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Why must the total increase in oxidation numbers equal the total decrease of them in redox reactions?
In my book, under the topic of redox reactions, it is given,

In all redox reactions, the total increase in oxidation number must be equal to the total decrease in oxidation number.

Is this a consequence of "Charges can never be created. The total charge of a system remains constant", or something else?
For the reaction,
$$\ce{
10 \color{red}{FeSO4} + 2 \color{red}{KMnO4} + 8 H2SO4 -> 5\color{red}{Fe2(SO4)3} + 2\color{red}{MnSO4} + K2SO4 + 8 H2O
}$$
(Reactants and products in red undergo oxidation or reduction)
For the above reaction, I tried whether the property stated above works or not.
The oxidation state of Fe and Mn in the reactants side are +2 and +7 respectively. The oxidation states of these elements in the product side are +3 and +2 respectively. The increase in OS is 1 and the decrease is 5.
Finally, the increase and decrease in oxidation states are not equal.
Kindly give an explanation for this fact.

In any balanced redox equation, there are no free electrons on either side of the reaction. So whatever electrons were gained by the one or several oxidizing agents must have come from the one or several reducing agents. The total number of electrons gained must equal the total number of electrons lost. Now think about what happens to oxidation numbers when electrons are gained or lost. And look over on the right side for examples of redox reactions, etc.

I am deleting my answer because I do not consider it to be anywhere near good enough to be called “high quality”, and that it the desideratum of this site.

Can you please explain how you arrived at the statement "Finally, the increase and decrease in oxidation states were not equal." You might have made a mistake calculating that.

Please add a citation for the quote you are using, otherwise this can be considered plagiarism.

@Martin-マーチン, The statement is from the book Physical Chemistry for JEE Mains and Advanced - Resonance DLPD. It's just a book from one of the coaching institutes in India.

@Martin-マーチン, The oxidation state of Fe and Mn in the reactants side are +2 and +7 respectively. The oxidation states of these elements in the product side are +3 and +2 respectively. The increase in OS is 1 and the decrease is 5. Clearly, both changes are not equal and hence I arrived at that particular statement that you have mentioned.

The 'total' is the key word in the given statement. You have to sum up all the oxidation states. Hence 10 times +1 is +10, and 2 times -5 is -10. The decrease therefore equals the increase. Thank you for editing in your thoughts, this makes the problem a lot more understandable.

@Martin-マーチン, Thank you for your reply. Got it, we need to consider the stoichiometric coefficients in calculating the change in oxidation numbers. One of the two issues clarified! Could you please tell the reason for this if possible?

So it is exactly what I said in my first comment and like @Martin-マーチン said: total means total. Each of the 2 Mn atoms formally gains 5 electrons, for a total of 10 gained. Each of 10 Fe atoms loses 1 electron, for a total of 10 lost. And, as Poutnik says, oxidation numbers are just a matter of formality. Basically a very useful accounting notion.

The reaction is with an advantage written in the simplified ion form:
$$\ce{5 Fe^2+ + MnO4- + 8 H+ -> 5 Fe^3+ + Mn^2+ + 4 H2O}$$
Total oxidation numbers $+17=+17$
The oxidation number is formally the net charge an atom within a molecule would have held, if we had broken all bonds and bond electrons  had sticked to atoms according the atom electronegativity.
Therefore, the constant total sum of oxidation numbers across all involved atoms directly relates to the constant total sum of valence electrons and to the law of charge conservation.


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Why are the equilibrium exponents stoichiometric when the rate law's are not?
I'm learning about rate law and equilibrium, and the textbook really hammers in that the exponents for the rate law must be determined experimentally - you can't just use the coefficients from the balanced equation. In a generic reaction aA + bB -> cC + dD:
rf = kf[A]w[B]x
rr = kr[C]y[D]z
It is not always the case that w = a, x = b etc.
Equilibrium is then described as the point that the forward reaction is occurring at the same rate as the reverse reaction. For the same reaction as above, we set the two rates to equal each other.
rf = rr
kf[A]w[B]x = kr[C]y[D]z
Kc = kf / kr = [C]y[D]z / [A]w[B]x
I'm told that the exponents here are based on the coefficients of the balanced equation, such that w = a, x = b, y = c, and z = d. This seems to be in direct opposition to what I read earlier about the rate law, as equilibrium is just built from two rates.
Why is this?

Interesting question, but it has been addressed nicely https://chemistry.stackexchange.com/questions/6243/how-is-it-that-the-equilibrium-constant-does-not-depend-on-the-mechanism. Basically, you have to distinguish something called as the elementary reaction where the rate constant is indeed based on the equation as written and the overall reaction.

Equilibrium is not necessarily "just built from two rates"; you don't actually need the concept of a rate to define an equilibrium. The approach you're taught is the simplest and most intuitive, which is why it's taught first, but it has its limitations. Refer to any physical chemistry / thermodynamics textbook.

@orthocresol Thanks. So a higher level definition of equilibrium is NOT "the point at which the forward and reverse rates are equal?" This is just something that is intuitive but already falls apart when trying to use rates equations to define equilibrium equations.

It only works when your forward and reverse reactions are elementary reactions, in which case the coefficients in the rate law are indeed the stoichiometric coefficients. However, the equilibrium constant can be proven to have this form (involving the stoichiometric coefficients), even when the forward and reverse reactions are not elementary. I am trying to find the correct links, it has probably been asked/answered here a few times.

I suggest reading https://chemistry.stackexchange.com/questions/75274/relation-between-chemical-kinetics-and-chemical-equilibrium which also (perhaps more directly) addresses your question.

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Can you heat something at alcohol bath?
I have a mixture of $\ce{Et2O},$ $\ce{EtOH}$ and water, and I want to distill off the ether. I am using a water bath so not much water is distilled over, as the boiling water is 100 °C, and the solvent has like 20 °C lower temp than the water bath working temp, I believe.
So, my point is, is it safe and reasonable to make a bath with 95% ethyl alcohol to prevent much $\ce{EtOH}$ evaporation without needing to closely keep track of the temperature?

No, it's not a good idea.

Ok, but why not?

Well, that depends if you try to make a safe experiment, or explosion.

It would be extremely dangerous to use a flammable solvent for a heat bath because any spill on the hot plate (or even worse, a Bunsen burner) could cause a fire.
If you want to distill off ether, it is by far better to just use warm water from a tap or a kettle, and replace your hot bath regularly.
Trust me, ether can spontaneously catch fire VERY easily ! I have been there!

I get you, but I am distilling about 100ml of ether in a really well ventilated area. I am not blowing my self. I get it though, it is a fire hazard...

Such distillation is done every day in great quantities in closed ( air free) stills ; they  are called refineries and chemical plants.

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How would you order alkyl groups of this compound numerically?

If akyl groups are ordered alphabetically and ethyl comes before methyl, would this compound be called 2-ethyl-3,4 dimethyl pentane? Since that would be less than 4-ethyl-2,3 dimethyl pentane, correct?

You picked wrong chain... It's like most common test trick ever.

How would you choose the right chain?

You need longest chain and you didn't think what really should be in this chain and not be a substituent.

Would it be 2,3,4 trimethyl hexane? Since the longest chain would start from C2H5 then, without CH3 because that is a substituent?

There should be no spaces in the name, though.


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Applying the constant number of electrons in DFT density optimisation
(Previous related question: Finding mathematically the ground state density in DFT)
I am studying the density optimisation procedure (in particular for Orbital-Free DFT) this thesis.
The derivative of the energy is given as:
$$\frac{\delta E[n(r)]}{\delta n(r)} = v(r) + v_i(r) + \frac{1}{2}(3\pi^2)^{2/3}\left(n(r)\right)^{2/3},$$
where $v(r)$ is the external potential and $v_i(r)$ is the Hartree potential, or electron-electron repulsion.
The chemical potential can thus be obtained after integration:
$$\mu = \frac{1}{V} \int_{\Omega} \frac{\delta E[n(r)]}{\delta n(r)}dr.$$
What confuses me is the next equation, which allows to find the density of the next step $k+1$:
$$n(r)^{(k+1)} = n(r)^{(k)} - t\left(\frac{\delta E[n(r)]}{\delta n(r)} - \mu\right),$$
where $t$ is a step size to help convergence.
If I understand this correctly, $\mu$ is an average variation of energy. The density at a given $r$ is thus varied by the difference between the derivative at that position and the average derivative.
$v(r)$ is a constant for a given position, and $v_i(r)$ depends only on the density, just like the kinetic term above (Thomas-Fermi in this case). It thus seems like the variation of density will never quite get to zero except if the density is exactly the right value to balance the external potential. This seems simplistic and wouldn't explain orbitals beyond the $s$ orbital.
In parallel, I am implementing a simple code for the hydrogen atom, but I can't get anything to converge nor give reasonable values; I must be doing something wrong somewhere.


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Which is more basic: oxaziridine or diaziridine?
Which is more basic: oxaziridine or diaziridine?

Oxaziridine's got one nitrogen and one oxygen as heteroatoms, while in diaziridine there are both nitrogens.

I don't think this is a bad question per se. However, since we have a homework policy, it would be nice if you could elaborate a bit more and add your thoughts or reasoning (besides stating the fact how many atoms are there) in order to prevent the question from being closed.


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Is it safe to boil water in a foil pan on an electric burner?
Is it safe to boil water in a foil pan on an electric burner?
Can the metal leach into the water?
I've heard some reports recently about avoiding aluminum foil due to it leaching into food.
Photos of the items in question, below:

Unless the food is acidic, Al won't leach much.

@IvanNeretin "much" ?

Not significantly. However this alu pot might just fail, depending on how hot the heater coil becomes. Also you canot properly hold it when hot. Good luck with your hands. My advice is to not do it. Totally inappropriate.

Personally, I would put a single layer of aluminium over each electric burner (just in case) and then the aluminium container. As long as you regularly check, there should be no problem. And for the aluminium ions entering your body, you may get much more of them from your antiperspirant (depending on which country you live in).

Never put aluminum foil or thin Al on a burner: it will melt onto the heating element, possibly causing uneven cool and hot spots afterwards. That is not a problem as long as the pan has water, though.

@Karl I've already been doing it dozens of times. The pan doesn't melt if you set it on low/med and it has water in it. My only question was regarding the safety.


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Is there a commercial liquid with refractive index greater than two?
I need a commercially available liquid with the following properties:

refractive index $n > 2$ at room temperature $(T \approx\pu{23 °C});$
transparent in visible region $(380$ to $\pu{700 nm});$
suitable for gap fillers in refractometers (gap between prism and stones);
not poisonous.

