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Best way to bond polyethylene to to aluminum?
I am a quality engineer looking for some chemistry expertise that can hopefully assist in solving a problem we are having at our company. We are a water filtration company and our location produce the filters used for our machines which are in restaurants and stores around the world. For simplicity, the filter is comprised of an aluminum can that we extrude, wash, add a polyethelyene coating (liner), then add the filter material and a cap with an O-ring. We often get reports from the field of filters that are leaking during use. Upon return, we open the filter up and see that the polyethelyene liner is not adhering to the inside of the aluminum can and has started to tear, creating a leak path past the O-ring.
Our current focus is on the cleanliness of the inside of the aluminum can. Much like anything else, the more dirt on the surface, the less likely anything will adhere to it. We are trying to improve cleanliness to increase adhesion.
Another thought I had was when we heat the cans to around 500F during the polyethelyene coating process, the contraction of the can after the thermal expansion when we quench is causing the liner to detach.
What I am asking from a chemist is, what can improve the adhesion of the polyethelyene to the inside of the aluminum can? Are we chasing the wrong thing with cleanliness? Is my thermal expansion theory completely off?

Maybe you should try to avoid quenching and to cool the pieces of aluminum coated with polyethylene as slowly as possible.

If you want to fix this completely, then you'd need different polymer altogether, I'm afraid. PE isn't something that's gonna stick to Al. It difficult to even dye, much less bind to stuff. On the other hand, if you used epoxy...

Have you tried anodizing the aluminum? You'd need to test, but the increased surface porosity and roughness should give the plastic a larger surface to which it could adhere.


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What is the real cause of burn-in effect on OLED(organic light emitting diode) displays and how can burn-in effect be solved?

In the picture above you can see how some logos have permanently burnt-in. I wonder if it has to do with the fact that the materials are "organic" as the name suggest, or because current oled displays use a combination of fluorescent and phosphorescent materials which causes the uneven aging.
"In OLEDs using fluorescent organic molecules only, the decay of triplet excitons is quantum mechanically forbidden by selection rules, meaning that the lifetime of triplet excitons is long and phosphorescence is not readily observed. Hence it would be expected that in fluorescent OLEDs only the formation of singlet excitons results in the emission of useful radiation, placing a theoretical limit on the internal quantum efficiency (the percentage of excitons formed that result in emission of a photon) of 25%.[4]
However, phosphorescent OLEDs generate light from both triplet and singlet excitons, allowing the internal quantum efficiency of such devices to reach nearly 100%.[5]"

"[O]rganic materials that make OLED TVs’ self-emissive displays could decay under prolonged exposure to static image elements," per https://www.whathifi.com/advice/oled-burn-in-should-you-be-worried-about-it-and-how-can-you-prevent-it . That is, actual chemical decomposition occurs from localized heating or other cause.

@DrMoishePippik The decay is usually photochemical. If there were localised heating, that would make it worse, but that isn't the primary cause.

The simple explanation is that some of the non-emissive pathways damage the emitting material
There are several ways to turn electrical energy into light. Old monitors and TVs used phosphorescent minerals that glowed when hit by electron beams. They suffer burn-in if the same image is displayed for an extended period.
OLEDs use a variety of organic materials (often specialised polymers, but sometimes small organic molecules) which glow when an electrical current is applied across them. They suffer burn-in too.
Some screen types do not. LCD displays use liquid crystals to "filter" light from a backlight source. They don't (usually) suffer burn in.
Why are they different?
LCDs switch polarisation driven by very small voltages and the process is not a chemical reaction in any sense, but is driven by small physical changes in how the liquid crystals are arranged in the liquid crystal layer. The energies involved are tiny and the rearrangement effectively completely reversible so no lasting changes occur even when the same image is held for a long time.
But, in the other two cases, there is a photochemical reaction occurring with much higher energies involved. This is more obvious in a CRT display driven by high-speed electrons hitting a phosphor. The energies are significant. What happens is that electrons in the phosphor structure are bounced to much higher energy levels and then emit light when they decay back to the ground state. But these reactions are never 100% perfect, and there may be a very small probability that, instead of simply falling back to the ground state and emitting light, a different reaction pathway disrupts the structure of the phosphor, damaging its future ability to emit the right colour of light. Even when these alternative pathways are very improbable, they can eventually (because they may be irreversible) damage the phosphor, yielding a burnt-in image in the parts of the screen that see the most action.
Exactly the same problem occurs in OLEDs but at lower energies (but then OLED materials are usually a lot more fragile than CRT phosphors). The photochemical reactions are driven by the injection of electrons or holes into the materials, bouncing some of the electrons in the material into higher energy levels which then decay, emitting light. But, again, the pathway the leads to the emission of light is not exactly 100% efficient, and the small number of alternative paths available to the excited molecules are often damaging to the molecular structure and slowly reduce its ability to emit the right wavelength of light. That the energy involved matters is illustrated by the observation that it is much harder to make good blue OLEDs (higher energy light=more energy required to excite the electrons) than red OLEDs (lower energy light...)
The bottom line is that the light emission from CRT phosphors or OLED compounds is a photochemical process and they are never perfectly 100% efficient, and the side reactions eventually damage the material.
Fixing the problem perfectly is impossible. But finding better materials (with better quantum efficiency or fewer side reactions) is an important task. Of course, other strategies are also possible. Screen savers were invented when CRT computer displays were common and work to prevent the same image being displayed for extended periods, reducing the scale of burn-in. Many OLED manufacturers have built in similar ideas to the software running their monitors and TVs. A common strategy is to regularly shift the image by a few (not very noticeable) pixels so the same pixel does not always see the same part of the image.
I suspect both the software and the choice of OLED materials have led to big improvements in the last decade as the burn-in problem seems much less severe now than it used to be.

Since light emission from OLED compounds is a "photochemical process" which is why burn-in occurs, does the same apply for microled displays which instead use inorganic leds made from gallium?

@AquaNeon it is certainly possible that micro LEDs will see similar effects. And there is little evidence that other LED-based displays see the effect, possibly because their simpler solid-state structures don't have so many opportunities for side-reactions. It might be a matter of degree where the effect is small enough not to be noticable in normal lifetimes.

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Why is boiling point of hydrogen greater than of helium?
If we compare the boiling point of hydrogen and helium using molecular weight criteria (both have London dispersion forces as intermolecular forces of attraction because both are non polar then the one which have which have higher molecular weight will have higher intermolecular attraction forces) then helium should have greater boiling point, but if we see the boiling point data for H2 and helium then we found that H2 have its boiling point as approximately 20 Kelvin while He will have approx 4.3 Kelvin.

Already addressed here: https://chem.libretexts.org/Bookshelves/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/01%3A_Atoms/1.8%3A_Interactions_Between_Helium_Atoms_and_Hydrogen_Molecules

Higher molecular weight is not the determining factor. Rather the number of electrons that could be polarized and the volume of space over which they may be polarized are the key factors in dispersion forces.
For species with similar structures higher molecular weight goes along with more or larger atoms, thus more electrons and greater polarizability; but "monatomic" and "diatomic" are not really similar structures. Compared with helium, hydrogen has as many electrons (two), and the presence of two atoms instead of one allows an opportunity for polarization over more volume. So hydrogen will have more dispersion forces.

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Why is the relationship between molar mass and boiling point for alkanes a square root relationship?
I plotted the boiling point of alkanes against their molar mass and found the following trend:
I expected the trend would be linear, since (based on my understanding of what I've learned in class so far) for non polar molecules like alkanes, the boiling point depends on London dipole forces acting between them, and these are proportional to surface area. As such, I expected that as CH$_2$ were added, the surface area of consecutive alkanes would increase linearly with their mass and thus the London forces and boiling point would also increase linearly with mass.
However, as can be seen in the graph above, the relationship between boiling point and mass seems to roughly follow a square root trend.
Why is this the case?

Just for fun, when you plot boiling point vs. number of carbon atoms what do you get? Hopefully a straight line. These mathematical musings currently do not have a solid base, however such relationships are good for predictions. However square root relationship is not universal. Boiling point or melting point is among the hardest things to predict.

Also read about "Correlation does not imply causation" on Wikipedia https://en.wikipedia.org/wiki/Correlation_does_not_imply_causation

Also consider this: surface area does not grow linearly with extra carbons. Try calculating the actual surface area and plotting that against BP.

Is the units of your molar mass g/mol? If so, where did you get data for boiling points of such large alkanes? In the range of hundreds to thousands g/mol, polymer phenomena starts to become relevant and one might expect very different behavior than for the lighter alkanes. For instance, polyethylene (alkane with MM > 1000 g/mol) will never boil, and instead degrades around 650 K. Similar effects might be relevant at the scale you're considering here, unless you've been able to rule that out.

@matt_black, can you please provide a reference for your statement? According to data from Zhao et al. (2003), the van der Waals surface area does grow linearly for linear alkanes  (at least up to $n=10$).


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Software for running dynamics simulation of polymeric and mixed biomaterial (e.g hydrogels)
I was working on the development of a soft solid hydrogel having polymeric chains and molecules of different compounds dissolved in Phosphate Buffered Saline solution. There is a sub-study regarding the interaction of two samples of the same hydrogel placed in contact with each other.
I want to study the changes in this interaction behaviour upon changing the ratios of constituent compounds in the hydrogel through dynamics simulations.
Can anyone please suggest the best or appropriate software or softwares for this purpose? (Several aspects or combinations can be considered including high precision and accuracy, visual simulations, ease of use, molecular and material dynamics, etc.)


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Possible experimental errors and how to improve them
I'm doing an experiment at home where I determine the relative atomic mass of magnesium by collecting the hydrogen produced from its reaction with hydrochloric acid. I ended up getting around 26, but as I know the relative atomic mass of Mg is around 24.3. So I would like to ask whether the following errors can be improved in any way:
1.When I removed the magnesium oxide coating on the surface of the magnesium ribbon, I knew it would affect the result because I couldn't remove all of them.Is there a way to remove all the magnesium oxide coating without having the magnesium metal react?
2.I used a gas syringe to collect the hydrogen. Would applying vaseline to the syringe and the place it connects with a tube actually work （as the syringe moves?)
3.I assumed that the molar volume was 24dm3 when it wasn't exactly room temperature. I looked online and saw an 'ideal gas law'. Could this be applied without having to add extra complicated apparatus to collect data? (pressure?)

Have you made a correction for the water vapor pressure in the collected gas ?

Um what is that?

Focus first on eventual conceptual errors (what and how to measure and calculate) before experimental errors (how well is was done).

If you want to use $pV = nRT$  in your experiment, you must correct your pressure measurements for the content of water vapor in  the gaz produced by hydrochloric acid aqueous  solution. Suppose the pressure measured on the classroom barometer is maybe $98 000$ Pa, and the temperature is $20°$C. Now, at $20$°C, the water pressure is $2340$ Pa, from the Tables. So the pressure $p$ to be introduced into the formula $pV = nRT$ is : $p = 98 000$ Pa - $2340$ Pa = $95 660$ Pa. With this pressure, the number of moles $\ce{H2}$ is : $n(\ce{H2}) =  pV/RT$,  with $p = 95660$ Pa.


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What is the mechanism for nitration of cinnamaldehyde?
The group added on to the benzene ring(-CH=CH-CH=O) should be an electron withdrawing group, yet the sources I found online suggest an ortho-/para- product rather than the expected meta product. Is there an explanation for this and preferably a mechanism for the reaction?

