text:
What precipitate is formed when a copper coin is dissolved in concentrated nitric acid?
The method being used.
I am currently carrying out a school project where I am determining the copper content of 1p coins. The coins used are British 1 pence coins dated before 1992 as after this year they were altered and no longer composed mainly of copper. In my experiment, the coin is placed in 16M nitric acid until dissolved and then titrated against sodium thiosulphate.
I have noticed that when the coin is dissolved in the nitric acid and made into a working solution with deionised water, a slight precipitate is formed and the solution appears cloudy which is not the case with the pure copper sample I have repeated the experiment with. The coins are made up of mainly copper with small amounts of zinc and tin so I am wondering whether this is causing the precipitate and affecting my results.
For both copper coins I have used, the percentage copper that I have calculated has been around 95% (online sources say coin is 97% copper). However when running the experiment with the pure copper sample my result was only 90% copper. While my results for the coins are fairly good, the result for the pure copper sample is far from ideal.
I will also add that the pure copper sample dissolved much more rapidly than the coin which could be a factor if the coin just simply isn’t fully dissolved.
I am looking for answers as to why this might be happening and any suggestions as to what could be changed.

Some precipitation of hydrated tin(IV) oxide? Copper and zinc should dissolve completely.

Just a point on notation: I think you mean working solution here: " and made into a standard solution with deionised water". A standard solution is one of known concentration. You have the opposite.

The textbook analysis of brass precipitates tin as a metastannic acid that was given an empirical formula. The procedure requires heating to dryness and redissolving copper and zinc.

BTW, how clean is the surface of the coin? Could it be foreign matter (e.g., organic residue from handling)?


text:
UV-Vis spectrum of 1,3-butadiene

Today I had the first lesson about UV-Vis spectrum. From the picture, the transition from HOMO to LUMOs of 1,3-butadiene depends on whether the structure is s-trans or s-cis.
My question is:

Why the π-π* transition from $\psi_2$ to $\psi_4$  is not allowed in s-trans conformation while it is a bit opposite for s-cis conformation
There is a slight change in wavelength of the transition from $\psi_2$  to $\psi_3$  in two conformation, while the transition from $\psi_2$  to $\psi^*_3$  remains the same value. Why is it so?

After Porphyrin's suggestion about symmetry of the transition, the only symmetry group that I could deduce is from the molecule itself, is there anyway to identify the symmetry of the orbital?

(1) the symmetry of the molecular orbitals between the ground state and the excited state in each point group, C2h for trans C2v for cis, determine if the transition is allowed for forbidden.  You need to know how to use point groups to work this out.

@porphyrin Its listed in the selection rule about the symmetry of the ground state and excited state. Is it the key to solve this problem?

yes. selection rules the symmetry properties of orbitals explains where these come from.

source: Stack Exchange

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

sort: votes


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I have fabricated a starch film impregnated with calcium carbonate, and the film is showing hydrophobic behaviour after addition of calcium carbonate
I have fabricated an extruded starch film, impregnated with calcium carbonate. After addition of calcium carbonate the contact angle of the film has risen suggesting a more hydrophobic surface. I have gone through various literature sources and the behaviour of calcium carbonate with water is very confusing. Some researchers have stated that calcium carbonate is hydrophobic and some have stated it is hydrophilic. Please suggest a relevant theory behind the increase in hydrophobicity of the starch films after calcium carbonate is added.

$\ce{CaCO3}$ itself is hydrophylic, though it might be less so than starch.
However, surface texture may make a substance hydrophobic. Simply forming a surface into nano-pillars can make it water-repellent. There are some "hairy" materials in nature so water-repellent that a layer of air is maintained even when fully submerged: the Salvinia effect.
Examine the surface carefully under high magnification, with various types of lighting (e.g., bright-field and dark-field illumination) to look for surface "fuzz". If you have access to an electron microscope, give that a try.
Perhaps your technique could be used to make a useful product, e.g., slow-release starch capsules for medicines... or not-so-useful, e.g., water-repellent pasta -- boil for hours and it stays dry.

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


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Why do some powders form powder-filled blobs in a liquid?
e.g., why does this happen? (I think in the picture, it's flour mixed with water. I've seen examples of this with many different powders mixed in liquids, usually water.)

What has happened at the boundary of the powder that made the powder-walls impenetrable to the fluid, so much so that the powder inside the blob is, by all appearances, as dry as if the liquid were not even present?
BONUS QUESTIONS:

What I think would be really cool, and impressive, is if someone can provide some mathematics to calculate, given a particular powder and a particular liquid at a particular temperature and at atmospheric pressure, the expected thickness of the blob-walls. (My intuition—unjustified for the time being—is that this thickness should not significantly depend on other factors than those I've just listed.)

Finally, one more piece of data associated with a slightly different line of inquiry: in my experience mixing powders and liquids, sometimes the blobs form and sometimes they do not. This seems to depend, among other things, on the manner in which the mixing happens. (For example, I have a collagen powder that I mix in my coffee every morning. I have noticed that the blobs do not usually form when I pour the powder on top of the coffee; in this case, the powder dissolves easily even without stirring. But when I pour the coffee on top of the powder and stir with a spoon, blobs galore!) This observation prompts the following question: what determines whether blobs actually form, or not?

Why was this downvoted?
Help me and show mercy. I'm a physicist not a chemist.
I posted it in the physics stackexchange before coming with my question over here, as someone suggested it might be more suitable here.

Soft matter physics are on topic in chem.se. Your question is perhaps bit hard to read, maybe you can make it a bit more succinct?

I can almost guarantee however that there is no simple property and formula to predict blob formation. ;)

There are multiple parts to this question (multiple sub-questions), and some of those parts I expect to be easier than others. I'll edit in order to prioritize the simpler questions, with the harder questions a "bonus". I know blob formation prediction is likely to be an exceedingly difficult sort of problem, but who knows.

Cross-posting is frowned upon. https://physics.stackexchange.com/questions/803963/why-do-some-powders-form-powder-filled-blobs-in-a-liquid

"Some powders" is terribly broad. Why flour does, not so much.

Lumps in a sauce are the bane of many a cook, and you can find out how to avoid this issue at seasoned advice SE (the cooking SE site).
What happens, taking bleached wheat flour as an example, is best explained as a step-wise mechanism, since this is a kinetic, and not mainly a thermodynamic problem (if you mix the components long enough you should arrive at a homogeneous composition such as a a nice liquid batter, although this may phase separate if left unattended for longer times).

Flour is poured on the surface of the water. The flour is composed of small particles containing a mixture of amorphous and semi-crystalline regions of starch (amylose and amylopectin) and protein (including gluten).
The bottom surface of the flour in contact with water is wetted. Absorbed water causes swelling at the water boundary as starch polymer chains dissolve and expand. Water can infiltrate the boundary but is limited by slow diffusion through the gel.
Gravity draws the gel into the less dense solvent, dragging down the dry solid atop. Capillary forces cause water to encapsulate the dry-wet bilayer and prevent dispersion of the dry solid, engulfing the whole in solvent. Step 2 repeats for previously dry surfaces.
Air remains entrapped in the dry-wet blobs, providing buoyancy. Sufficiently wet blobs might become entirely submerged due to density differences.

The main point to explain the heterogeneity of the mixture is step 2. Swelling of starch and gel formation slows penetration of the solvent. If you have made a batter before, you might notice that leaving it standing typically leads to it becoming more viscous. This is due to water penetrating the starch granules reducing the volume of free water and causing the polysaccharide chains to swell.
I'd add that there are two issues here: one is movement of water into and through blobs, the second is dissolution of starch. Both movement of water through blobs and dissolution of starch are slow but persistent. Dissolution of starch is however much slower because in addition to being larger, polymer chains within a gel are entangled.

The case of creamer/coffee is similar, though here the interface occurs on top of the creamer at the bottom of the cup. Why it doesn't occur when creamer is poured atop coffee might depend on mixing technique and vigour.

Interesting. So you're claiming the walls are not as impenetrable as they appear. Next time I observe this, I could let the blobs sit for awhile (rather than stirring until they disappear, as I usually do) and according to your explanation they should eventually become permeated with the liquid (no dry powder inside) left undisturbed. Thanks for your answer.

@DavidC. You may have to wait a long time. Gels can be fairly stable.

Upon review I have ignored the importance of crystalline or amorphous structures that do not readily dissolve. While these are barriers to water diffusion and impede dissolution, it is the gel that forms a layer surrounding the dry flour.


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What are some general chemistry textbook with good problems?
I’m currently in a General Chemistry course and am wondering if you guys can recommend me a textbook or resource where I can find good problems to work through and further my understanding. We’re currently using Oxtoby but I find the problems to be a bit straightforward.


text:
synthesis of Red Lead, Pb3O4
I have tried heating $\ce{PbCO3}$ at $\pu{400 ^\circ C}$ temperature to form $\ce{Pb3O4}$. After sometime of heating, only a layer of $\ce{Pb3O4}$ is made and the reaction does not complete. The substance stays white even after hours of heating.

More heat. 2. If a crust forms, it prevents oxygen getting in. Agitate and crush.

Crust means partial oxidation. Yes, more heat and more reaction time.