What is the chemical composition of such liquid? The RI liquids with related properties I found online usually only have $n = 1.81.$

@Persian_Gulf So a liquid with n>2 but which only transmits, say, 390nm-700nm is no good for you?  It must have transmission down to 380nm?  What is a tolerable absorption coefficient over than range?  n is also not constant over the range in any material.  Do you care what the index dispersion of the liquid is?

By any chance, does this question have to deal with gemology and OTL ("over the limit") refractive index measurements of gemstones? If so, please see this URL for how to modify a microscope to measure refractive indices above 1.81: https://www.gemsociety.org/article/measuring-gemstone-ri-1-81/ . Given the stated requirements (room temperature; 380-700 nm; transparent), the standard sulfur in methylene iodide is the practical solution for gemological and mineralogical testing with the standard methodology.

there is such a commercial product. immersion glass for the diagnosis of diamond defects, commercial name s255_1. its predominance is 2.41, the softening temperature is 130 degrees Celsius. we use such a drug.

Can you add a link to where someone can find basic information for this product?

So not a liquid at 23 degrees Celcius, as specified by the OP.

Taking the definition of refractive index $n = c/v$ into account, $n > 2$ means the velocity of light in the material $v$ is less than half than in vacuum $c,$ so there must be a strong interaction with the material.
One of the highest refractory index in liquids is $1.79~(\pu{20 °C})$ possessed by a solution of sulfur in methylene iodide $(\ce{CH2I2}).$ Liquids are less dense than solids and this explains why it's difficult to have a refractory index greater than 2. According to Anderson and Payne [1]:

Selenium monobromide, $\ce{Se2Br2}$, has a higher refractive index than that of any pure liquid hitherto recorded. Prepared by direct combination, the value for $\nu$ is $1.96±0.01$ rising to $2.02$ on exposure to the
atmosphere, owing to decomposition of the bromide, with separation and reabsorption of selenium.

From a more recent paper [2], selenium monobromide is confirmed as one of the liquid with the highest refraction index $(n > 2).$ However, diiodomethane is more easily accessible commercially.

A promising candidate solvent is the organic liquid diiodomethane $(\ce{CH2I2}),$ which is one of the liquids with the highest known refractive index values $(n = 1.74).$ While other high refractive index liquids exist (phenyldiiodoarsine $(\ce{C6H5AsI2})$ with $n = 1.85$ and selenium monobromide $(\ce{Se2Br2})$ with $n = 2.1),$ diiodomethane has the key advantage of being commercially available. In addition, diiodomethane is an excellent solvent, and many liquid formulations using salts dissolved in diiodomethane are reported to increase the refractive index and are even available commercially.

References

Anderson, B. W.; Payne, C. J. Liquids of High Refractive Index. Nature 1934, 133 (3350), 66–67. DOI: 10.1038/133066b0.
Laskar, J. M.; Kumar, P. S.; Herminghaus, S.; Daniels, K. E.; Schröter, M. High Refractive Index Immersion Liquid for Superresolution 3D Imaging Using Sapphire-Based Aplanatic Numerical Aperture Increasing Lens Optics. Appl. Opt., 2016, 55 (12), 3165–3169. DOI: 10.1364/AO.55.003165.

Ah, and why you'd think sulfur in any solvent would be transparent?

Important note, "hitherto recorded" refers to 1934, the year of the paper.

@Mithoron it is sulphur in CH2I2 not in any solvent!

To complement @blu_potatos's answer, there are some other contenders having refractive index greater than 2

Arsenic di/trisulfide and sulfur and/or selenium and/or mercuric sulfide in arsenic tribromide solvent having refractive index ranging from 2.0 to 2.07 (arsenic tribromide seem to attack the lead-glass prisms of refractometers, corroding metal, poisonous in nature and reactive toward some minerals)
phosphorus and/or sulfur in carbon sulfide solvent having R.I= 2.01/2.07 (carbon disulfide is highly volatile, very flammable and poisonous.
Solutions containing white phosphorus should be kept in the dark as
light causes the conversion of the white phosphorus to the red form,
which is more poisonous).
Selenium and/or sulfur in phosphorus solvent having R.I =2.2-2.5 (Solutions of selenium in phosphorus cannot be stored for long under water as they decompose).
Selenium in selenium monobromide solvent having R.I>2.02 (Selenium monobromide is unstable decomposing to selenium)
A ternary system of phosphorus-sulfur-methylene iodide having R.I upto 2.06 has been observed

\begin{array}{c|c}
\mathbf{component} & \mathbf{R.I} \\\hline
\text{P(solid) at 29°C } & \text{2.15}\\
\text{P(liq.) at 44°C } & \text{2.10}\\
\text{P-S-CH2I2 at 25°C } & \text{2.06} \\
\text{P-CH2I2(satd.) at 18°C} & \text{1.94}\\
\text{CH2I2 at 15°C } & \text{1.74}\end{array}
There are loads of other liquids having R.I value in between 1.5-2.0 which are stable and can be used commercially. See references
References

http://www.minsocam.org/ammin/AM40/AM40_398.pdf (1-4)
http://www.minsocam.org/ammin/AM21/AM21_245.pdf (5)
https://www.osapublishing.org/DirectPDFAccess/71A8545D-AFC6-6350-82F044C815273272_338928/ao-55-12-3165.pdf?da=1&id=338928&seq=0&mobile=no

Are you just listing the real part of a complex index of refraction of opaque liquids? Or is a solution of selenium in phosphorous somehow transparent at some wavelengths? (cf. Born & Wolf; Alkali metals transparent to UV? Cesium transparent to blue?)

Hmmm, none of these seem especially attractive to use in a classroom demonstration ...

@davidbak yeah, I won't recommend using those chemicals. They are nasty.

-1 Seriously, what does selenium dissolved in phosphorous actually look like? Are you sure it's not metallic and opaque, and you've simply dropped the imaginary part of the index of refraction? Which materials in your list can actually transmit light to any appreciable degree?

Solutions of phosphorous in carbon disulfide are not only smelly and toxic, but they're hideously prone to catching fire. Methylene iodide is carcinogenic. Selenium and its compounds are toxic and notoriously evil-smelling. Because we're playing around in the periodic table's bad neighborhood, high refractive index (RI) liquids are uniformly extremely unpleasant.
The RI of liquid iodine is 1.934 at 114 °C [1]. This might be the best candidate for conducting experiments. Transparency in the visible range (above 450 nm) is very limited, but at least the stuff won't incinerate you, kill you, or induce your neighbors to kill you.
Reference

Donaldson, A.; Caplin, A. D. The Optical Properties of Liquid Bromine and Iodine. Philos. mag. B, 1986, 54 (3), 231–239. DOI: 10.1080/13642818608239022.

In gemological testing of the refractive indices of gemstones, the upper limit of refractive index is $1.81,$ if a contact liquid is used in a conventional gemological refractometer. See, for example, The International Gem Society — How to Measure an OTL Gemstone Refractive Index. As noted in The Gemology Project — Refractometer, the two high refractive index contact liquids are

a saturated solution of sulfur in diiodomethane $(n = 1.79);$
a saturated solution of sulfur, diiodomethane and tetraiodoethylene $(n = 1.81).$

Not the most pleasant solutions, but much better than the exotic and far more toxic possibilities that

are so toxic that they are only used in specially equipped laboratories.

There is no such RI liquid that would satisfy all parameters requested by OP.

Besides the already mentioned negatives of 'Highly Toxic', 'Noxious', 'Flammable', 'Carcinogenic', 'Corrosive', it needs to be understood that in order to use such high RI materials, a suitably high RI Prism or Hemicylinder and appropriate scale are also required ( there were only ever a few such instruments ever produced - using, Sphene, Strontium Titanate and Diamond - none of which were made commercially ).
My first response to your question was - what is it that you are looking to test ?
If it is a natural ( or even synthetic ) OTL stone I would suggest a 'Digital' Refractometer ( ie. Reflectometer ), which although somewhat inaccurate, will get you close - then confirm it with an SG.
NM Aug, '22

Sucrose or glucose in H2O = approx 1.4-1.5 depending on composition and concentration
Glycerol = 1.473
Toluene = 1.5
Chlorobenzene = 1.524
Getting higher than that without very high flammability/toxicity/reactivity is hard

See https://en.wikipedia.org/wiki/List_of_refractive_indices for some approaching 2

If someone's looking for a refractive index greater than 2, a list of substances with refractive indices around 1.5 isn't very useful.

@Mark I have a feeling that answer telling no, it's a bad idea, even indirectly, is only appropriate here. Much better then then listing hideous and mostly opaque things, just because they're on some list.


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Atmospheric methane to carbon
I read an article by popsci that talked about huge fans forcing methane over a catalyst.
Now that got me wondering, can't we produce hydroxyl radicals with powerfull uv lasers blasting the sky, creating more hydroxyl radicals to break down the methane.
Would this be effective enough to help create enough hydroxyl radicals to help break down methane or would it be a waste of energy against too small an offset?

It's kinda mean to insta-fry unsuspecting birds and bats. ;-)

Arising questions: a) How much methane is in the atmosphere, b) how effective is a hydroxyl radical in destroying methane, c) how effectively can we produce them, d) can we actually increase the local concentration, e) how long does the effect last, f) does it make sense to e.g. increase the concentration overnight? If you have all that info, it's easy to put the numbers together and make a business case for it.

It would be a total waste of energy.
Remember that the Sun is sending orders of magnitude more energy than humanity can produce. A significant fraction of that is UV, so the Sun is already breaking up more Methane than humanity could do by adding laser diodes or whatnot.
We might be able to do more at the ground, where the Sun's UV is mostly filtered.
It would still be an enormous feat to merely reach the Sun's UV level.
However, there is a big reason NOT to do it at the ground: UV does not only break up methane, but also oxygen. Part of that will recombine to ozone, which is an irritant (in the medical sense, i.e. extended-time exposure will cause damage no matter what).
(I do not know whether the oxygen bonds are stronger or weaker than the methane bonds. If methane has the weaker bonds, one could use an UV wavelength below the oxygen energy. You'd still break up bonds of any atmospheric molecule that has weaker bonds than methane, such as pheromones and whatnot; it is entirely unclear what consequences that would have.)

The ozone argument isnt really one: Hydroxyl radicals are made via singlett oxygen, which is also resonsible for ozone generation. As long as your artificial light source is less intense than the sun, it wont create unhealthy ozone levels. Of course it likely also wont make enough hydroxyls to make a difference. ;)

Powerful ground lasers are surely inefficient. The radicals are created in close proximity, meaning they likely have a good chance to recombine. Also highpower lasers (or any strong light source) are generally not very efficient. A lot of small ones would be better.
I would mount simple UV LEDs on the underside of passenger planes, where they are switched on to illuminate clouds at night (there is no sense in doing it during daytime, obviously). If that's enough to lift the project over the "waste of time and energy" limit, I'm not so sure.

