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How is there still iron on earth?
Iron rusts and the earth is pretty old, so how is it that there is still iron left that has not oxidized(/rusted)?
I tried looking it up, and the amount of iron on earth is mind boggling, but is that it?
Is there simply enough iron that not all of it has oxidized yet?
Maybe it's just that well insulated inside the crust, or does it occur naturally so more is created all the time?

There is very little metallic iron on the surface of the earth. And most of what is there is because humans made it by smelting iron-containing minerals, which are abundant.

The problem with availability of elements is often not the oxidation state but the concentration (or the dispersal). It would not be economical to collect all the rust falling off of vehicles to turn it into elemental iron. There is a similar issue with other elements, e.g. those used in electronics (although some of them do get recycled). There are some "endangered elements" but iron is not one of them.

Although iron is ubiquitous (Iron is the most abundant element, by mass, in the Earth, constituting about 80% of the inner and outer cores of Earth.), most of it is either molten or is complexed and not naked. So, there is little bare iron that can rust.

You may confuse iron(1) as an element and iron(2) as the metallic form of iron(1). Rusting of iron(2) does not destroy iron(1), but converts it to iron(1) compounds like non-stoichiometric hydrated oxides.
Iron(1) of Earth is much older than Earth and the Solar system. But most of surface iron(2) is less than 100 years old. All but the one contained in rare meteorites was produced by men from iron(1) ores.
Iron(1) in Earth's mantle and Earth's crust occurs in oxidized forms in various minerals, ores and rocks. E.g. the primary mineral of Earth upper mantle is olivine $\ce{(Mg, Fe)2SiO4}$
There is also iron(2), together with nickel, in Earth core, at high temperature and extreme pressure, where is no rusting.

There was lots of elemental iron before the great oxygenation event that created all the iron oxide deposits we mine today. But that happened a bit over 100 years ago…

Sure, I have been implying that  It is usually being related to Snow ball Earth  era 2.5(?) bill. y. ago,  when CH4 GH effect  was being replaced by CO2 GH effect.

And majority even before that.

There can also be iron(1) formed from iron(2)-bearing minerals via serpentization. Essentially, the FeO displaced from mafic/ultramafic minerals by this process undergoes various redox reactions, including disproportionation to Fe3O4+Fe. Also Ni-Fe alloys are possible if nickel ions are in the minerals.

You have it backwards. For about 2000 years people have taken iron oxides  (primarily) and converted them into metallic iron and steel. So there is dramatically more metallic iron on the earth surface now than 2000 years ago. about 6 million tons per day of new metallic iron/steel is made on earth each day. Admittedly much of that steel is recycled steel. Some mills only recycle metallic steel. Certainly many tons of oxides, etc, develop each day but I can't guess what that amount is except it is much less than new production of new metallic steel

I must admit very poor guess of tonnage by me . Wikipedia lists about 6 million tons per day . I have changed the answer but the error was so large I thought I should note it.


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Buffer solution of NaH2PO4 and Na2HPO4
We have this exercise without solutions

From a 0.2 M $\ce{NaH2PO4}$ solution and a 0.2 M $\ce{Na2HPO4}$ solution, a buffer
solution with pH = 6.8 is to be prepared. The total concentration of
$\ce{PO_4^3}$- should be 0.1 mol l-¹. Calculate the volumes of the two solutions needed to prepare one litre of buffer solution.

So my understanding is that the $\ce{Na}$ will dissolve in the water. Also, both $\ce{NaH_2PO_4}$ and $\ce{Na_2HPO_4}$ are acidic, but $\ce{NaH_2PO_4}$ will be more acidic in this case because is has one more $\ce{H}$ than  the other. So if a buffer consists of those two substances, then $\ce{NaH_2PO_4}$ will be the acid and $\ce{Na_2HPO_4}$ will be the base
First, is this correct ?
What I don't really understand is

The total concentration of $\ce{PO_4^3}$- should be 0.1 mol l-¹.

Does this mean that $c(\ce{NaH_2PO_4}) + c(\ce{Na_2HPO_4})= 0.1 $ ? In other words, we would need to set up the Henderson Hasselbalch equation
$$pH = pK_a + \log{\frac{c(\ce{Na_2HPO_4})}{c(\ce{NaH_2PO_4})}} = pK_a + \log{\frac{c(\ce{Na_2HPO_4})}{c(\ce{Na_2HPO_4) - 0.1}}} $$
and then solve for $c(\ce{Na_2HPO_4})$, and then we would find $$c(\ce{NaH_2PO_4}) = 0.1 - c(\ce{Na_2HPO_4}) $$
We have one litre of buffer, so the number of moles is simply
$$n(\ce{Na_2HPO_4}) = 1 \cdot c(\ce{Na_2HPO_4})$$ respectively
$$n(\ce{NaH_2PO_4}) = 1 \cdot c(\ce{NaH_2PO_4})$$
and then the volume is found by dividing the number of moles by the original concentration of $0.2 M$
Is this how they expect us to proceed ? We have no solutions so I prefer to ask
Are there some other important things I need to be aware of ? The $\ce{Na}$ seems to dissolve in the water, but what happens to the different hydrogen atoms? Do they simply become $\ce{H+}$ and $\ce{HO-}$ ?

You need just consider $\ce{H2PO4^-(aq) <=> HPO4^2-(aq) + H+(aq)}$ // Total phosphates means the sum [$\ce{H3PO4}$] +[$\ce{H2PO4-}$] + [$\ce{HPO4^2-}$] + [$\ce{PO4^3-}$] // As both stock solutions are 0.2 M, the sum of their volume is 0.5 L for the final 1 L of 0.1 M solution. // Volume ratio of stock solutions follows Henderson Hasselbalch equation.

@Poutnik But in our case we only have $[\ce{H2PO4-}]$ and $[\ce{HPO4^2-}]$ ? The volume ratio follows from the Henderson Hasselbalch equation: $pH = pK_a + \log(\frac{\ce{H2PO4-}}{\ce{HPO4^2-}} ) \rightarrow 6.8 = 6.8 + log(\frac{\ce{H2PO4-}}{\ce{HPO4^2-}} )\rightarrow$ $ 0 = 0 + log(\frac{\ce{H2PO4-}}{\ce{HPO4^2-}} ) \rightarrow 0 = log(\frac{\ce{H2PO4-}}{\ce{HPO4^2-}} ) = \log{1} \rightarrow \ce{H2PO4-} = \ce{HPO4^2-} = 0.5 $ L

You have trace equilibrium amounts of H3PO4 and PO4^3- too, but negligible, several orders lower concentration. You have switched fractions. More of acid form does not mean higher pH.

@Poutnik Yes in fact I switched the fraction, my bad. What do you mean by more acid form does not mean higher pH ? Because it's a buffer, the pH value shouldn't change drastically ?

The note was related to the wrong fraction.

@Poutnik Ah okay, if you want feel free to post an answer ! Aside of the inverted fraction, are there other important things to know ?

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Difference between sodium aluminate and sodium metaaluminate
So I was studying Chemistry (Hydrogen - Uses, Properties and Preparation) from 2 different textbooks (based on Grade 9 syllabus, I.C.S.E. Board, India) and I found what seems an astounding anomaly to me.
In the two textbooks, the same reaction for liberating Hydrogen gas from a hot concentrated solution of $\ce{NaOH}$ and $\ce{Al}$ is given as follows:-
Textbook 1: $\ce{2Al + 2NaOH + 2H2O -> \underset{\text{Sodium metaaluminate}}{2NaAlO2} + 3H2 ^}$
Textbook 2: $\ce{2Al + 2NaOH + 2H2O -> \underset{\text{Sodium aluminate}}{2NaAlO2} + 3H2 ^}$
If the two reactions are the same, why and how is the name of the same compound ($\ce{NaAlO2}$, Sodium meta aluminate / Sodium aluminate) different? Secondly, what does 'meta' imply in Sodium metaaluminate?

It is rather $\ce{[Al(OH)4]-}$, respectively $\ce{[Al(H2O)2(OH)4]-}$

There are multiple cases in chemistry when prefix is used either explicitly either implicitly.

You may want to check: https://chemistry.stackexchange.com/questions/40506/what-do-the-prefixes-meta-ortho-pyro-mean-in-inorganic-chemistry?noredirect=1&lq=1 // Besides, if you look at the WP article of sodium aluminate, you can see  sodium metaaluminate as one of the other names.

Seems like just a nomenclature thing to me. Prefixes meta, ortho and pyro are being sometimes used when giving names of inorganic acids to indicate a ratio of hydrogen atoms to the central atom(s).
And so:

meta means an acid with a minimal number of hydrogen atoms in relation to central atom (given its oxidation state), e.g. $\ce{(HPO3)n}$ is metaphosphoric(V) acid
ortho is a prefix given to an acid which has emperical formula having an H2O more than its meta acid counterpart, e.g. $\ce{H3PO4}$ is orthophosphoric(V) acid
pyro denotes an acid formed by condensation of an ortho acid, so pyrophosphoric(V) acid is $\ce{H4P2O7}$

In case of a compound from your question $\ce{NaAlO2}$ can be treated formally as a salt of "$\ce{HAlO2}$" acid - which would be of meta type. However, this a nomenclature thing so this prefix may be skipped.

What appears to be an "astounding anomaly" to you is totally normal in chemical nomenclature and biological sciences in general. Remember is chemistry has been studied by millions of people for centuries. One cannot make them follow identical nomenclature rules, although general consesus is developed by IUPAC. Many chemical names have 10 or even more synonyms. If you study botany, a single plant may have plenty of radically different Latin names. Here are some more names (16) identified by SciFinder (Chemical Abstract Service, USA) for the same compound. I list only a few as examples. Now note that that some names may not the be standard ones.
In short, both texts are fine.
Aluminum sodium dioxide
Sodium aluminate ($\ce{Na2Al2O4}$)
Sodium aluminum dioxide
Sodium aluminum oxide ($\ce{NaAlO2}$)
Sodium metaaluminate ($\ce{NaAlO2}$)

Certainly yes, I seem to be too used to the standardization offered by Physics and many other sciences. But thank you very much for the answer!

Since physics heavily relies on mathematics, it is more consistent. However sign conventions in physics can also drive some students crazy!

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Assigning configuration of chiral plane - selecting of the pilot atom
I am little confused about when it comes to assigning configuration of planarly chiral cyclophanes - and I think it comes down to selection of pilot atom.
All sources I have came across say that the pilot atom is the first out of the plane atom and it should be selected in order assign configuration of chiral plane. However, these sources present slightly different strategies for selecting a pilot atom.
(a) One says that the pilot atom should be the one which is on the same side as the highest priority substituent which is causing desymmetrization the plane (i.e. the one that is closer).
(b) While the other source says simply that the pilot atom should be selected according to CIP priority rules.
In most cases it doesn't make any differences. Actually, it doesn't make a difference for any of the examples that I have came across. However, when thinking about it a structure came to my mind that brings conflict between these two strategies, it looks like this:

Seems to me that if we followed strategy (a), then the pilot atom would have to be a carbon from the top CH2 group and chiral plane would be assigned with Rp configuration. However, following strategy (b) means that oxygen atom gets the pilot status and chiral plane has to be assigned with Sp configuration. Only one approach can (if any of these) can be correct - could you tell me which one is it?
I have checked IUPAC recommendations (https://iupac.qmul.ac.uk/BlueBook/PDF/P9.pdf), but I didn't find an answer to my question.