According to your post, you have tried heating $\ce{PbCO3}$ at temperature $\pu{400 ^\circ C}$ to form Red Lead, $\ce{Pb3O4}$. You haven't give how long you have heat, but mentioned the substance $(\ce{PbCO3})$ stays white even after hours of heating. I think your starting material is wrong in this instance (it should be beneficial if you have given us what literature procedure you have been following).
According to studies done in Ref.1, your starting material should be Basic Lead White $\ce{\left(2PbCO3·Pb(OH)2\right)}$ instead of $\ce{PbCO3}$ itself. Accordingly,

The thermal decomposition of basic lead(II) carbonate $\ce{2PbCO3·Pb(OH)2}$ in static air under atmospheric conditions has been studied for the first time using a combination of thermoanalytical, X-ray diffraction and Raman spectroscopic techniques. New intermediate compounds of the type $\ce{4PbCO3·3PbO}$ and $\ce{PbCO3}$ have been observed in addition to previously reported ones of the type $\ce{2PbCO3·PbO}$, $\ce{PbCO3·PbO}$ and $\ce{PbCO3·2PbO}$. Post-decomposition studies have identified the presence of the tetragonal-to-orthorhombic phase transition in $\ce{PbO}$. Results are also presented for the decomposition processes in nitrogen and oxygen atmospheres. Oxidative decomposition studies have shown that $\ce{Pb3O4}$ forms in oxygen and in flowing air.

According to the work described in here and here, it is clear that the starting material for this conversion is better to be basic lead(II) carbonate, $\ce{2PbCO3·Pb(OH)2}$ and temperature should be around $\pu{475 ^\circ C}$ (the heating should be for longer time period).

References:

Dan A. Ciomartan, Robin J. H. Clark, Lachlan J. McDonald, and Marianne Odlyha, "Studies on the thermal decomposition of basic lead(II) carbonate by Fourier-transform Raman spectroscopy, X-ray diffraction and thermal analysis,"   J. Chem. Soc., Dalton Trans. 1996, 3639-3645 (DOI: https://doi.org/10.1039/DT9960003639).

the substance I have used is PbCO3 because when I overheat it, it turns to a dark substance (PbO2). and I have tested all of the reaction period such as 30mins ,1 hour . after a layer of Pb3O4 is made it turns to yellow when I continue heating

You should have included these observations in your question at first place. that is important to reader to give you their thoughts. It is still not too late to do so.


text:
Non-catalysed exhaust gas
I'm writing up an academic paper (medical) on in-car exhaust levels and their effect on health. So as to ensure detection of leak-points that are located further along than the catalytic converter, in the discussion I would like to recommend the testing of an exhaust gas that is not affected by passing through the catalytic converter. Would I be correct in thinking that toluene is one such gas, and if so, would it be a good marker gas to use? Many thanks!

If you are measuring the actual outputs from a real engine, then carbon monoxide would be a good candidate as it is a key target for catalytic converters and is easy to measure. Toluene would not be as the amount will vary hugely by fuel composition and is probably not a major feature in exhausts from modern engines anyway.

Hi matt_black, many thanks for your comment. As it happens the study measured precisely that - I put CO dataloggers in in a bunch of cars driven by laypeople - and the results were pretty worrying! Some of the highest readings, interestingly, fell very quickly as the engines warmed up. This made me think that the CO may have originated from leaks distal to the catalytic converter - which is why a dual test, of CO and also of a non-catalysed gas, would be necessary to detect all in-cabin leaks. You can read about the project here - https://www.airsafe.london.

While the possibility of leaks from the exhaust is worth exploring, it would not be my first or most likely hypothesis explaining the presence of CO inside the car. Just normal atmospheric circulation will lead to that. And while cold engines might produce more CO, cold catalytic converters certainly do so the profile of the gas over time inside the care doesn't need leaks to explain the pattern observed.

If you're interested in learning more about the many factors that influence in-cabin CO, Kaur, Niewenhausen et al 2009 is a good place to start - from there, read Alameddine et al 2016, Hachem et al 2021 and Lim J 2019. Then look at the field studies - Moreno T et al 2019, Hachem M et al 2021, and round off with the Lancet meta-analysis (Cepeda et al 2016). And then once my paper is out, you could even read that too. TLDR, cars do often leak.

(and much thanks for the practice in thesis defence - always helpful! have a good one)

(oops sorry - Harik G et al 2017 is the second field study, not Hachem)

A much better tracer for leaks than toluene would be helium.
Compared to toluene, helium is a noble gas and hence unlikely to react at the temperatures and presence of the catalysts in a converter.  (Different to toluene, it plausibly would pass unchanged the combustion process of an Otto, or Diesel engine, too.)  It is inflammable and non-toxic. Its monoatomic molecules are considerably smaller in size, than a toluene molecule and hence can pass smaller cracks and openings.  At ambient conditions, helium already is a gas .
Helium sniffing already is a technique introduced in the industry to check e.g. joined gas pipes and vacuum installation for leaks: an external probe aspires air for instance next to a fitting, while the tubing is flushed internally with some of helium.  The air aspired outside is submitted to a small mass spectrometer attuned to the very signal of helium and of high (often adjustable) sensitivity.  The technique equally benefits from the the low terrestrial concentration of helium (except a few known natural sources and artificial processes) and hence low background.

(image credit to TQC Automation & Test Solutions)
A little cart can house both the gas cylinder with the tracer gas and the necessary equipment:

(image credit to Leybold, which provides some additional information here)

Hi there Buttonwood, this is really helpful: very much appreciated.


text:
Determination of Calcium from a mixture of calcium and magnesium using Back Titration
I've some confusions in the experiment. Step 7 gives us the concentration of both calcium and magnesium (Ca+Mg combined), right? How do we find the concentration of magnesium only from back titration described in the next steps?
Can someone please explain.

What is the meaning of the end of step $3$ ? Namely, I repeat : dissolving $0.2$ g of ErioT in $15$ mL of triethanolamine and $5$ mL of absolute ethanol or in $20$ mL of the latter ? What is the meaning of this "or $20$ mL of the latter" ? Should I use $5$ or $20$ mL ethanol ?

@Maurice This step just describes the preparation of indicator solution which can be prepared either by dissolving 0.2g of EBT in a mixture (15mL triethanolamine + 5mL absolute ethanol) or by dissolving 0.2g EBT in 20mL of Ethanol only.

Welcome to CH SE site!
Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of essential parts.
// Optional: Formatting guides for texts and formulas/equations/expressions.

The general reactions are described very well here: https://chem.libretexts.org/Ancillary_Materials/Laboratory_Experiments/Wet_Lab_Experiments/General_Chemistry_Labs/General_Chemistry_Labs/Complexometric_Calcium_Determination_(Experiment)
Ca++ and Mg++ both form complexes with EDTA, Ca++EDTA and Mg++EDTA. The Ca++ complex is the stronger so Mg++ will not destroy the Ca++ complex. Both ions form complexes with the Eriochrome. Here the Mg++ complex is stronger and the Ca++ complex relatively weak such that the end point with Ca++ is vague and the end point with Mg++ is sharper. Both complexes with the indicator are red. When the indicator complexes are destroyed by EDTA excess the indicator is blue. Titration of the sample to the blue endpoint gives an approximate Ca++ amount, record this value. Add the excess EDTA and record the total volume exactly. The solution should be blue. All the calcium is complexed to the EDTA. Titrate the solution with the Mg++ standard. This converts the excess EDTA into the Mg++EDTA complex. As the end point is neared be careful. When the blue color fades all the EDTA is the Ca++ and Mg++ complexes. Record the volume. The total EDTA solution is their sum. The difference is the calcium.
The procedure should be more explicit in recording data.


text:
Is this a allyl bromide? If not what is it?
This molecule should be a allyl bromide:

but what about this one:

What is it?

2-Bromopropene?

As you correctly mentioned,
is allyl bromide.
It's IUPAC name is 3-BromoProp-1-ene.

The IUPAC name of this compound is 2-BromoProp-1-ene.
It is a homologue of Vinyl Bromide.


text:
Why can't the diene in a diels-alder also be a dienophile
why can't the diene react with itself like this

it does it just has a really low yield

... because it's a bad dienophile

It can and does given the right diene. The classic case is cyclopentadiene:

At room temperature, this cyclic diene dimerizes over the course of hours to give dicyclopentadiene via a Diels–Alder reaction. This dimer can be restored by heating to give the monomer.

Note the thermoreversibility of the reaction in this case, which opens up a relatively easy route to aromatic cyclopentadienide-ion rings.

Freshly distilled cyclopentadiene cracked from the dimer is neutral, and its dimer/product of the Diels-Alder reaction is uncharged, too.  Neither cyclopentadiene, nor dicyclopentadiene feature a planar cyclic array of $(2n + 2)$ $\pi$ electrons to qualify as aromatic. The large gain of stability if cyclopentadiene is deprotonated because the resulting anion/base then equally is an aromatic aromatic structure is separate, independent of cyclopentadiene's ability to dimerize quickly at room temperature.


text:
Why is 1-ethylcyclohexene more stable than ethylidene cyclohexane?
Both compounds yield the same product, ethylcyclohexane, when hydrogenated. However,
1-ethylcyclohexene has lesser heat of hydrogenation than ethylidenecyclohexane, hence more stable. I want to know why that is the case although both are trisubstituted. (I can only think of comparing them as cyclohexane ring vs cyclohexene.)
Please note: I am trying to learn Chemistry to teach my child. I left chemistry a long time ago and am now refreshing the topics.

In general, exocyclic double bonds are less stable than endocyclic double bonds for six-membered rings. Plus, since hyperconjugation stabilizes double bonds, one must note that the number of $\alpha$ H are more for 1-ethylcyclohexene and hence, it is comparatively more stable.


