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What is it about bleach that keeps Pharaoh ants away even after drying?
I posted this question here and one of the commentors thought that this Chemistry SE might be able to offer more sure answers.
My apartment high-rise has been afflicted with ants in the past few years.  The pest control technician says that it has become a common problem due to warming weather.  Various things have been tried, and I am leaving this in their expert hands (or rather, the apartment management is).  This question is not asking for further countermeasures.
When I find a place or pathway with many ants, I find that simply wiping with a damp cloth or vinegar doesn't keep them away for long.  Wiping is supposed to disrupt their pheromone trails, but I only find somewhat lasting effects if I wipe with undiluted household bleach.  I am pleasantly surprised that they stay away even after the bleach has dried.  After some internet research, I found that bleach leaves behind a salt residue when dried.  Is it the salt residue that keeps the ants away when wiping otherwise doesn't have lasting effect?
I've had another corroborative experience that leads me to suspect that salt residue has a repulsive effect on pharaoh ants.  I keep a large beer glass of very salty water on hand because of the dental benefits of salt water rinse.  I probably use too strong a solution, which likely has drawbacks (something I have yet to research).  It is kept on top of the fridge.  Around the glass is salt water staining.  I never see ants around it, even though ants like water.  I admit that this corroboration is not very strong, as I generally do not see ants on top of the fridge anyway.

This question is admittedly cross-disciplinary (chemical ecology/biology), but probably better suited for biology SE since they can discuss issues such as sensorial thresholds and range of chemical perception and interference, issues that have less to do with basic chemistry and more to do with neurophysiology/ecology/animal behavior.

However, you might want to explain what kind of chemical answer you are looking for (and I don't think "Is it the salt residue that keeps the ants away when wiping otherwise doesn't have lasting effect?" is a chemical question per se.

Try an experiment. Leave some of your strong salt solution to dry in an area occupied by the ants and see if they avoid it when it has dried.

Personally I use butter to convince the ants to stay away. It works. They hate butter.

@Maurice: Can you please share how you discovered this?  My online surfing indicates that they like fat, and at least one mention of butter.

@Waylander: I'll keep the dried salt solution in mind.  However, the ants show up in a food preparation area (kitchen counter, often near the sink), so I'm not keen on deviating what seems to work (bleach) for any amount of time.  The sample sizes per day are small (3 to 5 sightings per day, sometimes with application of bleach, often applied at night), so it may take a many-day controlled experiment to be sure.  And the will to avoid wiping down with bleach when I see the ants.

@ Buck Thorn: OK, thanks for your assessment of the question.  I just didn't know enough to mark the answer as correct in the Biology SE.  Would you have an opinion on that?  I feel that leaving it unmarked is irresponsible, as it could be the right answer.  I just don't know enough about the chemistry to judge.

I have not discovered the effect of butter. I just got it a couple of years ago in a museum where a real living ants hill was installed by specialists. There was plenty of people  around the nest. And some were asking questions. I remember a visitor asked the specialists how to repell ants from entering his home. The specialist mentionned the butter as an old receipt. I was an auditor, but I remembered it. Much later, I had the opportunity to test the trick at home, and I cut the ants way with a line of butter. It worked. The ants refuse to cross the line, even if it is thin.

Further web searching reveals that it depends on the ant. If what I read is to be believed, grease ants will love butter, but general ants won't because the the fat makes it hard to digest with their acid. The ant in question here is the Pharaoh ant, which has more of a sweet tooth. This page, however, implies that they like butter, as does this page. I'll play it safe for now and keep the butter idea for later.

Household bleach acts as a broad oxidizing agent of organic compounds, and this activity is rapid and usually enough to suppress the original activity of a pheromone. So my assessment would be that the answer at bio SE is correct. The residue is probably table salt. The bio question is, do the ants mind the salt, why or why not? See also: https://www.clorox.com/learn/does-bleach-leave-a-residue-dried-bleach-crystals/

In retrospect this post seems to contain plenty of chemistry. Feel free to request migration to chem SE at biology SE.

I did, days ago.  No one responded.  That's why I ended up cross-posting.  Thanks for the vote of confidence in the answer at the Biology SE.  I'll mark it as "the answer".

My bleach solution seems ill-advised, both because of the fumes and the adverse effects on laminate countertops.  It looks as if I will have to get comfortable again with ants on my food preparation surface.


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Electrons in Bohr's atomic model
Why electrons don't fall into the nucleus in bohr's atomic model?
Is there a reason or explanation or is just a consideration?

Is just a postulate, I think anyone can actually explain why.

Related: https://chemistry.stackexchange.com/questions/54914/stability-of-bohr-orbits

@JorgeBonifaz That was a postulate - quantisation of energy. Now it's quite more than that and learning quantum mechanics gives some explanations.

So you mean it "was" in the past? Or only that was in some way the base of future advances in quantum mechanics?

@JorgeBonifaz I mean Bohr just postulated it, but future advances changed this situation.

The site expects that you write explicit compact summary of your prior effort to answer the question, based on your knowledge and on searching for existing related  info or answers. It helps preventing  others to tell you what you already know or what you could easily find yourself, being answered many times in multiple ways.

As far as I know, it's a consideration that the electrons must describe circular orbits around the nucleus of the atom without radiating energy.
So this is because is a "postulate", since according to classical electrodynamics a charge with an accelerated movement must emit energy in the form of radiation.
To maintain the circular orbit, the force experienced by the electron (the Coulombian force due to the presence of the nucleus) must be equal to the centripetal force. So here it is:
Taken from Chemistry Libretexts - 6.2 The Bohr Model

$$k\frac{Ze^2}{r^2} = \frac{m_ev^2}{r}$$

Note that this model is only valid for hydrogen-like systems (only with 1 electron).

How does that relate with the "quantisation of energy"?

since according to classical electrodynamics a charge with an accelerated movement must emit energy in the form of radiation. - This is valid only for nonzero time derivative of acceleration vector. Linear constant acceleration does not lead to radiation, what is used in linear accelerators.

source: Stack Exchange

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

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Molecular orbital mixing in second period elements
Why do the $\pi_u(2p)$ and $\sigma_g(2p)$ diatomic MO mix within the second period elements as it's shown in the image below?

If we consider the OM from both 2s and 2p AO, we have the following diagram, which is only valid when the 2s and 2p are far in energy (O2 and F2). This is because as they have a higher Zeff, the 2s orbitals are more stabilized.
For lighter homo-diatomic molecules the AO are very near so they mix

This is a zero order answer but hope it is helpful.
The model that is commonly used to describe the MO is based on the results of expressing the MO as a linear combination of atomic orbitals (LCAO), thus obtaining the MOs and their energies. The contribution of each AO to a particular MO depends on its shape and energy. In general, the LCAO coefficients are optimised to minimise the energy of the molecule and its occupied MO.
Since the atoms of different elements have different nuclear charges and number of electrons, the magnitudes of the attractive nucleus-electron and repulsive electron-electron forces are different and, in consecuence, the energies of AOs of the same type for different elements differ. Thus, when the LCAO method is applied to calculate the MO of a homonuclear diatomic molécule, differences in the energies of the MO are observed.

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


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Excited states and emission lifetimes
I have some doubts about the nomenclature of the lifetimes. Are singlet and fluorescence lifetimes the same? As well as the triplet and phosphorescence lifetimes?
Also, can the triplet/phosphorescence lifetimes be measured by phosphorescence decay as well as by transient absorption of the triplet state by means of laser flash photolysis?

Yes, singlets fluoresce and triplets phosphoresce. So the singlet lifetime is the fluorescence lifetime and is the inverse of singlet excited state decay rate constant and similarly for phosphorescence.
Both singlet and triplet decay can be measured by flash photolysis more often nowadays this is called pump-probe spectroscopy or pump-probe transient absorption (or some variation of this) but it is the same technique as flash photolysis. It has kinetic (decay measured at one wavelength at a time) and spectroscopic (spectrum at sequence of times after excitation) variants.
If the excited state is, say, a quartet or quintet etc. we use the term luminescence.

Thank you for your clear answer. However, it makes me feel uneasy.  In some molecules the ground state is a triplet -e.g $O_2$ ($X ^3\Sigma_g^- $)- and the emission from its triplet excited states is called fluorescence taking into account that the transitions are allowed and the lifetime is short. Also the OH radical ($X ^2\Pi$) has fluorescence and the radical is frequently detected using Laser Induced Fluorescence (LIF).

I wonder if as a general statement (without specifiyng a context) this definition maybe an oversimplification, and I wonder if it misses the importance of distinguishing between an allowed (high-probability, fluorescence) versus  a forbidden transition (low-probability, fosforescence)?

The length of an excited state emission  has nothing, per se,  to do with whether it is fluorescence or phosphorescence, although phosphorescence is normally longer in the same molecule. Symmetry can also be involved in determining emission probability, thus there are many molecules whose singlet is longer lived than other types of molecules triplet state.

Completly agree,  Did not explain myself very well. I wrote a more detailed comment as an answer below.

Thanks @Achem and @porphyrin for your comments. I do not know if an Answer is the right format for this Comment but the comments do not allow enough character to develop my argument.
I think we are all in the same or close pages. My reservation is that we have to be a little nuanced when using this terminology, since otherwise it may leed to confusion.
I rewied this morning the the IUPAC's 3rd edition of Glossary of Terms Used in Photochemistry and look for the definitions of luminescence, fluorescence and phosphoresnce. I quote from the document:

Luminescence.
Spontaneous emission of radiation from an electronically excited species or from a >vibrationally excited species not in thermal equilibrium with its environment.
Fluorescence
Spontaneous emission of radiation (luminescence) from an excited molecular entity >with retention of spin multiplicity.
Phosphorescence
Phenomenologically, term used to describe long-lived luminescence. In mechanistic >photochemistry, the term designates luminescence involving change in spin >multiplicity, typically from triplet to singlet or vice versa.
Note: For example, the luminescence from a quartet state to a doublet state is also
phosphorescence.

