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How do you determine the NIE equation of a buffer?
For example, I have formic acid and sodium formate. How do I know the NIE(net ionic equation) equation is $\ce{HCHO_2 + H_2O <=> H_3O+ + CHO_2-}$ and not $\ce{HCHO_2 + OH- <=> H_2O + CHO_2-}$ ? If I added $\ce{HCl}$ or $\ce{NaOH}$, would that help decide which equation? Is a buffer always added to water? Why or why not?

Both reactions happen. The first with the equilibrium constant Ka, the second with the equilibrium constant 1/Kb=Ka/Kw. When I drink water, I do not add buffer, so buffer is not always added to water.
( Sorry, I cannot help myself, the question is formulated rather poorly and not clearly.)

For eventual writing and formatting of chemical formulas or equations, see how to use MathJax with mhchem extension

@Poutnik fixed it

For next time, note that \ce{} need not underscores to display subscripts in chemical formulas.

@Poutnik I don’t know about your water but my tap water contains hydrogencarbonate which technically is a buffering agent ;)

$\ce{HCHO_2 + H_2O <=> H_3O+ + CHO_2-}$ and $\ce{HCHO_2 + OH- <=> H_2O + CHO_2-}$ are both reactions that can happen.
If you added $\ce{HCl}$ (of high concentration), you added a strong acid, and there will be extra hydrogen ions. Therefore, it will drive your first reaction backwards due to the extra hydronium: $\ce{H_3O+ + CHO_2- <=> HCHO_2 + H_2O}$.
If you added $\ce{NaOH}$ (of high concentration), you added a strong base, and there will be extra hydroxide ions. Therefore, it will drive your second reaction forwards: $\ce{HCHO_2 + OH- <=> H_2O + CHO_2-}$.
Distilled water has no buffer added to it, and it is not a buffer itself because of extremely low concentrations of $\ce{H_3O+}$ and $\ce{OH-}$. Tap water might have buffers in it. Lakewater have buffers to prevent sudden changes in pH. For example, it might have limestone at the bottom of the lake, which serves as a buffer against strong acids. As an example, when acid rain falls, it brings with it sulfuric acid, and the buffers prevent the pH from dramatically decreasing, by the equation $\ce{H2SO4 + CaCO3 => CaSO4 + H2O + CO2}$. With the buffers, the fish living in the lake will not be affected by the sudden changes in pH, otherwise, they might stress or even die (but that's a question for Biology SE).


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Exactly how the subscripted numbers are given to the different types of boron sheets (borophenes)?
I was going to start a project on Borophene, and hence I was going through a review journal, 2D Boron: Boraphene, Borophene, Boronene, where they have said many types of borophenes had been proposed. They have been given different greek letters like α, β, γ (β12), δ and χ(chi) to classify different types and they also classified those types into different sub-types by using subscripted numbers like α1, α2, α3 or β1, β2, β3, β12 etc.
At first, I thought it was coming from the hexagonal hole density (η) term which is defined by -
η = (a number of hexagonal holes/a number of atoms in the original triangular sheet). But later I found that it's not the case (considering all the types).
My question is - how exactly these subscripted numbers are given to sub-classify the different types of borophene?
Different types of Borophene(name) and their properties

P.S. - All the numbers next to the greek letters will be in subscript.

All the subscripted numbers are to denote different phases.
In the physical sciences, a phase is a region of space (a thermodynamic system), throughout which all physical properties of a material are essentially uniform. Distinct phases may be described as different states of matter such as gas, liquid, solid, plasma or Bose-Einstein condensate.
Distinct phases may also exist within a given state of matter. As shown in the following diagram for iron alloys, several phases exist for both the solid and liquid states.
For Borophene, α, β, χ, δ and ψ are different types of structures based on their coordination number(CN) like,

α (CN = 5,6)
β (CN = 4,5,6)
χ (CN = 4,5)
δ (CN = 3,4,5)
ψ (Single valud CN)

Now each different structures can have more than one phases. When any particular type of structure has more than one phases, those are denoted by numbers.
For example, α has 7 different structures. Now the first one is named α. After that total of 6 phases are found. One of which is a very similar structure with α. Hence it was named α'. The difference between α and α' is that there is no corrugation in α whereas α' has a corrugation resulting in better stability than α. However, the other 5 phases are named accordingly from α₁ to α₅ based on their earlier prediction. You can find the different phase structures of α Borophene from the following image -
Ref:

https://en.wikipedia.org/wiki/Phase_(matter)
Xiaojun Wu, Jun Dai, Yu Zhao, Zhiwen Zhuo, Jinlong Yang, and Xiao Cheng Zeng. Two-Dimensional Boron Monolayer Sheets. ACS Nano 2012, 6, 8, 7443–7453. https://doi.org/10.1021/nn302696v

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What does heterogeneous mean?
We say that a homogeneous sample consists of the chemically same type of molecules whereas heterogeneous ones vary chemically, i.e. the molecules throughout are not same and also not uniform.
Why do we treat a (water + water vapour) system as heterogeneous in spite of the presence of the same kind of molecules? Also if there exists an equilibrium in the two phases would we still not call them homogeneous?
Does heterogeneity have a chemical basis or physical basis?
I do not seem to get a clear idea as to what defines heterogeneity. The definition has kept on changing since my high school and later on.

Any classification like chemical or a physical change, homogeneous or heterogeneous substance, will eventually fail at one point. From an analytical or physical chemistry's perspective, homogeneous substance is a substance, in a single phase, whose chemical composition is uniform. A heterogeneous substance has different composition from place to place like your garden soil.

Why do we treat a water+water vapour system as heterogeneous in spite of similar molecules present? Also if there exists an equilibrium in the two phases would we still not call them homogeneous?

It is a bad example to label water+ water vapor system as heterogeneous- but with respect to what? It is better to call it a two phase system. You have to define heterogeneous with respect to some property.
For example, if we take tap water in a glass, we can say the liquid (water) is homogeneous with respect to its chemical composition. Let us put some ice cubes in water, now it is a heterogeneous phase, although chemically it is all water.
More formally, from the International Union of Pure and Applied Chemistry's Gold Book:

homogeneity (in analytical chemistry): The degree to which a property
or a constituent is uniformly distributed throughout a quantity of
material. A material may be homogeneous with respect to one analyte or
property but heterogeneous with respect to another.

You can see the subtilty above in italics.

It is a bad example to label water+ water vapor system as heterogeneous- but with respect to what? WRT near any property but chemical composition. A liquid/gas fog or solid/liquid mash is very heterogenous.

General concept of heterogenity says the matter properties within the given region are location dependent ( within the given resolution and tolerance of the scenario ).
The change of these properties can be the gradual ( like a concentration gradient ), abrupt ( like a phase boundary ) or combination of both.
Practically speaking, matter is homogenous, if any its point-like sample has representative properties for the whole region, otherwise it is heterogenous.
So mixture of liquid propane and n-butane is homogenous, if its macroscopic properties, like composition, temperature or density,  are the same across the volume ( within the given tolerance ).
OTOH, a system containing just water in 2 or 3 phases is heterogenous, as a point-like sample from either of phases is not representative for the other phases.
See also wikipedia Homogeneity_and_heterogeneity  and Homogeneous_and_heterogeneous_mixtures

Your definition of homogeneity is wrong
Homogeneous systems are not, as you seem to assume, about the makeup or constituents of the system being uniform. Homogeneity is about the distribution of the components in the system being uniform. A solution of salt in water is homogeneous (if well mixed) despite having multiple types of molecule in it. If those molecules are evenly distributed, the mixture is homogeneous.
More examples: ethanol and water form a homogeneous mixture under all normal conditions as the two components are mutually soluble. Oil and water form a heterogeneous mixture and the best you can get is a fine dispersion of one liquid in the other with two distinct liquid phases. Air normally contains some water vapour as a homogeneous mixture. But, when too much water vapour is present, you get fog which is an heterogeneous suspension of small water droplets in air (which is saturated with water vapour). Water is present in both phases but in two different states, liquid and vapour.

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Why is oxygen sp2 hybridised in acetate ion
The oxygen atom with the negative charge on the acetate ion is said to be sp2 hybridised but I can seem to figure out how
I feel I may not have out much effort into this question so even a link to a useful website where I can learn how to find hybridizations for atoms other than the central atom will be much appreciated.
Chemistry is hard :(

The oxygen atoms in the acetate ion are identical. The fact that we draw them differently just reflects the inherent imperfection of our language.

Somehow I doubt that knowledge that both oxygen atoms are in practice rather sp hybridised won't help you. And that "are" part should be more like can be described using such approximation, if hybridisation is even used on occasion.

related: https://chemistry.stackexchange.com/a/9740/102629

Oxygen atom in the acetate ion is $\ce{sp^2}$ hybridized because as @Ivan Neretin said in comments, both the atoms are identical because the negative charge is in resonance with the $\pi$ bond of the other oxygen atom and thus it has to be $\ce{sp^2}$ hybridized to show resonance.


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How to explain the relation between atomic number and energy of orbitals?
I came across a certain line in my textbook that says:
"As the atomic number increases the energy of orbitals decreases"
Can someone please explain how this happens?
My explanation is: As number of protons increases, the increased positive charge pulls the electrons closer to the nucleus, thus reducing the size and energy of the orbitals (since orbitals with lower energy tend to be smaller and closer to the nucleus)
Is my explanation correct? If not can someone please explain it to me in simple words? I have just started getting into chemistry

It is valid, while there is a single electron(*). With more electrons, things get complicated by nucleus charge screening/shilding and mutual electron repulsion. E.g. the $\mathrm{2s}$ electron of $\ce{Li}$ atoms "feels" the effective charge somewhere between $\mathrm{+3e}$ as the true nucleus charge and $\mathrm{+e}$, the nucleus charge perfectly(**) screened off by the $\mathrm{1s}$ electrons. See Slater's_rules.

(*) There is important essential thing: the quantum effects. They  defining the allowed levels of electron energies, that go down for respective orbitals with the raising atomic number. If they had gone up, the stronger nucleus attraction would not have helped.
(**) - That would have happen, if 3D distribution of $\mathrm{1s}$ electrons occurance had been perfectly spherically symmetric AND not overlapping $\mathrm{2s}$ electron distribution, forming exclusive regions for $\mathrm{1s}$ and $\mathrm{2s}$ electron distribution..


