text:
Breaking/reducing water adhesion with an electric field
Does anyone know if the presence of an electric field can help to reduce water adhesion to a surface?

related   https://en.wikipedia.org/wiki/Electrocapillarity

On the question relating to breaking/reducing water adhesion with an electric field, I suspect, the answer is that magnetic fields can have an impact.
Support, here is a 2008 paper published in Journal of Applied Physics, The surface tension of water under high magnetic fields, to quote in part:

The surface tension of water-air interface was measured by means of a surface-wave resonance method under magnetic field up to 10 T. It was found that the surface tension increased linearly with an increase in the square of the magnetic field.

Of note, background on the concept of surface tension, per Wikipedia, to quote:

At liquid–air interfaces, surface tension results from the greater attraction of liquid molecules to each other (due to cohesion) than to the molecules in the air (due to adhesion).
There are two primary mechanisms in play. One is an inward force on the surface molecules causing the liquid to contract.[1][2] Second is a tangential force parallel to the surface of the liquid.[2] The net effect is the liquid behaves as if its surface were covered with a stretched elastic membrane.

So, per the cited paper, surface tension increased linearly with the square of the magnetic field. Per Wikipedia, surface tension relates directly to cohesion, as opposed to adhesion.
As such, the question likely correct cites increased magnetic fields as reducing water's adhesion (as it apparently relates to promoting surface tension/cohesion).


text:
What does *oxidizing* have to do with *hydrogen*?

An oxidizing acid is a Brønsted acid that is a strong oxidizing
agent. All Brønsted acids can act as oxidizing agents, because the
acidic proton can be reduced to hydrogen gas.

(source: Wikipedia)
But what does oxidizing have to do with hydrogen? Isn't it about oxygen?
Like how iron becomes iron oxide after prolonged exposure to oxygen in the atmosphere?
So why is hydrogen relevant?

Do you know the chemical definition of oxidation (and reduction)?

Moreover, generally speaking, oxidation has nothing to do with oxygen either.

i believed so. perhaps not. the definition seems to vary with the source. i assumed the acquisition of an oxygen atom was the crux of the concept.

is oxidation not contingent on the synthesis of an oxide?

@voices Well, reaction of fluorine with oxygen is fluorine reduction, not oxidation. Free oxygen is one of many potentially oxidation agens. Some of them contain oxygen, some do not. One of them is hydronium ion H3O+, which can e.g. oxidize metallic zinc to Zn^2+ ions.  Zn + 2 H2O+ -> Zn^2+ + H2 + 2 H2O

if it's only about the transfer of electrons, isn't that ionization (as opposed to oxidization)?

https://chemistry.stackexchange.com/questions/69798/an-introduction-to-oxidation-state-oxidation-number

Ionisation is a particular kind of oxidation, so does reaction with oxygen. Oxidation is increasing of oxidation number, what is a formal charge an atom would have, if all bonds were broken in such a way the electrons involved in bonds would stay  with more electronegative atom.

An oxidizing acid is a Brønsted acid that is a strong oxidizing agent.
All Brønsted acids can act as oxidizing agents, because the acidic
proton can be reduced to hydrogen gas."

The whole sentence is written in a very convoluted fashion, and it is partially wrong as well. Someone should improve this Wikipedia section.
All they are saying is that strong acids do react with many metals and oxidize them. In turn their acidic proton is reduced to hydrogen.
For example, metallic Zn reacts with dilute sulfuric acid to form zinc sulfate and hydrogen gas. Zinc (0) has been oxidized to Zn(II). Consequently, Zn(0) also reduced the hydrogen ions to hydrogen gas.
Why this statement is wrong? The reason is that it says ALL Bronsted acids can do this, which is simply an over-generalization. For example, benzoic acid would not produce hydrogen with zinc. It is sparingly soluble in water.
Independent of this point, sulfuric and nitric acid are strong oxidizers in concentrated form. Here it is not behaving like an acid. Sulfuric acid, if concentrated will oxidize copper to Cu(II), and itself get reduced to sulfur dioxide.

The definition of oxidation and reduction are

Oxidation is the loss of electrons or an increase in the oxidation state of an atom, an ion, or of certain atoms in a molecule.
Reduction is the gain of electrons or a decrease in the oxidation state of an atom, an ion, or of certain atoms in a molecule.

By this definition a $\ce{H+}$ ion can gain an electron by oxidizing a species, in this process since it has gained an electron we say it has been reduced.

All acids are oxidizing. They are all able to oxidize metals $\ce{M}$ whose redox potentials are negative with respect to hydrogen, like zinc $\ce{Zn} ~(E° = - 0.76~\mathrm V$), iron $\ce{Fe}~ (E° = -0.41~\mathrm V)$, and magnesium $\ce{Mg} ~(E° = -2.37~\mathrm V)$. The reaction produces some hydrogen gas $\ce{H2}$ and the metallic cation $\ce{M^{z+}}$. But usual acids like $\ce{HCl}$ or $\ce{H2SO4}$ cannot oxidize metals whose redox potentials are positive, like copper $\ce{Cu}$ or silver $\ce{Ag}$, at least at room temperature. That is why $\ce{HCl}$ and $\ce{H2SO4}$  are not called "strong oxidizing acids". For example : $$\ce{2 HCl + Fe ->Fe^{2+} + 2 Cl- + H2}$$ The $\ce{Fe}$ atom is oxidized from $0$ to $+2$, and the $\ce{H}$ atom is reduced from $+1$ to $0$. $$\ce{HCl + Cu -> no reaction}$$
On the other hand, nitric acid $\ce{HNO3}$ is able to oxidize metals whose redox potentials are positive, like copper $\ce{Cu} ~(E° = +0.34~\mathrm V)$ and silver $\ce{Ag}~ (E° = +0.80~\mathrm V)$. That is why nitric acid is called a "strong oxidizing acid". It is worth observing that the reaction does not produce hydrogen gas $\ce{H2}$. It produces nitrogen oxides like $\ce{NO}$ or $\ce{NO2}$. For example with copper $\ce{Cu}$: $$\ce{3 Cu + 8 HNO3 -> 3 Cu^{2+} + 2 NO + 4 H2O + 6 NO3-}$$ In this reaction the $\ce{Cu}$ atom is oxidized from $0$ to $+2$, but the $\ce{H}$ atom is neither reduced nor oxidized. It is the $\ce{N}$ atom that is reduced (from $+5$ to $+2$).

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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Why is an Octet Complete Cation more stable than a Resonance Stabilized Cation?
Our instructor has lately been teaching us about different types of cations namely-Classical, Non-Classical, Resonance stabilized (R.S.C) & Octet complete cation (O.C.C) & how O.C.C is more stable than a R.S.C .
One thing I did not get isn't donating lone pair like in the below given example too is resonance ,so how come these two phenomena are any different?
  (R.S.C)(O.C.C)& also, why is the latter structure more stable ?

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text:
Relationship between microplastics and wildfire smoke
My understanding is that:

microplastics are short fragments of previously long polymer chains;
wildfire smoke largely consists of incompletely burned organic materials such as wood. Wood burns by heat first breaking long polymer chains into small fragments, which then ignite (or not).

So ... it sounds like there is quite some similarity between (some parts of) wildfire smoke and microplastics.
What are the key differences? Does it boil down to nature's ability to recycle one but not the other because their composition is different because they came from different types of polymers?
This is a noob question. But I'm still curious :-)

The word polymer is very broad. Wood and plastics are very different. I wouldn’t begin to compare wood and plastic smoke. Both aren’t great for you health but I’ll take a camp fire over burning plastic any day

Smoke is mostly polymeric carbon. Polymeric carbon has quite a long lifetime in nature, google "terra preta". Charcoal. Microplastic is also polymeric. It is not necessarily bad. Depends on the exact type and place. Wood was not "recyclable" for the largest part of prehistory. That's why we have coal. Etc.

In addition to the importance of chemical composition, I would add that both smoke and microplastics are ambiguous terms in that they can refer to many size distributions.

The term «polymer» only refers to using small molecules (called «monomer», needn't be of one type) multiple times to build a larger molecule.  E.g., if you would use the small molecule thrice to get a larger one, it were a trimer, if four times, a tetramer to be more specific; but these were all examples of «polymers».  The term does not tell if the material were of industrial / artificial (like for a PET bottle), or biological origin (e.g., lignin, chitin, cotton, or kreatin).
Given the small size of molecules, even tiny particles known as microplastic consist of many polymer molecules each.  Like if the band of a tagliatelle would just be cut into smaller and smaller pieces and still count as a noodle.  The fine mechanical division  / chopping into smaller and smaller fragments only increases the accessible area per unit of weight.  Similar to a combustion engine, this material then spread out may yield a combustion more efficient with less smoke and soot than, e.g., a torch, or a candle. Not all polymers burn equally well, even if finely divided.
The «problem» with wood, compared to natural gas - chemically speaking - is that it consists of many different compounds.  Beside the lignin, there are a number of minerals, sugars, fluids, resins, etc. and water present in a tree; each with its own characteristics.  Some will catch fire quickly no matter what, like human hair.  Some need to be heated first so that fire nibbles them away.  Meanwhile others may be heated and then burn as gases, rather than as solid or liquid, sometimes liberated like in an explosion.  Some materials require much air (wind) per unit of mass to burn, or otherwise would just smoulder; so now equally soil, weather and topography of the terrain enter the the set of parameters yielding a typical wild fire with incomplete combustion and smoke.


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What is the duration of time needed if heating activated carbon at 100 degrees celsius to recharge?
I have 5 pounds of loose activated bamboo charcoal that I am using as an odor absorber.  When I need to recharge it, I would like to heat it to 100 degrees celsius.  How long does it need to remain at this temperature to recharge, that is, de-adsorb anything on its surface?

I'm afraid that the temperature to remove the odor agent depend on the odor components.
In the industry you will have a time/temperature protocol with three steps.
First near 100°C (a little more), to remove humidity from the activated carbon.
Then a temperature dependant of the adsorbed component. (around 400° commonly).
Finally above 700°C to reactivate the carbon (making sure is very porouse again).
If you have a lot of humidity you can restore a lot of the adsorbent power with 100°C. This is not uncommon.
There is not specific amount of time (depends on several variables), you can know by the weight profile.
If your adsorbed components have more afinity than 100°C. You will have to increase the temperature.
Best Regards.