Thus, both fluorescence and phosphorescence are examples of luminescence, independent of the spin of the states coupled in the transition.
Both triplet-triplet transitions (for example in O$_2$) and doublet-doublet transitions (for example in OH and NO) are examples of fluorescence. I am not familiar with any molecular transitions involving quartet or quintet states but that is my ignorance.
I guess that in some fields the molecular ground states are singlets and higher spin ground states and oddity, so I understand that the terms fluorescence, phosphorescence and luminescence can be used more loosely, even if it makes me unconfortable, but in that case I feel it is better to state in what context you are using the definition.

I think the definitions are self-consistent, a triplet to triplet in oxygen is fluorescence because the spin multiplicity is the same. I am sure there must be plenty of examples in atomic fluorescence. Like Na fluorescence is doublet to doublet.

Agreed. I meant molecular systems involving transitions between quartet or quintet states. I will edit my answer.

Regarding porphyrin's response and PAEP's reservation about oxygen emission from an excited triplet to a ground triplet being labelled as fluorescence, the issue can be addressed by defining fluorescence as the excitation and emission having the same spin multiplicity. This removes the condition of fluorescence as singlet to singlet transition and then it is not based on short lifetimes. In a similar manner, phosphorescence can be defined as emission from a different multiplicity from the ground state. IUPAC's gold book seems to silent on this topic but other IUPAC documents agree with above.
To the OP, you will have to look at the context. Singlet state does not necessarily imply fluorescence. All it means that all electrons are paired in that system. Anyway, these are man-made definitions/ labels. Luminesence to analytical chemists is a catch all term for emission of light from molecules or atoms by means of various excitations. It has no connotations associated with the spin multiplicity.


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Reaction of hydrogen peroxide with acetic acid and potassium permanganate
I conducted an experiment where I mixed hydrogen peroxide $\ce{H2O2}$, acetic acid $\ce{CH3COOH}$ and potassium permanganate $\ce{KMnO4}$ under vigorous stirring. The reaction hasn't stopped after 15 min and the colour is cycling between pink and colourless whenever I add more potassium permanganate.
What reaction is going on? I think despite a medium created with weak acid, acidified potassium permanganate is a much stronger oxidising agent and it should react with a lot of peroxide rather than the opposite.
But hydrogen peroxide and acetic acid react to form peracetic acid $\ce{CH3CO3H}$ (PAA). I don’t know if PAA reacts with potassium permanganate or something entirely different occurs.
I am concerned about the reaction and products. The mix smells strongly of acetic acid and most of the permanganate is still in there. Could it be the manganese oxide formed in the reaction being oxidised? Well first of all no bubbles the limiting reagent of the reaction was hydrogen peroxide and I used aqueous potassium permanganate. And by most of the permanganate was still there I meant I added half a teaspoon of permanganate and I used less than 1% acetic acid and 6.5 percent peroxide. I was trying an acid catalysed reaction between and oxidiser permanganate and and reductant peroxide. When I drained the solution (by excess of water)a most of the amount added of permanganate was in the solution with manganese oxide indication it reacted with the peroxide.

Try to better explain what you have done. In which order have you mixed the three components ? And how much of each ? How many times ? Did you use solid permanganate or its solution ? In this case which concentration ? The same question for hydrogen peroxide : have you used $30$% or $3$% hydrogen peroxide ? Did you use pure acetic acid, or an aqueous solution ? Which concentration ? What is the meaning of "most of the permanganate was still in there" ? Please explain thoroughly all what you have done
!

Please be careful, peroxide and permanganate are strong oxidizers.  Acetic acid is potential fuel.  But what you may be doing is oxidizing the peroxide (with permanganate).  Did you see bubbles?

The limiting reagent was not defined. Acetic acid is a suitable solvent for permanganate or chromate oxidations at low temperatures but still should be used very carefully. This idea of mixing peroxides with a strong oxidzer and a possible reductant sounds like potential bomb making. This person is playing with chemicals not "conducting an experiment"

The Kinetics of the Acid-Catalyzed Formation of Peracetic Acid from Acetic Acid and Hydrogen Peroxide has been studied in details as early as 1965.

Well first of all no bubbles the limiting reagent of the reaction was hydrogen peroxide and I used aqueous potassium permanganate. And by most of the permanganate was still there I meant I added half a teaspoon of permanganate and I used less than 1% acetic acid and 6.5 percent peroxide. I was trying an acid catalysed reaction between and oxidiser permanganate and and reductant peroxide. When I drained the solution (by excess of water)a most of the  amount added of permanganate was in the solution with manganese oxide indication it reacted with the peroxide.


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1,1-Diphenylethylene subproduct/impurity
I'm working with 1,1-diphenylethylene and I've observed that, if left at room temperature in the air atmosphere, a white insoluble powder forms. I'm unable so dissolve this powder in every solvent that I've tried (acetonitrile, acetone, ethyl acetate, hexane, water, etc), so I can't get an NMR spectrum. Moreover, I can't find any paper or information online about this subproduct/impurity.
Edit: The melting point of this powder is around 120-130 ºC.
1,1-Diphenylethylene has to be stored between 2-8 ºC, but the safety data sheet doesn't say that it has to be stored under an iner atmosphere. However, we have it stored under argon in the refrigerator and the impurity only appears when leftovers of 1,1-diphenylethylene are left outside in a vial, for example. Futhermore, I observed that 1,1-diphenylethylene is also oxidized to benzophenone when left outside.

Hmmm, a polymer?

Polymerisation seems very likely

Or possibly peroxide formation, An ATR FTir will give a spectrum that should help you. More information on purity, preparation method, storage details etc. would help us.  Does the compound arrive stored under N2?

@jimchmst I've added an edit with more information.

This may be of help.


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Terms used to describe the Pauli exclusion principle
This is a very simple question. Today, someone explained to me that the Pauli exclusion principle says that "no two electrons can occupy the same energy state". However, I believe that this language is not precise as two electrons can occupy the same energy state, allowing they have different spin. Rather it is the quantum state that cannot be the same.
So, my question is, is it wrong to say energy state?

I’d go one step further and argue that the term ‘energy state’ is not well-defined. If they meant that two electrons can’t have the same energy, then they should just say that (even though it’s obviously wrong).

The principle means that electrons cannot have the same set of quantum numbers and be in the same state. You realise this by saying that they must have different spin. (Electrons have two spin quantum numbers. Spin quantum number in your question means magnetic/projection/azimuthal spin $m_s=\pm1/2$, whereas the other electron spin quantum number is always the same and is $S=1/2$)

Depending on the level at which it is taught, you will find different statements, some of which are just wrong. Related: https://chemistry.stackexchange.com/q/111062/72973


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Has dilead hexahydride been synthesised?
Dilead hexahydride is listed on Wikipedia as a compound composed of two lead atoms, and six hydrogen atoms:
  H  H
  |  |
H-Pb-Pb-H
  |  |
  H  H

As the wikipedia page for this compound does not exist, and Google scholar, along with Google, show no results on it, it's suggested that this is an artificial chemical that might have to be created in a similar manner to plumbane (PbH4).
How is dilead hexahydride synthesised?
My secondary question is, what are it's properties/characteristics? (if any info can be obtained?)

Even PbH4 was only made about 20 years ago. I doubt diplumban is gonna be made anytime soon.

https://doi.org/10.1002/chem.200400525 predicts low $\ce{H3Pb-PbH3}$ bond dissociation energy, so let me state the obvious: diplumbane should be quite unstable. Wiberg's Anorganische Chemie, 2017, says $\ce{Pb2H6}$ is an unknown compound.

This blows lead based life forms out of the water! Is there still hope for silicon, germanium, and tin?

@jimchmst Umm, not exactly. Tetraethyllead was added to fuel, so I think there could be plenty of chain derivatives, just not much of binary hydrides, that are somewhat unstable even for silicone.

Plumbane: https://chemistry.stackexchange.com/a/162199/17368

Apparently diplumbane itself is not well enough known to appear  in amy casual Google search. According to andselisk in a comment citing Wiberg's Anorganische Chemie (2017), $\ce{Pb2H6}$ is unknown.
However, some substituted diplumbanes appear to be more stable. hexaphenyldiplumbane, $\ce{Pb2(C6H5)6}$, appears to have known spectra, including the complex pattern below for protons in $\ce{CDCl3}$ solvent1. The complexity of the pattern suggests the phenyl groups are not equivalent in this compound. Since lead (and other Group 14 elements) has larger atoms than carbon, the structure follows the substituted ethane model unlike "hexaphenylethane"; the same is true with other Group 14 elements besides carbon in place of lead.

Reference

John Wiley & Sons, Inc. SpectraBase; SpectraBase Compound ID=1WQFC6VQS6Z SpectraBase Spectrum ID=6esVxgAw9qV

https://spectrabase.com/spectrum/6esVxgAw9qV (accessed Oct 8, 2022).

This is not an answer to the question, while the observation of Pb2Ph6 is interesting, it does not answer the question about Pb2H6.

Apparently, protonated diplumbane $\ce{Pb2H7+}$ have been predicted to be energetically favorable to varying degrees.

@NuclearChemist please see my update. On this basis the claim should be resolved.

@NilayGhosh I saw this. One notable feature: the proton inserts itself into the lead-lead bond si that the protonated species no longer has a direct bond between the lead atoms.

The Pb2H7 ion is interesting, maybe it has a three centre two electron bond. But the species is clearly not something with a Pb-Pb bond. I reason that Pb2H7 could be reasoned to relate to MeOMe while Pb2H6 could be argued to relate to ethane.


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Titanium-free catalysts to decrease decomposition temperature of magnesium hydride
Magnesium hydride $(\ce{MgH2})$ is a metal hydride with a relatively stable structure, at least compared to other metal hydrides. Magnesium hydride splits at around 300 °C into magnesium and hydrogen. The decomposition temperature can be halved with a catalyst.
Unfortunately, these catalysts are often titanium-based (example A and example B), which does not make this a realistic way to reduce the temperature required, at least not if made for wide-spread production, as titanium is a rare and expensive metal.
What catalysts are proven and effective at reducing the temperature required to break magnesium hydride, that don't include titanium?