So if you would take a couple of those superlong solar powered self flying planes and patch them up with uv leds you could have a sustainable solution?

@Tschallacka No. You just make it mandatory for all planes landing in US or EU to have those and present a usage log. A kilowatt or two of electricity are basically free on a jet airplane. Much better scaleability.

At the risk of sounding too obvious, doesn't the sun already illuminate clouds with UV?

@immibis Well, only during the day. But you're right. The underside of clouds is likely the only place where this might possibly make sense.


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Deriving a Real Gas equation
I've read about the ideal gas law and the van der Waal's gas law. I know that the ideal gas law is horribly inaccurate for temperatures close to, but above the critical temperatures, so it can only be used for permanent gases like nitrogen, oxygen, etc. The van der Waal's equation is a little better in this regard, but also fails for temperatures below the critical one, notaby by the "van der Waal's loops".
I looked into how the van der Waal's equation and how it was derived. Reading the Wikipedia page about it, I find that van der Waal's assumed that gas molecules are hard incompressible objects that have a certain volume (evident by the constant $b$, defined as the volume occupied by one mole of molecules themselves), and these molecules attract each other (I couldn't find how these molecules attracted each other, like perhaps inverse square or cube?). I find the rigid volume assumption a little too generous. I also know that molecules attract each other at long ranges due to London dispersion forces, but these dispersion forces also act as repulsions when the molecules get too close. We can describe this by the potential energy of the two molecules:

Differentiating this curve (a.k.a Morse potential) should give us the force between two molecules. Now what we have is a single expression for both attraction and repulsion, rather than one for attraction and another one for repulsion (as hard spheres, in van der Waal's attraction). This single expression of attraction and repulsion gives the molecules some "squishy" nature which I believe is how real gases behave.
Naturally, I want to derive a gas equation using the Morse potential graph, rather than how van der Waal's has assumed. This is where I hit my roadblock. So far I have compared the derivation of the van der Waal's equation, and I have tried to make modifications as per the Morse potential's force law. Here are some conclusions I came to:

The original van der Waal's equation is $$(P+\frac{a}{V^2})(V-b)=RT$$ I believe the new equation that I plan to derive will not have any volume correction parameter, since with our assumption of how molecules are "squishy", there is no correct volume correction.

I plan to incorporate repulsions into the pressure correction factor as well. My equation should be of the form $$(P+K_0)V=RT$$ where $K_0$ is a factor that corrects pressure due to both attractions and repulsions.

I looked at the derivation of van der Waal's and found this

Next, we introduce a (not necessarily pairwise) attractive force between the particles. van der Waals assumed that, notwithstanding the existence of this force, the density of the fluid is homogeneous; furthermore, he assumed that the range of the attractive force is so small that the great majority of the particles do not feel that the container is of finite size.[citation needed] Given the homogeneity of the fluid, the bulk of the particles do not experience a net force pulling them to the right or to the left. This is different for the particles in surface layers directly adjacent to the walls. They feel a net force from the bulk particles pulling them into the container, because this force is not compensated by particles on the side where the wall is (another assumption here is that there is no interaction between walls and particles, which is not true, as can be seen from the phenomenon of droplet formation; most types of liquid show adhesion). This net force decreases the force exerted onto the wall by the particles in the surface layer. The net force on a surface particle, pulling it into the container, is proportional to the number density
$$C=\frac{N_A}{V_m}$$
The number of particles in the surface layers is, again by assuming homogeneity, also proportional to the density. In total, the force on the walls is decreased by a factor proportional to the square of the density, and the pressure (force per unit surface) is decreased by
$$a'C^2 = a'\left(\frac{N_A}{V_m}\right)^2 = \frac{a}{{V_m}^2}$$

So we can see that the pressure correction factor is described by considering the force experienced by the particle is proportional to the number density of the particle. This seems well and good for the van der Waal's equation, since it only considered attractions. In my case, I dont think it's so simple. Molecules closer to the molecule in consideration would repel it, and those further would attract it. I have no idea on how to incorporate this in terms of number density, even though I believe it's possible (denser gas would mean more attractions as well as more repulsions, and depending on the Morse potential is what would decide if actual pressure is more or less). This where I need help. How do I incorporate the Morse potential curve into the pressure correction factor?

Have you reviewed also other real gas models ? https://en.m.wikipedia.org/wiki/Real_gas

@Poutnik I have. But I wish to see if what I've written above can help me get an equation. I dont want the "best" gas equation, but hopefully just one that's more accurate than the van der Waal's equation.


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Testing for manganese in water
Is there a decent way to test for solvable manganese in water without purchasing a 150USD test kit? I have a fair chemistry set available (I use it to teach my kids chemistry), and would rather pay 50USD for something to add to it that can also be used for other things later than pay for some test that may not be that accurate anyway.
Our city water supply to our house produces a black slime that is likely from bacteria that feeds on iron or manganese. I’d like to start getting to the bottom of it. This is a first step.

Forget about iron and manganes (you will definitely find some). Your water supply produces some black slime, period. Could be a problem in your house, or in the street. Likely both, now. Somehow the pipe system is contaminated. Alert your city officials, this is just inaccetable. Btw., are you sure this is in the pipes, and not just in your faucet?

Unfortunately, for random strangers on the internet, there is perhaps no way situations like this can be adequately explained. Even the best answers would be nothing more than educated guesses, which is why I'm afraid I have to vote to close this question as "unclear".

@Karl yeah, it’s several faucets on two floors, unfortunately.

@M.A.R. - perhaps that’s true in this case. I don’t know, which is why I asked. But there are often simple options. Kjeldahl Digestion for nitrogen would be an example. A Prussian Blue test for cyanide would be another. Is there such a method for soluble manganese (quantitative)?

Best way is sending a sample for Atomic Adsorption Spectroscopy (AAS). You can get the concentration for $\ce{Mn, Pb, Fe, Cr, Ni}$, and other water goodies. Contact a local university or lab and get a price quote for a test.
Or get a flame spectroscope and DIY https://web.pdx.edu/~atkinsdb/teach/427/Expt-AtomicSpec.pdf


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Mixtures readily oxidzing organic matter
Are there other mixtures than piranha solution with similiar or better ability to completely and readily oxidize organic matter? Perhaps peracetic/performic acid?

Is it a practical question for a certain application (e.g. cleaning glassware), or just a search for the best oxidant ( there are plenty exotic compounds and mixtures such as $\ce{ClF3}$ or liquid ozone/oxygen)?

...and there is manganese heptoxide. I would recommend staying away from it.

In biophysics, when you want to destroy anything organic on a platinum probe(1), you usually use sulfochromic acid (mixture of chromic acid and concentrated sulfuric acid). I do not know anything stronger.
(1) A platinum probe is used just because it can handle being subjected to sulfochromic acid. Most other metals would not.

There are many, many dozens of substances and mixtures that can readily oxidize organic matter, in some cases explosively. The degree of completeness of the oxidation process is dependent on the specific circumstances, e.g., is the organic matter sucrose or a piece of wood?
That said, a few notable oxidizing substances (or liquid mixtures) are the various sulfuric acid family members (https://en.wikipedia.org/wiki/Sulfur_oxoacid), chlorosulfuric acid, fluorosulfuric acid, nitrosylsulfuric acid, white fuming nitric acid (WFNA), red fuming nitric acid (RFNA), hydrogen peroxide (30% or higher), $\ce{FOOF}$, $\ce{OF2}$, $\ce{ClF3}$, ‘mixed acid' (WFNA with 10-17% $\ce{H2SO4}$), $\ce{FNO3}$, $\ce{FClO4}$, $\ce{ClF}$, $\ce{BrF5}$, $\ce{F2}$, $\ce{O3}$, $\ce{NO2}$, $\ce{N2O4}$, $\ce{NOF}$, $\ce{F2NNO}$, $\ce{HNF2}$, $\ce{ClF5}$, and far more. And this does not even consider exotica such as alpha particles, plasma etching, ion milling and so on.
Almost all of the substances/solutions listed above are very dangerous and quite a few are either explosive per se or as close to it as makes no difference. Some are way too unstable to exist at room temperature.  In some cases, mixtures are possible and have even been made by intrepid people. In other cases, mixtures would be catastrophic.
I highly recommend Dr. John Clark’s book [1] if anyone wants to know more about what happens when extemely powerful oxidizing agents meet various potential rocket fuels. (This is the source of some of the oxidizing agents listed above.) Dr. Clark was a physical chemist with deep insider knowledge of the development of liquid fuel rockets.
[1] J.D. Clark, Ignition! An Informal History of Liquid Rocket Propellants, Rutgers University Press Classics, New Brunswick, NJ, ©1972 & 2017.


text:
Is "stainless" a bulk or a surface property of stainless steel?
In particular, if I cut a block of stainless steel in half, would the newly formed faces be stainless as well?

Yes, it is stainless steel throughout. Look up “stainless steel” in wikipedia to see lots of information about the many varieties.

"Chrome-plated" is sth that peels off, eventually.

The headline question and the text block of clarification appear to ask slightly different questions regarding the perceived properties vs the bulk metallurgy/chemistry/physics of the material. (an "if a tree falls in a wood and there is no one there to hear it, it only produces pressure waves" viewpoint conflict)

The bulk property is that any surface exposed to the atmosphere develops a surface layer that protects it from further corrosion. The mechanism protecting the material is a surface mechanism but the property that creates that surface is a bulk property.

It is usually a bulk property though you would need to know exact regulations for your country to be certain.
Stainless steel is steel (i.e. iron + a little bit of carbon) alloyed with another metal (usually chromium) which makes it resistant to oxidation by atmospheric oxygen, but not to strong acids (e.g. concentrated hydrochloric acid) or strong oxidizers.

Indeed, for example stainless steel feels like the perfect electrolysis electrode, until you find out it corrodes because of the O+ ions and releases highly toxic chromium compounds in the electrolyte.

Re strong oxidizers, bleach products often say that they're not to be used on stainless.

@David possibility because releases chloride when it breaks down and that attacks SS passive film.

Stainless steel isn't "melted steel". It is a solid now. It is an alloy. Being "melted" in the past has nothing to do with it stainless. It's the alloying with things like chromium that makes it stainless.

@Gimelist That is that SteffX meant. How do you alloy two metals together without melting them?

@Ingolifs easy - solid state sintering. SteffX's answer is misleading, suggesting that the "melting" part is important. It is one way to achieve it, but not the only one. And SteffX does not use the word "alloy" which is the critical thing here.

