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What is the state of matter of a substance if the melting point or boiling point of a substance is exactly at room temperature(25 degree Celsius)?

A substance can exist both in solid and liquid state at melting point, and A substance can exist both in liquid or gaseous state at boiling point. In the above image the state of matter is differentiated in solid, liquid or gas on basis of what the state of matter is the substance at room temperature (25 degree Celsius)(i.e. if the substance is at liquid state at room temperature (25 degree Celsius) than the substance is liquid). What if the melting point or boiling point of the substances is exactly at room temperature (25 degree Celsius) .it can exist both in solid or liquid state at 25 degree Celsius if it's boiling point is equal to 25 degree Celsius, and it can exist both in liquid or gaseous at 25 degree Celsius if it's boiling point is exactly equal to 25 degree Celsius. What will be the state of matter of that substance?

Well, it can be either solid or liquid or a mixture, depending on the heat content added.

The answer is: Temperature is not sufficient info to determine the state of matter at these cases.

You should look at  a phase diagram, (which shows pressure vs temperature) such as for water or benzene etc. If you fix just the temperature then it is possible for example, to be either in the gas or liquid phase. Only if  pressure and temperature are both fixed to be on a line in the phase diagram will, for example, a solid or liquid be in equilibrium.

It could be a gas. The physical state depends not only on the temperature but also on the pressure. And you can observe multiple physical states in the same system, even if at the same temperature and pressure. In the end, the cited statement is a bit too brief to be accurate. Room temperature, I think, is 20 degC now, not 25 degC.

This person should buy a bottle of tertiary butyl alcohol and observe it over the four seasons [global warming permitting] and occasionally open the bottle and take a sniff. If a liquid and solid are in equilibrium and there is a head space, this is a triple point. The triple point is a function of an inert gas pressure because there is an additional component in the Phase Rule.

There are two common ways of differentiating the state of matter of a sample, by directly observed properties, and by the nature of intermolecular (or interionic) interactions.
1. Properties
You can differentiate by whether the sample takes the shape of the container (assuming there is some gravity) and whether it has a definite volume or fills out the available space.
2. Interactions
Solids and liquids have interactions between particles, gases do not. In solids, the interactions are long-lived with the same partner, in liquids interaction partners change. In solids, sometimes it is difficult or impossible to decide what the particles are (e.g. diamond is one single molecule).
Assigning a physical state to a substance
You would test the physical state of the pure substance at atmospheric pressure and room temperature. This does not mean this substance always has that physical state, it depends on the circumstances (pure or not, interactions with other substances, etc.).
The elements are labeled as a single physical state in some periodic tables, using the rationale above. If you can't get a sample of pure substance (e.g. highly unstable, radioactive elements), then you can't assign a physical state. I don't think there is any element where the normal melting point or boiling point is so close to room temperature to make it ambiguous.

[OP] What is the state of matter of a substance if the melting point or boiling point of a substance is exactly at room temperature(25 degree Celsius)?

That depends on the specifics. Sometimes, you will have the substance in two or three physical states at the same time.
Here is a textbook problem and the model answer to illustrate the complexity:

Isopropyl alcohol is a colorless, flammable chemical compound with a strong odor. Its melting point is -89°C and it boiling point is 82.5°C. Is isopropyl alcohol a solid, liquid or gas at room temperature (25°C).
Solution: Since room temperature (25°C) is above the melting point of isopropyl alcohol (-89°C), but lower than its boiling point (82.5°C), it is a liquid at room temperature.

The answer is a bit incomplete. If the alcohol is stored with some air above it, there will also be some isopropanol in the gas phase. This is how we smell liquids - some molecules enter our noses in the gas phase. If store in an open bottle outside or in a building with good ventilation, all of the alcohol will eventually evaporate (in a building with bad ventilation, the open bottle will pose a health and fire hazard).

Heat is never distributed equally across an object, so if you have an object at 25 °C then that temperature is only an average measure of the energies of the particles which make it up. For example this means that in any liquid you have some particles with enough energy to become a gas, which is why water evaporates in the sun on 30 °C days. To answer your question then it doesn't matter if the object is sitting exactly at its melting/boiling point as some of the particles would have already changed state and others would not.
If you're asking more about characteristics then these are some examples I can think of:

Evaporated water just leaves and the remaining water acts like liquid
Water droplets which condense in clouds eventually stick to each other then fall down as rain
Most of the melting of an ice block is on the outside and flows off, however what is inside is trapped


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Should durations of time be considered to determine final significant figures of answer?
In preparation for AP Chemistry this upcoming semester I've been assigned a number of challenge problems. One of them is as following:

Suppose that polluted air has carbon monoxide $(\ce{CO})$ levels of $\pu{15.0 ppm}$. An average human inhales about $\pu{0.50 L}$ of air per breath and takes 20 breaths per minute. If the density of carbon monoxide is $\pu{1.2 g L=1}$, How many milligrams (mg) of carbon monoxide does an average person inhale during an $\pu{8 hour}$ period of time breathing polluted air? (Hint: $\pu{15.0 ppm} \ \ce{CO} = \pu{15.0L} \ \ce{CO}$ per $\pu{1.0E6 L}$ of air.)

I've set up the following stoichiometry to solve this problem:
$$\frac{\pu{8 h}}{1} \times \frac{\pu{60 min}}{\pu{1 h}} \times \frac{\pu{20 breaths}}{\pu{1 min}} \times \frac{\pu{0.50 L} \text{ air}}{\pu{1 breath}} \times \frac{\pu{15.0 L} \ \ce{CO}}{\pu{1.0E6 L} \text{ air}} \times \frac{\pu{1.2 g} \ \ce{CO}}{\pu{1 L} \ \ce{CO}} \times \frac{\pu{1000 mg} \ \ce{CO}}{\pu{1 g} \ \ce{CO}}\\ = \pu{86 mg} \ \ce{CO}$$
My question is regarding the number of significant figures for the solution, which I calculated as $86 g CO. Initially I assumed two sig-figs were to be used because of the value 1.2 g/L, but have since noticed that the given duration (8 hours) only has one sig-fig. So my questions breaks down to: are durations considered in the calculation of sig-figs? If so, would the correct answer be 90 mg CO as opposed to 86 mg CO?
TLDR;
When calculating sig-figs, do you consider the number of significant-figures for all given values, including durations, or just physical quantities?

The question isn't what units it's measured in (i.e. whether it's a time or a length or whatever), it's whether it's a quantity that's measured. The point is that something you measure will have an uncertainty, which is (crudely) reflected in terms of significant figures. I do think this question is really quite contrived, though, because arguably 20 is a measured quantity (I don't think humans take exactly 20 breaths per minute, not even on average, or else we'd have no need for clocks).

The $8$ hours is an integer so its value is exact, as a real number it would be  $8.000000\cdots$ so you use other numbers to determine the sig figs.

The question seems to be the ability to manipulate units and possibly to figure out what PPM really means not significant figures tho they matter. All the values are approximated o.5 L is low but might approximate exchange volumes, breaths are high error about -3 to +1 so lets say 10%. The error in time is interesting 7 of the hours are exact the eighth or start and stop are what affects accuracy, that is not important; the only thing that affects the accuracy is the volume exchanged and the measurement of breaths. This question is really an exercise in the folly of extrapolation. explain that

[OP] When calculating sig-figs, do you consider the number of significant-figures for all given values, including durations, or just physical quantities?

You consider all physical quantities including time or duration. Time has the unit of one of the base SI units, it is a quantity just as mass, volume and amount of substance.

[OP] Initially I assumed two sig-figs were to be used because of the value 1.2g/L, but have since noticed that the given duration (8 hours) only has one sig-fig.

Two significant figures for the result is reasonable.
There is ambiguity with numbers that don't have a decimal point (i.e. integers). Take the 20 breaths per minute. You would count to twenty, it can't be 20.1 or 19.9 (unless you take a population average or a personal average). However, without decimal point you could make the argument that 20 only has one significant figure. The 1 g in the conversion factor is intended as exact. I would think the 8 in "8 hours" is also intended as exact rather than 7 to 9 hours.
To resolve the ambiguity, you could write $\pu{2.0e1}$ instead of 20 breaths if the intent is to give two significant figures, or $\pu{0.8e1 h}$ if the intent is to show one significant figure for the duration.
In the absence of these explicit notations, you have to try to use common sense and hope that the person who writes the answer key also uses common sense.

Per a source: "Normal respiration rates for an adult person at rest range from 12 to 16 breaths per minute" and children likely more, and given the earth population distribution and where they live, sea level or higher, the value stated for breaths per minute is not likely precise. Link: https://www.hopkinsmedicine.org/health/conditions-and-diseases/vital-signs-body-temperature-pulse-rate-respiration-rate-blood-pressure#:~:text=Normal%20respiration%20rates%20for%20an,to%2016%20breaths%20per%20minute.

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What exactly happen when 532nm laser cause fluorescence in red objects

I hit the above plastic sliding toys with 532nm Laser in a Raman spectrometer and the following are the spectra of each. This is the blue sliding toy showing the right Raman for the plastic.

But when I hit the red sliding toy, there is no Raman signature detected, instead there is a broad Fluorescence.

What happen exactly at the molecular level when the red object get fluorescence from the 532nm laser but not the blue object?
And if I paint the red object with blue paint, would the fluorescence be suppressed producing the spectrum for the blue object?

The  laser excites both a Raman signal and fluorescence in both materials. In the blue dyed material whatever dye is used has a low fluorescence yield so that the Raman signal is seen on top of the fluorescence signal. In the lower figure the signal is about ten times bigger, and the fluorescence swamps the Raman signal. In this case the fluorescence yield of the red dye must be  large.
If you add paint and the laser passes through it to the material underneath then the sum of any Raman and sum of any fluorescence will be seen. Clearly, if the paint is so thick as to absorb all the laser only signal from this will be seen.

Can we say that in the blue dye, the gap between S0 and S1 levels where the electrons can get excited have more energy than the red dye that's why the green laser photons can't be excited from S0 to S1 electronic levels in the blue dye?

Not exactly. The laser excites both the dye in the blue and in the red material. The blue emission seems to have a max at about $550$ nm (its hard to tell as there are no values on your plot) and the dye in the red material at about $590$ nm.  Thus the blue dye has a larger S1-S0 energy gap than the red dye but both are less than the energy of the laser's photon. (i'm guessing that $532$ nm is off the left of either plot.)

If both the blue and red S1-S0 energy gap is less than the energy of the laser's 532nm photon. How come the green laser didn't cause muchg fluorescence in the blue dye when the S1 was reach and transitional down non-radiatively and then emitting fluorescence photon as it goes from S1 to S0 (like in the red dye)?