I would consider titanium neither rare nor expensive. Titanium is ninth most abundant element on Earth, and is much cheaper than other transition metals used as catalysts. My friend used to ride a bicycle with titanium frame (aquired about 20 years ago) before switching to carbon fiber, and he is no Rockefeller.

The only aspect somewhat affecting price is technologically demanding production of bulk metallic  titanium, due its high affinity of oxygen and high melting point.

From Shikin et al.[1], a recommendation to press the magnesium hydride with nickel powder:

We have studied the activation of magnesium hydride decomposition by means of its pressing with a catalyst. It is established that pressing leads to the formation of metal nuclei, which favor a decrease in the temperature threshold of magnesium hydride decomposition. The introduction of catalytic additives also reduces the temperature of dehydrogenation. The most effective in this respect was found to be the addition of nickel powder.

Reference*

Shikin, I.V., Elets, D.I., Voyt, A.P. et al.  (2017). "Activation of magnesium hydride by pressing with catalytic additives." Tech. Phys. Lett. 43, 190–193. https://doi.org/10.1134/S1063785017020262.

It's seems like the cheapest catalyst to ease this decomposition is $\ce{MgO}$ (Ref.1):

Other metal catalysts, which effect this decomposition are as follows:

Evidently, $\ce{MgO}$ seemingly better metaloxide than titanium oxides as well as the best metal catalyst so far, $\ce{Nb2O5}$.
References:

José-Ramón Ares-Fernández, and Kondo-Francois Aguey-Zinsou, "Superior $\ce{MgH2}$ Kinetics with $\ce{MgO}$ Addition: A Tribological Effect," Catalysts 2012, 2(3), 330-343 (DOI: https://doi.org/10.3390/catal2030330).


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Why can we add/substract/cross out chemical equations for Hess law?
Suppose I take the model example:
(1) $\ce{C + 1/2 O2 -> CO}$
(2) $\ce{CO + 1/2 O2 -> CO2}$
Adding both together:
(3) $\ce{C + CO + 1/2 O2 + 1/2 O2 -> CO + CO2}$
(=> $\ce{C + 1/2 O2 + 1/2 O2 -> CO2}$)
Now why is it generally ok to cross out CO on both sites? Clearly it's balanced with or without and it's not part of the reaction either but let's assume for a moment that this reaction (3) containing the reactants and products from (1) and (2) suddenly forms other products:
$\ce{C + CO + 1/2 O2 + 1/2 O2 -> C2O3}$
(Only hypothetically, there are certainly some reactions that then form something other than merely additively the same products), why can Hess law still be true than? The formation must clearly have another enthalpy.
Or do I always have to imagine with such manipulations that they are pure mathematical steps to the actual reaction of interest without the various reaction paths actually reacting in such ways?

Coal would burn, regardless if there is a stone laying aside at the beginning and the end, or if it is not there. // Get familiar with the concept of bystanders/spectators in chemical reactions, that are optional and unnecessary. BaCl2 + Na2SO4 -> 2 NaCl + BaSO4 is de facto  Ba^2+ + 2 Cl- +  2 Na+ + SO4^2- -> 2 Na+ + Cl- + BaSO4 , what can be written juyst as Ba^2+ + SO4^2- ->  BaSO4

When you add two equations such that the product from 1st equation is a reactant for the 2nd equation, instead of thinking of them "cancelling", think about the product from the first reaction is being used up for the 2nd one, thus being effectively "cancelled".

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

From your example, you were trying to use Hess's law to find the enthalpy change of the reaction $\ce{C + O2 -> CO2}$ from a two-step process:
$(1) \ \ce{C + \frac{1}{2}O2 -> CO}$
$(2) \ \ce{CO + \frac{1}{2}O2 -> CO2}$
As you said, adding these two processes combine to produce the overall reaction we are interested in as the $\ce{CO}$ on both sides would cancel. Now, what you are saying is that, why can't we just not cancel the two $\ce{CO}$ and then combine the $\ce{CO}$ and $\ce{CO2}$ to make a different hypothetical molecule $\ce{C2O3}$.
Sure, you can do that, and obviously, the overall reaction would have a different enthalpy, but the $\ce{C2O3}$ never appeared in our steps. You can easily fix this by adding an extra step as below:
$(1) \ \ce{C + \frac{1}{2}O2 -> CO}$
$(2) \ \ce{CO + \frac{1}{2}O2 -> CO2}$
$(3) \ \ce{CO + CO2 -> C2O3}$
Now if you combine them, you'd get what you had,
$\ce{C + CO + \frac{1}{2}O2 + \frac{1}{2}O2 -> C2O3}$
or $\ce{C + CO + O2 -> C2O3}$
However, we had to consider an additional step for that overall process which would be the source of the difference in enthalpy that you are pointing to.

Enumerated equations of chemical reactions, regardless of if they reflect really ongoing processes or not,, are just accountant inventories of atoms counts on both sides. They say in your example that numbers of atoms of carbon and oxygen on both sides are equal.
If there are the same terms on both sides, they can be subtracted as in any other mathematical equation.
If atoms counts on both sides are equal for 2 reactions, summing the sides is the equivalent operations and counts for both sides will be equal too.
Not there you cannot imply from the above reaction equations that CO and CO2 form C2O3. Such knowledge is outside of mathematical formalism of these equations.

Fundamentally, this is precisely because of Hess's law, which says that (as long as the reactants and the products are both in consistent standard state) the change in enthalpy depends only on the changes between the initial reactants and final products of the reaction, not on how the reaction actually happens (or even whether it can actually happen).
In particular, since a "reaction" like $\ce{CO -> CO}$ where nothing actually changes has $\Delta H = 0$, it follows from Hess's law that adding such a (non-)reaction to another reaction cannot change its $\Delta H$.  So the reactions $$\ce{C + O2 -> CO2}$$ and $$\ce{C + CO + O2 -> CO + CO2}$$ must have the same $\Delta H$, since they only differ by the addition of one $\ce{CO}$ molecule to both the reactants and the products.  According to Hess's law, it doesn't matter whether the $\ce{CO}$ actually participates in the reaction somehow or not: one molecule of $\ce{CO}$ goes in and one molecule of $\ce{CO}$ comes out, and that's all that matters.  Indeed, for all we know, that $\ce{CO}$ might be sitting in a sealed bottle on a shelf somewhere and might never even come anywhere near the other reactants!
It is, of course, extremely convenient that Hess's law happens to hold, since if it didn't, we might not even be able to define the concept of a standard enthalpy of reaction without having to consider details like the specific pathway the reaction takes and what other chemical species might be present and how they might affect the kinetics of the reaction.  But it does, and indeed that's exactly why reaction enthalpy is such a useful concept in chemical thermodynamics.

Ps. Of course, as others have already noted, a hypothetical reaction like $\ce{C + CO + O2 -> C2O3}$ would have a different $\Delta H$ than $\ce{C + CO + O2 -> CO + CO2}$, since it has different products: $\ce{C2O3}$ instead of $\ce{CO + CO2}$.  There is no contradiction here, since the reaction $\ce{CO + CO2 -> C2O3}$ has a non-zero $\Delta H$ that exactly accounts for the difference.

Now why is it generally ok to cross out CO on both sites?

Every chemical reaction tells you about the amount of reactant used up, and the amount of product made. To make things easier, we will consider a situation where exactly one mole reacts, e.g. a stoichiometric coefficient of "1" in front of a reactant means we use up one mole, and a stoichiometric coefficient of "1" in front of a product means we make one mole.
To understand why we can cancel CO on both sides, imagine we have a thousand moles of all species present (reactants and products). Each reaction will change the amounts in our stockpile. Because there is so much, we will never run out.
The first reaction adds a mole of CO to our stockpile. The second reaction uses up a mole of CO in our stockpile. The net effect is no change in CO. It does not matter whether reaction (1) happens first, second, or at the same time as reaction (2).

Or do I always have to imagine with such manipulations that they are pure mathematical steps to the actual reaction of interest without the various reaction paths actually reacting in such ways?

If the reactions are real, substracting or adding reactions is real, as well. When you add them in a one-to-one fashion, there is usually a physical reason. For example, if you start out with no CO at all, the second reaction can only use up as much CO (or less) as the first reaction produces.

Why can we add/substract/cross out chemical equations for Hess law?

When you substract a reaction, you are saying that it runs in the reverse direction.  In this case, your stockpile of products will diminish, and your stockpile of reactants will increase. Again, this might actual happen, or it might be artificial.
For Hess' law, it does not matter. The quantity you calculate with Hess' law, enthalpy, is a state function. Any path from A to B will result in the same enthalpy change, no matter whether real or fictitious. If you can measure the reaction from B to A, for example, you can also give the enthalpy for the reaction from A to B, even if you can't do the experiment.
UPDATE

[OP in comments] "If the reactions are real, substracting or adding reactions is real, as well." - That is a little unclear to me, although I can have two real reactions, but if I add them together, it can also happen that instead of the products merely being additive, another products are formed because, for example, a reactant from reaction a now suddenly prefers to bond with a reactant from reaction b. A+B->C and D+E->F, instead of A+B+D+E->C+F it is A+B+D+E->I+J, where A+D->I and B+E->J. That shouldn't be simply additive anymore because it doesn't produce the same products as the 1. and 2. reactions alone.

That is true in general. If I don't mix reactant, I don't get a reaction. If I add a third thing (maybe an acid or a base, for example), I might get a different product. Hess's law just says that once you describe a certain set of reactions with some reaction equations, the enthalpies add up in the same way that these component reactions do.