@Gimelist Ok so sintered stainless steel is indeed a thing. Regardless, melting remains the primary way of alloying two metals together, and this is the sense the OP quite clearly meant. "Melted with another metal" doesn't mean the OP thinks a metal would become stainless just through melting.

Why is chemistry dependent on each country's regulations?

Let's divide the steel world into two classes: 1) rusting steel and 2) stainless steel.
Rusting steel, in the presence of oxygen and moisture, will oxidize, forming hydrated iron oxides/hydroxides which have a greater volume than the original iron, and which have relatively little adhesion to the metal. They curl up and continue to expose bare metal, and so rusting iron/steel will continue to rust.
It was discovered in the early 1900s that addition of 12% or more chromium to iron would produce an alloy that oxidized very slowly, forming a very thin oxide layer that was adherent and did not continue to thicken. It resembled chromium in that respect, and the oxide was probably rich in chromium. Nickel was also found to increase the corrosion resistance of the alloy.
Now the bulk composition affects the surface oxide, but it is the surface that stains, or rusts - or doesn't. If you broke a block of stainless steel in half, in a vacuum, the fresh surfaces would be bare metal, active to many reagents. Exposure of the fresh surfaces to oxygen will oxidize them fairly rapidly (minutes, hours) to a passive state which resists many chemicals, like acids, especially oxidizing acids like nitric. (Interestingly, plain iron dissolves in dilute HNO3, but in concentrated HNO3 (>30%), the oxidizing power of the acid is so great that the surface of the iron is rendered passive: an oxide coat is produced which does not flake off and does not dissolve in the acid.)
Stainless steels are less resistant to chloride ion, which has a way to infiltrate the oxide layer and corrode the base metal. Stainless steels which have been brushed with a wire wheel made of ordinary steel will have a surface contaminated by tiny fragments of regular steel, which will rust. The oxide layer on the stainless steel body will be imperfect, and oxidation will progress thru and under the passive oxide until the whole stainless steel is corroded.
Since it is the surface of the steel which contacts the active reagent, it would seem that the surface (oxide) is the determining factor, but, of course, the bulk composition strongly affects the surface oxide layer. And the surface needs oxygen to resist further corrosion!

Btw, is the process with concentrated HNO3 you refer to, actually useful and used for some purposes?

I expect that nitric acid passivation of regular steel is not as durable as the passivation of stainless steel in air. But passivation of stainless steel with 30% HNO3 is common and there are ASTM specs for it. The HNO3 removes contaminant metals like Cu, Al, Zn and even regular iron from machining so the the passive film is as good as can be. Then it stays good unless chloride sneaks into the picture. But it is a neat demonstration!

"an alloy that oxidized very slowly, forming a very thin oxide layer that was adherent and did not continue to thicken."  That sounds like what happens to aluminum.

"Stainless" is not a specific definition. The stainless steel with the least alloy is $5\% \; \ce{Cr}$ ( grade 501) according to AISI (It can't be cut with an oxygen/acetylene torch-like regular steel). API considers $\ce{Cr :Mo}$ (9:1) as stainless for oil well tubulars. SAE consider $12\% \; \ce{Cr}$ as stainless (most modern auto exhaust pipe).
Stainless cutlery is also $12$ to $13\% \; \ce{Cr}$. Non-magnetic stainless starts at the proverbial $\ce{Cr :Ni}$ (18:8) (Grade 304 and a half dozen other grades). All stainless will resist corrosion in some environments and corrode in others. And you can make a pretty good income telling people which stainless they need in their specific environment.

Quick and simple:
Steel = iron + carbon (less than 2%; also called "forgeable iron")
Adding chromium (min. 12 %) makes it stainless. These chromium atoms are spread over the full volume of your block, also on the surface of it. There they create a thin layer of oxygen atoms. This layer makes the steel stainless.
So when you cut your block in half, a new oxygen layer needs to be created first, before you can expose your block to a corrosive environment – and this can take up to 48…72h. If the chromium atoms are covered by something else (casual steel or these colourful residues from welding), the oxygen layer cannot be developed in this area; thus you get a 'hole' there, where your steel is not stainless. That's why you never cut a stainless steel block with tools made of casual steel and always remove the welding residues.
BTW: Aluminium does the same; with the difference that it's oxygen layer is developed within milliseconds instead of hours.
Well, 12% chromium isn't that much. So your oxygen layer is not too 'stable'. For highly corrosive environments (saltwater, acid, etc.) you may need to add more chromium. Acid takes some oxygen atoms away, populates the surface and thus forms a similar hole in your layer as mentioned above.


text:
What is "super" in superphosphate?
This question is inspired from a previous question(marked unclear). I don't know the context of that question but I was intrigued by a statement:

Superphosphate is used instead of just phosphate because
superphosphate is a compound whereas phosphate is an ion.

How the name "superphosphate" describe a compound? Since it contains the suffix -ate, shouldn't it be considered an ion just like phosphate?
I googled "superphosphate" and it gave results about it being a fertilizer, its various types and its suppliers. Is it a type of phosphate fertilizer or is it just a trademark name? What is the significance of the word "super" in this context?
On some more searching, I came to know that the other name of calcium dihydrogenphosphate is calcium superphosphate. Is it the same superphosphate that we are talking about?  Does IUPAC recommends its usage?
If there are superoxide and superphosphate, are there any other ion containing the name "super" like supersulfate or supernitrate? The names seem to be too absurd/obsolete to be used. Searching for "supernitrate" gave 2 results: alibaba and super calcium nitrate which seems to be nitrogenous fertilizers. Searching for "supersulfate" gave me results of a type of cement(One example here). So, I think that the name "super" isn't bound to fertilizers only.
To clarify my questions:

What is the significance of "super" in superphosphate? Is it a real chemical name or a trademark name?
Is the same as superoxide?
Are there any other ions containing the name "super"?

In the bulk fertilizer  business there is a solution that contains more than 100 % H3PO4 ; I think it is 107% .  It is a gel that is kept hot (  about 150F) so that it will flow.  It is blended  at fertilizer plants to get custom fertilizer blends for specific locations.

The term superphosphate is really old, even well before the concept of atoms was proposed by Dalton. Therefore it is difficult to rationalize the choice of this terminology.
In the unabridged version of the Oxford English Dictionary, you can see the earliest usage dates back to 1798

Chemistry. A phosphate containing an excess of phosphoric acid; an
  acid phosphate. Now disused except in superphosphate of lime, calcium
  superphosphate: cf. sense 2.
1798   Philos. Trans. (Royal Soc.) 88 17   It was..Scheele who discovered, that the urine of healthy persons contains superphosphate,
  or acidulous phosphate, of lime.

Further information from the OED on the usage of "super" in chemical names confirms its use since antiquity. See antique examples

(b) Denoting the highest proportion of a component, esp. owing to a
  high oxidation state. Now chiefly archaic or hist., except in
  superoxide n.   and in the names of certain substances used in
  industry and commerce, as superphosphate n., supersulphate n. Cf. sub-
  prefix 4b(b).  (i) Categories » [1788   J. St. John tr. L. B. Guyton de Morveau
  et al. Method Chym. Nomencl. 107   New names... Acetite of lead.
  Ancient names... Sugar of lead, Super-acetated lead.] 1811   Jrnl.
  Nat. Philos. June 78   The aqua lithargyri acetati is a saturated
  solution of the proper acetate of lead,..it is an essentially
  different salt from the super-acetate of lead. 1913   Brit. Med. Jrnl.
  4 Oct. 875/1   Dr. Latham..used the superacetate of lead in
  consumption. 1979   A. J. Youngson Sci. Revol. Victorian Med. i. 18
  Acetate or superacetate of lead combined with opium was prescribed for
  haemorrhage of the lungs.

Indeed, I also found supersulfate and superacetate of iron in a paper published in 1835 so its usage has become obsolete.

Superphosphate is a name used in agriculture, therefore, it does not mean the same super in the superoxides. There are two important types of fertilizers in agriculture: Nitrogenous and phosphatic fertilizers. Superphosphate is an one of major segments in phosphatic fertilizers, which is described as follows in Ref.1:

The primary products of the phosphatic fertilizers industry are phosphoric acid, ammonium phosphate, normal superphosphate, and triple superphosphate. Phosphoric acid is sold as is or is used as an intermediate in producing other phosphatic fertilizers. Monoammonium phosphate is favored for its high phosphorous content, while diammonium phosphate is favored for its high nitrogen content. Normal superphosphate has a relatively low concentration of phosphorous, however it is used in mixtures because of its low cost. Triple superphosphate provides a high concentration of phosphorous, more than 40% phosphorous pentoxide.

Normal or ordinary superphosphate fertilizers (NSP) are produced by reacting phosphate rocks with sulfuric acid (for this reason, NSP retains its importance in wherever sulfur efficiency limits crop yields). Normal or superphosphate fertilizers are refer to the fertilizer material containing 15-21% phosphorus as measured as phosphorus pentoxide $(\ce{P2O5})$. NSP contains no more than 22% of available $\ce{P2O5}$ (Ref.1).
Triple superphosphate fertilizers (TSP) are produced by reacting ground phosphate rocks with phosphoric acid. Triple superphosphate fertilizers are also known as double, triple, or concentrated superphosphate. The phosphorus content of TSP is over 40% measured as phosphorus pentoxide $(\ce{P2O5})$, which is its major advantage over other phosphatic fertilizers (Ref.1).
Note: To my knowledge, there are no other names such as supernitrates in fertilizer industry or agriculture, to tell the least.
References:

Nicholas P. Cheremisinoff, Paul E. Rosenfeld, "Chapter 1: Industry and Products," Handbook of Pollution Prevention and Cleaner Production, Vol. 3: Best Practices in the Agrochemical Industry; 1st Edn.; Elsevier Inc.: Oxford, United Kingdom, 2011, pp. 1-24 (https://doi.org/10.1016/B978-1-4377-7825-0.00001-7).

According to the reference book I've given, phosphatic fertilizers are divided into three segments: phosphoric acid, ammonium phosphates, and superphosphates. However, I can't realize why it's called that name. It is clear, though, superphosphates is not a trade name. Also, triple superphosphates is the real superphosphates because of its high phosphate content; 2) Also, according to the book, there is no "super" outrageous fertilizer: Ammonium nitrate and urea are the most prominent ones.

"Super" in "superphsophate" probably refers to it being super-active, with its phosphorous more readily available than it would be in an ordinary phosphate.  Chemically this is done by combining calcium phosphate with sulfuric acid giving an acid salt that is more soluble than normal-salt calcium phosphate would be.  See here, for example.
Superoxide is not a analogous to superphosphate.  It is, as the OP may know, $\ce{O_2^-}$ instead of the normal oxide ion $\ce{O^{2^-}}$.  In simple salts it's stable only with alkali metal cations, but oxyhemoglobin has been considered as a superoxide complex.
I am not aware of any "super" ions other than superoxide, as superphosphate is just a catchy name for acid-sslt phosphates (see #1).  Feel free to edit with any corrections.