Because different types of molecules have different fluorescence quantum yields. This can depend on several factors such as energy gap to triplet, and /or presence of heavy atoms.

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Understanding Hydrocarbon Combustion
First ... I know almost nothing about chemistry.
So, for the past week or so, I've kind of been fascinated with this whole combustion thing (not in an evil way, of course). So, I've been reading, and youtubing etc. etc. What I have so far is below, some of which is also seen in this thread:
Mechanism of combustion of hydrocarbons
So I want to make sure the picture in my mind is right, so that I could reasonably explain it to ... say ... a 7th-grader. Let's take the (simple?) example of one molecule of methane CH4 and two oxygen O2 molecules.
So we'll put some energy "into" the methane and oxygen molecules, never mind where it came from. Now ... when we do that, the atoms vibrate (think of the bonds between the atoms as little springs). More energy ... stronger vibrations. If we put enough energy into them, the "springs" will "break" and the molecules come apart. Now we have 4 Hydrogens, 1 Carbon, and 4 Oxygens floating around. Think of this breaking-up happening to a small percentage of molecules among trillions and zillions of CH4 and O2 molecules.
Eventually the free atoms find each other and "fall" into deeper, stronger bonds of two H2O and a CO2. That's important: deeper, stronger bonds. Still the same number of atoms! Just rearranged. There are actually charts that show the energy levels, and for quantum reasons beyond my understanding and that aren't important at this level, there is a "well" of distances between atoms where the bonds are most stable. So remember when they fell into the new bonds? Well before they fell they had potential energy, that, after falling into each other, is now kinetic energy of those molecules vibrating. Eventually the new bonded molecules run into other molecules and transfer their energy to them, and the molecules that we just created settle down into the even deeper and stronger (more stable) energy states. The scales have to balance. Energy is never created or destroyed ... only changed.
Long story short: The energy given off by the atoms falling into each other to form the new H20 and CO2 molecules is greater than that needed to break apart CH4 and O2. That's why the reaction is exothermic, i.e. giving off heat. Eventually, this happens to enough molecules that they start bouncing around and vibrating faster and faster, that enough energy makes it to another set of CH4 and O2 molecules, and it happens again, and again, eventually consuming all the fuel.
Whew ... so did I get it right?

Now we have 4 Hydrogens, 1 Carbon, and 4 Oxygens floating around. - This does not happen, but at temperatures much higher than usual for fuel flames. Molecules and their fragments (radicals) exchange atoms, which replace weaker bonds for stronger ones, what is illustrated in the link you provided yourself.

Duplicate of  Mechanism of combustion of hydrocarbons

The whole story is pretty good, but there's one major thing I'd correct.
Molecules can undergo decomposition at higher temperatures (that was the springs stuff you were talking about) but that's not really what happens here. Think of it rather that heat is the energy of particles moving about, the hotter something is the faster the particles are moving. When CH₄ and O₂ are in a mixture they bump into each other a lot, however they're too stuck together for anything to happen unless they're moving a lot faster (like if you slowly place a lego creation on the floor it wont come apart, but if you hurl it against the wall it will). Now if there's enough heat (e.g. you put a match in a methane jet) and two O₂ molecules and a CH₄ molecule all collide then the atoms will all come apart, and if they're in the right orientation then instead of just reforming into their initial bonds they'll form into one CO₂ molecule and two H₂O molecule. Combustion has occurred.

"two O₂ molecules and a CH₄ molecule all collide then the atoms will all come apart," That's also not really how it works. COmbustion is a multi-step reaction and does not involve two O2 and one CH4 all coming together. Many different radical species are involved which react with each other in different ways, producing quite a few intermediates on the way to the final products


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Unequal bond lengths of resonating bonds in dinitrogen trioxide
Here is an image showing the resonance structures of $\ce{N2O3}$ molecule,
As the bonds numbered (2) and (3) are in resonance, they will have same bond lengths but each with bond order less than two. As the bond numbered (1) doesn't involve in resonance, it will have a bond order of two. Hence, the order of $\ce{N-O}$ bond lengths is (1)>(2)=(3).
But according to Wikipedia, all the three $\ce{N-O}$ bond lengths are of different lengths! Here is an image form Wikipedia showing the bond lengths and the bond angles in $\ce{N2O3}$,
So my question is why are the resonating bonds are of different lengths?

Bonds don't "resonate", and in dissymmetric molecule won't be equal

Consider the relative energies of the two resonance structures

Could you explain what you mean by and why you think "As the bonds numbered (2) and (3) are in resonance, they will have same bond lengths"?

@Mithoron if the bonds aren't in resonance, then there could be a significant difference in the bond lengths of bonds (2) and (3).

@Andrew I understand what you are saying. So according to bond lengths data, second resonating structure is more contributing than the first one. Can you explain, why?

The experiment describes what actually happens, not a theoretical resonance structure :)  So we must assume that there is limited rotation about the N-N bond otherwise the two bonds in the NO2 part would have the same length and same angles  by symmetry.  (Perhaps some rotation is happening as lengths are v similar and perhaps if the experiment was done at a lower temperature there might be a bigger difference in bond lengths and angles? )

The two oxygen atoms on the $\ce{NO2}$ nitrogen are not equivalent. The molecule is planar*, and one $\ce{NO2}$ oxygen (call it the "cis" oxygen) is closer to the third oxygen atom than the other $\ce{NO2}$ oxygen ("trans").
We then note that the oxygen atoms have nonbonding pairs around them, which tend to repel each other. So the cis oxygen will tend to form a stronger pi bond, pushing more nonbonding electron density onto the relatively remote trans oxygen.
*This planarity is due to hyperconjugation. See Halpern et al[1].
Reference

Halpern AM, Glendening ED (2007 Apr 21). "Ab initio study of the torsional potential energy surfaces of N2O3 and N2O4: origin of the torsional barriers." J Chem Phys.; 126(15):154305. https://doi.org/10.1063/1.2713756. PMID: 17461624.

What is the reason the NO group orientation is locked? Does  the N-N bond have partially pi-character too? As it does not seem that there is a sterical obstacle for the bond to rotate.

Halpern et al: "N(2)O(4) and N(2)O(3) exhibit strong hyperconjugative interactions of in-plane O lone pairs with the central N-N sigma* antibond. Hyperconjugative stabilization is somewhat stronger at the planar geometries because 1,4 interactions of lone pairs on cis O atoms promote delocalization of electrons into the N-N antibond. "


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Why are there so many metals in the periodic table?
Brief Background: I was studying about the 'classical electronic configuration' of the first 20 elements of the periodic table today and was bewildered by the fact that there are an equal number of metals and non-metals.

Question(s)
How on earth is it possible for 92 of the 118 discovered elements to be metals!? I've read a little about it here, but if I agree with this, how is it that Calcium essentially 'forgets' about its penultimate shell but Scandium and many others don't? And still, how is it possible that some electrons 'jump' from the penultimate shell to the valence shell to result in variable valency in such a case?
Holy grail
I hope that a conceptual answer awaits for me, but I may be fine with others as well. Specifically, what is the exact physical cause behind this phenomenon?

I think it's possible to argue that metallicity (or more generally, delocalized bonding) is more natural, and it's the non-metals which are unusual - why do we have a bunch of those?

Some non-metals have metallic states, see e. g. https://carnegiescience.edu/news/intense-conditions-turn-nitrogen-metallic

Simultaneous posting across various sites is generally discouraged, but if you do not obtain answers (or satisfactory ones) on one site it can be justified to consider another site.

@NicolauSakerNeto Can you please explain 'delocalized bonding'?

@Karsten Interesting! Never knew of this!

@ChinmayKrishna This might be a starting point to learn about metallic bonds, delocalized electrons and band structure: https://chemistry.stackexchange.com/q/25003

@Karsten when I read your post about nonmetals turning metallic I thought of silicon. All you have to do is melt it. The transition to a metal makes silicon contract more than water upon melting!

Very simply, metallicity is defined (other than in astronomy, where everything beyond He is a metal) by behavior: a shared sea (conduction band) of electrons, causing conductivity, reflectivity, etc.
Elements such as the alkali metals are "glad" to be rid of their lone electron to have a complete outer shell (well, to share with all atoms in the conduction band), and similar effects are observed for dropping an electron or two to complete an orbital.
However, other important effects are shielding, average distance from the positive nucleus, and relativity effects, all of which make outer electrons easier to drop off. For example, look down the column of chalcogens:

Oxygen is very quick to add an electron to complete its outer shell, definitely nonmetallic.
Sulfur is a bit less likely to grab an electron, but is still nonmetallic.
Selenium is a metalloid, shiny in some allotropes, and conducting electricity with the help of a photon or two. Clearly, the outer electrons are not firmly attached.
Tellurium is also a semimetal, more conductive than Se.
Polonium is a metal, conductive (0.40 µΩ⋅m at 0 °C)  and shiny (and incredibly dangerous).

In all these elements, the outer shells are somewhat similar, but the increasing size of that outer shell make them more metallic.

Well depending on who you ask,polonium is a semimetal.

@BruhMoments Maybe. But there are few chemists indeed who will volunteer to do the work with polonium to double check that idea.

@matt_black, Good point! See Derek Lowe on Po and other things he won't work with: https://www.chemistryworld.com/opinion/a-risk-not-worth-taking/4010881.article

No atom is "glad to " lose an electron or anything else for that matter. Electron loss is a brute force procedure call it atomic rape if one must anthromorphize things. The structure of the Periodic Table with a bit of QM explains it all nicely.

I will use this excellent comment as a starting point:

[Nicolau Saker Neto] I think it's possible to argue that metallicity (or more generally, delocalized bonding) is more natural, and it's the non-metals which are unusual - why do we have a bunch of those?

Definition and structure of metals
Mostly, we think of metals as solids, although the liquid mercury is commonly defined as metal as well. While the OP's question is about elements, alloys are classified as metals as well, even when they contain some non-metals like the carbon in steel.
So to be considered a metal, an element has to be in a condensed physical state. In a solid or liquid, the atoms will be close together; in an element, they will all be the same. Either the atoms will interact with each other in the same way (like in a metal), or they will form groups of strongly interacting atoms (called molecules) interacting less strongly with other groups (like liquid bromine or solid iodine). The former seems more "natural", given that the atoms are all of the same type. The absence of molecules and the abundance of nearest neighbors gives rise to the band structure of metals (and the delocalisation of electrons) that explains many of their properties.
Temperature and pressure

[OP] How on earth is it possible for 92 of the 118 discovered elements to be metals!?