"If the reactions are real, substracting or adding reactions is real, as well." - That is a little unclear to me, although I can have two real reactions, but if I add them together, it can also happen that instead of the products merely being additive, another products are formed because, for example, a reactant from reaction a now suddenly prefers to bond with a reactant from reaction b. A+B->C and D+E->F, instead of A+B+D+E->C+F it is A+B+D+E->I+J, where A+D->I and B+E->J. That shouldn't be simply additive anymore because it doesn't produce the same products as the 1. and 2. reactions alone.

That is true in general. If I don't mix reactant, I don't get a reaction. If I add a third thing (maybe an acid or a base, for example), I might get a different product. Hess's law just says that once you describe a certain set of reactions with some reaction equations, the enthalpies add up in the same way that these component reactions do.

Hess' Law is inferred from the nature of Enthalpy: It is a state function. This is derived through its equation,
ΔH=ΔU(or ΔE depending on how you represent the internal energy)+PΔV.
P, Pressure, and V, volume, are both state functions: their values are only based on the present state (for example you could use the present P/V/n/T values, assuming ideal gas, or using the Van der Waals equation to find P and V respectively). Similarly, change in internal energy is also a state function, since its value is irrespective of the pathway taken to achieve the change in internal energy (q+w is always constant).
There's an example in Atkins' Chemical Principles that illustrate this fact: Consider this: If you're rolling a ball, A, down a hill, it ends up bumping into another ball, B, making B roll up a part of the hill, gaining GPE, and A itself slows to rest at the bottom of the hill.
In this example, A does work w on B by allowing it to gain KE which is transferred into GPE as it climbs the hill, and A loses energy as heat q to the hill surface by frictional heating as it moves down the hill. Therefore, the total loss of Internal Energy of A, ΔU, is equal to q+w. If A had lost transferred more heat instead of work, it means the process' pathway has changed. However, we know that A must have lost an amount of energy equal to its initial GPE, which is constant.
Therefore, the value of Internal energy is irrespective of the relative pathway it takes to achieve it. As we get here, you might now have a clue about why this was brought up, because it's beginning to sound a lot like the statement of Hess' Law: The total enthalpy change of a reaction is irrespective of the path taken.
We know that ΔH=ΔU+PΔV. We also know that U, P and V are all state functions. Therefore, since enthalpy is the sum of these functions, we can conclude that enthalpy is also a state function.
Therefore, we only have to consider the net outcomes of a reaction when calculating enthalpy: the net reactants leading to the net desired products, which justifies the cancelling of chemicals on both sides.


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How do you achieve the exchange of reactants even though it is energetically less favourable?
Assuming that in a system the following pairs are energetically most favourable: AB and CD. I now heat the system until all bonds are broken. How do I achieve the following reaction: A+B+C+D -> AC + BD? What examples and methods can you think of?

It is case dependent. If there is T region where they are preferred, at least partially, the mixture may get quickly cooled to get the equilibrium frozen. It was used for reaction $\ce{N2 + O2 <<=> 2 NO}$ in electric arc before catalytic synthesis of ammonia.

Elementary type reactions will reach thermal equilibrium. The position of the equilibrium depends on the energetics and the temperature. The conditions are manipulated to give significant amounts of the material wanted and the reaction is kinetically stopped and the product isolated. There are many ways of doing this [most common is to simply remove a product]. A reaction can be quenched. The energy can be supplied in a specific manner such as photolytically, [photosynthesis], or by electrolysis, [batteries, plating]. There can be specific catalysis to separate competing reactions when the mechanisms differ enough. If a species is at a relative energy minimum there seems to be a method to characterize and possibly isolate it. I just looked up the half-life of Helium-5; it is 2x10^-21 sec. If that can be isolated anything can. When all else fails use a different method of synthesis! in your case A + C = AC and B + D = BD and be very careful how you mix them.


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Would a hypothetical Og2 +235 form a chemical bond?
In a hypothetical (?) Og2+235 we would have a simple sigma bonding orbital occupied by one electron (leading to a bond order of 1/2). But how to take into account the giant repulsion of the two nuclei? Generally how are nucleis taken into account for energy calculations from molecular orbitals in MO theory? So far I've always read about the electrons but very few about the nuclei in the system.
Same in He23+, half bond order but very large repulsion. My intuition says that He21+ should be more stable, even if both have the same bond order.
Where in the framework of MO theory is exactly that taken into account? How does nuclei repulsion change the MO energy scheme?

Oh, no, not again ;> For diatomics you get +2, tops. Some absurd charge is out of question.

Apply the common sense of classical electrostatics and you would see that the monotonous steep slope of potential energy of nuclei cannot be disturbed by very tiny bump caused by interaction of electrons.

https://chemistry.stackexchange.com/questions/106826/does-the-hydrohelium-dication-exist

@Poutnik And why is it then so often taught that bonds are formed when one simply takes mo theory into consideration and only looks at electrons? Nuclei repulsion is hardly mentioned in mo theory

Well, gosh, it's not like you need to consider dications all that often. Even multiatomic break up pretty fast.

@FlawC https://en.wikipedia.org/wiki/Born%E2%80%93Oppenheimer_approximation - calculating MOs first involves the assumption that that the nuclei are stationary (have fixed positions) and then optimising the electron wavefunction. The nuclear repulsion term is not ignored in the overall energy, but it is just a constant (for a given set of nuclear positions).

https://chemistry.stackexchange.com/questions/146481/why-is-oxygen-more-stable-than-oxygen-dication https://chemistry.stackexchange.com/questions/112087/why-doesnt-h%e2%82%84o%c2%b2%e2%81%ba-exist

@orthocresol ok so considering a system just with the "Og2 +235", at what distance apart would they be stationary and form a bond? I mean what would be the hypothetical bond length of such a compound?

@FlawC Short answer: never. Long answer: what you need to do is to plot the energy of such a compound as a function of bond length. (You can google 'hydrogen bond energy curve' for an example of what this looks like for H2.) Now, your curve for $\ce{Og2^{235+}}$ would be so biased towards large distances, and the 'dip' in the curve would be so small, that it can hardly be considered a molecule. To be technical, it wouldn't admit a vibrational state which is required for something to be considered a molecule. https://goldbook.iupac.org/terms/view/M04002 So there is no bond at any length.

That said, the original question isn't silly at all (although I guess you did choose a rather extreme example). I would love to write a proper answer, but it's very late where I am now. Feel free to ping me again tomorrow and maybe I can get round to it.

@orthocresol that would be great :-) I've played a bit with a electrostatic system, two positive charges (p) x apart and one negative charge (e) x/2 apart, basically I've found out that the term: -pe/(x/2)^2 + p^2/x^2 describing the electrostatic force on one (and therefore the other) nuclei becomes zero at a "e to p ratio" of 1:4, so relative any more proton in the nucleis than 4*N(e) leads to a net repulsion at any distance. But of course we deal with quantum mechanics and I can't assume a electron being in the middle. I don't know enough about the quantum mechanical "forces" :-(

@orthocresol But I assume at a far enough distance apart (still repulsive, but it approaches the x axis asymptotic) the repulsion get overcompensatet by a type of quantum mechanical force which comes from the fact that the system would probably have an even lower energy than if the nuclei went to infinity apart and no longer felt any repulsion at all? (So can those electron interactions at some very very large distance overcome electrostatic forces in my example?)

@orthocresol OK so basically the only force here is actually electrostatic, then the only way I can explain it is that the atoms stay together when they are far apart and the fact that the electron interacts by staying in the area between the two nuclei, the moment where the electron would be @ x/2 doesn't contribute to the cohesion, only those where it's much closer to one or the other nuclei. Like a child that tries to hold two trucks together that rolling apart, but does not have the strength to hold them together in the middle, it always runs from left to right and pulls one back again?

I'm afraid that here you've gone off on a tangent which I don't really understand any more. There are no special quantum forces, it's all electrostatics. The problem is that the electrons are not point charges. They're charge distributions. So it's rather more involved than just doing Coulomb's law.

@orthocresol That's what I thought while writing, but unfortunately I know absolutely nothing about the nature of charge distributions, hence the analogy with the child. But if you have time to write an answer that would make me happy :)

I would not bet on it. We apparently can't even get $\ce{He2^{3+}}$, let alone higher single-electron diatomic ions. From Wikipedia:

$\ce{He2^+}$ was predicted to exist by Linus Pauling in 1933. It was discovered when doing mass spectroscopy on ionised helium. The dihelium cation is formed by an ionised helium atom combining with a helium atom: $\ce{He^+ + He -> He2^+}$.[1]
The diionised dihelium $\ce{He2^{2+}} (1Σ_g^+)$ is in a singlet state. It breaks up $\ce{He2^{2+}->2 He^+}$ releasing 200 kcal/mol of energy. It has a barrier to decomposition of 35 kcal/mol and a bond length of 0.70 Å.[1]

Thus even with only two positive charges the helium dimer is prone to breaking up due to electrostatic repulsion. Although the dication above is metastable, we should expect the stability to only become worse with an additional positive charge and only one bonding electron instead of two.
A simple classical electrostatic calculation is instructive. Suppose you have a negative charge and two positive charges, the latter equidistant from the negative charge on opposite sides. The net force is attractive if all charges are one unit, but this net attraction is lost if we give two or more (and certainly 118!) units to both positive charges. The true quantum mechanical calculation is of course much more complicated, but can give only less favorable results because as the positive nuclear charges approach the electron cloud cannot stay fully between them. Thereby $\ce{H2^+}$ is predicted to be the only single-electron homonuclear diatomic ion that remains bound.
Cited Reference

Grandinetti, Felice (October 2004). "Helium chemistry: a survey of the role of the ionic species". International Journal of Mass Spectrometry. 237 (2–3): 243–267. Bibcode:2004IJMSp.237..243G. https://doi.org/10.1016/j.ijms.2004.07.012.