Is possible that names like perchlorate are abbreviations from former super- to per- ? Just thinking...

On 3, "feel free to edit".  If <2000 rep, I would accept such an edit.  On 2, potassium peroxide is stable but covered in the answer under "alkali metal cations".


text:
Coulombic energy of repulsion and exchange energy
In this Wikipedia link about the excited states of oxygen, this diagram is shown.  The left most molecular orbital diagram is the first excited state of oxygen, and the MO diagram in the middle is the second excited state of oxygen. It seems to imply that the energy of pairing is less than the exchange energy, and thus the first excited state exhibits the two electrons with opposite spins in the same MO, while the second excited state exhibits the two electrons with opposite spins in separate MOs.
However, in Miessler, Fischer & Tarr, Inorganic Chemistry, 5th Edition, this diagram is depicted, implying that the energy of pairing is greater and more unfavorable than the exchange energy.
These two diagrams seem to have different orders for the first and second excited states. Which energy energy state, having both electrons with opposite spin and in the same orbital, or having both electrons with opposite spin and in separate orbitals, has higher energy?

@porphyrin are you saying that the Wikipedia article is incorrect?

No, apologies, from memory I made a silly mistake, I stupidly confused singlet delta with singlet sigma. The singlet delta is the first excited state at 0.98 eV the singlet sigma at 1.6 eV above the triplet sigma ground state so the labelling in the Wikipedia article is correct. The second diagram does seem to imply that the singlet sigma is lower in energy than the delta but I don't follow why there are three lines in each, orbitals, energy levels?  It does not seem to make sense.

@porphyrin I think the three lines refer to an arbitrary degenerate $np_x$, $np_y$, $np_z$ orbitals; not necessarily the antibonding MOs in $O_2$.


text:
Can sulfuric acid itself be electrolysed?
We all could know by now that water can be electrolysed by adding some sulfuric acid.:)
But is it possible that the sulfuric acid itself becomes electrolysed when the concentration of the acid in the water gets higher ?
Would it give sulfur dioxide or sulfur trioxide ?

Rather hydrogen and peroxodisulphuric acid, with ozone as a byproduct. $$\ce{ 2 H2SO4 -> H2 + H2S2O8 }$$ https://en.wikipedia.org/wiki/Peroxydisulfuric_acid

@Poutnik Thank you for the useful link.  I wondered how the sulfuric acid clouds of Venus could be useful.

After some searching, I found this 1956 Thesis(Reference), which describes the production of peroxydisulfuric by electrolyzing concentrated sulfuric acid. The yield varied according to the concentration of the acid, temperature, current, and the nature of the electrode. In the Thesis, the chemist noted that discharged $\ce{HSO4-}$ united together to form peroxydisulfuric acid ($\ce{H2S2O8}$). It further stated that  oxidation of sulfate ions by anodic oxygen was responsible for the peroxydisulfuric acid formation. Also, hydroxyl ion at anode played an important role in the acid formation.
The reaction mechanism was confirmed by electrolyzing sulfuric acid and ammonium sulfate at various concentration at $\pu{16\!- \! 20 ^\circ C}$ using platinum electrode and the reaction occurred in accordance to the theory. It was also confirmed that amount of peroxidic oxygen increased with increase in ammonium sulfate. The specific gravity of sulfuric acid and the current density was varied and accordingly the highest yield of 78.4% was obtained. It was also noted that increasing the current density increases the yield up to a certain point as  peroxydisulfuric acid was found to convert into peroxymonosulfuric acid. At that moment, a small trace of hydrogen peroxide was noted.
So, the chemical species formed during electrolysis are $\ce{HSO4-}$, $\ce{SO4^2-}$, peroxide ion, protons, trace of hydrogen peroxide, and other side products.
Reference:
Sin-Chou Fan, “The electrolytic production of peroxydisulfuric acid using periodically reversed direct current and alternating current superimposed on direct current,” MS Thesis, Virginia Polytechnic Institute, 1956.


text:
Balancing the chemical equation using Oxidation Number Method

Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation for the reaction.

Please balance the equation using the oxidation number method. The skeletal ionic equation is:
$$\ce{MnO4-(aq) + Br-(aq) -> MnO2(s) + BrO3-(aq)}$$
This question is asked in Problem 8.9, NCERT Textbook Class XI, Pg. no. 267. My answer to this question is different from the answer given in the book. I'd love to know the correct answer to this question.

You may want tp write down the expected answer, your answer and your reasoning for it. It may prevent closing your question as "not elaborated homowork class question".

It's hard to pinpoint what the problem here is since I don't have the mentioned book to compare an answer with.
The simplest way is write down half-reactions for reduction (red) and oxidation (ox) processes once you've assigned oxidation numbers (denoted above the symbols of the elements which are participating in a redox reaction), then balance the number of the transferred electrons and the algebraic sum of both half-reactions will yield in a complete balanced redox reaction:
$$
\begin{align}
\ce{\overset{+7}{Mn}O4- + 2 H2O + 3 e- &→ \overset{+4}{Mn}O2 + 4 OH-} &\quad |\cdot 2 \tag{red}\\
\ce{\overset{-1}{Br}^- + 6 OH- &→ \overset{+5}{Br}O3- + 3 H2O + 6 e-} &\quad \tag{ox}\\
\hline
\ce{2 MnO4- + Br- + H2O &→ 2 MnO2 + BrO3- + 2 OH-} \tag{redox}
\end{align}
$$
As a sanity check, you can see that there are no protons $\ce{H+}$ participating in either of the reactions, and the resulting medium is indeed basic (slight excess of $\ce{OH-}$), which is in a good agreement with the conditions mentioned in the problem.


text:
Why do Ketones Have Lower Wavenumbers than Esters?
The average wavenumber for a ketone is about $\pu{1720 cm-1}$ and the average wavenumber for an ester is about $\pu{1740 cm-1}$. This, however, does not make sense, as the carbonyl group of an ester should have a greater single bond character than the ketone due to resonance from the adjacent oxygen atom. This greater single bond character should thus result in a lower wavenumber for the ester, but it does not. Is there an explanation for this?

For an undergraduate starts to learn IR spectroscopy, the stretching frequency of any $A-B$ chemical bond, $\over\nu$ (in $\pu{cm-1}$) can be calculated by using following equation:
$${\bar\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}= \frac{1}{2\pi c}\sqrt{\frac{k(m_A+m_B)}{(m_Am_B}}$$
where $k=\text{the force constant of the bond}=\text{bond strength}$, $m_A= \text{mass of atom  }A$, $m_B= \text{mass of atom  }B$, and $c= \text{speed of light}$. The equation shows $\bar\nu$ is at least depends on two factors, reduced mass $\mu$ and the force constant of the bond $k$.
The dependency on $\mu$ would be explained by the difference in stretching frequency of $\ce{C-H}$ ($\pu{\approx 3000 cm-1}$) and $\ce{C-D}$ ($\pu{\approx 2200 cm-1}$). The force constants of $\ce{C-H}$ and $\ce{C-D}$ are approximately equal.
The most important fact when compared to different carbonyl bond stretching frequencies is force constants of the bonds of interest. For example, ring strain in a cyclic ketone usually increases the $\ce{C=O}$ stretching frequency. That of cycloheptanone is $\pu{\approx 1702 cm-1}$, cyclohexanone is $\pu{\approx 1714 cm-1}$, cyclopentanone is $\pu{\approx 1747 cm-1}$, and cyclobutanone is $\pu{\approx 1783 cm-1}$. Therefore, we can generally conclude that the stretching frequency of the bond increase with the increase of the reactivity of carbonyl bond. Likewise, reactivity of carbonyl compounds in increasing order is: acid chlorides ($\pu{1780-1820 cm-1}$) > acid anhydrides ($\pu{\approx 1760 cm-1}$ and $\pu{\approx 1810 cm-1}$) > esters ($\pu{1730-1750 cm-1}$) > aldehydes ($\pu{1720-1740 cm-1}$) > ketones ($\pu{1705-1725 cm-1}$) > carboxylic acids ($\pu{1700-17205 cm-1}$) > acid amides ($\pu{1630-1680 cm-1}$). The reduction of double bond character in ester (average bond length of methyl acetate is $\pu{\approx 123.2 pm}$) is minimal but has a better leaving group than in ketones (average bond length of acetaldehyde is $\pu{\approx 123.1 pm}$) when subjected to react.

Using the equation above when isotopes are concerned, the force constant is equal between CD and CH because only a change in mass is involved so any change in frequency is solely due to changes in $\mu$.

But why does the leaving group ability explain the stretch frequency and the resonance effects?


text:
What is meant when people say a molecule has high multi-reference character?
I have read some other posts explaining what multi-reference and multi-configuration are with regards to the wave-function (such as What exactly is meant by 'multi-configurational' and 'multireference'?).
However, what does it mean for a molecule specifically to exhibit high multi-reference character? Does it just mean the results from calculations will be strongly dependent on if the level of theory incorporates multi-reference?

There are molecules, which are qualitatively treated wrong when only using one configuration (for example in the form of a Slater Determinant). This is often also called strong correlation, or dynamic correlation. This happens when you have partially occupied (near-)degenerate orbitals (like in transition metal complexes for example).

@Martin-マーチン , I think that maybe is not very clear that you mean by "dynamic correlation".  For clarification, those systems that are treated wrong when only using one configuration suffer of static correlation and have multi-reference character.

@user1420303 yes indeed, I misspelt. It should have been static correlation. If you have the time, maybe you could add an answer. I will unfortunately not be able to.

When doing a CI calculation based on Hartree-Fock orbitals, then usually the HF configuration has a very high weight in the CI vector (>90%). The next largest weight is then very small (for example only 1%), and there is a very large number of those very small configurations.
Such cases are called weakly or dynamically correlated. Because of the large weight, it is reasonable to optimize the orbitals for the leading configuration only and then run some truncated CI calculations (typically CCSD) on top.
If there are multiple configurations with similar (or same) weight, then it is no longer reasonable to optimize the orbitals for just a single configuration, because this will introduce a bias towards this configuration. The number of these configurations is rather small.
Additionally, those configurations are not necessarily related by Single or Double excitations, so CISD may not include all. However, they are commonly generated by an active space of just a few orbitals.
This is called strong or static correlation, and is typically found in systems with partially occupied close-degenerate orbitals, e.g. d orbitals in transition metals.
In FCI this orbital bias would not matter, because the basis is (numerically) complete. But in truncations like CISD or CCSD this leads to a qualitative error. Some important configurations are poorly represented (due to the bias), other are completely missing (due to the truncation).
A CASSCF calculation resolves the bias by considering all important configurations of similar weight (by chosing an appropriate active space) and balancing the orbitals to all of the considered configurations. Since only some configurations are required for static correlation, it is often feasible to optimize both (CI and orbital coefficients) simultaneously.
The term multi-reference then refers to a calculation where all (or most) of the CASSCF configurations are used to generate excitations from.


text:
What Is The Heat Of Combustion Of Diesel And How Do I Find It?
I need help in finding the heat of combustion of diesel, no where online can i find it, and my chemistry teacher does not know how to find it....