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How do I number the atoms in strychnine?
I know the basics of naming organic molecules, but this one is a bit more complex than I'm capable of. How can I number the atoms in strychnine?
Systematic IUPAC name: ($4$bR,$4$b$^1$S,$7$aS,$8$aR,$8$a$^1$R,$12$aS)-$4$b$^1$,$5$,$6$,$7$a,$8,8$a,$8$a$^1$,$11$,$12$a,$13$-Decahydro-$14$H-$12$-oxa-$7$,$14$a-diaza-$7$,$9$-methanocyclohepta[cd]cyclopenta[g]fluoranthen-$14$-one

related https://chemistry.stackexchange.com/questions/169432/iupac-naming-of-lysergic-acid-diethylamide-lsd-heterocyclic-core-and-its-locan

Nomenclature for organic compounds (Blue Book, see nomenclature references on the resources site) equally agreed on how structure formulae of polycylic compounds should be presented, and oriented.  For a couple of parent structures considered important enough, the numbering is defined, too.  While illustrating rule P-101.6.7, the Blue Book displays strychnidine as the plausibly most similar parent structure to strychnine itself (note C10 does not carry a carbonyl group):

(image credit to G. P. Moss' freely available, searchable compilation of the Blue Book, landing page, and link to the corresponding pdf of chapter P-10).
This depiction differs from the pattern you found / obtained by a computer program, for which bridge atoms (atoms belonging to multiple rings) are named with small letters (e.g., 12a) because the algorithm applied assigned a different stem systems (a motif of higher seniority), than by departing from strychnidine as parent structure.

For the numbering, one has to identify the largest ring to be oriented that vertically sketched bonds are as far as possible away of each other.  Subsequently, the orientation has to sequentially consider a) as many as possible rings are oriented along a horizontal line, b) as many as possible rings are in the top right quadrant, c) as few as possible rings are below a horizontal line, and d) as many as possible rings are above a horizontal line in an intermediate coordinate system.  The relevant set of rules in the Blue Book are e.g., P-25.3.3.1.1 and following, and P-25.3.2.3 and following.

Once this is set, one seeks for the position which is topmost, in the cycle most to the right hand side, the atom most to the left hand side which is member of only one cycle; from there, count clockwise within this stem system.  Rasmussen compiled a couple of examples, for instance

Among the molecule editors, the template library of the freely available ACD ChemSketch actually contains a set of alkaloid scaffolds already numbered, ready for copy-paste and further edit:

Rasmussen, S. C. The Nomenclature of Fused-Ring Arenes and Heterocycles: A Guide to an Increasingly Important Dialect of Organic Chemistry. ChemTexts 2016, 2, 16. https://doi.org/10.1007/s40828-016-0035-3 (open access).

thanks, i'm trying to apply this to the structure of strychnine but i need a little more. could you explain why C1 in morphinan is where it is and point me to C1 in strychnine?

@imrobert With reference to the Blue Book and a editor which has strychnidine as a structure closer to strychnine itself, I attempted to provide an entry into the nomenclature about fused organic rings.  It is complex.


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Phase Diagram Vertical Lines
We had to create a phase diagram of H2O and NaCl. My professor mentioned that the vertical line I’ve encircled is at the weight % of 61.9. I was just wondering if this is proven through experiment or is there a way to solve this?

Define ‘solve’? Phase diagrams were originally determined experimentally. Now, while their are various computational techniques most of them rely on experimentally determined points to make the computations ‘right’.

This was also what I was thinking--I was just unsure if it's possible to solve it without experimental data. Thanks so much for your answer! :)

The vertical line marked by you is a phase boundary.
At temperatures below $\pu{0.1 ^\circ{}C}$, anything to the right of this boundary is a mixture of variable composition consisting of solid sodium chloride, $\ce{NaCl}$, and sodium chloride dihydrate (hydrohalite). The two compounds coexist. The crystal structure of the later regularly contains molecules of water and is considerably less dense ($\pu{1.656 g/cm^3}$), than halite ($\pu{2.17 g/cm^3}$ [ref.]).

(structure model of hydrohalite, $\ce{NaCl * 2 H2O}$, as entry COD 4513223, based on Bode et al. Oxygen represented by red, hydrogen: white, sodium: purple, and chloride: green spheres.)
Left to this line, there is a liquid state of varying composition: solid $\ce{NaCl * 2 H2O}$ in a liquid aqueous solution of sodium chloride.
Bode, A. A. C.; Pulles, P. G. M.; Lutz, M.; Poulisse, W. J. M.; Jiang, S.; Meijer, J. A. M.; Van Enckevort, W. J. P.; Vlieg, E. Sodium Chloride Dihydrate Crystals: Morphology, Nucleation, Growth, and Inhibition. Cryst. Growth  Des. 2015, 15, 3166–3174. https://doi.org/10.1021/acs.cgd.5b00061.


text:
What is happenning with that single hydrogen in Phosphorous acid?
As an oxyacid, Phosphorous acid intuitively should have the structure

since the hydrogens should be on the oxygens.
However, the major form of this acid is actually

with the hydrogen attached to the phosphorous.
My question is, what fundamentally causes the second form to be more stable than the first?

Did you see and read the English article in Wikipedia?

@Buttonwood yes, I obviously did, and there was no information that answered my question as to why the H is bonded to P and not O. Or else I wouldn't be asking this question here lol

Related https://chemistry.stackexchange.com/questions/127414/why-does-noh3-not-exist https://chemistry.stackexchange.com/questions/85253/phosphorous-acid-vs-phosphonic-acid

Is it bond energies? Other tautomeric equilibria are driven significantly by bond enthalpies


text:
Why aren’t electrolyzers made like this(two inter-locking coils)?
The maximum current(based on how long you want your electrolyzer to last) is based on the electrode size according to my minimal research. The larger the electrode the more current is allowed. So why not make the electrodes as big as possible using the design in the illustration below (that I made in the camera app on my phone). Why don’t people do this?
Something I note is that comparatively to a dry cell since there are multiple electrodes you can share the voltage between many of them instead of sharing current. This concept can’t do that, but it doesn’t need to because it doesn’t need to share current either. Do what is the problem with this design? Why don’t people use it?

The risk of short circuit is high and difficult to control with such a spiral structure.

It's not unheard of: https://www.sciencedirect.com/science/article/abs/pii/S1359511315001944 It might also be related to ease of manufacturing.

The geometrical arrangement for galvanic cells and electrolytic cells is usually very different. The spiral arrangement is routinely used for the former, but not the latter.
For the electrolytic cells, there are preferred interleaved comb structures, with alternating plates A B A B.
This arrangement is easier to assemble and maintain. It is also easier to control the inter-electrode distance, preventing short circuits, and it is less blocking electrolyte motion.
In the case when products being formed on the cathode and anode are not to be mixed up, the electrodes are not even interleaved, but are kept in separate compartments.

I also suppose it has something to do with mechanical strength, where something like a spiral wound membrane element is fine. Since membranes are typically flexible, but electrodes are not as flexible and any mechanical failure will destroy the cell.

I found another reason it isn’t used, for anyone who may view this question in the future. Energy efficiency goes way down. Electrolysis only requires 2 volts to work. Anything more than that is really just an inefficiency. In a dry cell, it powers every nth plate and not power the ones in between, which causes the voltage to be shared between the plates. They do this because as the current goes up, so does the voltage based on resistance which means you have unnecessary voltage. However, in the spiral wound you can’t share the voltage which means you will be drawing unnecessary voltage and therefore unnecessary power for the electrolysis. The trade off is that you can go to much higher currents and have much more gas output.
Also a quick other reason is that is hard to compress the spiral wound uniformly to make sure there are no shorts, and it requires a lot of calculation to find the correct size for the container of the spiral wound for compression.


text:
Can phosphorous penta hydroxide [P(OH)5] exist?
I was looking at the oxyacids of Phosphorous and it is given that (in Wikipedia article of "Phosphorous acid"), some of these acids (Eg: $H_3PO_2$ and $H_3PO_3$) exist in a dynamic equilibrium between their keto form and enol form.
This is a pure speculation from my side but I want you to think about a molecule with $P(OH)_5$ molecule with each hydroxide in a trigonal bi-pyramidal geometry.
I feel like there could be two reasons to think about it's possibility:

Since 'P' have empty d-orbitals and there are lone pairs on the oxygen, it could form a type of back bond, thus making the P-OH bond a bit stronger.
Since there are OH groups, there could be a intramolecular H-bonding that could further stabilise the molecule.

Maybe this can form in this can form in such a reaction:
$$
\ce{H_3PO_4 + H_2O <=> H_5PO_5}
$$
where the water molecule could on the P=O bond.
So, is it even possible for such a molecule to exist?

No, it doesn't exist: https://chemistry.stackexchange.com/q/135208/17368

So it would be called Holophosporic acid. Is there a reason for why it couldn't exist?

Does this answer your question? Why is H5PO5 not called as phosphoric acid?

No. That question describes the nomenclature for this formula. I'm questioning whether this compound exists or not. And if it couldn't, why not?

The specific $\ce{P(OH)5}$ does not appear to be known, but high-pressure phosphates with pentacoordinate phosphorus have been observed.
Bykov et al.1 reported the first example, involving titanium(III) phosphate ($\ce{TiPO4}$), in 2016. Under ambient conditions this compound shows "classical" coordination in which the titanium shows polymerized octahedral coordination (blue) and the phosphorus tetrahedral (gold/orange), shown in the first figure below from the reference. Under 48 GPa pressure, both titanium and phosphorus increase their coordination numbers, giving the phosphorus coordination a polymerized trigonal-bipyramidal shape shown in the second figure (also from the reference above). The titanium has an irregular eightfold coordination. The 48 GPa pressure is too great to be realized within Earth but may be present in super-Earth exoplanets.