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The colors that we see
I have a question. When I was studying Coordination Compounds, I came across this para:

When a sample absorbs light, what we see is the sum of remaining colors that strikes our eyes. Observed and absorbed colors are generally complimentary.

Now, I have doubt that does sample word applies to all the objects of universe? I mean if we are, for example, painting an object orange, does it mean that we are actually painting blue and we are seeing it's complimentary color Orange?
Please clear my doubt on this.

@Arnav I suggest you take a look at how perception of color works for some "enlightenment". While I don't see anything wrong with the question, it seems off-topic to me for Chem.SE and better suited for Physics.SE.

Yeah or maybe just suggest a link where one can read about it and have it done with?

Yes to your final question. The point is that following your line I should have answered NO.

The OP is commendably curious, but this question is off-topic.

Thanks @user79161. I will post this at physics SE. Actually, I read about this in Chem, that's why.

These questions are fundamental, the best answered rather by searching than by querying/answering, as the former provides more complete and detailed answer. There are tons of information on internet  or in textbooks about this topic.

The question seems to be perfectly on topic in a Chemistry site. Now reading the entire thread it seems that a full answer is more time consuming that at a first glance. OP you should have fixed it now if you read my first ironic comment. What you say in the question is true. However color rendering can be achieved in different ways. For instance a red LED is not a red paint. In this case it is the LED itself that emits red, opposite to common case in which we see the light impinging an object minus the components that the latter has absorbed/diffused in other random directions.

See Additive color and Subtractive color

Complimentary colors are very polite, but complementary colors are the opposite.

Your problem is a much linguistic as chemical or physical
When you say you paint something "orange" you describe the visible colour you see as a result of applying the paint. It makes no sense to describe the "colour" of the paint as anything else.
The physical process involved in making the paint that colour is, indeed, absorbing the colour ranges complementary to orange. But it makes no sense to describe the paint as anti-orange. Unless you are colour scientist who needs to clarify how the object would look under illumination with a different mix of wavelengths. In which case you probably won't be unclear about the relevant terminology or processes.

That is (at least as formulated in the question) the point. Plus 1. It is also true that from just this point of view the question is off topic both here as well as in Physics SE.

In bold, it should be "as much linguistic as..."

@user79161 it should be than rather than as or a.

Suppose you illuminate with white light a blue object. At an atomic scale what is happening is the following:
Photons of all the visible spectrum frequencies are striking the object, and many of these are being absorbed by the object. This causes the electrons of the atoms to being excited and promote to higher energy levels. This state of the electrons are unstable so they decay with time generally to the ground state. In this way down, there are some "paths", some jumps between levels, that are very likely to occur which are characteristic of the particular atom. In our example, the atoms of the object must have a characteristic jump that emits photons in the frequency of blue.
To physically see an object, the object must absorb the light and then re emit it in a particular frequency to you. Glass for example do not absorb light (the photons do not get absorbed by its atoms), so they pass by and never come back to us. The glass therefore is transparent. A red object absorb light and then only re emits red photons,

Something like at the botton happens in part but it is not why we see a red object to be red. The first paragraph is totally wrong as it would imply absorbing red photons and emitting blu ones.

@Alchimista of course that a body can absorb red photons and emit blue ones. If for example a red object could absorb only red photons, it would be transparent for all the colors, which would travel throug it without being absorbed. Obviously this is not the case and maybe you are confused with the case, for example, of a pure sodium gas that when illuminated with white light, basically only absorbs a characteristic yellow and all the remaining photons would cross the gas. This is not what happens in a red opaque object.

"Of course... blue ones". This is blatantly wrong. Your answer implied up-conversion. True is that the colour in transmission can be even different than that seen by reflection. In short, it is not even true that "must absorb the light and then re emit". I am surprised by the number of down vote. The answer is totally wrong.

@Alchimista the electrons in the atoms do not "reflect" light in the classical sense, they absorb photons or do not interact with them, as they are bound to the nucleus, they cannot change their energy any way they want (obviously this is a simplification, but we are dealing here with the major effect). As you will disagree with this, please, explain how an atom "reflects" a photon (I mean, name the process).

do at least realise your sentences to be wrong? If you are after the microscopic quantum level mechanism of reflection, this is not what OP were asking for. Beware that it is right taking an atom (an atom does not reflect light by definition) that confuses you and/or your answer. There is nothing concerning atomic orbitals in reflection, including transition among them. Reflection is a properties you might even have at the interface of two both transparent clear materials. Finally beware that even a sample of gas, in which collective wavefunction aren't present, won't behave as you claim

In brief, a red opaque object visible only if illuminated is red because it absorbs blu, not because it emits red. More clear that this.... Case exist in which high energy photon results on fluorescence of course, but at least agree that we are leaving the OP trace. As far the question is of relevance, the answer is given above by Matt_Black. I am here just because I thought that your answer will let OP with the idea of knowing more, but that more is wrong.

@Alchimista I do not want to confront with you, I'm just trying to explain myself because I really think you point is wrong.

A red object for sure have spectral lines that lie in the red part of the spectrum. How couldn't it absorb and emit that frequency of light? if that object absorbs all the rest of the colours, what happens to the excited electrons? don't they emit light when they decay? I think that you must explain this very well for maintaining your position.

If you would like to read a good explanation of this absorption and re-emission process, see the answer from Selene Routley: https://physics.stackexchange.com/questions/137293/what-happens-to-photons-after-they-hit-objects of course I'm not inventing the wheel here.

Well I couldn't say that is particularly good. But at least I see that semantic can play a role in our diatribe. OK, let us call absorber everithing which is in reality interacting. In such a case every step of a photon-matter interaction can be called absorption. The point is that while this is clear in the answer you link, it is not clear in your answer. In other words, your answer suggest that reflection happens for photons that can be absorbed by the object, The level that you mention in the blu example are simply not there. Your mistake arises by seeing everything as a isolated atom

"A red object for sure have spectral lines that lie in the red part of the spectrum. How couldn't it absorb and emit that frequency of light? if that object absorbs all the rest of the colours, what happens to the excited electrons? don't they emit light when they decay? I think that you must explain this very well for maintaining your position".This is the source of your misunderstanding other answers. No, it is not true. An object that reflect blue does not need those levels, unless you bring in the discussion some virtual state or something really unstable. They will be at the surface in...

.. in case of perfect mirror. In no case that light propagate within the bulk, right because the material cannot absorb it. Your way of thinking basically negate reflection.

@Alchimista you can talk about reflection in a classical way: Snell's law, Frensel's equations, and so on. This approach, absolutely valid for a lot a issues, starts to fail if you dig into its nature, once you enter the quantum world. So I think when you try to understand how light interact with matter, you must talk about QM (I'm not trying to be "exquisite"). Here what we call classical reflection is the process of absorbing one type of photons and re-emitting exactly the same.

The metals are shiny because their "free" electrons can absorb and then re-emit practically all the frequencies (electronic band structure). If a metal could not absorb the photons, it would not be a mirror as you say, it would be transparent (the photons would never came back to your eye). That happens in glass, which is transparent for visible light but opaque for UV light (this photons have the adequate energy to excite the glass structure, thus, being absorbed).

Matter can absorb one colour and emit other. https://www.fnal.gov/pub/science/inquiring/questions/colorofatoms.html from Fermilab "In molecules, where two or more atoms share some of their electrons, the molecules can absorb light of one color and emit another color. This works whether the atoms are the same (eg two Nitrogen atoms) or different elements. Paints and dyes (the chemicals used to color clothes) are complicated molecules where the arrangement of the electrons make the molecules reflect specific colors very well."

From the same link, about reflection of light: " The part of an atom or molecule that "reflects" the light is the electrons on the outside of the atom. Now..the color comes in two steps..
Step 1: The electrons first absorb some of the light that hits the atom or molecule.
Step 2: The electrons which absorbed the light then emit (give out) some light "

You can put it as you wish. There must be an interaction,that is sure. The point is that your simple picture absorption - emission either negate reflection, as for tell me why the emission does not happen forwards in the bulk..., or negare bulk absorption. The absorber you refer to are quantum object at the surface. In such a sense a body that has them does not absorb. Your treatment mixes mirror with gases and it is messed up. I am perfectly fine with absorber on the surface of a mirror. The mirror doesn't not absorb.

Then it can be a linguistic friction between us. This does not make your question a better one, though we may substantially agree that an interaction of the type absorption/re-emission can take place at the external boundary of metals/mirrors. But if you take that photons can be absorbed/re-emitted at frequency where the absorbance is zero, that is wrong. We won't be able to record zero absorption! Which in turn invalidate the opening of your question. A red object does not absorb light and then re emit red photons. A red object absorbs blue photons, at least in this simple context.

@Alchimista little more I can say. For understanding this issue of colors you must use the quantum picture: photons are being absorbed and then re-emited for us to see the object, there is no other possible way. Assuming only the classical picture of reflecting EM wave lead to the well known errors one can read at the start of any quantum physics textbook. In this picture the classical reflection consist in the absorption of photons that excite electrons in the atoms, which then emit the same photons as they decay.

The classical absorption appears when an asborbed photon creates phonons in the material so the T raises, etc... (the zero absorbance you measure in a metal or a mirror, has nothing to do with the absorption and reemision of light, as you measure neat absorption when the light is finally not reemited, for example in a black body). But this is not what we are talking about, we are focus in the colors we see.

I don't understand very well why don't you accept that it is possible to absorb blue photons and reemit red ones as it is the same process that happens in fluorescence (the only difference is that the frequency of the photons are higher). The contribution to the final reflected ray will depend on the material, but this is certainly something that will happen, how couldn't be?

Under absorption, I consider what is really is absorbed. The absorption re-emission interaction you refer to dies not take place within the level you mention in your answer. The state you are referring to are either extremely short living, perhaps as much to be virtual ones, or are so called surface plasmon. Surface right because not the body nor its atoms or molecule can net absorb at that frequency. Again, it might be that the only difference between our two pictures is, as already have been the case with the question, a merely linguistic conflict. In the sense that you call absorption...

... what I call interaction. And absorbers what I call oscillators. Your mechanism is anyway messed up or mixing up different thing. Pure scattering (by emission or by Rayleigh type mechanisms) let us already see things. That is why we see spectral lines. Spectral lines of atoms are indeed lines which are  absorbed and emitted. But we see tham as lack of light! You omit that fluorescence is diffusive. You omit that generally the transitiond upwards predominate over those downwards, because of population of the levels. This said, of course we see what somehow arrives at our eyes.