On earth, indeed.
If an element is in the gas phase, we typically don't describe it as a metal. This includes hydrogen, oxygen, nitrogen, the halogens up to chlorine and the noble gases up to radon. Some of these elements (including hydrogen, oxygen, nitrogen, chlorine, and xenon) solidify and form metallic phases when under sufficient pressure. A paper on high-pressure forms of chlorine claims [[1]2]

Under sufficient compression, all molecular systems are expected to collapse into close-packed metals.

On the other hand, if we increase the temperature, most metals will vaporize (or even form a plasma). This shows that the number of elements we classify as metals depends on the ambient conditions we experience on the surface of our planet these days.
Allotropes
There are also elements that have allotropes, with one metallic and one non-metallic. A good example is tin, which has a metallic allotrope (white tin) and a non-metallic allotrope (gray tin), which is a powder and does not conduct electricity. Carbon comes in the form of diamond (non-conducting) and graphit (conducting). Boron has metal-like allotropes as well. Finally, metalloids are semiconductors, so their metallic character depends on temperature.
Chemical properties
This answer focused on physical properties. The key chemical property of metals is that they tend to form (monoatomic) cations. Taking away electrons from an isolated atom always costs energy (all ionization energies are positive). Interacting with a polar solvent (like water, made of nonmetals) or with anions (made of or at least containing nonmetals) offsets this cost in some cases, so we observe the formation of cations. This means that without non-metals, we would not be able to observed metals forming cations. To explain (or rationalize) which elements form cations or anions, we invoke the electronic structure of atoms, as you will find in any textbook. The introductory textbooks focus on main group metals and non-metals, yet the bulk of metals are transition metals and f-group metals, where things quickly become hand-waving (OP's "jumping" electrons).
Cause and effect

[OP] Specifically, what is the exact physical cause behind this phenomenon?

Sorry, other than pointing to experimental evidence showing that most elements have metallic properties under ambient conditions, and adding that this is well-described by theory (but the theory is complicated and the applications are subtle), I got nothing.
Reference

Dalladay-Simpson, P.; Binns, J.; Peña-Alvarez, M.; Donnelly, M.-E.; Greenberg, E.; Prakapenka, V.; Chen, X.-J.; Gregoryanz, E.; Howie, R. T. Band Gap Closure, Incommensurability and Molecular Dissociation of Dense Chlorine. Nat Commun 2019, 10 (1), 1134. DOI:10.1038/s41467-019-09108-x.

Can you please explain the theory behind it as well (in brief), mentioned in the subheading 'Cause and effect'?

@ChinmayKrishna Sorry, I can't. If anyone else can, chemistry instructors would all lose their jobs. The results are easy to show (e.g. on periodic table), the theory takes a couple of years.

The simplest view is if the bonding is due to overlap of the highest energy shells on the atoms then almost everything should be a metal as the resulting bands will be partially filled except when the atomic shell is full, such as the noble gases or alkaline earths. Thus the default position is nearly everything should be a metal and you have to explain why it isn't, not the other way around.

And as for why Mg and Zn and similar are metals, well the problem is more than one band is involved, see https://chemistry.stackexchange.com/questions/164694/what-theory-accurately-explains-metallic-bonding-in-beryllium/164703#164703

@Karsten regarding your tin can be a metal or non-metal, depending on the allotrope of tin. Are there elements that can be different states of matter, depending on the allotrope?

@NealConroy yes, different allotropes will have different melting and boiling points. For instance white phosphorous melts at around 317K, red at about 860K

The elements are defined by their "molecular structure", do they form covalent bonds. Covalent bonds are formed by elements with two types of valence orbitals, first: orbitals that are close to the nucleus [1s,2s] second: degenerate orbitals of the same energy level [2p, 3p, d and f to a lesser extent] that are not shielded from the nucleus as electrons are added. They form covalent molecular bonds that are strong and directional maximizing attraction of electrons to the nuclei and minimizing electron repulsion. This explains the nonmetallic behavior of hydrogen and helium, the less nonmetallic behavior of lithium and beryllium and the nonmetallic behavior of boron thru neon. As we progress down the periodic table the valence electrons, while possibly having similar shielding, are farther from the nucleus so molecular bonds are farther from the nucleus and are weaker. Rather than saying metallic character increases nonmetallic behavior decreases so eventually the only nonmetals left are iodine, possibly tellurium, and the inert gases; and they are all suspect.
Nonmetals usually have large ionization potentials and electron affinities; loss of electrons is difficult and gaining one is usually exothermic. This affects chemical properties and nonmetals tend to be involved with negative ions, anions. Finally! metals, since their valence electrons are more shielded, have lower ionization potentials and less exothermic electron affinities. The individual electronic structure is what is important. This means that metals in their lower oxidation states tend to form positive ions, cations, and in their higher oxidation states start to assume nonmetal properties forming complex anions; the borderline is the +3 ions. A question here might be why do ionic compounds form? why not heteroatom metals? A quick answer is electronegativity differences. What gets those electrons closer to the nucleus is what is important.
With the exception of hydrogen and lithium covalent bonding is weak for elements that have lower nonmetal properties; lets call them metals! They still bond together sometimes with immensely strong bonds. Their bonds involve mobile electrons evidenced by electrical conductivity and no or little nuclear motion. The bonding possibly involves attractions of nuclei for the electrons of surrounding nuclei added to the attractions of the electrons in conduction bands for the array of nuclei. There must be an explanation for the immense range in physical properties from mercury to osmium This is beyond the scope of my understanding.
There are more metals than nonmetals because the electronic configurations that result in nonmetal properties are very limited to the elements in the first few periods and far right groups in the periodic table.

In the array of nuclei of a metallic structure the better shielded the valence, especially the s, electrons are the stronger the attraction of nucleus A to the electrons of nuclei B,C etc. becomes than the repulsion of the nuclei. Someone must have worked out the QM implications for this.  There must be a reason for the strong attraction between positive ions in a close packed array. The more shielded the valence electrons are the more metallic properties.


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Tetra-tert-butylmethane

Does this molecule exist? I've been wondering about extreme steric hindrance and trying to find the limit of what can exist and what not. It appears to exist some basic data about it in PubChem, but does this really matter? Do all the molecules that are recorde in this site really exist? I haven't found anything more, and I would also like to know if there's any synthetic path to achieve it. Maybe it's impossible to know how to produce this molecule, but I'm sure that is possible to calculate the thermodinamic parameters or the relative stability.
Wikipedia also shows that it's an hypothetical compound and there are energetic studies in references:
...for tetra-TBM, a compound that has not yet been synthesized, the G3(MP2) results reported in this work should be reliable estimates...
So, does this mean that is possible to make it but it hasn't been  made yet?
https://pubs.acs.org/doi/pdf/10.1021/ci0497657
https://pubchem.ncbi.nlm.nih.gov/compound/14123361
https://pubs.acs.org/doi/full/10.1021/jp034879r

Even if it could exist, it would be hard to produce. The closes thing I'm aware of is tetra-tert-butyl-phosphonium, but the method used to produce it does NOT translate to nitrogen or carbon central atom. ref: https://doi.org/10.1002/cber.19801130440

related https://chemistry.stackexchange.com/questions/108219/largest-alkane-having-a-given-alkane-as-its-base-name

It is quite likely not to exist. Theoretical studies of possible alkane isomers show that this isomer and its neighbour with one fewer methyl group  are too crowded to exist never mind be synthesised: https://chemistry.stackexchange.com/a/27980/81

Something close to it, pentaerythritol, is used in paint resin formulas.

Henry Rzepa's blog has a post on this topic. According to him, the tetraisopropyl methane (formally bis(3,3-isopropyl)-2,4-dimethylpentane) has been made, but the tetra t-butyl "might never be achieved" and he provides an argument why: https://www.ch.imperial.ac.uk/rzepa/blog/?p=24159

http://dx.doi.org/10.1039/C5CP07005H terminal methyl groups are linked pairwise to benzene rings, but the core structure is the one from your question.


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Why is there a decrease in the total volume when NaOH dissolves in water?
When a solute is added to a solvent, the volume of the solution should be equal to the sum of the individual volume of the solute and solvent (i.e total volume), right? But why is there a decrease in the total volume when sodium hydroxide (NaOH) dissolves in water (H2O), given sodium hydroxide (NaOH) does not react chemically with water (H2O)?

Related: https://chemistry.stackexchange.com/questions/33058/why-there-is-no-change-in-water-level-when-salt-is-added/33063#33063

Does this answer your question? How does volume contraction in solvent mixing work?

When solving solid water into liquid water, the total volume also changes.

If you mix a bucket full of water and a bucket full of marbles, will you get two full buckets?

@eric-duminil I suppose if they don't react with each other, the volume would add up

A bucket "full of marbles" isn't really full, it has many holes. When you mix the two buckets, you might get a volume of about 1.75 buckets. The total mass should stay the same, though. And a bucket full of water is also full of tiny, empty spaces between molecules.

@eric-duminil I thought you meant the marbles are the same volume as the bucket of water

When dissolved into water, $\ce{NaOH}$ dissociates into $\ce{Na+}$ and $\ce{OH-}$. Charged ions interact strongly with water via ion-dipole interactions, which are stronger than the hydrogen bonds characteristic of water intermolecular interactions. These strong interactions pull water molecules into relatively dense localized hydration complexes.
The water in these complexes are denser than pure water, decreasing the volume of the water. Adding $\ce{NaOH}$ to water does increase the volume of the solution compared with the pure water, because the ions occupy some volume, but the total volume of the solution is less than the sum of volumes of pure water and solid $\ce{NaOH}$ due to ion hydration.

There is no law for volume conservation when dissolving a compound in another.
If "hetero-intermolecular" bonds are tighter than the weighted average of "homo-intermolecular" bonds, it leads to the overall volume contraction when they get dissolved. The same occurs for homo/hetero-intermolecular bond strengths and the energy outcome of mixing.
Analogical effects occur for ion-ion and ion-dipole interactions for ionic compounds being dissolved or formed.
Typical cases are dissolution of strong acids and alkali hydroxides. Another case is dissolution of ethanol and water.

(+1) I recall the demonstration of this effect using ethanol and water from my high-school chemistry class.

You can see it from the lens of intermolecular interactions. As answered earlier NaOH when dissolved in water creates $Na^{+}$ and $OH^{-}$ ions. Earlier it was just water which where hydrogen bonding (3rd strongest) was the force of attraction. After introduction of ions into water the force of attraction is ion-dipole (2nd most strongest force) which is much stronger as compared to the previous one.
Now higher intermolecular attraction means molecules tend to be near and more molecules exist in the same space (compared to earlier). Which mean higher density and lower volume.