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When we take baths, how does the water in the bathtub absorb only the dirtiness on our skin?
I'm 16. I study chemistry in college along with biology, physics, and maths, and I have a general interest in all of them.
I always found this interesting: sometimes when I left the bathtub, even when I forgot to use soap, the bath would still have a somewhat noticeable, murky tint in the water, indicating that something had dissolved in it, even though I didn't take the time to rub it off manually. When we get in the bath, how does the dirtiness (small amounts of mud, bacteria, pen markings etc.) dissolve in the bath alone without any of our skin dissolving as well so that we can leave clean? Maybe dead skin might dissolve, that would make sense, is there an adaptive mechanism on the skin involved in this sort of process of hygiene? The same thing would happen if I put any other object in the water with something on its surface and the object would come out clean when the thing on the surface dissolved.
How does the water break the attraction? Some sort of cohesive force, between the atoms/molecules of the thing (dirtiness) on the surface of the object and the atoms/molecules of the object (our bodies) and separate them to make it dissolve? And how can it differentiate between the cohesion of the skin and the dirtiness (what determines solubility, if both polar and non polar molecules on our skin will dissolve in the water)?

Usually taking a bath involves the use of soap to speed up the process; in fact, it may be hard to answer because baths without any soap may be outside most people's everyday experience, and that might be because water alone does not work as well as you describe it. Yes the tub got dirty - probably some fraction of the top most layer of dead skin cells fell off and floated to the edge of the water, but I do not think you were able to get actually clean (yes, it's a subjective term) without soap.

You can "get clean" using water alone. The next time you experience a summer thunderstorm, modesty permitting, strip and wash thoroughly. You will be cleaner and more refreshed than any soapy shower.

Water is called a universal solvent, and perhaps it is rather a miraculous solvent with unusual properties. It is able to "wet" a lot a surfaces and thus able to disrupt the adhesive forces. (Cohesion refers to sticking of like molecules only.)
However, the murkiness in water is not due to dissolution of dead skin or dirt dissolving in water, but rather it is a suspension (hence murky). If you had a microscope, you would rather see particles of dust, mud, soil, oils, dead skin etc if someone has not taken a shower for a while. Just pouring water does not remove these things from objects, mechanical action is also needed to dislodge them from the surface (e.g., dishwashers and washing machines use mechanical forces to clean items with water).
Since you appear to like science, next time do a Tyndall light experiment. Take that murky tinted water (as you call it), and fill it in a transparent glass tumbler. Pass light through it, and view that water at ninety degrees (to the light beam). If you see a light beam, it indicates really small particles floating in water. It is not a true solution.

The outer layer of skin is water-insoluble. According to this paper, there is a structure specifically made in skin cells before they die and become part of the outermost layer of the skin:

The outermost, cornified, layers of the epidermis are composed of terminally differentiated keratinocytes known as corneocytes. Corneocytes lack a plasma membrane and instead are encased in a structure known as the cornified envelope (CE). CEs consist of highly insoluble, cross-linked proteins with covalently attached lipids and are essential for the mechanical integrity and water impermeability of the skin.

So not dissolving in water is one function of the skin. It also protects the inside of the body from dehydration by preventing excess evaporation of water on the skin (unless you are hot, when sweating helps to maintain a lower body temperature, also important for survival).

[OP] How does the water break the attraction?

That depends on what is on the skin. If you are muddy, the components of the mud are easy to wash off with water because the components are polar, and water is polar. If you are covered in motor oil (which is non-polar), water alone will not be efficient to wash it off, but soapy water will work (with some scrubbing).

[OP] When we get in the bath, how does the dirtiness (small amounts of mud, bacteria, pen markings etc.) dissolve in the bath alone without any of our skin dissolving as well so that we can leave clean?

While the layers of dead skin cells are not water-soluble (or soapy water soluble), we nevertheless continually shed these former cells, and continually replace them. In fact, much of household dust is made of material we shed (e.g. it contains quite a bit of cholesterol, as described here).

When we take baths, how does the water in the bathtub absorb only the dirtiness on our skin?

It doesn't. When you take long bubble-baths, your skin will also lose oils and other substances (giving you "dry skin"). Some people use oils or lotions ("moisturizer") to replenish these after a long soak.


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Density of dissolved and non-dissolved salts
I am doing a calculation, which needs to be 100% correct. My question is as follows; Does the density of dissolved salt differentiate from the density of non-dissolved salts?
My specific problem is a solution of 75% potassium formate in water. When the salt is dissolved, the salt molecule is broken down to K+ and COOH-. This, at least in my mind have to affect the density of the dissolved ions from the density of crystalline salt (non-dissolved). This in turn, would mean that a calculation of the exact density can not be calculated from the density of pure non-dissolved potassium formate (@ 1.91kg/l).
Is there any table I can look up the density of the ions directly, or does anyone have any tips to point me in the right direction? Are this difference negligible? Or are my hypothesis just plain wrong? All answers are appreciated.

Would [chemistry.se] be a better home for this question?

https://www.sciencedirect.com/science/article/pii/S0017931018353067 has the data and fits you seem to want.

Thanks a lot for the answers and the reference. Yeah, I wrote a little quick there, we are using formic acid in the production for the formate :) The 20-degree reference is perfect, specifically, I'm looking for the values at 22C. My wording while looking for an answer may be the reason I was struggling, it's not easy having just a bachelor. Anyways, thanks a lot for the quick reply. Now I know where to ask when I'm stuck

This is not quite an answer but an edit to the question. I can see you haven't registered to this site. Please register and edit your question. This will merge your account.

Poutnik has nicely explained in his answer what you have to do to get reliable data. Unfortunately, tabulated density data of potassium formate is hard to found. However, I have found two reliable datasets (Ref. 1 & 2) with very agreeable data sets at two extremes (densities of 5-25% solutions in Ref.1 and that of 50-78% solutions in Ref.2, which was mentioned by Jon Custer in his comment):

Now, since the graph has good agreement of data $(R^2 = 0.9971)$, and using the straightline equation, you can find any density of given solution. Note that the data is for solutions at $\pu{20 ^\circ C}$.
References:

Åke Melinder, Thermophysical Properties of Aqueous Solutions Used as Secondary Working Fluids," Ph.D. Thesis 2007, Royal Institute of Technology, Stockholm, Sweden (ISBN 978-91-7178-707-1).
Tao Wen, Meng Wang, Yi Chen, Weifeng He, and Yimo Luo, "Thermal properties study and performance investigation of potassium formate solution in a falling film dehumidifier/regenerator," International Journal of Heat and Mass Transfer 2019, 134, 131-142 (DOI: https://doi.org/10.1016/j.ijheatmasstransfer.2019.01.031).

Better than "density of dissolved salt" is using the term "density of the salt solution", which if a function of the particular solution concentration.
As there is no law of total volume conservation, the volume of final solution may differ (is usually little smaller for salts) from the sum of volumes of solvent and (bulk) salt to be dissolved.
The density of a particular salt solution cannot be calculated from basic principles. It has to be determined experimentally, or interpolated from already known tabulated solution density data.
For common salts, there are often published tabulated data of solution densities for given mass concentrations or mass percentages.
If no solution density data are available, you can use as a very rough first approximation the calculated density from the total masses and total volumes of the solvent and the (bulk) salt. BUT, it is highly recommended to measure density (or calculate it from the mass and the volume) at least for one or two solutions close to solution concentrations of your interest.
Then, you can use an interpolation function, available in numerical mathematical methods, to interpolate density data.
Look also on the related question  Why water volume went up by almost the same amount after adding salt?

There is also no salt molecule $\ce{HCOOK}$. It is just a summary formula, expressing stoichiometric ratio of ions  $\ce{K+}$ and $\ce{HCOO-}$  1:1 in a salt crystal or its solution.


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Resolution in mass
I'm pretty confused about mass resolution. The resolution of my instrument is 1000 and I have two masses at 85.0639 and 85.0651. So, the difference is 0.0012.
I know that (M1-M2)=M/R.
But, I'm not sure if my spectrometer will be able to separate these ions.

It is not very clear what those "1000" means. As resolution is usually meant the smallest difference distinguished by device reading. E.g. the classical, old school  2-arm mechanical scales had resolution 0.0001 g.

My bad, I meant mass resolution.

From your Q it seems 1/1000 is the relative resolution of the device, so for relative mass 85, resolution is 0.085 > 0.0012 so it seems you are out of luck. // If it is wrong, that it is probably your fault due bad description of scenario.

related/helpful: Resolution in Mass Spectrometry

The mass resolution $R$ of a mass spectrometer is defined as
$$R = \frac{m}{\Delta{}m} $$
with $\Delta{}m$ the mass difference between two masses still be resolved, $\Delta{}m = m_1 - m_2$.  Thus, if your spectrometer specs detail out  $R = 1000$, then you record masses  of $m_1 = 100.0$ and $m_2 = 100.1$ as two separate signals.*
In your example, $m_1 = 85.0639$ and $m_2 = 85.0651$, or a $|\Delta{}m| = 0.0012$.  So you would need a resolution of
$$ R = \frac{85.0639}{0.0012} \approx 70887$$
to record these masses as separate signals. This number leans more toward a high-res spectrometer (e.g., to substitute a combustion analysis), than the simpler bench-tops routinely hyphenated at the exit of a HPLC.
* This omits how you define a peak in mass spectrometry.  For simplification, this back on the envelope computation assumes zero-width of the signals and no partial overlap.

+1, There is a non-ending argument about the terms spectroscopy vs. spectrometry :-) Your mass spectroscopy reminds me of my undergraduate analytical chemistry teacher held me outside the class and said that I was the only person in class who wrote mass spectroscopy in the exam. Why did you write that? It is mass spectrometry! I still see very established chemists arguing about it.

@AChem The inconsistency (twice spectroscopy, twice spectrometry) is fixed. With "The term mass spectroscope continued to be used even though the direct illumination of a phosphor screen was replaced by indirect measurements with an oscilloscope." from wikipedia, I agree, since its recordings no longer rely on emitted/captured electromagnetic radiation, MS is better off  as spectrometry. Google Ngram


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why does the single crystal ferroelectric have a different hysteresis curve compared to polycrystalline ferroelectric?
For the ferroelectric hysteresis curve, is there a reason why single crystal ferroelectrics have a 'hard' ferroelectric hysteresis curve compared to 'soft' ferroelectric hysteresis curve for polycrystalline ferroelectrics?