"I don't understand very well why don't you accept that it is possible to absorb blue photons and reemit red ones as it is the same process that happens in fluorescence (the only difference is that the frequency of the photons are higher). The contribution to the final reflected ray will depend on the material, but this is certainly something that will happen, how couldn't be?" It does happen. But this is fluorescence, which is not the only way a body is visible to us, and is even a relatively rare phenomenon in solids.

I love when in a interchange, the discussion finally gets distilled and more directed to the exact point of conflict. The process of absorption re-emission is not marginal, is THE process that it is actually going on for we could see. More precisely, it is the process that accounts for the classical reflection. You can get prove of this studying any spectrum of any atom. For example: if you illuminate  sodium gas with white light, and then you analize the colours of the beam after passing the gas, it seems the yellow line is missing,

but what really happens is that the sodium atoms are absorbing this colour and then re-emiting it in every direction, so the intensity you get in the direction of the white beam is greatly diminished. This has nothing to do with plasmons or virtual states. Please, if you have a reliable source that state this is a minor process, let me know. I can say by the way that thing like Raman or Rayleigh scattering indeed are minor in the matter we are talking about: "Rayleigh scattering usually has an intensity in the range 0.1% to 0.01% relative to that of a radiation source.

An even smaller fraction of the scattered photons (approximately 1 in 10 million) can be scattered inelastically, with the scattered photons having an energy different (usually lower) from those of the incident photons—these are Raman scattered photons" https://en.wikipedia.org/wiki/Raman_scattering. In regular objects it is very imprecise to say: "absorbs everything, and reflects red". Even in pure atoms their spectrum have peak with a little width. Regular object emit (reflect) a lot of colors and all that enormous combinations is what we call red.

They might have the most prominent peak around the red wavelengths, but certainly if you decompose their light you will see a rainbow with more intensity in the red part of the spectrum, etc... You will almost never get monochromatic colours with regular objects.This things that I'm talking about are the things I remember from my years at university studying a physics degree (quantum physics courses, particle physics, etc...) and the things I study now for my work as a science high school teacher. If you can point out with data a real mistake it would be very much appreciated.

Fich, all the above is certainly correct. The point is that if we see a gas, we see it by mean of its diffusion (proper scattering processes, like Rayleigh etc... , fluorescence, eventually phosphorescence) and not by absorption. But sn atomic gas is virtually invisible, except if you analyse the spectrum as you said. But atoms. You might well discuss the reflection as well as scattering in terms of abs/re-em from an oscillator. What is different is that the excited state created by the "abs" does not decay with time through "some paths*. No, it does not extend within the body because is

the corresponding state does not exist in the object. Again, is not really clear to me if all this has just a linguistic ground or is genuine. I might continue when I'll have time (if they don't stop us :))

@Alchimista there was indeed linguistic matter as at principle I did not get what did you mean by absorption (my net absorption). But anyway I think that we have reached a point that either one of us is right and the other wrong. I stated that a blue photon could eventually be reemited as red colour, you say this is wrong because the corresponding state does not exist in the body. This is not true because the spectrum of an ordinary object has a lot, plenty of lines.

I know this is this way for sure, but truth be told I never saw data not did any experiment to know the degree of importance of the effect (absorbing blue light and then reemiting red). If the effect is marginal, has little importance if it happens or not (for our discussion, as conceptually I think is important, marginal or not).

Finally I found a video that can settle down our discussion: https://www.youtube.com/watch?v=NJhuf0Um-Eo  See 0:50, he uses UV light, and then blue light to make orange, yellow and green liquids get their colours. This not happen when illuminated with red light as the photons have less energy. He gives an interesting data too in 7:44: "fluorescence is common among organic molecules". Why do you said that fluorescent is rare? (I do not know).

Re-reading the discussion I see I made the mistake of saying that an object can absorb red photons and emit blue ones. This is false as the red photons never can excite the atoms so much to re-emit blue ones (if an electron absorbs rapidly various photons of lower energy, it can reach states that could decay emitting higher frequencies, but this has to have a negligible probability). Evidently, you can only absorb in one frequency and then emit in the same or lower ones, not higher.

yes that is a point that is surely wrong but I did not put much weight on it. Though is related, as there won't be way to see a blue object when illuminating it with a blue lamp. Obviously isn't the case (contrast trouble omitted).

(Written before our last two comments above). I am not saying that a blue photon can't be absorbed than re-emitted as red. This is fluorescence, of course! But a body can be blue without fluorescence, and this is by far the most common case. Otherwise we would need UV lamp to light a room. The blue of the above body would be due to reflection and/or scattering. If you model the processes...

.... If you model the processes you..... might see them as absorption followed by re-emission, the point is that the emitted light is not that that leading to absorption / fluorescence in atom nor the one that would be emitted at the end of a decay process in molecules and solids.

No no, if a body has the transition to absorb the blue light, a great deal of it will be remitting transitioning down the same way (this is the classical reflection). But a significant part of this blue photons will produce other colours as well. In this video you can see how as you illuminate the bottom sea with blue light will produce an intense blue light being reflected (absorbed and then re-emitted in the same spectrum band-line).

But if you use a filter to filter the blue light, you will see all the colours. This is impossible to understand without the absorption re-emission process. https://www.youtube.com/watch?v=E56IuJumLAE This is purpose of my original answer: if an object appears yellow, the yellow we see is not only coming from the yellow that is being reflected (absorbed and then re-emited), there are contributions from other photons with higher energies that are decaying in such a way that eventually produce yellow photons as well.

I gave up. There is no way for water to absorb in the visible (except for overtones of the OH vibration giving, in long optical path, its blue-greenish hue. All the rest are reflected photons. Free to call them absorbed-re-emitted but they basically do not deposit energy into the body (except for vibrations, they can indeed move from surface into depth and warm the body). Another point it seems missed in your reasoning is that in molecules and even more bodies there are plenty of non radiative channels to relax. This is actually well known by all those aiming at fluorescent molecules...

..., which is already a task, able to retain fluorescence in solid state. You mixed up so many phenomena that I can't even answer if not by disparate points. Even in scattering, no the sky isn't blue because because UV transforms into blu. Blue is scattered. Calling absorber an oscillator, while can be adequate and correct in a context, can also cause both semantic as well as physical confusion.

ok @Alchimista. I'm gald we had the interchange. It served me to clarify some points (I even emailed an old university professor of mine to ask a few questions :)). Maybe we will have an encounter in another question haha bye!

one point is sure. I am not much into optics but I do agree that something can be called abs-reemit to justify interaction. This can solve the linguistic issue that was almost certainly present at beginning. Moreover I spotted this because of the supposed energy up conversion that was buried in your answer/comment. Other phenomena are relevant in context, surely we do not see object by Raman. Nor using laser as illumination source. Actually we should restart the discussion from scratch, possibly without thinking of atoms because we do not see them.


text:
Chromatography- why tM is separated from other peaks?
tM in Chromatography is the time of exist mobile phase. But, the mobile phase exist together with all the compunds that tasted in the process. So, why in the graph of signal to time we can see a peak separated from all the other ?
thanks,
Liat

You may need to study basics of chromatography.

"time of exist"—do you think the mobile phase disintegrates afterwards? "all the compunds that tasted in the process"—sounds like degustation, not chromatography.

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The dead time "disturbance" $t_M$, is not a proper peak per se in HPLC. It is a disturbance recorded by the detector when the injected sample solvent reaches the detector.
When you open your garden hose tap, it takes some finite time when you start seeing water out of the other end of the hose. Dead time is exactly like that. It is the time it takes for the sample solvent to travel from the injector, column and finally to the detector. This is the of time of exit as you call it.
Everything must elute after the dead time, because nothing can travel faster than the unretained sample solvent. There are a few exceptions but accept this as a fact for most of the cases.


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Why baking soda neutralize the flavor of coconut oil?
I discovered recently that adding baking soda to a recipe can neutralize the strong flavours of the coconut oil. Does anyone know how it is working and what is, in a simple way, the chemical reaction for this?
Thank you very much.

Seems like you're using unrefined coconut oil? From my memory baking soda could neutralise the lauric acid. Actually to convert unrefined to refined they put it through a process that has a sub-process of neutralisation to help reduce the strong flavour!

On the light of Desai comment OP you might specify it the flavor reduced was the coconut one or any other taste/note.


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Calculate Kc with dissociation degree
If I have the reaction
$$\ce{2HI (g) ⇌ H2(g) + I2(g)}$$
and know that at $T = \pu{448 °C}$ the dissociation degree $α = 0.2198,$ how do I calculate $K_c?$
I thought that it would be something like this:
$$
\begin{array}{cccCc}
\ce{&2HI &<=> &H2(g) &+ &I2(g)} \\
&2n && 0 && 0 \\
&2n(1 - α) && αn && αn \\
\end{array}
$$
Then
$$n_\mathrm{tot} = 2n(1 - α) + 2nα = 2n$$
$$p(\ce{HI}) = \frac{2nα}{2n} = (1 - α)p_\mathrm{tot}$$
$$p\ce{I2} = p(\ce{H2}) = αp_\mathrm{tot}$$
$$K_p = \frac{(αp_\mathrm{tot})^2}{((1 - α)p_\mathrm{tot})^2} = \frac{α^2}{(1-α)^2} = \frac{0.2198^2}{(1 - 0.2198)^2} = 0.0794$$
and then from $K_p$ I could calculate $K_c:$
$$K_c = \frac{0.0794}{(8.314 \times 721)^0}$$
I think the last step is where I'm going wrong since it shouldn't be raised to zero. The correct answer is $K_c = 0.01984.$

At $\alpha=1$, i.e. at full conversion, you errorneously suppose $p_{\ce{H2}}=p_\mathrm{tot}$ and $p_{\ce{I2}}=p_\mathrm{tot}$. Should be
$p_{\ce{I2}} = p_{\mathrm{H2}} = \frac{\alpha}{2} * p_\mathrm{tot}$

You don't need gas laws, temperature or $K_p.$
You overcomplicated the problem: $K_c$ can be found by following its definition with an RICE table:
$$
\begin{array}{lccccc}
&\text{R}\ce{&2 HI(g) &<=> &H2(g) &+ &I2(g)} \\
&\text{I} & 2c_0 && 0 && 0 \\
&\text{C} & -2\alpha c_0 && \alpha c_0 && \alpha c_0 \\
&\text{E} & 2(1 - \alpha)c_0 && \alpha c_0 && \alpha c_0 \\
\end{array}
$$
For the sake of consistency, I denoted the initial concentration with $c_0$ and dissociation degree as $\alpha.$
By definition, the equilibrium constant on a concentration basis $K_c$ can be found as follows:
$$K_c = \frac{[\ce{H2(g)}][\ce{I2(g)}]}{[\ce{HI(g)}]^2} = \frac{(\alpha c_0)^2}{(2(1 - \alpha)c_0)^2} = \frac{\alpha^2}{4(1 - \alpha)^2} = \frac{0.2198^2}{4(1 - 0.2198)^2} \approx 0.0198$$


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Why does a wooden spoon creates bubbles when put in hot oil?
This question might be a bit weird, but I just asked myself why a wooden cooking spoon creates bubbles when put in oil at about 170°C. This is a common trick to find out, if Oil has the right temperature for frying.
My idea is, that the water in the wooden spoon starts to evaporate, but why does this just start to happen when the Oil reaches approx 170°C? Why does this not happen at 100 or 120°C ?