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What does "shape of hybrid orbitals" mean?
My teacher and textbook distinguish between the $sp$,$sp^2$ and $sp^3$ hybrid orbitals' shapes by saying that the first has linear shape, the second has trigonal shape, and the third has tetrahedral shape. Are they saying this because in the case of $sp$ hybridization, since 2 different orbitals hybridized to form this orbital, the total number of hybridized orbitals will also be 2(stated by MO theory)?  So since 2 bonds are created, the shape of molecule will be linear. Similarly in $sp^2$, there will be 3 hybridized orbitals therefore 3 bonds, thus trigonal shape. Is this what it means to say that the $sp$ has "linear shape"?
(Meaning that the "shape of the hybridized orbital" is just the same as the molecule formed when that hybridization takes place.)
Also, won't this not consistently be true when lone electron pairs are involved as based on VSEPR theory, their repulsions affect the geometry of the molecule? Or is it that the shape of hybrid orbitals is the same despite lone pairs because sometimes lone pairs themselves are inside hybrid orbitals?

Well, the shapes of the orbitals are alike as they come from linear combinations of the same orbitals. However, they coordinate in different ways. More precisely, the coordination determines how the organisations will combine.

Related: https://chemistry.stackexchange.com/q/82133/72973


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Why does Hartree-Fock overestimate the energy gap?
Everybody knows that Hartree-fock overestimates the energy or band gap, but it is difficult to have a clear explanation of the origin of this large deviation from the experimental values.
I know that this may come from the bad description of the unoccupied states or directly from the exchange interaction which is used to correct the gap in DFT with hybrid functionals. So my question is to know if there is a clear mathematical, physical or chemical explanation of origin of these large gap values obtained with the Hartree-Fock approximation.

I suggest trying to fully understand Hartree-Fock, especially its shortcomings. Have a look at correlation energy and try understanding post-HF methods. Its mean field approximation and with it is physically not an accurate choice.

I will try to provide a "intuitive" explanation.
The HF approach seek to find the best set of orbitals that minimizes the energy of the ground state. This procedure leads to a set of optimized occupied orbitals that, in the context of solid state theory, form the valence bands, and a set of unoccupied orbitals forming the conduction band. These unoccupied orbitals are also refer to as virtual orbitals and are "only" a by-product of the orthogonalization procedure. They have no real "physical" meaning.
The energy gap between valence and conduction band in the solid state case is related to the energy difference between the highest occupied molecular orbital (HOMO) and lowest unoccupied (or virtual) molecular orbital in the molecular picture. In addition to electron correlation, the orbital relaxation effects, both not taken into account by the HF approach, contribute to the discrepancy in the calculated energy gap. While it is not straightforward to predict in which direction the electron-correlation effects will shift the energy gap, relaxation effects will more likely lead to a stabilization of the excited state, hence decreasing the energy gap.
One can summarize it as follow: the virtual orbitals that form the valence band are not optimized within a HF calculation, hence their energy is higher, leading to a larger band gap in the solid state case.


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Do polyatomic ions have oxidation numbers?
In a polyatomic ion, the sum of the oxidation numbers of all atoms is equal to the overall charge on the ion. However, does this polyatomic ion have an overall oxidation number?
We also know that the oxidation number of a monatomic ion is the same as the charge of the ion, does the same apply for a polyatomic ion?

Related: https://chemistry.stackexchange.com/questions/9516/charge-of-polyatomic-ions-doesnt-equal-the-charge-of-ions-it-consists-of ...https://chemistry.stackexchange.com/questions/87485/formal-charges-on-polyatomic-ions ... https://chemistry.stackexchange.com/questions/58548/how-can-one-determine-the-charge-of-a-polyatomic-ion

Oxidation number has a meaning for atoms or monoatomic ions. It has no sense for polyatomic ions. A polyatomic ion has a charge. It does not have an oxidation number, although all its atoms do have an oxidation number

"although all its atoms do have an oxidation number" and their sum should be equal to the charge of the ion.

TL;DR: A commentary on Maurice's answer.

We also know that the oxidation number of a monatomic ion is the same as the charge of the ion, does the same apply for a polyatomic ion?

This has one major gap in understanding. An oxidation state is defined for an atom, not a molecule. Look at the definition stated in the gold book[1],

Oxidation Number
In English is largely synonymous with oxidation state, and may be preferred when the value represents a mere parameter or number rather than being related to chemical systematics or a state of the atom in a compound.

Oxidation State
Gives the degree of oxidation of an atom in terms of counting electrons. The higher the oxidation state (OS) of a given atom, the greater is its degree of oxidation. Definition: OS of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds.

Both of these definitions have one thing in common. It talks about the state of an atom not a molecule. Hence, there is no such thing as an overall oxidation number.
References:

The IUPAC Compendium of Chemical Terminology. ** 2019. DOI: 10.1351/goldbook.

If you consider the "overall oxidation number" of a polyatomic ion to be equal to the sum of the oxidation numbers of the atoms that make it up, then the overall oxidation number of a polyatomic ion is equal to the charge on the ion.
For example:
Nitrate has 1- charge.  oxidation # for N plus 3(-2) for the three oxygens must equal 1-.  For that to be true oxidation number of N must be +5.
Sulfate ion has 2- charge.  Oxidation number for S plus 4(-2) for the four oxygens must equal 2-.  For that to be  true the oxidation number for S must be +6

Overall oxidation number is something that does not exist. Oxidation number is the state of an atom within a molecule, and doesn't talk about the whole molecule. See Maurice's answer

This answer probably gave the OP the connection between charge and sum of oxidation states they were seeking. However, making up new terms in not a long-term strategy for communicating within a discipline. I think just giving the example and illustrating the the sum of oxidation states is equal to the charge would have been a better answer (of course, feel free to edit to improve the answer if you agree).


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Can chemical potential be directly measured like this?
I have a system of a pure ideal gas $a$, which I call 'the reference', with a known $P^{ref}, V^{ref}, T^{ref}$ and $n^{ref}_a$. I know the chemical potential formula for an ideal gas, so can calculate how it changes as its variables change.
I place it next to an unknown system with unknown chemical potential $\mu_a$. This system must me "much bigger" than my reference so that it isn't disturbed by their interaction.
I make the separation a fixed membrane permeable only to $a$, and put both in contact with a bath of $T = T^{ref}$.
I wait for the systems to reach equilibrium, and measure $P^{ref}, V^{ref}, T^{ref}$ and $n^{ref}_a$, which let's me calculate how $\mu^{ref}_a$ changed.
Thermodynamics say that at equilibrium $\mu^{ref}_a$ must have become equal to the unknown $\mu_a$. So I just measured the difference between the unknown $\mu_a$ and the initial reference $\mu^{ref}_a$.
This device then let's me directly measure chemical potential (relative to a reference).
Is this correct? Can chemical potential be directly measured like this?
I believe this is completely analogous to how thermodynamic pressure is measured, (relative to a reference like atmospheric pressure), by connecting two systems through a movable wall and letting it reach equilibrium.
I was thinking maybe ph meters are an example of this. Osmotic pressure tubes too.

Only if the mixture and the pure component in the reference state behave as ideal gases.

@ChetMiller I'm not sure I understand what you mean. In my example the reference system is a pure ideal gas that always remains pure and ideal. The unknown system can be anything and behave in any way, but its so big that nothing significant will happen to it.

If  you are describing an experimental procedure to measure the chemical potential of a gas, at higher pressures the gas is not going to be behaving as an ideal gas (both the pure gas and the gas mixture).  So, for the pure gas, you need to take into account non-ideality.


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Finding mole fraction from molarity and density

A solution of H2SO4 with concentration of 2,325 M at 15 °C has a density of 1,140 g/mL. Find the molar fraction.

I know what molar fraction is, but I really don't understand how to get it from the data I get. I tried calculating all other things like mass/mass ratio or molality and rearranging them to find some sort of relation between the data, but I couldn't find any way of solving for n(solution) or anything like that.
Am I missing something?

Follow the steps to calculate molality, but calculate the amount of solvent instead of the mass of solvent.

Write particularly what you have tried, as the calculation is very simple and it is difficult to miss it.

@Poutnik i don't remember exactly what i tried because i erased everything that didn't work from my document, to save space. i can say that i'm probably missing something because i'm pretty much new to the topic, so all i've tried so far is stringing all various data i had together to try and find an equation that would allow me to find the molar fraction without any unknown variable, or having them all cancel out, but i always get one or two. what is the calculation you're talking about?

My point was, for homework-like questions, it is expected to provide evidence of history of your effort to understand principles leading to the solution. No explicit effort looks exactly like no effort. Considering the fact the solution is very trivial.....

You already know H2SO4 molar amount in 1 L. From that you get its mass in 1 L. Density provides total mass of 1 L. Subtracting gets you the mass of water, from which you get water molar amount. You need molar masses of H2O and H2SO4 (approximate integers or precise from looking them up.)


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Analog of chemical accuracy in vibronic energy calculation
Is there any analog of chemical accuracy in vibronic energy calculation? I understand that the unit is /cm however, probably, I shouldn't expect a given number in that unit as the analog.
I am guessing the gold standard for vibronic energy accuracy will vary depending on the size, shape, and symmetry. Whatever vibronic energy accuracy helps the goal of achieving  chemical accuracy would be the target accuracy.
Is it the right way of thinking?

It may not be very clear what exactly you ask about. Standard of energy transition? Wavenumber standard? Measurement accuracy?

Given that chemical accuracy is often associated with roughly 5 kJ/mol (of about 1 kcal/mol for the believers of the ancient scriptures), I'd say 400/cm is a good guesstimate. http://wild.life.nctu.edu.tw/class/common/energy-unit-conv-table.html

Well 1 kcal/mol (4.184kJ/mol) is the gold standard for electronic structure calculations. But here we are talking about vibronic which refers to vibrationally resolved electronic structure. In simple language in addition to electronic transitions we are looking at vibrational transitions too. So I would expect the accuracy constraints should be more stringent (Since we are looking at different portions of electronic transitions).


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Dissociation of Water?
My teacher told me that 1 in 553 million water molecules dissociate.
However, since water has a pH of 7, shouldn’t 1 in every 10,000,000 water molecules dissociate?

You have a concentration of 10e-7 mol/L for pH 7 (at the usual conditions and approximations). How many molecule of water are there in one litre?

[. . .] water has a pH of 7, shouldn’t 1 in every 10,000,000 water molecules dissociate?