I regret that the OP did not provide a link to the o,p-nitration of cinnamaldehyde. HotLasagna's surmise that m-nitration should apply is well-founded. I did a reaction search on Chem. Absts. for the nitration of cinnamaldehyde without success. Because cinnamaldehyde is a vinolog of benzaldehyde, the latter should serve as a model for the former. The literature confirms the preferred m-nitration of benzaldehyde.
Continuous flow nitration[1] ($\ce{HNO3/H2SO4}$) of benzaldehyde gives high yields of the o:m-isomers (1:4) without mention of p-nitrobenzaldehyde. Two additional methods employ trifluoromethanesulfonic anhydride (triflic anhydride) in conjunction with tetramethyl ammonium nitrate or ethyl ammonium nitrate. In the former case[2], the o:m:p ratio was 31/63/6 while in the latter study[3] the o:m:p ratio was 27:70:3.
Contrary to the rational suggested by @ M. L., the determining factor is (would be) the destabilization of the positive charge by the aldehyde and vinylogous aldehyde groups in 1 and 3, respectively, whereas the m-nitration resonance structures 2 and 4 avoid this problem. (Not all of the canonical resonance structures are shown.)

A. A. Kulkarni, V. S. Kalyani, R. A. Joshi and R. R. Joshi, Org. Proc. Res. & Dev., 2009, 13, 999.
S. A. Shackelford, et al., J. Org. Chem. 2003, 68, 267.
G. Aridoss and K. K. Laali, J. Org. Chem. 2011, 76, 8088.

Nitration of Cinnamaldehyde with acetyl nitrate giving o product  http://www.orgsyn.org/demo.aspx?prep=CV4P0722  The references to the Org Syn mention this procedure giving a mixture of o and p  Davey and Gwilt, J. Chem. Soc., 1955, 1384. No mention is made of m product

@Waylander: Well, so much for Chem Absts. and reasoning by analogy! The Davey/Gwilt paper does not mention which nitrocinnamaldehyde is formed---at least to my reading of it. The yield is of o-nitrocinnamaldehyde is modest after recrystallization. I wonder what other isomers might be in the mother liquors. What do you think about an internal delivery of -NO2 by some derivative of the -CHO and intermediate isomerization of the double bond to cis? A 7-membered TS and perhaps too far fetched.

An LCMS of the reaction mixture would certainly be interesting. I suspect that different regioisomer mixtures are obtained from different conditions. The acetyl nitrate conditions could possibly give rise to the intermediate you suggest.

@user55119: I agree with you. I think they were looking for only o-nitro product and got that. Other 50+% should be down the drain! :-)

Deciding how a group will affect an aromatic ring in electrophilic substitution comes from analyzing the possible resonance structures. With cinnamaldehyde, we can draw the possible forms of the para (which should be the same as ortho) and meta addition:

Clearly, the para/ortho configuration allows for an additional resonance form to take place utilizing the group added to the benzene ring. This higher stability of the resonance with para/ortho addition is likely why electrophilic addition to cinnamaldehyde favours those two configurations.


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Which energy increases when compression work is done on a gas?
When a gas is compressed using external pressure (piston-cylinder assembly) we say work is done on the gas and it increases the internal energy of the gas.
Where does this work mainly go ? Does it increase the Kinetic energy of the molecules (work energy theorem), does it increase the potential energy of the molecules since we are doing work against internal conservative intermolecular forces of the gas, or it goes to something else ?
Also can temperature of a gas be increased by work only ?

What happens depends upon whether the change is done isothermally or adiabatically.
Suppose the gas is ideal then the internal energy $U$ depends only on temperature and is independent of the volume or pressure (by definition) and therefore for an isothermal expansion/compression $dU=0$ and so by the first law $dq=-dw$ where the work is $dw=-pdV$ and for an ideal gas $p=RT/V$. As the internal energy and temperature is unchanged, so is the kinetic energy. In a real gas at low pressure intermolecular interactions can be ignored and this behaves as if an ideal gas. (If this is not the case then $p$ would have to be given by, say, the van-der Waal equation which greatly complicated any calculation, but some work will be needed to compensate for the intermolecular forces).
In the adiabatic case no heat leaves or enters the system (i.e. no heat exchanged with surroundings) and therefore and expansion or compression is done at the expense of the internal energy and the temperature will change. The change in internal energy is given from the first law as $dU=dw$. On compression the temperature rises. If the gas is one of atoms then there is only kinetic energy which increases. In a gas of molecules, kinetic energy also increases but additionally rotational /vibrational energy levels can be populated according to the Boltzmann distribution.
The original experiments by James Joule (1843, in a brewery in Manchester, England) showed that mechanical work results in an increase in temperature; "Wherever mechanical force is expended, an exact equivalent of heat is always obtained."

Where does this work mainly go?

It speeds up the gas particles as the collide with a wall moving toward them. A quantitative description (for a monoatomic or diatomic gas) is given here:
https://en.wikipedia.org/wiki/Adiabatic_process#Example_of_adiabatic_compression
The temperature of the gas increases, so the pressure goes up more than the compression ratio when the process is adiabatic.

[..] does it increase the potential energy of the molecules since we are doing work against internal conservative intermolecular forces of the gas, or it goes to something else ?

The particles are far away from each other, so intermolecular forces can be neglected (in an ideal gas model, there aren't any except for the interactions with the wall).

Also can temperature of a gas be increased by work only?

Yes, imagine a giant stir bar bumping into the gas (shaft work - this is more commonly invoked for liquids, see this historical account, but should work for gases as well).


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Absorption wavelength of nickel (II)
One of my students is trying to study a nickel(II) solution, prepared from nickel(II) sulfate, which is green. In solution it is also green, thus I would expect it to absorb red light. Here is a test run in a colorimeter:

with each signal corresponding to 430, 470, 565 and 635 nm, respectively. The absorption spectrum of nickel (II) sulfate is shown below:

What I don't understand is why 470 nm shows the greatest absorbance in the colorimeter, if in the UV-vis spectrum it quite clearly shows it should absorb more at 635 nm?

Looks rather like max at 400 nm, not 470 nm. Be aware of difference between radiometric ( objective ) and photometric ( eye sensitivity aware ) parameters. Eyes are much more sensitive to some 650-720 nm than to 400 nm. Absorbance at region of low eye sensitivity is less significant, as its influence on subjective light attenuation is low.

What is the meaning of the absorbance measured versus the time ? The time !!! Does it change with the time ?

It doesn't really change with time, no.
@Poutnik, max is indeed at about 400nm, which is why I would expect the colorimeter to absorb more at 430m, not 470. The colorimeter spectrum shows a higher absorbance at 470nm.

The spectrometer irradiates each wavelength alone so the green light is always transmitted, close to 100%T or zero A. The absorption bands are measured and are used analytically. The colorimeter passes the complete visual spectrum so what you see predominately are the wavelengths that are passed by the sample aka its color. This means that increasing concentrations of the material pass selectively more light of its color and the color becomes more intense. More green more nickel. The intense color means that the absorption bands, the light that you do not see, are relatively strong.  If you shine only green light on it in a darkened room All concentrations would look the same, possibly they would all look white since the transmission is 100% and our brain can interpret 100% as white.
A story. I was going to bleach a photograph using potassium dichromate, a bright orange solid. I weighed some out on a watch glass and went into the dark room placed the watch glass on the counter turned on the orange safelight and turned off the room lights and looked for the dichromate. It WASN'T there! There was a watch glass with a bright white chemical in it.  Quickly turned the room light on and there was the orange chemical. One must be very careful when working with selective frequency or monochromatic light sources such as Lasers. What you think you see may not be what is really there.


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Is 25% aqueous ammonia solution considerably less safe than concentrated HCl
I plan to sterilize and clean some deer bones for display in my amateur chemical lab. The needed concentration of ammonia is only about 2,5% but it would be significantly cheaper to buy more concentrated solution, and dilute it myself. I have worked with NaOH, and concentrated HCl before, but I'm wondering will this be a bigger safety concern, and should I risk it just to save some money. I will obviously be working outside, with a fan running and wearing proper PPE (faceshield, googles, nitrlie gloves and a lab coat)

Well, no, I wouldn't say so. Why ask that?

Even HCN can be safe and water dangerous, depending on circumstances. Concentrated ammonia with proper handling of liquid volatile chemical is rather annoying due its famous smell than dangerous. I would suggest to partially dilute it as an intermediate stock solution.


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What is mol % and how do you calculate the mass of a compound needed to be a certain mol % relative to another compound at a particular volume?
I know that there is some stoichiometry involved, but I am lost here. I know you need the molecular weight of both compounds and that's about it?

Is your mixture undergoing a reaction or not?

Always try to find answers on your own in offline and online resources, before asking.  It will give you more than just waiting to receive the answer.  And it will not force your  question to look like if you have not done so.

Assuming we have a binary mixture, or equivalently, a mixture that only involves two different components $A$ and $B$, the total mass of the mixture is simply the sum of the individual masses:
$$m=m_A+m_B$$
The same logic applies to the total amount of the mixture:
$$n=n_A+n_B$$
When you divide the mass of $A$ and $B$ (separately) by the total mass of the mixture, you obtain the mass fraction of $A$ and $B$ in the mixture:
$$X_A=\frac{m_A}{m}\quad\quad X_B=\frac{m_B}{m}$$
When you divide the amount of $A$ and $B$ (separately) by the total amount of the mixture, you obtain the mole fraction of $A$ and $B$ in the mixture:
$$Y_A=\frac{n_A}{n}\quad\quad Y_B=\frac{n_B}{n}$$
In a binary mixture, the sum of fractions of $A$ and $B$ is equal to 1:
$$X_A+X_B=1$$
$$Y_A+Y_B=1$$
The molar mass of a binary mixture can be calculated with either the mass fractions or mole fractions, and the molar masses of each component:
$$M=\left(\frac{X_A}{M_A}+\frac{X_B}{M_B}\right)^{-1}$$
$$M=Y_A\;M_A+Y_B\;M_B$$
Finally, if you want to convert a mass fraction into a mole fraction or vice-versa, you can use these formulas:
$$X_A=Y_A\;\frac{M_A}{M}$$
$$X_B=Y_B\;\frac{M_B}{M}$$
$$Y_A=X_A\;\frac{M}{M_A}$$
$$Y_B=X_B\;\frac{M}{M_B}$$
Note that the "mol percent" you're referring to is the mole fraction multiplied by 100, and neither fraction (mass or mole) depends on volume, assuming no reaction is taking place.
You should be able to solve your problem by using these formulas.


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Lattice energy as a factor that helps determine which charge the atoms take on?
I do not quite understand the following line of reasoning:

Why does Na not form $\ce{Na^{2+}}$ ions? To obtain $\ce{Na^{2+}}$  ions, the first and second ionization energy must be applied: 496 + 4563 = 5059 kJ/mol. The lattice energy of an imaginary "$\ce{NaCl2}$" would have to be in the same order of magnitude as for $\ce{MgCl2}$ (-2525 kJ/mol).

So far everything is understandable...

This amount would not be sufficient for ionization, the removal of an
electron from an ion with noble gas configuration costs too much
energy.
Why can magnesium form $\ce{Mg2+}$ ions? The magnesium atom reaches the noble gas configuration when two electrons are donated. The sum of the first and second ionization energies for magnesium is considerably lower than for sodium: 738 + 1450 = 2188 kJ/mol. The lattice energy of -2525 kJ/mol is high enough to provide the energy needed for ionization.

