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What does it mean and what are the implications of individual electrons being identical?
In Jeremy Harvey's Computational Chemistry, I found the following statement:

[...] all electrons need to be treated as being very strictly identical to each other, so the wavefunction cannot treat individual electrons differently.

If you could explain what the author is saying here, and what the implications are. I can't make sense of this paragraph/section. An image of the section can be found here.

You can probably look up Slater determinants in some other textbook of your choice (preferably one with a longer explanation, because the Oxford Primers can be a bit terse). More fundamentally, the antisymmetric requirement stems from the Pauli principle (or the spin–statistics theorem), and is often covered in QM textbooks under the guise of 'identical particles' or 'indistinguishable particles'. That's plenty of keywords for you to hit the library!

Thank you for replying. My aim is to go through this entire textbook as it seems like a good book to start with. My goals are to familiarise myself with computational chemistry so I can have a career in this field. I'm not very mathematically able (I am currently reading Maths for Chemistry Monk and Munro). I have looked at other books e.g. Szabo and Ostlund, Young. but found them difficult to comprehend. I'll have a look at these search terms, thank you.

What level of education are you currently at?

I have just completed an MSc Chemistry (June). I have some experience in computational chemistry (my final project was computational chemistry). However, during my MSc, this quantum chemistry stuff was done in one lecture and I think it was based more on application.

Ah, ok! I was going to say that it's easier to get into comp chem by doing it, rather than diving headfirst into the theory. But it sounds like you're at a stage where you probably should be getting into the theory if you want to pursue it further. I don't have any concrete suggestions for books, but if you aren't fully comfortable with QM, I'd probably suggest to brush up on that first.

Doing it has helped me understand things a lot better. I have a long way to go though! Looking forward to it. Thank you.

The Wikipedia article on identical particles explains what is meant by identical particles being indistinguishable:

There are two methods for distinguishing between particles. The first method relies on differences in the intrinsic physical properties of the particles, such as mass, electric charge, and spin. If differences exist, it is possible to distinguish between the particles by measuring the relevant properties. However, it is an empirical fact that microscopic particles of the same species have completely equivalent physical properties. For instance, every electron in the universe has exactly the same electric charge; this is why it is possible to speak of such a thing as "the charge of the electron".
Even if the particles have equivalent physical properties, there remains a second method for distinguishing between particles, which is to track the trajectory of each particle. As long as the position of each particle can be measured with infinite precision (even when the particles collide), then there would be no ambiguity about which particle is which.
The problem with the second approach is that it contradicts the principles of quantum mechanics. According to quantum theory, the particles do not possess definite positions during the periods between measurements. Instead, they are governed by wavefunctions that give the probability of finding a particle at each position. As time passes, the wavefunctions tend to spread out and overlap. Once this happens, it becomes impossible to determine, in a subsequent measurement, which of the particle positions correspond to those measured earlier. The particles are then said to be indistinguishable.

It also has a section on exchange symmetry that might help to see the bigger picture. Wavefunctions describing fermions (which includes electrons) have to be antisymmetric with respect to exchanging two identical particles in the wavefunction. As the article points out, this is an empirical observation. There are other particles (bosons) that don't have that requirement and have symmetric wavefunctions instead:

[...] symmetric and antisymmetric states are essentially unchanged under the exchange of particle labels: they are only multiplied by a factor of +1 or −1, rather than being "rotated" somewhere else in the Hilbert space. This indicates that the particle labels have no physical meaning, in agreement with the earlier discussion on indistinguishability.

This requirement for antisymmetry is one way of formulating the Pauli principle. If two particles had the same (single-electron) wavefunction and the same spin, swapping them would make no difference, breaking the requirement.

What I have understood now is that electrons are identical (indistinguishable) when not measured according to QM because we are looking at wavefunctions. Thus, the equation
Big Psi (r1, r2 ... rn)= Small Psi1(r1) * Small Psi2(r2) * ... Psi n (rn)     is modified to accommodate the identical. So we get the equation in the picture      Big Psi (x1, x2, x3, x4) = -Big Psi (x1, x2, x3, x4)       where the negative symbol is for the antisymmetric and instead of multiplying we have combined all the psi into one overall wavefunction to show the identical (Pictured Equation). Thank you.

Thank you for linking the articles and the information! This makes a lot more sense.

@PassionFruitChemistry You are most welcome. I spent some time trying to improve the quality of the question. Please think about a title that lets others find the information, too, and type the relevant quotes so they become searchable (I would have OCRed the entire page but the quality of the photo was insufficient to do that). If you feel like this answers you question fully, feel free to accept this answer. If not, you are free to edit your question to clarify what the point of confusion was. I had to guess, but judging by your comment, I guess my guess was pretty good...


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Equilibrium Bond Length & Diatomic Molecules
Must every diatomic molecule always have an equilibrium bond length? That is, is there always a distance between two nuclei such that the Coulomb attraction between the electron and protons counterbalances perfectly the electric repulsion between the protons?

I wonder what the alternative would be. Would the atoms come closer together, feel repulsion, move apart, feel attraction, move together? How do we even measure the distance of two particles that are too small to see?

Hmm, for polyatomic molecules this isn't true - there are molecules that don't keep one equilibrium position - fluxional molecules.

Taking the English maybe too literally one could just about read "an equilibrium bond length" as "a single equilibrium bond length", i.e. can we devise a system where there are multiple minima? Otherwise I can't think what physically the question could actually mean.

@IanBush There could be hysteresis.

@IanBush Yeah, that's what I mention. One could say it's a matter of temperature, but methanium is fluxional no matter how low it is.

Another option are molecules that are very far from equilibrium, perhaps excited or transient ones.

Please note the question specifies that it's about  diatomic molecules but then seems (in the second sentence) to be more generally about any two nuclei. Which one is it?

@Karsten Funny you should mention hysteresis, that was what was at the back of my mind - in a talk presented on Friday I saw a nice case of hysteresis in a phase transition in a metal organic framework driven by multiple minima in the potential energy surface.

@Mithoron the more I think about it the more I think the most likely scenario in a diatomic is a weakly bound system where the BO approximation is breaking down - i.e. at elongated bond lengths an excited but still bound state becomes lower in energy than the "ground state". Not my area, no idea if it is observed.

[OP] Must every diatomic molecule always have an equilibrium bond length?

The bond length depends on "what the electrons are doing". For example, $\ce{H2}$ and $\ce{H2+}$ have different bond lengths (the latter is a molecular ion, so maybe outside of the scope of the question). Also, if you are looking at an excited electronic state, the bond length will be different from that of the electronic ground state. It also depends on the temperature, and with that, the vibrational state. The atoms will always oscillate. For higher vibrational states, the average bond length will be longer (anharmonic oscillator).

[OP] That is, is there always a distance between two nuclei such that the Coulomb attraction between the electron and protons counterbalances perfectly the electric repulsion between the protons?

The main repulsive force for 2nd period and heavier atoms is the not repulsion of protons, but the quantum-mechanical effects keeping electrons apart (Pauli exclusion), see What repulsion keeps non-hydrogen atoms at a distance: between inner shells or between nuclei?. Aside from that, the question is

whether there is always a distance where the atoms experience neither a repulsive nor an attractive force (i.e. a potential energy minimum),
whether there is a single minimum, and
whether the atoms arrange such that they are in the minimum position.

The first part is easy to answer - it depends. For atoms that form a given molecule, there is a potential energy minimum different from infinite distance. When two atoms can't form a molecule (at the given pressure and temperature), there is not.
The second part is easy to answer (with no) for molecules with more than two atoms. There are examples where multiple arrangements of atoms in space give rise to a potential energy minimum (rotation around single bonds and other conformers, various types of isomers). I am not aware of any examples for diatomic molecules (if the number of electrons is constant and we are in the ground state) where there is more than one minimum.
The third part is easy to answer, again. When there is kinetic energy available, molecules will not always be at equilibrium distance.
Lastly, here is a definition of equilibrium bond length for computational chemistry:

The equilibrium bond length for a diatomic molecule is defined as the separation that minimizes the total electronic energy; we can identify this as the minimum on our potential energy curves.

Source: Molssi

I think that your last definition from Libretexts is not quite correct. Thermal motion does not come into it. It is the uncertainty principle that determines the minimum zero point energy and so the range of possible bond lengths that on average we may call the equilibrium length. Stronger bonds do not always have to correlate with bond length, I'm sure exceptions exits. An excited state can have a shorter bond length than a ground state, e.g. NO

@porphyrin I edited the answer, trying to address your points.

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Why Hg doesn’t form hydroxide?
In my textbook it was written that if you put alkali solution into Hg2+ solution you won't get Hg(OH)2 but you get HgO.
And this is because Hg has high electronegativity and this makes polarization of O-H in virtual Hg(OH)2 big enough to make dehydration of Hg(OH)2.
I couldn't  understand the reason well. Dehydration seems like a kind of neutralization because it emit H2O and makes salt(HgO), but neutralization between the same thing(in this case Hg(OH)2 )is usually not favored (reaction like 2HCO3-->H2O+CO2+CO32-  is not favored.) Why Hg(OH)2 dehydration favored ? Electronegativity of Hg is 2.0(Pauling) and 2.2(H) so it is smaller in Hg actually.

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Keep CH SE Q titles in plain text.

The core 5d orbitals of Hg are relatively close in energy to the valence oxygen orbitals and repel them.

@KanghunKim how does that explain formation of HgO over Hg(OH)2? In both of them oxygen is binding to Hg.

An Hg can barely mind binding to one oxygen(HgO is actually only kinetically stable)- it just cannot withstand binding to TWO oxygens.

Suppose that you get an upset stomach to apples. Eat one apple and you're barely trying not to throw it up(but it gets wasted as diarrhea at some time). This is HgO. Eat two apples and you throw half up(Hg(OH)2 -> HgO + H2O), later you excrete the other half as diarrhea. Not a metaphor that agrees with the table, but you get the idea.

@KanghunKim Do you mean, after one bond is made between a Hg and an O atom, Hg cannot make energetically stable bond with one more O atom? But I don't understand the reason for it well.

One Hg-O bond: only kinetically stable

Two Hg-O bonds: not even kinetically stable

Could anybody explain the difference between , say Ca(OH)2 and Hg(OH)2?

I thought since Hg is lower in ionization tendency and Ca is higher in ionization tendency, Ca2+ is stable as an ion with OH-, Hg2+ deprives e- from OH- and  form covalent bond with O emitting H+. Is this reasoning valid?