Reference

Bykov, Maxim; Bykova, Elena;  Hanfland, Michael;  Liermann, Hanns-Peter; Kremer, R.; Glaum, Robert; Dubrovinsky, Leonid; Van Smaalen, Sander. (2016). "High-Pressure Phase Transformations in TiPO4 : A Route to Pentacoordinated Phosphorus." Angewandte Chemie (International ed. in English). 55. 10.1002/anie.201608530.


text:
Find the pH of a solution obtained by mixing 100 ml 0.1m Na3PO4 and 100 ml 0.1 M NaH2PO4. Given that H3PO4: [Κα₁ = 10^-4, Κα₂ = 10^-7, Каз = 10^-11]
So, my query is that since NaH2PO4 is a weak acid, and Na3PO4 is a basic salt, wouldn't they react? Another theory was that since NaH2PO4 is a weak acid and Na3PO4 is a salt of strong acid and NaH2PO4, so they would form a buffer. So, I thought Henderson-Hasselbalch equation would work.
therefore, pH = pKa2 + log 1
pH = 7?
Neither does this answer make sense nor does it match with the textbook answer of 9(assuming the textbook is correct). So it would be much appreciated, if someone could clear this up for me, thank you.

The solution mix is equivalent to 200 mL 0.1 M Na2HPO4. Both cases contain the same concentrations of ions, reaching the same equilibrium from different starting points.

A good question title should be simple, unique, resembling a magazine article headline or a book title. It should be in plain text, summarizing the topic, optionally formulated as a question. It should  avoid being a click bait, saying too much or too little.

Your solution is simply a $0.1$ M solution of $\ce{Na2HPO4}$. And its pH is the average value of $p\pu{\mathrm{K}_{a2}}$ and $p\pu{\mathrm{K}_{a3}}$, which is $\frac{7+11}{2} = 9$.

I understand your solution but could you explain the logic behind taking an average of pKa2 and pKa3? That's kind of where my deficiency is.

See the CH SE Post or search using the string site:stackexchange.com pH amphoteric salt solution or site:libretexts.org pH amphoteric salt solution


text:
Change in pH/acidity of vinegar after adding aqueous NaCl
There's a question in my (grade 10) textbook asking to determine what will happen to the pH/acidity of vinegar if a solution of sodium chloride in distilled water is added to it drop by drop. The answer in the back says the pH will remain unchanged because $\ce{NaCl}$ will dissociate into $\ce{Na^+}$ and $\ce{Cl^-}$ ions, which are neither acidic nor basic.
This explanation doesn't really make much sense to me; wouldn't the water present in the aqueous solution of $\ce{NaCl}$ dilute the ethanoic acid in vinegar to increase the pH/reduce the acidity? Or, at least, wouldn't all the ions formed after dissociation interact with each other to change the pH in some other way?

The question may have different answers, depending on how deep through a rabbit hole you want or are expected to go.  I assume for the 10th grade, you can say the pH will not change. // The advanced answer would say it depends on which activity coefficient of respective ions decreases with the ionic strength faster.

The pH increases, as a more diluted solution has a lower oxonium ion concentration, as you pointed out. However, this dependence is not linear in any way

P.S.: and yes, diluting of vinegar by water or NaCl solution does increase pH, as there is lower acetic acid concentration.

For future reference: for the body of questions, answers, and comments, chemistry.se offers to use mhchem as a comfortable method to add chemical equations and report numerical values (\pu{}) including a non-breakable space.

In summary, it depends on the initial concentration of the solution.

Your thoughts are absolutely correct, when being absolutely precise. If one diluted vinegar (5% ethanoic acid) with an equal quantity of plain water or salt water, there certainly would be a noticeable pH change.
However, if one has a moderate size sample, perhaps 100 mL, and adds salt water drop-by-drop, it's doubtful even a precision pH meter or very sensitive pH indicator would resolve the small change. After all, a drop is defined as 0.05 mL, about 1/2,000 of the original sample.
The answer is open ended. What if the original sample were only a single drop, 50 µL?

Thanks for the clarification! Is it safe to assume that the ions won't react with each other to form any new compounds that may change the pH?

No it is not safe to assume anything. Each situation should be analysed.


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If a compound has a strong and weak minus I group, will the stronger group always have a δ- charge?
In a compound like acrolein (prop-2-enal or $\ce{CH2=CH-CHO}$, will the oxygen in the  $\ce{-CHO}$ group always have a $\delta$- charge?

Acrolein does have three resonance structures as shown below, in which on one of them, oxygen has a $\delta$+ charge.

How stable is the third resonating structure? Is it viable to even consider it a resonating structure, since $\ce{-CHO}$, a decently strong minus I group has a $\delta$+ charge, and $\ce{-CH=CH2}$, a very weak minus I group has a $\delta$- charge?
This doubt stemmed from the following question, about the resonant structures of acrolein (apologies for the low quality):

It's a multiple correct MCQ, and the answer given is $ABC$, but as per the resonating structures of acrolein, it should be $AC$, as B and D, both should be valid resonating structures of acrolein.

Hmm, as per my knowledge, the third resonating is of much higher energy and thus doesn't contribute much to the overall stability of the molecule. The reason being empty orbital, $\ce{O^+}$, in $\ce{sp^3}$ hybridised oxygen, which is very unstable.

That's a good logic, but is it really fair to say that the third structure entirely does not exist simply because it is unstable, it could still have a low resonance contribution.

The problem with the third structure is not the positive formal charge on oxygen as such, but leaving the oxygen short of its octet. Electronegative elements generally have their octets filled in stable molecules or ions, both organic and inorganic.
On the other hand, once the octets are filled electronegative atoms often back-donate their "nonbonding" electron pairs and may develop a positive overall charge through this mesomeric effect. Examples with oxygen include carbon monoxide, pyrilium rings, and the cationic intermediate formed when a hydroxy or alkoxy group activates electrophilic aromatic substitution.
See this question for a rather unexpected example involving chlorine as the -I function.


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Why does current remain constant in a conductor having non-uniform cross section for constant voltage supply if I=neAvd?
I am confused in this question:

A copper wire of non-uniform area of cross-section is connected to a d.c. battery. The physical quantity which remains constant along the wire is …

At first I thought this way:
Since $I=neAv_\mathrm d$
Where $I$ is current, $n$ is number density of free electrons, $A$ is area of cross section and $v_\mathrm d$ is average drift velocity of electrons in the conductor.
So from this relation one can easily see that current is directly proportional to area of cross section. So current should increase with increase in area of cross section. But it wasn't the explanation.
Its answer is given as current, that is current remains constant, and the explanation is provided as:
"Current is defined as charge flown per unit time, it should remain same along any cross section as there is no accumulation of charge in conductor. Hence current will remain constant."
I don't understand when to employ $I=neAv_\mathrm d$ and when to apply definition of current.

At the constant current, A.vd is constant. It has direct analogy with water flowing through a pipe with variable diameter. A narrower tube means faster flow.

Yep, just what the comment above said. It's the principle of continuity in fluids, analogously applied for current.

Also, isn't this question more suitable for physics stack exchange than chemistry stack exchange?

Even if chemistry is in this or that way involved in majority of domains from astronomy to psychology, for most of these domains there is better to ask somebody else than chemists. This is such a case, even if trivial enough so chemists can answer it as well.

You have forgotten to consider that $v_\text{d} \propto \frac 1A$.
A place with a narrower wire means there is stronger intensity of electric field, causing faster electron drift motion, compensating for smaller cross-section area.
Initially, when such a wire is just being attached to an external voltage, the intensity of the field is the same along the wire. But, as it causes smaller current at the wire bottleneck, there is being formed very tiny excess and deficit of electrons at the respective narrow place ends.
This deviation causes additional electrostatic field, making electrons moving faster, leading to the same current as for the rest of the wire.
You probably know the Ohm law
$$U=RI,$$
what is the relation of the current and the voltage (potential difference). But this is its integral, macroscopic form.
The Ohm law has also its equivalent, differential form, valid locally at any wire point.
$$\vec E [\pu{Vm-1}]=\sigma [\pu{\Omega m}] \vec j[\pu{Am-2}]$$
where sigma as material resistivity (generally a tensor, but effectively a scalar for isotropic materials) is the proportionality constant between vectors of current density $\vec j$ and electrostatic field intensity $\vec E$.

Similarly, for the water pipe analogy mentioned in the comment, the narrow pipe passage causes a local increase of hydrostatic pressure gradient, that causes pushing the same water flow even through smaller pipe cross-section.

The current is the number of electrons crossing a given section of the conducting wire per second. It does not depend on the importance of this section or of the position of this section along the wire. This wire can be thin or wide. The number of electrons crossing any part of the circuit or any section of the wire per second is always the same. One mole electrons crossing any section of the wire per second is $96500$ Ampere, whatever the diameter of the wire.


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Proper batch adsorption mechanism for heavy metals
I'm a college student with scope on environmental science, specifically the adsorption of heavy metals in aquatic media.
Currently, I suffered from serious issue of batch adsorption mechanism, which is hindering me from obtain a better result of iron adsorption on my adsorbent. The problem is (1) separation of iron ions to form iron complex with water in low concentration (20 ppm) and in less acid medium (pH > 3) [1][2] and (2) pHPZC value of my adsorbent is at 3.46.
I conducted the adsorption mechanism by diluting stock solution of iron (100ppm or 1000 ppm) to a lower concentration test solution, 20ppm (20mL in volume & pH at 3.46). The addition of adsorbent is at 25mg with a series of adsorption time (10-60min), including magnetic stirring (250rpm). Lastly, I filtered the solution with Whatman filter paper no. 42 (2.5 micron in pore-size), then I sent it to AAS (Atomic Absorption Spectroscopy/Flame method) for trace iron determination. But I still can't manage to obtain a decent result data.
Is anyone notice a mistake above, which lead to indecent data?
Does anyone kindly provide me with a proper mechanism?
Thank you for your kind response.