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What reaction would take place on an electrode in a secondary cell if it were charged and the other electrode removed?
I'm reading a book that covers the basics of electrical theory and I'm at a chapter that discusses the concept of a secondary cell. The author includes the redox reactions that take place in a primary cell when it is charging/discharging.
The secondary cell is comprised of an anode (composed of PbO2) and a cathode (composed of Pb) submerged in a solution of H2SO4. As it discharges both electrodes are plated with PbSO4 until they are fully covered, at which point the reactions that produce a charge for the battery cease and the original H2SO4 is now diluted with H2O.
My question focuses on the charging aspect of it not the discharging (because I have an easier time framing my question from this perspective) but I want to preface this aspect with two quotes the author said earlier in the chapter:
First quote: "From previous information, we found that H2SO4, when charging or discharging a cell, dissociates into hydrogen H+ ions and sulfate SO4= ions."
Second quote (this quote comes from a section describing a different cell): "The zinc element or electrode, when immersed in the acid, starts to dissolve, leaving its electrons behind and the zinc goes into solution as Zn++ ions. This action stops very shortly, as the zinc plate becomes negative and ceases to throw off positive ions. The amount of zinc thus dissolved is very minute. If the negative zinc plate is connected externally by a wire to the copper plate, an electrical current will flow from the negative zinc plate through the external wire to the copper plate (positive). As the negative charge of the zinc plate is thus removed, more Zn++ ions go into the solution"
As the original secondary cell I mentioned recharges the PbSO4 plating reverts to the composition of it's respective electrode and the diluted H2SO4 reverts to it's original solution. The reactions that take place when the cell is recharging as quoted from the book:
Negative Plate: PbSO4 + 2H+ + 2e --> Pb + H2SO4
Positive Plate: PbSO4 + SO4= - 2e --> Pb(SO4)2
I presume that each plate is a half reaction; the negative plate is a reduction half reaction; the positive plate is an oxidation half reaction; I believe these half reactions can be combined into the full reaction: 2PbSO4 + 2H+ + SO4= --> Pb(SO4)2 + Pb + H2SO4. Please correct me if I'm wrong about what these half reactions are or about how I combined them; I have only just recently learned about redox reactions in order to better understand this chapter.
Now for the question this has all been building up to; if the positive plate were removed during the recharge process would the half reaction that takes place on the negative plate become the half reaction: PbSO4 + 2H+ + SO4=  --> Pb + H2SO4 + SO4= for a short period of time until the solution becomes negative with SO4= anions and ceases to throw off H+ cations (similar to what happens in the second quote and considering what happens in the first quote)?
If this is the case, then I don't know how the difference of charge between the reaction and product side is possible. The only thing I can imagine is that the SO4= is just a spectator and does not participate in the actual reaction.

PbO2, not Pb(SO4)2

$Pb(SO_{4})_2$ breaks down in the presence of $H_2O$ to arrive at $PbO_2 + 2 H_2SO_4$

Rather $\ce{Pb^4+ + 2 H2O -> PbO2 + 4 H+}$  ( using \ce{} ).  There is no $\ce{Pb(SO4)2}$

The sulfate of the electrode must bond with the hydrogen in the end to restore the solution for further charge and the electrode itself. Also Pb does not react with $H_2O$ during electrolysis. How did you get to Pb from PbSO_4 (what both plates are coated with)?

And: $\ce{PbSO4(s) + 2 H2O -> PbO2(s) + HSO4- +  3 H+ + 2 e-}$

$\ce{Pb^2+ + 2 e- -> Pb}$ respectively $\ce{PbSO4(s) + H+ + 2 e- -> Pb(s) + HSO4-}$.

Note that primary cells are not supposed to be charged. Secondary ones are.

Sulphate (hydrogen sulphate) ions are spectator ions, aside of reversible forming of PbSO4(s), as you correctly   assumed.

Half reactions do not depend on the circuit being closed. If the circuit is open, the electrode potential converges to its equilibrium one, where the half reaction in ongoing by the same rate in both directions.

May I have your thoughts on 2 things? First thing: Are you saying that it's impossible for the sulfate anions to react with the positive plate when it's oxidized during the recharge process? Right now I'm thinking that it's possible for both reactions to take place when the positive plate is oxidized; one possibility is that the negative sulfate ions (from $H_2SO_4$) bond with $PbSO_4$ to arrive at $Pb(SO_4)_2$; the other possibility is that the negative oxygen ions (from $H_2O$ like you said) bond with Pb from $PbSO_4$. If it's not possible for the sulfate ions to bond, then my book is wrong

Let us continue this discussion in chat.

Second thing: When you say that half reactions do not depend on the circuit being closed I assume you mean the half reaction on the negative plate would remain the same if the positive plate were removed (correct me if my assumption is wrong).

Isn't that what I did with my extrapolated reaction of what would happen if the positive plate were removed? The only difference between my extrapolated half reaction and the one given is that the negative sulfate ion doesn't participate in the reaction so it's on both the reactant and product side. I only included the sulfate ion to illustrate that without the positive plate it wouldn't have anything to bond with (assuming that a negative sulfate ion could bond with the positive plate at all).


text:
comparing electronegativity for row elements and size of the atom for column elements
I have read that while comparing atoms in a row we compare their electronegativities to know which atom would better stabilize a negative charge on it, and for atoms in a column we compare their sizes. Why is that so? Like, for example if we consider Nitrogen and Oxygen (which are same row elements), why can't we say "as Nitrogen atom is bigger in size, it can better stabilize a negative charge rather than the Oxygen atom" (as Nitrogen has more room for stabilizing the charge), but we say it is the Oxygen atom which stabilizes the negative charge as it is more electronegative than the Nitrogen atom?

The difference in size is not very large across the row.

An oxygen atom has smaller radius than nitrogen, because of bigger nucleus charge, what contracts orbitals.
Oxygen has higher electronegativity than nitrogen, as p electrons have mutually  low nucleus/kernel screening coefficient, so present or incoming electrons are in oxygen atom  attracted by higher effective nucleus charge ( plus force is bigger due smaller radius. )
For more, see Slater's rules

Yeah, you have noticed right thing. Well, remember one thing nitrogen has half filled state so it's more or less inert in comparison with it's neighbours i.e oxygen. So it has no need to take electrons because if it takes it that will break it's maximum spin multiplicity which will result a little distabilised state . Where as oxygen is more electronegative has affinity towards electron so it going to take an election but remember the specific example you asked has a annoying point that is (2- ) charge or capturing 2 electrons will create a massive distabilised state because of its small size and inter-elctonic repulsion.


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Is the precedence of functional groups used in IUPAC naming arbitrary?
There seems to be a priority order that has to be followed while naming an organic compound with more than one functional group , according to IUPAC rules. I want to know if there is a basis on which this priority list has been made, or is it just arbitrarily ordered.
I have tried to connect the dots but failed, first i thought it may be the s-character of the atom which is in direct link with the parent chain, doesn't work out, then about their atomic weights, that too, failed.
Is there some order among this chaos?

Generally, languages are older than their dictionaries. Therefore, the first dictionaries are usually descriptive; i.e. they describe the language. Later there may also be prescriptive dictionaries; i.e. they are more concerned about a standard language and prescribe the proper usage and spelling of words.
The same principle applies to chemical nomenclature. The first chemical names are older than any systematic nomenclature rules. The first nomenclature rules tried to describe the already existing chemical names, as explained for example in an early version of the rules used in Chemical Abstracts [1]:

We have not tried to invent a new system. Our aim has been to follow existing usage as far as it could be made fairly consistent, choosing what appeared to us good practices and rejecting bad, and introducing new features only when some very positive advantage was to be gained.

In particular, this descriptive approach also applies to the seniority order of functional groups:

With compounds of mixed function there must be a definite understanding as to which is the chief one for purposes of naming. After considerable experimenting we have adopted the following order of precedence, based on existing usage:
(“Onium” compounds), acid (carboxylic first), acid halide, amide, imide, aldehyde nitrile, ketone, alcohol, phenol, mercaptan, amine, imine, ether, sulfide (and sulfoxide and sulfone).

Note that this list is already very similar to the current IUPAC recommendations.
Remarkably, however, the International Union of Chemistry at the meeting held in Liége (1930) could not decide to include this seniority order in their reform of the nomenclature of organic chemistry yet [2] (The Liége rules were the predecessor of the IUPAC nomenclature of organic chemistry.):

The Committee intended at one time to establish an order of precedence of functions (to determine which should be the principal function in a compound containing two or more different functions), but later gave up the idea.

Nevertheless, the 1965 IUPAC rules included such a list [3]:

’Onium and similar cations
Acids: in the order $\ce{COOH}$, $\ce{C(=O)OOH}$, then successively their $\ce{S}$ and $\ce{Se}$ derivatives, followd by sulfonic, sulfinic acids, etc.
Derivatives of acids: in the order anhydrides, esters, acyl halides, amides, hydrazides, imides, amidines, etc.
Nitriles (cyanides), then isocyanides
Aldehydes, then successively their $\ce{S}$ and $\ce{Se}$ analogues; then their derivatives
Ketones, then their analogues and derivatives, in the same order as for aldehydes
Alcohols, then phenols; then $\ce{S}$ and $\ce{Se}$ analogues of alcohols; then esters of alcohols with inorganic acids*; then similar derivatives of phenols in the same order
Hydroperoxides
Amines; then imines, hydrazines, etc.
Ethers; then successively their $\ce{S}$ and $\ce{Se}$ analogues
Peroxides

Patterson, A. M.; Curran, C. E. A System of Organic Nomenclature. J. Am. Chem. Soc. 1917, 39 (8), 1623–1638.
Patterson, A. M. Definitive Report of the Commission on the Reform of the Nomenclature of Organic Chemistry Translation with Comments. J. Am. Chem. Soc. 1933, 55 (10), 3905–3925.
Nomenclature of Organic Chemistry (Definitive Rules for Section C. Characteristic Groups Containing Carbon, Hydrogen, Oxygen, Nitrogen, Halogen, Sulfur, Selenium, and/or Tellurium). Pure Appl. Chem. 1965, 11 (1–2), 1–260.

This clears up everything ! I learnt much more than what i asked for, and although the comment instructions tell me to not write thanks, I can't help myself , so , a sincere thanks to you!