The filled core 5d orbitals of mercury(II), were they valence, would be antibonding with respect to the filled valence 2s and 2p orbitals of oxygen(-II).
For the lighter elements this effect, which technically exists in every species with filled core orbitals, could be ignored- however it is not that straightforward for the heavier elements, especially the later-period post-transition metals, due to relativistic effects that stabilise the valence s orbital and destabilise the core d orbitals.
The only ways to remedy it is to either form strong covalent bonds (using the low-energy unfilled 6s orbital of mercury(II)) to increase the bonds with the ligands, or to form dispersive interactions (using the relatively high-energy filled 5d orbitals of mercury(II)) with the ligands to decrease the antibonds with the ligands. The former requires high-energy ligand HOMO's, while the latter requires diffuse ligand HOMO's- oxygen(-II) has neither. (Sulfur(-II) has both, which explains why HgS is one of the most insoluble substances in water)
In short, HgO is unstable due to the "unfortunate" combination of core-valence repulsion (effected by relativistic effects) and poor bonding overlap. Of course, similar effects occur in ZnO and CdO as well, but HgO is the most unstable of these, in fact being the only thermodynamically unstable of the three, due to the very fact that relativistic effects get stronger when the element gets heavier.
All this is about how mercury(II) literally wants to "expel" oxygen(-II) when forced to eat it. Eat one oxygen(-II) and mercury(II) tries to keep it down(eventually expelling it as diarrhea)- eat two oxygen(-II)'s and mercury(II) ends up immediately throwing one of them up(no comments shall be made on the fate of the other one). This explains the kinetic (never mind the thermodynamic) instability of Hg(OH)2.

Could you explain what you mean by the first paragraph? Molecular orbitals are bonding or antibonding, not atomic ones. Do you mean higher in energy? If so that's not the same as antibonding.

"Were they valence", some of the Hg 5d orbitals would end up being the predominant parts of filled antibonding orbitals.

@KanghunKim can I ask you if you know a good book which explains correlations between occupied orbitals, periodic properties and reactivity?

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How do we know the ionic charge of iron or elements such as iron on the periodic table?
How do we find iron's ionic charge? I was taught that we use roman numerals but I don't see any on the basic periodic table of elements (I'm using this one). So how do we find iron's charge of Fe 2+ or 3+? Again I thought it was the column that told us this but that doesn't seem like the case because iron's column numeral is VIII.

Elements don't have "ionic charge". They can exhibit oxidation states and coordination numbers. While inorganic compounds are often treated like they were made of ions, that's hardly true. It's at best a lame approximation that easily outlives its usefulness.

It's not simply the column that determines the ionic charge of an element, but also other factors, such as row and with what other elements it's combined.
Epediaa states, "valence refers to the ability of an atom to be combined with another atom [which may vary, ed.] whereas valency refers to the maximum number of electrons that an atom can lose or gain in order to stabilize itself [a single value, ed.]." So an element has a specific valence, depending on its group (e.g., C, 4 or Xe, 0), but may have multiple values for valency, such as in $\ce{FeCl2}$ and $\ce{FeCl3}$, as you state
There are charts showing the most common valences, but take them with a grain of salt. For example, xenon, Xe, has a filled outer shell, ostensibly a valence of zero,  yet over the last 50-or-so years, a number of Xe compounds have been discovered, such as $\ce{XeF2}$, $\ce{XeF4}$, and $\ce{XeF6}$.
So there's no simple rule for learning the number of electrons an atom may gain or lose in a compound (and using electricity, for example, it's possible to remove electrons well below the valence shell), though the number of electrons in the valence shell is unique.

So if it were FeCl3, that would mean that it's Fe3+ because Cl3 is -3, so Fe has to cancel with positive three, thus Fe3+. Am I Correct? and wasn't the simple rule that if it is the first three columns excluding transition and the elements with carbon-silicon etc., then it (+) and otherwise it's (-) because they gain one more due to them having an imbalance.

Yes, Fe+3 in FeCl3, and Fe+2 in FeCl2. The first few rows tend to follow general rules, but consider carbon monoxide, or all the different oxides of nitrogen. https://en.wikipedia.org/wiki/Nitrogen_oxide


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Are substituents placed alphabetically in the name of a cycloalkane?
I named this molecule 1,5-diethyl-2-methyl-4-propylcyclohexane. Is this correct? Or should we somehow use the size of the substituents?

Yes, simple prefixes (like methyl, ethyl, and propyl) are arranged alphabetically disregarding any multiplicative prefixes. Any multiplicative prefixes are inserted later and do not alter the alphabetical order.
The corresponding actual wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows.

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Therefore, the name for the compound that is given in the question is (1S,2S,4S,5S)-1,5-diethyl-2-methyl-4-propylcyclohexane.


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Intense purple colour of litmus paper in contact with calcium oxide
When conducting an experiment to identify some common chemicals, I tested
crystal/powder samples of the chemicals with moist litmus paper. When I put a piece of blue paper on some $\ce{CaO}$ powder, much to my surprise, it turned an intense, bright purple (A). When I used a piece of red litmus paper later to test it, it turned a purplish blue as expected (B). Below roughly shows the colours I saw.

I have been unable to figure out what caused the intense purple colour, and why the same does not happen for red litmus paper.

What is the meaning of the eight long, thin and colored rectangles, which do not seem to be related to the letters R, G and B, printed before each line of rectangles? And what is the meaning of the eight numbers varying between $5$ and $232$ and corresponding to each rectangle ?

Yes, I left them out in a new edit. The values aren't too important for this question indeed.

@maurice FYI on those color rectangles. The first three are red, green and blue components in the rgb color solid. If there is 102 blue, let us say, the color would be 40% of the way from a relatively dark reddish or greenish color (0 blue) to a brighter, predominantly blue color (full blue component, which is 255). The last rectangle shows how bright the color is overall, from 0 for black to 100 for a fully brightened hue. Formally this number is the max rgb component divided by 255, as a percentage.

@ Oscar Lanzi. The colored rectangles have been removed now.

Still thought you'd like to know. Actually the red, green and blue numbers are still there as the letters and digits you still see, which are hexadecimal representations of the original numbers.

you also discovered how they test for water in petroleum storage

You may have overreacted the litmus paper. Such overreactions are possible when indicators are exposed to strongly basic (or strongly acidic) conditions.
The complex organic compounds that make up typical pH indicators may undergo multiple stages of acid-base reaction, which could then impart more than one color change. Perhaps the best known example is with phenolphthalein. We all know what happens when the indicator, typically added inits colorless "weakly acidic" state, is exposed to a basic solution. But:

Between strongly acidic and slightly basic conditions, the lactone form (HIn) is colorless. The doubly deprotonated (In2-) phenolate form (the anion form of phenol) gives the familiar pink color. In strongly basic solutions, phenolphthalein is converted to its In(OH)3− form, and its pink color undergoes a rather slow fading reaction1 and becomes completely colorless when pH is greater than 13.

So if we were to expose phenolphthalein to let us say, a one molar sodium hydroxide solution to get the beautiful pink color, we would instead see the pink "unexpectedly" fade away. Milk of magnesia is actually better optimized for turning phenolphtahlein pink, because of the lower pH to which magnesium hydroxide solutions are limited.
So it likely is with litmus. Adding a powerful base such as calcium oxide to an already "deprotonated" blue litmus may cause an overreaction similar to what Wikipedia reports for phenolphthalein. Here, too, using the less soluble but still sufficiently basic magnesium oxide or hydroxide should avoid the overreaction.
Cited reference

Kunimoto, Ko-Ki (February 2001). "Molecular structure and vibrational spectra of phenolphthalein and its dianion". Spectrochimica Acta Part A. 57 (2): 265–271. Bibcode:2001AcSpA..57..265K. https://doi.org/10.1016/S1386-1425(00)00371-1. PMID 11206560.

(+1) Nice answer! I used to do the calcium metal in water lecture demo, using a 2 L beaker with about 100 mL of water in it. I had a little phenolphthalein to the water initially and the resulting turbid solution, after the very exuberant reaction of calcium with water, was bright pink. But the color faded away, “softly and suddenly”, over a period of a few minutes. Then I learned about that further reaction you discuss.

Thanks Ed. As an example of magnesia being better optimized, magnesium metal will give an enduring pink color, which would be more prominent if warmed. Also be sure to avoid exposure to CO2 from air.

@EdV see https://www.sciencephoto.com/media/568914/view/reactivity-of-magnesium-in-water for the color imparted by magnesium metal with phenolphthalein. The left one is warm water, the right one cold.


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Store Palladium Chloride in Solution
I have one gram of palladium chloride ($\ce{PdCl2}$) and I want to use very small quantities of it, in milligrams somewhere. It is not possible for me to get my hands on a scale that can measure that low quantity. When I shall be dissolving it in HCl solution during use.
So what if I prepare its solution in HCl and then use that solution by measuring its volume in a syringe and calculating how much $\ce{PdCl2}$ there is in the solution? I have seen 5 wt% $\ce{PdCl2}$ in 10 wt% HCl solution available on the internet. So I take 54 mL of 37 % HCl and dilute it to 200 mL to get 10 % HCl. Then add 1 g of $\ce{PdCl2}$ to it. Then I can say that for every mL of the solution, I have 5mg of $\ce{PdCl2}$.
How much shelf life will this solution have? What kind of container should I use to store it? Or is there a better way to store it in the solution and measure small quantities of it?

I remember during undergrad storing Pd catalysts in a brown glass jar in the fridge. If making up 0.25g, you could use a brown 50 ml dropper bottle or something similar. As for shelf life, I would suggest making up 0.25 g, dating the jar and checking to see if the reaction stops working at any point. To measure the the amount, use a syringe and needle.

Was it Palladium Chloride in HCl that you stored in the brown glass jar?

No, Pd2dba3 came in a brown glass jar.


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Crude palm oil DOBI analysis
I'm performing Crude Palm Oil DOBI spectrum analysis using uv-1900i UV-VIS spectrophotometer at a range of 269nm and 446nm but it's only giving peak absorbance at 446nm but no reading at 269nm,what could be the problem? Please note that I took 0.4g of CPO and dissolved to 100ml mark volumetric flask using iso-octane, which was also used as the blank (iso-octane)

What is your cuvet material?


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Why can't galavanic cells run without an electrolyte?
If all that needs to happen in a galvanic cell is transfer of electrons from one electrode to another, why can't we just join them directly? What is it that the electrolyte does that keeps the cell running that doesn't happen without it?
I actually tried to ponder about this myself and came up with my own theory, i just need someone to confirm/correct my theory.
Considering a Daniel cell, copper electrode is dipped in let's say copper sulphate and connected to zinc electrode which is dipped in say zinc sulphate solution, a salt bridge is also setup, what happens is that copper having a higher standard potential starts pulling electrons from neutral zinc and the copper ions being positive start getting greedy and deposit on the electrode for those electrons, meanwhile zinc gets ionized and starts dissolving in the electrolyte, the free sulphate ions in copper side start jumping into the sea of zinc ions on the zinc side(technically its just one sulphate ion nudging other sulphate ions in a sequence which go and form zinc sulphate on the zinc side). This is what I think happens in a galvanic cell.
Now, since in dry form if the electrodes are just connected directly, the copper will be stronger on both ends of the circuit and there is no current as a result.
Edit: There has been a suggestion of a different already published answer, but it doesn't give a clear explanation enough, if someone could give a clear step by step journey of the workings of a galvanic cell and functions of its key components, I shall be grateful. Thank you.