I don't know, but you might find "Algorithmic Analysis of Cahn–Ingold–Prelog Rules of Stereochemistry: Proposals for Revised Rules and a Guide for Machine Implementation" useful in clarifying unclear CIP cases.


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Experimental evidence for zwitterions
If I have a diprotic acid that is +1 positively charged in its fully protonated state, I can figure out the apparent equilibrium constants by titration with base. The net charge will be neutral after losing one proton, and -1 after losing the second proton.
Examples would be the hydronium ion or the amino acid glycine at a pH < 2.
How could you determine whether the single-protonated state is a zwitterion (glycine in neutral aqueous solution) or a molecule without charged functional groups (glycine in neutral polar aprotic solvent, or water at neutral pH)?
Is it necessary to do an experiment, or is there a simple rule to figure out whether a zwitterion is expected?

Well, in principle there will always an equilibrium between at least four species (zwitterion, cation, anion and uncharged molecule), which makes it seem difficult to characterize. I'm also interested to know how reliable of a picture you can get with simple rules, and I'm especially curious about whether they can compensate for solvent effects.

Inspired by this closed question: https://chemistry.stackexchange.com/q/89937/72973

One has to consider nature of both protonated groups. In case of glycine, there is no reason to assume R-NH3+ is more acidic than R-COOH. OTOH, it is funny to assume H3O+ dissociate to a zwitterion.

Related: Charge distribution in molecules

That is a very difficult problem worth a PhD project in physical chemistry. Intially, I thought one could try capillary electrophoresis at different pHs and if the analyte travels with the electrosmotic flow marker at a certain pH, one can say a zwitterion has existed (or a neutral form) by inference. However, you query is about distinguishing the neutral state and the zwitterionic state. One has to think about spectroscopic techniques that rely on dipole moments. Since you are interested in aqueous phase, gas phase microwave spectroscopy of amino-acids is out of question (there are papers on it). Since zwitterion ions can possess large dipole moments as compared to their neutral counterparts, two remaining ones are infrared and photoelectron spectroscopy.
Zwitterion formation in hydrated amino acid, dipole bound anions: How many water molecules are required? J. Chem. Phys. 119, 10696 (2003); https://doi.org/10.1063/1.1620501

We utilize the facts that zwitterions possess very large dipole
moments, and that excess electrons can bind to strong dipole fields to
form dipole bound anions, which in turn display distinctive and
recognizible photoelectron spectral signatures. The appearance of
dipole-bound photoelectron spectra of hydrated amino acid anions,
beginning at a given hydration number, thus signals the onset of
greatly enhanced dipole moments there and, by implication, of
zwitterion formation.

IR would be finiky with water, but there are reports that water bound to zwitterions has a different vibrational frequency
Stability and IR Spectroscopy of Zwitterionic Form of β-Alanine in Water Clusters, J. Phys. Chem. B 2019, 123, 20, 4392–4399, https://doi.org/10.1021/acs.jpcb.9b00654

We perform an experimental and computational study on the number of
water molecules needed for zwitterion formation of β-alanine. Our
density functional theory investigation reveals that a minimum of five
water molecules are required to form and stabilize the zwitterion. A
characteristic connecting water molecule located between the COO– and
NH3+ groups is found to enhance the stability. This water molecule is
also involved in a characteristic infrared active vibration at ≈1560
cm–1, which is slightly shifted with the number of surrounding water
molecules and is located in a spectral region outside of water
vibrations.

Additionally, I assume measurement of permitivity spectrum of protic and aprotic solvents, with and without glycine, could confirm the zwitterion nature of glycine solution in water.

Poutnik, Have you checked relevant literature? I didn't, but that is an interesting idea.

I have not, was just wondering.....

Would NMR be capable of distinguishing zwitterions vs neutral molecules? As an example, one would expect 3 different peaks for glycine's neutral form vs 2 for the zwitterionic form on proton NMR, right?

@user73910, I do not know. In solution there is perhaps very, very little neutral glycine. NMR is fundamentally an insensitive technique. Feel free to do a literature search on Google Scholar.


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white vs light pink color of MgO
Recently I ordered some high purity MgO, which is light pink in color. However, the old powder I was using was white. MSDS also suggests that it should be white. Can anyone explain the reason of two different colors of MgO of same purity.
Thanks in advance
PS- I suspected it to be due to absorption of moisture earlier, but it is not the case. Even on heating at 800 C, does not change the color.

There is a proverb saying "When two are doing the same, it is not the same."

I guess, impurity.

Could be. but they both have the said purity of 99.95%. Anyways, thanks.

I've seen bright blue KCl and yellow NaCl. A few ppm of something (not sure it even counts an impurity) can do miracles.

99.95% ... Imagine how 500 mg Fe2O3 in 1 kg MgO could look like.

Small amount can go a long way with color. Cobalt blue glass may be colored by as little as 0.02% cobalt oxide. From the link: >Small amounts of cobalt (around,to a ton of glass). Deeper blues are obtained by adding up to ten pounds of cobalt oxide per ton of glass.

@Poutnik, I was suspecting iron as well. It is not very hard to detect iron by simple tests.

It is possible, just, that it might be due to band gap states in the MgO crystals themselves. Such states are very common.


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Gold Chloride as electroplating solution?
In some (maybe questionable) pedagogical texts at least, gold chloride solution is presented as an electrolyte in the electro deposition of gold.
In practice however, it seems that cyanide based solutions are used instead. Please excuse my ignorance, this is not my field of study, but could gold chloride be actually used in practice as an electrolyte?
The application is electroplating 304, 316 stainless steel and cast iron ornaments.

It would be rather H[AuCl4]. https://en.wikipedia.org/wiki/Chloroauric_acid

Note that generally, electroplating is easy. Quality electroplating (durability, shine) is difficult. It took a lot of experimenting to determine the right metal compound, auxiliary components and current density to achieve desired results. Many solutions for electroplating were initially kept as trade secrets.

Electrolysis of gold chloride solutions (or $\ce{HAuCl4^-}$)  produces a rough, dusty and pulverulent gold deposit, which is not really adherent. To get a bright and continuous gold deposit, electrolysis must be carried out with gold cyanide, because in gold cyanide solutions, the concentration of $\ce{Au^{3+}}$ ions is extremely low. And this rather low concentration is necessary to get a nice bright, shiny and continuous gold plating.

The necessity of keeping the ionic concentration as low as possible is well explained in the document published on December 23, about silver plating.

This one:  why-is-silver-cyanide-used-for-electroplating

Electroplating is one of the true black arts of chemistry…


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how helium can be compressed if its compressibility factor is alway greater than 1?
I was studying real gases and found that H2 and He have Z>1, that means they cannot be compressed even at high pressure as their graph is always upwards meaning Z>1 always. Then how come the balloon seller have compressed helium in his cylinder?

$$Z= \frac{pV}{nRT}$$ Where do you see not being able to be compressed for Z>1?

Hello, But Z is also equal to Vreal/Videal

That is obvious. It just means you can compress helium by given pressure slightly less than ideal gas. ( unless pressure and Z are too high - then not slightly ).

It also means you need for the same volume just slightly higher pressure than for ideal gas.

You have an incorrect interpretation of the compressibility factor $Z$. $Z$ being greater than unity does not mean that it cannot be compressed. All gases can be compressed (that is a part of the definition of this state). Even liquids can be compressed under extreme pressures. All it means that it's volume will not decrease like an ideal gas under pressure.

If $Z$ > $1$, it means that, when compression this gas,  $pV > nRT$. This is the only meaning. It does not mean that the gas cannot be compressed.


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Radical chain reactions
I have some questions about radical chain reactions. Let's take an example:

And here are my questions:

How can we define in the above case at which level the reaction 2) is located (initiation or propagation)? I understand initiation as: creation of new radicals and propagation as: left and right with equal number of radicals. My difficulty comes from the fact that for reaction 2), there is both the creation of a new radical, and an equality of the number of radicals. What confuses me is that I have the following example in my course:

We can see that there is radicals used as reactant in one of the initiation step, as well as in the propagation step. How to decide which is the right choice in this case ?

In a reaction like this, can we both use the pre-equilibrium hypothesis (pseudo-equilibrium due to quasi-stationarity of intermediate reactant) and the quasi-stationary state -for radicals- hypothesis?

Regarding the pre-equilibrium hypothesis, at what stage can we consider a pseudo-equilibrium and calculate an equilibrium constant? Only at the step of creating the intermediate reaction? Or also at the stage of its disappearance?

Concerning the hypothesis of quasi-stationarity of the radicals, can we equalize the speed of appearance and disappearance of the radicals only between the initiation and the termination, or is it also possible between initiation and propagation and/or between propagation and termination?

Many thanks in advance.
PS: my apologies if I'm not using to best words for the hypotheses, English is not my native language. I tried to find the corresponding words on internet.

Take new as more. Propagation, aside of keeping radicals, is regenerating cycle, producing one of its reactants, like here the ethyl radical.

Your step with $k_{init2}$ is technically a propagation step ( one radical in one radical out) but the reaction is given as 'fast' so that $\ce{CH3^*}$ is consumed faster than it is produced and we may assume $\ce{[CH3^*]}\to 0$ and takes no further part in the reaction. The chain carrier is thus $\ce{C2H5^*}$.


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How many E/Z isomers are possible for vitamin A?
If E/Z is a type of stereoisomerism, then you should be able to use the formula $2^n,$ where $n$ is the number of stereogenic centres. How is it that there are 16 possible E/Z isomers according to the answer in my textbook, but there are no stereogenic centres in all-trans-retinol?

E/Z is not about centres at all.

See E/Z notation

A molecule may include

any natural number including zero of double bonds (and for each of them, individually, one can check if their configuration can be described as E, Z, or if this label is not applicable here [e.g., ethene])
any natural number including zero of stereogenic centres (either R, or S)
any natural number including zero of helical axes (either P, or M)

These three properties are independent of each other.

The molecule contains four conjugated double bonds $\ce{C=C}$, which are all drawn in the E (trans) disposition. And there is no chiral carbon atom, so no R or S atom. But one, two, three or all of the double bonds $\ce{C=C}$ may be in the Z (cis) disposition. This produces $2^4 = 16$ different molecules which are all isomers. They are EEEE, ZEEE, EZEE, EEZE, EEEZ, then EEZZ, EZEZ, EZZE, ZEEZ, ZEZE, ZZEE, then EZZZ, ZZZE, ZZEZ, ZEZZ, and ZZZZ.