No specific reason. At 220, the bubbles become very strong. At 120, you barely see them.

Thanks for your answer. So my hypothesis that the bubbles are created due to the evaporation of water is true?

It may not be just free water evaporation, but also cellulose dehydration or even starting overall wood pyrolysis.

@Poutnik Wood pyrolysis seems unconvincing. I think cellulose dehydration is a better probability.

@desai Well, try to let organic material in an oven at 170 deg C for longer time. It may be slow, but... The famous novel 451deg F means 232 deg C.


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Radial wave-functions and spherical harmonics of p-orbitals in a certain hypothesis
I know that the s-orbitals of H atom are spherically symmetrical. But the p,d or f orbitals aren't. So, the H-atom needn't necessarily be spherically symmetrical, as the wave-function would be a linear combination of all the orbitals. But, in a certain hypothesis, if I assume the H-atom to be spherically symmetrical, what could I possibly comment on the specific orientations of $p_{x}$, $p_{y}$ and $p_{z}$ orbitals in space? The radial and angular wave-functions might/mightn't change for the p-orbitals, but how to comment on their spatial orientations? Can someone help me out in this?

Why take care of the $p$ orbitals of the hydrogen atoms ? It is like its $d$ and $f$ orbitals. They are empty. So why discuss about  their spatial orientation and of their linear combinations ?

Okay, but what about the excited states? Is it unnecessary altogether to look at $p$ or $d$ orbitals, or even $f$ orbitals in H atom?

Why take care of the excited states of the H atom in chemistry ? They last less than a nanosecond.

In "real life", the hydrogen atom has its only electron in the s orbital. And unless it gets to an excitated state, that's the only thing that matters (whether you're studying the energy of the atom or the bounds it can make in terms of energy or symmetry).
Nevertheless, the H atom is likely to have a spherical symmetry, so any of its p-orbitals will be orthogonal to the others, so their linear combination is likely to keep the overall spherical(-ish) symmetry of the atom.
Here is a representation of such orbitals :

(Reference)
Hope this helps!


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Stability of an atom in absence of EM field
According to Bohr model of atom, electrons move up an energy level in presence of EM field and emit a photon moving down the level.
In complete absence of any external EM field, shouldn't the electron collapse into the nucleus since it would keep losing energy as photons but wouldn't gain any?
(I do know (a little) about quantum mechanics and also that Bohr model is severely lacking. I was wondering if Bohr model had explained this phenomenon.)

Yes it should, so Bohr atom is impossible. End of story.

I'd suggest learning some basic quantum mechanics since this observation was one of the key anomalies that led to its development.

No, Bohr's model hadn't explained this phenomenon(historically). But during that time the idea of 'Quantum ' hadn't been yet applied to matter. By taking the ad hoc assumption that electrons can occupy only some specific orbits in which the angular momentum has specific values(quantised) and by assuming that in these energy levels it does not radiate energy(another ad hoc assumption), Bohr basically explained the major features of the atomic spectrum which was a great feat at that time since there was no classical explanation for the discrete nature of atomic spectra.

This collapsing electron orbit issue was a stumble stone for the last major classical Rutherford model of an atom. According to the classical electrodynamics, an electron accelerated by a central force should emit radiation and finally collapse on the nucleus.
The Bohr model of an atom was the first major quantum atomic model, addressing this electron collapsing issue. It has been postulated that just particular energy levels are allowed, where de Broglie matter wave of an electron forms a standing wave along its orbit.
The Bohr model was significant progress in its time, predicting many observations. But it was later abandoned for its various shortcomings. The major issue was addressing the electron orbital angular momentum and problems with its zero value ( the quantum number $\ell=0$ ).
It was replaced by the quantum model based on the Schroedinger's wave equation, that came with the concept of orbitals.
Later came the Pauli(Schroedinger-Pauli) equation addressing spin and then models involving Special relativity as Klein-Gordon equation or more famous Dirac equation.
The consequences of the Dirac equation are predictions of antimatter before it was observed ( specifically of a positron ) and natural integration of phenomena of quantum spin, that was until than rather an ad-hoc addition to non-relativistic models.


text:
Electrolysis of brine solution gives sodium hydroxide , chlorine and hydrogen gas but catner kellener cell it forms amalgam
Sodium hydroxide is generally prepared
commercially by the electrolysis of sodium
chloride in Castner-Kellner cell. A brine
solution is electrolyzed using a mercury
cathode and a carbon anode. Sodium metal
discharged at the cathode combines with
mercury to form sodium amalgam.
My question is, why sodium ion discharge on the cathode and why not hydrogen ion?

Please please explain Why Sodium-ion discharges on cathode instead of hydrogen ion???????

Discharge potentioal of sodium ion higher than the H+ ion

True, but there is a lot more Na+ than H+.

What is the question here? You have made a statement but not a question.

Na+ ions are discharged in preference to H+ ions due to high over voltage.  see this site:  http://www.citycollegiate.com/sblock3.htm

Find useful reading about overpotential. the overall effect is combination of this, low activity of H+ and very high activity of Na+ in concentrated NaCl solution.

When nobody understands a scientific phenomena, we give it a name. Here the name is overpotential. Here the hydrogen is said to have a big "over potential" on mercury cathode. That is a bright and remarkable way to hide our ignorance.

MOHIT SINGH GAUR - Please edit your question to include your first two comments, then delete those comments. We are not use to reading comments to find the basic question.

1-On electrolysis of brine (NaCl),  we get dihydrogen gas, Chlorine gas, and NaOH.

1-On electrolysis of brine (NaCl),  we get dihydrogen gas, Chlorine gas, and NaOH. But on electrolysis of Brine solution by using carbon anode and mercury cathode, we get chlorine gas at anode and sodium metal at the cathode which makes amalgam.  Why this happen???/

I’m voting to close this question because it's not a question, rather a collection of words.

I like what Maurice said "When nobody understands a scientific phenomena, we give it a name. Here the name is overpotential. Here the hydrogen is said to have a big "over potential" on mercury cathode. That is a bright and remarkable way to hide our ignorance."
This is little on the extreme side but people have spent their life on studying the hydrogen evolution reaction. It is a kinetic issue. It was one of the favorite topics of electrochemists. As one can see in the figure, each electrode evolves hydrogen at a different potential in water. Using mercury in the Castner-Kellner cells, allows sufficient negative voltage that can reduce sodium ions even in the presence of water. Additionally, mercury can form amalgams with many metals. Sodium is not an exception. The moment sodium is reduced at the mercury surface, it is dissolved in the mercury pool, which is periodically tapped off.
Figure taken from: An Electrochemistry Experiment: Hydrogen Evolution Reaction on Different Electrodes in the Journal of Chemical Education.


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What's the meaning of "wavefunction must be finite"?
In quantum mechanics it's important that the wavefunction has to be finite everywhere, including infinity. It's a criteria for acceptable wavefunction. What exactly does this statement mean? And how it's associated with square integrability?

And the square integrability is related to the fact the probability of electron being somewhere is certainty. Integral of the $|\phi|^2$  over all the space is equal to 1.

The probability has to be 1 over all space i.e. $\int \psi^*\psi dv=1$  which means that the 'particle' has to exist somewhere, but if the wavefunction was infinite at any point this integral would  not be possible and so the wavefunction would not make physical sense.

A function can indeed approach $\infty$ and still be square integrable, but the s orbital isn't an example of that!

A broader point regarding functions that are finite:
A function is finite if it never asigns infinity to any element in its domain. Note that this is different than bounded as $f(x):\mathbb R \to \mathbb R \cup\{\infty\}: f(x)=x^2$ is not bounded since $\lim_{x \to \infty}=\infty$. However, $f$ is finite since it does not assign $\infty$ to any real number.
This is a generalisation of the comments written above, not specifically quantum related, but a more general point on mathematical functions.

Your question is "Why does wave function must be finite"? Actually, that is not exactly true. Wave function can be infinite, as long as its follows this
$$\int_{-\infty} ^\infty \psi^*(x)\psi(x) dx=1$$
What's the example? Dirac delta function is one of them.
For most cases, it is true that wave function of a bounded system must be finite for the simple reason that a wave function has to be normalizable  ($\int_{-\infty} ^\infty \psi^*(x)\psi(x) dx=1$).
If it was infinite in some places (for example $\psi(x)=\frac{N}{x}$), we can try to integrate them over all space, and will get infinity
$$\int_{-\infty} ^\infty \frac{N^2}{x^2} dx=\infty$$
And, that is not just make no sense, it is also not physical.

Aside the requirement for the normalization, is not a wavefunction having a local singularity potentially violating the Heisenberg uncertainty principle?

I am not sure, but that do make sense. But the Fourier transform of Dirac delta function is just complex exponential. So, basically, the "uncertainty" in the momentum wave function is infinite, making the $\Delta x \Delta p$ become an indeterminate form.

And, eventually, are not there other mathematical requirements for wavefunction properties, like being differentiable in its domain? I guess that in context of being solution of differential equation, it looks like a mandatory requirement. Note that I am not skilled enough in math to decide myself.

(1) A distribution with $|\psi^2|=\delta(x)$ gives you an exact coordinate and a completely unknown momentum, so no violation of Heisenberg here. (2) A $\psi$ function with local singularity does not have to be like delta function. (3) There is no requirement for $\psi$ to be differentiable everywhere; indeed, $s$ orbital at $0$ is not.