For a rough start you can look at https://en.wikipedia.org/wiki/Energy_density#Table_of_energy_content and also https://en.wikipedia.org/wiki/Diesel_fuel#Fuel_value_and_price (about 45 MJ/kg) but the thermodynamics of combustion may make a complete answer more complicated than that


text:
Hemiacetal formation mechanism

The question is asking which reagents can be used to carry out this reaction. I get the L (PCC) is used to turn the primary alcohol into an aldehyde. However, I believe that is not the only reagent needed as the tertiary alcohol -OH must become -O- for the reaction to be SN2. A strong nucleophile is needed to carry out the formation of the SN2 reaction that forms the ring as the aldehyde is only primary (thus SN2 needed). I think that J, the tertbutoxide or S (NaOH) will deprotonate the OH group on the tertiary alcohol for the reaction to proceed.
Why is only L needed and not L and J (or S)?

The presence of acid in L make the smooth hemiacetal formation.

There's no SN2 process here. A nucleophile attacking an electrophile doesn't necessarily mean SN2; it's only SN2 if the electrophile is an alkyl halide (amongst other conditions).


text:
Constant concentrations in solution
I'm a high school student considering the concentration of solutions, and whether the concentration is constant at the bottom or towards the surface. Even then, how would I know that the solution is consistent and not more concentrated in places compared to others? Is there any visual way or method for identifying non-consistent concentrations in solution i.e. Copper (II) sulphate, Iron (II) sulphate

If the solution is well mixed (stirred or shaken, etc.), then the concentration will be uniform throughout. But a solution of $\ce{CsCl}$, in an ultracentrifuge, is a different matter. Drop a few grains of potassium permanganate into a beaker of water and you can vividly see that the dissolution is neither nearly instantaneous nor uniform. But stir or shake the contents of the beaker, and you get a uniformly optically dense solution.

A solution by definition is a homogeneous mixture, so its "concentration" is same throughout the bulk. Let us have a look as to how one would prepare a solution of a given concentration, say in 1 L. We would weigh out a known weight of a solute and dissolve it  ~ 500 mL of solvent, with mechanical stirring or shaking. We would add more solvent until the total volume is 1 L. We would stir it sufficiently, again, until we believe that the stirring has made the solution uniform.
a) If it a colorless solid, like NaCl, there is no visual way to tell if the mixing is complete and the solution is uniform. You have to stir it very well.
b) A colored solid, like copper sulfate or iron sulfate is more or less visible if it is not homogenous. Different colors in the bulk.
If you are still not (mentally) satisfied, you can take portions of the solution from surface, middle and bottom. Measure their refractive index and they should be identical. If they are not not, the mixture needs more stirring.
Edits:
Some people may wish to differentiate microscopic vs. macroscopic homogeneity regarding the definition of solutions. IUPAC does not offer such a distinction. For an analytical chemist or any researcher interested in measuring or making known concentrations of solutions- a solution has to have identical concentration throughout the bulk. Ultimately, it will boil down to semantics or the problem of "how many angels can land on the head of the pin" - a debate which has no end! en.wikipedia.org/wiki/…

+1 for the answer given by @M. Farooq, but one little thing: it is very easy to think that stirring is (or has been) adequate, especially with colorless solutes, just as it is easy to think that a deck of playing cards has been adequately shuffled to randomness, whatever that really means. If in doubt, stir (or shuffle) more!

An additional comment, though OP shall be satisfied by the actual answer above. I am not sure that homogeneity of a solution refers to its composition (concentration) or just to the absence of phase boundary. I mean, out of lab glassware size, as in atmosphere, oceans or just a tall reservoir, we should still see that as solutions, in spite of density and concentration gradients.Right?

Solutions refer to homogeneity in composition. In a strict sense air or oceans are not exactly solutions because their composition varies in the bulk as a function of depth or height.

This is not true - definition is about microscopic homogeneity - there's one phase not some tiny droplets. There's nothing about perfect mixing, which doesn't even exist.


text:
Electrons distribution of oxygen atoms in ozone
While dealing with the Lewis structure for ozone, most of the explanations show the electron distribution around the 3 oxygen atoms with a single and double bond.
But if you count the number of valence electrons of the middle oxygen atom, it will add to 5 valence electrons only, but the number of valence electrons of an oxygen atom is 6.
Similarly, the oxygen atom with a single bond in ozone, is being shown with 6 valence electrons plus one single bond, which accounts for 7 valence electrons, which does not tally with the expected number of 6 valence electrons.
Can someone please explain why this discrepancy has been permitted? Or what did we miss to consider?

Related: https://en.wikipedia.org/wiki/Ozone#Structure

You may want to check up the concept of "formal charge". There is no rule that would require the O to have 6 electrons. It can be less or more, depending on the molecule.

The ozone molecule is described with two resonant structures(as in figure) in which all the oxygen atoms respect the octet rule. The oxygen in the middle has a positive charge because has only 5 valence electron  and the oxygen with the single bond has a negative charge because it has 7 valence electrons.
 In the real molecule both the bonds have the same length so there isn't one single bond and one double bond. Also both the terminal oxygen atoms have the same partial negative charge instead of being one neutral and one negative(-1).
The picture below is more realistic of the situation:

In the electrostatic potential map you can see that the two lateral oxygen are equivalent:


text:
Storage of Halogen
I learned that we should store Bromine in brown coating glass bottle. But, I heard that halogen liquid is so reactive that it can melt even glass. So, I want to know the practical storage method of halogen.

Which halogen? Fluorine would eat glass; bromine definitely wouldn't.

The "halogen liquid that can even melt glass" could be chlorine trifluoride, which can indeed react with glass and just about every other oxide you can think of.  Fortunately, your bromine is not that stuff.

What do you mean by "melt"? Melting suggests a phase transition, has little to do with chemistry between whatever halogen you are asking about and the vessel, and is essentially a physical process.

I assume liquid halogen to be bromine. From here:

Glass, ceramic nickel or lead containers are suitable for bromine.
Lead-lined steel tanks can be used. Only highly fluorinated plastics
will resist corrosion. A free space of 8-10% by volume should be left
in the container.

Also see: https://patents.google.com/patent/US3375077

We know now that bromine can be stored safely in glass (with appropriate glass handling, of course).  Now for the "brown" part.
Bromine can break down on exposure to ultraviolet light, a reaction shown here as part of an alkene bromination.  Uncontrolled, this process could release reactive atoms; these may degrade surrounding materials and the products then contaminate the bromine and the surroundings.  The brown glass commonly seen in chemical storage bottles is designed to screen out this radiation.


text:
Why is the rate inversely proportional to the square root of temperature in Grahams Law of Diffusion?
In my book it is given that:

The general form of the Grahams Law of Diffusion can be stated as follows when one or all of the parameters are varied:
  $$\text{rate} \propto \frac{PA}{\sqrt{TM}},$$
  where $P$ - pressure, $A$ - area of the hole, $T$ - temperature, $M$ - Molecular weight.

How can rate be inversely proportional to the square root of temperature? I feel that on increasing temperature it must increase, but how can the rate decrease?
Is the formula in my book correct or incorrect?

It helps to rewrite
$$\text{rate} \propto \frac{PA}{\sqrt{TM}}$$
by assuming the ideal gas law holds, as follows:
$$\text{rate} \propto \frac{RA}{V_m}\sqrt{\frac{T}{M}}$$
where $V_m$ is the molar volume or inverse of molar particle density.
Written this way it is clear that if the particles occupy the same volume, increasing their temperature increases the rate at which they effuse through the apperture.


text:
Number of planes with maximum atoms

Number of planes in $\text{SiH$_2$FCl}$ which contain maximum atoms but all must be different?

I am interested in the above problem. I predicted the answer to $10$ which indeed is $5\choose 3$ because maximum number of points in a plane will be $3$ since the structure is tetrahedral. But the book writes an answer $7$ to the above problem!! Any help will be highly appreciated.

I predicted the answer to 10 which indeed is $5\choose 3$ because maximum number of points in a plane will be 3 since the structure is tetrahedral.

Yes, 10 is the number of sets with three atoms, and each set defines a unique plane.

But the book writes an answer 7 to the above problem

3 of those (HHSi, HHF, HHCl) don’t count, so 7.


text:
Permanent storage container with electrode for electroplating
I'm not a chemist at all but I am looking for a specific type of container which I'm sure must exist but I don't know it's name.
We want to store non-toxic (non-cyanide) gold solution for electroplating in a small (10-20ml) transparent container that has an inbuilt, permanent electrode for the plating process:

But I can't find anywhere to buy such a thing and I don't know if it has a specific name. The electrode would have to be made of a material which will not react with the gold solution, so as not to contaminate it.
Why do we want this specific setup? We want to gold plate fine wires which are inserted into the solution from the top of the container, these are fragile and expensive which is why I would rather avoid having a temporary electrode inserted into the solution each time. We have a small amount of gold solution at a time and it is expensive so I would rather store the solution in the plating container permanently. The container really needs to be transparent so we can see that the plating wires are in the solution.
Any help would be massively appreciated,
Thanks,
M.


text:
Electrolysis of molten sodium chloride
Can somebody explain me the electrolysis of molten $\ce{NaCl}$ via electrode potential?
Reduction potential of $\ce{Cl-}$ is -1.36 V, while sodium metal is -2.76 V, but still sodium is reduced and not $\ce{Cl-}.$

Cl- can't be reduced. What is dead may never die, they say. Same story here.

@andselisk, random capitalization is a typical writing problem among South Asians. I see this everyday. There is no concept of capital letters in the alphabetical system and definite articles.

@M.Farooq Oh, thank you, I wasn't aware of that. How interesting.

The misconception you have in the question is that the concept of tabulated electrode potentials do not apply to molten salts. The values in general chemistry textbooks have been calculated in water under specific concentrations. No need to extrapolate electrode potentials to molten states.