Rhetorical question: Where would dry electrodes take the energy from? Daniell cell takes it from reaction $\ce{Zn + Cu^2+ -> Zn^2+ + Cu}$.The respective electrode reversible half-reactions act as charge pumps, similarly as electronic charge pumps are used in chip to generate various voltage levels.

What is the Purpose of an Electrolyte in a Galvanic Cell?

"join them directly": you can do that. It's called a short circuit. ;)

Most textbooks or online educational sources give such explanations. Study them and come with what you do not understand. See e.g. https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/17%3A_Electrochemical_Cells

If all that needs to happen in a galvanic cell is transfer of
electrons from one electrode to another, why can't we just join them
directly?

Obviously, this is only a partial explanation; not only must electrons be transferred, but there must be chemical changes at the electrode surface as well.
If you were to join dry copper and zinc plates, there will be a development contact potential. This is explained very well in Wikipedia Galvani potential. However, one cannot extract useful work just from this transient contact potential. So this arrangment is useless as an electrochemical cell.
In order to draw continuous current from a cell for doing external work say in a bulb, there must be a chemical change. This means that the electrode material must ionize, say if you have a Zn electrode, leading to the formation of $\ce{Zn^{2+}}$ ions. As you know, charged ions alone do not have a long lifetime. For example, if you wanted to create a doubly ionized zinc ion directly from zinc, you would probably require the surface temperature of the Sun in the gas phase. However, presence of a solution makes ionization easy. A solution phase stabilizes ions.
Alternatively, one also requires a solution and a salt bridge because air is a poor conductor of electricity. All you need is a medium for holding ions in an electrochemical cell, and for that purposes an aqueous solution in Daniell cell is a good idea. Also, note that ordinary "dry cells" in the market are not dry at all. The internal stuff is very moist.
A final advice is to avoid anthropomorphizing non-living objects. Copper ions are not not greedy like humans, sulfate ions don't nudge and copper is not strong like a superman :-) These labels create problems in proper scientific explanations.

@ AChem. What do you mean with the sentence "sulfate ions don't nudge" ? In the Merriam-Webster Dictionary, the word "to nudge" is defined as : "to push gently (as with the elbow) in order to seek attention". So what ?

A cell $\ce{Zn/Cu}$ contains two separate compartments, one with a $\ce{Zn}$ plate dipped in a $\ce{ZnSO4}$ solution, and then another compartment made of a copper plate dipped in a $\ce{CuSO4}$ solution. If they are separated, nothing will happen, even if the the metal plates are connected by a copper wire.
Matter of fact, during the first microsecond, some zinc atoms will produce some $\ce{Zn^{2+}}$ ions in solution and electrons in the outer wire, by the reaction $$\ce{Zn -> Zn^{2+} + 2 e^-}$$ And the electrons will travel to the copper plate and discharge some $\ce{Cu^{2+}}$ ions according to : $$\ce{Cu^{2+} + 2 e^- -> Cu}$$ But these two reactions will only last a couple of microseconds and then stop. The reason is that the zinc solution contains now some positive charges, so that it becomes more and more positively charged. This supplement of positive charges will prevent new zinc atoms to produce new positive ions (and new electrons). The same thing happens on the copper plate, where arriving electrons will destroy copper ions, without removing $\ce{SO4^{2-}}$ ions. The copper solution becomes more and more negatively charged, which repels new electrons and prevents them from coming to the copper plate. The cell does not work.
In order for the cell to work, it is necessary to get rid of the positive charges due to newly created $\ce{Zn^{2+}}$ ions in the zinc compartment. And to get rid of the sulfate ions from the copper compartment.
The only way to do it is to make up a liquid bridge between the two compartments. This bridge is made of a solution of non reacting ions, like $\ce{Na+ + SO4^{2-}}$. With this setup, the positive ions in excess in the first compartment attracts the negative ions in excess in the second compartment, and they may join in the bridge, so as to establish electric neutrality everywhere. The cell is now working properly.


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Is there any calcium carbonate analogue drawing carbon from dry air?
Is there a nonliving (wood does not count) substance or process that accretes a solid material by withdrawing carbon from the air, in a manner analogous to calcium carbonate being accreted from seawater in biorock/seacrete?

The alteration of concrete is an example.

There is no carbon in the air. Air contains some carbon dioxide, but carbon is not carbon dioxide. Carbon is a solid stuff. Carbon dioxide is a gas. Carbon dioxide can be removed from air by bubbling in a suspension of (insoluble) calcium carbonate in water. The reaction occurs according to the equation :  $$\ce{CaCO3  +  CO2  +  H2O  ->  Ca^{2+} + 2 HCO3^-}$$


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Why does the coordination of alkaline earths by graphene oxides increase as a function of the increasing ionic radius of the species?
The literature describes that the affinity of graphene oxides towards alkaline earth metals presents a positive correlation with their ionic radius (for example, Klímová et al. (2016) and Sófer et al. (2014). Why does that happen?
The main results of the articles cited above are summarized below:
q (mmol/g): 0,17 (Be), 0.19 (Mg), 0.19 (Ca), 0.54 (Sr), 1.52 (Ba) (Klímová et al. (2016)
wt. % in GO: 0.11 (Mg), 0.26 (Ca), 0.54 (Sr), 2.19 (Ba) (Sofer et al. (2014).
KLÍMOVÁ, K.; PUMERA, M.; LUXA, J.; JANKOVSKÝ, O.; SEDMIDUBSKÝ, D.; MATěJKOVÁ, S.; SOFER, Z. Graphene Oxide Sorption Capacity toward Elements over the Whole Periodic Table: a comparative study. The Journal of Physical Chemistry C, v. 120, n. 42, p. 24203-24212, 14 out. 2016. American Chemical Society (ACS). http://dx.doi.org/10.1021/acs.jpcc.6b08088.
SOFER, Z.; WANG, L.; KLÍMOVÁ, K.; PUMERA, M. Highly selective uptake of Ba2+ and Sr2+ ions by graphene oxide from mixtures of IIA elements. Rsc Adv., v. 4, n. 51, p. 26673-26676, 2014. Royal Society of Chemistry (RSC). http://dx.doi.org/10.1039/c4ra02640c.
Thanks in advance!

The articles you reference are behind paywalls and so not available to many on this site, Please summarise in more detail the content of the articles in the question. In particular please quantify "greater ability" - what precisely do you mean by that?

@IanBush, thanks for the reply. I'll summarise the main results obtained by the paper with the amount of each alkaline earth metal adsorbed by GO: q (mmol/g): 0,17 (Be), 0.19 (Mg), 0.19 (Ca), 0.54 (Sr), 1.52 (Ba) (Klímová et al. (2016) / wt. % in GO: 0.11 (Mg), 0.26 (Ca), 0.54 (Sr), 2.19 (Ba) (Sofer et al. (2014). I used the term "ability" in the sense of having greater affinity, which means that GO better adsorbed species presenting greater ionic radius.

Perhaps because coordination sites have limited mobility - bigger cation can access more of them?


text:
Is it appropriate to call an element an amphoteric substance?
In most of Japanese high-school textbook it is written that Al,Zn,Sn and Pb are amphoteric element(in Japanese 両性元素）because they react both with strong acid and base.
For example Al reacts like below.
$\ce{2Al +2NaOH +6H2O->2Na[Al(OH)4] +3H2}$
$\ce{2Al + 6HCl-> 2AlCl3 + 3H2}$
It is true that these are reaction with acid or base with Al, but in these reactions Al is not acting as a base or an acid. Al is acting as a reductant.
https://goldbook.iupac.org/terms/view/A00306
In above IUPAC definition, it is written that "A chemical species that behaves both as an acid and as a base is called amphoteric.".
Since Al acts not as an acid or a base but as a reductant, I thought it is inappropriate to call the Al element itself as an amphoteric substance, although its oxide and hydroxide is amphoteric. Or are these  reaction actually a kind of acid-base reactions?

They obviously use the term "amphoteric element" as an element reacting with both acid and bases. That somewhat overlaps with the element ability to form amphoteric compounds, typically  oxides and hydroxides. But many compounds that are not usually considered amphoteric, becomes amphoteric in extreme conditions.

"Amphoteric element" is definitely a term used in the past but I think its usage has become obsolete. https://pubs.acs.org/doi/pdf/10.1021/ed008p2126

You are right, there is a problem in the textbook's description. Elements are not amphoteric, it is their oxides which are amphoteric. Zinc oxide, aluminum oxide, lead oxide will all dissolve in strong bases or strong acids. In the bases, we get correpsonding anions containing the metal which are called zincate, aluminate, and plumbate respectively. In the acids, we would get their corresponding cations.

Thank you! It seems Japanese ministry of education must learn more about chemistry.

It should be specified that : An amphoter is a substance which reacts with acids and bases, provided there is no change of oxidation numbers in the reaction.


text:
How many multivalent elements are there?
According to this video there are 20 multivalent elements, and from the periodic table, I see 20, but according to multiple websites, including this, it says there are only 5. What am I not understanding? Are there 20 or 5? I want to know so I can memorize them. Also, are there any that aren't transition metals that are exceptions, i.e. non-metal multivalent? If so, may you please list them in your response?
Again talking about multivalent elements because I am learning ionic compounds, and these are part of it. Also, please keep high-level vocabulary to a minimum, I know some science vocabulary but i am only in grade 10.
Edit: I would like to add that i am referring to how many metals have roman numbers, i.e like Fe which has 2+ and 3+ charge both. are there more than 5 of these or are there 20?
To anyone with the same question: this article (thanks to Achem) and the article listed here article are very useful.

You should not memorize all the periodic element's oxidation states randomly. Nobody will expect you to know all the elements. Learn the valency common elements which are mentioned in your grade-10 books. It is quite easy to see a pattern. Group I elements will exist as (+), group II elements (2+), etc. The transition elements show multiple valencies.