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Thermal properties of chemical compounds
I have a chemistry related question, I'm doing a research on some inorganic salts (KOH, KCl) and similar compounds and I'm struggling finding the thermal properties like (thermal conductivity, specific heat, coefficient of thermal expansion, kinematic viscosity, density, latent heat of fusion, melting point), I could find some but not all, can you help me in this

You can check the MSDS of the compounds.

MSDS will not have the information, except the melting point and density if they were liquids. Please clarify your question and give a context of your question. Do you have access to SciFinder? How can solids like KOH and KCl have kinematic viscosity?

Tell others where you were searching, not to repeat your actions.

Some of the data you want to access are available for free, others are available on purchase of reference materials and subscription of databases typically seen in research libraries and institutes.  Get in touch with your supervisor and libraries close to universities; it depends on their contracts with the publishers if access already is granted by your affiliation with a research group, or permissible as a walk-in user.
This being said, for some chemicals you find properties listed in the property boxes on wikipedia (example KOH), or in the NIST Chemistry WebBook (again example KOH).  A handy (somewhat handy) reference equally is the CRC Handbook of Chemistry and Physics you either access on-line (requires a subscription, some groups/schools/libraries have), or browse through the heavy print edition.  (With their virtual library card, there are a couple of editions on archive.org one can borrow online for 14 days, too.)
As @AChem already indicated in a comment, you would complement the fact retrieval with a consultation of a professional database.  In addition to SciFinder (by the American Chemical Society), I recommend Reaxys (by Elsevier) because it inherited so many data of the Gmelin handbook.  As your search includes thermal expansion, viscosity, etc. possibly your research is in solid state chemistry (e.g., KOH as a flux/melt, example); well, then ask for Springer Materials (because of the tables and phase diagrams once compiled in Landolt-Börnstein).  All three electronic databases require a paid subscription.

Although I like Reaxys (for the same reasons you mentioned Gmelin and Beilstein inherited data), but even many Tier-1 universities in USA do not have access to it-maybe it is too expensive. Also, SciFinder has recently changed. It has become SciFinder-n, after incorporating millions of abstracts from Chemische Zentralblatt (among other changes).

@AChem This pattern may happen anywhere, to any database. My awareness of walk-in user as a legal term is personal experience: compile my questions & of colleagues on paper, take the train, query Thieme's Science of Synthesis in a different school, with clearance -- because thanks to encouragement by Zass and Braendle (ETHZ), the methods of Houben-Weyl print were scanned and linked to modern references. So while there was training in Beilstein Crossfire & STN already as undergrad in France, the catalysis folks in the US couldn't believe how to work without SoS on PC, only few knew Reaxys.

Interesting, I think Reaxys is more popular in Europe. I hope one day major German works will be (machine) translated into English. MT has become quite reliable. A typical US graduate student does not even know about these names: Houben-Weyl, Beilstein, Gmelin etc.

Yeah; «you know Beilstein, you were in Europe; you know SoS, you were in the US» And as SciFinder is so expensive, our school holds 2 FTE (max), though already participating in a consortium with other schools. Access to Reaxys equally is more often constrained now (once access all across campus, now only to the groups), too. Are ACS' activities still all under IRC 501(c)(3), as are the Ivies? For some time, Elsevier's ROIs exceeded those of Apple (well in the double digit percentage); not only for Reaxys, but incl. for journal fees hurting all from the west (like UC Berkely) to East Europe.

@AChem The Beilstein English <-> German dictionary, by the way: http://web.stanford.edu/group/swain/beilstein/index.html (open access)

This is sad how publishing houses are charging like crazy. They don't pay a penny to review or write articles with the exception of encyclopedia chapters. They offer about $250 for a single contribution.

Let us continue this discussion in chat.


text:
Effect of magnesium hydroxide MgOH formation in water on pH (Why more basic?)
Elemental magnesium reacts in (hot) water to form magnesium hydroxide and hydrogen gas:
$$\ce{Mg_{(s)} + H2O_{(g)} -> Mg(OH)2_{(s)} + H2_{(g)}}$$
As indicated for example by phenolphthalein turning pink (Youtube video), this increases pH. Why?
I would assume that magnesium hydroxide as a strong base dissolves back to yield hydroxide anions, but those are previously taken away from the water, so isn't the net effect neutral? My assumption is that magnesium makes water dissociate at higher rate than it would be by itself, and that formation of hydrogen gas binds/removes protons from the solution, while the hydroxide dissociates back to the solution, therefore net making it basic. Does that add up or is the main reason another?

$\ce{Mg(OH)2}$ is "strong weak base", but just slighly soluble, with equilibrium $\ce{Mg(OH)2(s) <<=>[H2O]Mg^2+(aq) + 2 OH-(aq)}$ with trace amounts of   $\ce{Mg(OH)+(aq)}$ and $\ce{Mg(OH)2(aq)}$

I would like to copy your assumptions as printed in the text (in italics) and then add my comments and answers in the text (but not in italics), as one goes along. Let's go !
I would assume that magnesium hydroxide as a strong base dissolves back to yield hydroxide anions This is right.  but those are previously taken away from the water, so isn't the net effect neutral? What does it mean ? The hydroxide anions remain in water, at least at low concentrations, when the solution is not yet saturated in $\ce{Mg(OH)2}$. Why do you think that they are "taken away from the water" ? My assumption is that magnesium makes water dissociate at higher rate than it would be by itself, Well ! Hot water is more dissociated than cold water, but this is natural, and not due to magnesium or any other solute. and that formation of hydrogen gas binds/removes protons from the solution, while the hydroxide dissociates back to the solution, therefore net making it basic. Does that add up or is the main reason another? This is right. The solution gets more basic.


text:
Are physical space filling molecular modelling kits still available anywhere?
In the 1980s when software for drawing 2D chemical structures was in its infancy and most chemists didn't have access to the workstations that could convert 2D to 3D structures, many labs had physical molecular modelling kits that could be used to build 3D space filling models of molecules.
Stick and ball kits were common then and are still common, but the 3D CPK kits seem to have disappeared as soon as most desktop chemical drawing software was able to reproduce pictures of space-filling models.
For those too young to remember this era, most 2D and 3D drawings were done by hand (a tedious task even for 2D structures that used physical templates so your benzene hexagons were not distorted).
But despite the ease of getting high quality 2D or 3D pictures of molecules directly from desktop software, there was something fulfilling about having a physical 3D model you could hold in your hand.
Is there any current source of those 3D CPK kits? Or, is there any accessible way to produce their equivalent physical models by, for example, 3D printing the models generated by desktop software?
PS this almost deserves a "nostalgia" tag.

Note, CPK (for  Corey–Pauling–Koltun) is a colour scheme, i.e., a convention to symbolize atoms of carbon in black, oxygen in red, nitrogen in blue, etc. which took inspiration on the colour scheme in the wired models by Dreiding.  Wikipedia cites Chemical News

"Hofmann, at a lecture given at the Royal Institution in April 1865 made use of croquet balls of different colours to represent various kinds of atoms (e.g. carbon black, hydrogen white, chlorine green, 'fiery' oxygen red, nitrogen blue)."

From your description, I assume instead of a kit for ball-stick models, you seek one which is instead is spacefilling.  If we assume atoms as spheres, then sections of spheres are called calotte, the same term used in architecture and about the skullcap.  On occasion you see them (e.g., ebay), smaller ones still commercialized:

(image credit: Aug. Hedinger GmbH & Co. KG, Stuttgart/Germany)
The set depicted (55 atoms made of a polymer, 30 struts, and a tool to loose the struts) is sold for EUR 32.50 (plus taxes and shipping).
Larger in dimension (intended for demonstration in a lecture) are the ones by VSN:

(image credit VSN).
made of wood (Hofman like), with spheres of 3.5 cm (H), 6 cm (C, N, O), or 8 cm (Cl, S) in diameter.  You purchase per atom.

If you don't want to disassemble them, these models may be printed in 3D, too:

(image credit: shapeways.com)
I have no affiliation with either corporation.

Good clarification that CPK is a colour-scheme (though the kits were often casually called "CPK kits"). And a great link for a source that can create physical models: thank you.

@matt_black Today, it depends a bit on the software, too.  If you run Jmol, for example and use its terminal (File -> Console) load $glucose; followed by spacefill on; you get the van der Waals model.  Yet select all; color rasmol; yields a lighter gray on the carbons than select all; color cpk; (the default color scheme), one may select individual atoms/atom types, etc too.

By the way, Jmol equally exports the scene with write example.wrl; (VRML 2.0) or write example.stl; a 3D print shop accepts as input format.  CCDC's community edition of the Mercury has a similar export function (compared to Jmol, Mercury is more resource hungry for HDD and RAM): a brief how to, and some of the yearly contests (including polychrome models, too).

Avogadro will also export VRML 2.0 with colors. I've used Shapeways to print the structures in full color.

As the Wikipedia page mentions, Koltun patented an extension of the modeling kits developed by Corey & Pauling -- not just the color scheme, the models themselves. While CPK today is best known as a color scheme, there were actual kits sold as "CPK models" (for example), and that was a common "genericized" name for the kits in the 70s and 80s. (And apparently Harvard Apparatus still sells "CPK" branded kits today.)

@GeoffHutchison Given the publications around blender e.g., in JChemEduc (https://doi.org/10.1021/acs.jchemed.1c00515, or https://doi.org/10.1021/acs.jchemed.7b00549 to mention just two) I welcome if programs leave the "chemistry only" bubble. Despite all advances in computing, a physical model -- either by kit, or by 3D print -- still has its value.  For molecules, as for molecular surfaces (example Hirshfeld surfaces, https://pubs.rsc.org/en/content/articlelanding/2009/ce/b818330a -- but no, CrystalExplorer no longer exports vrml).

Is there any current source of those 3D CPK kits?

You have to distinguish space-filling and semi-space filling, see below:

For semi-space filling, many ball-and-stick models also have pegs instead of sticks to connect atoms without extra space between them, like this Molymod Biochemistry set:

It is more difficult (and more expensive) to find space-filling model kits, but they still exist, e.g. this HGS model kit:

History
Space-filling models are associated with the CPK initials because Corey and Pauling published a home-grown non-commercial space-filling model in 1953, and later Corey, Pauling and Koltun made them available commercially.

They were sold by a company called Ealing, and searching for "Ealing CPK" might lead you to a vintage model such as this one:

I'm far more interested in the "proper" space filling kits with VdW atomic radii. The semi-space filling kits look like a terrible shonky compromise to me.