To answer this question, what does pH really mean?
Taken from Chemistry Libretexts - pH and pOH, pH is defined as follows,

The pH of a solution is therefore defined as shown here where $\ce{[H3O+]}$ is the molar concentration of hydronium ion in the solution:
$$\mathrm{pH} = -\log \ce{[H3O+]} \tag{14.2.2}$$
(emphasis mine)

This means at pH $7$, there are $\pu{10^-7 mol}$ of $\ce{H3O+}$ in $\pu{1 l}$ of water.
This means that $\pu{10^-7 mol}$ of water dissociate. But now how many moles are there in $\pu{1 l}$ of water?
There are a total of $\pu{55.49952 moles}$ in $\pu{1 l}$ of water. Therefore $1$ in every $\pu{554995200 molecules}$ of water dissociate to be exact. This is assuming $\pu{0.9998395 g/cm3}$ to be the density of water though.

pH is defined for activity a(H+), not concentration. IUPAC Gold Book ref.. Libretexts used the simplification for education reasons. // Molar amount with 4 valid digits cannot lead to molecule count with 9 valid digits.

@Poutnik The textbooks are also lazy about units when they do that. It makes more sense to write $$\mathrm{pH} \approx -\log \frac{[\ce{H3O+}]}{\pu{1 M}}\tag{ for dilute solutions}$$

@Karsten Sure, as arguments of logarithms ( and most of other functions)  must be unitless, as activity is, being defined as $\mu = \mu_0 + RT\ln{(a)}$.


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Regioselectivity of the Hydroboration and Epoxidation of Limonene

Could someone help me out with the most likely major products for each reaction?
1 eq. of reagent is used for both reactions, so my initial thought was that both processes would only occur at the C=C double bond on the 6-membered ring as that one is more substituted and therefore more electron rich compared to the C=C bond at the bottom. However, the C=C bond at the bottom is less hindered so I was wondering if maybe sterics was a more important factor when considering regioselectivity for either one or both reactions.

Sterics is the deciding factor in hydroboration which occurs at the least hindered carbon, thus you get the primary alcohol from limonene. The hydroboration of limonene is discussed here
MCPBA selectively oxidises the more substituted alkene source here so with limonene the selectivity is the reverse of the hydroboration and the cyclohexene double bond is epoxidised.

Thanks for your answer. Is there a reason why sterics is the deciding factor for hydroboration but not for epoxidation? Especially as BH3 (less bulky) is used instead of BHR2 as the reagent.

@SashaJ  Does this answer your question? https://chemistry.stackexchange.com/questions/74005/monoepoxidation-of-1-2-dimethylcyclohexa-1-4-diene-using-m-cpba

That's very helpful, thank you for sharing. I think I understand why epoxidation occurs at the more substituted electron rich bond, however I am still unsure why hydroboration does not follow the same selectivity. Isn't BH3 also acting as an electrophile, so shouldn't it also be attracted to the more electron rich bond? I get why sterics would be the deciding factor if the hydroboration reagent used was bulkier, but in this case it isn't.

@SashaJ  I cannot immediately answer this. I think it deserves to be a new question.


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How to turn a copper/zinc penny battery rechargeable?
I just saw the copper+zinc+vinegar battery also known as "penny battery" and I thought it would be perfect to make at home since I don't know anything about chemistry.
In the wikipedia page linked it is said that the process is irreversible, so the battery can't be recharged.
Searching a bit I've seen here that some company patented their Copper/Zinc rechargeable batteries (patent link).
However, I couldn't make heads of its structure, I'm not trying to make a industrial level battery, just a homemade simple thing.
In order to make it rechargeable, I just need to put the zinc and copper inside their own separator (like in lead batteries in cars or ion-lithium batteries, I believe) and fill it with vinegar, right? Or there is something else to it?

This is just a variant of the lemon “battery”. It is not rechargeable. Also, look up the Daniell cell.

You can benefit greatly by getting this and working through it: K. Schmidt-Rohr, "How Batteries Store and Release Energy: Explaining Basic Electrochemistry", J. Chem. Ed., 95 (10) (2018) 1801-1810. The Zn and Cu Daniell cell is addressed in great detail. TL; DR Cohesive energy differences are the major factor in explaining the behavior of this famous galvanic cell.

Now cross posted at EE SE: https://electronics.stackexchange.com/q/633077/223146.

On the other side of the coin, you could use two lead "slugs" to make a rechargeable cell.


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What is a good strategy for fully quantitative X-Ray Diffraction on coal fly ashes?
I am having some trouble with fully quantitative phase identification of coal fly ashes using x-ray diffraction (XRD). Here follows some basic information about the experiment:
--
Instrument: Bruker D8A
Software: TOPAS5
Database: COD (But can also use PAN ICSD if need be)
Sample: Coal fly ashes from a Western Australian coal power station combusted at around 900$^\circ$C.
The goal: How to accurately fit an XRD profile to the pattern and accurately quantify the phase composition (i.e. crystalline and amorphous wt%s) of the fly ashes. For example, I'd like a good fit of the XRD pattern so that I can get an accurate output phase composition of the fly ash along the lines of (as an example):

2% Hematite
4% Magnetite
8% Mullite
16% Quartz
70% Amorphous

--
I have used TOPAS before for easier and idealised samples (i.e. crystals and simple amorphous structures using the internal standard method) and have fitted the curves great. However, I'm having trouble conceiving of an analytical strategy to analyse and quantify the crystalline and amorphous material in coal fly ashes; a "real world" sample. Refer to the picture below where I've prepared the sample in a way to minimise preferred orientation and with a 20 wt% corundum internal standard and have done phase ID to find quartz, hematite, magnetite, mullite and corundum (from the standard) along with allowing for 1/X background and zero error, absorption and LP corrections:

The goal here is to input the phase composition of the fly ashes (i.e. the quantified crystalline and amorphous wt%s) into a mathematical model (the "mix designer") to synthesise geopolymers, without these wt%s the geopolymers I make don't have good consistency in their measured durability properties (i.e. compressive strength, rheology, viscosity etc.) nor quality assurance and control. Note that procedures for this have been developed in literature (i.e. https://www.sciencedirect.com/science/article/pii/S0016236110003789), but unfortunately, the detailed XRD method employed is not usually provided.
Does anyone have any experience or learnings here and can provide some insight or knowledge? I have done some literature review on this but a lot of the fly ashes mentioned are very different due to differing parameters like location, coal power station combustion conditions, source coal geology and ash content (etc.). XRD of coal fly ashes is a niche and an uncommon area of materials science, accurate quantification is even more uncommon.

I might be ignorant here, but after subtraction of the identifiable patterns, what do you expect to be able to analyse and quantify?

".. mix design and synthesis geopolymers .." Can you elaborate on that a bit? I haven't got the foggiest idea what that sentence means. Typo? Missing word?

I've made some edits to the question for clarity. In short, I input the crystalline and amorphous wt%s (i.e.  16% quartz, 8% mullite, 70% amorphous etc.) into a mathematical model (the "mix designer") to make geopolymers, without these wt%s the geopolymers I make don't have good consistency in their measured durability properties (i.e. compressive strength, rheology, viscosity etc.) nor quality assurance and control. So a good quantitative XRD strategy is crucial to this process.

Hm, and why do you think that model, which you cannot test, because you cannot get the numbers, accurately describes the material properties of a geopolymer made from fly ash? ;) Sorry for playing the devil's advocate here.

I think XRD is probably the best standard analytical method to use here, but it's also very probably totally insufficient. Fly ash is very likely far too complex and inhomogeneous to be sufficiently described by its XRD pattern. Just my gut feeling.

All good, I believe it provides a good indicator because of literature such as https://www.sciencedirect.com/science/article/pii/S0016236110003789 (among others). Fly ash is certainly very complex and inhomogeneous, but the latter complication could be improved upon with good statistics. It will never be homogenous, but a margin of error could be implemented in the model to have some form of quality control/consistency of results (etc.).

I should add also that other tests, such as XRF, ICP, PSD and BET (etc) are done along with the fly ash XRD.

I am the author of that paper.
Reach out on linked in if you would like to discuss in detail. The bot deleted my previous message when I sent my linked in address. You can also get more information my PhD thesis available on the Curtin university library website. It probably has the what I refined and in what order if I did a sequential process.
Like in the paper you need XRF and Quant XRD.
The fluorite internal standard was used to not overlap and minimize micro-adsorption mismatch.
A very important thing to do is fit the mullite with a le-bail fit to get the lattice parameters, then find the most appropriate mullite as it is a solid solution. Then you can use the correct mullite in the Rietveld refinement. There is a set of Mullite's which cover the range I covered in 9 Australian fly ashes, I think it was Ban and Okada (1992). Fitting the minor phases is also important to accurately quantify the amorphous fraction.
Then as per the paper you calculate the composition of the amorphous fraction using Oxide fractions from XRF minus oxide fractions calculated from XRD.
This works ok for many of the Australian fly ashes. In subsequent work we saw that fly ashes with amorphous Si/Al >> 2 did not work out well as Geopolymer target Si/Al ~= 2, in retrospec is was probably an issue with the commercial sodium aluminate product that we were using. Industrial byproduct sodium aluminate worked fine.
If I was given a new fly ash I would follow this procedure and make about 9 paste mixes with a variation of Si/Al and Na/Al ratio to confirm most appropriate mix. If I was in process control I would use the above process to flag when there is likely going to be an issue to investigate.
Ban T, Okada K. Structure refinement of mullite by the Rietveld method and a
new method for estimation of chemical composition. J Am Ceram Soc
1992;75:227–30

I've since gotten better at XRD and GP mxi designing and have been doing a lot of this it's been great

Hi @Ross Williams, amazing response thank you! I tried adding you a few months ago and messaging but it was locked. I can't seem to PM you here on SE but my name is Ramon S on Linkedin, please reach out as I can't message you and couldn't get your contact through Arie :) Would LOVE to have a coffee and chat !

So when using TOPAS the fits look much better and I've learned a lot. My refinements now are nowhere near as heinous as the pics in my original question 7 months ago!


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Disposal of Metal Ion Solutions After Flame Test
I just bought a flame-test spectroscopy kit to use with our small homeschool group. It includes calcium chloride, potassium chloride, strontium chloride, and lithium chloride. I also bought a bottle of copper chloride.
After we do the lab, how can I safely dispose of the solutions? Can I just dilute the solutions and flush them? Or do they require a Hazardous Material Pickup?
Thanks in advance!

The copper chloride might be the largest hazard, depending on amount. It is a poison for plants and corrosive for your pipes. How much did you buy? Here is a related question showing how it "eats up" aluminum foil (while being reduced to elemental copper, which is less of a hazard and present in the solid waste stream anyway): https://chemistry.stackexchange.com/a/128099/72973

The American Chemical Society has a series of safety videos. This one might be a start for discussing safety with your group. It does not directly address hazardous waste disposal, but it does mention safety data sheets (SDS), which also contain disposal information.