I don't understand why the lattice energy released has to be sufficient for ionization (at all why any ionization should even take place here). I wonder in general where the connection is.
One is the heat release when positive and negative ions in the gas state are combined to form a crystal and the other is just the energy to knock out electrons. Why does the lattice energy have to be high enough now to apply the first, second etc. ionization energy?
(In this section and before in the book the entropy or Gibbs energy was not introduced yet, the author himself says: "The correct quantity to be considered here is the free enthalpy instead of the enthalpy. However, we do not commit a big mistake if we calculate with enthalpy in this case".)
That's the information I'm taking out:
$$\ce{Na -> Na^{2+} + 2e-}  ~~~ \Delta H = \pu{5059 kJ/mol} \\ \ce{Na^{2+}(g) +2Cl-(g) -> NaCl2 }  ~~~ \Delta H \approx \pu{-2525 kJ/mol}$$
Then it is said that the amount of the lattice energy is not sufficient for the ionization of Na. This is true, after all 2525 kJ < 5059 kJ and now? Why should the lattice energy be used for this? I don't know what the author wants to make clear now, the jump to the original question why Na does not form $\ce{Na^{2+}}$ ions does not work for me. Why should the lattice energy even be responsible for the ionization of Na?

The enthalpies for the two steps, ionization and lattice formation, must add up to a sizable negative number for the atoms to associate from the gas phase and overcome an entropic penalty.

Have you read about Hess's law and the Born-Haber cycle? https://en.wikipedia.org/wiki/Born%E2%80%93Haber_cycle

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.

Note there was an error in second equation (Cl should be Cl-) and state of matter is not specified in the first (presumably all are gases).

It might help if you tell us what question you think is trying to be answered by the text you quote . Also where is it from?

@IanBush The section is from the textbook "Mortimer, Basiswissen der Chemie, 13th edition, chapter 7.6 - Types of ions". I think that the author wants to explain with the help of the lattice energy why Na does not exist as Na2+. But I don't understand where exactly he means that, Na2+ can exist with enough energy, does he mean why it doesn't exist in an ionic lattice? (1/2)

Do I then have to imagine it as "dominos" (in an environment of only one Na+ and Cl- and the rest is Na and Cl is enough if once Na+ + Cl- -> NaCl becomes and thereby releases energy which can ionize all others and which then again become NaCl and so on)? (2/2)

The author is trying to explain why $\ce{NaCl2}$ doesn't form given that  $\ce{MgCl2}$ does, where Mg is just one electron away from Na on the periodic table. The reaction being considered is
$$\ce{X(g) + Cl2(g) -> XCl2(s)} \tag{1}$$
Thermodynamic arguments (having to do with the stability in terms of energy of products relative to reagents) or kinetic arguments(having to do with energetic barriers to reaching the products from the reagents) can be used to explain why compounds don't form as expected.
Here the author uses a thermodynamic explanation. Given the apparent similarity of $\ce{NaCl2}$  and $\ce{MgCl2}$ the failure to form $\ce{NaCl2}$ is traced to the energetic cost of forming it.
And why is $\ce{Na^{2+}}$ so costly to form? Because sodium has a large second ionization energy.
How do you do energy accounting to show this? You use Hess's law, which allows splitting a reaction into smaller steps through intermediates. Since enthalpy H is a state function, the path that takes you from reagents to products doesn't matter. To obtain the change in enthalpy for the overall process just add the changes for the individual steps.
In this case the steps are
$$\ce{X(g) + Cl2(g) -> X(g) + 2Cl(g)}\\\ce{X(g) + 2Cl(g) -> X^{2+}(g) + 2Cl-(g)}\\\ce{X^{2+}(g) + 2Cl-(g) -> XCl2(s)}$$
The first step is the atomization of $\ce{Cl2(g)}$, the second the ionization of the atoms in the gas phase, the third the formation of the lattice. The total enthalpy change for reaction (1) at top is just the sum of the enthalpies for these individual steps.
The enthalpies for the steps, atomization, ionization and lattice formation, must add up to a sizable negative number for the atoms to associate from the gas phase and overcome an entropic penalty.
There is some subtlety to this. An ionization step is introduced as a step on the way from gas phase atoms to the solid as a way to explain the stability of ionic solids and an additional justification for the proposed structure of such solids as composed of lattices of ions. You might have skipped that step and gone from the gas phase atoms to the lattice, without incorporating an ionization step, but that would not have provided much insight into the fundamental question, why does sodium behave differently?
Born-Haber cycles are a method of accounting (a "book-keeping" tool) but also have explanatory power, in that like an accounting sheet they can point out where there might be important steps responsible for differences between processes. Ionization does occur, and a lattice does form, but the events might not necessarily occur in the order suggested by a Born-Haber cycle. These events may occur simultaneously, for instance. In that case, the energy for the simultaneous event is the sum of the energies for the step-wise events.
There is also the issue of assuming that the lattice enthalpy for formation of the hypothetical $\ce{NaCl2}$ and $\ce{MgCl2}$ are equivalent. You have no way of proving this without further experiments (or computational chemistry), but it is consistent with the fact that in practice $\ce{NaCl2}$ crystals are not observed.
In practice the formation of salts by mixing gaseous ions would be difficult to carry out (although clusters might be studied this way). So this is primarily a conceptual problem that demonstrates the importance of ionization energies in determining the relative stability of ionic compounds, and it also illustrates the usefulness of Hess's law.
It might just be possible to make $\ce{NaCl2}$  under the right circumstances, just not at room temperature and pressure (or maybe not?).

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NaCl2 would be unstable to NaCl + Cl

@jimchmst Yes. The question is how this can be explained in terms of the 2nd IE of sodium.

I suppose deltaH, deltaS, delta G, Keq could all be calculated, and one could determine if there were enough Na and Cl on Earth for a molecule to exist at equilibrium.  The answer to the ultimate question [last sentence in the post]is that the lattice energy has nothing to do with the ionization energy. It's responsible for what happens when the positive and negative ions are brought together to form a compound.   An Na++ ion and electron at infinity are perfectly stable as long as nothing comes close to the Na++. Think of Na++ as an almost alpha particle. Born-Haber cycle is not a mechanism.

@jimchmst Yes, I added more to emphasize that the Born-Haber method is a book-keeping tool.

My equation was unbalanced [by someone].  NaCl2 [were it ever possible] would compose to NaCl + Cl atoms not Cl- ions.


text:
How do I predict the spectrum when a magnetic field is applied on a specific transition from a term symbol to another?
More specifically the question pertains to physical/quantum chemistry, and to the transition 2D (with J = 3/2) -> 2P (J = 1/2)
where the numbers before the letters are multiplicities (these are term symbols.) likely an application of the zeeman interaction, but I don't understand how.
Thank you.

In a weak magnetic field the $J$ levels are split into $m_j=\pm1/2$ for $J=1/2$ and $m_J= 3/2,1/2,-1/2,-3/2$ when $J=3/2$. Transitions between these $6$ levels follow the $\Delta m_J=0,\pm1$ selection rule.  As both states are doublets the $\Delta S=0$ selection rule is also satisfied and the $\Delta L=1$ also between P and D. The splitting between the $J=1/2$ and $J=3/2$ sub-levels (fine structure) is not quit the same so the spectrum should have six lines probably split into three sets of doublets.


text:
Purification of compound containing dicarboxylic acid and amine
The prep-LC in my lab is broken and I need to urgently purify my compound which contains two carboxylic acid groups, one amine, and one amide bond. This compound was synthesized with the scheme and it did dissolve well in the reaction solvents, which are DMF and water, but it does not seem to dissolve in any other organic solvents... I am having a difficulty in finding out a suitable TLC eluent system because of this issue. Can anybody please advise me what to do?

Do you have access to reverse-phase TLC plates?

@Waylander I guess I can buy. I am struggling to find a correct solvent system that can dissolve the compound.

Have you tried these systems for very polar compounds
1 - 10% NH 4OH in MeOH solution as your polar solvent: try 1-10% of this mixture in dichloromethane.
2-  EBAW: EtOAc/butanol/HOAc/H 2O in a 80/10/5/5 ratio


text:
How do you calculate percentage of an ion in a non stoichiometric compound?

In an experiment lanthanum $\ce{^{57}La}$ was reacted with $\ce{H2}$ to produce the non stoichiometric compound $\ce{LaH_{2.90}}$. Assuming that the compound contains $\ce{La^{2+}}$, $\ce{La^{3+}}$ and hydride ions, what is
the % of $\ce{La^{3+}}$ present in $\ce{LaH_{2.90}}$ ?

I tried to find the percentage but I got a quadratic equation which has imaginary roots. Any help would be appreciated. I am aware of how to find the percentage of an element in a compound but it gets kind of fuzzy when we get to non stoichiometric compounds but I can understand the basics.
Here is what I tried:
Let the number of $\ce{La^2+}$ ions in $\ce{LaH_{2.90}}$ be x then the number of $\ce{La^3+}$ ions should be (2.9-x)^2 considering that the number of hydride ions are equal to the total number of La ions be it 2+ or 3+. Hence the quadratic expression emerges that x+(2.9-x)^2 = 1 which has no roots. I may be wrong so please do tell me if my assumption is incorrect.

What compound would $\ce{La^{2+}}$ form? What about $\ce{La^{3+}}$?

@BuckThorn the question should be read as LaH2.90 where 2.9 is in the subscript of H

I think you make this more complicated than need be. You can't have LaH2.90 in practice (only as a stoichiometric formula) so it has to be a mixture of LaH2 and LaH3.

@BuckThorn well when we come to practicality all I can say is it wasn't my idea.. it was given by my teacher from general chemistry by Zumdahl & Decoste

http://dl.iranchembook.ir/ebook/General-Chemistry-604.pdf
Q187 chapter 3.

What I mean is that you can't in practice split a hydrogen atom to make 0.90, right? So the question you should ask is, what La ion can form LaH2? Which one can form LaH3?

Well I assume you are familiar with the metal vacancy defects so I think LaH2.90 doesn't mean in particular for one molecule but as a mass phenomenon

Yes but you need to satisfy electroneutrality locally too, otherwise there really is no point to the question.  The problem presumes the existence of the two lanthanum ions in the lattice, and electroneutrality is satisfied through the particular stoichiometry.

If you do not say what you have done, others cannot say what you have eventually done wrong.

[OP] Let the number of $\ce{La^2+}$ ions in $\ce{LaH_{2.90}}$ be x then the number of $\ce{La^3+}$ ions should be (2.9-x)^2 considering that the number of hydride ions are equal to the total number of La ions be it 2+ or 3+. Hence the quadratic expression emerges that x+(2.9-x)^2 = 1 which has no roots.

The charge of the lanthanum ions is equal to the charge of the hydride ions. Your statement that the number of lanthanum ions is equal to that of hydride ions is incorrect.
Starting with the charges is more difficult, but possible. Let $c$ be the charge of the +2 ions in the formula unit. $2.9 - c$ will be the charge of the +3 ions. I can get the stoichiometric coefficient of the separate ions by dividing by 2 or 3, and these have to add up to one:
$$\frac{c}{2} + \frac{2.9 - c}{3} = 1$$
Multiplying by 6:
$$3c + 2(2.9 - c) = 6$$
Solving for c:
$$ c = 6 - 5.8 = 0.2$$
So $\ce{La^2+}$ contributes 0.2 of the charge, and $\ce{La^3+}$ 2.7 of the charge.
Of course, calling $x$ the stoichiometric coeffient of the $\ce{La^2+}$ ions gives an easier derivation, without fractions. You get
$$ x * 2 + (1 - x) * 3 = 2.9 $$
which gives $x = 0.1$.
Here are three ways to write the result:
$$0.1 \ce{La^2+ + 0.9 La^3+ + 2.9 H-}$$
or
$$\ce{\overset{+2}{La} _{0.1}\overset{+3}{La}_{0.9}H_{2.9}}$$
or
$$\ce{LaH2.9LaH3}$$
(with thanks to BuckThorn).