In the second quotation, there is a line which says "After considerable experimenting" , does this refer to experimentation as in with the substances, or with the existing literature and its usage?


text:
What (stored-as-solids) acids/liquids can clean corroded copper?
I'm looking for a liquid that can remove copper corrosion (oxides and basic carbonates - the black and green stuff) off copper piping.
The corrosion will be located on the inside of the pipe, in hard-to-reach areas, so ideally I should be able to just pour the liquid in, leave it for a while, and just flush it out with water. It will be practically impossible to scrub the corroded areas.
Additionally, the piping is brazed or soldered together with an unknown filler metal.
I've read around, and it seems people commonly use acetic acid (vinegar) or hydrochloric acid for this. Is this correct?
However, I'm interested in if any acids (or other liquids) that can be stored as solids can be used for this instead. Mainly because it's easier to store and transfer solids than liquids. (As far as I'm aware, acetic and hydrochloric acid are only readily available in liquid forms)
Some common solid acids that seem to be effective on rust (iron oxide) that I can buy off Ebay include citric acid, oxalic acid, and sulfamic acid. But will these also be effective on copper corrosion, and keep the byproducts in solution?
Additionally, will I need to use any corrosion inhibitors, to prevent the acid for eating away from the raw metal itself? If so, what should I use, and what acids are they compatible with? I'd like to avoid anything that would leave phosphates/silicates/etc coated on the metal.
TL;DR: What (stored-as-solids) acids/liquids can clean corroded copper, that will leave the byproducts in solution? Do I need to use any corrosion inhibitors?
Note: I do not have access to chemical suppliers, so only things commonly available on Ebay and the like please. Additionally, I do not have fume hood, so no toxic fumes too.

Google is your friend. When typing "Copper cleaning" on Google, you will find a lot of different recipes for cleaning oxidized copper surfaces. In the majority of these processes, vinegar is used, usually with some slat. There are plenty of videos about each recipes. And I can tell you, that vinegar will not corrode any other metal (tin alloys usually) used for soldering. Of course, sulfuric acid may also do the job.

Yes, I'm aware of vinegar (as mentioned in the question), but I don't have any so I would have to go out and buy some anyways. In which case I would rather buy something stronger like HCl instead.
But I also want to know if there are any stored-as-solid alternatives readily available

See https://patents.google.com/patent/US5468303A/en

Try sodium bisulfate. It is cheap: used to lower pH in swimming pools. Readily available.

Per a source to generally clean copper metal surface, to quote:

A few hard rainstorms may clean the surfaces sufficiently to start operations. However, it is always advisable, particularly if the coloring is to be done immediately after installation, to go over the surface with a commercial chemical metal cleaner. Cleaners of the trisodium phosphate type should be satisfactory. Avoid cleaners which leave a coat of oxide on the copper surface.

However, assuming one has already a basic copper carbonate, the application of an acid, like HCl or more friendly acetic acid, is likely suitable, especially with the addition say NH4Cl or NaCl to increase the activity coefficient of the acid (so less strong acid may be employed).
However, a review of the cited source notes that these reagents are the very agents suggested to create a patina coating (which can be characterized as electrochemical based corrosion) occurring in the presence of air/oxygen.
So, do limit air/oxygen exposure by immersing the copper surface to be cleaned in say a HCl/NH4Cl mix.
Also, employing the concept of a sacrificial anode, experiment with performing the latter cleaning in the presence of a more anodic metal (say a small piece of Aluminum metal in acetic acid and NaCl), especially when the requirement of limiting air exposure could be difficult to implement. This may prove to be actually beneficial in avoiding surface
oxidation.

Sulfamic acid is a standard for cleaning pipes. Typical use is to remove blockage by carbonate and rust scale, not green film on copper. But it will clean copper. Why do you want the remove the film ? Do you have a corrosion problem that you want a protective coating for ? Solder is lead + tin, or just tin if it is newer, after the lead hysteria started. Caution , do not use any ammonia compounds on copper- severe corrosion potential.

Two common acids that can be used to clean corroded copper are:
Vinegar (Acetic Acid): Vinegar is a weak acid that can effectively dissolve tarnish and corrosion from copper surfaces. Soak the corroded copper item in vinegar or create a paste with vinegar and salt to gently scrub away the corrosion.
Lemon Juice (Citric Acid): Lemon juice contains citric acid, which can also be used to clean corroded copper. Apply lemon juice to the corroded area and use a cloth or sponge to gently scrub away the corrosion.
Always exercise caution when working with acids, even weak ones, and wear appropriate protective gear such as gloves and safety goggles. Additionally, ensure proper ventilation when using these cleaning solutions. For more information us on - https://goo.gl/maps/CzTRuLaAi5h1Mxe46


text:
how to obtain potential curves from database?
I'm sorry if this is a really super basic question. I am looking at the potential curves of a diatomic molecule in a book , and to plot them. I google the molecule and got to the NIST website, there I get diatomic mass and constants for each state,  such as:

Te -     minimum electronic energy (cm-1)
ωe  - vibrational constant – first term (cm-1)
ωexe -     vibrational constant – second term (cm-1)
ωeye - vibrational constant – third term (cm-1)
Be -   rotational constant in equilibrium position (cm-1)
αe -   rotational constant – first term (cm-1)
γe -   rotation-vibration interaction constant (cm-1)
De -   centrifugal distortion constant (cm-1)
βe -   rotational constant – first term, centrifugal force (cm-1)
re -   internuclear distance (Å)
Trans. -   observed transition(s) corresponding to electronic state

Great, I should be able to use these to plot the potential energy curves for each electronic state somehow right? there is no formula i can see that here, is it just morse potential? what happens if the state is unbound?
Please help me understand how to plot the potential curves. Thanks!

You could try the  method in this paper. https://www.sciencedirect.com/science/article/abs/pii/009784859500016L

Im sorry, that doesn't answer the question. and that paper is from 1995.  mathematica (which I dont have) is probably very different today. the paper related to RKR potentials, why is that appropriate for for this question?

The fact that it is Mathematica is unimportant. The equations for the classical turning points of the potential for the potential are given in terms of the parameters you quote. I coded this up in Python (free) quite easily. If you cannot see equations due to pay wall I will put them in an answer

yes, I can't access that paper. Thanks for the clarification. I'd very appreciate if you can help with posting the equation in your answer.

The RKR method determines the classical turning points $r_{\pm}$ of the potential energy based on knowing experimentally determined spectroscopic constants. ( Rydberg, 1931, Z. Physik. v73,376, Klein, 1932 Z. Physik. v76, 226, Rees Proc. Phys. Soc.(Lond.) 1947, v59, 998.)
The energy equation takes the usual form. Parameters all in cm$^{-1}$, distances in m, reduced mass $\mu$ in u, $h$ in cm$^{-1}$s.
$$E_v=\omega_e(v+1/2) - x_e\omega_e(v+1/2)^2 + y_e\omega_e(v+1/2)^3 + z_e\omega_e(v+1/2)^4 +\cdots$$
The vibrational quantum number $v$ should be considered as a variable that has values from $-1/2$ to whatever positive value is required.
The derivative of the rotational energy wrt. $J$ at $J=0$  is also needed,
$$B'(v)= B_e-\alpha_e(v+1/2)+\gamma_e(v+1/2)^2+\cdots$$
The function representing the turning points is
$$r_\pm= \frac{f(v)}{2}\left (\sqrt{1+\frac{4}{f(v)g(v)}}\pm 1 \right)$$
where functions $f, \, g$ are integrals. These functions 'blow up' at the upper integration limit $x \to v$ so a correction is added to prevent this. The basic equation is
$$f(v)=\frac{h}{\pi\sqrt{2\mu}}\int_{-1/2}^v\frac{dx}{ \sqrt{E(v)-E(x)}}$$
and the modified one
$$f(v)=\frac{h}{\pi\sqrt{2\mu}}\left(\int_{-1/2}^{v-\delta}\frac{dx}{ \sqrt{E(v)-E(x)} }+ 2\sqrt{\frac{\delta}{Q_v}}\right)$$
where $\delta$ is small $\approx 10^{-5}$ and
$$Q_v=\omega_e - 2x_e\omega_e(v+1/2) + 3y_e\omega_e(v+1/2)^2 + 4z_e\omega_e(v+1/2)^3 +\cdots$$
The function $g$ with correction is
$$g(v)=\frac{4\pi\sqrt{2\mu}}{h} \left( \int_{-1/2}^{v-\delta} \frac{B'(x)}{\sqrt{E(v)-E(x)} }dx+2B'(v)\sqrt{\frac{\delta}{Q_v} } \right) $$
A set of values for CO is
$\mu = 6.85841,\; \omega_e= 2169.8135,\; x_e\omega_e=13.2883,\; y_e\omega_e=0.010511,\; z_e\omega_e= 0.000057$
$ B_e= 1.9312808,\; \alpha_e= 0.0175044,\; \gamma_e= 0.000000548$.
Plot energy $E(x)$(cm$^{-1}) $ vs $r_\pm(x)$ (pm) is shown. The value at $x=-1/2$ should be  $r_e=103.14$ pm the equilibrium bond length with energy of zero. As values are based on cm$^{-1}$ a scaling of $10^{14}$ is needed to go to pm.

I used Jupyter notebooks (via Anaconda) and python 3 to do the calculation using built in integrator from numpy/scipy and matplotlib to draw the graph. All are free and very easy to use.


text:
Why is aluminum oxide resistant to corrosion?
It is common knowledge that aluminum is relatively resistant to corrosion due to its tough oxide layer. The ultrathin (1 nm even) layer is impervious to water and oxygen. What I can't figure out is, why? What makes $\ce{Al2O3}$ so strong compared to $\ce{Na2O}$ or $\ce{MgO}$? My theory is that the aluminum oxide molecules are tightly bound together due to the high charge density of aluminum ions, and this forms a physical barrier where water and oxygen molecules can't come through. This will explain why this only applies to aluminum, as its small size and 3+ charge should make its charge density amongst the highest known. I can't find anything corroborating this claim, though.

While waiting for an answer you may want to search for "passivation" and either spelling of aluminium on this site, as there are many relevant anwers, for instance: https://chemistry.stackexchange.com/questions/125691/will-aluminum-continue-to-corrode-after-its-surface-is-oxidized/

Also Al is not unique in this regard.