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UV-Vis spectrum of alkane
My reference books stated:

Methane's maximum absorption wavelength is 125 nm, ethane's is 135 nm. The longer the carbon chain, the larger the absorption wavelength, it is 165 nm for n-octane.

Well, I have tried to explain this based on organic chemistry, specifically hyperconjugation. However, it is still an unexplainable problem for me.

There is no simple explanation for this behavior.  Increasing $\lambda$ is not just predictable with increasing carbon number. For example, structure of the molecule plays a role too. Hope this would help you.

Yes, the paper helps. Since the question is mainly for hypothesis about interaction for linear alkane chain. I'll take other factors into account when it comes to more carbon atoms


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Which of these molecules have stereoisomers?
Problem:
Which of these molecules have stereoisomers?
Solution:
(The molecules with a red X beside them are the ones with stereoisomers.)

Question:
What is the method used to identify which molecules have stereoisomers? I can see the commonality between the three molecules with stereoisomers being that the two atoms that can exchange side are right in the middle if that makes any sense. But is that the correct way to think?

ii and iv have optical isomerism and iii has geometrical isomerism, both come under stereoisomerism.

Stereoisomerism is a type of  isomerism in which molecules have same molecular formula and structural formula (sequence of bonded atoms), but differ in orientation in space.
Stereoisomerism is of two types:

Conformational : The type of stereoisomerism in which isomers can be converted into one another by free rotation about $\sigma$ bond.

Configurational : It is of two types-

Geometrical Isomerism
Optical Isomerism

(Note: When in questions, stereoisomerism is mentioned, it's self understood that configurational stereoisomerism has been asked, unless mentioned specifically.)

Geometrical Isomerism (GI)
Geometrical isomers are compounds having same molecular and structural formula, but different geometry across restricted rotation system.
Conditions for GI:

restricted rotation system (like $\pi$ bonds)
Distance between terminal atoms or groups must be different in both isomers.

Optical Isomerism (OI)
Optical isomers are compounds having same molecular and structural formula, but different optical behavior.
Conditions for OI:
[First check for Asymmetric/Chiral Centre (CC)]:

If CC Present -- Compound would surely show OI.
If CC Absent then:
Check for Plane of symmetry(POS), Centre of symmetry(COS) & Alternating axis of Symmetry(AAOS)/n-fold axis of improper rotation symmetry.

If all three are absent then compound would show OI otherwise not.
(For details on symmetry elements, click here.

Now Answering your question:
Compound(i) has a $\pi$ bond, but both the groups on oxygen are same (i.e. Lone pairs) hence no GI. It has POS as shown, so no OI.
Compound(ii)has two chiral centers as shown, so it shows OI.
Compound(iii) has a $\pi$ bond, and the distance between terminal groups is different as shown, hence it shows GI.
Compound(iv) has a chiral center as shown so it shows OI.

Hope helpful.

Thank you so much. And thank you for making the added effort of drawing the molecules. Very helpful :)


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Analysis of a sample with several Na compounds
I have a sample that is composed of NaOH, NaNO3 and NaCl in unknown proportions and I would like to determine the approximate proportions. I can determine the melting point of the combination.
I was thinking that I could determine the NaOH percentage by using a pH meter and making an aqueous solution - finding the amount of pure NaOH that it would take to get the same pH in the same volume  of water. My assumption is that the other two components won't affect pH.
That still leaves the other two. I could fiddle with the percentages until the melting point is matched, but is there an easier way that is not too expensive and requires only limited equipment and knowledge?

A simple titration with standard acid will measure NaOH or the pH. A Mohr titration [or potentiometric since you have a pH meter] with silver nitrate will determine chloride. The nitrate makes up the difference or you can buy a nitrate test kit.  More sophisticated methods would be ion chromatography or ir or Raman analysis to  determine nitrate. I like the two titrations and a Hach test kit. If you have a differential scanning calorimeter the mp test might work.

@jimchmst Thanks. The silver nitrate approach seems simple enough. That looks a lot like an answer...


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Aqueous electrons in electrochemical cells
In the context of a zinc-copper electrochemical cell, zinc atoms at the anode are oxidized, releasing electrons and forming zinc ions that dissolve into the solution ($\ce{Zn(s) -> Zn^{2+}(aq) + 2e-}$). This process leaves electrons behind on the electrode. Given the high reactivity of electrons, it's intriguing that they don't also "dissolve" into the aqueous solution like the zinc ions. Considering the potential for radical formation if electrons were to interact with the solution, how is it that electrons are prevented from entering the solution? Is this just a case of textbooks leaving out a loose thread in the story of electrochemical cells?
As a side-note, I can imagine a metal plate under water onto which ultraviolet light is shone. I would guess that surface electrons could absorb the light and become free in the water. So I don't see why would a similar effect not sometimes occur from the bombardment of the electrode surfaces by molecules (rather than light) in the electrochemical solutions?

Zinc doesn't simply release electrons. This process is endothermic. The first and second ionization energy is 9.4 eV and 18.0 eV, respectively.

If zinc does not release electrons in water, alkaline metals can, but these electrons reacts immediately with water according to :  $\ce{2 e- + 2 H2O -> 2 OH- + H2}$

Forming aqueous electrons is energetically highly disadvantageous and occurs at  a highly negative potentials, followed in typical time scope of  hundreds of microseconds by reducing water to hydrogen. Interaction and reduction of H+(aq ) occurs at higher potential, so metals get dissolved in acids.// To the side note: Thermal energy of molecules is typically 100 times smaller than energy of photons of UV/VIS light.

Please avoid cross-posting. If you have trouble getting an answer then you might try another site.

Is there a reason for de-accepting? I see no alternate answer. I will gladly answer questions.

Because the zinc in your example is the anode, the electrons are actually not released into the elecrtolyte. They are pulled through the external circuit to the cathode, in your case copper, and it is the cathode that releases the electrons into the electrolyte.
In the case of a water-based electrolyte the electrons do not get very far before they react with the water to form hydrogen gas, or perhaps with metal ions in the electrolyte (including, possibly, zinc ions coming from the anode) to reduce and deposit the metal. Liquid ammonia solutions, on the other hand, are less reactive towards electrons. They are often used as media for electrolytically generating blue-colored solvated electrons in solution.

To the first part, disconnect the two halves of the cell so we have just zinc (or copper if you prefer) in a solution. In equilibrium there will be atoms leaving the electrode and returning to it at the same rate. Do they leave electrons behind when they do so?

@Kris hard to answer. Atoms of metallic elements dissolced into water are likely to react, releasing hydrogen, possibly even before the electrons would have become solvated.


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Electrolytic Substrate modification to enhance corrosion resistant of reactive metal
So, I was working on a project lately. The project I was working on involves the utilization of inexpensive, non-toxic, and inert electrodes. Unfortunately, the only substrates I have access to are copper, aluminium, and iron plates. Now, my assumption was to coat these substrates with inert, inexpensive metal oxide articles like lead dioxide and manganese dioxide. Clearly though, the lead dioxide won't be an option as it can be potent, leaving me with one option.
I tried heating the freshly HCl etched iron plate while applying a thin layer of manganese nitrate. However, as it formed, the plate started rusting and then gradually darkened. I was confused for a moment, so I tried cooling the plated substrate and applied 30% hydrogen peroxide to make sure we have got manganese dioxide, not magnetite. Upon testing, the misty gas generation confirmed the test.
The next step was a corrosion test. In the experiment, I placed the control and the newly plated electrode in 20% brine with a 1 amp current draw for 180 sec. After examination, they both formed hydroxides with different oxidation states II and III, respectively. Consequently, this disproved my hypothesis.
Question:
What went wrong in the process?
Is there any pre-treatment that I could perform?
Are there any chances or possibilities for any of the remaining substrates?
Are there any inexpensive accessible articles that can be utilized instead?
NB: I planned to use the plated the substrate in electrolysis without it being involved in the actual reaction. Therefore, I did not want to just prevent it from corrosion caused by oxygen but prevent it from further corrosion caused by electrolysis

You should absolutely separate the treatments of the electrodes. Anodes and cathodes do not behave in a similar way in electrolysis. Probably only anodes must be protected against corrosion. But I don't know if it is possible to prevent anodic oxidation of usual metals like copper, aluminium or iron. The best way (and maybe the only way) is using a noble metal like platinum as an anode.

An anode reaction is essential to the cell therefore the anode must either be involved in the reaction or be inert and a conductor to supply electrons to the anode reaction. Painting the electrodes is a good way to stop electrolytic reactions. That is corrosion prevention. What is your reaction? An anode must be less reactive. Two choices: use a less reactive material or chose an anode reaction that is more easily oxidized.

@jimchmst my intention was creating an anode that is both inert and allows circulation of electron .Certainly, I never wanted an anode that involves on the chemical reaction or completely insulated electrode that doesn't circulate electron.