If the title length is comparable to the body length, the former is too long and the latter too short.

Crystals of polycrystallic materials do not have uniform orientation of their crystal lattices and it is well known crystallic materials have usually anisotropic properties.

@Poutnik - and being ferroelectric requires the lack of a symmetry center

Keywords to complememt «ferroelectric hysteresis»: Curie temperature, Néels temperature, Weiss domains. And conceptually, there are similarities between domains within a single crystal (note that twinning may occur) , individual crystals, or large grains.

@JonCuster Nice to see an echo of crystallography here.


text:
How do solutions act as solvents?
For example:

Silver chloride, $\ce{AgCl}$, dissolves in dilute aqueous ammonia, so this can be used in qualitative analysis, after reacting $\ce{Ag+}$ ions with $\ce{Cl-}$ ions, to differentiate the white chloride-containing precipitate from a similarly colored cream bromide-containing precipitate ($\ce{AgBr}$) or a yellow iodide-containing precipitate ($\ce{AgI}$).

The ammonia is already dissolved in a solvent (presumably water), therefore the ammonia molecules are sparse and separated in the solution (because low concentration, hence "dilute"), how does the $\ce{AgCl}$ dissolve in an already dissolved solution - where the ammonia is not concentrated-in-one-space-enough to have something 'dissolve' in it (shouldn't it dissolve in the original solvent instead?)
Or do the polarities of the molecules (eg. $\ce{H2O}$ and $\ce{NH3}$) in the new solvent 'combine' somehow to make an even more polar solvent and therefore overcome the dipoles in the polar solute, making it dissolve? If this were case, why is more concentrated ammonia needed to dissolve $\ce{AgBr}$ (less polar than $\ce{AgCl}$ - hence shouldn't need as much concentration of ammonia (extra polarity) to dissolve it)? How do polarities of molecules in solvents combine?

In case of AgCl and ammonia solution, it is not true dissolution by solvent, but chemical reaction, forming soluble [Ag(NH3)2]Cl. AgBr has very limited "solubility" in ammonia and AgI is not soluble
.

For future reference: chemstry.stackexchange allows you to edit chemical formulae with mhchem.  Use it in questions, answers, comments; but because it is something special not all browsers understand well enough, do not use it in the title of a question.


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What is the charge of amino acid in alkaline conditions?
This problem refers to the protein buffer in a living system.
"If the blood pH becomes alkaline, there is a release of a proton from the
$\ce{NH3+}$ ion, which takes the $\ce{NH2}$ form."
I understand that the charge of $\ce{NH2}$ must be negative alone and neutral in an amino group. However, the subject of the protein buffer is very briefly touched upon in the text our lecturer gave us and it doesn't detail the reactions in different $\mathrm{pH}$.
I think that since $\ce{NH3+}$  loses a positive proton to balance the blood pH during alkaline conditions, this must mean it becomes neutral. The problem is: I'm not sure if $\ce{NH2}$  is actually in a group and I suspect that it might be alone, which necessitates that it carry a negative charge.
This is my current understanding of the problem.

See here: https://chemistry.stackexchange.com/questions/170512/how-to-determine-the-charge-of-amino-acid-at-certain-ph.

Protein buffer refers to the buffering action of amino acid side chains of proteins. The “$\ce{NH3+}$” is more accurately $\ce{R-NH3+}$, becoming $\ce{R-NH2}$ when losing a proton. Other side chains have $\ce{R-COOH}$ or $\ce{R-COO-}$ groups, also helping to buffer.
$\ce{NH2-}$ does not occur in aqueous (biological) systems, it is too strong of a base. However, it does occur in other solvents as part of chemistry in the synthetic lab.

@MuAlShammary No, $\ce{R-NH3+}$ has a positive charge, and it goes away with the proton, yielding $\ce{R-NH2}$. The nitrogen atom in the latter has a formal charge of zero, with two electrons in the lone pair, and three electrons involved in the three single bonds.


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Why does increasing the concentration of an electrolyte decrease the mass deposited when electroplating?
I am in the process of writing a lab where the manipulated variable is the concentration of the electrolyte (zinc sulfate heptahydrate) and the responding variable is the mass of zinc deposited on a copper cathode. I electroplated each manipulation at a constant current. Hence, I found that as the concentration increases, the voltage and the mass deposited decreased.
I have researched a lot and have not found an explanation to why mass deposited decreases when current is controlled. Please help if possible!

Did you see zinc collected at the bottom of the beaker?

I did observe a few small zinc sheddings in the electrolyte solution after electroplating.
However, my procedure involved creating a concentrated 1.0 mol/L solution and diluting it for my manipulations with smaller concentrations. Does this impact the relationship?

Depositing Zinc metal requires 0.76 V.  It may be possible that too concentrated an electrolyte solution was too conductive to have a large enough voltage difference near the cathode.  So you may try various concentrations and also experiment along the axis of variable voltage.  This work should yield an excellent report and graphs even if it does not succeed.

@RobertDiGiovanni How exactly would a too conductive solution cause a lower faradaic efficiency? My guess would be these sheddings are significant or you're having a higher partial current for the HER, how exactly a high ZnSO4 causes this to happen I am not sure of.

Just a gut feeling.  "Constant current".  V = IR.  I hope it's DC.  If so, the solution (and apparatus) may be simply grounding the current.  I would increase voltage.  Using a DC transformer (like those used for heating mantles) may help.

Thank you for your replies! I will definitely consider them in my lab writeup.


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Is there any special meaning to the angle between two peptide planes?
Take a look at the following image:

Is there any special meaning to the angle between two peptide planes (marked with red arrow heads)?
If so, why isn't a specific name (like phi, psi, etc.) given?

Hmm, I'm pretty sure that angle is just $\phi - \psi$ (or $\psi - \phi$; I guess it depends which way round you define your angle, but my point is that it's not a 'new' value).

There is an angle defined by three consecutive alpha-carbon positions, and pseudo-torsion angles defined by four consecutive alpha-carbons (for course-grained modeling). Because this is a 3-dimensional structure, the angle shown in the picture is unrelated to the angle of intersection between the peptide planes (which is about 90 deg for a alpha-helical segment and maybe 50 deg for a beta strand).

@orthocresol It is more complicated. Plus, the angle shown is ill-defined (is it a proper angle, or a dihedral, and does it depend on the position of the amide proton - which is ill-defined itself).

I stared at it again, and to be honest, I still think it is some kind of combination of $(\psi, \phi)$; but I'm fully willing to accept that I'm missing something, so I'll defer to you. I don't have a proof, it's just a strong intuition, which may be wrong. FWIW, my reading of it is this: if the peptide units are indeed planar, then the angle between them is the angle between the two normal vectors (= $\arccos(\vec{n}_1 \cdot \vec{n}_2/|\vec{n}_1||\vec{n}_2|)$). There's an ambiguity in that $\cos(x)$ is symmetric about 180°, but that can be resolved through careful definition (I think).

@orthocresol If the angle at the alpha carbon were 180 deg, then I would agree. In the conformation shown, the peptide planes are almost coplanar. If you rotate both planes by 90 deg (so that there are far outside of the plane of the paper), they would have an angle closer to 60 deg. The figure can be found here and is ambiguous about the direction of rotation (what are we keeping constant and what are we rotating).


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Are there xenon compounds with two xenon atoms bonded directly to each other?
I've seen so many compounds containing xenon, and xenon was mostly the central atom, and if we had 2 xenons they would never bond directly to each other. I was curious if it is possible to have for example 2 xenons in a compound that are bonded directly to each other?

As you said yourself, we haven't seen any examples yet.

From a quick search, https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7435756/ explains: "Noble gas (or aerogen) bond (NgB) can be outlined as the attractive interaction between an electron-rich atom or group of atoms and any element of Group-18 acting as an electron acceptor."

One example is the $\ce{Xe2+}$ cation. Quoting from the abstract of Formation of the dixenon cation, $\ce{Xe2+}$, in fluoroantimonate(V) media by oxidation–reduction methods: spectroscopic properties of the ion

A green species has been generated both by oxidation of gaseous xenon
with a dioxygenyl salt and by reduction in antimony pentafluoride
solution of the $\ce{XeF+}$ cation, e.g. with water or a suitable metal; the
Raman, u.v.-visible and e.s.r. spectra of the product are all
consistent with the identity $\ce{Xe2+}$, representing the first homonuclear
noble-gas cation to be characterized in the condensed phases.


text:
What does it mean when there is a term in parenthesis in an electronic configuration?
This may be a silly question, but I have not yet found an answer and I don't really know how to word my searches. I have been looking at the NIST Atomic Spectra Database (and also atomic physics papers) and I see configurations of excited states that have terms in parenthesis in them and I don't know what it means. For example, I might see a configuration for N II
2s2p2(4P)3p
and the term is listed as 3So.
What would the 4P term in the middle of the configuration mean? When would you write a configuration like this?

At least to me it isn't really clear what you are asking. It would probably be beneficial to view this in context. Can you provide a citation for a paper or a link to the database.


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

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

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

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


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

The text says:

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

On the other hand, the research article

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

also mentions and repeatedly uses the word restraint.

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

I have two questions:

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

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

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

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

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

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

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

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

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

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

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

Could you add some references in support of your answer?

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


text:
How is Clausius statement violated for reversible heat engines that are assumed to have different efficiencies?

I'm trying to verify that all reversible heat engines have equal efficiencies by using this diagram. Here, the second heat engine (HE2) is assumed to has higher efficiency than the first heat pump (HP1). Since W2 > W1, some work produced by the HE2 can be used for other work, which is denoted as W2 - W1. This concludes that the overall engine will be extracting heat energy from the low temperature reservoir and completely convert it to work (W2 - W1 = Q2 - Q2'), which violates Kelvin-Planck's statement.
What I'm having trouble is, I don't get how Clausius statement (heat doesn't flow spontaneously from cold to hot regions) is violated here? Can someone help?


text:
Band gap voltammetry sensors
I'm new in electrochemistry and I am leading with modification of sensors and analytes. I would like to know why the analyte voltammetric signal should be between the band gap of the modifier. If I run a DPV passing through the band gap limit do I lose my electrode? or modification?