@Alchimista - true, but only 3 elements have a self-limiting oxidation layer so it is close to unique...

@JonCuster ... chromium, titanium, and nothing else?

@Karl - aluminum, silicon, and beryllium (or boron - I really don't deal with those). Neither Cr or Ti self-limit.

@JonCuster Ah, you wrote "element", I read "metal". ;-) What exactly do you mean by "self-limit", i.e. why does that exclude Cr and Ti?

Self passivating is the more proper term perhaps. A clean silicon surface exposed to air grows a passivating oxide layer. It stops somewhere around 1.5nm in thickness and does not grow any further (under steam at 1000C you can grow it thicker). Similarly aluminum in air gets a 5nm thick oxide film, and it stops growing. Oxide layers on most other metals will keep growing under STP conditions, it just slows down. But you wait a few years and it will be thicker. Not silicon or aluminum.

You might also look up the chemical resistance of sapphire or alundum.


text:
Are crystallographic omit maps only used with molecular replacement?
(This is in the context of macromolecular X-ray crystallography.)
From Lamb et al. 2015:

Model bias is the result of how maps are calculated: because the
phase estimates [of the structure factor] for the calculation come from the model, maps will
tend to show electron density for an atom in the model whether it is
truly there or not. A simple omit map is normally a difference (Fo −
Fc) map calculated after omitting specific atoms from a model, like a li-
gand or a functionally important loop.

From this the purpose of an omit map is rather clear.
But what I don't understand is when omit maps are used w.r.t. the source of the phases. From what I can tell, if the phases have been determined experimentally (e.g., using MAD), there cannot be any extraneous electron densities in the atomic structure model. This drives me to the conclusion that omit maps are used when the source of the phases is molecular replacement (I presume this is what the "model" alludes to).
Is this correct? Are omit maps only used with molecular replacement?
Edit: I am also confused with regards to how an omit map actually works. If the phases are what determine the final electron density irrespective of the amplitudes used, how does deleting a part of the atomic model, getting the phases from that and recalculating a new map (with these phases and the experimental amplitudes) actually help?
Edit 2 (from comments): I got the impression that if we don't have the phases (say for a part of a model), it just wouldn't get built even if we have the amplitudes for that part, right? Or is it that in this case (of omitting a part), there would be an electron density (because the amplitudes are still there), just that it would be featureless?!
I would be very glad to receive some clarifications and/or pointers.

What kind of macromolecular system is this about? And I'm confused about your "phases". Phases of a signal/wavefunction, and/or amorphous/crystalline phases?

Biomolecular (protein) crystals, I guess. And by phases I was referring to the relative phases of the the diffracted X-rays that together make up the diffraction pattern from the crystal (so I guess these are signal phases then?). I apologize if my terminology is confusing, I am new to this (+ information overload, so I am confused myself).

[OP] ...if we don't have the phases (say for a part of a model), it just wouldn't get built even if we have the amplitudes for that part, right? Or is it that in this case (of omitting a part), there would be an electron density (because the amplitudes are still there), just that it would be featureless

For each reflection, the structure factor $F_{hkl}$ gives the magnitude and phase resulting from interference of all atoms in the unit cell (see e.g. this slide show). This is unlike other methods (like NMR) where signals originate from given parts of a structure, i.e. can be assigned to an atom or groups of atoms. In crystallography, every atom contributes to a given diffraction spot. On the other hand, a given atom contributes to every structure factor in the data set. In this respect, crystallography is a "holistic" method.
If you are familiar with the 3D fourier transform and the math behind it, you can see this immediately:
$$ F_\vec{h} = \int_{cell} \rho(\vec{r}) e^{2 \pi i \vec{r} \vec{h}} d \vec{r}$$
In the formula, $ F_\vec{h}$ refers to complex structure factors (so a combination of amplitudes and phase) in reciprocal space indexed by (hkl) or $\vec{h}$. On the other hand, $\rho(\vec{r})$ refers to the electron density at a location x,y,z or $\vec{r}$ in the unit cell, i.e. in direct space. So the electron density (and the coordinates) "live" in direct space, and the diffraction pattern and calculated phases of the structure factors "live" in reciprocal space (see e.g. this slide).
Omit map
The omit map will show features even in regions where no model has been built yet, or where the model is omitted from calculating phases (as long as the remainder of the model is decent and the diffraction data is decent). For a pictorial demonstration, see Kevin Cowtan's cat and Manx

In this picture, a cat was used to calculate structure factor amplitudes, and a Manx (cat without tail) was used to calculate structure factor phases. Then, they were combined and Fourier-transformed. The resulting (2Fo-Fc) electron density shows the tail at lower signal strength than the remainder of the cat, and there is some noise. The difference between the perfect image of a cat and the observed image is caused by the phases lacking the information of the tail. (See this slide for the entire pictorial explanation.)
Use of omit maps

[OP] But what I don't understand is when omit maps are used w.r.t. the source of the phases.

To reiterate the other answer, omit maps may be used whenever the phases of the structure factor are based on an imperfect model of the structure. If parts of the electron density are difficult to interpret, the model in this part is left out, and a new density is calculated. This density is not biased by a potentially incorrect model in this area, so it might help to correctly build that part. (Here is a slide on model bias.)
With the computer power available nowadays, you can also omit every part of the structure in sequence and piece back together the entire electron density (if you are not sure which part of the model is incorrect). See e.g. https://www.phenix-online.org/documentation/reference/composite_omit_map.html
Omit maps can be calculated no matter where initial phases came from. In the final stages of model building and refinement, the phases are usually obtained exclusively from the model, even if you start with experimental phases.
Additional aspects raised in the chat

[OP in comments] is the reason why omit mapping works because of the fact that every atom contributes to a diffraction spot?

Yes, exactly. The phases of each structure factor are almost correct, and the amplitudes are experimental (so unbiased but they do have experimental error, e.g. noise). It is worse to have a bad part of the model compared to leaving that part out all together.

[OP in comments] So even if the phases from that particular part of the structure itself are missing/omitted, there are still phases from elsewhere that do contribute to the electron density at that location. So if the feature were a model artifact, these other contributive phases wouldn't exist, and the omit map would show zero density there. Is this correct?

No, the language you are using is incorrect. There is no one-to-one correspondence of spots in the diffraction pattern and locations in the unit cell. Intensities and phases are indexed by (hkl) or $\vec{h}$, i.e. which plane of reflection they originate from. Locations in the unit cell are designated by x, y, z or $\vec{r}$. Every atom contributes to each diffraction spot. Every diffraction spot contributes to the calculated electron density in a given location.
Here is a way to say it: Even if the phases lack the information from the omitted part of the model, they are still good approximations of the phases of the actual structure. What really carries the information is that the structure factor amplitudes are derived from the diffraction pattern, which of course reflects the real structure. The amplitudes carries less information than the phases, but as long as you don't introduce incorrect information via the phases, that weak signal from the intensities is good enough to build or correct the model

[OP in comments] So if the feature were a model artifact [...], the omit map would show zero density there. Is this correct? Or is it more that an omit map does not "delete" the density in question so much as it "un-defines" it?

If you put atoms into the model in a region that is "empty" (filled with solvent) in the real structure and calculate electron density based on that, there will be some electron density (with lower signal) looking like the structure of the added atoms. If you remove those atoms again, the artifact will disappear again. If you replace an correct part of the model with incorrect atomic positions, the density will be a mix of both. Depending on the quality of the diffraction data (especially the resolution limit), the density will reflect the correct or the incorrect model more. In the scenario of removing a spurious part of the model, you could say it deletes the density, in the scenario of removing an incorrect part of the model, you could say it corrects the density.

Thank you for this very detailed answer. I am still trying to wrap my head around this, so I have a question for you: is the reason why omit mapping works because of the fact that every atom contributes to a diffraction spot? I'll go read all the links you've included. I think I am getting it, but I still haven't had that "a-ha" moment yet.

@Dunois Yes, exactly. The phases of each structure factor are almost correct, and the amplitudes are experimental (so unbiased but they do have experimental error, e.g. noise). It is worse to have a bad part of the model compared to leaving that part out all together.

Ah ah, so even if the phases from that particular part of the structure itself are missing/omitted, there are still phases from elsewhere that do contribute to the electron density at that location. So if the feature were a model artifact, these other contributive phases wouldn't exist, and the omit map would show zero density there. Is this correct?

No, the language you are using is incorrect. There is no one-to-one correspondence of spots in the diffraction pattern and locations in the unit cell. Intensities and phases are index by (hkl) or $\vec{h}$, i.e. which plane of reflection they originate from. Locations in the unit cell are designated by x, y, z or $\vec{r}$. Every atom contributes to each diffraction spot. Every diffraction spot contributes to the calculated electron density in a given location.

Here is a way to say it: Even if the phases lack the information from the omitted part of the model, they are still good approximations of the phases of the actual structure. What really carries the information is that the structure factor amplitudes are derived from the diffraction pattern, which of course reflects the real structure. The amplitudes carries less information than the phases, but as long as you don't introduce incorrect information via the phases, that weak signal from the intensities is good enough to build or correct the model.

Thank you for the explanation and also pointing out where I am mistaken. I think I was trying to formulate the essence of what you've stated, albeit with very poor language (as you pointed out). But is my reasoning regarding the model artifact aspect of this nevertheless correct? Or is it more that an omit map does not "delete" the density in question so much as it "un-defines" it? (So that we can correct/rebuild it.)

Let us continue this discussion in chat.

@Buttonwood and Karsten Theis have provided a lot of information and feedback to me through past discussions (especially the former). That taken into account, I will go ahead accept Karsten Theis ' answer given its immediate relevancy to the post itself and utility to fellow beginners who are likely to stumble upon this thread; I think this answer can help people glean insights from the others here.

Experimental phasing and molecular replacement provide initial phases used to start model building. Later on (refinement, validation) phases are calculated from the model.

If the phases are what determine the final electron density irrespective of the amplitudes used

Both amplitudes and phases are used to determine the electron density.
If you calculate the phases omitting part of the model, the electron density in this place won't be biased by that part of the model. (It may be biased by the absence of the model in this place).

I apologize if this is a dumb question (and if I am running in circles), but I got the impression that if we don't have the phases (say for a part of a model), it just wouldn't get built even if we have the amplitudes for that part, right? Or is it that in this case (of omitting a part), there would be an electron density (because the amplitudes are still there), just that it would be featureless?!