Im aware of that just ask extra pratice. could you please answer my question though? I know the (+) and (-) tendencies aswell

Both videos/websites are wrong. There are more than 20 elements which have multiple valency. All transition metals will show multivalency. All lanthanide and actinide elements show multivalency. Count how many transition elements are there in the periodic table. It is certainly more than 20.

no!!! That's not what i meant. I meant how many transition metals are there, that have more than one roman numeral sign / charge like Fe which has 2+ and 3+ i.e Iron(11) or Iron(1). How many metals have that, could you please list them so i can write them down as notes. I know that many transition metals have more than one atom, im just wondering for the roman numeral in specific

This is exactly what I am trying to tell you, there are about 38 transition elements and they will show multivalency. So both websites are wrong. There are more than 20 multivalent elements. Fe(III) and $\ce{Fe^{3}}$ is the same thing.

So there are 38 elements that have roman numerals or more than one charge like Fe(III) and Fe3 Please confirm, that is what your saying

Read this link: https://www.americanchemistry.com/chemistry-in-america/news-trends/blog-post/2018/discover-the-world-of-transition-metals#:~:text=There%20are%20total%20of%2038,%2C%20Manganese%2C%20Zinc%20and%20Mercury.

Thanks for helping me so much!!! I didnt know that the 38 transition elements all had roman numeral charges like that. Guess i aint memerozing or writing them down lol

You are not supposed to memorize those 38 elements.

Elements with a single valency are the minority.

Valency is not always a clear-cut concept in chemistry. Also, beyond all the d and f elements (which Achem mentioned), many p elements and even H is multivalent (H- is clearly different than H+).

I'm surprised those fraudulent papers saying there are only 20 or 5 multivalent elements got published—illiteracy to the highest level.

The elements that aren't multivalent in my understanding are alkali metals, alkaline earth metals, fluorine, and noble gases. Correct me if I am wrong.

@BruhMoments All the alkali metals except Li have at least M+ and M- to my knowledge (https://en.wikipedia.org/wiki/Alkalide , https://chemistry.stackexchange.com/questions/109846/unusual-oxidation-states-of-alkali-metals/109854#109854). Also to my knowledge the alkaline earths except Be form monovalent compounds involving M-M bonds (https://pubs.rsc.org/en/content/articlelanding/2011/dt/c0dt01831g#!divAbstract). Ar, Kr, Xe and Rn all form well defined compounds (https://en.wikipedia.org/wiki/Noble_gas_compound), Xe and Rn in multiple oxidation states.

Almost all of them, maybe?

haha i was so dumb back thrn! Thanks to everyone who assisted me


text:
Converting structures into Fisher projections
I am really struggling with converting pyrano/furanosides into their Fisher projection counterparts. Especially when they are drawn like in the shown projection. When do you know to connect the anomeric carbon? Do you always need to show the connection like in the attached photo?
With this in mind, please could someone walk me through how to draw isopropyl beta-D-mannopyranoside or a similar structure as a fisher projection ? Or point me towards a resource that could help?

Who said these two representations are equivelent?

My apologies I should have been more clear. I am not saying these are equivalent. I am giving an example of how a pyranoside is drawn occasionally and asking how do you know when to draw it in that form vs its acyclic? Then asking for help applying that reasoning to converting isopropyl-d-mannopyranoside into a fischer

First of all, mannose is an aldohexose. Your compound is a pentose, which has two anomeric carbons. You may need to get the correct structure for isopropyl $\beta$-D-mannopyranoside (is this the correct name?).

Perhaps the first thing is to know the person's name. It is Emil Fischer. In the left hand structure, all OH's ar equatorial except for the one at C2, which is axial. You need to work on stereochemistry in 6-membered rings and carbohydrates. I'll post some useful links in a separate Comment. Oh! Welcome to ChemSE!

Abigail: Try these links. DrawingCyclohexanes:
http://ursula.chem.yale.edu/~chem220/chem220js/StudyAids.html#Conformation
http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/cyclohexane.pdf
Carbohydrates:
http://ursula.chem.yale.edu/~chem220/chem220js/StudyAids.html#carbohydrates

This page gives a satisfactory structure for Isopropyl $\β$-D-mannopyranoside with correct number of carbons.

Thank you both for these very helpful resources! I feel much better equipped to approach this problem, and other similar problems in the future!

Abigail: I think that you should have enough information from the links (vide infra) I provided to get you through most of what bothers you. It takes time to get these issues straight. Structures 1 represent the isopropoxy pyranosides of D-glucose ($\ce{C5}$-(R)). Assuming a chair conformation in 1a, all of the groups alternate, i.e.,up,down,up, etc., around the ring. All of the substituents are equatorial. Structure 1b is the chair conformation of 1a. The isopropoxy group is in the β-configuration, meaning that it is above the plane of the ring as drawn in 1a and 1b. The Fischer projection of this D-glucopyranoside is shown in structure 1c. By convention, the isopropoxy group of the β-configuration is drawn on the left side of the Fischer projection.
As @Mathew has noted, your left hand structure is a pentose. But if the hemiacetal OH were replaced by -$\ce{CH2OH}$ of the same configuration, then structures 2 would be the same as structures 1 except for the configuration at $\ce{C2}$. The $\ce{C2}$ epimer of D-glucose is D-mannose wherein the -OH group at $\ce{C2}$ is axial.

Now your right hand structure has some issues. It is a heptapyranoside as shown in structure 3a given the configurations in your structure with the addition of the configuration at the anomeric carbon based upon what your structure displays. Structure 3b is the chair conformation of 3a. Again, by convention, when the methoxy group is drawn on the right of the Fischer projection, it means that the  methoxy group has the α-configuration. In keeping with the Fischer convention, structure 3c is the correct representation with the correct configurations.

Drawing Cyclohexanes:
http://ursula.chem.yale.edu/~chem220/chem220js/StudyAids.html#Conformation
http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/cyclohexane.pdf
Carbohydrates:
http://ursula.chem.yale.edu/~chem220/chem220js/StudyAids.html#carbohydrates

@Abigail if this answer helped you, please accept the answer by clicking on the checkmark.


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How are photons converted into thermal energy?
I know when light strikes an object it excites electrons into higher energy states. I know thermal energy is basically just a lot of atomic and molecular kinetic energy. Does the excitation of the electron when struck by light cause a molecule to vibrate or make atoms move faster?

Atoms do not vibrate. Atomic bonds or molecules do. Electron excitation is part of thermal energy and redistributes toward translation, vibration and rotation energy, respecting quantisation discrimination.

Quantum excited states, rotational, vibrational, electronic, are heat content only when in thermal equilibrium. Absorption of radiation does not raise the temperature until the excited states deactivate by collisions, emission etc. transferring the quantized energy into random velocity of molecules and thermally equilibrated quantized states.

When light hits an object, different things can happen. The light can be reflected or absorbed. When visible light is absorbed, it will usually excite electronic states.
This excitation typically does not last long. As Poutnik mentioned in the comments, the excitation energy can dissipate in the form of vibration, rotation or translation. It can also be emitted or partially emitted, either quickly (fluorescence) or slowly (phosphorescence). The energy can be transferred to another electron (exciton transfer) or result in breaking a bond (photochemistry).
When the energy ends up as thermal energy, the mechanism of dissipation is described as coupling (e.g. vibrionic coupling) or non-radiative decay. I am not aware of a simple conceptual explanation of how this happens other than a coupled oscillator.


text:
Does carbon dioxide come out of solution on freezing?
Years ago in a lab where I was working someone had the excellent idea of cooling down their can of coke by pouring liquid nitrogen over it. The can exploded and resulted in quite a mess.
My question is twofold:
(a) Does $\ce{CO2}$ really come out of aqueous solution when water freezes?
(b) If so, why and under what conditions?
I would have expected it to form some kind of clathrate.

It would be combination of expulsion of dissoved CO2 and volume expansion of ice. The clathrate seems to need high pressure > 45 bar.

@Poutnik But why would the CO2 be expelled? And given that ice expands, surely there would be more space for CO2?

Space inside the ice structure is unsuitable for CO2. Very few things are soluble in solids.

@IvanNeretin Clathrates

The problem is to know the maximum value of the internal pressure that can exist inside the can before it explodes. Who knows its value ?

@DirkBruere They are not solutions, though. They are different solids with different crystal structure.

Wikipedia - CO2 hydrate/clathrate-  phase diagram

What happens is more a kinetic phenomenon than a thermodynamic one, and more a physical reaction than a chemical one. The carbon dioxide is first introduced into the drink under several atmospheres pressure, so it becomes a supersaturated solution. When the water is cooled and forms ice crystals, the ice crystals act as nuclei for the carbon dioxide gas to  "boil". This converts its latent potential (fugacity) to actual pressure, which explodes the can.

I suspect that the can explodes from the ice expansion rather than gas pressure? I'm fairly certain that would be the case if the freezing were done slowly. Not sure about a flash freeze. Similarly, in a slow freeze, the growing ice crystals (100% H2O) cause the concentration of all solutes to increase in the remaining liquid volume, thus increasing the equilibrium CO2 pressure in the gas phase. But again, I'm not sure what happens in a flash freeze where rapid amorphous solid formation might be more common than actual crystal growth.

Usually there is free space above the liquid in a soda can, so ice expansion has decreased effect

Hmmm. It becomes supersaturated after the can ruptures or when enough ice is formed. Cooling down a closed can before freezing makes it undersaturated as solubility increases.

It is supersaturated from the start. The carbon dioxide gas is introduced under pressure. Also note what happens when a chilled carbonated dring is poured onto ice (without the can).

A CO2 bubble would be even better centre than ice particle, if it was supersaturated. But it is in equilibrium at the start at given pressure when closed, as gas solubility is proportional to gas pressure. Pouring is very different situation. It gets supersaturated when the can or bottle of carbonated drink gets opened, because pressure and therefore solubility suddenly drops.

Perhaps we have different Q understanding. I think the can was closed, you may think the can was open. Otherwise I have no other explaination for your evaluation.


text:
Are there any aromatic systems that don't contain any rings?
I'm curious if there are any aromatic systems that don't contain any rings. I just wonder and want to know if there is any exception to the aromatic system.
I know the definition of aromaticity is that it must have a ring but my friend told me there are some aromatic molecules that don't have a ring.

The definition of an aromatic system is that it has one or more rings

That depends on what kind of aromaticity you'd talk about. What is commonly know as aromaticity, precludes something like that. That are some extensions of the concept on other systems, though.

https://chemistry.stackexchange.com/questions/32354/what-is-y-aromaticity-is-the-trinitromethanide-anion-aromatic

The guanidinium cation wants to be aromatic, definitions be darned!