The linked answer shows how I reacted aluminum foil with the solution containing copper ions, resulting in copper metal powder and aluminum ions in solution. That is your disposal method: react your copper chloride solution with an excess of aluminum foil, until no green color remains in solution, then toss it all away: it is harmless. This assumes you do not have large quantities to deal with.

Thanks Kartsten and Ed! The lab calls for a solution of about 8.5g of copper chloride in 50mL or water. I love the aluminum foil idea because that would be a fun lab in and of itself to do with the group. Is my solution volume doable with that technique?


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Molecular collisions in an ideal gas
(The equation of ideal gas is $pV=NRT$. If p = 1 atm, N = 1 mole, T = 0°K, and R = gas constant, then volume = V = zero. Hence, the volume of an individual molecule of ideal gas is zero).
An individual molecule of ideal gas is assumed to have zero volume. The molecules of ideal gas are assumed to be dimensionless points. Then how do the dimensionless points collide with each other in accordance with the kinetic theory of gases?
I assume that the individual molecules of the gas have non-zero volume such that they are able to collide with other molecules or the wall of the container. If a molecule has zero volume (i.e. a dimensionless point), then how can it collide with other molecules (how can points collide with each other)?

They don't collide for the reason you have given, they are point particles. You don't need to have interatomic collisions to derive the ideal gas law. The wall is modelled as a continuous surface, even point particles can collide with that.

@IanBush But they have to mutually collide to achieve thermodynamical equilibrium and Maxwell-Boltzmann speed distribution. // The idea is zero-like as negligible volume compressibility-wise, but non-zero volume collision-wise. BTW 0 K is not achievable.

No they don't, not in the ideal gas model. Thermodynamics says what the final state will be. It says nothing about how that state is reached. That is kinetics. c.f. forbidden transitions in Quantum mechanics.

@IanBush Well, the ideal gas model has both thermodynamic and kinetic parts/aspects. I do agree that for the state equation itself, the former is sufficient.

@Poutnik The speed distribution could work via the wall.

Mean free path is a great concept: https://chemistry.stackexchange.com/q/111532/72973

@Karsten It probbaly could.  // AFAIK, there is said if the mean free path is (significantly?) longer than container size, the vacuum-like gas does not reach equilibrium. Or the temperature does not make sense for such a gas. But I may remember it wrong. It may be just for low molecule count.

The first sentence is not relevant to what follows, or at the very least not correct. You can also assume that the pressure = 0 if T=0. This makes sense: at T = 0 the molecules are at rest and do not collide with the container. The volume of the container can be anything under these conditions, assuming it is rigid.

@Karsten The Boltzmann distribution doesn't require collisions. which is what the speed distribution really is. It is just how classical particles occupy the available energy levels at a given temperature. The ideal gas model has no interparticle collisions.

@IanBush A devil's advocate could object with the case of ideal gas sphere, kept together by gravity. Such a gas would never reach thermal equilibrium.

It's not an ideal gas anymore - you've included inter-particle interactions (gravity)

The ideal gas is a 'model' of how an actual gas behaves. It is still used because it agrees with many different types of experiments on gasses, but its limitations are also recognised.
The initial assumptions are that molecules are represented by perfect and elastic spheres of infinitesimal size, these particles travel in straight lines and strike the walls of any vessel in which they are contained. By being elastic means should they collide with one another no energy is lost, i.e. no energy is retained within a particle so that the total kinetic energy after collision is the same as before collision, but their direction and hence momentum changes.
On collision ideal gas particles impart momentum to any wall thus producing pressure. The collision with a wall is elastic so no energy is lost. The incident and reflected angles are the same. To calculate the pressure ($p$) we do not need particle to particle collisions and all molecules can travel at the same speed then
$$p=\frac{2\cdot \mathrm{kinetic\; energy}}{3\cdot V}$$
where $V$ is volume. Even if there were collisions this would not affect the pressure as this is caused by collisions with the walls.
As kinetic energies are additive so are pressures, this means that the pressure of mixtures is the same as that of each gas taken separately which is Dalton's Law. If the volume changes it does so in inverse to the pressure which is Boyle's law.
If two gasses at different temperature are allowed to mix, say by removing a partition between them, they will diffuse into one another and at the same time the temperature will eventually become uniform throughout the gas. This can only happen if collisions between particles occur, so this is an essential assumption of the ideal gas model, unrealistic though it may seem for particles of an infinitesimal size.  Once the absolute temperature is introduced the kinetic energy can be shown to be $3RT/2$ which produces the ideal gas law via the previous equation.

The zeroth law says that two bodies in thermal equilibrium with a third will be at the same temperature. I don't see the requirement for direct collisions when you mix gases.

@Buch Thorn Suppose that I have a vast volume of gas so large that the walls are unimportant and I excite just the centre part with a focussed laser, without collisions I will have the gas at two different temperatures persisting for ever. Does this happen in practice?  Or I could have two gas clouds in space at different temperatures that collide but retain their individual temperatures. Does this happen?

Yes, I see what you mean. Two mixed gases at different T would be strange indeed. But I suppose something like it could be generated. The excited focal point would dissipate, but it need not do so into a uniform gas, since there is no wall to ensure mixing. The universe is a good example. It is not uniform and the absence of a boundary to keep things mixed is probably why (ok, I actually assume this, but I would like to hear the counterargument).

Gas clouds in space? they would be at a T low enough to prevent dispersion.  A heated gas mass expands radially to equalize pressure or gravitational energy.  To see what happens compare the planets in the solar system in relation to the sun. The collisions with the wall are more complicated ; the wall must also have a 2mv change in momentum. The wall is superfluous; all one needs is a bunch of molecules with the same average KE; when they hit they bounce.

@jimchmst yes, but you are bringing gravity into the story, and there is no comparable force at the atomic scale (other than electromagnetic, but we assume an ideal gas). The point of the wall is that it leads to exchange of energy (in hindsight mixing is not the key point, but is important too given the counterargument that started the discussion).

Gravity is the container. My point is that to transfer momentum there must be a loss of energy.  If the collision is elastic the change in momentum of both particles is equal and momentum and KE are both conserved. This does not require a wall but simply molecules at the same temperature and the same average MW [if the MWs differ diffusion happens]. If there is a wall the molecular motions in the wall must have the same momentum change as the gas molecules. since the velocities etc. differ there is an energy exchange. Supposedly this is how solar sails work. I cannot figure it out.

I think we would be much better off were the ideal gas concept never been mentioned and PV simply be limiting example of actual gas behavior. Why learn something to be unlearned later


text:
Does the MS oxygen peak at m/z = 16 correspond to $\ce{O+}$ or $\ce{O2^{2+}}$?
In a mass spectrum, while annealing with $\ce{O2}$ of a sample in UHV, I observe peaks at $m/z = 16$ and $32 (= 16 \times 2)$. While the latter is $\ce{O2^+}$, I'm unsure what the peak at $m/z = 16$ is. I lean toward $\ce{O2^{2+}}$ because of the molecular orbital diagram.

I wonder what may be the eq. constant $\ce{O2^2+ <=> 2 O+}$...

Most likely the signal is predominantly or entirely $\ce{O^+}$.
The question is whether two $\ce{O^+}$ ions will bind to each other forming a diatomic ion in the gas phase. A rough energy balance calculation suggests they will not.
The observed bond length in the singly charged $\ce{O2^+}$ ion is $112.3$ pm. Now suppose two $\ce{O^+}$ ions, each with a charge equivalent to one electron, are brought together to this internuclear distance. The standard Coulombic repulsion calculation gives about $1.24$ MJ/mol.
This might be overcome if the covalent binding energy is sufficiently strong, but the isoelectronic and similarly sized $\ce{N2}$ molecule has a bond energy of only about $0.94$ MJ/mol. So the covalent binding energy expected for the proposed $\ce{O#O}$ triple bond in the gas phase would be insufficient to overcome the repulsion between two positive charges, and we should expect the proposed $\ce{O2^{2+}}$ ion to be less stable than a pair of single $\ce{O^+}$ ions. Therefore the m/z signal is most probably due predominantly or entirely to $\ce{O^+}$.

That looks incorrect - there's no reason to think distance would be so small. Charge would obviously make the bond much longer.

Well, peroxide is a thing, and IIRC even dihelium dication is metastable, so I think you need evidence to say it doesn't exist.

What we call peroxide is not so in the gas phase, I have only ever seen it in condensed media intertwined with counterions.

Yeah, yeah, I even started to look for second affinity for O2, but hey, it's you who decided to answer ;D

@mithoron do you agree with my modifications? The energy balance still seems to favor the monitoring ion to me.

There is paywalled https://www.sciencedirect.com/science/article/pii/S1387380614000189 but I checked some other articles and the dication is most likely indeed repulsive, and dimerisation unlikely. Oh well, I learnt bit about interesting topic ;)

This is a very tricky molecule. A single m/z value cannot tell you whether this is singly ionized or doubly charged on a low resolution mass spectrometer. One would look at mixed isotope patterns of oxygen-17 and oxygen-18 molecules. A doubly charged molecule would then appear at half-integer values (17+18)/2 =17.5. Without any mixed isotope enrichment experiment, a real mass spectrum will be noisy, so it is hard to tell anything with ordinary oxygen which is dominated by a single isotope.


text:
Regioselectivity in oxo-Diels-Alder reaction
I have one doubt about the dimerization of methacryloyl chloride into the product 3:

The dimerized product was observerd around 10-15mol% in different commercial samples of methacryloyl chloride, as it is explained in this research paper:
https://doi.org/10.1016/j.tet.2014.07.019
My question is why the product 3 is observed and no the other possible dimer X:

Taking into account the partial charges of the different atoms, would it not be more logical to obtain the X dimer regioselectively?
Thanks!

The ionic mechanism does not work as you have discovered. It is a "tail-to-tail" dimerization best handled as a radical reaction.  This may help:

Thanks for answering, first of all! Sorry for my ignorance, but I don't get the radical mechanism and why it is a radical reaction instead of normal Diels Alder.  Then, I've found next paper 10.1039/d0ra10084f talking about that it is unclear the origin of the regioselectivity of dimerization of acrolein (similar to the dimerization of methacryloyl chloride) and they do not comment on the possibility of being a radical reaction. Sorry but I am a mess now jaja!

I'm not implying that it is radical mechanism but rather treat it as one that allows you  to get two stabilized radicals that couple.


text:
Use the first law of thermodynamics to derive the pressure as a function of T, U, V and N
I am struggling with a thermodynamics question given
$$S(U,V,N)=C_VNK_\mathrm b\ln U/U_0+NK_\mathrm b\ln V/V_0$$ (where $U_0$ and $V_0$ are the reference energies and volumes) and the first law of thermodynamics which is $\mathrm dU=T\,\mathrm dS-p\,\mathrm dV+\mu\,\mathrm dN$. The question asks to use the two expressions above to derive the pressure $p$ as a function of $T$, $U$, $V$ and $N$.
Would anyone be able to give me a hint as to where to start with this question?