Re «space-filling vs semi space-filling» in the visualization by Jmol/JSmol: the command to Jmol's command line select all; spacefill on; yields the same result as select all;, followed by spacefill 100%; (100% of the van der Waals-radii). And so offer spacefill off; and spacefill 0% a similar result.  Intermediate dials as spacefill 22.2%; are permitted, too (somewhere close to Jmol's default).

I got a pile of CPK atoms on ebay. Sadly, the bonds had become brittle and useless over the decades. Harvard Instruments was able to supply new bonds, at roughly 100 times the original price of 2 cents a pop. Yep, that's right.
There's a pile for sale on ebay right now: https://www.ebay.com/itm/204536539768. The bonds will almost certainty be of no use though, just so you know.

Was there ever an answer to the question of whether there is a modern producer of CPK model parts?  I've recently obtained a very large set of parts that was sitting in the back of a storage room in our chemistry department, potentially for many years.  Unfortunately, the connector pieces are made of a softer plastic that over time degrades and becomes very brittle, so when you try to take these apart the connectors shatter.
Back in the late 1990s when I was a graduate student at UCLA, we had a similar old set, but at that point I was able to order a bag of probably 100 of the connector pieces, so someone was still producing these parts.  Unfortunately I don't remember who produced them, and Googling is not helping me.
Bruce

The modern way is to print them yourself, e.g. https://www.morressier.com/o/event/623377e0b300ee00119b311f/article/6234a1c8818a915252b81c5f


text:
Compound of C+Li+F possible?
Sorry my chemical knowledge is not very deep so some of what you assume is common knowledge will probably be new to me.
For a game i am trying to make i wrote a simple atom / compound simulator where i try to build compounds with simple covalent and ionic bonds but in a very basic sense without taking anything into account beside the number of electrons
so far my approach was to implement rules and every time i the simulator spills out a new compound i google if really does exist (did not find a comprehensive list of existing compounds, please link if you have one) and if it does not i dig deeper to understand the rules better
so after implementing dativ bonds i got a compound of C, Li and F where
Li forms an ionic bond with C and C forms a covalent bond with F and F helps C out with a dativ bond to reach the 8/8
so far i only found CF5Li which would suggest that this compound can exist in some way but i am not sure if the dativ bond between C and F is "real" or my implementation is beyond flawed because its much more complicated (like for example that F does have "free" electrons but there is no overlap between the missing electrons orbits of C and the available ones of F)

The problem is that rules based on valency can only get you so far. You can generate lots of compounds that formally satisfy all atoms' 'desired' valency (or octet), but don't make sense IRL because they would immediately decompose to a more stable arrangement. I'm guessing you are a programmer - this is like trying to generate working programmes just using the syntax rules of a language: you can generate valid programmes, but the vast majority of them wouldn't make any sense.

You might think CF3Li should exist but it is unstable with respect to LiF and difluorocarbene

@Waylander indeed, i was already thinking about some kind of "bonding priority" because i am also not able to create O3 or HNO3 , with your comment you gave me one more reason to dig into that

@orthocresol i am indeed a programmer, do you think its possible to define whats stable and what is not with enough knowledge ?

In principle, the answer is yes, you can - however, the amount of 'knowledge' you'd have to encode is huge, because determining stability (computationally) essentially involves solving physics equations (e.g. Schrödinger) from the ground up. Naturally, the less knowledge you provide it, the more errors you will run into. One sensible compromise might be to restricting the elements to a set of well-behaved, organic-ish elements, namely C, H, O, N; it's often possible to use lower-level heuristics (i.e. based on bonding patterns, not quantum mechanics) to make a guess as to their stability.

Unfortunately it is not yet possible to get an algorithm which is able to say if a given structure does exist. Let's take for example the molecule H-N=O. It should exist, by the octet rule, but it does not exist. Why ?

@Maurice Nitroxyl does not exist?

@orthocresol i see, thank you very much for your input. restricting the elements will not be an option for me because the whole goal is to have a "good enough" representation of the "real rules" and than create fictional elements / atoms that not exist but follow the same rules for bonding but I appreciate your suggestion non the less

Generalising chemical rules across all possible chemistry is extremely ambitious. The vastly simpler problem of enumerating all possible simple hydrocarbons (just hydrogen and carbon with single bonds and no loops) is already very hard despite being nearly adressible using graph theory. See this question for some of the detail of that problem.

@Braunbaer. HNO does not exist. If somebody did succeed in synthesizing it, it gets immediately transformed into $\ce{N2O + H2O } $

@Maurice Not in gas phase https://en.wikipedia.org/wiki/Nitroxyl and it hardly is on topic under this question.

@Mithoron. You are right. HNO is not on topic under this question. It was just here to illustrate that it is not always possible to foresee the stability of a structure by counting its electrons.

You want to blend $\ce{C}$, $\ce{Li}$ and $\ce{F}$ into one molecule and state "found" $\ce{CF5Li}$ as a solution.  From this I take, there is no constraint on the numbers of atoms as long each element is present by one atom, or multiple.
With 3,3,3-trifluoropropyne there is an entry on PubChem beyond of living in silico only, with a CAS registry number 661-54-1 (reference). The compound is literature known since 1950,(1) described with as "a gas of boiling point c $\pu{-48^\circ{}}$ with an odour reminiscent of that of acetylene".  Though rather on the expensive side for such a small molecule ($\ce{C3HF3}$, $M = \pu{94.04 g/mol}$), it is commercially available; for \$US 1,010.00 MilliporeSigma sells you $\pu{10 g}$ in a cylinder.
Attempting chemical overkill on paper, one could deprotonate this ethyne with butyl lithium (BuLi), formally creating $\ce{C3F3Li}$:

From a practical perspective, however:

the reaction has to be carried out at a temperature of $\pu{-48 ^\circ{}C}$, or below, to retain 3,3,3-trifluoropropyne in the liquid phase.  Likely, a reaction temperature considerably lower will better dissipate the heat of reaction.

butyllithium is a base way stronger, than typically required to abstract a protons attached to an ethyne.  With the presence of a conjugated, electron withdrawing trifluoromethyl group ($\ce{-CF3}$), sodium hydride often used for such a reaction equally would be multiple orders of magnitude stronger, than necessary.
The byproduct of butyl lithium's reaction as a base is butane.  Wikipedia's corresponding property box lists a  boiling point of pure $n$-butane at atmospheric pressure in the range of $\pu{-1 \ldots +1 ^\circ{}C}$, so this doesn't leave the reaction mixture.  This possibly could hamper attempts to isolate the envisioned product, $\ce{CF3C#CLi}$.

Haszeldine already described the preparation of compounds where hydrogen was replaced by a metal, $\ce{CF3C#CCu}$, $\ce{CF3C#CAg}$, and $\ce{(CF3C#C)2Hg}$ with copper, silver and mercury where "the first acetylides decompose quitely on gentle heating and violently when heated rapidly."  Maybe $\ce{CF3C#CLi}$ is similar, i.e. unstable.  Given the formation of e.g., tetramers for butylithium, the notation of $\ce{CF3C#CLi}$ as an isolated molecule might not be the correct representation of the structure of the reaction product, either.

This doesn't sound like an easy synthesis.

Because you mentioned your background leans more toward computation, than chemistry:  the systematic generation of molecules in silico stays a field of ongoing research.  Challenges include the identification of efficient algorithms (and their implementation) to shuffle the atoms to build chains, and rings.  Then you need quality checks to retain molecules which are chemically plausible e.g., by count of valences (how many bonds an atom typically engages), or ring strain. Last but not least, you equally want to screen these structures for useful chemical and physical properties to identify interesting candidates for drug design, or material science -- still before performing any chemical synthesis in the lab.
Thus, examples like MAYGEN and OMG, or Reymond's massive compilations GBD-11, GDB-13, and GDB-17 can only provide a glimpse.

addition past acceptance of the answer: User @Mithoron suggests in a comment pentafluorobenzene as an alternative, i.e. implies a reaction like

This would be much more affordable ($\ce{C6HF5}$, $M = \pu{168.06 g/mol}$) per gram, as well per mole as the same source as above
currently sells $\pu{5 g}$ for \$US 40.00.  In addition, the analogue phenyl lithium (lacking the fluorine atoms) is readily available and used a chemical reagent solution.  Thus, the synthesis departing from pentafluorobenzene likely is less troublesome than the one on 3,3,3-trifluoropropyne, too.
(1) Haszeldine, R. N. Synthesis of 1:1:1 Trifluoropropyne. Nature 1950, 165, 152–153; doi 10.1038/165152b0
(2) Peironcely, J.E., Rojas-Chertó, M., Fichera, D. et al. OMG: Open Molecule Generator. J Cheminform 2012, 4, 21; doi 10.1186/1758-2946-4-21 (open access); http://sourceforge.net/p/openmg
(3) Yirik, M.A., Sorokina, M. & Steinbeck, C. MAYGEN: an open-source chemical structure generator for constitutional isomers based on the orderly generation principle. J Cheminform 2021, 13, 48; doi 10.1186/s13321-021-00529-9 (open access);  https://github.com/MehmetAzizYirik/MAYGEN
(4) https://www.gdb.unibe.ch/downloads/
(5) What are the GDB-13 criteria for "synthetically accessible organic molecules"?

Darn, that's just the example I though of earlier, sneaky telepaths!

@Mithoron Give me some rope here, did I touch a taboo? This aside, PubChem's entry 21122483 about 5,5,5-Trifluoro-1,3-pentadiyne could be equally be used here, $\ce{CF3-C#C-C#C-H}$, which is accessible enough to record microwave spectra (entry Springer Materials), just as e.g., 7,7,7-Trifluoro-1,3,5-heptatriyne $\ce{CF3-C#C-C#C-C#C-H}$ possibly amenable to similar chemistry.

snort Kidding :D That's a pretty obvious example, no wonder you'd think of it too!

BTW pentafluorophenyllithium also exists, and many others, I guess.

@Mithoron $\ce{C6HF5}$ is a starting material much less exotic than $\ce{CF3-C#C-H}$; with reference to you, the idea was added. Thank you.


text:
Dehalogenation of vicinial halides via Sodium Amide vs Sodium Ethoxide
I noticed that vicinial dihalides can undergo elimination to create two alkenes or an alkyne. These are the following reactions I've seen.

From Solomons & Fryhle 10th Edition Organic Chemistry Problem 13.16

from E2 reactions and Alkyne synthesis on LibreTexts
Is it just the heat that makes a difference here? (i.e. could you get an alkyne to form with sodium ethoxide under cold conditions), or is it a combination of the difference in heat and reagent?
A similar post about the dehalogenation of gem dihalides mentioned the relevance of heat (especially that reactions with sodamide typically lack it), so I'm fairly certain that heat does distinguish the two outcomes, but I'm not sure if there is something special about sodamide as well.

Write formulas with hydrogens and see if you get it then.

Thanks for pointing out my mistake. I have fixed the question so that it actually makes sense.

Umm, I'm not so sure you got it. If you want to compare reaction with amide and etoxide, then first example isn't great, because you simply couldn't get the triple bond.