The word "aromatic" is most commonly used for systems that do have rings, but an enhanced resonance stabilization with delocalized $\pi$ bonding can occur without rings whether we apply the label "aromatic" or not.
Carbon dioxide is actually the first of a series of linear molecules having the formula $\ce{C_nO2}$ or $\ce{O=C=C=...=C=O}$, in which all the pairs of neighboring atoms are joined by double bonds and thus we have a linear molecule.
If we draw out the molecular orbitals of this structure we discover that the molecule will have a closed-shell structure when there are $4n$ $\pi$ electrons (not $4n+2$), and carbon dioxide with eight ($n=2$) has all the hallmarks of a system with extra $\pi$ electron stabilization and delocalization (see here for a discussion). Similarly, the $12\pi$ member carbon suboxide, $\ce{C3O2}$, formally has cumulated double bonds: but its carbon-carbon and carbon-oxygen bond lengths are shorter than those in ethylene and formaldehyde respectively. Carbon suboxide is prone to polymerization but can be kept as molecular $\ce{C3O2}$ under carefully controlled, dark conditions. By contrast, ethylene dione, $\ce{C2O2}$, with an unfavorable electron count for forming a closed shell, decomposes on a nanosecond time scale.

Resonance isn't, in general, enough for aromaticity. Did someone invent something like Y-aromaticity, but for cumulenes?

@mithoron I haven't quite figured out the difference between enhancement of resonance stabilization with an optimal number of electrons but reduced stabilization with a wrong electron count, and aromaticity. Do tell.

@mithoron how about this: a chain of atoms has linear aromaticity if it (1) is cumulatively pi-bonded, (2) is linear or nearly so, (3) has all atoms in the chain sp hybridized, (4) contains 4n cumulated π electrons. The most stable linear carbon dioxides, plus some other elatively stable cumulated species such as nitrous oxide or allenide ion, all follow these rules. Makes as much sense as Y aromaticity to me.

Oh, well, you didn't really call something like that aromatic in the answer, so it's just that it is kinda a propos post, I guess. And yes, some Y-aromaticity makes about as much sense. You could expand the answer though.


text:
Where do lactone groups occur in graphene oxides and how are they identified using FTIR measurements?
Usually, the literature (for example, Serodre et al. [1]) shows the occurrence of lactones in graphene oxides through potentiometric titration measurements. In the nanosheets, do these groups occur in defect regions (such as carboxylic acids and carbonyls) or in the basal plane (such as hydroxyls and epoxide groups)? Furthermore, in the FTIR measurements, how can we identify lactones?
Reference

Serodre, T.; Oliveira, N.; Miquita, D.; Ferreira, M.; Santos, A.; Resende, V.; Furtado, C. Surface Silanization of Graphene Oxide Under Mild Reaction Conditions. J. Braz. Chem. Soc. 2019. DOI: 10.21577/0103-5053.20190167.


text:
Oxidation of Methanol - how to write the half reaction
I am trying to write the half reaction equations for:
$$\ce{CH3OH + O2 -> HCOOH + H2O}$$
The primary goal for me here is to find out which reactant is oxidising without using prior knowledge of alcohol reactions.
So I wrote
$$\ce{CH3OH -> HCOOH}$$
and balanced the non hydrogens and non oxygens.
I then added $\ce{H2O}$ and $\ce{O}$ to the sides appropriately
I calculated the charge of the alcohol by adding up the individual ionic charges of each atom in the molecule
Each side ended up with the equal charge of -2.
However I had no need to add any electrons into the equation to balance it out.
So how can I figure out whether it is oxidising if it doesn't indicate electron transfer? Is it some fundamental mistake in my method? Also, do alcohols have 0 net charge when dissolved in water?

If you don't respect oxygen's O's, chemists will be upset. See if this helps for starting: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Chemical_Reactivity/Balancing_Organic_Redox_Reactions

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Keep CH SE Q titles in plain text.

This problem is not as easy as it looks, because there are two sorts of oxygen atoms. Some change their oxidation numbers, and some not. The first half-reaction is an oxidation of the reducing agent $\ce{CH3OH}$ which produces $\ce{HCOOH}$ on the right-hand side. The second half-reaction is a reduction of the oxidizing agent $\ce{O2}$ which produces $\ce{H2O}$ on the right-hand side. The two half-equations are : $$\ce{CH3OH + H2O -> HCOOH + 4 H+ + 4 e-}$$ $$\ce{O2 + 4 H+ + 4 e- -> 2 H2O}$$If you make the sum of these two half-equations, the $4$ e- and the $4$ $\ce{H+}$ disappear, and the final equation is : $$\ce{CH3OH + O2 -> HCOOH + H2O}$$


text:
Rationalisation of stabilising dispersion interactions in HgS via QM
It is well-known that mercury(II) sulfide, HgS, is extremely inert- for example, it has one of the lowest solubilities of nearly all known compounds in water(check Wikipedia). There are two well-known ways to rationalise this interness; one is HSAB theory, in this case that the LUMO of the mercury(II) cation is relatively low (soft acid) while the HOMO of the sulfide anion is relatively high (soft base), and the other is that the core 5d electrons of Hg(II) are relatively high in energy, sufficiently high enough to "dispersively interact" with the diffuse electrons of S(-II).
However, the conventional quantum chemical method CCSD(T), and to a lesser extent DFT, do not explain the latter in a satisfactory manner. All the former does is to include the excited states of HgS, and the crucial functional of the latter is not even known in closed form.
My question now follows- how does one explain the dispersion interaction of (say) two filled orbitals via excited-state-including, so-called "correlated" wavefunctions methods, and/or via the exact DFT functional?


text:
Can an equilibrium be reached when there is an excess amount of one reactant over the other reactant?
For example, consider a net reaction between ferrous nitrate and potassium thiocyanate:
$$\ce{Fe^3+(aq) + SCN-(aq) <=> (SCN)^2+(aq)}$$
Can an equilibrium be reached if there is a much greater excess amount of one reactant over the other:
$$[\ce{SCN-_{(aq)}}] \ll [\ce{Fe^3+_{(aq)}}],$$
or
$$[\ce{SCN-_{(aq)}}] \gg [\ce{Fe^3+_{(aq)}}]$$?
If not, is it okay to consider it as an irreversible reaction in these cases?

Equilibrium constant when adding more of a reactant

Does exist a solution for $$K=\frac{x }{ (a-x)(b-x)}$$ for the real $K \gt 0$ and  real $a \gg b \gt x \gt 0$? Yes, it does. // You must mean ferric/iron(III) nitrate, not ferrous/iron(II) nitrate.

Whatever the initial and the final amounts of reactants and products, the formation of $\ce{[Fe(SCN)]^{2+}}$ ion is always an equilibrium. If a strong complexing solution, like EDTA, is added to the red solution containing $\ce{[Fe(SCN)]^{2+}}$ ion, whatever the relative amounts of the remaining $\ce{Fe^{3+}}$ and $\ce{SCN-}$ ions, the red color disappears, which means the formation reaction is reversible.

Not necessarily reversible the complex is destroyed. Reversibility requires the same mechanism in both directions.

The reversible reaction is $\ce{Fe^{3+} + SCN- <-> [Fe(SCN)]^{2+}}$. It does not imply that the formation of the complex of $\ce{Fe}$ with EDTA is reversible.


text:
Which one reacts first with NaOH; Triglycerides or FFA?
I want to saponify Crude Palm Oil (CPO) with NaOH. The CPO itself contains 97,35% of triglycerides and 2,5% of free fatty acids. Which one reacts first with NaOH?

The free acids will react first with the NaOH to form the salt (soap)

If there is still some FFA, there is no NaOH yet. If there is already some NaOH, there is no FFA anymore and NaOH can finally start the long awaited and slow saponifying.

Triglycerides require nucleophilic displacement of the glycerin component, whereas FFAs require only deprotonation. The latter is much faster.


text:
Hund’s Rule - Please help explain below using Hunds rule?
Why is it easier to excite a 2s electron in a carbon atom to a 2p orbital than it would be to do the same in nitrogen?

No own attempt to answer it?

Yeah I do have my own attempt, obviously Hunds rule states that max multiplicity = max unpaired electrons = lowest energy. In carbons case you have 1s2 2s2 2p2, with the electrons in the p orbitals occupying 2px and 2py respectively, with up spin, maximising multiplicity. This is not the case for nitrogen (2p3, paired electrons,higher energy). Meaning that the energy required to excite an electron in the carbon atom from 2s to 2p would be less than it would be for nitrogen, in regards to hunds rule

I just wanted to get a rough idea of other peoples takes on it, note the question I’m asked is centred around hunds rule

The question is too simple, because nitrogen's 2s electrons have higher Zeff. Comparing carbon and nitrogen: when you increase 1 proton and 1 electon, the electron has some shiending effect, so while the effective positive charge increases 1, the effective negative charge can't reach 1. In conclusion, it's easier to promote 2s>2p in carbon than in nitrogen atoms.
Also, this video might help you:
https://www.youtube.com/watch?v=dXOyNaxhc4A&list=PLavaRHHaRimXdmzRB21Ew9abP66YvxKzh

There's a missing step in your logic: ionisation energy corresponds to the removal of the outermost electron (which is 2p in both cases), but here OP is asking for 2s -> 2p promotion.

Oh sorry, I didn't read it well.

The question I’ve been asked at university is asking to explain it using Hund’s Rule


text:
Getting magnesium chloride gunk off a wood shelf
I have a situation.  Some magnesium bath salts got left on a wood shelf in my bathroom in a bag that was not closed.  Being hydroscopic, it sucked up a ton of water, and then spilled onto the wood.  Of course I can try to dry it off but it never will really, and it will just suck in more water later.
Is there something I can do to react the magnesium chloride with a relatively harmless chemical to make it no longer hydroscopic, and hopefully then I can clean it up?

warm water will dissolve it completely so you can mop up the solution, wash the wood down with more water to ensure it is all removed.

I thought about dilution, but it's a large area, inside a cabinet.  My last resort I guess.

Please chose a meaningful title. Please do not repeat tags within the title. In this case, you should also read the tag description; applied-chemistry is misapplied (pun intended) in this case.

"Chemical processes and theories being used in practical and experimental applications"   Getting gunk off my shelf isn't a practical application?

I think the use of "applied chemistry" here is intended to be as one would use the term in Department of Applied Chemistry, in which case, no, it would not apply (pun intended) to your question. For your use case, I recommend "everyday-chemistry".

Converting it to something else would lead to either another mess either cleanup up anyway.

You've raised a valid point. We should probably work on the description of this tag. I apologize for being harsh with my words; it really isn't that obvious.