Half of success of obtaining useful answers is providing sufficient relevant context along the questions.


text:
Are configuration interaction (CI) and configuration interaction method (CI-method) the same thing?
I am trying to study molecular quantum mechanics and I am quite confused because the material that I'm reading has a separate section called "Configuration interaction" and another separate section called "Configuration interaction calculations".
In the CI section, it is explained that in $H_2$-molecule, if the basis set consists of 1s-orbitals of the hydrogen atoms ($\phi_A$, $\phi_B$), the two electrons can occupy the molecular orbitals (MOs) according to four different configurations:
$1\sigma_g(1) 1\sigma_g(2)$
$2\sigma_u(1) 2\sigma_u(2)$
$1\sigma_g(1) 2\sigma_u(2)$
$2\sigma_u(1) 1\sigma_g(2)$
The last two configurations can form antisymmetric and symmetric linear combinations:
$\frac{1}{\sqrt2}(1\sigma_g(1) 2\sigma_u(2) - 2\sigma_u(1) 1\sigma_g(2))$
$\frac{1}{\sqrt2}(1\sigma_g(1) 2\sigma_u(2) + 2\sigma_u(1) 1\sigma_g(2))$
The four possible total wave functions are the first two configurations multiplied by spin wave function and the linear combinations multiplied by spin wave functions. The term symbols associated with these wave functions are  $^1\Sigma_g$, $^1\Sigma_g$, $^3\Sigma_u$, $^1\Sigma_u$, respectively. As you can see, two wave functions have the same symmetry, and as the internuclear distance $R \rightarrow \infty$, the states of the same symmetry converge to the same dissociation limit. This means that the energies of the states of the same symmetry never cross at any $R$ due to the non-vanishing matrix element between the states in the Hamiltonian. Due to the off-diagonal matrix elementrs, configurations of the same symmetry mix. This is called the Configuration interaction.
Now, the other section is the Configuration Interaction calculations, where it is explained that CI-method is a method to approximately solve the electronic SE. In CI-method, the exact wave function for any electronic state can be expressed as a linear combination of all possible n-electron Slater determinants arising from a complete set of spin orbitals. This is called the CI wave function. In practice, e.g. CISD is used, where only singly and doubly excited determinants are used in the calculation.
I kind of understand both of the things explained above, but I am very confused since I'm not sure if the two things explained above are the same thing? Or are they completely different topics but have the same name? Can someone explain, are these the same thing? Or is it that the CI-method just takes the Configuration interaction into account?

CI is within its limitations an exact theory. CISD is truncated and will have shortcomings. The latter is an approximation to the former. So the method and the theory are definitely not the same thing.


text:
Relative Concentrations of Reactants and Products at Equilibrium
I cannot understand what makes a reaction at equilibrium lie far to the right (large K) or far to the left (small K). What exactly makes the product or the reactant concertation larger than the other when the reaction producing the products (forward reaction) and the reaction producing the reactants (reverse reaction) have the same rate, and rate is change in concentration over time.

When I was reading to find an answer, I came across the definition of the equilibrium constant as a ratio of the rate constants of the forward reaction to the rate constant of the reverse reaction. Does that have some effect on determining the concentration which reaction species will be greater at equilibrium, given that at the end of the day, the rates of the reactions are the same. If the rate constant does have an effect, then what is the factor determining how large the rate constant at a given temperature is.

I just can't wrap my head around it, if the rates of the reactions are the same, (and obviously the initial concentrations of the reaction species have no effect on K), then there must be another factor affecting the concentrations of the reactants at equilibrium, but what is it?

The point is, the rates of the forward and backward reactions are equal at and only at the reaction equilibrium. If for $\ce{A <=> B}$ and $[\ce{A}] = [\ce{B}]$ isf the forward rate  (much) bigger than the backward rate, then K>1 ( K >>1 ) and vice versa. $$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = k_f[\mathrm{A}] - k_b[\ce{B}]$$ and similarly for $\ce{A}$, assuming the reaction is elementary.


text:
How to choose the solvent gradient for the flash column chromatography of a complex mixture?
When performing a flash column chromatography, it is advised to select an eluent in which the desired compound has an $R_\mathrm{F}$ of about $0.25$ in TLC.
However, if I want to separate and isolate all the compounds present in the mixture, or I do not know which is the desired product, it is advised to use an eluent gradient for the column. But what eluent should be selected to start with? Maybe an eluent in which the compound with the highest $R_\mathrm{F}$ (least polar in direct phase chromatography) has $R_\mathrm{F} = 0.25$? What should I consider when performing such chromatography?
I do a gradient from hexane to ethyl acetate until everything is separated. Is there a more efficient method that saves time and solvents?

That would be good start since you don't know which product is the desired one. To save time and chemicals, I'd buy an authentic sample and follow through.

I would do a complete gradient from 100% A to 100% B where A is something non-polar (hexane maybe) and B is something polar (e.g. EtOAc) to ensure I separated everything, and then a final wash with 10% MeOH in EtOAc


text:
Is cleaning silverware with tinfoil and salt dangerous?
Premise: I'm not a chemist and I've already had a look at this and this but they don't fully answer my question.
I have read that you can clean silverware using tinfoil, salt and water, though from what I understand, $\ce{H2S}$ is (or could be?) freed. Again, from what I understand, this a dangerous compound to breath. Now, is it still safe to clean silverware using this method? Is the amount of $\ce{H2S}$ freed dangerous?
I don't want to come across a "bleach + vinegar situation", which is why I'm asking here.

Many things are unsafe if you do them incorrectly, unfortunately. If you perform the reaction in a well ventilated place you reduce the risk effectively to near zero. The answer presumably  also depends on the amount of silverware and the state of the silverware.

I've used this method several times. You'll need to make sure your silver is in very good contact with the tinfoil so wrap it quite tightly. I do not recall ever smelling H2S from the system

The amount of sulfur in the tarnish is microscopic. 2. Much of it reacts with the Al, and is not released as a gas. Consider that there is cyanide bound in almonds and apple seeds. Eating them in reasonable amounts is usually not fatal. There have been exceptions.


text:
post-treatment of silicon surface after silanisation
I recently attempted to make a silicon oxide surface hydrophobic using trimethoxyoctylsilane and methanol as a solvent in ambient air conditions. I left the silicon in the solution for 24 hours, but after rinsing the surface with ethanol and pure water, the surface is still not clean. (I even rinsed it with methanol, water, and ethanol.)I have two questions:

Will I receive a hydrophobic surface if I reduce the silanisation time from 24 hours to just 5 minutes? I read somewhere that a maximum of 5 minutes is enough to achieve a hydrophobic surface.

How can I effectively clean the silicon after silanisation to receive a clean surface?

Surface should be treated before silanization.


text:
Calculating the enthalpy of reaction at non-standard concentration for reactions in aqueous phase
Standard conditions for a chemical reaction in the aqueous phase, for instance, a chemical reaction in solution, consider that reagents should be at a 1 M concentration. I was wondering how to calculate the $\Delta$H for other concentrations. Let's say we have the next reaction:
$$\ce{2HCl(aq) + Na2CO3(aq) -> 2NaCl(aq) + CO2(g) + H2O(l)}$$
It resumes to:
$$\ce{2H+(aq) + CO3^{2-}(aq) -> CO2(g) + H2O(l)}$$
To determine the $\Delta H^0$ we can perform:
$\Delta H^0$ = $\Delta H^0_{f}$ $\ce{CO2(g)}$ + $\Delta H^0_{f}$ $\ce{H2O(l)}$ - 2$\Delta H^0_{f}$ $\ce{H+(aq)}$ - $\Delta H^0_{f}$ $\ce{CO3^{2-}(aq)}$
Which results in: - 3 kJ/mol
But this is assuming standard conditions and thus 1 M concentrations for the reagents in the aqueous phase. How can I calculate the $\Delta H$ for other concentrations, for instance, when mixing 100 mL of 0.2 M $\ce{HCl(aq)}$ with 50 mL of $\ce{Na2CO3(aq)}$ 0.3 M?

Using Hess law and tabulated or measured enthalpies of dilution.


text:
Why is nucleus coupling neglected to explain electronic states but LS coupling only rises because of?
Here, with LS coupling I mean coupling between total orbital momenta and S total spin momenta and treating eletrons in light atoms with Russell-Saunders approximation. I found a bit paradoxical that LS coupling happens because of a positive nucleus charge Ze - and its dependence follows Z^4 - but the coupling between the spin momentum of the nucleus and the eletrons' is alltogether neglected in at least this approximation. My reference is Hollas, Modern Spectroscopy, 4th Edition, pg. 208, cap. 7 (Electronic Spectroscopy). That is, why is nucleus spin momentum weak?
Also, what are the evidences that support the efficiency in Russell-Saunders approximation for lighter atoms (ll coupling strong, ss appreciable, j weak) and jj- coupling for "a few states of heavy atoms" (j strong, jj appreciable, ll and ss weak)?
It is curious that the extent of the coupling j apparently inverts from light to heavy atoms, given these 2 approximations.