As far as the difference in reactions with these reagents in general, I'm pretty sure it's a duplicate.

https://chemistry.stackexchange.com/questions/101339/can-vicinal-dihalide-produce-conjugated-alkene-in-strong-base

https://chemistry.stackexchange.com/questions/48347/what-is-the-major-product-in-the-reaction-of-2-3-dichlorobutane-with-sodium-amid

https://chemistry.stackexchange.com/questions/23966/why-can-syn-periplanar-e2-elimination-happen-in-molecules-like-vinyl-bromides-bu

Oh, whatever... Etoxide is simply not strong enough (at least in solution), amide is.

Judging by the references in comments, it seems possible the free electron pair in an amine adduct could influence the adjacent carbon, forming an aziridine intermediate.  Elimination would form the triple bond.  Ethoxide may not be able to do the same after it replaces one of the bromines.

@RobertDiGiovanni I hope nothing I linked there says it. That's just elimination! Ethoxide eliminates to double bond, amide can react further, because it's a friggin' strong base.

@Mithoron and it's friggin' hot too!  That, and a bowl full o' nucleophiles strongly suggests an equilibrium somewhere.  Br- is literally swamped with sodium amide and ammonia.  I'm seeing a solvent effect and, perhaps not an aziridine closed ring, but the free nitrogen electron pair (as an intermediate) interacting with the adjacent hydrogen on the nearest carbon (4 membered complex and/or solvent effect).  I'm seeing, perhaps a concerted reaction leading to double elimination.  Just a thought.

The intramolecular concerted mechanism yields 2 ammonia and the triple bond.


text:
Are A1g modes reduced at 180 degrees polarisation in Raman spectra?
I have read that A1g modes are identifiable in Raman spectra by looking at cross-polarised Raman spectra and seeing peaks that strongly reduce in intensity. So essentially, the intensity of A1g modes is at a maximum at 0 degrees (or XX polarisation where the first is the incident laser light polarisation and the second X is the polarisation of the backscattered light) and a minimum at 90 degrees (YX polarisation configuration).
I have some Raman spectra where I'm looking at A1g modes by varying the polarisation of the incident laser across 360 degrees and I saw intensity minima at 90 degrees and 270 degrees as expected but I also saw a dip in the intensity at 180 degrees (i.e. -XX configuration). My question is, should A1g peaks be reduced at 180 degrees due to being antiparallel to the polarisation of the backscattered light? Or might it be the case that the dips in intensity are due to something else as A1g modes are totally symmetric under all symmetry operations?

In solution the theoretical ratio of parallel to perpendicular polarised Raman intensity (measured relative to laser polarisation) is $\rho=3/4 $ if the  mode is non-totally symmetric, and $0\lt \rho<3/4$ if it is totally symmetric, and closer to zero the more highly symmetric the molecule is. Also the totally sym. vib. often is the most intense. $180$ deg should be be the same as 0, so it seems that your signals may have another source. (The intensity is over the integrated band, not just peak)


text:
Chemistry of a proposed Capsaicin extract
Background:
I’d like to make an extract of capsaicin using the following ingredients: Capsicum powder, Fractionated coconut MCT oil, apple cider vinegar, and saline. The idea is for the capsaicin to undergo absorption from the capsicum and into the oil to create an extract. The apple cider vinegar will be used as an emulsifier so I can mix the solution with saline. I chose these specific ingredients because they work well with my particular use case, but let me know if I should use a different solvent and or emulsifier (I'd like it to be food safe, and I'd like to avoid alcohol.)
Proposed process:

Combine MCT oil and Capsicum powder and heat to around to around 250° F and let cool
Filter out the leftover solids using a coffee filter
Combine with apple cider vinegar and saline.

Questions (in order of importance):

Could this combination of ingredients react to create unwanted chemical byproducts?
Is capsaicin soluble in Fractionated coconut MCT oil, or will I need to use a different kind of oil?
Will this proposed process actually “extract” the capsaicin from the capsicum powder, or is additional processing required?
How should I determine the ratio of ingredients?
How long should I keep the oil & Capsaicin over heat?
I used the term “extract” to describe this, but is it actually a decoction, infusion or something else?
Any other tips, suggestions, warnings?

I know just enough about chemistry to know that I should probably ask questions before giving this a go.

Some of this might be answered at https://cooking.stackexchange.com/ As for your questions, I would answer: (1) depends. Why are you concerned about chemical byproducts? For safety, taste, other? (2,3) probably. Run tests. (4) Run tests (5) As little as possible (6) extraction is the right term; the other are cooking terms (7) test test test; chemistry is largely an empirical science. Also, google works wonders: search for "oil solubility capsaicin"

See for instance: https://www.researchgate.net/publication/263619725_Effect_of_vegetable_oil_in_the_solubility_of_capsaicinoids_extracted_from_capsicum_Chinense_Bhut_Jolokia


text:
Why water volume went up by almost the same amount after adding salt?
Seems like it's a well-known fact that adding salt to water will not raise the water volume by much (e.g.: Why there is no change in water level when salt is added? )
I've done this experiment at home and saw unexpected results:
Setup:

Start with 500 ml of tap water in measuring cup (500 grams)
Measure 50 ml of iodized salt in measuring cup (80 grams)
Add salt to water, stir until dissolved

Result: around 540 ml (give or take) of salt water in measuring cup weighing 580 grams
Weight was measured with kitchen scale, volume measure by eye, but ~40 ml increase in total volume seems way too much.
Followed up by replacing salt with sand, volume went up by about the same amount.
Video: https://www.youtube.com/watch?v=rLTn5jKqipU
What am I missing here?
p.s.: the commenter pointed out that my salt volume is off, I should have looked up the density to go by weight rather than by measuring cup notches. Going by weight, I've added 37 ml or salt, making the result event closer. Volumes were judged by eye in food measuring cups, not exactly the proper lab equipment.

With the density of NaCl at 2.16 g/cc, 50 ml should weigh 108 g. Bottom line is there is an error on your part somewhere.

@ToddMinehardt I guess the error was that instead of 50ml of salt I've added 37ml, which makes the result even closer. As I've said, I used food measuring cups, so volumes are not too precise, that's why I've used weight to keep me honest. Bottom line, water volume went up by roughly the amount of salt added. I'm sure you have salt and water handy, give it a go.

@ToddMinehardt forget the experiment, how much should the water volume go up if I add 50ml of salt to 500ml of water? Is there a reference table for something like this? I'd expect there to be one.

Why, you can look up the solution density tables.

How do you measure the volume of the salt ? If you measure it by filling a graduated glass column, you obtain the volume of the salt plus the volume of the air between the salt crystals. This measured value is NOT the volume of the salt. The volume of the salt cannot be measured. It must be calculated from the mass and the density.

You have water volume.
Calculate salt volume from its mass and density.
Calculate their total volume before dissolution.
Calculate solution mass concentration in mass %.
From tabulated data or online calculators, obtain its density.
Calculate solution volume from its total mass and density.
Subtract from it the total volume of separate water and salt.
You have the volume difference.

Here is the simple formula to calculate the difference:
$$\Delta V = V_\mathrm{solution} - V_\mathrm{solute} - V_\mathrm{solvent} = \\
\frac{m_\mathrm{solute} + m_\mathrm{solvent}}{\rho_\mathrm{solution}} - \frac{m_\mathrm{solute}}{\rho_\mathrm{solute}} - \frac{m_\mathrm{solvent}}{\rho_\mathrm{solvent}}$$
For our particular $\ce{NaCl}$ case:
$$\Delta V = \frac{m_{\ce{NaCl}} + m_{\ce{H2O}}}{\rho_\mathrm{solution}} - \frac{m_{\ce{NaCl}}}{\rho_{\ce{NaCl(s)}}} - \frac{m_{\ce{H2O}}}{\rho_{\ce{H2O}}}=
\\
\frac{m_{\ce{NaCl}} + V_{\ce{H2O}}\cdot \rho_{\ce{H2O}}   }{\rho_\mathrm{solution}} - \frac{m_{\ce{NaCl}}}{\rho_{\ce{NaCl(s)}}} - V_{\ce{H2O}}$$

$V$ is volume [mL]
$\rho$ is density [g/mL]
$m$ is mass [g]

Volume changes during dissolution are common, usually contractions. $\ce{NaCl}$ solution volume decreases, as water molecules as electric dipoles are attracted to ions $\ce{Na+}$ and $\ce{Cl-}$. As the result, ions and molecules in solution are packed in average better than salt and water apart.
It can be somewhat illustrated on mechanical macro analogy of 2 jars with marbles and sand. If mixed together, their total loose volume decreases as they better utilized the unused space.
There happened an error in obtaining density for given salt mass fraction, which is smaller than originally provided, therefore the volume of solution is larger.
The volume contraction $\pu{10.2 mL}$ is less than 1/3 of the (real) volume of added salt $\pu{36.9 mL}.$
Using the link at the solution density (or other density data sources), you can predict volume contraction for any salt/water mass ratio (within salt solubility)
Here are calculates values for you particular case:

System
Quantity
Value
Unit

Water
Volume
500
mL

-
Density
0.9982
g/mL

-
Mass
499.1
g

Salt
Volume
36.9
mL

-
Density
2.17
g/mL

-
Mass
80
g

Total
Volume
536.9
mL

Solution
Massconcentration
13.815
%w/w

-
Volume
526.7
mL

-
Density
1.09947
g/mL

-
Mass
579.1
g

-
Volumedifference
-10.2
mL

https://www.youtube.com/watch?v=rLTn5jKqipU the level clearly went up way more than 3 ml, what am I missing?

I have fixed the error due somehow wrong value of tabelated solution density. Volume contraction is less than 1/3 of volume of added salt.


text:
How to dissolve beeswax/carnauba wax to spray
I’m trying to make an edible hydrophobic coating for a tart shell. If anyone knows of a way to lower the viscosity of an edible wax enough to make it sprayable I’d really appreciate it.
I have a paper here that explains that it could be done but it doesn’t state the solvent used
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7435775/

Because of the specific tag baking and tart, I speculate sibling seasoned advice may provide an audience with more (practical) expertise here.

Also, since random organic solvents might not be food-safe, this question should not be here due to possible health concerns. That said, if you do succeed in waterproofing that tart, you might snack on it submerged: https://www.youtube.com/watch?v=5UILp88D3XQ

In Section 2.2 of the paper cited, the method and solvent are specified:
"...soybean wax suspension was prepared through mixture of 1 g of soybean wax and 50 mL ethanol solution followed by heating at 65º C for 3 min." The superhydrophobic character of the coating described in the paper may not be necessary for the tart shell (see below for examples). Extra care must be observed when following this recipe, for the flash point of ethanol is only 14º C, but this method has the advantage of easy evaporation and safe consumption of the carrier solvent.