Here is one possible solution that first should be tested on a small area as it may damage the wood imbued with magnesium chloride.
The proposed path to removing $\ce{MgCl2}$ is based on the reaction between aqueous $\ce{MgCl2}$  (or $\ce{MgSO4}$) with some dissolved $\ce{Na2CO3}$ that is commonly available as Washing Soda. The resulting created so-called Magnesium Alba salt can be physically removed followed by treatment with an acid (lemon juice should suffice).
I also recommended this reaction as a quite surprising demonstration as it forms a massive water-loving basic magnesium carbonate hydrate that may actually consume all the water in your vessel! The further addition of the acid results in its complete dissolving.
Here is a historical reference from Atomistry.com, to quote, in part:

Magnesia alba is commercially prepared as a white powder by precipitating magnesium sulphate or chloride with sodium carbonate. Magnesia alba levis results from precipitation in the cold, and magnesia alba ponderosa from precipitation in the heat. Both the light, bulky form and the heavier variety are used as laxatives in medicine, in dentistry, and for toilet purposes. They ignite respectively to the light and heavy oxides of magnesium. The light carbonate, " when examined under the microscope, is found to consist of amorphous particles with numerous slender prisms intermixed.

And further:

Magnesia alba is only slightly soluble in water (the solution has an alkaline reaction), but is soluble in acids and in solutions of ammonium salts...Its composition is usually said to approximate to that of hydromagnesite, Mg(OH)2.3MgCO3.3H2O, but analyses of its composition were never very consistent, and it seems quite certain that its composition is very variable.

I hope this helps but, at least, it should make an interesting experiment.


text:
How to minimize sample dispersion during spotting for paper electrophoresis?
I am separating highly polar amino acids and their derivatives using paper electrophoresis. This requires pre-wetting a piece of filter paper with electrophoresis buffer (in this case, carbonate-bicarbonate buffer) and depositing spots of samples on the wet paper.
When I do so, the molecules in the samples instantly disperse on the wet paper, and form fairly wide spots (wider than on dry silica/cellulose TLC). This makes it hard to separate closely related molecules.
How can I minimize the dispersion of the molecules during the spotting step on wet paper?
I tried to make the samples slightly viscous by adding large amounts of poly(ethylene glycol) or glycerol, but it did not seem to make any difference in terms of spotting size. Concentrating the samples to reduce the loaded volume is not a viable solution, as I am already spotting very small volumes (1-2$\mu$L). Are there tricks used for example in TLC that could apply there?

The classic way is to start with a dry stationary phase, spot a small volume, let the solvent evaporate, spot another small volume. This does not work if the protocol calls for wet paper. Another way is to minimize the volume applied while maximizing the sensitivity of detection (i.e. not look for color in daylight but fluorescence under UV light, or use a specific dye to stain after you run and dry the TLC).


text:
Why did The Action Lab say that solutes that don't dissolve as well in hot water are exothermic?
Somewhat recently, The Action Lab made a video called "Why does this powder only dissolve in cold water?" , saying the powder which was calcium acetate dissolves in cold water better than in warm water.
He also said that gases get less soluble in hot water than in cold water, which is true, by demonstrating it with a cold and hot soda can, noticing that the hot soda can bubbled up more violently than the cold one.
But the real point is what he said:

"All exothermic reactions slow down as it gets hotter."

A video reacting to The Action Lab from That Chemist, in which he says :

"His rationale in the video-which is wrong is that because it's exothermic when it's dissolved, if you add heat it becomes less soluble."

Sodium hydroxide when dissolving in water is exothermic, but it still gets more soluble as it gets hotter.
And not only that, but he had also said that dissolving sugar in water was a chemical change, even though it is strictly physical.
In my opinion, what I think That Chemist said was true, but correct me below.

The action Lab is wrong. The reason is that his knowledge of chemistry is weak and he leapt to a plausible conclusion based on too few examples (and din't check his conclusion by asking people with better chemistry knowledge).

okay, I'll bite. They're both wrong.
The Action lab is specifically incorrect in conflating kinetics (reaction rate) with thermodynamics (equilibrium). Pretty much all reactions go faster with higher temp, but in the case of reversible exothermic reactions, the reverse reaction increases in rate more quickly than the forward reaction, so the final equilibrium state has more reactant remaining.
Where Action Lab is correct and The Chemist is wrong is in the application of Le Chatelier's principle to heat. The van't Hoff equation shows clearly that the change in the equilibrium constant for a reaction with temperature is dependent on the enthalpy change of the reaction (and nothing else):
$\ln {\frac {K_{2}}{K_{1}}}={\frac {\Delta _{r}H^{\ominus }}{R}}\left({\frac {1}{T_{1}}}-{\frac {1}{T_{2}}}\right)$
So if $\Delta_r H^\circ < 0$ and $T_2 > T_1$, then $K_2 < K_1$.
Where The Chemist is correct is when he points out that the rection can change as the temperature increases, which can affect this result. In the van't Hoff equation, we've assumed that $\Delta_r H^\circ$ is constant over the temperature range of interest. This isn't always true, but in most cases is a good approximation.
Strangely, after pointing this out, he makes exactly that mistake himself. In his counterexample of NaOH increasing in solubility with temperature, he wrongly assumes that the published enthalpy of solution of NaOH is the relevant value in his experiment. That value unfortunately is not applicable to this situation.
For any compound the "enthalpy of solution" is defined for the case of infinite dilution. So for NaOH, that means adding solid NaOH into pure neutral water in such a large volume that the added NaOH does not appreciably change the pH. In that case, there are two reactions going on:
(1) The separation of the NaOH lattice and hydration of $\ce{Na+}$ and $\ce{HO-}$ ions: $$\ce{NaOH(s) <=> Na+(aq) + HO-(aq)}$$
This reaction actually has a positive enthalpy change.
(2) The reaction of hydroxide ion with hydronium ions: $$\ce{HO- + H3O+ <=> 2H2O}$$
This reaction has a large negative enthalpy change.
In the case of infinite dilution (pH=7), both reactions are occurring, so the large negative enthalpy change of (2) results in a net negative enthalpy change, ie the reaction is exothermic.
However, when we get to the point that the solution is saturated, the pH is extremely high and there is very little hydronium left to react, so reaction (2) contributes much less, and the positive enthalpy change of reaction (1) dominates, making the reaction endothermic. (More accurately, the amount of hydronium is so low relative to hydroxide that the decrease in concentration of hydronium necessary to restore Kw for a given amount of added hydroxide is relatively very small.) Since the maximum solubility is determined by the behavior of a saturated solution, it is this enthalpy change that determines the change in solubility with temperature. That means that NaOH exhibits the expected behavior of an endothermic salt dissolution, ie increasing solubility with temperature.


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How do salts affect pH?
Since pH measures the effective proton concentration, it is the concentration of a very specific type of cation. I know if salt reacts with water, it will affect pH. But my question is, how does the presence of salt affect the pH in general in aqueous solvents, especially for, say, a protein where there are many titratable protons? Can pH and ion concentrations be considered to be totally decoupled from one another in such cases?

See ionic strength,  activity coefficient, Debye-Hückel equation and pH


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Will potassium nitrate and petroleum jelly work as a rocket propellant
I saw a post on a online forum that potassium nitrate and petroleum jelly could be mixed to create a rocket propellant. What is the reaction that would take place?

Yes, a bit crude as a rocket propellant, but it works as igniting the petroleum jelly in air leads to the associated thermal decomposition of the potassium nitrate liberating oxygen:

$\ce{2KNO3 (s) → 2KNO2(s) + O2 (g)}$

where the generated oxygen accelerates the burning.
The petroleum jelly itself is not particularly volatile, so the dangers of an explosion are low.
In general, however, I would recommend exercising considerable care when working with a fuel and oxidizers mix.

@ AJKOER. 1) Volatility and risk of explosion are not necessarily related. 2) your equation corresponds to what happens when heating pure $\ce{KNO3}$. When mixed with carbon containing substances $\ce{KNO3}$ looses all its oxygen atoms. For example, the famous black powder is a mixture $\ce{KNO3 + C + S}$ that reacts according to $$\ce{2KNO3  +  3 C + S -> K2S + 3 CO2 + N2}$$

Good points Maurice, hence my cautionary note on handling such compositions, especially in larger amounts.


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Which is more acidic : 2,6- dihydroxy benzoic acid or 2-hydroxy benzoic acid?
This post address the acidity of ortho and para hydroxy benzoic acid. However, what effect is there if there are two hydroxy groups ortho to carboxylic group?
I know hydrogen bonding is also at work but won't the two ortho substituents would decrease the acidity by highly destabilising the conjugate base due to steric hindrance and repulsion; as there are three big groups near to each other?
[P.S.:My worksheet claims that ortho di-hydroxy benzoic acid is more acidic(Option-A is correct in below figure). Also, since about the mono-hydroxy substituted benzoic acids are discussed already on the site, so my question is regarding between only  di- and mono- hydroxy benzoic acids.)

Thanks!

The highly electronegative O atom of the carboxyl group (which by itself in an aliphatic compound would be highly stabilised by equivalent resonating structures) forms very efficient H-bonds with the fairly acidic phenolic hydrogen atoms. The ring formation provides enough stability to favour release of a proton.

The increased acidity can also be explained by the First Ortho Effect which basically states that steric hindrance forces the carboxyl group to move out of the plane of the ring which essentially stops resonance with the phenyl ring, thus inhibiting the slightly destabilising crossed-conjugation:

P.S. I could not find a suitable source for the $\ce{pK_a}$ of $2,6$-dihydroxybenzoic acid. I would be grateful if someone adds them in the comments.

But sir I asked for 2,6-dihydroxy benzene mainly because mono hydroxy acids are already been asked. Anyway, I hope you will expand your answer later . Thanks!

What exactly would you want me to add?

That why dihydroxy one is more acidic than mono-hydroxy one.

2,6 -dihydroxybenzoic has a pKa of 1.64 (computed by MarvinSketch)

"The increased acidity can also be explained by the First Ortho Effect ". Isn't the A value of the hydroxyl group less?


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Why an irreversible isobaric process won’t satisfy the above given equation in the image above

Here, my doubt is , why is option D incorrect?( the ans  key says A,B,C )...
in isobaric process, the work done is same for both reversible and irreversible processes, then why isnt D correct as well?)

Welcome to CH SE site!
Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of at least essential parts.
Suitable formatting can be done according to formatting math/chem expressions/equations.

Why do you think the term in parenthesis describes the external pressure in an irreversible process?

That doesn't look like constant pressure for me...

In an irreversible process, the equation of state for the gas is not satisfied (due to the presence of viscous stresses).  The equation of state is valid only for thermodynamic equilibrium or for a reversible process.  So use of the VDW equation of state for the pressure is not valid for an irreversible process.


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Can a weak acid ever fully dissociate?
From what I understand, the pH of the solution the weak acid is added to determines the extent that the weak acid will dissociate. But would there be a pH which would allow the weak acid to fully dissociate?