True, but there is a much deeper misconception which (IMHO) has to be dealt with first. I mean, the basic idea of what can and what can't be reduced.

I think the student is connecting reduction potentials as "reduction only" potentials.


text:
Using an oxidizing agent to improve combustion of a fuel
If I were to add an oxidizing agent like dilute Potassium permanganate (aq) to a potential fuel such as plastics (polystyrene etc), would the combustion of this fuel be more complete ie: less sooty flame, brighter/hotter flame etc?

In general, yes (remember self-ignition experiment when a cotton wool soaked in glycerin ignites after a while when sprinkled with solid potassium permanganate). However, you should account for leftovers ($\ce{K2CO3}$, $\ce{MnO2}$) which would add to soot, and if you want to go for a hotter flame, you probably want to look at stronger oxidants.

There is no "if". Unless you add an oxidizing agent, there will never be a flame in the first place. So everybody who starts a flame, without much trouble and possibly without giving it much thought, is already adding an oxidizing agent, and a pretty powerful one: $\ce{O2}$.
With good chances, whatever else oxidant you wanted to add will be worse than oxygen in one or more regards. It will probably be weaker. It will probably carry many times more weight per electron transferred (surely the case with $\ce{KMnO4}$ even you use it solid, and twenty times more so if you dilute it with water; if you thought of diluting it with gasoline, then it just won't dissolve at all). It will produce much more byproducts, maybe including solid ones (again, surely the case with $\ce{KMnO4}$). In short, there are good chances that it will make the flame more sooty and less bright (if maybe more colorful), and the combustion less complete.
Short of these issues, the idea is fine. If you blow fluorine into your flame, I bet it will burn a good deal brighter. The possibility of your survival through this, however, is another story.

The answer would be incomplete without mentioning the two cases when other oxidants, despite not being fluorine, might have an edge over oxygen:

Lower temperature (colder than flame, that is), when kinetics matters a great deal. This covers the case mentioned by andselisk, and also all other cases when you can run a controlled redox reaction in a beaker without much interference from $\ce{O2}$. These, however, are beyond the scope of your question if you want a flame.
Tight packing. Solid oxidants generally beat gaseous oxygen in that you can fit much more of them in a small volume. So if you mix the said permanganate with I'd rather not tell you what, you'll get yourself a decent bomb. The reaction will certainly get more noisy, if not less sooty.

So it goes.


text:
Why exactly is a fluoride anion a stronger base than iodine anion?
First off, I've learnt that stronger acids produce weaker conjugate bases (through Brønsted–Lowry acid–base theory).
Then I looked at the $\mathrm{p}K_\mathrm{a}$ values of $\ce{HF}$, $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$ and came to the conclusion that $\ce{HI}$ is the strongest among them and that explains why $\ce{I^-}$ is the weaker base.
So far so good.
But then I remembered that $\ce{F^-}$ is in fact the strongest oxdizing agent out there.
So shouldn't that mean that $\ce{F^-}$ ion is more stable than $\ce{I^-}$? And hence does not give away its electrons easily and therefore is a weaker base?
Where am I going wrong here?
I have a little bit of gut feeling that this is because of the large size of iodide ion which can hold the negative charge better than the tiny fluoride ion. But I have found nothing to back this up.

I would rather say F- is the weakest oxidizing agent, as there are similar complementary rules for redox pairs as for acid-base pairs.

There is no connection whatsoever between the "strength" of the acid base pairs and the redox pairs. Your gut feeling is basically right, but dont drag the redox chemistry in to the argument.

If the fluoride anion, $\ce{F^-}$, were an oxidizer, than because it would removes at least one electron from an other atom, ion, or molecule.  Yet, as fluoride anion, $\ce{F^-}$ already possess completely filled s- and p-orbitals.  And as member of the second period (counting the one with H and He as the first one), the octet rule typically is observed strictly.  There use of d orbitals (like for sulfur, one period later) isn't typical either.  So, while while the fluorine molecule ($\ce{F-F}$) is capable to oxidize all up and including oxygen, the same by fluoride ($\ce{F^-}$) is unclear.

Related: https://chemistry.stackexchange.com/questions/34818/why-is-hf-the-least-acidic-hydrogen-halide

You are mixing several concepts in your question: being an acid/base is very different to being an oxidizer/reducing agent

According Lewis acid base theory where , electron pair acceptor are acid and vice versa. So ,as reducing agent donates electron it might seem to be as a Lewis base. I think this what op is trying to tell

First off, I've learnt that stronger acids produce weaker conjugate bases (through Brønsted–Lowry acid–base theory).

That is correct.

Then I looked at the $\mathrm{p}K_\mathrm{a}$ values of $\ce{HF}$, $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$ and came to the conclusion that $\ce{HI}$ is the strongest among them and that explains why $\ce{I^-}$ is the weaker base.

I don’t agree with that explains; but the observation is correct, iodide ($\ce{I-}$ is the weakest base and $\ce{HI}$ is the strongest acid.

But then I remembered that $\ce{F^-}$ is in fact the strongest oxdizing agent out there.

No. The fluoride anion is not an oxidising agent. If it were, it would take up electrons to become a fluoride dianion ($\ce{F^2-}$), inserting an electron into the 3s level—not happening.
What you probably meant is fluorine (the element; the diatomic molecule) being the strongest oxidising agent out there. That might be correct or it might not (I am not up to date with extremely strong oxidising agents) but it is not a statement I could dismiss at first glance. Certainly, fluorine is a stronger oxidising agent than iodine.

So shouldn't that mean that $\ce{F^-}$ ion is more stable than $\ce{I^-}$?

Thermodynamic stability is sometimes a tricky beast, but in general a fluoride ion is more stable than an iodide ion which is prone to oxidation to give iodine or iodate. This is connected to the oxidising ability of fluorine (strong) versus iodine (weak), not to the strengths of the acids, though.

And hence does not give away its electrons easily and therefore is a weaker base?

The conclusion is invalid. Not wanting to give away one’s electrons means not wanting to get oxidised. As mentioned, fluorine is a strong oxidising agent so to reduce fluorine to fluoride is a favourable process. But acid-base chemistry has nothing to do with gaining or losing electrons: it is an entirely electrostatic process and builds on different principles.
The high oxidising ability of fluorine can be thought of as a side effect of its high electronegativity: it has a tendency to strongly attract electrons since it has a highly positively charged nucleus that is shielded by only two core and seven valence electrons. Adding an eighth electron is comparatively easy. For iodine, there are four core shells and a valence shell between the nucleus and the incoming electron; this shielding effect can be thought of as reducing electronegativity and oxidising ability.
The high acidity of $\ce{HI}$ and low basicity of iodide is a direct consequence of its size: the negative charge is distributed across a much larger volume meaning it is more stabilised. In fluoride, the negative charge is confined to a much smaller volume meaning that positive charges such as a proton are attracted stronger. This is the underlying reason for the observed acidities/basicities.

Second period nonmetals form much stronger bonds with hydrogen than their heavier congeners and so tend to bind more strongly with protons than we might otherwise expect.  Not only is fluoride ion a stronger Bronsted-Lowry base than chloride ion (and heavier halides), so are hydroxide ion versus hydrosulfide ion and ammonia versus phosphine.  The last pair has the property that phosphine is much more highly flammable, which can be traced to the phosphorous-hydrogen bonds being easy to break while nitrogen-hydrogen bonds are more robust.


text:
Why is the valency of the Calcium (Ca) Bohr model, set to 2 electrons, instead of 1 electron for stability?
This is an Ascii table for the elements on the 4th period (row) of the periodic table, and distribution of electrons in each orbit.
+---------+----------+---------+---------+----------+---------+
|         |          |         |         |          |         |
+---------+----------+---------+---------+----------+---------+
| Element | Atomic # | Shell 1 | Shell 2 | Shell 3  | Shell 4 |
| K       | 19       | 2       | 8       | 8        | 1       |
| Ca      | 20       | 2       | 8       | 8 (9?)   | 2 (1?)  |
| Sc      | 21       | 2       | 8       | 9        | 2       |
| Ti      | 22       | 2       | 8       | 10       | 2       |
| V       | 23       | 2       | 8       | 11       | 2       |
| Cr      | 24       | 2       | 8       | 13       | 1       |
| Mn      | 25       | 2       | 8       | 13       | 2       |
| Fe      | 26       | 2       | 8       | 14       | 2       |
| Co      | 27       | 2       | 8       | 15       | 2       |
| Ni      | 28       | 2       | 8       | 16       | 2       |
| Cu      | 29       | 2       | 8       | 18       | 1       |
| Zn      | 30       | 2       | 8       | 18       | 2       |
| Ga      | 31       | 2       | 8       | 18       | 3       |
| Ge      | 32       | 2       | 8       | 18       | 4       |
| As      | 33       | 2       | 8       | 18       | 5       |
| Se      | 34       | 2       | 8       | 18       | 6       |
| Br      | 35       | 2       | 8       | 18       | 7       |
| Kr      | 36       | 2       | 8       | 18       | 8       |
+---------+----------+---------+---------+----------+---------+

I know all elements must have 4 orbitals due to the period (row #) of 4, and the outer most orbital (or shell), 4, is the valence.
Cr and Cu, have their valance set to 1, instead of 2, because, the electrons closer to the nucleus, requires less energy to sustain itself, and is thus more stable, but at the same time the each orbit (or shell) must have at least 1 electron on it. In regards to Cr and Cu, I thought there would be a precedence for orbit 3, the orbit preceding the valence, to be more full and within 18 or less electrons.
This forum post (link), hosted on a college's server, unfortunately fails to address the issue with Ca.

You should use the standard electron configuration nomenclature instead of numbering "shells" and "orbitals". The configuration of Ca is $[Ar] 4s^2$, chromium $[Ar] 3d^5 4s^1$, copper $[Ar] 3d^{10} 4s^1$

@Karl instructor went straight to drawing the Bohr model, rather than familiarize us with SPDF filling patterns and the square notations.

Well, Bohr is only step one. It seems you have progressed now, keep it up!