Edible waxes can be emulsified and dispersed in water and are widely used to coat fruit and vegetables. Spraying an aqueous solution onto a tart shell could be adjusted so as not to overwet the shell, but some experimentation will be necessary. Wax emulsions are commercially available from many sources, such as Hexion (Ref 1) and Lubrizol (Ref 2). More commercial sources are listed in Ref 3.
Or, with a bit of experimentation, a food-grade wax (e.g., carnauba or polyethylene) and a food-grade emulsifier can be emulsified with intense stirring in hot water (to melt the wax), then cooled to stabilize the formulation.
Cooking and baking are specialized branches of chemistry, but you can eat your mistakes.
Ref 1. https://www.hexion.com/en-US/chemistry/wax-emulsions
Ref 2. https://www.lubrizol.com/-/media/Lubrizol/Coatings/Coatings-Literature/Wax-Additives-Product-Guide---22-122162.pdf
Ref 3. https://www.thomasnet.com/products/fruit-vegetable-coating-wax-93060200-1.html

I never tried rubbing carnauba wax on a tart and brushing it to a glossy shine, but one other idea came to mind:  food grade ethyl acetate.
Carnuba wax could be dissolved in ethyl acetate and sprayed on, with a few more minutes in the oven at elevated temperature to evaporate the solvent.
Ethyl acetate is approved for use in foods as a flavor additive and should be safe in low concentrations.
Carnuba wax could also be softened with heat and applied to tarts in a rotating tumbler.  This is how it is added to the surface of candies to make them shine.
It should be cautioned that adding excessive amounts of wax may affect the palate feel and texture of the tart.  Finding a waterproof wrapping may be better.  However, if feeling a bit adventurous, one might consider a chocolate coating as well.

Wouldn't a tart fall apart if tumbled?

Hmm, I assume the approve as flavour additive is not supposed to be extended for bigger amounts, used for spraying. As there are no harmful nor harmless chemicals. Their amounts are.

@Poutnik surprisingly, ethyl acetate (ester of ethanol and acetic acid) has an LD 50 comparable to neat alcohol. (One might get drunk and have a hangover at the same time).  Other waxes could be tried.  Cocoa butter had me curious.

It does not contradict what I have said. Food acceptance involves also long term influence, not covered by LD parameters. I do not have handy conditions of its usage, but I doubt it would be unconditional.

@Poutnik Aye, and we're the right track with waxes and solvents.


text:
Water Electrolysis Efficiency
During my water electrolysis with NaOH, I measured the voltage and current, and the volume of gases (hydrogen and oxygen) displaced at every 30-second interval. 10g NaOH in 2L water for 8 minutes.
Then, I was able to calculate the rate of reaction of the gases in terms of volume per unit time. Dividing the rate of reaction by power gives the efficiency of the trial (how much oxygen or hydrogen are produced per unit energy). As I vary the voltage from 5V to 10V and then to 15V, there is a decline in efficiency (less gases produced for each joule).
What is the reasoning for this? My hypothesis is that while the rate of production is directly proportional to current, as I increase the voltage, both current and voltage increases which causes the power to increase at a quicker rate. Ex. $P=IV$, then we increase $V$ to 2$V$, $I$ therefore increases to 2$I$ due to $V=IR$. Therefore $4P=2I \cdot 2V$ where the rate of reaction only increases by a factor of 2.
Or is energy lost to heat? But this will not fully account for the large decline as shown in the following picture.

One easier way to measure the amount of products formed (albeit not 100% correct*) would be to indeed measure the charge that passed through the circuit as Poutnik mentioned.
In an electrochemical system there are different processes which all lead to losses. Not all of them follow Ohm's law $IR=V$. First there's a minimal potential required by the Nernst equation (approximately 1.23V at STP), the equilibrium potential.
However there are also kinetic losses, at some small finite potential over the equilibrium potential one will find that the current is still approximately 0. This is because the rate of charge transfer (kinetics at the electrode) at the electrode-electrolyte interface is determined by the applied potential as well. The dependence of this rate on potential follows an approximately exponential course: $i = i_0e^{\eta_{kinetic}/b}$ with $\eta_{kinetic}$ called the 'kinetic overpotential' which is the applied potential  minus the equilibrium potential (and the potential of the solution $\eta_{kinetic}=U_{applied} - U_{equilibrium} - U_{electrolyte}$).
The next loss is the conduction of the electrolyte, which follows ohmic behavior. So: $\eta_{ohmic} = I*R_{electrolyte}$, which is the "Ohmic overpotential".
Finally, we need the hydroxide/water to get to the electrode and O2 and H2 out the solution. This is mass transport and is governed by the Nernst Planck equation and there is not a simple expression for this loss, but only starts being significant when you go to very high currents. At some point you will find that no matter how much potential you apply no increase in current happens, this is due to mass transport losses (as water, H2 and O2 transport is independent of the potential). One gives this also an overpotential: $\eta_{transport}$.
The total current-potential behavior is thus not straight forward, these simple 3 phenomena  already are quite complex and I have not described each contribution in enough detail. For more information I would definitely start looking at electrochemistry textbooks. In general the current is some complex function of the potential, and thus the efficiency at different rates (i.e. currents) is different: $I=f(U_{applied})$ and $efficiency=\frac{U_{applied}}{U_{equilibrium}}$.
Finally, all losses end up as heat.
*(Current going to side reactions or double layer charging of your electrodes)

Thanks, but is there any theoretical data (methods to find the supposed rate of reaction to power ratio) to compare with my experimental results? As you mentioned, this is extremely complex, so is calculating the theoretical ratio feasible?

You could start with measuring the resistance due to the electrolyte, you can measure the conductivity and then also the distance between the two electrodes. This would give you the resistance due to the electrolyte which will probably be a large part of your losses. If you know the surface area of your electrodes and the material you could also find kinetic parameters for the kinetic overpotential.

The amount of products is determined by passed charge $Q = \int_t{I \cdot \mathrm{d}t}$, not by spent energy $E = \int_t{U \cdot I \cdot \mathrm{d}t}$.
The relation of the charge $q$ and amount of substance $n$ is $q = znF$, where F the the Faraday constant $F = e \cdot N_\mathrm{A} \approx \pu{94685 C mol-1}$ and $z$ is the number of electron per particle.
Majority of energy even for $\pu{5 V}$ is wasted. Even worse for $\pu{10 V}$ and $\pu{15 V}$. You need not a source of high voltage. You need a low voltage source with low internal resicence, able to provide high current.
Theoretical voltage needed to electrolyze water is $U_\mathrm{theor} = \pu{1.23 V}$. Practical needed voltage lays in range $\pu{1.5 - 2.0 V}$, due kinetic reasons, depending on used electrodes, solution, geometry of the cell and required production rate.
The power being lost (as heat) $$P_\mathrm{lost} = ( U - U_\mathrm{theor})I \tag{1}$$
and the efficiency $$\eta = \frac{U_\mathrm{theor}}{U} \cdot 100\ \%  \tag{2}$$
As first (linearized) approximation:
$$\eta = \frac{U_\mathrm{theor}}{U_\mathrm{theor} + RI} \cdot 100\ \%  \tag{3}$$
where $\mathrm{R}$ is the Ohmic resistance of the electrolytic cell. Note that $\mathrm{R}$ is not constant, but depends on current, temperature, mixing, geometry, solution and other factors.
There is trade off between the electrolysis efficiency and the production rate. The maximum efficiency is near zero production rate and decreases with the rate.

You are right. But you should not use informations which are difficult to understand like "thrown out of a window" or "hard source" or "soft source", or "absolute loses" or "relative loses", or "minimal loses of both kinds means also the minimal rate".

What is difficult to understand? Those are frequently used terms. But OK, I have rephrased the considered parts.


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Is there an international standard for rating the danger of chemical elements?
I'm writing a small little tool (something like an interactive periodic table) and I wanted a good "guesstimate" of the danger of certain elements, to visualize across said table.
I could go and guess by manually looking up the physical / chemical / health / other effects of every element, but that would take an obscene amount of time and effort, not to mention it wouldn't be very accurate / precise.
My question is, is there a standard or "list" recognized internationally as a good scale of the danger of certain elements?
I know of the NFPA 704 Hazard Classification diamond, and while that would be perfect for my use, it doesn't classify every element, mainly just chemical compounds + a few noteworthy elements.
For example, I could find an NFPA 704 for just about every major compound of arsenic, but not pure arsenic (the element), which would be a pretty obviously dangerous element.

GHS.

You are asking for an international standard or an internationally recognized list; however, the NFPA diamond you give as an example is neither.

@DrMoishePippik as I mentioned in the question, I was specifically looking for a system of measuring safety as a whole.. simplistically. I also specifically mentioned that I wanted to avoid having to look up the safety data sheets / other measurements manually for every element by hand.

Assessing the safety of an element if pretty futile, it only makes sense for specific compounds. Chlorine gas is dangerous and toxic, but PVC (a polymer containing chlorine)is pretty inert and table salt (NaCl) is safe and a vital ingredient for life. How to rate the element makes little sense given those facts.

@matt_black That... that logic makes no sense. Im assessing the danger of elements on their own, not their compounds. NaCl might be safe, but Na and Cl on their own can be very dangerous. You can very much so rate an element and give it safety regulations / standards, as the answer to this question has already proven.

You can check the European ECHA Database (or the corresponding original CLP Regulation), which uses the Globally Harmonised System (GHS) managed by the United Nations, if it contains the relevant elements.
For arsenic, you would find that it's

Toxic if swallowed (H301)
Toxic if inhaled (H331)
Very toxic to aquatic life (H400)
Very toxic to aquatic life with long lasting effects (H410)

thanks, this will work for what I need. Ill have to estimate a kind of scoring system to actually visualize it, but this is a good start.


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Why does a symmetric stretch mode not have an imaginary frequency?
Currently studying potential energy systems and our professor asked a question at the end of a lecture and I can't wrap my head around it.
He said, "Picture a 2D PES for the simple reaction of H+H2 -> H2 + H. The direction of the symmetric stretch mode would be to the top right corner where both distances between the hydrogens in the transition state are large. This mode does not have an imaginary frequency. Why is that?"
I can't think of a reason that it would be an imaginary (negative) frequency since the TS must already have one for it to be a TS. So my only thought is that the TS already has one and it cannot have two.
Is my thought process too basic? What am I missing here?

Does the symmetric mode contribute to the transition state? Isn't the dissociation path equivalent to a high amplitude assymmetric stretch?

The path over the transition state is like an inverted parabola so has a negative frequency, i.e. its an upside down harmonic oscillator if you like. But orthogonal to this is a second potential like a Morse potential  and this has a normal (positive) frequency, just as is the case of a saddle point. Buy a pack of Pringles if you want to see 3D saddle points :)
You can see this in the energy contour plot calculated for H+H2 below (using the LEPS potentials). The hydrogen molecule is vibrating as it approaches the H atom and the product is similarly vibrating. The diagonal is the symmetric stretch at the saddle point ( circle) and the asym. stretch is along the reaction path, shown  as a short line at right angles to the other. The transition state is located at the highest point between the two valleys on approach and retreat and this has the shape of a saddle as you can see from the potential.