What are your thoughts so far?

Start by writing $dS=\left(\frac{\partial S}{\partial U}\right){V,N}dU+\left(\frac{\partial S}{\partial V}\right){U,N}dV+\left(\frac{\partial S}{\partial N}\right)_{U,V}dN$

It seems that the entropy function is for a thermodynamic system of a pure fluid. So let me:

Define $ s \triangleq S/N $ to have the entropy per molecule.
Multiply both sides for Avogadro's number $N_A$ to have the molar entropy $N_As$ in the left-hand side and the $R$ constant in the right-hand side.
If you don't mind, I will abuse the notation a bit, and also call this new guy also $s$.
Eliminate the $C_V$ that appears in your equation, it seems to be a typo, and also it is not dimensionally correct.

The molar entropy then is
$$ s(u,v) = R\ln\left(\dfrac{u}{u_0}\right) + R\ln\left(\dfrac{v}{v_0}\right)\tag{1} $$
We can also put the molar internal energy in terms of the molar entropy and molar volume
$$ u(s,v) = u_0 \exp\bigg[\dfrac{s-R\ln(v/v_0)}{R}\bigg] \tag{2} $$
The first law for a closed system and a pure fluid yields
$$ du = -Pdv + Tds \tag{3} $$
For an isentropic process, we take the appropriate partial derivative for Eq. (2) and see what the 1st law tells us
\begin{align}
  -P &= \bigg(\dfrac{\partial u}{\partial v}\bigg)_s \\
  -P &= u_0 \exp\bigg[\dfrac{s-R\ln(v/v_0)}{R}\bigg]\bigg(-\frac{1}{v}\bigg) \\
  -P &= u\bigg(-\frac{1}{v}\bigg) \rightarrow P = \dfrac{u}{v} \tag{4}
\end{align}
We could end here but we should continue. For an isochoric process, we take the appropriate partial derivative for Eq. (1), and see what the 1st law tells us
\begin{align}
  du &= Tds \\
  \dfrac{1}{T} &= \bigg(\dfrac{\partial s}{\partial u}\bigg)_v \\
  \dfrac{1}{T} &= R\left(\dfrac{1}{u}\right) \rightarrow u = RT \tag{5}
\end{align}
Combining Eqs. (4) and (5)
$$ \boxed{P = \dfrac{RT}{v} = \dfrac{nRT}{V} = \dfrac{Nk_BT}{V}} $$
This is, the entropy function they gave you is that of an ideal gas.


text:
Schlögl model - Stationary States far from Equilibrium
The Schlögl model can be represented as follows:
$$
A + 2X \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} 3X \, ,
\\
X \underset{k_4}{\stackrel{k_3}{\rightleftharpoons}} B \, .
$$
The chemical reactions can be written in ODE form:
$$
\dot{C}_{X} = k_{1}C^{0}_{A}(C_{X})^{2} - k_{2}(C_{X})^{3} - k_{3}C_{X} + k_{4}C_{B}^{0} \, ,
$$
where we fixed the concentrations of $A$ and $B$ to the initial ones $\left[(C_{A} \equiv C_{A}^{0}) ;  (C_{B} \equiv C_{B}^{0})\right]$.
The concentration of stationary states does not evolve, therefore $\dot{C}_{X} = 0$:
$$ k_{1}C^{0}_{A}(C_{X})^{2} - k_{2}(C_{X})^{3} - k_{3}C_{X} + k_{4}C_{B}^{0} = 0
\\
\iff (C_{X})^{3} - \frac{k_{1}}{k_{2}}C^{0}_{A}(C_{X})^{2}  + \frac{k_{3}}{k_{2}}C_{X} - \frac{k_{4}}{k_{2}}C_{B}^{0} = 0
\\
\iff k_{2} = \frac{a}{(C_{X})} - \frac{k_{3}}{(C_{X})^{2}} + \frac{b}{(C_{X})^{3}} \,
$$
where $a = k_{1}C_{A}^{0}$ and $b = k_{4}C_{B}^{0}$.
In order to find the stationary states, we proceed the following manner:
$$
\frac{dk_{2}}{dC_{X}} = 0 \iff C_{x} = \frac{1}{a}\left(k_{3} \pm  \sqrt{k_{3}^{2} - 4ab}\right)
$$
If $k_{3}^{2} > 4ab  \, \, (1)$, there is 3 possible values and if $k_{3}^{2} < 4ab \, \,(2)$  there is one possible value.
Reichl's book, "A Modern Course in Statistical Mechanics", says that the Schlögl model, away from equilibrium, has multiple stationary states, which seem to represent situation (1) and in equilibrium seem to be (2). However, why does (1) represent the model far from equilibrium and (2) the model in equilibrium?

If $k_3^2 < 4ab$, the expression under the root is negative. The square root of a negative number cannot be calculated, as it is not a real number.

There are two stable equilibria. There are details and a clear discussion in  MIra et al. J. Chem. Educ. 2003, v80, p1488, also a v. thorough explanation in D. Gillespie, 'Markov Processes' publ Academic Press 1992.

to continue. You have to solve the cubic equation to find the deterministic values for to equilibrium values, or numerically integrate for the time profile, ,but then $k_1$ is divided by $2$ and $k_2$ by $6$ to relate the stochastic calculation to the usual rate equation. This is because the propensities can be approximated when the number of molecules becomes v. large. The solution to cubics is v messy but can be done numerically using python/numpy root solver.

Ilie et al give a numerical example that has two steady states. Here are the parameters (ignore the propensities, I think they are for the stochastic models):

with $C_A$ set to $\pu{1E5}$ and $C_B$ set to $\pu{2E5}$. Depending on the value of the initial concentration of intermediate, you get one of two possible steady states (figure from the same paper, I did not numerically solve myself):

In our context, the y-axis shows the concentration of the intermediate X. If you want to know whether the system is capable of reaching equilibrium (with fixed concentrations of A and B), use the criterion
$$ \frac{C_B}{C_A} \stackrel{?}{=} K = \frac{k_1 k_3}{k_2 k_4}\tag{3}$$
For the given set of parameters, this is not the case, so it is not at equilibrium (also, you would not expect two possible steady state concentrations of X if it were at equilibrium).

[OP] ... there is 3 possible values...

That seems too much for a quadratic equation. According to
Vellela and Qian (2008, doi:10.1098/rsif.2008.0476),

Because the ODE form of Schlögl's model is a cubic, there can be one, two or three steady states for a given set of parameters.

The OP does start out with an expression containing concentrations cubed, so unless some terms cancel out during differentiation, I would expect a different sort of solution for the steady state concentration.

[OP] However, why does (1) represent the model far from equilibrium and (2) the model in equilibrium?

You can check by testing one of the following:
$$ \frac{C_X}{C_A} \stackrel{?}{=} \frac{k_1}{k_2}\tag{4}$$
or
$$ \frac{C_B}{C_X} \stackrel{?}{=} \frac{k_3}{k_4}\tag{5}$$
as long as (3) is also true. I'm not attempting this because I have trouble understanding the derivation and the result given by the OP. Specifically, I am puzzled by the assumption that

[OP] $$
\frac{dk_{2}}{dC_{X}} = 0$$

Why would the steady state concentration be independent of one of the rate constants? That is counter-intuitive and requires some explanation.

[my comment] There are two stable equilibria. There are details and a clear discussion in MIra et al. J. Chem. Educ. 2003, v80, p1488, also a v. thorough explanation in D. Gillespie, 'Markov Processes' publ Academic Press 1992.

You have to solve the cubic equation to find the deterministic values for the equilibrium values, or numerically integrate to get the time profile, but then $k_1$ is divided by 2 and $k_2$ by 6 to relate the stochastic calculation to the usual rate equation. This is because the propensities can be approximated when the number of molecules becomes v. large.
Using the values given in the answer by Karsten the polynomial to solve is
$$\displaystyle ax^3+bx^2+cx+d=0,\quad a=k_2/6, b=-k_1A0/2, c=k_4,d=-k_3B0$$
which has roots at $566.9, 247.6,  85.5$. The middle value is a 'barrier' that the deterministic equations (i.e. normal rate eqns) cannot cross. (When the parameters as such that they produce only one real root then only one equilibrium will exist).
You can see the three roots in the plot below as horizontal lines.

The blue horizontal lines are the roots and the grey lines the rate equation solutions starting at X0 values $0, 50, 150, 240, 260, 450, 700$ which you can see in the plot.  You can see how the population is 'attracted' to the upper or lower root but never cross the barrier root at $247$.
The stochastic calculation is very different as there is a chance to rise or fall to an equilibrium value (a root of the cubic) and the barrier can be crossed or recrossed as at each step as there is a random probability of going one way or the other.
The second plot shows this and I have chosen to illustrate that it behaves much as the deterministic plot does, but not all do as can be seen the the third plot.

Having multiple equilibria is of course a result of having a cubic equation. Thus this is unrealistic chemically as a termolecular reaction has such a small chance of occurring that most termolecular reactions are found experimentally to involve two steps.