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Electronic Configuration of Atomic Numbers 20-28 and 29 (Calcium - Nickel) and Copper

I have always been confused as to why elements of atomic numbers 20-28, i.e., Calcium to Nickel have 2 valence electrons in their outer shell. Starting from Potassium, why is one electron present in the valence shell, making the configuration 2,8,8,1 and why doesn't it become 2,8,9 ? Same for Calcium, why does another electron join the Valence shell making the configuration 2,8,8,2 ? For scandium, 2,8,9,2 ? And in Copper, 2,8,18,1 ? why does one electron go from being in the valence shell to going in the 3rd shell?
What makes Chromium's and Copper's electronic configuration so weird?
Why does this happen?
So, in general how is this weird behavior defined?
Thanks.

Does this answer your question? How come electrons in the d-subshell are considered valence electrons in transition metals?

Need a simpler explanation. I am a student, hence I couldn't understand what the conversations mean.

In accordance to Aufbau's principle, the subshells with the lower energy will be first filled with electrons.
Assuming that you are a student and are not aware of subshells and orbitals yet( By looking at the way you wrote down the electronic configuration), I believe that you will learn about it in a higher grade.
I will still try to explain in simple terms.
The atom has shells. These shells are divided into subshells which are further divided into orbitals. The orbitals can be considered to be boxes in which only two electrons can fit. The subshell is a collection of these boxes and the shell is a collection of subshells. The first shell has only one subshell and two electrons can fit in it.
The second shell has two subshells of different sizes and a total of 8 electrons can fit in it.
The third shell has three subshells and 18 electrons (maximum) can fit in it and so on.
The electrons are filled in the lower energy subshells first. The ones closer to nucleus have less energy and electrons are filled in the first shell first then the second and then the third. However while filling the third (which has 3 subshells) we fill the first and the second just fine and in sequence (this sums up to 8 electrons in that shell).
Now the third subshell in the third shell has a problem. It can accomodate 10 electrons in total but the problem is that it has much higher energy than the next shell's first subshell. So the electron rather prefers to enter the higher shell as in potassium leaving the third subshell unfilled and with only 8 electrons.Two electrons can be filled in the first subshell of the fourth shell. The second subshell of the fourth shell now has higher energy than the third subshell of the third shell.So the electron now prefers to enter the third shell once again as in scandium. The same pattern continues. '
ps CHROMIUM AND COPPER POSSESS EXCEPTIONAL ELETCRONIC CONFIGURATIONS DONT GET ANGRY WHEN YOU LOOK AT THEM :> we can talk in comments if you want to know why though

Thanks a lot, my confusions have been solved to a great extent, and I completely understood what you wrote. Why do Chromium and Copper posses exceptional electronic confs? Also, why is this post getting so many downvotes?

I answered your question and i suggest you edit your question and add this question as well. As for the downvotes, the question (as well as the answer) may sound somewhat amateur.

Chromium has atomic number 24 and an electronic configuration of 2 8 13 1 but according to the rule i mentioned it must be 2 8 12 2. The electron fills the 3rd shell's 3rd subshell to give the electronic configuration 2 8 12 2 however this electronic configuration is not so stable so, an electron jumps from the 4th shell's first subshell to the 3rd shells 3rd subshell. Because half filled subshells are more stable than those which are one less than half filled.
This is why it happens in simple terms:
The 3rd shell's 3rd subshell is a large one. It has 5 orbitals or boxes (remember, each can accomodate maximum of two electrons only ).You fill all the boxes first with one electron and then only after that you pair them (this is a complicated topic again). For example: After Calcium 2 8 8 2 the 3rd orbital's 3rd subshell starts filling up. It has 5 boxes. The first box gets filled with an electron. You get scandium. The second electron is filled in the second box and you get titanium (remember pairing will not be done first).The third electron goes in the third box giving vanadium. And finally its time for the 4th electron to go in the 4th box for chromium leading to a problem. As now the 3rd subshell of the 3rd shell is one less than half filled (all boxes filled with one electron) the electron jumps from the 4th shells first subshell ( The fourth shell's first subshell has only one box and right now it has 2 electrons) to the third shell's 3rd subshell as its better to keep both the subshells half filled than to keep one hungry.The real reason lies in a concept called exchange energy and I suggest you read about it yourself.
Similarly in copper (Z=29) the expected atomic configuration would be 2 8 17 2 but again one electron jumps from the 4th shell's first subshell to the 3rd shell's 3rd subshell giving 2 8 18 1. In this case the 3rd shell's 3rd subshell gets fully filled with electrons and is the most happy (even more than the half filled one).


text:
Can you tell if a HOMO to LUMO transition is allowed just by picturing the MOs?
Someone told me that they said the HOMO to LUMO transition is allowed in benzene and naphthalene just by looking at the MOs and without using the irreps for each orbital and the character tables?
How can you do this? I thought you need to use the character and product tables to determine whether a transition is allowed (non-zero transition dipole matrix) Is there a method in doing so involving the symmetry of the lobes of each orbital?

You will always have to look at the symmetry each of the MO's with the point group to determine its symmetry species. You also need to know the sym. species for the x,y,z transition moment vector, so five sym species in total.  Of course if one  is the ground state it is always totally symmetric so you know this just by looking and as its totally symmetric can be ignored.  (btw the lowest energy transition in benzene is forbidden so whoever told you this has not got it quite sorted yet)

@porphyrin so does the HOMO-LUMO transition in benzene give a zero transition dipole matrix element? If so, why?

In principle why not? Remember ultimately all symmetry arguments are are a more general version of "an even function times an odd function is an odd function, hence the integral over all space of the product is zero". Consider a simple case Hydrogen 1s to Hydrogen 2s - it's easy (at least to me) to see that the transition dipole matrix element for x, y, or z is zero by the odd/even argument - and if you can visualise more complicated cases I don't see why you couldn't see if in certain cases an integral is zero or not simply by "looking"

yes it is forbidden, which is why with complicated MO's we need to use symmetry, and yes the transition moment is zero. The weak uv spectrum observed is due to coupling with a nearby state of the correct symmetry to effectively 'borrow' some intensity. Called Herzberg-Teller coupling or intensity stealing. See J. Steinfeld 'Molecules & Radiation' for a clear discussion, p270-276.


text:
Is it true that inorganic compounds cannot be synthesized by living things?

Inorganic means not organic. In biology, they are compounds such as water, minerals, acids, bases and salts that are found in the structure of living things but are not living things or are not produced by a living thing.

https://tr.wikipedia.org/wiki/%C4%B0norganik_bile%C5%9Fik
According to the information on wikipedia, inorganic compounds cannot be synthesized by living things. So why is water an inorganic compound? Water can be produced by cells as a result of respiration.
Edit: It turned out that the information in Turkish wikipedia was wrong. This information is given correctly in English wikipedia.

Edit 2: The website below contains information similar to that in our textbooks. I quote:

The chemicals that living things produced had been called organic since they came from organisms. Those that were obtainable from non-living things had been called inorganic, meaning “not organic”. This was the fundamental boundary that defined organic from inorganic compounds.

Inorganic compounds were thought of as compounds that were not derived from organisms. They may be derived, for instance, from geological systems, e.g. sediments and ores.

This belief had long been held for many centuries until Friedrich Wöhler (1800 – 1882) disputed it with empirical evidence from his experiments. In one of his experiments, he found out that urea, which was once thought to be produced only by living things, could be produced from inorganic precursors. He discovered in 1828 that urea could be chemically produced from salts potassium cyanate and ammonium sulfate. This is considered a crucial turning point that later led to the rise of modern organic chemistry.

Here he clearly provides evidence that organic compounds can be produced from inorganic compounds. There were people who explained this in the questions and answers. However, he argues that inorganic compounds are obtained from inanimate objects. He even argues that the word inorganic got this name because it is produced from inanimate objects.
https://www.biologyonline.com/dictionary/inorganic-compound

Last Edit: I have compiled the information I collected and corrected here. In addition to the answers below, you can check this text to convey accurate information. We know that organic compounds are no longer produced only in the nature of living things, but can also be produced in laboratory environments. We know that inorganic compounds can be produced by both living and non-living entities or in the laboratory.

The basic structure of all living things consists of inorganic compounds as well as organic compounds called carbohydrates, lipids, proteins and nucleic acids. Organic contain compounds carbon and hydrogen elements in their structure. In addition to these elements, organic compounds may also contain elements such as oxygen, nitrogen, phosphorus and sulfur. (CO2 is one of the rare inorganic compounds containing carbon.) Scientists believed that organic compounds were completely synthesized by living things, but urea, which is organic in the laboratory, was produced by inorganic compounds. It has been revealed that organic compounds can be obtained in laboratory environments other than living things. The main source of organic compounds is living organisms. Inorganic compounds can be found in the structure of living things and in nature, they can be produced by living things or in laboratory environments, their main source is minerals in nature. Inorganic compounds are compounds that do not contain carbon as a major element. These compounds participate in the structure of living things. It can pass through the cell membrane without being digested.

The premise of the title question is wrong. They synthetize inorganic compounds like H2O or CO2 routinely.

The title I gave is quoted from wikipedia information. If you pay attention, it is stated this way in the text I quoted.

If Wikipedia says nonsense, then it is still nonsense.

@IvanNeretin Well, can I ask one more question? Is the element oxygen inorganic? I'm sure it's not organic, but can we call it inorganic?

Inorganic compounds can be synthesized all right. Water is inorganic. Carbon dioxide is inorganic. Elemental oxygen, for that matter, is inorganic too. The distinction between organic and inorganic compounds is mostly arbitrary.

Thanks for the answer, I understand very well but; https://socratic.org/questions/is-oxygen-an-organic-or-inorganic-molecule There is information saying that oxygen is not inorganic.