(I just noticed that the pic has been clipped on the right shouid read H+H2.)

Below is a sketch of the atomic arrangements when considering a linear attack.

The blue diagonal represents the symmetric stretch. The reaction coordinate (dotted line) takes the lowest possible path from reactant (top left) to product (bottom right). If I stretch the atomic arrangement, I make it worse, and if I scrunch the arrangement, I make it (much) worse. So this is like a vibration in a molecule, with a minimum somewhere in the middle.
On the other hand, starting from the transition state ("‡"), the asymmetric stretch makes one bond longer (the two atoms that will no longer be bound) and one bond shorter (the two atoms that will have a bond in dihydrogen). So there is a maximum in the middle, and both directions have lower energy.
Another way of saying it is: The "middle" hydrogen "wants" to make a single bond, not two or zero. The asymmetric stretch gets the transition state there, the symmetric stretch doesn't.
For an actual potential energy surface, see e.g. https://people.uleth.ca/~roussel/C4000foundations/slides/07PES.pdf

So is the reason that the symmetric stretch mode doesn't have an imaginary frequency due to the minimum it has? Because imaginary frequencies signify a maximum somewhere on the PES?


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Do formed galvanic cells counteract electrolysis?
I was wondering about the following:
Suppose we carry out an electrolysis with constant applied voltage. During the process, a galvanic cell should be formed by the reactants which are formed during the electrolysis. This galvanic cell should counteract the electrolysis in some way, if it gets discharged.
However is there any way to quantify this matter? I only found the comment on this website:
https://www.galvanotechnik-for-you.de/lexikon/electrolysis/
Any help would be appreciated!

Consider rechargeable cells.

Yeah I have thought of that, the discharge only happens really when the electrolysis is completed or halted

This is a very good question I don't know why it has close votes. The proper term you should be searching for is called back-emf (emf=electromotive force). It was known long time at least a century ago that if you start electrolysis of water, stop it temporarily, the ammeter will show current moving in the opposite direction for a very short time. This was attributed to the formation of gaseous products at the electrodes, that formed a galvanic cell i.e., adsorbed oxygen and hydrogen want to for form water again. There are many reasons for back-emf and the formation of electrolytic products is one of them.
I assume you can read German. One of the finest works is 2-volumes on Electrochemistry History is by Ostwald (Elektrochemie, ihre Geschichte und Lehre). It is freely available from the Internet Archive. The English volumes also exist. Check that reference for the development of these ideas. Many electrochemical engineering textbooks will have more details on back-emf.

Hi! Thanks for the reply, and yes, back-emf is the keyword here :)

Consider an electrolytic cell, consisting of identical half cells with copper electrodes and the same electrolyte - e.g. $\pu{0.1 M}$ $\ce{CuSO4}$ solution, separated by a salt bridge with the same solution.
Any external nonzero DC voltage will start an electrolysis. As $\ce{Cu^2+(aq)}$ will be depleted at cathode and formed at anode, there will be formed DC voltage of galvanic cell, acting against the external voltage.
$$U = \frac{RT}{2F} \ln{\left(\frac{a(\ce{Cu^2+aq},\mathrm{anode})}{ a(\ce{Cu^2+aq},\mathrm{cathode})}\right)}$$

For general cases, electrolysis shifts redox systems at each electrode, the one at cathode obtaining more negative equilibrium potential, the one at anode obtaining more positive equilibrium potential.
The consequence is there is progressively needed higher voltage to keep the constant current. There is also higher threshold voltage (equal to the cell equilibrium voltage) as the electrolysis needs voltage higher than this. It was routinely observed e.g. at charging of rechargable cells.

When net electrolysis happens then, at the cathode, the rate of reduction is (much) higher than the rate of oxidation a vice versa at the anode. Without the external voltage and without electrode galvanic connections, both rates are equal with the zero net effect.
By other words, the rate or reduction exponentially grows with decreasing electrode potential, while the rate of oxidation exponentially grows with increasing electrode potential. Both until it hits other limitations, like ion diffusion. At given scenario, an electrode establishes such a potential that leads to particular value of the net current.

Is this only the case when both cells are identical or is this the general case?

Yes and no. It depends on what you mean by "this".

Let me rephrase a bit: Consider electrolysis of a Zn/Cu system. If we start with a 1M solution of zinc sulfate and 1M copper sulfate, we need >1.100 V to start electrolysis.  After a while, for example zinc ions are at 0.5 M and copper at 1.5 M and the cell voltage is at 1.114V. But as long as the voltage is always greater than the decomposition voltage, only electrolysis happens. For example, the system could not discharge (reverse reaction) while this voltage is applied, is that correct?

It is and is not correct - see the A update.

Poutnik, How can the rate of reduction be not the same as rate of oxidation. For each electron leaving, there must be one electron entering the electrolytic cell?

How can the rate of reduction be not the same as rate of oxidation - On the same electrode.

You mean in the absence of any external current.

As reversible processes, that are on macro level at static equilibrium, are at dynamic equilibrium on atomic/molecular/ionic level.

Or absence of potential drift, at which both rates just converge to the same value.

Yes indeed, but that is not my point. I am trying to confirm if you meant that the rate of reduction and oxidation is the same on the same electrode, in the absence of any externally applied potential.

At absence of external current or forced potential AND constant electrode potential, both nonzero oxidation and reduction rates on the same electrode are equal.

Btw shouldn’t it be the other way around? Rate of oxidation should increase when the potential gets lower (more negative)?

No, it should. Forced low potential means raised chemical potential of electrons, forcing reduction by  accepting electrons.  And vice versa.


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Why is the inductive effect possible?
Firstly, I must preface that I am a Biology student. To elaborate my question, I was looking for the reason why carboxylic acids are more acidic than alcohols, despite both compounds containing —OH groups. From my understanding it was because of the inductive effect, an example of this effect is CH3COOH, shown in the figure below:

The inductive effect in CH3COOH occurs because of the electron deficiency in the carbon atom, C (which has a partial positive charge, δ+). This electron deficiency on the carbon atom is present due to the more electronegative double bonded oxygen atom, O. However, the O single bonded to the C is also more electronegative, as a result, the electron density of the single bonded O moves to the C in order to 'counteract' the δ+ on the C, leaving the H 'less strongly attached' to the O (and therefore, the molecule), thus, increasing its acidity.
Now, in an alcohol there is only one oxygen atom, see the figure below:

Because there is no second O attached to the C, which would cause it to become more positive, the O does not become less positive, by sharing electrons to counteract the positive C. Thus, the hydrogen atom is 'more strongly attached' to the O which is more negative compared to the single bonded oxygen atom in the carboxylic acid.
My question is, why does the carbon atom in the carboxylic acid (first figure) counteract the change and increase its pull for the electrons of the single bonded oxygen atom, instead of becoming more positive?

Compare acidity of these 2 pairs:  $\ce{Cl-O-H}$ vs $\ce{O3Cl-O-H}$ and $\ce{R-CH2-O-H}$ vs $\ce{R-CO-O-H}$. Carboxylic acids and perchloric acid have stronger polarization of $\ce{O-H}$ bond by oxygen induction and better delocalication of anionic charge, what stabilizes the anion.

This thing with acid is rather describing mesomeric effect in a misleading way.

That's an interesting reasoning you give here.
The point is, the inductive effect is not the major factor explaining the acidity of a compound.
Electronegativity will give insights on the position of electrons along the bond, explaining that the hydrogen atom can be detached by a base in both cases (the lengths of carbon-oxygen bonds also need to be considered to understand the interaction between oxygens and carbon).
Yet, what makes a carboxylic acid more acidic than an alcohol is the stability of the conjugated base : a carboxylate ion, thanks to the $\ce{C=O}$ bond on the same carbon as the $\ce{C-O^-}$, is stable by mesomeric resonance (the lone-pair gained by the oxygen atom releasing the hydrogen can 'move', the direct consequence is that both carbon-oxygen bonds will have the same length, between $\ce{C-O}$ and $\ce{C=O}$).
However, the alcoholate will not have this resonance, thus will be less stable (see this link for a longer development on that, and images).
I don't know if this is what you wanted to know, do not hesitate to tell me otherwise.

Firstly, I always want to know new things. Secondly, thanks Thomas I appreciate the depth, as a second year Biology student (with one completed introductory chemistry course) I tried my best to correctly interpret your answer and I have read the link, though I lack the proper foundational knowledge required for me to understand the terms and concepts presented in the LibreTexts page.

Nonetheless, from my understanding, the two O atoms in the carboxylate ion (R−C−O₂⁻) (conjugate base of carboxylic acid) can distribute or 'spread out' the charge of -1, whereas the -1 charge is locked on the single O atom in the alcohol ion (R−C−H₂−O⁻) (conjugate base of alcohol acid). Therefore, carboxylate is less polar and more stable than the alcohol ion, since H⁺ will be less attracted to the less negatively charged single bonded O atom in carboxylate, compared to the more negatively charged O atom in the alcohol ion.

However, I have no idea why 'charge delocalisation' and resonance occur. From what I understood, according to resonance theory, the carboxylate can have two structures with a -1 charge on either oxygen atom (no idea how the electron hops from one O atom to the other), these structures are averaged resulting in each oxygen atom having a charge of -0.5, for example. If the single bonded O atom in one structure of carboxylate had the -1 charge then wouldn't carboxylate be just as stable as an alcohol ion?

Obviously, my lack of chemistry background is involved here. Lastly, I do not understand how this answers my initial question on the inductive effect.

Thanks for your comments @Simon. I didn’t have any clear idea of what your chemistry exactly is, yet I’ll try to give more precision.

“ I have no idea why 'charge delocalisation' and resonance occur”: it is a matter of bonds and electrons. Compared to a single bond with electrons following the inter-atom axis, the double bond is a conjunction of both a single bond, and an extra pair of electrons less strongly fixed in between atoms. That explains why these two extra electrons can move more freely from one bond to another if they can (standard bond number rules still apply). Hence, the lone pair giving the -1 charge to the alcoholate can form a double bond only if the other bond gives -1 to the other oxygen.

To answer your question, what is observed in the carboxylic acid is explained by the fact that mesomeric effect is stronger than inductive effect. Hence the increased acidity and observed results.