Nucleus spin momentum is not weak. It's magnetic momentum is, being reciprocal to mass. Angular momenta interact with each other via their magnetic momenta.


text:
Calculating concentration quotient and determining direction of reaction for equilibrium involving gases
I'm currently working on a chemistry problem involving chemical equilibrium, and I'm having some trouble figuring out how to calculate the concentration quotient $Q$ and determine the direction in which the reaction will proceed.
At a certain temperature, when nitrogen dioxide $(\ce{NO2})$ is in equilibrium with nitrogen monoxide $(\ce{NO})$ and oxygen $(\ce{O2}),$ the equilibrium constant is $\pu{0.50 M}.$ A container with a volume of $\pu{0.50 dm3}$ initially contains $\pu{0.020 mol}$ of $\ce{NO2},$ $\pu{0.10 mol}$ of $\ce{NO},$ and $\pu{0.010 mol}$ of $\ce{O2}.$
Now, I need to calculate the concentration quotient $Q$ to determine whether the system is at equilibrium or not. If $Q$ is greater than the equilibrium constant, the reaction will proceed to the left, and if $Q$ is less than the equilibrium constant, the reaction will proceed to the right.
However, I'm not sure how to calculate $Q$ for this particular system. Could someone please walk me through the steps involved? Also, if you could let me know which direction the reaction is likely to proceed based on the value of $Q,$ that would be great.

You have merely just verbosely described the task in the hand. But you have not described what you have tried to do. Like learning what is the an equilibrium constant,  a reaction quotient, what is their relations and what the 0.50 M is hinting to you.


text:
What is the energy of activation of uncatalyzed ATP hydrolysis reaction?
If we ignore all the biological significans for a while, ATP hydrolysis, the reaction:
ATP + H2O -> ADP + Pi
is still a chemical reaction that must go through transition state to yield product. This is no doubt a nucleophillic substitution reaction that, according to all the paper I've found, occurs by SN2 mechanism. But what's totally missing is the energy of activation of the uncatalyzed reaction.
There are a lot of paper on the mechanism of ATP hydrolysis in the active site of this enzyme and that motor. There are a few paper on uncatalyzed reaction and various nucleophile. A few paper does estimate the activation energy of ATP hydrolysis by various enzyme and motor proteins. But I haven't found a single paper estimating the Eact of the uncatalyzed reaction.
My failure apart, what is the energy of activation of this reaciton without the aid of any enzyme or other catalyst? Did nobody make any estimation?

https://pubs.acs.org/doi/pdf/10.1021/ja01098a020


text:
polymerized Dimethylbutadiene or "methyl rubber"
"Methyl rubber", polymerized Dimethylbutadiene, is the first synthetic rubber, but it falls behind polyisoprene having inferior shear strength & oxidizes more easily.
I'm looking into the isomeric composition of these methyl rubber products. Particularly, I'm looking into varieties mostly composed of isomers cis-1,4 & trans-1,4. Typically, ones that have been polymerized with Anionic Chain Polymerization. Have the physical properties of these varieties more of cis-1,4 & trans-1,4 been characterized?


text:
Zinc organic-water solvent extraction (Hydrometallurgy)
Consider a biphasical system consisting of an aqueous phase and an organic phase.
Consider the following equation:

$\ce{Zn^2+(aq) + 2HR(org) <=> ZnR2(org) + 2H+(aq)}$

Let [ $Zn^{+2}$ ]$^{0}$$(aq)$ = 0,1 $\pu{mol L-1}$ be the initial concentration of ionic zinc in the aqueous phase.
Let $\ce{[HR]}$$^{0}$$(org)$ = 0,5 $\pu{mol L-1}$ be the initial concentration of phosphinic acid in the organic phase.
Consider that the volume of the organic phase $\ce{V2}$ is twice the volume of the aqueous phase $\ce{V1}$ ($\ce{V2} = \ce{2V1}$ )
Consider that the equilibrium constant of the equation above is $K = 10^{-3,259}$
Let $D = \frac{[Zn^{+2}]}{\ce{[ZnR2]}}$ be the distribution coefficient of the equilibrium above and $E = \frac{100D\ce{V2}}{D\ce{V2}+\ce{V1}}$ be the extraction coeficcient. (note that since $\ce{V2} = \ce{2V1}$ , $E = \frac{100D}{D+0.5}$ )

Given the information above,plot the graphic of E versus the ph of the aqueous phase.
Suggestion: Use $\ce{V1} = 1L$ and $\ce{V2} = 2L$

My attempt at it:
Assuming $\ce{V1} = 1L$ $\ce{V2} = 2L$ and assuming that $\ce{n0,n1}$ are the initial mols of ionic zinc and phosphinic acid respectively,we obtain that $\ce{n0} = 0.1 mol$ and $\ce{n1} = 1 mol$.Also assume that $x$ is the quantity (in mols) of $\ce{Zn^2+(aq)}$ that reacted.Noting that we are analysing the equilibrium for each $\ce{[H+]}$ ,after we draw the reaction table we obtain:
$K = \frac{2x\ce{[H+]^{2}}}{(0.1-x)(1-2x)^{2}}$ and $D =\frac{0.5x}{0.1-x}$ .If we consider $2x <<<< 1$ we obtain that $K \approx \frac{2x\ce{[H+]^{2}}}{(0.1-x)(1)^{2}}$ $=>$ $K \approx 4D\ce{[H+]^{2}}$ and with that we obtain a relation between D and the ph of the aqueous phase.After that I was able to plot the graphic of E versus the ph.
Did I answer it correctly?Any help/correction would be kindly appreciated.Thank you in advance.

Ph is probably pH. See also Upright vs italic
(element and unit symbol upright)  and $n = \pu{1 mol}$ as $n = \pu{1 mol}$ // See CH SE META post about policy for "Am I right?" questions


text:
What are the differences between the different Rate Equations?
I currently am about to finish general chem 2 so my knowledge is still basic.
There are different kinetics and rate equations and im not quite sure I know the differences.

Breaking rate of a reactant: ${\frac{\ce{-[R]}}{Δt}}$.

Formation rate of a product: ${\frac{\ce{[P]}}{Δt}}$

$v$ = $\ce{-\frac{1}{a}}\frac{\ce{d}\ce{[A]}}{\ce{d}t}
= \ce{-\frac{1}{b}}\frac{\ce{d}\ce{[B]}}{\ce{d}t}=
\ce{\frac{1}{p}}\frac{\ce{d}\ce{[P]}}{\ce{d}t}
=\ce{\frac{1}{q}}\frac{\ce{d}\ce{[Q]}}{\ce{d}t}$

Rate = ${k\ce{[A]}^\ce{{n}}} \ce{{[B]}^\ce{{m}}}$.....

If I remember and understand correctly, equations 1 and 2 are in regards to the initial concentration, so they're the rate at the beginning of a chemical reaction... I think..
If so, does that mean equation 3 is for any given point in time, just insert the concentration at given time?
Can anyone please explain what are the differences and when to use what?
Thanks!

Welcome to CH SE site!
Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of at least essential parts.
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Note that, simplified, anything in italic is considered as a variable or scientific constant.  // For more details: Upright vs italic

You should mention that your expressions are only valid in an equation written $\pu{a \mathrm{A} + b \mathrm{B} -> p \mathrm{P} + q \mathrm{Q}}$

For reaction
$$\ce{a A + b B -> p P + q Q}\tag{1}$$
is
$$r_\textrm{A} = -  \frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t}\tag{2}$$ resp. $$r_\textrm{P} =   \frac{\mathrm{d}[\ce{P}]}{\mathrm{d}t}\tag{3}$$
the consumption rate of the reactant $\ce{A}$, resp. the production rate of the product $\ce{P}$.

$$r = -\frac{1}{a}\frac{\mathrm{d}\ce{[A]}}{\mathrm{d}t} = -\frac{1}{b}\frac{\mathrm{d}\ce{[B]}}{\mathrm{d}t} = \frac{1}{p}\frac{\mathrm{d}\ce{[P]}}{\mathrm{d}t} = \frac{1}{q}\frac{\mathrm{d}\ce{[Q]}}{\mathrm{d}t} \\ = \frac{r_\textrm{A}}{a} = \frac{r_\textrm{B}}{b} = \frac{r_\textrm{P}}{p} = \frac{r_\textrm{Q}}{q}\tag{4}$$
is the reaction rate, being the absolute value of the reaction rate of any reactant or product, divided by the respective stoichiometric coefficient.

$$r = k[\ce{A}]^a[\ce{B}]^b\tag{5}$$
it the reaction rate expressed by the reaction kinetic constant and concentration of reactants. Note that this is true only for single step elementary reactions, otherwise things get complicated by reaction chains and loopbacks.
If there is equation $y = x$ and $y = z$, then $x = z$, because we can substitute $y$ in the latter by $x$ from the former.
Therefore:
$$r = -\frac{1}{a}\frac{\mathrm{d}\ce{[\ce{A}]}}{\mathrm{d}t} = -\frac{1}{b}\frac{\mathrm{d}\ce{[\ce{B}]}}{\mathrm{d}t} = \frac{1}{p}\frac{\mathrm{d}\ce{[\ce{P}]}}{\mathrm{d}t} = \frac{1}{q}\frac{\mathrm{d}\ce{[\ce{Q}]}}{\mathrm{d}t} = k[\ce{A}]^a[\ce{B}]^b \tag{6}$$

You can determine

the reaction rate from the reaction rates of reactants or products,
the reaction rate from the concentrations of reactants or products,
the reaction kinetic constant $k$ from both rates and concentrations.

Thanks, it was trivial mistake.

Thank you so much.
So just to make sure, the above equations are used to express the rate of a reaction by using different data, i.e.  stoichiometry or concentration for example.
So when im asked to calculate the initial rate, I can use equations 4 and 5, and when asked to calc concentrations at specific time, I can use equation 5 (or 4, but i'd need the difference in both time and concentration)
Thanks again!

It is like the equation z=x.y . If you know z, you know x.y.  If you know x and y, you know z. If you know z and ý, you know x.

The second equation is the definition of reaction rate. The symbol used for rate sometimes is $v$, or it is written out as rate. The third equation is the rate law, telling us how the rate changes with concentration (at a given temperature).
The first set of equations is unfamiliar to me, and don't look like a rate as given.

the first ones are the average rates of breaking and forming of components.
So as Poutnik commented, are the reaction rate and rate law equal?

The quantities "rate" and $v$ are equal. The rate law is a different equation than the equation defining rate.