So what? They say so, I say otherwise. Whom do you trust is up to you.

Both organic and inorganic compounds can be created with or without involving living things. // All elements as elements are inorganic. Organic elements are meant as the element involved in organic compounds, not as different element quality. // C is inorganic. C in CO2 is inorganic C as CO2 is inorganic. C in glucose is organic C as glucose is organic. P in DNA is organic P as DNA is organic. P in H3PO4 is inorganic as H3PO4 is inorganic.

@Poutnik So all elements alone are inorganic. However, when they are in compound form, since the properties of the elements change completely, isn't it right to interpret the compound as inorganic or organic, rather than the element? I understood what you said. If a compound is organic, you mean that the elements it contains are now organic for that compound.

@IvanNeretin I want to believe you. But the only doubt remains. In biology, we show photosynthesis as the production of organic matter from inorganic substances. and as a result of this synthesis, oxygen gas and glucose (organic compound) emerge. If the oxygen is inorganic, not only organic glucose but also inorganic oxygen gas will be produced as a result of photosynthesis. This can happen, of course.

"organic element" = element in organic compounds. It is shotcut for properties of compounds, not elements. Like there is organic C or M or P in waste water and inorganic C  and P or N as carbonates, ammonia, nitrates, nitrites or phosphates. It is still the same element.

Living things very much can synthesize inorganic compounds, and nonliving things can synthesize "organic" ones (so we have to interpret the latter carefully when we find them off Earth).

@OscarLanzi Living things and non-living things can both form organic compounds, right?

Yes indeed. For example, methane found on Mars could have cone from a mineral process called serpentuzation. That's why news stories I read a year or two ago about this methane did not include claims that we had found life.

My question has been answered but I don't understand why I got downvoted. So I was restricted from asking questions. However, I didn't ask about homework. Or the question I asked does not contain any shortcomings. That's why I was banned from asking questions for 4 days.

When dealing with confusing or suspicious information, always check multiple alternative offline/online sources before asking. Aside of Wikipedia, search on sites like stackexchange.com or libretexts.org. there is plenty of useful info. You may want to prefer EN wikipedia as there is more content and details about most of topics.

As humans are breathing biosynthetized $\ce{CO2}$ and $\ce{H2O}$, both being inorganic compounds, it negates the title question premise.
The term organic has at least three distinguished meanings:

Organic (compound) in the organic chemistry context means a compound with molecules containing carbon, with enumerated exceptions as below.
Organic (element) in context of chemical analysis means an inventory term about the forms of element occurence. E.g. if water sample contains 40% of all carbon as organic carbon, it means that 40% of carbon is present in structures of organic molecules, or even in inorganic compounds contained in living thing tissues. The key is, it is not floating around as simple inorganic molecules/ions/particle, but is captured.
Organic (biologically natural) in context of biology, life style, nutrition means being produced by natural processes involving living things and processing without involving explicit man-made chemicals not present in living things. Involving chemicals produced by living things can be questionable.

Wikipedia - Organic_compound - Definitions_of_organic_vs_inorganic

For historical reasons discussed below, a few types of carbon-containing compounds, such as carbides, carbonates (excluding carbonate esters), simple oxides of carbon (for example, CO and CO2), and cyanides are generally considered inorganic. Different forms (allotropes) of pure carbon, such as diamond, graphite, fullerenes, and carbon nanotubes are also excluded because they are simple substances composed of only a single element and therefore are not generally considered to be chemical compounds.
The word "organic" in this context does not mean "natural."

Initially, scientists thought organic compounds were unique to living things. Inorganic compounds could be created by whatever way.
Later, the boundary got complicated, starting with synthesis of urea that was considered as organic compound. The original criteria is not used any more.
Currently, as organic compounds are considered all compounds containing carbon, with short list of enumerated exceptions like carbides, simple carbon oxides, carbonates, metal carbonyls, cyanides, cyanates and few other simple compounds.
To say it explicitly, being an organic molecule has no direct relation to being created by living things, being matter of chemistry convention.
Inorganic water and organic acetic acid can be synthesized both in living things and by chemical reactions excluding living things

Notes to comments:
A used terminology mismatch is the cause of most miunderstandings.
What is said on the provided table is partially true, what is the most serious kind of errors, mistakes or misinterpretations ( and of lies too, but that is another story).
The matter categorization organic/inorganic in a biological way and a chemistry way, they do not exactly overlap. There are not 2 categories but 4 categories.  chem/bio-organic, chem-organic-bio-inorganic, chem-inorganic-bio-organic, chem/bio-inorganic.
In chemistry context, a compound being organic or not depends on what the compound is, not on what is the compound origin.
Example:

Is Organic(biology)
Is Organic(chemistry)
Example

Yes
Yes
Proteins, DNA, monosacharides, aminoacids, nucletides, lipids

Yes
No
Biogenic $\ce{H2O, CO2, NH4+, NO, HCO3-, HPO4^2-}$ within living things

No
Yes
Man-made organics, like pesticides, chemical weapons, plastics

No
No
Materials not containing carbon, product of nature processes or human activity

Saying Organic substances are synthesized by living things, unlike inorganics ones., is true just partially, for blatantly obvious cases.
And it seems misinterpreted.

It focuses on the fact substances like proteins, DNA, sacharides and fat are formed in living things only.
It does not try to say living things do not synthesize inorganic substances found in nature.

[OP] If living things cannot synthesize inorganic compounds, why is H2O
an inorganic compound?

Living things are able to synthesize inorganic compounds.

[OP] According to the information on wikipedia, inorganic compounds cannot be synthesized by living things.

The English Wikipedia article does not make this claim. If my Turkish were better, I would edit the article you cite to improve it. I encourage anyone else capable to do so.

So why is water an inorganic compound? Water can be produced by cells as a result of respiration.

The historic definition of "organic compound" was a compound that could not be synthesized in a lab, and one that wasn't found in environments that lack life. "Inorganic compound" was the opposite - compounds you could make in a lab, irrespective if they are also made by metabolic processes. If I had more time, I'd draw a Venn diagram.

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed.

No.
Long ago there was the believe that organic and inorganic matter was inherently different, because organic matter was believed to contain "live force", which was believed to be some substance or energy.
For Chemistry, at least, this is not true.
But it resulted in the naming of organic chemistry and inorganic chemistry.
Not sure if the definition changed since I heard it, but organic chemistry is anything with a covalent C-H-bound, and inorganic chemistry is everything else.
There is up to now no evidence of anything like "live force" or "vital energy".


text:
Is Polymerized lithium niobocene a possible compound?
I was doing some research on different types of body armor and found a fictional compound by the name of Polymerized LiNb. I wanted to find what the structure of this compound looked like so it could be recreated. But to my surprise, nothing is on the internet about it. So I was hoping that someone up here could tell me how it could be formed and polymerized and if it could be used in the piezoelectric effect.

Nothing is on the internet about it? You should brush up your google-fu ;)

If it was real, our world probably wouldn't be what it is. Don't you think?

The Li-Nb phase diagram is quite simple - there are no known compounds in the system. Solubility measurements indicate that the heat of mixing is on the order of 100-300 kJ/mol which is quite large (particularly considering that both Nb and the high temperature phase of Li are both bcc). The two want nothing to do with each other...

Niobocene is a sandwich compound of Nb. I imagine one could call plenty of organometallic polymers in similar fashion, but the name itself is vague and incorrect.


text:
FTIR data: Is there a way to obtain stat significant differences observed between groups in LD-PCA?
Besides checking the loadings and choosing peaks of interest, there is a way to point out from the LD-PCA or PCA the differences between groups that are significant?
All ideas are very much appreciated.

For ease of access (subscribers of the page of all levels of study) as well as to attract future interested into the topic (newly submitted tag chemometrics) I suggest you either explain once the abbreviations used, or link them to an accessible reference.


text:
Why NaOH(s,l) + CO2(g) → Na2CO3(s)? Why not NaOH(g) + CO2(g) → Na2CO3(g)?
Why doesn't $\ce{NaOH(g)}$ react with $\ce{CO2(g)}$ to produce gaseous sodium carbonate $\ce{(Na2CO3(g))}$:
$$ \ce{2 NaOH(g) + CO2(g) → Na2CO3(g) + H2O(g)} \tag{3} $$
Why don't we see any gaseous $\ce{Na2CO3}$ at high temperatures (above 800)?

Check their boiling points. You will probably have difficulties to find it for the carbonate.

Most carbonates decompose, to some extent at least, at high T.    A quick look at MPs BPs and guessing temperatures indicates that there could be some gas phase reaction. As temperatures cool rapidly gas phase reactions should minimize.

@Poutnik curiously, Wikipedia reports neither a boiling point nor a decomposition point for sodium carbonate.

@OscarLanzi I assume it is not a surprise.

You are looking at a high entropy situation (high temperature, gas phase, and likely low vapor density as sodium hydroxide does not achieve one atmosphere vapor pressure until 1388°C). Under these conditions the proposed equilibrium
$\ce{2NaOH + CO2 <=> Na2CO3 + H2O}$
tends to favor more gas molecules with fewer atoms apiece, which is seen to be the left side (the sodium compounds, even if ionic, exist in the gas phase as neutral clusters of said ions).
Upon condensation of the sodium compounds the gas phase entropies are more nearly balanced, with only one three-atom molecule remaining gaseous on either side, and enthalpy effects that favor forming the carbonate come to the fore.
Incidentally, sodium hydroxide is one of only a few metal hydroxides that actually make it to the gas phase with significant vapor pressure. Most metal hydroxides, even with the lighter alkali metal lithium, decompose by emitting water vapor, due to those sane entropy effects, before they would become volatile.