I mean, come on. Take any square-integrable function with infinite singularity at some point. Plug it into $\hat H\psi=E\psi$. Now choose your $V(x)$ so as to make this true. It is as simple as that.

Nitpick: V(x) and boundary conditions

@IanBush Provided that we decide to have some, yes.

No boundary conditions, no particular solution - a general solution maybe, but you can't just neglect the boundary conditions if you want a specific square integrable function as a solution.

But I can have a specific particular solution square integrable on $\mathbb R^n$ with no boundaries.

You still have a boundary condition as how that function behaves as it approaches infinity, typically that it tends to zero - this is the usual boundary condition when solving the molecular Schrodinger equation. I also do solid state stuff where periodic boundary conditions may be more appropriate - and they will give you very different solutions.

Free particle with wave function of $e^{ix}$ is a valid solution for $V(x)=0$, and this function does not vanished at infinity.

Indeed @Tensor . Commonly seen as a boundary condition when solving scattering problems. I didn't say the wavefunction had to disappear at very large r. I just said that it is a common boundary condition, being the appropriate one for when solving for bound states of the molecule. Other problems have different boundary conditions.


text:
Strength of the hydrophobic interaction
How strong is the "hydrophobic force"?

Hydrophobic interactions are weak interactions but can have greater strength than hydrogen bonds. I find the strength of the hydrogen bond in literature to be near 1 pN. Finding the strength of the "hydrophobic bond" has proven to be a bit more elusive. I would guess that there is a hydrophobic interaction strength spectrum (dipole strengths and such), but would also guess there is a fundamental order of magnitude of strength of the hydrophobic interaction.
Hydrophobic interactions result from the hydrophobic effect and manifest in measurable attractive forces. For example, the Polyhydroxyethylmethacrylate (pHEMA) backbone has dipole induced interactions, hydrophobic interactions, and hydrogen bonding; each of these interactions being identifiable in molecular position on the backbone (aliphatic regions being identified as hydrophobic sites).
To date, i have found a measurement of such resulting attractive forces in a specific interaction to be near 15 pN (oil-oil droplet AFM approach)[1]. However, I can also find literature that discusses forces in avidin-biotin  complex which are dominated by hydrophobic interactions; resulting attractive forces range from 80 to 250  pN [2].
[1] Tabor et al, 2013; DOI: 10.1021/jz402068k
[2] Moy et al, 1994; DOI: 10.1126/science.7939660

With this not being an actual force you can have some trouble here...

"The hydrophobic force describes the attraction between water-hating molecules (and surfaces) that draws them together, causing aggregation, phase separation, protein folding and many other inherent physical phenomena"
it is not a fundamental force of nature, but it is a widely recognized intermolecular force.

You may have success if you look into surface tension and consider the area of the hydrophobic interfaces you are interested in.

Get better resources https://chemistry.stackexchange.com/questions/37287/basis-for-the-hydrophobic-effect https://chemistry.stackexchange.com/questions/135081/how-does-the-entropic-force-arise-within-the-hydrophobic-effect

I refer my information from a text book titled Intermolecular and Surface Forces; Israelachvili 3rd edition. I think your comments have been a bit patronizing, and if you refer to the question i say hydrophobic interactions and use quotations around "hydrophobic force". i realize this is not perfect, but it is a phenomena despite the etropic underpinnings.

Well, I think there's a need for clarification - so-called hydrophobic effect in more of a phenomenon then force. There are obviously various interactions between induced dipoles, quadrupoles etc. but do you ask about them, then?

hydrophobic interactions result from the hydrophobic effect and manifest in measurable attractive forces. the attractive force (per interaction) apparently depends on a lot of things, which is the basis of my question. induced dipoles are not required in hydrophobic interactions but can be another sort of interaction in a molecule. for example, the Polyhydroxyethylmethacrylate (pHEMA) backbone has dipole induced interactions, hydrophobic interactions, and hydrogen bonding; each of these interactions being identifiable in molecular position on the backbone (hydroxyl groups being hydrophobic).

I have edited the question to reflect my above comment, as it might clear up what i mean and it provides an example

The type of "force" in question does not have the classical meaning as "derivative (or gradient) of energy wrt position". This is what makes providing an answer far from trivial. Say you establish that the free energy of two states of your polymer in solution differ by some amount. What do you use as "distance" to compute a force?

An ideal answer would identify the questions shortcomings, discuss these shortcomings, but also recognize and identify ranges of forces in measurable hydrophobic interactions. I think the comments so far have been extremely useful and valid, but being bogged down by these characteristics to the point where an answer is unachievable I don't think is productive.

I have edited the question to loosen language on the origin of the forces cited in the final paragraph as comments have pointed out important notes on resulting attraction forces in hydrophobic interactions.

You are right that this is a force; it is not a "conservative" force, see eg https://en.wikipedia.org/wiki/Force#Nonconservative_forces

The hydrophobic force does not exist. It is a phenomena resulting from the following considerations.
In principle, liquids should be like gases, perfectly miscible in all proportions. This is the case for non polar molecules, like hydrocarbons. This is not the case for liquids where molecules are hold together by hydrogen bonds, like water. In these liquids, adding solutes requires to break hydrogen bonds. It is acceptable if the solute is able to form new hydrogen bonds, like with small alcohols, polyols and sugars. But if mixing with another substance will break too many H-bonds, Nature refuses to make up this mixture, as if there was a hydrophobic force. Of course, this force does not exist. It is simply a consequence that mixing very different substances would cost too much energy for breaking all H-bonds of one liquid, without forming back new ones.

I might ask: is there such a thing as a "frictional" force? Or an "elastic" force (eg in a rubber band)?

The elastic force is a result of literally stretching bonds in a solid, friction is super complicated.. probably similarly complicated to forces in hydrophobic interactions


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Comparing of melting point of dibromo p-xylenes

How to compare the melting points of the above isomers? I know that melting point depends upon the packing of the molecule, more efficient the packing, higher is the melting point. Symmetry in a compound also increases it's melting point. But all the above compounds have symmetry in them and I can't figure out how to compare their melting points.

I and III have 2 planes and 1 axis. II has 1 plane and 1 axis

@Maurice So what should we conclude from that?

Based on «the better (i.e., more regular) packing, the higher the melting point»,  my speculation the melting points increase along III -> I -> II.  This extrapolates*) from the packing of atoms in metals.  While none of the three molecular structures is a sphere, in my perception, compound II is the most sphere like of the three.  And neither one of the three has strong hydrogen bond donors and acceptors to increase the strength of the intermolecular attractive interactions (except H-B). *) Of course, extrapolations always come with the danger to potentially lead to a wrong conclusion.

@Buttonwood: Thank you for your comment. But how are you perceiving the 2nd compound as more sphere like? Can the melting point data be checked from somewhere?

@Buttonwood I think you should read about pi-stacking

I am thinking of an answer as for at a first glance this series has everything to conjure against rules of thumb. It would be nice if someone has access to the mp of compound I.

@Mithoron I join Alchimista's comment, «everything [can be used] to conjure against rules of thumb», and substitute sphere like by more round.  I can't quantify this.  But the polymorph of II accessible in the COD (preview CCSD) looks more herring bone, than $\pi$-stacking, don't you think, too?

@LightYagami If prepared as compound, and if characterized by recording a melting point, this property would enter the primary literature (e.g., the article), then a database like Scifinder or Reaxys (both commercial).

@Buttonwood Looks like T-shaped stacking.

@Buttonwood yes more global symmetric or round... But we are all on the same meaning. Also do not debate about parallel vs herring bone arrangement much. The latter is normally the case with aromatic molecules this size. It is still of stacking led by a pi system and still flat regular symmetric molecule favour that. I really need the mp of I otherwise I would write something more "didactic" than an answer somehow on point... A true one does not exist, or better, one can't do it.

@Mithoron I couldn't tag you in the above comment.


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Name of toxic liquid product that can be cleared with flocking agent
I'm writing a paper rpg scenario, and even if we don't need to be 100% accurate about chemistry and physics in such games, one of my players has an advanced level in chemistry and I don't want him to cringe at the situation (and I can't ask him, obviously, or he'll be spoiled).
So, I need a river to be polluted with an invisible and odorless product, that slowly destroys the fauna. As it happens in Siberia, I thought about tritium first, but I don't think it's easy to clean. So I thought about a product that could be cleaned with a flocking agent. Problem is, when I was searching for "products reacting to flocking agent" on Google I didn't find any name of product that I could use.
So, which invisible, toxic and odorless product react to flocking agent to the point the river is cleaned in one or two months maximum (and eventually produced in the 1980's) ?
Thanks a lot, and sorry if it's not the SE to ask this question.

If your impurity is tritium, it can not be purified, whatever the method used. If your impurity is another substance, it can be purified by adding aluminum sulfate. Aluminum sulfate is slowly hydrolyzed in dilute solutions. And this phenomena produces a light and insoluble substance called aluminum hydroxide $\ce{Al(OH)3}$. The particles of this substance look a bit like snow flakes and have the property of fixing impurities on its surface, between its branches. So after a couple of hours, this aluminum hydroxide will be deposited on the ground of the river or the lake, with the impurities fixed inside its particles.

I've seen that Pyrogallol is treated with aluminium hydroxide and seem to have the wanted properties (odorless and toxic), it colors the water in purple apparently but I'll say it's really diluted ^^ thanks a lot!


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How to mitigate salicylic acid saponification?
I started to make soap a few months ago and came up with a few formulas, one of it contains salicylic acid as a exfoliant/anti-acne agent, I've just realized that since salicylic acid is, well, an acid, and sodium hydroxide a base they will react together to form sodium salycilate, which doesn't have the required properties.
I add the salicylic acid and the other agents at (really) thick trace, when the saponification reaction is quite advanced but not complete, actually I couldn't find any data on how far the reaction has completed at this stage (if anyone knows this is always good to know).
Is there any way (besides waiting for reaction completion) to keep the NaOH reacting with triglycerides instead of starting reacting with the salicylic acid?
Is this even necessary?
I mean I know some reactions will be more likely to form a product or another depending on the amount of one of the reactants, like the reaction between salicylic acid and NaOH will give either sodium salicylate or disodium salicylate depending on how much NaOH is available (how is called this phenomenon by the way?), so maybe the unreacted NaOH will be that low at the time I add the salicylic acid it won't react with it at all or so little that there's nothing to be worried about.
What do you think?