Leaving out part of the model results in lower intensity of the electron density in that region, and in noise throughout the unit cell. You might get some bias with the omitted region has a peculiar shape (like leaving out atoms whose x-coordinate is between 0 and 0.1 in fractional coordinates - but who would do that?).

The following doesn't fit into a comments box.

From the perspective of small molecule crystallography (say, ShelX), once you solved your model completely or completely enough to engage refining, you may recognize reflections leaving the straight line in the Fobs vs. Fcalc map.  Within reason, it is possible to omit these outlier reflections from the .hkl file and still obtain a refinement which is both chemically reasonable, and converges.  Not necessarily designed for macromolecules, but one of tutorials for Olex2 showcases this for a small complex.
Independent to the intent to remove  a few outliers, it is possible to constrain structure solution and refinement to reflections within a certain range of resolution, even if there are recorded data passing beyond this interval.  This equally falls into the category of omission of data.

The second perspective is when experimentally recorded electron density can't be nicely attributed using the «standard» approach to solve a structure.  Orientationally disordered solvent molecules may be an example for this to happen, or long flexible alkyl chains lowering crystallinity of a sample.  The old way is to use Platon's squeeze instruction (again, small molecule perspective, solution and refinement with ShelX, reference) to keep pockets of electron density in Fobs for those you do not attempt (any more) allocation to specific atoms at specific positions.  Eventually, you document a model lacking a few atoms, including a statement that such a map was used and how many electrons are in these pockets (application in a paper, a small molecule tutorial).  The visualization of the structure model highlighting these pockets and writing a .cif file including the reduced .hkl help others to understand what you did encounter.
For small molecule samples, there are libraries like DSR with preset geometries about molecules known to possibly yield such problems.  If there is enough experimental evidence, their input, complementary to squeeze's  maps, may be very helpful.

To the first bullet: You are talking about omitting reflections. In the omit map, part of the model are omitted from phase calculation in preparation for calculating maps. Omitting reflections is done in macromolecular crystallography as well, to assess bias via calculation of R$_\mathrm{free}$. This is part of assessing model bias, but only distantly related to the question the OP posed.


text:
Under what conditions does Real Gases show Ideal Behaviour?

MY textbook says : Real gases show Ideal Behavior under low pressure ,high temperature.

So I wanted to know what happens to real gases at high pressure ,low pressure
Do they also show ideal behavior under those conditions?

In order to make comparisons you have to specify other conditions i.e temperature and volume. so do the other conditions remain same or not?

When real gases are at high temperature, the kinetic energy prevents any gas particles from interacting via intermolecular forces. With low pressure, the gas particles are separated enough that the intermolecular forces are sparse, therefore, giving rise to the ideal behavior since ideal gases are defined as non-interacting particles.
When real gases are at high pressure or low temperature, they deviate a lot from ideal gases.
If you compress a gas enough, you can induce a phase change.

The van Der Waal equation is $$\left(p+\frac a{V_\mathrm m^2}\right)(V_\mathrm m-b)=RT$$
Here $V_\mathrm m$ is molar volume. When pressure is low and temperature is very high, we can qualitatively say that the molar volume will be very large. Due to this the volume occupied by the molecules (given by $b$) becomes insignificant.
The pressure is low but the molar volume $V_\mathrm m$ is very large and thus the term $\frac a{V_\mathrm m^2}$ comes out to be much much lower than the already low pressure.
$$p+\frac a{V_m^2}\approx p$$ and $$V_\mathrm m-b\approx V_\mathrm m$$ Thus the gas shows ideal behavior at low pressure and high temperature.


text:
Counting electrons in Ozonolysis
I can't understand the electron count made on this page about Ozonolysis. (http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/ozonolysis/Oz.html)
This is the specific part of the article:

From my calculations the oxidation of an alkene to an ozonide is a 4 electrons oxidation and not 6e, in fact the two carbons in the double bond loses 2e each from bonding two oxygens.
Anyway I agree with the 4e oxidation from alkene to two carbonyl compounds. Where am I wrong?
Thanks in advance.

You misstated what you read. Formation of the ozonide from an alkene is a 6-electron oxidation. The dimethyl sulfide is oxidized to dimethyl sulfoxide by 2-electrons meaning that the ozonide is reduced by 2-electrons. So alkene to two carbonyl compounds is a net 4-electron oxidation. Welcome to ChemSE. Good choice of citation!

Thank you for your comment. In other words I can't understand why ozonide formation from alkene is a 6-electron oxidation (instead of 4-electron that I calculated).

I don't know how you count. If you accept that alkene --> 2 carbonyls is a 4e oxidation and DMS --> DMSO is a 2e oxidation then alkene --> ozonide must be a 6e oxidation. Read the link on oxidation levels in your citation.

Yes, I understand. But if the problem was to calculate the electrons number involved in the oxidation of an alkene to form an ozonide (as in the link on oxidation levels in the citation), how would you do?
P.S.: So not deducing that the number is 6 from indirect path

The oxidation levels link only assigned charges to a Grignard reagent but used partial charges in all other examples to determine the polarity of the bonds. Assigning charges to every atom in an organic compound (sounds like homework) is not a productive exercise. You want the relative change. The method used is direct. Let me say that I am quite familiar with the intent of the author of the website you have cited. If you want, I will expand on the explanation with an Answer. https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange.

If you want to demonstrate that the addition of ozone to an alkene is a 6-electron oxidation of the carbon framework, here are two methods. To assign a charge to the carbon atoms of a neutral compound (sum of all charges equals zero) treat all hydrogens as +1 and all oxygens as -2, then do the math. Each carbon in ethylene (1) has a calculated (in red) value of -2. Molozonide 2 has carbons with a charge of +1 but, for convenience, they are written as zero and +2. Already one can see that 1 $\rightarrow$ 2 is a 6-electron oxidation [-4e $\rightarrow$ +2e]. Collapse of the molozonide gives formaldehyde (3) of carbon charge zero and zwitterion 4 with carbon charge +2. These species recombine to form ozonide 7 that has the same carbon charge as the molozonide 2. That the zwitterion 4 is the equivalent of formic acid 6, follow the steps 4 $\rightarrow$ 5 $\rightarrow$ 6. Thus, both the molozonide 2 and ozonide 7 are equivalent to one formaldehyde and one formic acid.
Alternatively, oxidation of ethylene (1) to ethylene glycol (8) [-4e $\rightarrow$ -2e] is a 2-electron oxidation and a second 2-electron  oxidation [-2e $\rightarrow$ 0e] affords two moles of formaldehyde. Oxidation of one of the formaldehydes to formic acid [0e $\rightarrow$ +2e) is the third 2-electron oxidation. Both pathways demonstrate that the oxidation of ethylene to formaldehyde and formic acid are 6-electron oxidations.


text:
Why does ozone have a dipole moment different than 0
First of all, I know that the reason for this is the bent structure of the molecule.
However, we were taught in my class that the dipole moment of a molecule is equal to the vector sum of the dipole moment of each bond in the molecule.
But ozone only has bonds between two oxygens, so why is the dipole moment of each bond not equal to zero?

It can't really be understood without orbitals.

@IvanNeretin Please elaborate, we have learned about orbitals (the basic stuff), and I wonder if it's outside the scope of my course.

Well, then you must know what is $sp^2$. Also, you surely have seen the structure which looks like $\ce{O=O+-O-}$, only bent.

Because it's not just oxygen atoms. The oxygen atom in the middle has two oxygen atoms bonded to it and the ones on the end have one. Because the environment around the atoms are different, they are different. Specifically, they have different electronic properties so there is a dipole moment.

The electrons are not shared equally among the oxygen atoms.  The central oxygen atom alone donates three electrons to the covalent bonding whereas the other two oxygen atoms combined donate the other three (there is a 3-center, 4-electron pi bond in which two of the rlectrons are shared between just the end atoms).  Since the central oxygen atom is providing an excess of bonding electrons, it becomes positively charged.  The other oxygen atoms, contributing fewer electrons to the bonds and retaining more "lone pairs", take the compensating negative charge.
Trisulfur is a similarly polar elemental species, with a bent structure and unequal electron sharing similar to ozone.  Sulfur dioxide and disulfur monoxide, the latter found along with trisulfur on the Jovian moon Io, have a similar unequally-shared-electron structure with sulfur in the middle.  But we take little notice in the latter two cases because the sulfur, being less electronegative than oxygen, would be positively charged anyway.

However, we were taught in my class that the dipole moment of a molecule is equal to the vector sum of the dipole moment of each bond in the molecule. But Ozone only has bonds between two Oxygens, so why is the dipole moment of each bond not equal to zero?

Chemical education is in such a sorry state. Why do they teach chemistry like chicken or eggs story? Did the chicken come first or the egg? I have yet to find a single general chemistry book which even mention how the dipole moment is measured (even in one line). If you want to pursue science, instead of thinking about imaginary fairy tales, ask the teacher how would you experimentally measure the dipole moment of a gas? How do we know that a given gas has a permanent dipole moment?
The dipole moment of gases is mostly determined by microwave spectroscopy. In chemistry, experiment usually comes first (unlike chicken and egg). So if you shine microwaves on ozone, ozone would show a rotational spectrum. This is a signature that yes ozone must have a dipole moment. Now you have to consider a consistent electron distribution which would lead to a permanent dipole moment.
Similarly, electron diffraction experiment will tell you that ozone molecule is not linear. When it is not linear and it is rather bent, and yet it has a dipole moment, now people have to think of the electronic distribution.

Quite a tangent and not wholly relevant to the question asked. In fact, I’m arguing you’re not answering the question at all.

In such cases, as I always suggest, write a separate answer if you feel there is room for improvement or have a better explanation, which you have done shortly.

It is common practice at a high school or first year undergraduate level to determine whether bonds are in any way polar (and, by extension, whether a molecule can be a dipole) just by looking at the atoms on either side of the bond. If you do that for ozone, you will obviously come across two $\ce{O-O}$ bonds and would have to assume that these bonds are unpolar.
A more sophisticated on-paper analysis (and one that is rarely done in a lecture setting even at higher university levels) would not only take the atoms on either side of the bond but the entire fragments. It would recognise that the $\ce{O-O}$ bond is, in fact, an $\ce{O-[O2]}$ bond, where one side has a single oxygen atom but the other side a two-atom fragment. The mere presence of a third, distant atom means that the two sides of this bond are not equivalent and thus this bond must be ever so slightly polar.
Symmetry considerations then lead to understanding that the bond on the other side has exactly the same feature; thus, these two slightly polar bonds are even in strength but not antiparallel. The overall dipole moment loosely corresponds to vector addition of the two polar bonds as you would expect; naturally, it falls right into the axis and planes of symmetry as theory expects.