Thanks for clarifying about the expression

In an imaginary scenario where we could dissociate:
$$\ce{LaH_{2.9}->xLa^{+2} +yLa^{+3}}+2.9H^{-}$$
In terms of charge neutrality, we would need to satisfy:
$$2x+3y-2.9=0$$
In terms of stoichiometry, we would need to satisfy:
$$x+y=1$$
So we would get the following system of equations:
$$2x+3y=2.9$$
$$x+y=1$$
Solving it:
$$x=0.1$$
$$y=0.9$$
So there would be 10% of $\ce{La^{+2}}$ and 90% of $\ce{La^{+3}}$ out of all the $\ce{La}$ available.
The resulting dissociation reaction would be:
$$\ce{LaH_{2.9}->0.1La^{+2} +0.9La^{+3}}+2.9H^{-}$$
The total amount of species would be:
$$n=0.1+0.9+2.9=3.9$$
So the composition of each species in $\ce{LaH_{2.9}}$ would be:
$$X_{\ce{La^{+2}}}=\frac{0.1}{3.9}=0.02564=2.564\%$$
$$X_{\ce{La^{+3}}}=\frac{0.9}{3.9}=0.2308=23.08\%$$
$$X_{\ce{H^{-}}}=\frac{2.9}{3.9}=0.7436=74.36\%$$

These percentages are probably not what the textbook wanted. They probably want to know what percentage of the metal has one or the other charge.

@Karsten The wording of the problem is not clear enough to conclude that, but either way, both have been answered.

The odd-numbered questions have the numerical answers (i.e. without the work) in the appendix.

You're correct. However, either OP misrepresented the problem in the textbook or someone who made an edit did, since the question posted here adds "present in $\ce{LaH_{2.9}}$" at the end of the statement, which is not said in the textbook statement.

You are also correct. This is the full wording in the document the OP referenced in the comments: "Lanthanum was reacted with hydrogen in a given experiment
to produce the nonstoichiometric compound LaH2.90. Assuming that the compound contains $\ce{H2, La^2+, and La^3+}$, calculate the fractions of $\ce{La^2+ and La^3+}$ present."

I suppose then that since additional context was added here, the composition %'s are somewhat of a form of extra credit.

If you take $10$ $\ce{LaH2}$ and $90$ $\ce{LaH3}$ molecules, it makes a total of $100$ $\ce{La}$ atoms. But the number of $\ce{H}$ atoms is equal to $10·2 + 90·3 = 290$. This corresponds exactly to the formula $\ce{LaH_{2.90}}$


text:
How to estimate the mobility constant and the surface energy parameter in the Cahn-Hilliard equation?
As a mathematician, I want to simulate phase separations with the Cahn-Hilliard equation
$$
\frac{\partial c}{\partial t} = M \Delta \bigg(\frac{\partial \mu}{\partial c} - \kappa \Delta c \bigg),
$$
where $c$ is the molar fraction, $\mu$ is the chemical potential (given in the unit $\text{energy}/\text{length^2}$), $M$ is the mobility constant, $\kappa$ is the surface energy parameter and $\Delta$ is the Laplacian. (Since my simulations are in 2D, the units are correspondingly chosen with the unit length of $1\mu\text{m}.$)
Question 1: How do we get reasonable (good) estimates for $M$ and $\kappa,$ let's say for a certain mixture such as $\text{Na}_2\text{O-SiO}_2$? Are there standard references?
Literature and Problem: In the paper of Kim and Sanders (2020) the surface energy parameter is estimated by
$$
\kappa = \frac{\mu - \text{common tangent}}{(\partial c / \partial x)^2}
$$
while the mobility is expressed as
$$
M = \frac{D \cdot N_A}{\partial^2 \mu / \partial c^2}
$$
where $D$ is the diffusion coefficient (given in $\text{length}^2/\text{time}$) and $N_A$ is the number of particles per unit area (i.e. per $\text{length}^2$). According to these definitions, the corresponding unit for the mobility is
$$
\frac{\text{length}^2}{\text{energy} \cdot\text{time}}.
$$
(length is e.g. $\mu\text{m},$ energy is e.g. $\text{kJ},$ time is e.g. $\text{min}$). For the unit of the surface energy parameter we have just
$$
\text{energy}.
$$
The surface energy parameter estimate seems to have the correct unit due to
$$
\bigg(\frac{\text{energy}}{\text{length}^2} - \text{energy}\frac{1}{\text{length}^2}\bigg).
$$
However, the estimate of Kim and Sanders for the mobility constant seems to be incorrect, because
$$
\frac{1}{\text{time}} \not = \underbrace{\frac{\text{length}^2}{\text{energy} \cdot\text{time}}}_{\widehat = M} \cdot \underbrace{\frac{1}{\text{length}^2}}_{\widehat = \Delta} \cdot \underbrace{\frac{\text{energy}}{\text{length}^2}}_{\widehat = (\frac{\partial \mu}{\partial c} - \kappa \Delta c)}.
$$
This former question was already concerned with the units of the Cahn-Hilliard equation where the mobility $M$ was determined to have the unit
$$
\frac{\text{length}^4}{\text{energy} \cdot \text{time}},
$$
which in fact would lead to the correct unit.
Question 2: Do I misunderstand something?

You multiplied by an extra $\frac{1}{length^2}$

Yes, but I have to do that. The ''extra'' $\frac{1}{\text{length}^2}$ comes from the Laplacian in front of the bracket in the Cahn-Hilliard equation.

I edited the corresponding displaymath-line to make the origin of the ''extra'' $\frac{1}{\text{length}^2}$ clear.

I think i've figured it out, but are you sure $c$ is molar fraction and not concentration?

I am not completely sure. But according to some papers where I learnt the thermodynamic relations to that equation, I guess(ed) it's ''always'' molar fraction.

Do you know for a fact that the units of chemical potential given are $\frac{energy}{length^2}$? It usually has $\frac{J}{mol}$ units

The chemical potential is usually normed by the molar area $A_m$ (unit $\text{length}^2/\text{mol}$), i.e., we consider $\frac{1}{A_m} \mu$ and hence $\frac{\text{mol}}{\text{length}^2} \cdot \frac{\text{energy}}{\text{mol}}$

I think $N_A$ is what is causing the problem. The only way I can manage to get dimensional equivalence is when I ignore its units.


text:
Usage of thermobaric explosives to combat forest fires
This is a general question but in a thermobaric explosion I know the bomb uses all the present oxygen in the reaction. Could you please give an example of one of these reactions? Even if it isn't exactly what happens during one of these explosions an example would help. Also, could one of these bombs help stop forest fires because it takes away the oxygen which the fire needs for fuel? Could a combustion reaction potentially be a thermobaric reaction?

As simplification, thermobaric bomb is like spreading volatile fuel in large volume of air (partly evaporating, partly making aerosol) and then igniting the explosive mixture. In a way, it is the same process as the one occuring in car combustion engines, just in a gigantic volume and in open air. An alternative can be disperging of finely powdered combustible substance. See also Dust/Fuel-air explosion.
If liquid n-alkane is used, the reaction will be
$$\ce{2 C_{n}H_{2n+2} + }(3n+1) \ce{O2 -> 2n CO2} + (2n+2)\ce{H2O}$$
Big problem with fire extinction would be scalability of procedure.
Another issue would be that such explosions may ignite what is not being combusted yet at the explosion border regions. As their heat waves are more intense and last longer than when high explosives are used. Be aware fuels contain several times greater specific energy (in $\pu{MJ kg-1}$) than explosives.
It is like impossible to properly dose the fuel to spend like most of oxygen. Its natural content in air is about 21%. The threshold for sustained burning of organic matter is typically 17% (mostly 16-18%).

If there is at a particular place too little fuel, there would be still oxygen for continuing the fire and explosion can intensify it.
If there is too much fuel, it would burn later when there is enough oxygen again.


text:
The differential rate law applies to only one single species
In my book it was written that

The differential rate law applies to only one single species, so unless we have equimolar amounts of each reactant or make an approximation using an excess of one, we cannot apply the differential rate law.

What does this statement mean?  Can anyone please explain?

Do not confuse the applicability of the differential rate law and ability (easily or at all) to solve respective differential equations. There are many cases in chemistry and physics  where the law and it's equation fit well, but it is not possible to solve it analytically.

Which book is that?

For a reaction at constant volume of the form:
$$\ce{A +B->C}$$
If the rate of this reaction follows a power law relationship, then:
$$-r_\pu{A}=-\frac{\mathrm{d}C_\pu{A}}{\mathrm{d}t}=k\;C_\pu{A}^{\alpha}\;C_\pu{B}^{\beta}=f(C_\pu{A},C_\pu{B})$$
Since both $C_\pu{A}$ and $C_\pu{B}$ are functions of time $t$, you won't be able to integrate -$\frac{\mathrm{d}C_\pu{A}}{\mathrm{d}t}$.
However, if we use an excess of $\pu{B}$, then the change in $C_\pu{B}$ will be negligible compared to the change in $C_\pu{A}$ after the reaction is finalized, and the following approximation becomes valid:
$$C_\pu{B}^{\beta}≈constant$$
This constant can be grouped up with $k$, which is another constant:
$$k'=k\;C_\pu{B}^{\beta}$$
So we would get:
$$-r_\pu{A}=-\frac{\mathrm{d}C_\pu{A}}{\mathrm{d}\pu{t}}=k\;C_\pu{A}^{\alpha}\;C_\pu{B}^{\beta}≈k'C_\pu{A}^{\alpha}≈f(C_\pu{A})$$
So the rate would only depend on $C_\pu{A}$, and you can integrate it:
$$\frac{-\mathrm{d}C_\pu{A}}{C_\pu{A}^{\alpha}}=k'\;\mathrm{d}\pu{t}$$
Conversely, when no excess of either $\pu{A}$ or $\pu{B}$ are used, but equimolar initial amounts are, we can integrate regardless.
Assuming $\pu{A}$ is the limiting reagent, and constant volume, pressure, and temperature conditions, the concentration of $\pu{A}$ and $\pu{B}$ in terms of conversion are defined as:
$$C_\pu{A}=C_{\pu{Ao}}\;(1-X)$$
$$C_\pu{B}=C_{\pu{Ao}}\left(\frac{C_{\pu{Bo}}}{C_{\pu{Ao}}}-X\right)$$
Since we have equimolar amounts, it follows that:
$$C_{\pu{Ao}}=C_{\pu{Bo}}\implies \frac{C_{\pu{Bo}}}{C_{\pu{Ao}}}=1\implies C_\pu{B}=C_{\pu{Ao}}\;(1-X)\implies C_\pu{A}=C_\pu{B}$$
So the rate would be:
$$-r_\pu{A}=-\frac{\mathrm{d}C_\pu{A}}{\mathrm{d}t}=k\;C_\pu{A}^{\alpha}\;C_\pu{B}^{\beta}=k\;C_\pu{A}^{\alpha+\beta}$$
And you could integrate the resulting expression if orders are known or reaction is elemental:
$$\frac{-\mathrm{d}C_\pu{A}}{C_\pu{A}^{\alpha+\beta}}=k\;\mathrm{d}t$$
In conclusion, if either of the two conditions are used, and you know the values of $\alpha$ and $\beta$, you can integrate these differential equations to obtain a function of $C_\pu{A}$ vs $t$.

for further reference: upright vs italic

There is possible the parameterization , expressing both concentrations of A and B as functions of the single parameter. Still, more complicated solving than the simple case of pseudo first order.