After following Mithoron's link, I familiarized my self with the writing on the following ChemGuide site (link).
I went into Excel spreadsheet, and had each 'sub orbital' it's own column, and came up with the following.
+---------+---------+----+---------+----+----+---------+-----+
| Shell 1 | Shell 2      | Shell 3           | Shell 4       |
+---------+---------+----+---------+----+----+---------+-----+
| 1s      | 2s      | 2p | 3s      | 3p | 3d | 4s      | SUM |
| 2       | 2       | 6  | 2       | 6  | 0  | 1       | 19  |
| 2       | 2       | 6  | 2       | 6  | 0  | 2       | 20  |
+---------+---------+----+---------+----+----+---------+-----+

This is what the overall table (link) looks like in excel; I can't attach the image directly to my post.
While googling, to get a clear understanding of S,P,D,F I came across ThoughtCo's article (link). This let me know the stability precedence, i.e. electron distribution, more clearer, as follows:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f
4s is filled before, 3d. 3s and 3p, can hold a combined maximum of 8 electrons.
The following (link) has a diagram with downward/diagonal arrows, showing how one can better remember the stability precedence.


text:
Which is more acidic HOF or HOCl?
My attempt to the question:
We know that more polar the $\ce{O-H}$ bond, the more acidic is the compound. According to electronegativity, the $\ce{O-H}$ bond of $\ce{HOF}$ should be more polar than $\ce{O-H}$ bond of $\ce{HOCl}$ since electronegativity order is $\ce{F} > \ce{Cl}$. But we should not miss another factor! That is the backbonding of  2p electrons (lone pairs) of oxygen with the 3d vacant orbital of chlorine. This is not possible in $\ce{HOF}$ due to the absence of vacant orbitals. Backbonding reduces (maybe, I do not know) electron density in oxygen and therefore polarity of $\ce{O-H}$ bond increases.

For future reference, if deploying Latin script, do not skip spaces after the period.  This renders recognition where a phrase ends and a new one starts easier.

The 3d orbitals in 3rd row p block elements are not available for bonding, see any number of questions here, e.g. the first answer to https://chemistry.stackexchange.com/questions/13949/what-allows-sulfur-and-phosphorus-to-expand-their-octet. Thus there is no back bonding.

If we compare the HOF and HOCl
1) In HOF , the oxygen atom cannot do backbonding since fluorine has no vacant orbital while in HOCl it can do hydrogen bonding it can do backbonding which reduces electron density on oxygen by a slight extent( Remember 2pπ-3pπ backbonding is not very much effective due to size difference of 2p and 3p orbitals) . But yeah this factor makes HOCl bond more polar .
But there are two more important factors-:
2[)INDUCTIVE EFFECT]-  among oxygen and fluorine , fluorine is more electronegative , so it partially pulls the sigma bonded electrons towards itself giving a  delta+ charge to oxygen .
While in HOCl , O is more electronegative than Chlorine so the opposite mechanism happens and O gets a slight negative charge.
So according to this HOF has increased bond polarity than HOCl.
3) This factor is the most important and sufficient to answer almost any acid comparison . Compare the stability of the ion formed after releasing H+ ( formally called as the conjugate base of the acid).
So the question left is simply the comparison between OF- and OCl- ,
 The first two factors again work here but another important factor is the 2p-2p electron repulsions( due to smalls ize of sigma bond the electrons in the orbitals of O and F repel each other. This factor  in itself is quiet dominating) present in OF- makes OF- less stable than OCl-
So according to me HOCl must be more acidic than HOF.
Another important thing i wanted to draw attention to is that HOF is not the common acid that can be kept in laboratory. It is quite reactive and reacts vigorously even at 0°C , measuring its acidity is quite diffcult . I also searched online but couldn't find any data concerning Ka values for ionisation of HOF.
I hope this helps.

Fluorine has exactly the available orbitals of chlorine (one s, three p) so either both or none of the two can back-bond based on an orbital discussion.

According to me, $\ce{HOF}$ is more acidic than $\ce{HOCl}$.
As the $\ce{O-H}$ bond length decreases, the acidic nature of the compound increases.
In $\ce{HOF}$ , the $\ce{O-H}$ bond length is $\pu{96 pm}$ while in $\ce{HOCl}$, the $\ce{O-H}$ bond length is $\pu{97 pm}$ and so $\ce{HOF}$ is more acidic than $\ce{HOCl}$.


text:
Could coating silver inside of a cuvette be used for SERS?
I understand that molecules adsorbed to a coinage metal surface experience plasmonic enhanced Raman shifts and I have a spectrometer that I would like to perform SERS measurements with but do not have a Raman microscope/probe. I do have a light-tight cuvette cell and I am wondering if I could coat the inside of a cuvette on two sides with silver to obtain SERS spectra of liquids? This is a standard 1x1 cm cuvette and the SMA cable going to the spectrometer has a long pass filter installed for the correct wavelength.
Most of the literature I have read points to using a suspension of silver nanoparticle colloids mixed with the analyte liquid within the cuvette to obtain SERS spectra, but I have not been able to find any papers that use what I am proposing.

@EdV This paper on ResearchGate suggests it may be possible: "(SERS) of 4-mercaptopyridine (4-mpy) adsorbed on HNO3 etched silver foil, chemically deposited silver films (silver mirror) and silver colloids were measured".

I did an independent study with Prof. J Alan Creighton, one of the discoverers of the surface-enhanced Raman effect, during an exchange year at the University of Kent at Canterbury. The goal was to make a pyridine-silver sandwich complex to compare the molecular spectrum with that of the surface-bound pyridine. Unfortunately, this does not give me any extra insight into your question...

It really depends. In a back scattering configuration a simple aluminium foil folded around the sample holder can enhance raman spectra of thin films. It really depends on the sample. Most of the liquid probed would be far away from any silver, that's why you mentioned nanoparticles

from theory this should work what you have scatched here. When it comes to SERS a main point is, that you got a surface which is rough enough that you can create mini hot-spots for the electric field. This will give you the enhancement.
Maybe you must put a little bit of work in you set up. The scattering is needed to arrive at the detector. maybe you try a Raman probe or do a other angle for the wall with silver on it.
Best regards!


text:
Supposed low temperature of calcium carbide synthesis
I came across an interesting reaction on allreactions.com under silicon tab:
$$\ce{SiO2 + 5 C(coke) + CaO ->[\pu{800–1000 °C}] Si + CaC2 + 3 CO}$$
Unfortunately the website doesn't provide sources and I wanted to read further about this reaction, particularly about the reaction time. If anyone here could point me to another source I'll be very grateful.

Liu et al. in 2012 [1] showed experimental data that lead me to believe this to be inaccurate. Two relevant informations from their findings:

$\ce{SiO2}$ and $\ce{Al2O3}$ do not react with coke at temperatures lower than $\pu{1460 °C}$, but they do react with $\ce{CaO}$ to form $\ce{Ca3SiO5}$ and $\ce{Ca2Al2O5}$, respectively […]
Note that the behaviors of $\ce{SiO2}$ and $\ce{Al2O3}$ observed in this work do not agree with some literature, which reported that both $\ce{SiO2}$ and  $\ce{Al2O3}$ reacted with coke at temperatures of $\pu{∼ 1200 °C}$.

Also, in Fundamentals of Metallurgical Processes: International Series on Materials Science and Technology [2, p. 174] it is written that calcium carbide is not stable in the presence of silicon and undergoes the reaction:

$$\ce{2 SiO2 + 2 C + CaC2 <=> Ca + 2 Si + 4 CO}$$

References

Li, G.; Liu, Q.; Liu, Z. $\ce{CaC2}$ Production from Pulverized Coke and $\ce{CaO}$ at Low Temperatures—Reaction Mechanisms. Ind. Eng. Chem. Res. 2012, 51 (33), 10742–10747. DOI: 10.1021/ie300671w.
Coudurier, L.; Hopkins, D. W.; Wilkomirsky, I. Fundamentals of Metallurgical Processes, 2nd ed.; Hopkins, D. W., Ed.; International series on materials science and technology; Pergamon Press: Oxford [Oxfordshire]; New York, 1985; Vol. 27. ISBN: 978-0-08-032536-1, eBook ISBN: 9781483162409, DOI: 10.1016/C2013-0-03773-5.

@MathewMahindaratne (cc Nilay Ghosh) Please note that using mhchem for denoting references like $\ce{^{[1]}}$ is just evil. Square brackets are used in Markdown markup, hence you might want to escape them with \[ or \], but not by abusing MathJax. Also, keep in mind commonly used Latin words and abbreviations are never italicized, and we prefer to use ACS citation style (or, a least, keep your style consistent).

@andselisk: I agree with you, specifically with ~\ce{[]}$` to show the references. But I didn't have a choice because Nilay already have done that for first reference. I was just afraid I'd ruin the formulation and go with the his method. I have seen suire bracker with references with other experience bloggers. Sorry.

@MathewMahindaratne No need to apologize, this is just a reminder/suggestion, not an accusation (sorry if my comment sounds so). Figuring out a mishmash of edits isn't an easy job, so thank you for participating in posts improvement!

@andselisk: I have a suggestion if you can go with it. When I'm editiong I can't see how I'm doing because edited version is not visual until I submit it. Can you do something to visualize the text while doing editing. That would be really helpful. Thanks for encouragement. ;-)

@MathewMahindaratne Hmm, probably a side-by-side view would be more useful for you; there are some solutions mentioned in Editor to use on PC with windows to format answers?, such as using SOX or VSCode. I have an ancient square monitor with plenty of vertical space, so I just scroll back-and-forth a little when I want to check what my scribble renders to.


text:
What's the difference between atomic orbitals and energy levels?
If I'm drawing a Bohr model of Neon, the middle is the nucleus, the first energy level contains 2  electrons. The second level contains 8 electrons. 8+2= 10, Neon's atomic number. But then what are the atomic orbitals, like s, p, d, and f. Where do they come in? Are they completely different from the energy level?

Things like s, p, d, and f never come into the Bohr model.

There is always a number before the letters $s, p, d, f$. It tells you which Bohr shell youre in.

Neon's atomic number. But then what are the atomic orbitals, like s,
  p, d, and f. Where do they come in? Are they completely different from
  the energy level?

Atomic theory was developed in stages almost a century ago. Bohr's model is the simplest version of hydrogen atom. General chemistry textbooks write rarely mention the Sommerfeld's model from which the concept of orbital angular momentum originated on elliptical paths (recall that a circle of Bohr's model is a special ellipse). Sommerfeld model model explained more features of the hydrogen spectrum. After Sommerfeld the next person was Schrodinger who proposed the H atom in terms of probabilities of finding electrons rather than fixed orbits.
In short, s, p, d, f come into play in more advanced models of Sommerfeld. Bohr was the first one to talk about electronic configurations for multielectron atoms.
Sommerfeld' model

I'm sorry, so what you mean is: Neon's electron configuration, [He] 2s2 2p6, has nothing to do with me drawing 2 electrons on the first energy level, then 8 on the second energy level. These things aren't related. Is that what you mean? I'm sorry, I just want to clarify.

No need to be sorry, these are genuine questions which kept the best minds in physics busy. The picture of 2 electrons in the first shell and 8 in the second is a primitive picture. This picture is not enough to explain the experimental spectrum. [He]$2s^22p^6$ is an advanced version of that primitive model.