Sorry to say but this is also hand waving and not much different from what the rest have stated.

@M.Farooq I don’t see any other answer stating anything about how to a priori derive at the two oxygens being different without invoking additional premises (such as a positively charged central oxygen). My method even works if you assume $\ce{^.O-O-O^.}$ as a structure for ozone. But then again, you do you.

Jan, The OP is basically learning about dipole moments. I think Oscar Lanzi provided the classical explanation which the OP apparently understood.

The hybridisation of the central oxygen atom is $\mathrm{sp^2}$. The major contribution in the dipole moment of ozone is due to  the lone pair in the directional orbital of the central oxygen atom.
The orbitals containing valence electrons of other two oxygen atoms has more directional property. They unsuccessfully tend to cancel the dipole moment created due to lone pair on central atom .

You act as if the central atom were the negative end of the dipole.  It isn't.


text:
I need help finding several immiscible liquids
I am trying to fill a vial with as many immiscible liquids as possible so that they form many separate layers. I want to do this as a way to visualise how liquids with different refractive indexes flow past one another.
TLDR Is there some list or way of finding out which transparent liquids (at STP) I can mix in a container which won't mix together or react with one another?
The liquids need to:
a) Be insoluble with eachother even if container is shaken.
b) Be at least as transparent as olive oil or better.
I have seen stuff like this done with: maple syrup, honey, milk, dishwashing liquid, water, oil, rubbing alcohol and lamp Oil.

However, most of these substances aren't transparent like I need (I'm willing to use any liquid that's  more transparent than Olive oil even if it has a slight hue). I'm also not sure if these substances would react with each other if the container is shaken.
I don't mind mixing soluble substances together in order to change a particular layers properties (such as density, viscosity, refractive index, or surface tension).
From my trial and error tests, I've found that the more types of liquids I add, the higher the chance that a liquid will cause a reaction between two previously unreactive substances.
Apart from common household liquids like cleaners, water, vinegar, bleach, etc, I have the following liquids at my disposal (all of which are transparent in form).
Kerosene

Kerosene (Paraffin Oil)
Lamp Oil
Glycerin
Liquid Paraffin (Mineral Oil)
Coco Betadine
Methyl Salicylate
Transmission Fluid
Micellar
Dishwashing Liquid
Motor Oil
Isopropanol
Mineral Turpentine
Methylated spirits
White Spirits
Acetone
Ammonia
Sugar Soap
Lemon Juice
'Clear Glue' (like alcohol based
PVA glue)
Castor Oil
Linseed Oil
Grapeseed Oil
Eucalyptus Oil
Propelyne/Dipropelyne Glycol
Glucose Syrup
Vegetable/Olive/Sunflower Oil
A-B Epoxy
Hydrogen Peroxide 6%
Poly Sorbate

If there's anything else that's reasonably common (that I could get ahold of on eBay, supermarket or hardware store, etc) that would help with this please mention it!
I really hope you can help me figure out what liquids I can use!

The liquids in your example are mostly layered not immiscible. For example water, milk and dishwater soap are miscible. So do you just want layers, which are unstable, or truly immiscible phases.

It's far more difficult to make these layered liquids stable to mixing. It's unlikely you'll be able to do more than four layers without using toxic/dangerous materials. The safer liquids I can think of for this experiment would be perfluorocarbons ("fluorinert") and polysiloxanes (liquid silicones). These are fully immiscible with water, and can be fully immiscible with some kinds of hydrocarbons.

Glycerol has a density $1.263$ g/mL. It is miscible with water in all proportions. So by changing the ratio glycerol:water, you can make a large number of mixtures having all possible densities between $1.263$ and $1.000$ g/mL. And all these mixtures are made of polar molecules. On the other hand, you may make up all sorts of mixtures with non polar substances like dichlormethane (density $1.320$ g/mL) and cyclohexane (density $0.77$ g/mL) in different proportions. These non polar substances are miscible in all proportions. These mixtures can have densities varying between $0.77$ and $1.32$ g/mL. And they will never mix with glycerol-water mixtures.
So try to make $1:9$, then $2:8$, then $3:7$, etc. mixtures of glycerol-water, then of dichloromethane-cylohexane. Measure their density. Or simply try to mix them just to see which one is the heaviest.
You may get as many different mixtures as you want it, that can be superimposed in your container. You may also choose ratios which are different from the suggestion $1:9$, $2:8$, etc.
As you may have trouble seeing the separation between the layers, you should add a droplet of ink in all aqueous mixtures.

Thanks for the responses, so if I can somehow prevent the layers from mixing, I could theoretically have an infinite amount layers stacking on top of eachother with just 2 oil based and 2 water based liquids mixed in different ratios. Maybe I could get around this by surrounding each liquid layer with a thin film or surfactant to stop layers mixing together? Perhaps if the fluids have a low enough viscosity, I could contain the liquids in a ball shaped container to prevent them from mixing by being tilted?

Considering the limited stable options, and since I have suitable safety equipment available, I might try adding both Mercury and Galistan (room temperature liquid metal that's immiscible with Mercury) into the container to add additional stable layers.

Of course you can add mercury. But it is not colorless, as expected in you first demand.


text:
Wiki’s explanation of the rate of chemical reaction
Could you please help me understand the Wiki’s explanation of the rate of chemical reaction?

It says: the rate of the chemical reaction is directly proportional to the product of the activities or concentrations of the reactants.

I can’t wrap my head around this statement. Could anyone explain it in other words please?

Simply put, if your reaction is A+B, then it is [A]*[B].

Still not ringing a bell... lets say, that the sum of concentration of reactants is 200 mmol/l. How is this value relevant to the rate of reaction?

It isn't.$\mathstrut$

If I understood it correctly, this is applied in kinetic reactions: v=k*[A]*[B]....does the explanation only mean that the speed of reaction depends on the conc of reactants because when they are higher, the reaction rate is higher? or am i missing something?

Not only that, but it says how exactly rate depends on concentrations.

Is it basically saying that concentrations of reactants is gonna influnce the speed of reactions in reactions with higher order number (order 0 isnt affected)

The quoted statement is only true if the reactants are directly reacting in an elementary step.

Wiki ≠ Wikipedia. Also, the URL to the target article is missing, and questions like this are usually getting closed as opinion based since it is essentially a request to digest studying material. This corresponds to the cognition, not chemistry.

The rate of a reaction, or how fast reactants become products, depends on the collisions of particles. This is because when a particle collides, the reaction occurs (provided that it has enough KE). So, if we increase the amount of particles (the concentration), we will have more collisions and therefore the reaction will go by faster.


text:
Do enzymes that digest ᴅ-glucose react with ʟ-glucose?
ᴅ-Glucose is common in nature and ʟ-glucose is synthesized in the lab.
I know that humans can't use ʟ-glucose in their aerobic pathways because it doesn't match the active site of the enzyme, but why can't our cells use ʟ-glucose?

You can't put a left shoe on your right foot. Same thing here. Enzymes are chiral too.

Alice speaking to Kitty. ‘How would you like to live in Looking-glass House, Kitty? I wonder if they’d give you milk in there? Perhaps Looking-glass milk isn’t good to drink—' I have always felt that Lewis Carroll was aware of the mirror image world. Is milk (lactose) on the other side comprised of L-glucose and L-galactose? Is that why Alice questions whether it is good to drink? @Ivan Neretin

Wikipedia has a pretty good answer; https://en.wikipedia.org/wiki/L-Glucose. It includes one species able to digest L-Glucose, and the information that the taste is the same (which makes you wonder about those chiral taste receptors).


text:
If I hard boil eggs in a pressure cooker, will I get the same results regardless of altitude?
This seems like it should be obvious but for some reason (my chemistry is very rusty) I'm having a hard time articulating why the hardboiled egg recipe in an instant pot should yield the same results regardless of altitude.
Questions:

Would I get the same results at altitude using the pressure cooker?
Why? or why not?

Example recipe using pressure (via damndelicious.net)
INGREDIENTS:
1 cup water
6-12 large eggs

DIRECTIONS:
1. Add water to a 6-qt Instant Pot®. Place metal trivet into the pot. Gently add eggs on top of the trivet.
2. Select manual setting; adjust *pressure to high*, and set time for 3-7 minutes.* When finished cooking, quick-release pressure according to manufacturer’s directions.
3. Cool eggs in a bowl of ice water for 1 minute. Drain well and peel.

NOTES:
3 minutes: Soft, just set whites and runny yolks
4-5 minutes: Medium, firmer whites and jammy yolks
6 -7 minutes: Hard, firm whites and fully set yolks

The question is what exactly sets the pressure? Having an Instant Pot myself, I believe the pressure is set by the temperature of the heating element, so the pressure inside the pot should be independent of the pressure outside the vessel.

A traditional pressure cooker uses a weight that blocks the opening. In that type of setup, the pressure of the pressure cooker is relative to ambient, not an absolute setting.

@Andrew - the Instant Pot does not use a weight. High vs low pressure is just a setting on the control panel. And the safety valve does not set the pressure. Just different equipment.

Instant Pot uses a mechanical technique to control the pressure, described here: https://instantpot.com/portfolio-item/how-electric-pressure-cookers-work/

According to their website, the manufacturers say "The Ultra and Max models of Instant Pot come with the ability to adjust pressure settings to account for altitude."
So, yes, the timings will be affected by altitude, even in a pressure cooker.  This is probably because the pressure inside the cooker is controlled by some method that uses the difference in pressure between the inside and the outside.  Decrease the external pressure and that will decrease the internal pressure.
There is a USDA factsheet on high altitude cooking: https://www.fsis.usda.gov/shared/PDF/High_Altitude_Cooking_and_Food_Safety.pdf
The size of the eggs will probably have an even bigger effect on the cooking time though, with larger szed eggs taking longer to cook.  Since you're aiming for hard-boiled eggs though, it shouldn't be too difficult.  Your "recipe" suggests 6 - 7 minutes and even making a very generous allowance for altitude and egg size, 10 minutes should be fine.  If you're aiming for hard-boiled eggs, it would be difficult to over cook them.
Soft-boiled eggs would require much more precise timing and that can be best achieved by, and I have to say this, eggsperiment.