How do you know that sodium salicylate has not the same anti-acne property ? You say that you "add the salicylic acid and other agents". What are these agents? What is a thick trace ? Now $\ce{NaOH}$ will react immediately with salicylic acid. The saponification rection with oils is much slower.

Hi Maurice,
I couldn't find any data on a anti-acne property of the sodium salicylate so as far as I know sodium salicylate isn't an anti-acne agent.
Perfume, vitamin E and exfoliating walnut powder are the other agents I add
The thick trace is when the mixture of oils and NaOH consistency starts to thicken then it hardens in a few minutes/hours depending on the oils used in the recipe, chemically I guess it corresponds to the stage of the reaction where most of the triglycerides have been replaced by fatty acids sodium salts.
Thank you

What salicylic acid or sodium salicylate enters the human body, they are soon in contact with the plasma whose pH is fixed, and does not change. Both substances are then transformed into the same mixture salicylic acid + salicylate ion corresponding to the pH of the plasma, whatever their origin, acid or salt.


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Assigning carbon-13 NMR for ethyl 2-oxo-2H-chromen3-carboxylate
I have a molecule and an NMR spectrum, and I need to assign the NMR peaks to each of the carbons to the molecule:

I know that the 14.19 ppm and the 61.94 ppm correspond respectively to the carbon 12 and 11. However, I struggle for the other peaks especially between 116.76 and 148.51 ppm. I just know that it corresponds to the aromatic ring. Finally, I think that the 156.65 ppm and 155.15 ppm are respectively the carbon 9 and 10.
I also have the 1H NMR for this molecule:

The name of the compound and your numbering are not consistent.

It is, for all intents and purposes, impossible, or at least extremely difficult to assign a bare 13C NMR of a moderately complicated molecule with no other information. You should at the very least get a 1H spectrum and HSQC, and perhaps after that a HMBC for your quaternaries. In the absence of that, the best you can do for this compound is to circle a bunch of peaks in the 110–160 ppm region and mark them as "aromatic".

In addition to @orthocresol comments, I'd say $\pu{155.2 ppm}$ is definitely not $\ce{C}$-10.

No, the 1H alone isn't enough: you need a HSQC too, or else you can't work out which proton belongs to which carbon. Given that you have five quaternaries which won't show up on a HSQC, I'd actually upgrade the HMBC from a "perhaps you need it" to a "you definitely need it", if you want to unambiguously assign every 13C peak.


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Graham's Law of Effusion when Looking at Samples with Equal Mass?
So, I had a question on a chemistry test about two gas samples, one of O2 and one of N2, that had equal masses. We were given the time of effusion for the sample of N2 and then asked to calculate the time of effusion for O2. My initial reaction to the problem was to think that the nitrogen would effuse faster, as the effusion rate of nitrogen is clearly faster than that of oxygen. However, after doing the problem and adjusting for the number of mols of each sample, the oxygen had a faster time than the nitrogen. I got this question wrong (the "right answer" on the answer key had a time of effusion for oxygen that was slower than that of nitrogen). This doesn't make sense to me. When I asked my professor about it, they replied that we were supposed to use the equation t(N2)/t(O2)=sqrt(MolarMass(O2)/MolarMass(N2). I have my doubts that this is correct when looking at gas samples that have the same mass but not the same mols. It would seem that O2 should have a quicker effusion rate. Because r1/r2=sqrt(Mm2/Mm1), according to Graham's law, the ratio of Mm2/Mm1 is always going to be greater than ratio of r1/r2, so it would seem to me that the number of particles that need to diffuse should play a larger role in effusion time than how fast the individual particles are moving, at least when looking at samples of equal mass. Am I wrong here?

Please ! Be serious ! $\ce{O2}$ and $\ce{N2}$ do not have the same mass !

Please read the question more carefully. I am talking about SAMPLES of N2 and O2 that have the same mass but different mols. I am not saying that O2 and N2 have the same mass.

Effusion is how isotopes of say UF$_6$ can separated. The rate of efflux of a gas with mol wt $m$ is $\rho\sqrt{RT/(2\pi m)}$ where $\rho$ is the gas density. Thus at the same density gasses undergo effusion as $1/\sqrt{m}$.  (To be accurate this eqn. applies only if the mean free path is 10 times the size of the efflux pinhole.)

You need to be more precise with the terms you use, you are mixing up rates of effusion and times of effusion. This is a trick question, gases with lower molar mass will always effuse faster as per Graham's law


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Getting close to zero percent of particles with energy greater than an activation energy of 101kJ/mol using the Maxwell-Boltzmann energy distribution
I experimentally obtained the activation energy of a reaction to be approximately 101kJ/mol. I was then going to use the value for the activation energy to see how the number of particles with sufficient energy to undergo a successful reaction changes with temperature using the Maxwell-Boltzmann energy distribution. However, I noticed that the graph already approaches zero close to 10kJ/mol. I attached a photo of the equation and graph.

Am I doing something wrong?
I am certain that the activation energy of the reaction I am studying is around 100kJ/mol and that a reaction really does occur.
Another question, what does the y-axis really represent?
I graphed the proportion of particles with sufficient activation energy vs. temperature in Kelvin and I found that it increase exponentially. So exponentially that it seems almost vertical. Here is a picture of it. I could also have made a mistake though.

So what if only $1\over1000$ of all molecules are able to react at any given moment? We'll wait a few moments longer, that's all.

With typically $10^{10}$ collisions /second  and 1/1000 with enough energy is still a huge number of collisions and will be a pretty fast reaction.

Oh right, thank you both for your answers. I got the proportion of particles with sufficient energy but the values that I got were to the order of 10^{-52}. Is that unusual? @porphyrin , Ivan Neretin

Well this is a tiny, tiny number, and so small that even with a mole of reactants you will have to wait, and wait, and wait...

I see. Is that number usual? Actually, are activation energies of around 100kJ usual?

Large but not unusual. I calculated a fraction $\approx 7\cdot 10^{-22}$  at 280 K using the energy equation as given by Wikipedia and 100 kJ/mol, but you should check my calc I may have made an error.

I see. Perhaps I used the wrong method? I did the integral of f(x) with the activation energy as the lower boundary over the integral of f(x) lower boundary, 0. Could I please have the link to that Wikipedia page? I can't seem to google it.

Nevermind, sorry. The values increased exponentially after 100K. The fraction I calculated with an Ea of 101 was $1.08*10^{-18}. I suppose that would be more "normal." It's quite interesting how it increases so suddenly. I edited the original post to add the graph of my results. My equations could be wrong too, though.

@porphyrin Is not 1 degree of freedom energy distribution more appropriate than 3 degrees of M-B distribution ? Like the Ke for motion perpendicular to direction to the other molecule do not really contribute much to the reaction activation, if the frontal attack to overcome potential barrier is expected. ( But I am too long out of Academia to be a good theorist. )

@Poutnik, yes that would be better. The effect is probably small, however,  as the exponential will dominate.

See hyperphysics : Velocity Distribution in One Dimension  $$f(v)=\sqrt{\frac{m}{2\pi kT}} \exp {\frac{-mv^2}{2kT}}$$ with the maximum at $v=0$

You might find it helpful to replot the lower plot with a log scale on the y-axis and a narrower range of T on the x-axis.

@Andrew Yes!! That was extremely helpful! Thank you so much!


text:
Right conductivity of protein sample for Protein A purification of human IgG?
A sample from a cell culture has some IgG and this is to be purified using protein A chromatography. Based on the chromatography handbook from GE-Healthcare/Cytiva, the protein sample should have "physiological" pH and conductivities. This corresponds to a pH between 6-8, but it is not clear what the conductivity sould be. Is there a ballpark value for this?
Also in general, the series of buffers for Protein A chromatography including a wash step looks something like this

Binding buffer (physiological conditions) to equilibrate the column
Protein sample to load the column
Binding buffer (same as above) to wash impurities while retaining the protein
Elution buffer to unload the column

Does the protein sample have to have the same pH and conductivity as the binding buffer? If so, it has to be adjusted after coming from a bioreactor. What happens if the solution from the bioreactor already has a very high conductivity above that of the binding buffer?


text:
Direction of valence bond and Unsöld's theorem
Is it true that the atomic orbitals of separate atoms already have directionality, which ensures the directionality of the valence bond? Or another words, what is the real spatial distribution of the electron density in the atom, obtained in the one-electron approximation?

Doesn't this contradict Unsöld's theorem that states the square of the total electron wavefunction for a filled or half-filled sub-shell is spherically symmetric?

I think you might be getting confused by the way those orbitals are drawn. Yes, they have directionality, but they aren't long and narrow like that. For example, a p orbital is more like two big spheres, such that all three p orbitals collectively make a big sphere. That's why if all three of them are half-filled or are all completely filled, the net density is spherical.

@Andrew Ok, if we look at the d orbital along z-axix. It differ from y- and x-orbitals. But an atom does not know about any axis, so, what a real angular distribution of electron density in atom with d - orbital?

A half-filled subshell means one electron in each of the orbitals, so it doesn't matter that the dz2 orbital looks a bit different from the others. What matters is the sum of all five d orbitals. Look up the equations for the five hydrogen atom d orbital wave functions in spherical coordinates. Square each one to get the electron density. Add all five together and simplify the result. You'll see that all of the angular dependence disappears and the result is only dependent on r, ie it's spherical.

@Andrew "You'll see that all of the angular dependence disappears and the result is only dependent on r, ie it's spherical." --- yes, I did it $(2\ell + 1)/4\pi |R(r)|_{nl}^2$. But I'm interested in the direction of valence in this regard. For example, to which end should a two hydrogen atoms join another d-valence orbital atom?


text:
Disposal of Silver Halides dissolved in Ammonium Hydroxide
I was reading the Lab Manual for Chemistry prepared by the NCERT . While going through the the test for Chloride ions (pg 88) I came across the following Hazard Warning :

The solution obtained after
dissolving AgCl/AgBr/AgI
precipitate in ammonium
hydroxide should be
acidified with 2M HNO3
and should be discarded
quickly to avoid serious
explosion.

How does it cause a serious explosion?