In mathematical sense (no molecule left)? Or in scientific sense (too little left in given context to be detected or to matter) ? // Note that even strong acids do not fully dissociate in the former sense, as their Ka is very high, but not infinite.

A weak acid is in equilibrium as $\ce{HA + H2O \rightleftharpoons  A^- + H3O^+}$ so the position of equilibrium can be altered by adding or reducing acid or base concentrations just as for any equilibrium. If you want close to 100% dissociation, probably unlikely but 80% may be possible depending on equilibrium constant and realistic concentrations. What is possible depends on the weak acid used and you will have to calculate this.

At what temperature? At sufficiently high temperature, everything can be considered dissociated.

Not even "strong"  acid dissociate all that completely https://chemistry.stackexchange.com/questions/143459/do-strong-acids-actually-dissociate-completely

If the acid is not too weak, we can come close to full dissociation by making it very dilute. See https://chemistry.stackexchange.com/questions/118826/strength-of-bases-and-acids/145644#145644.

Strictly speaking, no acid is fully ionized in water solutions. Even strong acid solution has traces of a not dissociated forms, as their acidity constants are very high, but not infinite.
For all practical purposes, we consider the acid as fully dissociated if the not dissociated form is negligible for given purpose or context.
All strong acids can be considered fully dissociated in not too concentrated water solutions. It is obvious the solution with molar ratio $\ce{2 H2SO4}$ : $\ce{1 H2O}$ cannot be fully dissociated, as a part of the acids has no water available to be protonated.
For weak acids, if there is required e.g. 99.9% (99.99%) of dissociated form, the requirement is fullfilled at $\mathrm{pH} \ge \mathrm{p}K_\mathrm{a} + 3(4)$

Yes there would, simply add strong base in sufficient quantity.  The pH curve will show the weak acid buffering until it is fully depleted, then a sharp rise in pH.
This would be under the assumption that the weak acid and its deprotonated form are both fully soluble.  This is essentially how soap is made from fatty acids.

pH = pKa + log (salt/acid)

One can see as the strong base depletes the weak acid, pH becomes the negative log of the (proton) dissociation constant of either the solvent or the protonated form of the strong base!
The former weak acid simply becomes a salt in that solution.

Even if you add that much base it's still not completely deprotonated. And if you add enough to make it so, it's not an acid but a salt in a solution of the base.

@Mithoron "even when you add that much base"?.  It's not quantity, it's strength.  The concentration of acid form is essentially zero, especially considering solvent effects, once sufficient strong base is added.

In water an acidic solution contains a [H3O+] > sqrt Kw so any weak acid no matter how dilute will not be 100% ionized. The product of hydronium ion and conjugate base will always be too high. To make a weak acid strong a more basic solvent is necessary. [Please ignore for now the small amounts remaining because Keq is not infinity.]

@jimchmst One can look at it this way, adding strong base to the solvent changes the pKa of the acid, reducing its concentration to essentially zero.

Did I not say a more basic solvent?


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Arranging the following elements in the increasing order of their oxidising power
I was given the following equations and asked to arrange them in increasing order of their oxidizing power:

$\ce{Co^{+3} + e- -> Co^{+2}}$ with an electrode potential of $+1.81\, \mathrm{V}$

$\ce{Pb^{+4} + 2e^{-} -> Pb^{+2}}$, with an electrode potential of $1.67\, \mathrm{V}$

$\ce{Ce^{+4} + e^{-} -> Ce^{+3}}$ ,
with an electrode potential of $1.61\, \mathrm{V}$

$\ce{Bi^{3+} + 3e^{-} ->Bi}$. with an electrode potential of  $0.20\, \mathrm{V}$

as electrode potential is the same as reducing potential,  the greater a substance's reduction potential, the lesser its oxidsing power.
So I Got: $$ \ce{Bi^{+3} > Ce^{+4} > Pb^{+4} > Co^{+3}}$$
however, my book says that this is wrong. Why is this wrong?

What does your book say? Your order is correct. Co(III) is a powerful oxidizer and bismuth is almost nil.

my bad, I wrote it the other way round, why is that wrong? Why is this order correct?

Now that you have edited the order, then you have to re-visit the concept of electrode potentials and its relation to oxidizing power. It will be a 30 minute lecture, I were to explain it. And also learn the use of the symbols < and > properly. In short, re-visit the textbook explanation.

A higher reduction potential correlates to higher oxidation power. Think about a redox reaction: one of the reagents gets reduced (gains electrons), and the other gets oxidized (loses electrons). The more one reagent favours reduction (higher electrode potential), the more strongly that reagent can act as an oxidizing agent on another reagent.


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Erythritol oxidation using acidified KMnO4
I have recently experimented with erythritol oxidation my primary aim was to make glyoxal. However regarding the nature and structure of erythritol several other side-products were also observed one of them which I have confirmed to be oxalic acid using FeCl3 solution which precipitated ferric oxalate which was kind of dirty green in color. carbon dioxide was also seen to be evolved which was most probably from the decarboxylation of beta-keto acid which on paper seems to be a non-planar derivative of dioxobutanedioic acid

however due to the nature of the compound erythritol several other oxidation products are also possible which among which glyoxal seems to be present in minimal or negligible amounts

notice the structure of the compound which has bonds which extend both inside and outside the plane.  My question is what other compounds can be formed in the reaction? and if possible how to detect them and is it possible to make and extract a compound like dioxobutanedioic acid or glyoxal from erythritol oxidation with common oxidizing agents found in a lab setting.

Dioxobutanedioic acid would be hydrated.

i think that it would be decarboxylated as its a beta keto acid but please elaborate what you mean

:/ I you got it, it would be in gem-diol form, like glyoxal itself.

i have anticipated that but there is a large possibility of formation of byproducts like oxalic acid which i have confirmed to be present in solution since i want a fairly pure solution of glyoxal i wanna know about other byproducts


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Can metals have both Fracture Toughness and Elasticity properties?
Fracture Toughness is the resistance of a material to cracking and Elasticity is the ability to bend without breaking. I want to know if there is any metal or metal alloy that can easily bend without breaking and still stay strong against force coming towards it and even other materials colliding with it. This could be useful to make safety equipment which protect people and are still not too uncomfortable. (For example a helmet which protects a builder even when sharp bricks fall on it and bends but prevents damage and does not break meaning that sharp pieces wont go inside the person's head, or another example is a place where people can go when there is a natural disaster to stay safe and because of it not fully bending the people who rescue the survivors can see that it is not fully bent meaning that there is someone inside).

The bending can be useful for things like shaping it back together when it is deformed and so it could also be used for armour which would not take hours to wear.

What I am saying is that the material only bends and deforms instead of cracking but still absorbs and stops any force that could cause something behind it to be damaged.

To what extent? Without actual values of what you consider "fracture tough" and "elastic", the question has no meaning.

Ask a swordsmith?

There are plenty of varieties of steel (alloys of iron) that combine fracture toughness with a decent amount of elasticity. The control of these properties involve manipulating the material structure using processing methods and other ingredients in the alloy. Worth reading JE Gordon's "The New Science of Strong Materials" for lucid overview of the details and history.

Well the one which could be suited for this conditions could be steel. Which is normally an alloy of iron and carbon and have great modulus of elasticity. Even various types of steel are used in making armours. Other options could be osmium, being the densest element it also posses really good elasticity and decent amount of fracture toughness but it's a rare element and so, an expensive choice. So, the best choice is to go with alloys of steel having both  high fracture toughness and great modulus of elasticity.

It sounds like you are interested in helmets; modern helmets are all polymer (eg. polycarbonate) or FRP as for as I know. Steel is a "no brainer" for it has high fracture toughness and elasticity. But it may be heavy for a helmet and steel helmets are no longer used in industry because of the electrical shock possibility. I expect this is an OSHA rule.


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How does having a basic stationary phase help the separation of acidic/basic compounds from non acidic/basic ones?
I have to reason whether silica or alumina would be a better choice of stationary phase for separating a mixture of pyridine and cyclohexanone with column chromatography.

I would expect both pyridine and cyclohexanone to be slightly polar (silica because it has the $Si(OH)$ groups and alumina I suppose because it has a metal attached to highly electronegative Oxygen), but the main difference I'm getting between the two is that pyridine (according to Wikipedia) is considered to be slightly basic, whereas cyclohexanone is neither really acidic or basic. So I suppose alumina is somehow a proton acceptor?
Another Google search shows that pyridine has a $pK_b \approx 8$, which doesn't seem particularly basic to me, but I suppose it's enough.
My intuition would tell me that since alumina is basic it would be best for separating basic pyridine from non-acidic/basic cyclohexanone. But I'm having some confusion wrapping my head around about why using an acidic or basic stationary phase would be helpful (or perhaps it's not helpful and I'm heading on the wrong track -- but even then I'm confused in justifying how it wouldn't be helpful). This brings me to my questions:

How would having an acidic or basic stationary phase be useful for chromatography? Chromatography, as I understand it, relies in things adsorbing and sticking to the stationary phase for different retention times, whereas acid/base properties just have to do with proton transfer. If alumina is a basic stationary phase, I suppose it could be accepting protons from the solution -- but how it would this lead it to adsorb and retain one of pyridine vs cyclohexanone stronger? I would guess that maybe having a protonated stationary phase leads to the basic compound (pyridine) competing for those protons on alumina and somehow the pyridine ends up stuck to the stationary phase where all the protons are ... but I'm not sure if this would be logical.
How does pH of the stationary phase affect the quality of separation? Retailers seem to be selling alumina stationary phases in various pH ranges: pH 6-7.5 (called neutral); pH 4-5.5 (called acidc...); pH 8-9.5 (called basic). How could the pH range which the alumina is kept affect its separation properties? I'm thinking of things in a "Henderson-Hasselbalch amino-acid titration way," but I suppose alumina may be more protonated at lower pHs than higher ones ... does this affect it's separation properties?

See if this helps: https://www.dynamicadsorbents.com/alumina/what-is-alumina/

@Achem The source says "active sites used for separation consist of acidic alumina ions, basic oxide ions and polar Al-OH groups" but also says " the hydroxyl groups at the surface are responsible for the selective retention of adsorbed materials."

The latter statement implies alumina's OH groups are responsible for separation quality. But how do the acid/basic properties of alumina affect its separations of other acidic/basic compounds (if that is the case at all). What's the "mechanism" for how acid/base properties would help a stationary phase? That's what I'm confused on because I don't understand how proton transfer leads to adsorption.

"Retailers seem to be selling alumina stationary phases in various pH ranges: pH 6-7.5 (called neutral); pH 4-5.5 (called acidc...); pH 8-9.5 (called basic). " That rings my absurd detector. What the heck is that even supposed to mean in such a solid?