There is language elsewhere on their site that says they explicitly increase cooking times for altitude, indicating that, as you said, the pot creates a pressure difference in pressure-cooker mode, hence the internal pressure, and thus the cooking times, will be affected by ambient pressure:  "there is no indication the altitude has been adjusted. However, you may note the Cook time will be adjusted once the altitude has been updated. For example, depending on your altitude, a 00:30 minute Cook cycle may be automatically adjusted to 00:34 minutes when it begins cooking."

eggspiriment!!!!

I actually prefer soft boiled. I'm at sea level now. I guess the thing to do is perfect my recipe here and then take my pressure cooker to altitude and try it there. ...or zoom with my friend who has the same cooker and lives at altitude.

An ideal pressure cooker is an isolated vessel in which what goes on inside is independent of what happens outside. It allows the temperature of what you put in the cooker to be raised until the vapor pressure reaches a set point (a pressure higher than the pressure at the standard boiling point of water). Since cooking time depends only on the temperature of the water it should remain the same at different altitudes provided the inside of the cooker is pressurized to the same level.
However, pressure cookers are usually fitted with safety valves (or some system that regulates the pressure to prevent accidents). A safety valve prevents the pressure within the vessel from climbing to a dangerously high level which might result in a failure. The safety valve is actuated whenever the pressure difference between interior and exterior exceeds a safety threshold, reducing the internal pressure back to a safe level. The pressure inside such a cooker does depend on the external pressure (it cannot be higher than the difference deemed safe).
As an interesting aside, the Wikipedia writes that

The pressure cooker was invented in the seventeenth century by the physicist Denis Papin, and works by expelling air from the vessel, and trapping the steam produced from the boiling liquid inside. This raises the internal pressures and permits high cooking temperatures.

In fact, what Papin was first to do was to routinely incorporate a safety valve into his pressure cooker designs. The modern pressure cooker would not be safe without such mechanisms.

When we cook something inside a pressure cooker, we are able to manipulate the pressure inside of the pressure cooker. This is useful, because we know that changes in pressure can cause changes in boiling point. Let me explain.
Boiling Point In Terms of Pressure
A chemists definition of boiling point is the point at which the vapor pressure of the liquid overcomes the atmospheric pressure. The vapor pressure increases as we increase the temperature of a liquid.
Let's say that we were to boil a pot of water at sea level, where the pressure is very high. It would require a lot of heat to get the vapor pressure high enough to overcome the atmospheric pressure. Now let's say that we boil a pot of water at the top of Mt. Everest. Since the air pressure is so low, it would require less heat to overcome the atmospheric pressure and therefore boil the water. Thus, a pot of boiling water at 0 ft above sea level is hotter than one at 100000 feet above sea level.
Now that you understand the basic concept of boiling, lets investigate how a pressure cooker works. By manually increasing the pressure inside the container, we are able to cause the water to boil at a higher temperature. The outside atmospheric pressure may be different, but that doesn't matter, because the pressure cooker is sealed and the pressure is kept constant inside.
Hope this helps!

When water evaporates the hotter molecules leave and they take some heat with them. Which cools down what's left in the pot. You keep heating your pot and more molecules evaporate, which in turn cools down the pot more. When you just started to heat water - the amount of heat absorbed by the water is greater than the cooling effect of evaporation and the temperature keeps rising. But once you get to boiling - these processes balance each other. You keep heating, but evaporation keeps cooling down the pot at the same rate. Hence no net increase of temperature.
But if you were to trap the evaporated molecules in the pot? They wouldn't leave and thus they wouldn't cool down the pot. So you'd be adding more heat and there won't be evaporation to cool it down. The temperature will keep rising and rising.
When you cook in normal conditions the air particles keep hitting water (they exert pressure) which controls how many water particles are able to evaporate. At high altitudes the pressure goes down and so the water evaporates more readily (and thus cools the pot down faster). The goal of pressure cooker is to trap some of the evaporated water and thus keep the pressure high.
PS: Physics SE is a better place for such questions.


text:
When writing the net ionic equation, if one of the products ionizes, what is the most appropriate way to account for this in the answer?
Take these two practice problems and their solutions from Ebbing (8th ed) that involve completing the molecular equation, then writing the net ionic equation:
4.42a: $\ce{Ca(OH)2(aq) + 2H2SO4(aq) -> }$
soln:
$$\ce{Ca(OH)2(aq) + 2H2SO4(aq) -> 2H2O(l) + Ca(HSO4)2(aq)}$$
$$\ce{OH-(aq) +2H+(aq) +SO4^{2-}(aq) -> H2O(l) + HSO4-(aq)\\ \text{ (ionizes to } H+ + SO4^{2-})}$$
4.42c: $\ce{NaOH(aq) + H2SO4(aq) -> }$
soln:
$$\ce{NaOH(aq) + H2SO4(aq) -> NaHSO4(aq) + H2O(l)}$$
$$\ce{OH-(aq) + H+(aq) -> H2O(l)}$$
In 4.42a the authors do not ionize the product, only note to the side that it does ionize. In 4.42c the authors do ionize the product and account for this in the net ionic equation.
In general, is one practice preferable or more correct than the other?

Please note that I did not intend to answer my own question. After reading some of the answers, I feel it is necessary to provide clarification for future readers. Hopefully, my answer is enlightening.
Let us begin with a point of interest, reading again from Ebbing:

Phosphoric acid is an example of a triprotic acid. By reacting this
acid with different amounts of a base, you can obtain a series of
salts:
$$\ce{H_3PO4(aq) + NaOH(aq) \rightarrow NaH_2PO_4(aq) + H_2O(l)}$$
$$\ce{H_3PO4(aq) + 2NaOH(aq) \rightarrow Na_2HPO_4(aq) + 2H_2O(l)}$$
$$\ce{H_3PO4(aq) + 3NaOH(aq) \rightarrow Na_3PO_4(aq) + 3H_2O(l)}$$

And now, let's consider the problem.

Problem 4.42 Complete the right side of each of the following
molecular equations. Then write the net ionic equations. Assume all
salts formed are soluble. Acid salts are possible.
a. $\ce{2H_2SO_4(aq) + Ca(OH)_2(aq) \rightarrow} $

To give us insight into the nature of this reaction, let us first consider a complete neutralization reaction for calcium hydroxide and sulfuric acid and see what that looks like.
$$\ce{1H_2SO_4(aq) + 1Ca(OH)_2(aq) \rightarrow CaSO_4(s) +2H_2O(l)}$$
Now, let us again consider our original problem in terms of the complete neutralization. We divide the coefficients by two.
$$\ce{1H_2SO_4(aq) + \frac{1}{2}Ca(OH)_2(aq) \rightarrow \text{ incomplete neutralization}} $$
This helps give us an idea of how to predict the products for the molar ratio provided. We know the acid salt formed is soluble as we are told this in the problem statement.
$$\ce{2H_2SO_4(aq) + Ca(OH)_2(aq) \rightarrow Ca(HSO_4)_2(aq) + 2H_2O(l)}$$
For the net ionic equation, we consider that $H_2SO_4$ is a strong acid and will ionize completely in solution and $Ca(OH)_2$ is a strong base and will dissociate in solution to be present entirely as ions.
$$\ce{4H^+(aq) + 2SO_4^{2-}(aq) +} Ca^{2+}(aq) \ce{+ 2OH^-(aq)} \rightarrow  Ca^{2+}(aq) \ce{+2HSO_4^-(aq) + 2H_2O(l)}$$
Eliminate the spectator ions.
$$\ce{4H^+(aq) + 2SO_4^{2-}(aq) + 2OH^-(aq) \rightarrow 2HSO_4^-(aq) + 2H_2O(l)}$$
Reduce coefficients.
$$\ce{2H^+(aq) + SO_4^{2-}(aq) + OH^-(aq) \rightarrow HSO_4^-(aq) +  H_2O(l)}$$
Note that $\ce{HSO_4^-}$ ionizes to $\ce{H^+}$ and $\ce{SO_4^{2-}}$. Substitute and reduce.
$$\ce{H^+(aq) + OH^-(aq) \rightarrow H_2O(l)}$$
We accept either form of the answer, though if to go with the former you must notate the species that ionizes, otherwise is incorrect.
$$\ce{2H^+(aq) + SO_4^{2-}(aq) + OH^-(aq) \rightarrow H_2O(l) +  HSO_4^-(aq) \text{ (ionizes to } H^+ \text{ and } SO_4^{2-}\text{)}}$$

Preferable is the correct one. 4.42a is wrong. There is no way to have HSO4- after neutralisation, but SO4^2- before it.

Note that molecular equations are not generally correct, as ions in solution do not belong to particular counterions. E.g equimolar solution of KOH and NaCl Is equivalent to the equimolar solution of KCl and NaOH.

Note that for H2SO4 mass concentration > 1 g/L (about  [H2SO4] >0.01 M, pH<2 ), the majority of H2SO4 exists as HSO4-. For pH=0 it is about 99%. So you cannot expect initial H2SO4 contains SO4^2-, but after partial neutralization by Ca(OH)2 there are suddenly HSO4-. The latter are there all the time.