See these questions: https://chemistry.stackexchange.com/questions/43441/is-tollens-test-safe .... https://chemistry.stackexchange.com/questions/67395/mirror-reaction-avoiding-silver-nitride ... https://chemistry.stackexchange.com/questions/122666/will-disodium-edta-effectively-destroy-tollens-reagent

In the test for Cl- the chloride ions are converted into AgCl by the following reaction,
Cl- + AgNO3
→
NO3- + AgCl
which is then dissolved in ammonium hydroxide by the formation of diammine silver chloride by the following reaction:
AgCl + 2NH4OH
→
[Ag(NH3)2]Cl + 2H2O
On standing, the diammine silver complex subsequently breaks down to Ag3N.
The standard free energy of Ag3N is about +315 kJ/mol . Hence it is unstable and decomposes explosively to metallic silver and nitrogen gas .
2Ag3N → 6Ag + N2
So the diammine silver complex has to be destroyed after use so that it does not break down into Ag3N.
One way of destroying the complex is by adding nitric acid.
[Ag(NH3)2]Cl + 3HNO3
→
AgNO3 + 2NH4NO3 + H2O
So in all reactions involving the formation of diammine silver complex, the complex must be destroyed using acids after the purpose.
The complex is formed when AgCl, AgBr(sparingly soluble)and AgI(the complex will be formed only in trace amounts) are dissolved in NH4OH . The comlex is also formed in the well-known Tollen's test. Tollen's reagent is [Ag(NH3)2]OH. Ammoniacal Silver Nitrate is also used in Fontana–Masson Stain and also in silver mirroring(such as inside insulated vacuum flask).
The Wikipedia page on Silver Nitride
A reported accident

Actually, the mechanism of $\ce{Ag3N}$ creation (and also the associated products, Silver amide $\ce{AgNH2}$, and the imide $\ce{Ag2NH}$), have not been definitely accounted for, either by accident investigation reports or per a recent review of the literature, on possible paths to explosive Silver nitride.
I have ideas of a wider chemistry spectrum based answer which accounts for the apparent role involving $\ce{Ag2O}$.
To begin, I cite the more common background description, per a government source, to quote:

Silver nitride is an explosive chemical compound with symbol Ag3N. It is a black, metallic-looking solid which is formed when silver oxide or silver nitrate is dissolved in concentrated solutions of ammonia, causing formation of a silver-amide or imide complex which subsequently breaks down to Ag3N.

So, of interest is the cited reaction:
$\ce{Ag2O(s) + 4 NH3 + H2O ⇌ 2 [Ag(NH3)2]OH}$
where I would expect any Silver oxide formed from the reverse reaction of dissolving Ag2O in ammonia would then likely create a high surface area (with enhanced reactivity) $\ce{Ag2O}$. It should also importantly be noted that the reaction, itself, is reversible (see as a source, "Second year college chemistry" by William Henry Chapin, page 255), to quote:

As might be expected, the silver-ammonium complex dissociates slightly into its constituents as indicated by the equation

$\ce{[Ag(NH3)2]+ ⇌ Ag+ + 2 NH3}$

This is a reversible reaction, very much like the ionization of a very weak acid or base.

So, once any Ag2O is formed (by say moving the equilibrium to the left from a possible loss of water on standing in an open vessel or upon addition of dry alcohol), a radiation source could further accelerate the process (see, for example, Ag2O as a New Visible-Light Photocatalyst: Self-Stability and High Photocatalytic Activity). To quote from the abstract:

Ag2O is unstable under visible-light irradiation and decomposes into metallic Ag during the photocatalytic decomposition of organic substances. However, after partial in situ formation of Ag on the surface of Ag2O, the Ag2O-Ag composite can work as a stable and efficient visible-light photocatalyst.

As such, the introduction of appropriate light, or other irradiation paths (like X-rays commonly associated with radical formation) may be instrumental. In fact, here is an interesting pertinent radiation study: The X-ray activated reduction of silver (I) solutions as a method for nanoparticles manufacturing.  To quote from a passage discussing an associated radiation treatment:

The next step was aimed at eliminating glucose-containing ammonia solutions. The reactions taking place in these solutions resulted sometimes in the formation of precipitates with unstable composition. These precipitates contained considerable amounts of silver nitride Ag3N, which formed in some ammonia solutions of the silver salts and exhibited explosive properties in a dry state [15]. The results from analysis of the deposits obtained for three selected types of solutions are presented in Table 1.

As the irradiation of glucose is a path to radicals, my proposed mechanics incorporates an ostensible promotive radical pathway for the allude to creation of $\ce{Ag3N, Ag2NH and AgNH2}$, which also appears to be accelerated in the presence of $\ce{OH- and Ag2O}$ in the presence of ammonia.
The proposed reaction path is:
$\ce{NH3 + H2O ⇌ NH4+ + OH-}$
$\ce{NH4+ ⇌ NH3 + H+}$
$\ce{[Ag(NH3)2]+ ⇌ Ag+ + 2 NH3}$  (per above)
In the further presence of light:
$\ce{Ag+ + hv -> •Ag + h+}$
Source: See Photodecomposition and Luminescence of Silver Halides
$\ce{OH- + h+ -> •OH + hv}$  (same source as above)
$\ce{OH- (aq) + hv -> •OH + e- (aq)}$
As a reference, see: Flash photolysis in the vacuum ultraviolet region of sulfate, carbonate, and hydroxyl ions in aqueous solutions
$\ce{NH3 + •OH -> •NH2 + H2O}$
From a related work: Kinetics and Mechanism of the Reaction of •NH2 with O2 in Aqueous Solutions
$\ce{NH3  + •H <=> •NH2 + H2}$
As noted in this work.
$\ce{•NH2 + e- -> NH2-}$
$\ce{Ag+ +  NH2- <=> AgNH2}$
$\ce{AgNH2 -> Ag2NH + NH3}$
$\ce{AgNH2  + Ag2NH -> Ag3N + NH3}$
Note, my suggested pathway is largely promoted from the recognition of Ag2O as a recent (article was 2011) visible-light photocatalyst with the possible introduction of contributing radical based reactions.
In summary, I am suggesting chemically diverse mechanisms, involving both light and radical based reactions, as a more likely explanation.

In layman's terms, silver would have to be in the form of an oxide to set the creation of silver nitride, when dissolving in ammonium hydroxide. It is a simple tradeoff... the silver oxide takes on the nitrogen from the ammonia and donates it's oxygen to the hydrogen. Then you get water and silver nitride. Without silver being bonded to the oxygen, silver will not form an explosive compound when ammonium hydroxide is added to the reaction.

Welcome to Chemistry Stack Exchange. Please visit here - https://chemistry.stackexchange.com/help/how-to-answer - to see how you can improve this answer.


text:
Crucible design for melting alumina
I'd like to melt alumina in a slow, controlled manner.
So I thought a graphite crucible with a tungsten wire wrapped around it for heating would be a good start.  To reduce heat losses, I put that crucible inside an outer crucible.
Question 1 - can I get away with Al2O3 for the  outer one, given that the radial temp profile will be something like 1/r .  If not Al2O3 then what - Al2O3 would be nice as it has the lowest thermal conductivity that I can find of the refractory materials I looked at (W,SiC, C, BN)  .  And should I fill the annular space with e.g. YSZ powder (the quartz fiber I used to use for this kind of thing wont handle 2000C - is there some other fiber or 'wool' that can be used?)
Finally, can I zets the melt from above with microwaves conducted thru a waveguide - will the electrical  conductivity of the graphite be trouble or maybe help matters due to eddy currents?

A possible concern with the graphite crucible is contamination, sublimation and ignition. Would you be better off using zonal melting and horizontally directed crystallization. per http://www.crystaltechno.com/Al2O3_en.htm , which does not contact a crucible, with contamination issues? See also https://www.universitywafer.com/sapphire-wafers-al2o3.html

I'll read up on those methods - the crucible method seems more easily controlled than the Verneuil flame method - but I suppose crucible contamination may be possible. I don't know if there's any reactions expected bet. liquid Al2O3 and graphite. I don't think either material will ignite as both the Al2O3 and graphite are  quite stable (and the Al2O3 of course is already 'burnt').  In principle  I guess I could use a tungsten  crucible (wow, even platinum is already liquid by 1770C)

Having become aware the Pt is also liquid at my temps of interest, how do I measure temperature - the Pt/PtRh thermocouple clearly won't suffice...maybe there's a W/W+something thermocouple ...yup there it is , tungsten/tungsten+rhenium , up to 3000C

If platinum thermocouples won't do it , the next step is optical/ infrared.  /Thorium oxide fiber will tolerate your temperature .

Tungsten oxidizes at high temperature so must be in an inert atmosphere .Like little brother molybdenum , the oxide has some vapor pressure ; makes 316 SS a poor choice for high temperatures.

As for my first question it looks like 1800C is the upper use limit for alumina crucibles so I suppose I'll be use a graphite-in-graphite double wall.

Is zirconium a possible crucible material? I see that most use tungsten or molybdenum crucibles in industry.


text:
Stirring bar cleaning
I have a small question concerning the stirring bars cleaning. I use mine in BOD test wastewater for 5 days and I want to clean them so my question is can I soak them overnight with sodium hypochlorite (Javel) or it can affect their magnetic ability or something?

The magnetic properties of a stirring bar are not affected by any chemical treatment like cleaning with hypochlorite.

I found sodium in liquid ammonia got them really clean!

@Maurice Erm … this lab had a stirring bar lose its magnetic property. I suspect the outer teflon or whatever layer died a slow and painful death and then something acidic affected the metal core. So true in principle but exceptions exist ;)

No. No chemical treatment can change the magnetic state of an iron piece. The only thing that could change the magnetization of an iron bar is the heat (above the Curie point) and the time.

Is this really a question from someone doing chemistry?

Alchimista thanks for your kind reply I was asking because it's the first time I soak them in Javel for 24 hours, anyway learn from the above how to reply to people question gently ;)

@Maurice I believe that you mean no chemistry happening around a magnet can change its magnetic properties. Obviously isn't the case if the magnet react itself.

@ShafikSaad sorry. I apologise if it turns impolite. Nothing was aimed at you. Rather the question would have been not great but more sounding if formulated as "can a prolonged treatment in eau de Javel modify the Teflon coating of a stirring bar?". I thought to delete it but then a reader could think of a more personal impolite attack, and that was no in my scope. So I let it there. Mods might consider removal(s) for the cleanness of the page.