That is true. I had initially expressed both concentrations as functions of conversion $X$, but I thought that could confuse OP, so I just left it as is.

I see, I can agree with that.

If the reaction has the second order rate law $dA/dt=-kAB$ with initial concentrations $A_0,B_0$ then by stoichiometry $A-A_0=B-B_0$ and then
$$\frac{dA}{A(B_0-A_0+A)}=-kdt$$
which can be integrated after expanding as partial fractions as it is a function only of one variable, $A$.
Alternatively using the reaction variable method let $x$ be the amount reacted at time $t$ then $\displaystyle \frac{dx}{dt}=k(a-x)(b-x)$, which can be integrated when $a,b$ are initial amounts, to give an equivalent result to that above.
These solutions would seem to indicate that the statement in your book is not quite correct.


text:
Variation in the energy of the $\ce{\sigma_{2pz}}$ orbital
Is there a specific reason why the energy of the $\ce{\sigma_{2pz}}$ orbital varies relative to the  $\ce{\pi_{2px}}$ and $\ce{\pi_{2py}}$ orbitals, depending on the atomic number of the atom?

You should look at diagrams that are not utterly misleading. The should be s-orbitals, too.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/08%3A_Advanced_Theories_of_Covalent_Bonding/8.4%3A_Molecular_Orbital_Theory

@Martin-マーチン - unfortunately, the orbital diagram in that link for $\ce{Be2}$ is incorrect. See, for example https://ch301.cm.utexas.edu/section2.php?target=imfs/mo/secondrow-diatomic-mos.html

See also https://chemistry.stackexchange.com/a/34834/5017

Orbital mixing is the general answer. Molecular orbitals with the same symmetry, like $\ce{\sigma_g(2s)}$ and $\ce{\sigma_g(2p)}$ can overlap. The result is that $\ce{\sigma_g(2s)}$ will decrease in energy while $\ce{\sigma_g(2p)}$ will increase in energy. The same thing happens for the antibonding orbitals with $\sigma_u$ symmetry.

sp-mixing: Left diagram shows the MO orbital diagram for s and p orbitals without sp-mixing, while the right diagram shows one possible configuration after considering sp-mixing.
As the effective nuclear charge increases, the sp mixing is smaller due to the larger energy difference between the $\ce{2p}$ and $\ce{2s}$ orbitals. Therefore, for $\ce{O2}$, $\ce{F2}$ and theoretical $\ce{Ne2}$, you see the characteristic molecular orbital diagram shown on the left. However, with diatomic elements like $\ce{N2}$ and theoretical diatomic molecules with lower atomic numbers such as $\ce{C2}$ and $\ce{B2}$, you get more orbital mixing. That means the $\ce{\sigma_u(2p)}$ would increase more in energy, which is why you may see the right molecular orbital diagram representing certain diatomic molecules.

The left diagram is - if not completely and utterly wrong - at the very least incomplete.

@Martin-マーチン - the left diagram omits the 2s orbitals, but I have seen diagrams like this which focus on the 2p alone because of this unusual orbital energy ordering (e.g., in $\ce{N2}$. This answer is +1 from me.

@GeoffHutchison I was fully aware of this. But how do you explain orbital mixing, if you omit half of the originals that mix?

@Martin-マーチン - I'm not justifying the diagram. I think the whole valence picture is necessary. So I completely agree with "incomplete" but I disagree with "completely and utterly wrong" .. I think that goes too far.

@GeoffHutchison You are right and I have overreacted. I'm too easily annoyed by misleading depictions like the one in the question. However, this shouldn't be the place to comment this and it wasn't helpful. Especially given this being the answer. I'll be more careful in the future.


text:
Calculating required mass of reactant to produce a specific product
I want to know how many grams of $\ce{C}$ (carbon) are required to create $\pu{58 grams}$ of $\ce{Na2S}$, given
$$\ce{Na2SO4 + 2C -> Na2S + 2CO2}.$$
First I calculated molecular mass of $\ce{Na2S}$ which is $\pu{78.04 g/mol}$.
Then I divided to get the amount of substance: $\frac{58}{78.04} = \pu{0.743 mol}$.
The ratio between $\ce{C}$ and $\ce{Na2S}$ is $2 : 1$, so the amount of substance of $\ce{C}$ is $0.743 \cdot 2 = \pu{1.486 mol}$.
Now I multiplied the amount of substance of $\ce{C}$ with the molar mass of $\ce{2C}$ to get the mass: $1.486 \cdot 24.022 = \pu{35.69 g}$
I submitted this result and got a wrong answer. I don't really understand what I did wrong here. What am I missing? (Assume no limiting factor).

You made a mistake in the last calculation. Stoichiometric coefficients are not included when calculating mass of a species from amount (moles). They're only used when converting amount of product into amount of reactant or vice-versa. In short, your answer divided by 2 is the right answer.

You don't multiply by the molar mass of $\ce{2C}$. You simply multiply by the mass of one carbon as you already accounted for the ratio in the previous step. Note that $\ce{2C}$ doesn't imply two carbons bonded to each other. It only represents the stoichiometric ratio in the reaction.

I was not aware that stoichiometric coefficients are not included when calculating mass of a species from amount. Now I know. Thank you guys for the answer

You should always use symbols and write actual equations with them, before you put in numbers. Never ever skip the units. You'll catch about 99% of the mistakes you make with such approach.

Echoing Martin's excellent comment in answer form, I think it is very helpful to write out these units conversion problems in "railroad tracks" format.
The only trick with chemistry and stoichiometry when it comes to unit conversions is that there are often multiple species whose mass or moles (or volume or density etc) we are tracking.  You'd better write down species labels in your units.  And when you do this, a mol of Na2S is different than a mol of reaction; the two mole units do NOT cancel out if you take the ratio of one to the other.
The stoichiometry of the particular reaction you're interested in is what lets you convert from moles of one species to another.  So your particular reaction is telling you that 1 mol of Na2S is equal to 1 mol of reaction.  But if you had a different reaction, the "units conversion" between moles of one species and moles of another could be different.
For this particular problem, I'd use the "railroad tracks" method to write out the solution like this (n.b. this is hard to do in MathJax; apologies if I do anything wrong):

You start off with 58 g of $\ce{Na2S}$ and the idea to convert it to moles is a good one.

$$\frac{\pu{58 g \ce{Na2S}}}{1}\frac{\pu{mol \ce{Na2S}}}{\pu{78.08 g \ce{Na2S}}}$$

Once you have moles of $\ce{Na2S}$, you can convert to moles of "reaction" using the stoichiometric coefficient of $\ce{Na2S}$ in your reaction, which in this case for $\ce{Na2SO4 + 2C -> Na2S + 2CO2}$ is 1.
$$=\require{cancel}\frac{\pu{58 g \ce{Na2S}}}{1}\frac{\pu{mol \ce{Na2S}}}{\pu{78.08 g \ce{Na2S}}} \frac{\pu{mol reaction}}{\pu{mol \ce{Na2S}}}$$
And once things are on a "moles of reaction" basis, you can easily "convert" units to the species that you care about, in this case, carbon.

$$=\require{cancel}\frac{\pu{58 g \ce{Na2S}}}{1}\frac{\pu{mol \ce{Na2S}}}{\pu{78.08 g \ce{Na2S}}} \frac{\pu{mol reaction}}{\pu{mol \ce{Na2S}}}\frac{\pu{2mol C}}{\pu{mol reaction}}$$

Lastly, you can convert from moles of carbon to grams.

$$=\frac{\pu{58 g \ce{Na2S}}}{1}\frac{\pu{mol \ce{Na2S}}}{\pu{78.08 g \ce{Na2S}}} \frac{\pu{mol reaction}}{\pu{mol \ce{Na2S}}}\frac{\pu{2mol \ce{C}}}{\pu{mol reaction}}\frac{\pu{12 g \ce{C}}}{\pu{mol \ce{C}}}$$

The final step is to check that all units cancel except the output units you desire, in this case grams of carbon.

$$\require{cancel}=\frac{58 \cancel{\pu{g \ce{Na2S}}}}{1}
      \frac{\cancel{\pu{mol \ce{Na2S}}}}{78.08\cancel{\pu{ g \ce{Na2S}}}}
      \frac{\cancel{\pu{mol reaction}}}{\cancel{\pu{mol \ce{Na2S}}}}
      \frac{2 \cancel{\pu{mol \ce{C}}}}{\cancel{\pu{mol reaction}}}.
      \frac{\pu{12 g \ce{C}}}{\cancel{\pu{mol \ce{C}}}}$$
$$=\left(\frac{58}{78.04}\frac{2}{1}\frac{12}{1}\right) \textrm{g}\;\ce{C} =\textrm{17.84 g C}$$


text:
Benzene Identification in a liquid?
how can we identify the presence of benzene in a liquid? How to find the percentage of benzene in a liquid? Are there any feasible tests that could be done to find it?
Honestly, all we have is a chemistry lab with the bare minimum necessities.

Do you have access to a GC-MS?

Maybe GC is enough !

I have no access to a GC -MS. Is there any way to do this without them?

What do you have access to? and what are the liquids to be analysed? and what are approximate concentrations? Benzene has distinct uv, ir-Raman, nmr spectra, can be separated by gc or lc and possibly by azeotropic distillation. Learn about these various methods and see if what you have is appropriate. Then get some professional help perhaps starting with some chemistry courses.

Generally, there are no simple and cheap tests for it without instrumental analysis. Naive ideas like Add this and a color forms if the substance is present work for very limited number of cases.
Probably the easiest way is to contact some environmental lab or water treatment company lab as an external private customer, if they could do such an analysis for you. But it may cost up to (hopefully lower) hundreds of dollars. The best if you have a friend working there, the price could be addressed by alternative payment.
I worked in such a lab 15 years ago and trace analysis of aromatic hydrocarbons in water were done routinely.


text:
What is white powder on lead? How to remove?
What is the white powdery substance that has formed on lead airgun pellets stored in an oak cabinet for 35 years? How can it be removed safely?  The lead could possibly be alloyed with some antimony, the cabinet and pellets have been in a damp basement for over 35 years. The cabinet appears to be made of oak (perhaps the tannins from the oak acted on the pellets).  Here is a picture of the pellets in increasing degree of "corrosion." What is the white substance and can it be safely removed?  Thank you!

I would guess it is lead oxide. It is dangerous, but so is lead itself. Get rid of it completely. Bring it to a local business which can safely dispose of it.