We consider mixing of $\ce{H2SO4(aq)}$ and $\ce{Ca(OH)2}$ in molar ratio 2:1.
Molecular equation version:
$$\ce{2 H2SO4 + Ca(OH)2 -> Ca(HSO4)2(aq) + 2 H2O}$$
Eventually if $[\ce{Ca^2+}][\ce{SO4^2-}] = [\ce{Ca^2+}] \cdot K_\mathrm{a2} \cdot \frac {[\ce{HSO4-}]}{[\ce{H3O+}]}> K_{\mathrm{sp,}\ce{ CaSO4}}$ ($\mathrm{p}K_\mathrm{a2}=1.99$):
$$\ce{Ca(HSO4)2(aq)  <=> CaSO4(s) + H2SO4(aq)}$$
Ionic equation version:
Separated solutions:
\begin{align}
  \ce{H2SO4(aq) + H2O &-> H3O+(aq) + HSO4-(aq)}\\
  \ce{HSO4-(aq) + H2O &<<=> H3O+(aq) + SO4^2-(aq)}\\
  \ce{Ca(OH)2(aq) &-> Ca^2+(aq) + 2 OH-(aq)}
\end{align}
More exactly:
\begin{align}
  \ce{Ca(OH)2(aq) &<=>> Ca(OH)+(aq) + OH-(aq)}\\
  \ce{Ca(OH)+(aq) <=>> Ca^2+(aq) + OH-(aq)}\\
\end{align}
Mixing solutions with 2:1 molar ratio:
$$\ce{2 H3O+(aq) + 2 HSO4-(aq) + Ca^2+(aq) + 2 OH-(aq)-> 4 H2O + 2 HSO4-(aq) + Ca^2+(aq) }$$
And after elimination of spectator ions:
$$\ce{2 H3O+(aq) +  2 OH-(aq)-> 4 H2O }$$
Then, further ionizations of the bisulphate, as $\ce{[H3O+]}$ decreased:
$$\ce{ HSO4-(aq) <=> H3O+(aq) + SO4^2-(aq)}$$
Again, eventually if $[\ce{Ca^2+}][\ce{SO4^2-}] = [\ce{Ca^2+}] \cdot K_\mathrm{a2} \cdot \frac {[\ce{HSO4-}]}{[\ce{H3O+}]}> K_{\mathrm{sp, }\ce{CaSO4}}$ ($\mathrm{p}K_\mathrm{a2}=1.99$):
$$\ce{Ca^2+(aq) + SO4^2-(aq)  <=> CaSO4(s)}$$

I am not sure where your solutions come from but 4.42a isn't correct.  The products are calcium sulfate and water, and not, as Poutnik points out, $\ce{HSO4-}$. So the molecular formula should finish like this:
$\ce{Ca(OH)2(aq) + H2SO4(aq) -> CaSO4(aq) + 2H2O(l)}$
Calcium sulfate is actually pretty insoluble so would normally appear as a precipitate but the 5th edition of Ebbing that I looked at does say to treat all salts as soluble for this question. The ionic equation for the reaction should be this:
$\ce{2OH-(aq) + 2H+(aq) -> 2H2O(l)}$
Your solution to 4.42c is correct. So, that is what is intended.

In excess of acid, there will be hydrogen sulphate on the left and on the right, until CaSO4(s) starts precipitation, what is not much probable, considering Ca(OH)2(aq) at the beginning. pKa2=1.99

$$\ce{Ca^2+ + 2 OH- + 2H3O+ + 2HSO4- -> Ca^2+ + 2 HSO4- + 3H2O}$$

@Isobutane, the solutions come from General Chemistry: Solutions Manual, Ebbing, 8th ed.


text:
Why are these H the most acidic?

I don't quite understand why these two hydrogens are the most acidic. From a hybridization perspective, these are connected to a $\mathrm{sp^3}$ carbon (I thought the more s character, the more acidic). So why is it not any of the other hydrogens (connected to $\mathrm{sp^2}$ carbon)?

Consider the structure of phenol, how stable is the phenoxide ion?

The two hydrogen atoms are acidic, but things change after one of them is lost. But I think that the question was aimed at the acidity of either of the hydrogen atoms and they are equivalent, so let's see why:

As you can see in the given figure that after abstraction of $\ce{H+}$ the conjugate base can resonate to attain aromaticity. It is this stability that makes the hydrogen atom acidic. Now to remove the second $\ce{H+}$ the story changes, but I would be leaving out that discussion as it has not been asked in the question.

If the answer smoothens your path of learning chemistry, do take a moment off to accept it.


text:
Is using oxidized aluminium baking forms safe?
My apologies in advance if this is not the best forum for this kind of question.
Recently, I've bought a few non-anodized aluminium bread-baking forms, such as this one:

As you can see, there is already some discoloration on them when they come from the shop. My knowledge of chemistry is rudimentary but I presume it is an oxidized aluminum layer (do correct me if I am wrong here).
Now, my wife decided it is a good idea to wash these forms in a dishwasher. The forms became even darker, becoming rather gray over their entire area. There are a lof of posts on various forums and blogs telling you to not put aluminium cookware in dishwashers, and most of them go on about how to remove the oxidized layer.
For my purposes, I do not care how it looks. I would like to know whether it is safe to bake bread in those (post-dishwasher and without removing the oxide layer). There is very little information about this, as the online discussions revolve mostly about removing the gray layer. I've read someone saying that the oxide layer can even be beneficial as it might prevent aluminium leaking into food. This does make sense, presuming that the oxide itself is not harmful and does not leak. Could someone comment on this please? Any reference to scientific literature would be a bonus.
As a backstory, the reason I bought these forms is because they are traditionally used for making specific sorts of rye sourdoughs in the Baltic region. So I'd like to give them a go before buying more modern (and more boring) stainless steel forms.
Thanks in advance!

https://chemistry.stackexchange.com/questions/141001/why-is-aluminum-oxide-resistant-to-corrosion just to say that Al object always have a layer of oxide onto them. All the discussions about cookware discoloration etc are basically based on aesthetic. There are not really existing ruined pans etc, at least they perfectly keep their function. We don't give health advices, this is to say that your form is substantially the same independently of the look. Anodization is just a way to control the film of oxide. It forms anyway and stain or discoloration is just matter of optical phenomena.

Here is an answer to the title question :  the alumina layer is safe. Even if you swallow a bit, this oxide will not react with the acid in the stomach, and the whole stuff will go through the intestine without being modified.

@Maurice unfortunately I can not under vote a comment. I invite OP to read my comment as the answer is it in there. Plus the obvious fact that if something is in use as cookware since long time there cannot be evident acute or long term effects by just ruining it's appearance.

@ Alchimista. Sorry ! I do not understand your comment. All I have said is that aluminum oxide is not soluble in the dilute HCl solution in the stomach. What do you have against it ? Of course, your comment is valid, but it is not in the same domain as mine.

@Maurice because I can't see how subnanoparticles of Al oxide won't dissolve in the stomach. The fact remains that we know it is safe to use Al cookware. All the rest, can eventually shorten of a good looking pan. That will keep functioning for years of mishandling. This is even no to say that we don't get Al intake by the cookware, but just to put the issue in the right perspective.

In accord with this forum's policy on not providing health advice, I  will just comment on the chemistry surrounding any created Al3+ interacting with the superoxide radical anion (which is incidentally, present in the human body). As a source, here is a 2014 article by Tushar Kanti Das, et al, published in Archives of Neuroscience, available here.
The authors explicitly references, in Figure 4, “Formation of Aluminum Superoxide Semi reduced Radical Ion and Aluminum Superoxide Complex (43)”, with described reactions proceeding as follows (also adopting notation and water complexing from yet another article):
$$\ce{[Al(H2O)4](3+) + O2•− <-> [Al(O2•−)(H2O)4](2+) }$$
$$\ce{[Al(O2•−)(H2O)4](2+) + Fe(3+) --> O2 + [Al(H2O)4](3+) + Fe(2+)}$$
And, in acidic conditions:
$$\ce{[Al(H2O)4](3+) + O2•− + H+ <-> [Al(O2•−)(H+)H2O)4](3+)}$$
$$\ce{[Al(O2•−)(H+)H2O)4](3+) + [Al(O2•−)(H+)H2O)4](3+) --> 2 [Al(H2O)4](3+) + H2O2 + O2}$$
where the above chemistry suggests a role for Al3+ in possibly assisting with the subsequent production of miscellaneous powerful radicals associated with a Fenton reaction  as sourced from an iron ion presence (as in blood).
Now, as to the possible originating source of Al3+ itself, in the particular context of baking with Aluminum pans, and further washing said aluminum ware in a dishwasher also employing Baking Soda, is that the latter NaHCO3 on warming in aqueous solutions, apparently will attack Al metal, commencing with the reaction:
$$\ce{ NaHCO3 + H2O + Heat --> NaOH + H2O + CO2 }$$
And further, to quote Wikipedia:

Sodium aluminate is also formed by the action of sodium hydroxide on elemental aluminium which is an amphoteric metal. The reaction is highly exothermic once established and is accompanied by the rapid evolution of hydrogen gas. The reaction is sometimes written as:
$$\ce{ 2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2 }$$
however, the species produced in solution is likely to contain the [Al(OH)4]− ion or perhaps the [Al(H2O)2(OH)4]− ion.[6]

This problem with Aluminum ware is well known (see warnings, for example, here, and also here).
I hope this, in part, advanced chemistry presentation nevertheless is of assistance in answering your question "Is using oxidized aluminium baking forms safe?".

As I mentioned in the question part, my knowledge of chemistry is quite rudimentary (from a limited secondary school program some 30 years ago). Would you care to summarize your reply in key take-aways in layman's terms? I totally get the policy of not providing advice on health matters, though I hope that calling out obvious toxins in still within the policy. I think the key question is whether dishwasher added something on top of the "normal" oxidation process in the course of use. These forms have been used for half a century in multiple countries, so normal use should be OK.Cheers!

I believe the last reference I cited (repeated here https://dishwashingsecrets.com/aluminum-in-dishwasher-toxic/ ) does a fair job in expressing and addressing your concerns. Note: this reference uses bold letters on the point of  using dishwasher safe detergents. Cheers!

To continue my comment, surprisingly, the cited reference even brings up the case of muffin pans!

I am afraid it does not address my concerns at all. It deals with whether it is safe to put aluminium cookware in the dishwasher (e.g., to avoid discoloration). It does not say anything if the oxide/discoloration itself will affect the food. Aside from "I think baking cakes [...] in aluminum dishes are acceptable". The question was whether oxidation produced by the dishwasher is OK. There is ample scientific evidence regular aluminum cookware is fine. Also, judging by its level and style of English, the source does not inspire confidence, I'm afraid.

I am referring to the 2nd referenced cookware article which does make an opinion-based health assertion, not necessarily mine, or in accord with this forum. Note: unlike copper cookware toxicity, no need to be afraid of being immediately poisoned!  At worst, it is more of a longer-term opinion of possible ascribed health problems, which you may (or never) experience. Certainly, one can always more safely use steel, but it does not distribute heat well. If you are concerned or which to be very cautious, use NEW coated cookware and discard quickly if the coating is damaged. Cheers!