@Mithoron I really don't know. It confused me as well which was part of the reason I asked here to get some more information. You can look at the hyperlinks and investigate if you wish.

The responses I've gotten so far make me think that the choice of stationary phase between silica and alumina is largely empirical and that we don't have a mechanistic understanding of how this works. Given the link from @AChem and other things I read online about alumina being a good stationary phase to separate amines, I'm going to assume it's better suited to separate pyridine, which I guess is something like a tertiary amine.
It feels pretty sad to say that this is empirical and not understood mechanistically, so please correct me if I am wrong about that.

Well, afaik surface Al could coordinate nitrogen or oxygen - being somewhat Lewis acid in places where coordination number is low. By this reasoning, if there's a lot of such centers, then better ligand would get slowed down.

pH of alumina
Alumina for chromatography is not "just" simple aluminum oxide.  It is made by heating aluminum oxide particles in a carbon dioxide stream at very high temperatures:

When aluminum hydroxide is screened for particle size and heated in a carbon dioxide stream at about 900°C, it is converted to individual particles of aluminum oxide that are coated with a thin layer of aluminum oxycarbonate (with the approximate formula: $\ce{Al2(OH)5]2CO3@H2O}$. The alumina particles are between 70-290 mesh (50-200 mm), and most are "approximately 150 mesh".  Water content and alkalinity are then adjusted by washing with dilute acids.
So the high-temperature (900 °C!) treatment with $\ce{CO2}$ alkalizes the alumina surface by forming "aluminum oxycarbonate".  These alkaline surface groups can be removed by washing with acid.  Without washing, the alumina will give a basic pH on addition to water.
Why is this useful for chromatography?
Altering the proton affinity of the solid (aka stationary) phase in chromatography can alter the way it interacts with analytes.  I'd suppose that since pyridine is a weak base, it exists in most solutions in equilibrium with some amount, perhaps a very small one, of pyridinum ion, and that these pyridinium ions would interact much more strongly with a basic surface than would non-ionic species.
However, it's worth noting the same document I excerpted above cautions against using basic alumina with ketones like acetone.  It can cause polymerization reactions.  Whether the same thing happens to your cyclohexanone may be concentration dependent.


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Why doesn't hydrochloric acid oxidize tin to tin(IV) chloride?
Diluted hydrochloric acid oxidizes tin to $\ce{Sn^2+}$ ion:
$$\ce{Sn + HCl(aq) -> SnCl2(aq) + H2(g)}$$
Why is oxidation not complete? $\ce{Sn^4+}$ has closed shell and is more stable.

Neither Fe being dissolved in HCl form Fe^3+. Additionally, SnCl4 is a liquid with covalent bonds, there would not be Sn^4+. There is no shortcut for studying chemistry of elements.

@ Poutnik.  Anhydrous $\ce{SnCl4}$ is a liquid, and it is decomposed in water. But if produced by a redox reaction in aqueous solution, $\ce{Sn^{4+}}$ is hydrated, and its evaporation may produce hydrated compounds like $\ce{SnCl4·4H2O}$.

@Maurice  There is no real M^4+(aq) ion in water, the field is too strong for that. Neither the famour Ce^4+ ( cerimetry ) is true Ce^4+(aq), being a dimer with bridges, not handy if oxo- or hydroxo- ones. // BTW, I have not said SnCl4(aq) remains in the same form as SnCl4(l).

Look at the standard electrode potentials  here
The std electrode potential for Sn/Sn(II) is -0.14, the std electrode potential for 2H+/H2 is 0 (by definition) so the reaction of Sn + 2H+ to Sn(II) + H2 is energetically favourable.
The std electrode potential of Sn(II)/Sn(IV) is +0.15 so H+ going to H2 is not a strong enough oxidant to oxidise Sn(II) to Sn(IV).

The reaction $\ce{Sn + HCl}$ produces $\ce{SnCl2}$ and $\ce{H2}$. If by chance or by accident some $\ce{SnCl4}$ or $\ce{Sn^{4+}}$ is present in solution, it will be reduced by the reaction : $\ce{Sn^{4+} + H2 -> Sn^{2+} + 2 H+}$.

With such small differences in potentials there possibly are measurable amounts of oxidation to Sn(IV) but a see-saw reaction?

If both concentrations of $\ce{Sn^{2+}}$ and $\ce{H+}$ are $1$ M, Nernst law states that the equilibrium concentration of $\ce{Sn^{4+}}$ is about $0.003$ M. It is small, but it may be considered as measurable.


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Does ozone decay in pure water?
I am trying to understand ozone decay in water and its rate. Some say ozone has a first-order decay at high pH and second-order kinetics at low pH.
Does ozone decay in pure water? Or does it decay because it reacts with organic matter or other minerals in water?

Based on academic work by Bjorn Dahlgren, “On the aqueous reactions of the aminyl radical with molecular oxygen and the superoxide anion" where, per his summary in Table 2.1, details applicable radical reactions, including those associated with ozone, all importantly, with the original source reference and associated reaction rates, from which I selectively cite the following reactions:
$\ce{•O2- + O3 → •O3- + O2  (1.5 × 10^9)}$
$\ce{O3 + •OH → •HO2 + O2 (1.1 × 10^8)}$
$\ce{•HO2 → H+ + •O2-  (1.35 × 10^6)}$  (pKa 4.88)
$\ce{•O3- + •OH → OH- + O3 (2.55 × 10^9)}$
which, in my opinion, constitute the likely reactions of interest in the decay of ozone as could occur in natural waters (containing dissolved oxygen and transition metals) implicitly capable of inducing formation of the superoxide radical anion ($\ce{•O2-}$) and the hydroxyl radical ($\ce{•OH}$).
In strong sunlight, photo generation of electrons combing with $\ce{H+}$ creating the hydrogen atom radical from natural photocatalyst (like dissolved organic matter, DOM) may also occur leading to further pathways:
$\ce{H+ + e- → •H  (2.3 × 10^10) }$
$\ce{•H + O3  -> •HO3  (3.8 × 10^10)}$
$\ce{•HO3 -> O2 + •OH  (1.1 × 10^5) Slow}$
Note, in alkaline conditions, only a very limited reaction:
$\ce{OH- + O3 → •O2- + •HO2 (70) Extremely Slow}$
which I would cite also for the distilled water decay of ozone (absence transition metals, DOM and dissolved oxygen).
So, to answer the question does ozone decay in pure water, yes, but apparently, per the reaction above, slowly. Clearly, per the cited reaction rates possible in natural water, the decay can be greatly accelerated with the creation of transient reactive oxygen species including superoxide that can participate in Fenton (with iron, and Fenton-like with other transition metals) associated radical attacks on organic matter in the presence of dissolved oxygen and minerals.
I hope this helps.

Ozone decomposes to OH radicals in pure water.  Factors influencing the decomposition of ozone in water are temperature, pH, environment and concentrations of dissolved matter and UV light. A more detailed discussion is here

thank you so much for your answer. does it decompose because its simply unstable? doesnt it react with something? and in pure water, what happens to the OH radicals? there will be no free ions in water.

@Hana Ozone decomposes because it is unstable. The OH radicals are highly reactive and will find soemthing react with if only each other forming hydrogen peroxide. If you are happy with the answer please accept it.

How can I accept it?

@Hana there is a check for acceptance beside the question


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Why the density of liquid decreases and density of vapour increases as we approach towards the critical state/point
Pressure and temperature have opposite effects on density. As temperature increases at constant pressure density decreases and as pressure increases at constant temperature the density increases (and vice versa). The change in density for liquids is smaller compared to gases.
But when we talk about the density change when both pressure and temperature increases, how can we say that the density of liquid decreases as we go towards critical point and the density of the gases increases as we go towards the critical point. Since pressure and temperature have opposite effects on density.

For $a,b \gt 0$:
$$\frac{\Delta a}{a} \gt \frac{\Delta b}{b} \implies \Delta (\frac ab) \gt 0$$
$$\frac{\Delta a}{a} \lt \frac{\Delta b}{b} \implies \Delta (\frac ab) \lt 0$$

If there are 2 mutually opposite effects, the net outcome depends on what effect is stronger.

Imagine the phase diagram of the compound, figuratively going along both sides of the boiling curve toward the critical point.
For the liquid, the density decrease due thermal motion and formation of nano-cavities is much higher than the density increase due pressure compression. Therefore, the net density change is negative.
For the vapour, the density increase due pressure compression is much higher than the density decrease due gas thermal expansion. Therefore, the net density change is positive.

Why is it that for vapour, the pressure compression results in higher density decrease than the increase caused by thermal expansion? I haven't heard about nanocavities and stuff will have a look into that.

Read carefully:  For the vapour, the density increase due pressure compression is much higher than the density decrease due gas thermal expansion.

Ah I meant the other way. Got it right in my head but wrong in typing.

I see.......  Been there, done that. :-) // Nanocavities get significant when T is approaching Tc. If you think about it, liquid properties have to converge to gas properties somehow.


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Can vinegar be used to clean wood (a porous surface) after it has been cleaned with germicidal bleach?
I recently cleaned moldy air vents and hardwood floor with diluted germicidal bleach water (1/3 cup bleach to 1 gall). I would like to go back in with vinegar to make sure the mold stays gone, mostly as peace of mind for me. Is it safe after 24+hours to clean with vinegar? I’m also concerned that the areas may not be completely 100% dry from the bleach mixture.

Bleach and vinegar do not go well. You do not want chlorinated acetic acid products.

I’m wondering if the bleach being diluted and applied over 24 hours ago if it’s okay to spray with vinegar

Sodium hypochlorite (bleach) is a solid, it does not evaporate. It is sold in solution. It decomposes slowly. Give it a week.

Actually, AChem is confusing calcium hypochlorite (a solid) with aqueous sodium hypochlorite as commercially prepared from the electrolysis of aqueous NaCl. Solid NaOCl exists but very unstable (as in explosive).

I am not confusing bleaching powder with sodium hypochlorite. Both are solids! Since sodium hypochlorite is unstable at room temperature, it is solid as a stablized solution at high pH. Yes, it is made electrolytically. In short bleach does not evaporate from surfaces.

Solid NaOCl is not sold as a commercial bleach due to its extreme instability. One only prepares dilute aqueous solutions!

Further,  aqueous NaOCl rapidly decomposes to just NaCl in solutions exposed to light, dust,  and transition metal-rich tap water, so it does not stick around.

More precisely, boiling down aqueous NaOCl is a path to NaCl and some NaClO3. This is a commonly known (but inefficient) method for chlorate preparation.

I’ll wait a week before using vinegar on the wood. I’m very very novice and ignorant with chemistry but thank you for the information and assistance! Pesky mold.