It is probably lead carbonate $\ce{PbCO3}$, because it is white, and lead(II) oxide is yellow, and lead(IV) oxide is brown


text:
What is mAh/g Specific capacity?
In battery chemistry I've been reading about specific capacities of various electrochemical cells as $\pu{mAh/g}$. For example, in one article it says the specific capacity of the system is $\pu{75 mAh/g}$ at $\pu{1.7 V}$.
Does that mean the a $\pu{1 kg}$ battery would give 75 x 1.7 or 127.5 Watts of energy?
Or is it some "electrod-ic" "cathodic" or "anodic" capacitance. In that case how do we calculate how much Wh/kg of power we would get effectively from a battery?
High Coulombic efficiency aluminum-ion battery using an AlCl3-urea ionic liquid analog electrolyte

The battery exhibits ∼99.7% Coulombic efficiency and a substantial rate capability, with a cathode capacity of $\pu{73 mA g-1}$ at $\pu{100 mA g-1}$ ($\pu{1.4 C}$).

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

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.

The unit mA-h/g means milliampere-hours per gram mass. However, since mA-h means milliampere-hours, that is, providing a stated current for a stated time, it cannot be compared with watts, a unit of power. Comparing mA-h and W is like comparing apples and barium. An automobile lead-acid battery might be rated 12 V at 60,000 mA-h (60 A-h), meaning it can provide about 20 amperes for 3 hours, or 10 A for 6 h. That is, 12 V * 60 A-H = 720 watt-hours (W-h). Each of the six cells in the battery have the same current rating, but at just 2 VDC, each stores 120 W-h.
To try to understand the question you might want to ask, learn about dimensional analysis. You can't measure area in fortnights, nor measure mass in furlongs.

The parameters with units $\pu{mAh}$ and $\pu{mAh/g}$ express ability of a system to absorb, hold and provide back the particular electric charge. $\pu{1 mAh}$ is a non-SI unit of charge, equivalent to $\pu{3.6 C}$.
The value usually relates to the whole cell package, but as in the paper, it  can relate to one of the electrodes as well, evaluating the electrode charge storage efficiency.
Assume this illustrative case:

Decription
Value

The cathode specific capacity
$\pu{75 mAh/g}$

The anode specific capacity
$\pu{50 mAh/g}$

The desired capacity
$\pu{300 mAh}$

The needed cathode mass
$\frac{\pu{300 mAh}}{\pu{75 mAh/g}}=\pu{4 g}$

The needed anode mass
$\frac{\pu{300 mAh}}{\pu{50 mAh/g}}=\pu{6 g}$

The electrolyte mass
$\pu{6 g}$

The supporting non-active material mass
$\pu{6 g}$

The case mass
$\pu{8 g}$

The total mass
$\pu{\pu{4 g} + \pu{6 g} + \pu{6 g} + \pu{6 g} + \pu{8 g} = \pu{30 g}}$

The cell specific capacity
$\pu{300 mAh} / \pu{30 g} = \pu{10 mAh/g}$

Quantity
SI unit
Unit relation
Calculations

Charge
$\pu{1 C}=\pu{1 A} \cdot \pu{1 s}$
$\pu{1 mAh} = \pu{3.6 C}$
$\pu{75 mAh} = \pu{0.075 A} \cdot \pu{3600 s} = \pu{270 C}$

Specific charge
$\pu{1 C/kg}$
$\pu{1 mAh/g} = \pu{3.6 C/g}$
$\pu{75 mAh/g} = \pu{270 C/g}$

Voltage
$\pu{1 V}=\pu{1 J} / \pu{1 C}\\=\pu{1 W} / \pu{1 A}$

Energy
$\pu{1 J}=\pu{1 C} \cdot \pu{1 V}$
$\pu{1 mAh} \cdot \pu{1 V} \\= \pu{1 mWh} = \pu{3.6 J}$
$\pu{75 mAh} \cdot \pu{1.7 V} = \pu{270 C} \cdot \pu{1.7 V}= \pu{459 J}$

Power
$\pu{1 W}=\pu{1 J} / \pu{1 s}\\=\pu{1 A} \cdot \pu{1 V}$
$\pu{1 mW} \\ = \pu{1 mWh} / \pu{1 h} \\ = \pu{3.6 J} / \pu{3600 s}\\=\pu{1 mJ} / \pu{1 s}$
$\frac{\pu{75 mAh} \cdot \pu{1.7 V} }{ t \  \pu{s}} = \pu{\pu{\frac{459}{t}} W}$

Specific energy
$\pu{1 J/kg}$

$\pu{75 mAh/g} \cdot \pu{1.7 V} =  \pu{459 J/g}$

Specific power
$\pu{1 W/kg}$

$\pu{75 mAh/g} \cdot \pu{1.7 V} / t\  \pu{s} = \pu{\frac{459}{t} W/g}$

Let's solve your problem properly, using SI Units.
The specific capacity of $\pu{75  mAh/g}$, expressed in SI unit, is : $$\pu{75 mAh/g = 75*0.001 A*3600 s/g = 270 C/g}$$ If the battery has a nominal voltage of $\pu{1.7 V}$, it means that it is able to produce a total energy which is not expressed in watts (as mentioned in the original text), but in joules. This total  energy is $$\pu{270 C/g  *  1.7 V  = 459 J/g}$$ If the battery weighs $1$ kg, it is able to deliver a total energy of $\pu{459 kJ}$. It is not a power (as mentioned in the original text). No. It is an energy.


text:
Why is parafluorophenol is more acidic than phenol?
I have learnt that the +Mesomeric effect (+M effect) of fluorine at para position dominates its -Inductomeric (-I effect) nature. If we see that the acidic strength (with the help of stability of their conjugate anion then in parafluorophenol fluorine) will exert its +M effect at the negative charge of oxygen, then that's why the stability the of parafluorophenoxide ions should be less than normal phenoxide ion. Hence, the acidic strength of the phenol should be greater than parafluorophenol, but the actual acidic strength of parafluorophenol is greater than phenol.  Why is this?

From the experimental pKa tables in a fairly authoritative reference, phenol and 4-fluorophenol have indistinguishable acidities (both compounds have pKa 18.0 in DMSO and 9.95 in water). It would be interesting to compare the gas-phase acidities to remove the solvent dependence, but regardless of those last data, it's clear that no hand-waving argument would be sufficient to explain the precise ordering of acidities for these compounds,

@Nicolau Saker Neto: I think the value of pKa for 4-fluorophenol is a mistake. The referenced article ( J. Org. Chem. 1984, 49, 1424) did not reported a value for this compound. However, it reported the values for 3-fluorophenol and 3-chlorophenol, which are 15.88 and 15.83, respectively (remarkably close). It also reported the value for 4-chlorophenol (16.74). So, I assume the value for 4-fluorophenol should be around that (about 100 times acidic than phenol).

@MathewMahindaratne That is an interesting find, though it's puzzling that the link I gave also reports the pKa in water, compiled by a different author (R. Williams instead of Reich and Bordwell), and yet once again the two compounds are indistinguishable.

@Nicolau Saker Neto: The values I gave here are from original article. They have some values in water and in gas phase as well. I think the link you sited may have made a mistake. :-)


text:
Why is it recommended to not store commercial sodium silicate solutions in direct sunlight, what are the effects of doing so and can it be modelled?
I regularly purchase aqueous sodium silicate solutions from commercial providers for my studies and note that a storage recommendation for all different grades (i.e. differing $\ce{SiO2 / Na2O}$ ratios) is to "not store in direct sunlight" with no explanation as to why this is the case. I tried asking the provider why this is a recommendation but didn't get a substantial answer apart from "it turns a weird [brown-like] colour after a while and is not really usable after that".
I believe the issue here is something to do with the polymeric character of sodium silicate - with the formula given as $\ce{Na_{2x}Si_yO_{2y+x}}$ or $\ce{Na2O_x SiO2_y}$ - which would inevitably cause issues with sunlight cleaving parts of the sodium silicate polymer backbone and creating a heterogenous mixture (i.e. with different parts of the aqueous solution having polymers of different chemical formulas) which would be a QAQC (quality assurance and quality control) nightmare for using sodium silicate solutions in commercial developments. Adding to this, I believe the poor usability probably comes from precipitation of something and have found (if useful) that sodium silicate solutions with increased levels of $\ce{Na2O}$ can cause solution precipitation.. however this precipitation is a white gel-like substance (which I believe comes from the solution approaching the saturation concentration of $\ce{NaOH}$) and nothing "brown-like".
My Questions are:

Does anyone have experience/insight here with storing sodium silicate solutions in direct sunlight?
What predictive mathematical and/or empirical models can be used and/or made to assess how much sunlight will cause QAQC issues with sodium silicate?

These silicate solutions come in plastic bottles, and plastic does degrades with sunlight. Storing away from sunlight is a common warning on reagent bottles, because plastic can also become brittle. Sodium silicate should be UV insensitive. This browning is hard to believe is due to sunlight.

The brown substance is NOT a compound of silicon and sodium. All sodium silicates are colourless. It must be due to some organic impurity which is sensitive to light.

Sodium silicate may attack the paper liner inside some bottle tops, causing the brown coloration, but, as others state, it should be relatively insensitive to light. BTW, never store in a glass container, which it does attack over time.

Along with the issues with the container, it seems that photolysis from the sunlight can also degrade the sodium silicate by changing the chemistry of the bonds and thus impacting its QAQC utility in experiments (etc.).


text:
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.


text:
Why does silver react preferentially with chlorine instead of chromate?
Mohr's method of precipitation titration relies on the fact that silver reacts preferentially with chloride ions instead of chromate ions in solution. When I searched up why this is true, the only response was that the $\text{Ksp of }\ce{AgCl}$ ($1.77\times10^{-10}$) is greater than the $\text{Ksp of }\ce{Ag_2CrO_4}$ ($1.2\times10^{-12}$) seen here : https://www.answers.com/earth-science/Why_does_silver_nitrate_combine_first_with_chloride_ions_in_water_and_not_with_potassium_dichromate.
However this doesn't make sense to me since shouldn't the more soluble a compound is indicate that it is less likely to react and form a precipitate meaning that by this logic shouldn't $\ce{Ag_2CrO_4}$ be preferential not $\ce{AgCl}$? Could someone explain what is wrong with this reasoning and how to actually explain why silver reacts preferentially with chloride ions?

The idea of $K_{sp}$ is just the equilibrium constant for a dissolving reaction of a solid in liquid.
$$aA_{s}\leftrightharpoons cC_{aq} + dD_{aq}$$
$$K_{sp}=[C]^c[D]^d$$
But, another concept is the solubility $s$ of a substance. As you calculate the solubility of a salt you should take into account the stoichiometry of the reaction.
$$AgCl_{s}\rightleftharpoons Ag^+_{aq}+Cl^-_{aq}$$
$$K_{sp}=[Ag^+][Cl^-]$$
$$s^2 = K_{sp} \rightarrow s = \sqrt{K_{sp}} = \ce{1.33·10^{-5}}$$
On the other hand for the silver chromate reaction we get:
$$Ag_2CrO_4\leftrightharpoons 2Ag^+_{aq}+\ce{CrO_4^-_{aq}}$$
$$K_{sp}=[2s]^2[s]=4s^3\rightarrow s = (K_{sp}/4)^{1/3}=6.69· \ce{10^{-5}}$$
And we get that the silver chloride is less soluble than the the silver chromate due to the stoichometry of the two reactions.


text:
Why does NaCl form when it requires 1.52 eV of energy?
An atom of sodium has one 3s electron outside a closed shell, and it takes 5.14 eV of energy to remove that electron. The chlorine lacks one electron to fill a shell, and releases 3.62 eV when it acquires that electron.
This means that it takes 1.52 eV of energy to donate one of the sodium electrons to chlorine when they are far apart.