Thank you very much for your help! Can I make another question about this? Or should I make a new post?

probably a new post unless it has a short answer that can be put as a comment

My question would be about the stability analysis of the Schlögl model (or reaction models in general); is it better to make a new post?


text:
Why can water act as a base under acidic conditions in organic chemistry mechanisms?
In Organic Chemistry mechanisms, I've commonly seen water getting protonated to complete a mechanism, and I'm getting a bit confused on how water can easily get protonated.
From the autoionization of water, I've learned that water is in equilibrium with $\ce{H3O+}$ and $\ce{OH-}$, but it has an equilibrium constant of $K_\text{eq} = 10^{-14}$. So more often than not I suppose, water should not be protonated and the $\Delta G^\circ$ at standard state for protonating/deprotonating water would be rather large and positive.
In the mechanisms I've seen, the protonation of water happens in some sort of acidic conditions. For example, the mechanism of the oxidation of alcohols with $\ce{NaOCl}$ (from Wade "Organic Chemistry" 9th Edition pg 508; I made some annotations to illustrate my question)

In the second step of "Formation of an alkyl hypochlorite derivative," I see water come in and deprotonate the oxygen. The way I'm seeing it, such acid/base reactions can only happen if they result in the formation of weaker conjugate acids and weaker conjugate bases. So does that mean the intermediate they show which gets deprotonated is a stronger acid than the $\ce{H3O+}$ that is formed (and I believe $\ce{H3O+}$ has a $\mathrm{p}K_\mathrm{a}=0$ so that intermediate would have a $\mathrm{p}K_\mathrm{a} < 0$)? And the alkyl hypochlorite that is formed is a weaker base than $\ce{H2O}$ (and I think $\ce{H2O}$ has a $\mathrm{p}K_\mathrm{b} = 14$ so that alkyl hypochlorite would have a $\mathrm{p}K_\mathrm{b} > 14$)?
Something bothers me about thinking that the intermediate is a stronger acid than $\ce{H3O+}$ and the alkyl hypochlorite is more basic than water, and I'm not sure I'm thinking about this in the right way. If acidic conditions make water more basic and more likely to be protonated into $\ce{H3O+}$, how come it starts taking protons from the intermediate in this mechanism instead of just taking protons from the acid that induced acidic conditions (which I think is $\ce{CH3COOH}$ in this case)? Is that intermediate in the mechanism really more acidic than the $\ce{H3O+}$ that forms?
And why do the acidic conditions of lots of protons dissolved in solution lead to more protonation of water -- wouldn't the dissociation of the acid (that's causing acidic conditions in the mechanism) lead to a greater amount of $\ce{H3O+}$? So shouldn't that shift the equilibrium in (2) to the left, leading to less protonated water?
$$\ce{CH_3COOH + H_2O <<=> CH_3COO^- + H_3O^+ (1)}$$
$$\ce{2H_2O <<=> H_3O^+ + OH^- (2)}$$

From the autoionization of water, I've learned that water is in equilibrium with $\ce{H3O+}$ and $\ce{OH-}$, but it has an equilibrium constant of $K_\text{eq} = 10^{-14}$.

The value of $K_\text{w} = [\ce{H+}][\ce{OH-]} = 10^{-14}$ is, on its own, meaningless. You could obtain that by setting $[\ce{H+}] = 1$ and $[\ce{OH-}] = 10^{-14}$, which actually corresponds to a case where quite a lot of water is protonated. You could also obtain that by setting $[\ce{H+}] = 10^{-14}$ and $[\ce{OH-}] = 1$, which means that very little water is protonated at all. The product being $10^{-14}$ isn't what matters; it's the individual concentrations which matter. An 'acidic' solution would lean far more towards the former than the latter.

The way I'm seeing it, such acid/base reactions can only happen if they result in the formation of weaker conjugate acids and weaker conjugate bases.

This is true, but only from a thermodynamic perspective, i.e., it only applies to the net reaction, or the equilibrium position. At any single point in time, protons are flying around all over the place. If it's the only way the mechanism can go forward, it's not illogical to protonate something, even if the equilibrium position doesn't favour it outright.
The point is, even if you only have a tiny concentration of a reactive intermediate, it's more relevant than a large concentration of unreactive stuff.

If acidic conditions make water more basic [...]

It doesn't make water 'more basic'; you're confusing the basicity of water with the position of the equilibrium.
If you're sitting on a seesaw, and someone heavy comes and sits on the other side, sending you flying upwards, it's not because you got lighter.

And why do the acidic conditions of lots of protons dissolved in solution lead to more protonation of water -- wouldn't the dissociation of the acid (that's causing acidic conditions in the mechanism) lead to a greater amount of $\ce{H3O+}$? So shouldn't that shift the equilibrium in (2) to the left, leading to less protonated water?

This is a common misconception. The issue is that in a neutral solution of water, there is not much room for that equilibrium to go to the left.
$$\begin{align}
\ce{HA + H2O &<=> A- + H3O+} \tag{1} \\
\ce{2H2O &<=> H3O+ + OH-} \tag{2}
\end{align}$$
At pH 7, before you add any acid, you have $[\ce{H3O+}] = [\ce{OH-}] = 10^{-7}$. Now, if you add $\pu{0.1 M}$ of some acid to the solution, sure, you can push this equilibrium of $(2)$ 'to the left'. But by how much? There's only $\pu{10^{-7} M}$ of hydroxide for this new acid to react with. Even if you assume that all of this hydroxide reacts up, it's barely enough to make a dent in all the extra acid you added.
For all intents and purposes, then, you can ignore the autodissociation of water. The concentrations involved in this reaction are completely insignificant when compared against the concentrations involved in the other reaction, namely the acid dissociation, $(1)$.

This is why the rate of reactions often is exquisitely sensitive to pH - for every pH unit, a reaction step might be 10 times faster or slower. Here, the acidic pH helps with the protonation of the hypochlorous acid, but makes the deprotonation to the alkyl hypochlorite more difficult. In solution, you have to choose a single pH as a compromise. Enzyme, on the other hand, are capable of protonating one location while deprotonating another by immobilizing acids and bases in specific positions of the active site.

So then what I'm getting is that we have an acidic solution from dissociation of $CH_3COOH$ but that doesn't really shift the equilibrium of water much, so those acidic conditions don't impact whether water can be protonated? And while protonating water at an individual level may be energetically unfavorable (just to confirm: is just the protonation of water step by itself energetically unfavorable?) or kinetically slow (as @Karsten explained), in the context of the mechanism as a whole it can happen because the entire reaction is favorable?

And the fact that this reaction needs acidic conditions doesn't really have to do with the protonation of water/deprotonation to the alkyl hypochlorite step (if anything, I suppose having acidic conditions makes protonating the water more difficult)? But the fact that we have acidic conditions is really just helping to protonate to hypochlorous acid?

Also, is the reason we need to use $H_2O$ as the base instead of $OH^-$ because we have an acidic solution and some of the very little $OH^-$ has been neutralized (although I guess the amount neutralized is very small)? Even if we didn’t have an acidic solution, would there even be enough $OH^-$ to use as a base at neutral pH?

In general, the steps that lead to a successful reaction are rare. However, these are the only steps  we show, not the many collisions that don't even give the first step, or those that almost make it to completion but then go back to the starting point.

Instead of comparing or even considering dissociation of water into H+ and OH-, the key is consider relative reactivity of the the components in the equation to each other.
Protonation on NaOCl creates a leaving group of H2O, allowing Cl+ to "oxidize" or "bleach".
Remember these are equilibrium reactions.
In the second equation, ClOH2 will favor binding to an alkyl group because the alkyl group contributes negative charge a bit more than H.  Water simply acts as a leaving group, and actually CH3COO- can be the proton acceptor in both steps, though it is listed only in the second step where the ketone is formed and Cl- is the leaving group.
Finally, HClO is a stronger acid than H2O because Cl is more strongly electron withdrawing than H.

Would the downvoter please step up and explain why they feel this answer is incorrect.

When you hover over the down-vote button, it says "this answer is not useful", it does not say "incorrect". I think there is a lot of good stuff in your answer. One individual, with the right to vote anonymously, did not think what you had (or what you had written before the edit) was a useful answer. Wasn't me (you have to trust me on that because voting is anonymous)....

Appreciate that.  It would be really helpful if the downvoter would leave a comment so I can improve my answer.  I only did this stuff for a living for more than 20 uears.


text:
What layer is formed on Titanium when soaked in Hydrogen Peroxide?
A few pieces of TI-6AL -4V ELI ASTM F136 Titanium jewelry were left to soak in a small cup of 3% household Hydrogen Peroxide.
After about a week of soaking, all of the Titanium had acquired a smooth, dark grey/black coating. In the process, the Hydrogen Peroxide acquired a yellow tint.
I am familiar with Ti color anodizing, but this is a color that, as far as I know, is impossible to obtain with traditional anodizing methods. It more closely resembles industrial anodizing and the grey/silver color they obtain.
What process is taking place?
What is the grey coating made of?
Is it just oxidized titanium?

the yellow color is from peroxotitanic acid and the grey color is titanium dioxide

@SafdarFaisal Thank You! That is exactly what I wanted to know.

I have some papers that show the reactions.. If you want i'll write an answer at some point


text:
Why does the rate of change of atomic radius in the second period change so drastically?
I was reading in my textbook Chemistry Part I, Textbook for Class XI
by NCERT, ed. January 2021 that:

The atomic size generally decreases a period across as illustrated in Fig. 3.4 (a) for the elements in the second period. It is because within the period the outer electrons are in the same valence shell and the effective nuclear charge increases as the atomic number increases resulting in the increased attraction of the electrons to the nucleus.

I understood what they've tried to say, but when I looked at Fig. 3.4 (a) that they provided:

I noticed that the rate of change for the atomic radius becomes less drastic as I move across the period from left to right, albeit there's an anomaly: the slope of the graph between C and N is less steep than the slope of the graph between N and O.
What explains this?
To reiterate, I understand how the atomic radius changes across a period, and I also understand why it does so. What I don't understand, however, is why the rate of change of atomic radius changes? Why is the graph not linear?

Why do you say the graph should be linear? E.N differences are not linear.

That's exactly what I'm asking: why's it (not) linear? It doesn't help when I ask a question and I get a reply to it saying why I asked the question.

It should help you to ask yourself why you have asked this question and not another. Like, "What users want is not always what they need.". // See also "effective nuclear charge" in context of Slater's rules.

@Poutnik thanks for the link. I'm new to the site and still learning the rules, so thanks once again for bearing with me.

For linearity, especially in context of nonlinear interaction, must be very good reason. So more proper question than "Why is not it linear?" would be "Why should it be linear?"

Within a given period, one might expect such a graph to be of roughly the same shape as the one-electron Bohr radius $r=\frac{53}Z$ pm, which is nonlinear and follows from the basic force equations of the Bohr model. Of course, the presence of additional electrons will perturb this basic structure to some extent, but there is no reason for it to become linear.

This question might be best answered conceptually, rather than mathematically:
Lithium has but 3 protons in its nucleus. Adding a fourth, to make it beryllium, increases the nuclear charge by 33%. But if you add yet another proton, to make it boron, you are increasing the nuclear charge by 25%. And so on, down to the change from oxygen to fluorine, an increase in the nuclear charge (really, the proton count) of only 12.5%. This accounts for the decline in rate of change that you observe.

If this explanation is true, then what accounts for the jump that you would have observed if your graph had considered Period 3, and 4, and so on?

Remember that your graph only considers atoms within the same period. And now, I ask you to think about atoms in different periods.
This effect is explained by the fact that, when you move to a different period, the outermost electron (which determines the atomic radius) is forced to occupy an entirely separate principle quantum level. And, to get more mathematical in my answer, recall that the positions of the principle quantum levels varies as the square of the principle quantum number.

If this explanation is true, then the rate of declination in atomic radius should decrease as you consider each period in turn.

And that is exactly what is observed.

And if this explanation is true, then you should observe almost no change in atomic radius as you march across the Periodic Table, by the time you get to uranium.

Seems to be what the data shows.
Thanks for asking. I hope this helps.