode of Zn metal, a high area cathode of Copper metal, all in an electrolyte of copper ions. The anodic corrosion of the zinc is now observably accelerated due to the now very favorable ratio of the low surface area zinc anode to the high surface area black Copper cathode.
Further, electrochemical reactions, in general as long as there is some reagent concentration presence, are NOT driven by the relative concentration considerations as occurs with standard chemical reactions (for the current reaction, see graph of reaction rate here depicting a flatting).
So apparently, there are perhaps several reasons including an electrochemical underpinning, as to why to preferentially employ dilute $\ce{HCl}$ along with impure $\ce{Zn}$ metal.


text:
What is the difference between these two formulas for pH of weak acids?
In the German Wikipedia article on $\mathrm{pH}$, I found the following formula for calculating the $\mathrm{pH}$ of weak acids (which are there defined of having $4.5 < \mathrm{p}K_\mathrm{a} < 9.5:$
$$c(\ce{H3O+}) = c^\circ\cdot\sqrt{K_\mathrm{a}\cdot c_0/c^\circ}\label{eqn:1}\tag{1}$$
I am a bit confused about this as in school, I learned that the $\mathrm{pH}$ (or more precisely, the concentration of $\ce{H3O+};$ one would have to take $-\log c(\ce{H3O+})$ to get $\mathrm{pH})$ for weak acids is
$$c(\ce{H3O+}) = \sqrt{K_\mathrm{a}\cdot c_0}\label{eqn:2}\tag{2}$$
I see that these formula are somewhat similar, but they aren't identical. The Wikipedia formula \eqref{eqn:1} includes $c^\circ,$ of which I could not find out what it is (it also appears in the equations for strong and very weak acids).
So I have two questions:

What is $c^\circ$?
Are these formulae different and if yes, how?

That's not how it works. Until you know the meaning of every term in a formula, there is no formula, just some random heap of letters and math symbols. So there is only one formula here.

@IvanNeretin agreed. Not understanding the second term was probably my main motivation to ask this question

Then why didn't you read that article and find it out? True, they never say that quite explicitly, but one might deduce that $c^0$ is some kind of activity coefficient. If we disregard it altogether (that is, consider $c^0=1$, as people often do), then the formula (1) becomes the same as (2).

@Jonas Your confusion probably arises from improper notations (which I edited for clarity). It's not "$c$ to the power of zero" (c^0) which  indeed mathematically leads to $c^0 = 1$ and makes little sense in this context. It should be either $c^\circ$ (c^\circ) or $c^⦵$ (c^⦵), which is standard concentration.

The purpose of this $c^\circ$ is to ensure dimensional consistency. Keep in mind that Wikipedia can be edited by anyone, although the content is often of very good quality, sometimes the volunteer writer misses some points and assumes that the reader might be aware of his/her symbols. In your link, the writer does not explicitly define $c^\circ$.
The German version is ensuring that the equilibrium constant is dimensionless and all the quantities which are mathematically operated upon are dimensionless numbers. Therefore $c^\circ$ must be 1 mol/L, if $c_0$ has molar units.
$$c(\ce{H3O+}) = c^\circ\cdot\sqrt{K_a\cdot c_0/c^\circ} \tag{1}$$
Note the quantity under the square has been made dimensionless. In order to attach a unit to the square root term, $c^\circ$ has been multiplied.
If you want to dig deeper, there is something called quantity calculus, which I quote again from my previous answer

Are the units of mole of oxygen molecules the same with the units of mole of nitrogen molecules?
Calculus here is not the integration / differentiation, but rather the
Latin calculus implying a method of calculation.
There is a very nice article "Quantity Calculus: Unambiguous
Designation of Units in Graphs and Tables" by Mary Anne White in the
Journal of Chemical Education. Please read this if you are seriously
interested. Search on Google Scholar and it is free to download from
there.
In quantity calculus Each physical quantity as the product of a
numerical value and a unit:
physical quantity = numerical value × unit
This approach was introduced by British scientists and many leading
physicists used it. Now there is there is nothing less or nothing
more. Therefore your ambiguity arises from introducing another factor
such as "oxygen" or "nitrogen". The unit mol does not know whether it
belongs to oxygen or nitrogen.
As explained in the comments, suppose we write L symbolizing the
height of a tree, then I can only write, L = 10 m. For mathematical
purposes, I will not introduce "tree" anywhere in this equation. The
tree is already incorporated in L (in your mind) but not in the
mathematical equation. One can also write L/m =10. Now you have a pure
number on both sides.

Just a dimensional unit, then. So I was wrong. Well, good to know.


text:
Why don't atoms with expanded
My title may be confusing, so let me explain. Let's take the molecule SO2 for example...

Above, you can see that the Sulfur atom has 6 electrons from its original valence shell involved in bonding/in a lone pair. However, if you add up these bonds, the atom has 10 electrons, not 8 (which disobeys the octet rule). I was told this is because sulfur has access to the d orbital of electrons. That makes sense, but why does it not get a 2- charge if it is getting 2 more electrons in it's valence shell?
Thanks!
Will

It is not getting any extra electrons. Why would it? Where from?

If this puzzles you, you can remove one of the two bonds between S and O and put it entirely on the oxygen atom which becomes negatively charged. The sulfur atom, having lost one electron, will then become positively charged, but it is now surrounded by 8 electrons.

This is because covalent bonds are when electrons are SHARED BETWEEN THE ATOMS, not all given to 1 atom (that's an ionic bond). The sulfur may gain a partial charge (positive in this case), because more electronegative atoms hold electrons closer, but never a full charge.