What causes the system even to brake through his metastable state to finally reach the stable state?
Shouldn't it even cost +5.14eV according to that chronological order? I don't understand why one can say here that it costs 5.14eV - 3.62eV when 5.14eV must be applied first:
I've found a guy on yt and he explains that the activation energy must be 1.53 eV. (https://youtu.be/np7GcXTYKlI?t=396) [@06:36]
But again, I wonder why it's not 5.14eV as he clearly statet that first it requires 5.14eV to strip down that electron from Na and only afterwards those 3.61eV releases due to Cl gains that electron.

Well, if you knew that NaCl in gaseous phase is molecular then it would save you the trouble of writing all this, I guess. Yes, it's just NaCl molecule (or its dimer), no magical ionisation or whatever.

https://chemistry.stackexchange.com/questions/14174/what-is-sodium-chloride-like-in-gas-state https://chemistry.stackexchange.com/questions/2511/will-gaseous-ionic-compounds-be-free-moving-ions?noredirect=1&lq=1

Ionic pair could be better term, as the bond is more ionic than covalent.

You are only considering the ionisation potentials. A lot of the net energy of the reaction of Na and Cl comes from the large energy released when two opposite charges come together. And we observe that release when Na burns in Chlorine gas. Also the starting point is not usually Cl atoms but dichlorine molecules.

The interaction potentials of the two cases you propose are dramatically different, because the Coulombic interactions are dramatically different. In one case you are talking about the interaction of two particles with opposing charge, in the other of two neutral particles:
$$\begin{align}\ce{Na(g)  ... ... Cl(g) } ~~~~~~ & \textrm{very weak electrostatic interaction}\\ \ce{Na+(g)  ... ... Cl-(g) }~~~~~~& \textrm{strong electrostatic interaction}\end{align}$$
In the first case you can essentially ignore interactions between the particles until they are within a distance of the order of the sum of their van der Waals radii. In other words, until very close the particles behave as in an ideal gas. If the neutral particles collide they might share electrons and bond covalently or, somewhat more likely, exchange an electron to form ions and an ionic bond. The actual bond that may form will be intermediate to these extremes and closer to ionic.
In the second case the particles attract strongly even at fairly large distances due to the $r^{-1}$ Coulombic potential. This is a very unstable situation. The attraction may lead to a collision if the particles do not fly too fast and too far apart.
A bond formed during a collision will be stable provided the collision energy was not too great, otherwise the particles will bounce off each other. They might then retain any charges gained through the collision. These topics are the subject of gas phase molecular dynamics. See for instance Ref 1.

When the neutral Na and Cl pairs come close to each other, potential energy is released, but why is it now energetically more favorable when Na transfers its electrons with this released energy to the Cl? Wouldn't the potential energy be even more negative if they came closer and didn't transfer anything?

Since the potential between uncharged particles changes very little as you bring them closer, whereas that of oppositely charged ones decreases much more rapidly, a transfer from the neutral to the charged potential energy curves becomes increasingly favorable as the particles get closer. Eventually the change in the attractive potential between the charged particles compensates for the cost of transferring an electron between the atoms and the transfer becomes viable and may occur spontaneously (for instance through quantum tunneling). This is illustrated in the following figure (distance is in units of Bohr radius):

Various sites including this one report that this happens when the internuclear distance falls below 9.4 Å.
However, I have no answer for what is the probability of an electron transfer between gas phase sodium and chlorine as a function of interatomic distance. In essence that is what a large part of your post boils down to.
References
(1) Šimsová née Zámečníková, M.; Gustafsson, M.; Soldán, P. Non-Adiabatic Dynamics in Collisions of Sodium and Chlorine Atoms and Their Ions. Phys. Chem. Chem. Phys. 2022, 24 (41), 25250–25257. https://doi.org/10.1039/D2CP03361E.


text:
Cell notation for simple voltaic cell
We can make a simple voltaic cell consisting of Cu and Zn electrodes dipped in an aqueous H2SO4 solution for demonstration purposes. I am curious to know whether this has a constructible cell notation or not, since I have not come across one in any book so far.
Also, the cell reaction generates H2 gas which makes me doubt that an electric double layer involving H+ ions is formed faster than that involving Cu2+ ions from a Cu electrode. It is also  For the reason that , we see no detectable blue colouration around Cu when the set up - simple cell is in operation. Is that right?
Moreover, does the acidic solution have a liquid junction potential?

This is simply a variant of the well-known lemon cell, i.e., lemon ‘battery’.

It is cell Zn/Zn^2+  H+/H2.  Cu acts  as pseudo inert electrode there. Cu^2+ would be reduced first, before H+, but there are no Cu^2+ ions.

"There are no Cu²+ ions." Does this mean double layer of Cu/ Cu²+ is negligible? It is little unconvincing, because Cu can dissolve in H2SO4 unlike pt/C

Well...this notation apparently lacks symbols for junction potential/anode cathode separation etc... Is it possible to write like this?


text:
How to define percentage by volume for ideal gas and Why percentage by volume is equal to percentage by mole?
Why do we define percentage by volume for an ideal gas in a closed container in spite of that we know the ideal gas will occupy the volume of the whole container?
I have seen many people who prove that % by volume is equal to the percentage by mol for an ideal gas (at constant temperature and pressure conditions). They follow the following procedure...
PV = nRT
V = nRT/P
(let RT/P = K as a constant as pressure and temperature are constants)
V=Kn
as volume is directly proportional to the number of moles and hence % by volume is equal to the percentage by mole...
But the doubt is here, when we write PV=nRT for a gas in a closed container then we write volume as the volume of the container then where do we define % volume for gas?

It is easy to remember that for ideal gases,
$$V_{\text{const }p} = \sum_{i=1}^{n}{V_i} = \sum_{i=1}^{n}{\frac{n_iRT}{p}}\tag{1}$$
$$p_{\text{const }V} = \sum_{i=1}^{n}{p_i} = \sum_{i=1}^{n}{\frac{n_iRT}{V}}\tag{2}$$
It can be also easily seen that
$$\frac{V_i}{n_i} = \left(\frac{RT}{p}\right)_\text{p = const}\tag{3}$$
$$\frac{p_i}{n_i} = \left(\frac{RT}{V}\right)_\text{V = const}\tag{4}$$
And the molar, pressure and volume ratios:
Constant pressure:
$$\frac{n_i}{n} = \frac{\frac{pV_i}{RT}  }{  \frac{pV}{RT}} = \frac{V_i}{V}\tag{5}$$
Constant volume:
$$\frac{n_i}{n} =\frac{\frac{p_iV}{RT}  }{ \frac{pV}{RT}}=\frac{p_i}{p}\tag{6}$$

Considering variables without subscripts represent total state functions (mixture), and variables with subscript $i$ represent the state functions of a single component $i$ in the mixture, at constant pressure and temperature for an ideal gas, according to Amagat's law, we have:
$$PV_i=n_iRT$$
$$PV=nRT$$
Dividing the first equation by the second:
$$\frac{PV_i}{PV}=\frac{n_iRT}{nRT}$$
Cancelling all equal terms:
$$\frac{V_i}{V}=\frac{n_i}{n}$$
Defining the volume fraction of $i$ and mole fraction of $i$, respectively:
$$\upsilon_i=\frac{V_i}{V}$$
$$x_i=\frac{n_i}{n}$$
It follows that both fractions are equal:
$$\upsilon_i=x_i$$
A second way of showing they are equal is using the relationship between both fractions:
$$\upsilon_i=x_i\frac{M_i}{M}\frac{\rho}{\rho_i}=x_i\frac{C}{C_i}=x_i\frac{\bar{V_i}}{\bar{V}}$$
Where $\bar{V}$ represents molar volume, $M$ molar mass, and $C$ molar concentration.
The molar volume of an ideal gas is equal to that of the mixture at constant temperature and pressure, so we get:
$$\frac{V}{n}=\frac{V_i}{n_i}\implies\bar{V}=\bar{V_i}\implies\upsilon_i=x_i$$

when you wrote PVi = nRT then why did you use total pressure instead of partial pressure i.e it should be PiVi = nRT and from that after dividing the two equations and applying dalton's law of partial pressure on gaseous mixture you will get Vi = V.

Amagat's law uses total $P$, while Dalton's law uses total $V$.


text:
Purpose of a pH Problem
I was doing a multiple choice question as follows.
The pH scale measures the

Strength of an acid
Concentration of hydrogen ions
Strength of hydrogen ions
Concentration of an acid

The answer is 2 but why is 1 not sufficient. The way I understand the pH scale is that it measures how acidic or alkaline a substance is. So does it not measure the strength of an acid?

Low concentration of stronger (like sulfuric) acid can have higher pH than high concentration  of weaker (like acetic) acid.

@Poutnik Could you explain that statement briefly? Why is this the case?

Concentration of H+ ions and therefore pH depends on acid concentration (as their source) and degree of their production (depending on acid strength).

Strength ≠ concentration for acids and bases. A stronger acid has a lower pKa, regardless of concentration. I think the confusion comes from the colloquial use of "strong/strength" to refer to relative concentrations.

pH is approximated as $\mathrm{-log\ce{[H+]}}$. It specifically has the $\ce{[H+]}$ or concentration of $\ce{H+}$ term, so pH is a direct measure of that.
$\mathrm{p}K_\mathrm{a}$ is a better measure of the strength of an acid. Some acids are stronger than others, and dissociate more in water. While, yes, more acidic species produce more $\ce{H+}$ ions, thus having lower pH, there isn't a fully direct relationship.

Rather pH=f(acid concentration, acid strength)

pH is not defined as $-\log[\ce{H+}],$ rather as $-\log a(\ce{H+}).$ See https://doi.org/10.1351/pac198557030531. Note the proper typesetting for $\mathrm pK_\mathrm{a}$.

@andselisk While that is truth, sometimes simplified truth is more helpful at certain education levels. Like W=F.l in elementary school, later vec F . vec l in high school, later int(vec F . vec dL) in college/university. // E.g. standard school formulas for pH almost never deal with activities.

@Poutnik Fine, but then don't lie by saying "defined as", at least say "approximated as". Yes, I'm aware of eigenchris — What Is Momentum? (joke video).

@andselisk Imperative don't lie assumes lying, what is not confirmed. I have not said such a thing.

@andselisk Good point, I've only dealt with activity coefficient a few times and just got too used to that 'approximated' form for pH.

@andselisk But as a figure of speech I agree.  BTW a nice video.