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Electron orbitals
Can electrons be found anywhere within the space described by a 3D orbital "90% of the time" (as stated in my textbook)? But that would mean they can be found right next to the nucleus or in the space of a lower energy level "90% of their time" (since the spheres and other shapes overlap starting right next to the nucleus).
We also know that electrons cannot jump from one energy level to another without absorbing or releasing energy and that they maintain a specific average distance from the nuceleus as denoted by a 2D depiction of their electron shells in the form of concentric circles (also written in the same textbook).
Even if they only skimmed the surface of the 3D spheres or dumbell shaped orbitals, the latter, e.g. a 2p orbital, still starts close to the nucleus while it should be farther away from the nucleus than a 1s orbital.
PS Thank you for your answers. But please elaborate on them since I did not understand at all. I just finished high school but am extremely curious and confused by the new information we are studying, and I want to understand...

Note that even the Bohr model, when modified to adapt electron orbital angular momentum quantization, stopped using the original constant radius of circular electron orbits. It was replaced by elliptical orbits. The the whole Bohr model is already for long time obsolete after creation of the quantum atomic model, based on Schrödinger wave equation. It is used only for education purposes in context of history of science.

Yes. The electron has a small but not zero possibility to stay quite near the nucleus, and for example at a smaller distance than the traditional radius of the first 1s orbital. The probability of finding an electron is a number that looks a bit like the deviation of a vibrating rope fixed on the nucleus and going to an infinite distance, if all its points are attracted by the nucleus. Hmmm ! Beg the pardon from all real scientists !! It is not like this. It just looks like.

Can electrons be found anywhere within the space described by a 3D orbital "90% of the time" (as stated in my textbook)?

Yes, they can.

But that would mean they can be found right next to the nucleus or in the space of a lower energy level "90% of their time" (since the spheres and other shapes overlap starting right next to the nucleus).

Yes, it means that. There is intense overlapping of orbitals.

Particular points of 3D space around a central force do not belong to any particular energy level, being it around an atom nucleus or a star, because the sum of potential and kinetic energy must be considered.
By other words, the electron energy level does not determine the electron distance to nucleus. It only determines the probability of being at that distance.
The direct analogy is in planetary orbits around the Sun. The Earth is about $\pu{150000000 km}$ away from the Sun on near circular orbit with orbiting speed near $\pu{30 km s-1}$. But Earth could have at the same distance higher speed up to $\pu{42 km s-1}$, being at its perihelion (being the closest to the Sun on an elliptic orbit). Or a much lower speed, being at its aphelion (The most distant point of an elliptic orbit).

Consider classical gravitational analogy of the probe orbit. The mechanical energy alone(+) of the probe determines the farther distance from the planet the probe can reach. The closest distance(+) is not limited ( but by the radius of the planet ).
Similarly for orbitals. Electrons from all s orbitals can all occur near nucleus, as their zero orbital angular momentum is  no limitation. 6s electrons of gold  move at relativistic speed there, leading to orbital energy shift and color of gold.
(+) -  It can be further limited by the conservation of the probe orbital angular momentum.

If I got it right, from the example of Earth at its perihelion and aphelion, the Earth increases its speed close to the sun and decreases it away from the sun so that the sum of potential and kinetic energies remains the same... Similarly electrons increase their speed when they come close to the nucleus and decrease it away from it. Is that right? Also considering 1s and 2s orbitals... do they overlap or does 2s begin where 1s ends? Because in the former case an electron would positively be jumping to a lower energy level for most of its time which is not possible...

But if the lower distance is not limited, why would electrons need to release energy to jump to a lower energy level...

What I wanted to point out it could have higher or lower energy than it has now and still be at some moment and with some speed at the same distance as is Earth now. All ( or near all? ) orbital overlap.

Jumping between energy levels is not jumping between 2 different places.  It means the original and new orbital overlap at that point, both with significant probability density at the3D region where the electron is. The energy change is via kinetic energy change ( together with angular momentum change)

Ok so when an electron enters a lower energy level it might increase its kinetic energy or as we know, it might release energy to stay in that shell..? Both things are possible?

Lower energy levels have lower kinetic energy in the same place than higher energy levels. OTOH, lower energy levels have higher average kinetic energy. In these aspects it is the same as in classical gravity.

Well, now what you have written does not make any sense.

Then I will delete my first comment. I am sorry for being rude because you did try to help. I just got carried away because I have been trying to understand this for so long and.. k tc!

Classical analogies about simple orbits don't describe electron behaviour well
Thinking about electrons as having "orbits" is rarely helpful as their behaviour only makes sense in a quantum mechanical (QM) approach. In QM it is meaningless to talk about position and velocity separately and you can't know both precisely at the same time.
Orbitals describe the probability of finding electrons in a particular region of space. And, indeed, in some types of orbitals (eg 1s) there is finite probability of the electron overlapping the nucleus (2p orbital have a node there so this isn't true for them): this matters for some other quantum effects involving nuclei and electrons. But the probability described by the shapes of orbitals is not uniform, some areas "inside" the area shown in most pictures have a higher probability density than others.
So saying that "they maintain a specific average distance from the nucleus" is not really correct: That is a classical analogy that is a poor description of what electrons do. The average distance is a vague way to describe the average of the probability cloud of electron location, but the idea that a distance is "maintained" contradicts the quantum picture.
And the picture involving orbits where being further from the nucleus is "higher energy" is far too simple. The energy is a function of the whole probability distributions, not the specific average distance. The electron location clouds of different orbitals do overlap.
Classical analogies about electron behaviour are not usually helpful. Unfortunately, QM pictures are often not simple until you have reached a certain ability to the complex mathematics behind QM. Until then "fuzzy clouds of probability" are the best you can do.

Classical analogies are not intended to describe electron behavior. They just point out some shared principles, while omitting the difference. Otherwise, they would not be analogies, but identities. Both work with potential and kinetic energies.

Thank you matt_black.

@Poutnik That might not be the intent of classical analogies, but that is how many untutored people read them, causing confusion.

The essential thing is to know the general purpose of any analogy.

[OP ...] that would mean they can be found right next to the nucleus

They are close to the nucleus at times. The nucleus is tiny compared to the dimensions of a bond (or an atomic radius), so the electron can come pretty close. When it comes too close, it can get captured (electron capture, for certain radioactive nuclei), or gets scattered (by forces that only act at very short range and are usually ignored in chemistry because these are very rare events).

[OP ...] or in the space of a lower energy level "90% of their time" (since the spheres and other shapes overlap starting right next to the nucleus).

The most powerful way to explain this is by treating electrons as a wave (you might have heard orbitals mentioned together with wave function). Just as we can hear multiple musical notes at the same time in the same place, electrons can occupy the same space. The technical term is superposition of waves. Below are two standing one-dimensional waves, one with a single node and the other with four nodes, superimposed (amplitudes are added).

Even though the two waves are using the same space, they are independent of each other, and you can separate them mathematically (because they have different wavelength and frequency). Our ears do the same thing with music, taking a single input stream and separating out the different pitches of musical notes and overtones.
The waves describing electrons are more complicated (they are three-dimensional and have an exponential decay away from the nucleus rather than ending at a defined position), but the idea of superposition still holds. Here is a depiction of the 3p wave function (a snapshot, not showing the time-dependency like I did above), showing various nodes (zero probability) and that there is no defined boundary to where electrons might be found (exponential decay means it gets very unlikely for large distances from the nucleus).

Source: https://mathematica.stackexchange.com/questions/32378/is-there-something-like-densityplot3d-to-visualize-atomic-orbitals

Thank you so much! This was very helpful... However, I didn't quite understand what you meant by "three dimensional waves," what they are or how to visualize them.. Could you please elaborate on that? Does that mean an electron vibrating in .. a certain way? Or.. the 3D shape of the electron itself? Does the three dimensional nature of these waves have anything to do with the three dimensional space occupied by the electrons?

You lost me with "exponential decay away from the nucleus...." too. Didn't understand the last paragraph actually..

@Falak Just to be sure, have you read Wikipedia: Atomic orbital ? Or similar searched for on Libretexts.org or hyperphysics.phy-astr.gsu.edu

@Falak 1 dimensional waves are like the vibrations of a violin string. Most people are very familiar with those. But 2D surfaces also have vibrational patterns, but more complex ones (search for Chaldni patterns to see examples). 3D waves are even more complex: and the type represented bay electron orbitals are known as spherical harmonics which is also worth searching for. Different orbitals are, basically, spherical harmonics.

3D waves are hard to visualize - we run out of dimensions. Sound is a good example, though, and there are some cool effects like echo chambers or the Doppler effect you can appreciate without extra math. The deeper questions you are asking, though, require quite a bit more math for solid answers. BTW: The square of the wave function is proportional to finding the electron in a given spot, so there is a direct relationship.


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Is determination of nuclidic mass by nuclear reaction currently state of the art for any nuclides?
In Section 4.9 of Linus Pauling's General Chemistry Book (Dover 1988 edition, a copy of the WH Freeman and Company 1970 edition) he describes the principle behind the determination of nuclidic mass by nuclear reaction, giving the following beta decay reaction as an example for the determination of the mass of $\ce{{}^{40}_{19}K_{21}}$:
$$\ce{{}^{40}_{19}K_{21} -> {}^{40}_{20}Ca_{20}^+ + e^-}$$
I was wondering if nuclidic masses determined this way remain state of the art (the origin of the currently accepted value of the mass) for any nuclides? More generally,  how important is the technique currently?

The canonical source for masses, the Atomic Mass Evaluation 2020 part 2 (https://iopscience.iop.org/article/10.1088/1674-1137/abddaf) shows in column 5 of Table I ("The 2020 Atomic mass table") the 'origin' of the evaluated mass, and many indicate using various decay paths. Interestingly, the example above is not one of them...

I would expect, considering broad electron-antineutrino energy spectrum, that beta decay would not be the best way for precise measurements.

@Poutnik - sure, for measuring a single decay. You measure a bunch and get the full electron energy spectrum and you have a quite reasonably precise measure. Yes, the alpha channel will be more precise with less data.


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Reaction of BTNA, DMSO, and phosphate?
I prepared a solution of n-benzoyl L-tyrosine p-nitroanilide (BTNA) in dimethyl sulfoxide for use in an enzymatic assay. However, mixing it with a solution of phosphate buffer produced a color change for unknown reasons. Here is what I did:

A 1.4mM solution of BTNA in DMSO was prepared.
5mM phosphate buffer was prepared. This was done by dissolving dipotassium phosphate in distilled water and adjusting the pH to 7.59 by addition of hydrochloric acid.
Into a cuvette was added 0.4mL of the BTNA solution, followed by addition of 0.1mL of the phosphate buffer solution.

When the BTNA solution and phosphate buffer were mixed, the mixture took on a yellow color. I tried mixing the BTNA solution with distilled water without any phosphate added to see if, perhaps, the act of mixing water with the BTNA solution is responsible for the color change. In this case, the mixture did not change color.
My question is thus: Why did mixing 0.4mL of BTNA solution in DMSO with 0.1mL of 7.59 phosphate buffer result in the mixture taking on a yellow color, whereas mixing BTNA solution with water did not result in a color change?
Sidenote: None of the solutions were tested for contaminants.