Alchimista your apology means so much thank you for your kidness and yes that was my question usually I soak my bars when I arrive to work and take them out before leaving but I forgot last time and I was affraid if the coating will be damaged... Thank you again for your apology and wish you the best :)


text:
Which of the following can be a source of error in Spectrophotometry? Mark the wrong answer
a) Differences in solvent
b) Acid base equilibrium (differences of pH)
c) Parasitic or stray radiation
d) Monochromatic radiation
I have some doubts about pH in spectrophotometry. I'm not sure if it is correct but here is what I thought:
a)Differences in solvent. It can be a source of error because disociation, reaction or asociation of the analyte with solvent can generate absorption products different of the analyte.
b)Acid base equilibrium. I don't know why pH is a variable to take into account in spectrophotometry. I thought maybe the case of the determination of phenolphthalein that  depends of pH medium. Spectrophotometry doesn´t work when phenolphtalein is in acidic medium. So pH is a multiplicative interference in this case.
c)Parasitic radiation. It is a source of error because  it is radiation that doesn't come from the analyte and it generates signal.
d)Monochromatic radiation. I think that is the wrong answer because it is what we need to avoid big changes in the absorbance as a function of the wavelength.

The worst source of error is parasitic radiation, because they cannot be understood or corrected. The monochromatic radiation is a good way of making spectrophotometric measurements. Change of acidity and change of solvent can be corrected. But parasitic radiation produces errors that cannot be corrected.

I doubt title and body form the exact test question as personally I won't be able to answer. My major activity is spectrophotometry so I am really surprised by the question, at least out of a chapter context.

It is a very open ended question, but first three choices will cause errors. Regarding your pH issue, you are right, it is multiplicative error. Take the example of phenolphthalein, as you say that its color (=absorption spectrum) is different in acidic or basic medium. So pH change can shift the lambda max of analytes, hence the cause of error.
Phenolphthalein still absorbs light in the acidic medium. I cannot find phenolphthalein spectra, but let us take phenol red indicator, you can see drastic changes in absorbances at the lambda max as a function of pH.
If by some random chance, you were measuring at the isosbestic point, then pH will not affect absorbance but in all other cases pH will change the A values drastically.


text:
Will a titanium piercing tarnish if submerged in sulfur water (hot spring)?
I have tried to research whether titanium piercings will tarnish if one submerges in sulfur hot springs. Will titanium piercings react to the sulfur water?

No.$\ce{~~~~~}$

I understand OP is new to the site so I would like to point to the elephant in the room. Your question received downvotes because of one of the reasons mentioned on this page: https://chemistry.stackexchange.com/help/how-to-ask

Neither Hydrogen sulfide nor sulfur dioxide will corrode titanium at ambient temperatures.

We are not actually addressing the properties of pure titanium, but an alloy thereof.

Grades 1 to 4 are considered to be unalloyed.. they have increasing levels of oxygen , nitrogen and similar interstitials that have little affect on corrosion resistance; They are increasingly stronger with less ductility. Jewelry would logically be made of them. Grade 2 is most common , difficult to imagine a jewelry producer going to the expense of other grades . And of course Grade 5 would need to be heat-treated - way beyond a jewelers skill.

Wikipedia is good but they do have some poor listings. The one for Inconel is very misleading.

Actually, per my edited answer, a source marketing such rings, clearly recommends rinsing off the ring after normal seawater exposure, which suggests an electrochemical promoted corrosion issue here, especially in a highly salty mineral bath (also with a sulfur presence).

The question actually relates to Titanium piercings. Per Wikipedia on the topic of Body Piercing Materials, to quote:

Titanium body jewelry is often manufactured in either commercially pure grades 1 to 4, grade 5 TI6AL4V alloy or grade 23 Ti6AL4V ELI alloy.

where the keyword is "alloy".
So, rephrasing the question, could a Titanium-based alloy be subject to electrochemical corrosion in the presence of hydrogen sulfide?
My answer is a possible yes, and I suspect the person asking the question has already confirmed this via direct observation.
This article: Corrosion characterization of titanium alloys by electrochemical techniques appears to confirm potential corrosion issues in even a lab constructed physiological solution.
I suspect, aqueous $\ce{H2S}$, would be even more problematic, and likely could result in visible electrochemical based corrosion.
[EDIT] Did find an interesting suggestion on how to care for your Titanium jewelry, particularly rings here, to quote:

You should also rinse the salt off of your ring after swimming in the ocean.

So, on the question as to "whether titanium piercings will tarnish if one submerges in sulfur hot spring", a source marketing such rings, recommends rinsing off your ring after exposure to a salt-rich solution. Note, salt is well known to generally promote electrochemical based corrosion and, not even further considering a sulfur presence, there appears to be a potential issue here (else it would not be recommended!).


text:
Quantum Internal Efficiency
I'm reviewing a homework problem where we're given a hypothetical organic semiconductor which has a fluorescence efficiency of $40$% and a phosphorescence efficiency of $60$%.
From theory, I know that, the internal quantum efficiency is given by: $$\large\eta\small_{EL}=\large\eta\small_{PL}\cdot\large r_{st}\cdot\gamma_{cap}$$
where $\large\eta\small_{PL}$ is the photoluminescence efficiency, $r_{st}$ is the fraction of singlets over the total number of excitons formed and $\gamma_{cap}$ is the exciton formation number.
We assume that that simple spin-statistics arguments are valid to calculate the fraction of singlets or triplets generated by electron-hole mutual capture, and that the hole-electron capture efficiency is one.
I need to calculate the quantum internal efficiency of this diode with the additional assumption that intersystem crossing is negligible.  I am not sure where to start.

If intersystem crossing is not possible it means that once singlets and triplets are formed  they do not interconvert to form the other species.


text:
Le Chateliers principle and increase in temperature
If you increase the temperature, the endothermic reaction is favored. The Le Chatelier's principle states that the endothermic reaction is favored in order to minimize the effect of an increase in temperature. Endothermic reactions absorb heat so the temperature must, in a way, remain constant in that system. So that way, the equilibrium constant is never changed due to a change in temperature as the temperature doesn't change. I know I went wrong somewhere but can someone explain where. Please don't include equations like the ones with Gibbs free energy and Van't hoff equations and stuff as we didn't learn all of that in school yet. Please make the explanation as intuitive as possible.

Are you confusing the reaction being endothermic or exothermic with the system being isothermal or adiabatic?

You may perhaps have an idea endothermic reaction absorbs near all provided thermal energy. It is not like that, it absorbs just a small part.  ...in order to minimize the effect...  is misleading. "to decrease the effect" is more correct.

Heat is not a species in the reaction. It does not enter into the equilibrium expression. Therefore, you cannot apply Le Châtelier's Principle.

The right way to do this is to understand the Van't Hoff equation.
Barring that, I can provide some intuition, but it'll have to be a bit hand wavy.
The way I think about this is that at higher temperatures, the intrinsic preference for reactants or products becomes less. That's because the thermal energy of the system is becoming more relevant than the energy difference between reactants and products.
For an exothermic reaction, when we increase the temperature, we'll reduce some of the preference for the product, thereby pushing the reaction backwards.
For an endothermic reaction, when we increase the temperature, we'll reduce some of the preference for reactant, thereby pushing the reaction forwards.
Of course, it's pretty hard to truly justify these kinds of quantitative relationships without the use of proper mathematical tools. In this case, the relationships are most clearly captured in the Van't Hoff equation.


text:
Are all protein tetramers considered to be "dimers of dimers"?
Is every tetramer thought to be a dimer of dimers? Because even if every subunit is unique in structure, it could be a heterodimer of heterodimers? Or is the term "dimer of dimers" reserved for something more specific like symmetric tetramers?

No, only if it's justified.

A dimer is an aggregate of two identical subunits, not necessarily connected in a symmetric fashion! "Heterodimer" is a somewhat unusual term. And a "dimer of dimers" must actually fit that description. $\ce{A + A3 <=> A4}$ does NOT, obviously. ?!

The jargon in my grad-lab was that a dimer of dimers has D2 symmetry while a tetramer could also refer to C4 symmetry. All of these would be homotetramers. Hemoglobin is a good example of a heterotetramer. The 4 protomers all have the same fold but slightly different sequence. See e.g. https://proteopedia.org/wiki/index.php/Tutorial:How_do_we_get_the_oxygen_we_breathe and click on the "light-blue chains" link.

https://www.biorxiv.org/content/10.1101/431288v1.full

Logic
A dimer of dimers implies that there are four units (2 x 2 = 4) somehow organized into two pairs. To be meaningful and not arbitrary, the units within pairs should somehow be related in a different way to each other than units in between pairs. Units within pairs should probably also connected in some way.

Is every tetramer thought to be a dimer of dimers?

No, not by that logic. The four branches of a multiplication sign ($\times$) don't look like dimers of dimers because of the four-fold symmetry. On the other hand, it would make perfect sense to talk of a dimer of dimers for the protein quaternary structure depicted below because the protomers of like color have a different (and larger) interface than the one of unlike color.

Source: https://www.biorxiv.org/content/10.1101/431288v1.full
If a structure containing 4 units always falls apart into the same 2 pairs of units, you could also speak of a dimer of dimers. So if a complex ABCD always falls apart this way
$$\ce{ABCD -> AB + CD}$$
but not e.g. that way
$$\ce{ABCD ↛ AC + BD},$$
speaking of a dimer of dimer would make sense.
Jargon
The jargon in my grad-lab was that a dimer of dimers has D$_2$ symmetry while a tetramer could also refer to C$_4$ symmetry (or some other arrangement lacking 2-fold axes). Strictly speaking, all of these would be homotetramers but you could use this for protomers that are unequal but highly similar. Hemoglobin is a good example of a heterotetramer. The 4 protomers all have the same fold but there are two sequences, represented twice. You could summarize this by saying hemoglobin has an $\alpha_2 \beta_2$ quaternary structure. See e.g. this Proteopedia page and click on the "light-blue chains" link. The symmetry could be described as dimer of dimers, even though logic might tell you not to. Certainly, there is no evidence for hemoglobin falling apart into separate $\alpha_2 $ and $\beta_2$ parts.

Because even if every subunit is unique in structure, it could be a heterodimer of heterodimers?

Yes, if there is a reason to group the four units into two pairs in a specific way.

Or is the term "dimer of dimers" reserved for something more specific like symmetric tetramers?

There is no central agency that regulates how scientists talk and write about their objects of interest. So you will find inconsistent language until someone writes a textbook or review paper from which most people adopt the usage of a technical term.