Suppose we have two different aqueous solutions, one being a strong acid and the other being a weak acid:
$$\ce{HCl(aq)<=>H+(aq) +Cl-(aq)},\quad \pu{K_a}=10^6$$
$$\ce{HCOOH(aq)<=>H+(aq) +COOH-(aq)},\quad \pu{K_a}=10^{-3.75}$$
The strength of an acid is a measure of to what extent it is dissociated into its constituent ions. The dissociation fraction $\alpha$ gives us an idea of this, and it is defined as:
$$\alpha=\frac{x}{C_{\pu{Ao}}}$$
Where $x$ represents the equilibrium concentrations of $\ce{H^+}$ ions or conjugate base $\ce{B-}$ ions, assuming no initial amounts of either are present, and we can make the approximation:
$$\pu{pH}=\pu{-log}\;{C_{\ce{H^+}}}\implies x=10^{\pu{-pH}}$$
The initial concentration $C_{\pu{Ao}}$ can also be set in terms of $x$:
$$\pu{K_a}=\frac{x^2}{C_{\pu{Ao}}-x}\implies C_{\pu{Ao}}=\frac{x^2}{\pu{K_a}}+x$$
Substituting in $\alpha$:
$$\alpha=\frac{10^{\pu{-pH}}}{\frac{10^{\pu{-2pH}}}{\pu{K_a}}+10^{\pu{-pH}}}=\frac{1}{\frac{10^{\pu{-pH}}}{\pu{K_a}}+1}$$
We can observe that:
$$\pu{K_a}\to\infty\implies \alpha\to1$$
$$\pu{K_a}\to0\implies \alpha\to0$$
If we plot $\alpha$ vs $pH$ for $\ce{HCl}$ and $\ce{HCOOH}$, we get:

Now, let's suppose someone were to hand you an aqueous solution of an unknown acid with a known $\pu{pH}$ value of 2.
You might be tempted to conclude that since the $\pu{pH}$ is low, the acid must be a strong acid like $\ce{HCl}$ that is completely dissociated ($\alpha=1$); however, as shown in the plot, it could very well be a weak acid like $\ce{HCOOH}$ that is partially dissociated ($\alpha=0.017$).
In other words, $\pu{pH}$ alone is not a reliable criterion for gauging the strength of an acid. If it is coupled with $\pu{K_a}$ however, it can be.


text:
Rates of a chemical reaction
Consider the reaction between Calcium Carbonate and Hydrochloric Acid. I understand the reaction rate can be increased by increasing the concentration of reactants. However I can't seem to understand the intuition behind it. If we were to increase the quantity of Hydrochloric Acid, would it not also increase volume of the reaction having no affect on the reaction rate. Furthermore does increasing both reactants increase the rate of the reaction? Increasing both would increase the volume meaning their would be the same concentration of Calcium Carbonate per unit volume of Hydrochloric Acid. So shouldn't the reaction rate not be affected.

Rates of heterogeneous reactions are always very tricky, as there is many unknown variables.

Quantity and concentration are different things. 2L of 36% HCl does not have higher concentration than 1L. But 1L of 36% HCl has higher concentration than 1L of 20% HCl.

Please try to use meaningful yet concise titles that reflect as much as possible what the question is about without opening it. Note that chemical names are common nouns and follow corresponding capitalization rules.

Equations of reaction rates are based on statistical reasoning about collision frequency, which is supposed proportional to the reaction rate at given temperature.
Imagine
$\pu{1 mmol/L}$ of $\ce{A}$ and $\pu{1 mmol/L}$ of $\pu{B}$ forms $\ce{C}$ at rate $\pu{1 mmol L-1 min-1}$ by a simple bimolecular reaction:
$$\ce{A(aq) + B(aq) -> C(aq)}$$
$\pu{2 mmol/L}$ of $\ce{A}$ and $\pu{1 mmol/L}$ of $\pu{B}$
would lead to doubled $\ce{A}$ and $\ce{B}$ collision frequency and doubled reaction rates.
$\pu{2 mmol/L}$ of $\ce{A}$ and $\pu{2 mmol/L}$ of $\pu{B}$
would lead to quadrupled  $\ce{A}$ and $\ce{B}$ collision frequency and quadrupled reaction rates.
Therefore:
$$\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} = k [\ce{A}][\ce{B}]$$
Rates of heterogeneous reactions are always very tricky, as there is many unknown variables. Such reactions of gases or liquids with solids complicate everything. As quantity more or less analogic to concentration, we could use the parameter:
$$\text{"surface density"} = \frac{\text{"total surface area of the solid"} }{\text{"(total) volume of suspension"}}$$
assuming suspension is evenly distributed in the volume. It would start failing for bigger chunks, sitting on the bottom, on at sedimentation of not well stirred mixture.
Note that this parameter has non trivial dependence of the reaction quotient (progress). It does not decrease linearly, as particle sizes decrease, particle surface/mass ratios increase and smaller particles react relatively faster.
If we imagine $\ce{B}$ is solid and reaction $\ce{A(aq) + B(s) -> C(aq)}$, it could be like:
$$\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t} = k [\ce{A}] \cdot \text{"surface density of B"}$$
But $k$ is not really a constant (for given $T$) here, as it depends to many factors, e.g. the way or intensity of mixing, or forming other phases lile a gas. There remain many unknown variables, mostly related to diffusion in heterogenous environment and surface layer phenomena. That makes models and rate equations reflecting reality semi-empirical at the best.


text:
Relation between symmetry of a reciprocal vector and lattice planes perpendicular to it
What is the relation the symmetry of a high-symmetry point in the first Brillouin zone and lattice planes perpendicular to it? Are the two symmetries equivalent?
I have this question because I want to "visualize" the symmetry operations of a high-symmetry point (a vector w.r.t. $\Gamma$ point) in the first Brillouin zone in direct space. I think the point group of a wave vector is equivalent to the point group that leaves the lattice planes perpendicular to the wave vector invariant. But I haven't find a solid proof for this hypothesis.
Below is a proof by myself:
Let $K_m = (h, k, l)$ denotes a high-symmetry point in the first Brillouin zone, $P_\alpha$ a symmetry operation of the point group that leaves $K_m$ invariant, $R_n$ a lattice vector with its end point on a $(hkl)$ plane. By definition of reciprocal lattice, we have
$$(P_\alpha K_m) Rn = 2\pi N_1$$
where $N_1$ is an integer. Since a point group operation on a constant is still a constant, we act $P^{-1}_\alpha$ on each element of the above equation, and we get
$$P^{-1}_\alpha (P_\alpha K_m) (P^{-1}_\alpha R_n) = 2\pi N_1 = K_m (P^{-1}_\alpha R_n)$$
Therefore, $P^{-1}_\alpha R_n$ is also a lattice point on the same $(hkl)$ plane. Since the same proof can run in the other direction, the point group of $K_m = (h, k, l)$ is equivalent to the point group which leaves the $(hkl)$ lattice plane invariant.
Is my proof correct? I doubt it since only the magnitude of $K_m$ is not considered, while $K_m$ in the same direction while of different magnitudes can be of different point groups. For instance, in rutile TiO2, the point group of high-symmetry point X (0.5, 0, 0) is $D_{2h}$, while the point group of (0.25, 0, 0) is $C_{2v}$.
The point groups (little co-group) of the two high-symmetry points were queried from Bilbao Crystallographic Server

I suspect you will be more likely to get an answer to this in the matter modelling group - it has a more mathematical outlook and periodic systems are more commonly looked at there. It is also dealt with by the code CRYSTAL (https://www.crystal.unito.it/index.php) when it applies symmetry adaption to the reciprocal space representation of the wavefunction, but that's a part of the code I have never really got my head around - full disclosure, I am an author of CRYSTAL.

What is the relation the symmetry of a reciprocal vector and lattice planes perpendicular to it? Are the two symmetries equivalent?

The meaning of symmetry in reciprocal space is unclear. If you are just talking about the location of lattice points in reciprocal space, every point on this lattice has the same symmetry. If you are also talking about the intensities (like a diffraction image), all symmetry elements include the origin, so other points lack point group symmetry.
However, you can ask the question whether a point k in reciprocal space lies on a symmetry element, i.e. whether applying the symmetry elements results in different k-vectors or the same.

Let $K_m = (h, k, l)$ denotes a high-symmetry point in the first Brillouin zone [...]

The Brillouin zone refers to a part of reciprocal space that only includes a single reciprocal lattice point (the origin). Conventionally, $h, k, l$ are used as indices for reciprocal space points, i.e. whole numbers. Here whole numbers would not make sense. If the numbers are not whole, no lattice planes in real space are associated with the point in reciprocal space, so the definition of $R_n$ does not make sense, and $N_1$ would not be an integer.

Is my proof correct?

As I don't understand the first part of the proof, I find it difficult to tell.
This paper introduces the concepts underlying the data of the Bilboa database that are now referenced in the question.

Thanks! The symmetry of a reciprocal lattice vector is referred to the little co-group of a high-symmetry point in the first Brillouin zone, which can be queried from Bilbao Crystallographic Server. The $D_{4h}$ is actually a typo, it is $D_{2h}$ point group from Bilbao Crystallographic Server.


text:
Why does XeO3 not hybridize d orbitals?
Apologies if this is a dumb question. In my chemistry class we are discussing hybridization, and an example that came up for $sp_3$ hybridization was the compound $\ce{XeO3}$.
However, this seems counterintuitive. While the lewis structure of the compound is

and by inspection we see that there are 4 groups of electrons, the three double bonds and the lone pair which implies $sp_3$, from my understanding, the ability of xenon to form compounds relies solely on the fact that Xenon can use its $d$ orbitals to bond.
Why are the $s$ and $p$ orbitals implicated in the bonding and not the $d$ orbitals?

https://chemistry.stackexchange.com/questions/67352/about-dative-covalent-bonding-and-expanded-octet-rule

@Mithoron From the first link you shared, can I basically interpret this as $\text{Xe}$ bonding with $\text{O}$ in a way where xenon sends an electron to oxygen but not vice versa, so the bonding only inolves the s and p orbitals?

Yes, the bonds can be considered dative. Also there's hardly any involvement of d orbitals in hybrids in every compound you'd think of as "hypervalent".

https://chemistry.stackexchange.com/questions/167501/bonding-in-hypervalent-molecules

BTW titles shouldn't have mathjax because of searchability issues.


text:
Diffusion vs mobility and the Einstein relation
This former question is about the mobility constant $M>0$ in the Cahn-Hilliard equation. To determine the value of such a mobility for a simulation study, Kim and Sanders (2020)
use the formula
$$
M = \frac{D \cdot N_A}{\partial^2 G / \partial c^2},\qquad \text{(*)}
$$
where $D$ is the diffusion coefficient, $N_A$ is the number of particles per unit area, $G$ is the molar Gibbs free energy and $c$ is the molar fraction.
Question: Where does the formula (*) come from? Does this formula have something do to with the Einstein relation?


text:
Enthalpy and Bond Energy Sign Convention
I read that a negative enthalpy of formation means an exothermic reaction. Does that mean this reaction is true? The enthalpy of formation for $\ce{H_2O}$ is -285.8 kJ/mol.
$$\ce{2H2(g) + O2(g) -> 2H2O(l) + 285.8*2 kJ}$$
What about the enthalpy of reaction though? Is it also negative for exothermic reactions, like the one above? I think it's calculated with: $\Delta H$(Rxn) = Sum of Heat of Formation of Products – Sum of Heat of Formation of Products.
What about bond enthalpy (bond energy)? I know the calculation is different:
$\Delta H$(bond energy) = Sum of Bond Energy of reactants – Sum of Bond Energy of products, so are positive bond enthalpies exothermic (like in the example above)?
Finally, for Gibbs free energy, $\Delta G=\Delta H-T \Delta S$, is enthalpy in the equation for enthalpy of reaction since it's the change in enthalpy?

Enthalpy of product formation = enthalpy of reaction (if normalized to the same amount), if reactants are elements in their standard state, as the enthalpy of formation of the latter is conventionally zero.

When writing $\ce{2H2(g) + O2(g) => 2H2O(l) + 285.8⋅2 kJ}$, the sign plus (+) before $285.8$ shows that the reaction is exothermic, and that this amount of heat is getting out of the system. But it also implies that $\ce{\Delta H_{reaction}}$ is negative, and equal to  $- 285.8·2$ kJ !