Your equation implies bot $\ce{NaOH}$ and $\ce{Na2CO3}$ have to be the gas state. Well! It is hard to obtain. $\ce{NaOH}$ is gaseous at temperatures above $1500$°C. And $\ce{Na2CO3}$ does not exist in the gaseous phase. It melts at 851°C, and is decomposed at higher temperatures.


text:
1,3-butadiene conformation difference
I am aware that there are 2 well-known conformations of 1,3-butadiene which is s-cis and s-trans conformation.

The s-trans conformation is clearly more stable than the s-cis conformation whose interaction of two hydrogens destabilizes. Based on the picture in Shira's question, the energy difference of the HOMO and LUMO in s-cis is lower than s-trans (calculated via wavelength). Why is that so? Could it be the contribution of the mentioned hydrogen interaction?

https://chemistry.stackexchange.com/questions/44053/relative-rates-of-two-acyclic-dienes-in-the-diels-alder-reaction

Much thanks for your suggestion, but I don't think think that I have the answer for the question


text:
Does ionisation change the physical/chemical properties of an atom?
From what I have learned, the number of electrons on the outermost shell (called the valence electrons), determines the chemical properties of an atom, this is why elements in the same group have similar physical/chemical properties.
I was wondering, if you had a neutral sodium atom (Na), and you ionising it by making it lose an electron ($\ce{Na+}$), it now has a total of 10 electrons, and a full outer most shell. Does the $\ce{Na+}$ ion have the same chemical/physical properties as a neutral Neon atom and the rest of the elements in the noble gas group, if not how are they similar/different.

Atoms and their ions have totally different chemical and physical properties. Apart from the mass and the number of protons in the nucleus, there is no similarity between the properties of neutral atoms and the properties of corresponding  ionized atoms. For example, the behavior of a given atom in contact with another one changes if the colliding atom is ionized or not.

You're correct in that there are similarities in the spectra of $\ce{Na+}$ and $\ce{Ne}$. That is, in an isolated ion or atom of $\ce{Na+}$ and $\ce{Ne}$, there is a similarity in the mathematical relationship amongst the energy levels.
Meghnad Saha and Protap Kischen Kichlu, in the Indian Journal of Physics, are quoted as stating, "that when we take a number of successive elements, which, on account of loss of one or of more electrons by electric discharge, are reduced to have the same electronic structure, [e.g., $\ce{Na+ and Ne}$, ed.], then the frequencies of the corresponding lines of these elements arising from the same transition form an arithmetic progression, provided the transition takes place between levels having the same total quantum number."
This seems to me to be one of the fundamental studies in quantum mechanics, though not as easy to find on the internet as it should be. You can read an article by Saha and Kichlu, Extension of the Irregular Doublet Law to Complex Spectra, at arXiv.org.
However, it is more than a quantum jump to assume that ionized sodium and neutral neon behave in a similar chemical fashion, or that they have similar physical properties. Well, other than providing useful illumination via the neon lamp and the sodium vapor lamp. But then we're back to spectra, aren't we?


text:
How is the electronic configuration of elements determined practically?
How is the electronic configuration of elements determined in practical? Theoretically, we use certain rules (Aufbau principle, Pauli exclusion principle, and Hund's rule) to define the electronic configuration of elements. There are some exceptional cases and some explanations for them. This is what I know about electronic configurations so far.
I wonder how we know the electronic structure of any element practically and how the exceptional configurations became known to us? I mean the electronic configuration of copper, according to rules, should be [Ar] 3d⁹ 4s² but in actual it is [Ar] 3d¹⁰ 4s¹. How did we determine this?
Similarly, how do we know that the electronic configuration of tungsten is [Xe] 6s² 4f¹⁴ 5d⁴? It could have been 6s¹ 5d⁵ (there is possibility). I'm not looking for the reasons for these exceptional configurations but a detailed explanation of the practical methods to determine the electronic structure of elements.

See https://chemistry.stackexchange.com/questions/31473/how-are-electronic-configurations-determined-experimentally#:~:text=There%20are%20lots%20of%20techniques,EELS%20spectroscopy%2C%20ELNES%20and%20EXELFS. for example


text:
How to explain oxidizing power from Molecular Orbital Theory
I am trying to explain the oxidizing property of $\ce{Fe^3+}$ in $\ce{H2O}$ and $\ce{CN^-}$ which is $\ce{[Fe(H2O)6]^3+}$ and $\ce{[Fe(CN)6]^3-}$.
The electrochemical reactions along with their standard reduction potentials are:
\begin{align}
  \ce{Fe^3+(aq) + e-        &<=> Fe^2+(aq)} \qquad
  E^\circ = \pu{0.77 V} \tag1 \\
  \ce{[Fe(CN)6]^3-(aq) + e- &<=> [Fe(CN)6]^4-(aq)} \qquad
  E^\circ = \pu{0.36 V} \tag2 \\
\end{align}
and the higher the standard reduction potential, the higher the equilibrium constant. I deduce this from
\begin{equation}
  \Delta_\mathrm{r}G^\circ = -nFE^\circ = -RT \ln K \tag3
\end{equation}
Which could also give the result that $\ce{Fe^3+}$ in water oxidizes stronger than when the ion is in cyanide medium. I want to ask whether we can deduce the same way by using the molecular orbital of two complexes mentioned above?

In this picture, instead of water, chloride ion could be used. So I am thinking of predicting oxidizing strength using the energy of HOMO and LUMO. Could I do this?
My thought was that the $t_\mathrm{2g}$ in cyanide complex is lowered in energy than that of the chloride complex. So, if the electron was to be accepted, then the cyanide complex is prone to accept better. However, the reduction potential speaks otherwise.

@ananta-This one is correct, and it can also be said that if you are looking at standard oxidation potentials, higher values mean lower reducing power (from the mirror image view)

I am a bit confused right now. If I read correctly, then wikipedia is wrong, is it not?
https://en.wikipedia.org/wiki/Standard_electrode_potential
It is stated "The larger the value of the standard reduction potential, the easier it is for the element to be reduced (gain electrons); in other words, they are better oxidizing agents."

I reconsidered my comment, and the standard reduction potentials are indeed an indicator of the oxidizing power. Higher potential means the moiety gets reduced easily which means it some other moiety gets oxidized easily.


text:
How do you identify an organic base?
Problem:

Solution:
The molecules with red crosses underneath them are considered organic bases.
Question:
From my understanding an organic base has an atom with a lone pair of electrons. Doesn't molecule (ii) have a lone pair of electrons and should thusly also be considered an organic base? If I am using the wrong method to identify organic bases, what other method should I use?

Nora, your understanding is correct. just remember that there are degrees of basicity. Amines are stronger bases than amides. An available electron pair makes a base, all that is needed is an appropriate acid. Check this for intra molecular acid base reactions with strange bases. https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/rearrang.htm

Organic bases are just organic compounds which act as bases. These primarily include amines.
Compound two is an amide. Amides are more acidic than amines, and they very rarely act as organic bases. They aren't classified as such.
The lone pair is not much available in an amide due to resonance, hence it acts as a base only when a strong acid is present. On the other hand, amide might act as an acid, donating proton. As $\ce{H+}$ comes out of the nitrogen in an amide, the negative charge is delocalized and hence the conjugate base is stabilized.
PS. This doesn't mean that an amide is much of an organic acid, either. They are both very weak bases and very weak acids.

You contradict yourself not a base! a very weak base!

@jimchmst By using the word very weak base I mean to say that it can't be treated as a base just like sodium hydroxide can't be treated as an acid. The words acid and base are often relative and interchangable in situations like these. Ethanol for example is more inclined towards the basic side and so it is a weak acid. It is just a matter of words.

If it is a matter of words lets use them correctly. There are degrees of basicity and it depends on the acid it is reacting with.  Add Al powder to a strong hydroxide solution to see how acidic it is. As for ethanol it definitely is an organic base but does not qualify under your answer or the tone of original question. Again you contradict yourself.

That is what I mean. The terms acid and bases are relative and depend on the reagent added.So I used the word very weak base instead of "not a base" at the end

Organic base has an atom with a lone pair of electrons. This statement is not true considering the fact that the lone pair of the compound may be in resonance.
Consider the example of a carboxylic acid. They are organic bases under your definition as the oxygen has a lone pair. The second question is similar to a carboxylic acid.
Consider the image:
The first situation shows amide acting as a base.This leads to a highly unstable structure.
The second situation shows amide behaving as an acid. This leads to a stable structure.
Hence as @TheFrightOfTheNight pointed out,the amides are acidic and so can't be called organic bases.


text:
How to deduct the concentration of chloride in a silver/silver chloride electrode, given the cell notation
I have this cell notation:
$$ \ce{Zn|ZnCl_2(0.05M)||AgCl(s)|Ag} \tag{1} $$
What is the physical shape of this cell? I know $\ce{AgCl}$ has very low solubility, and the problem has no concentration on the right side. Is it an aqueous solution, and they just forgot to write the concentration of $\ce{Ag+}$ and $\ce{Cl-}$? Or is it really a solid with no aqueous solution?
Can it have a porous disk or salt bridge if it is non aqueous?
The half reactions are:
oxidation: $$\ce{Zn -> Zn^{2+} + 2e^-}$$
reduction: $$\ce{2AgCl(s) + 2e^--> 2Ag(s) + 2Cl^-}$$
So, when calculating the reaction quotient:
$$Q=[\ce{Zn^2+}][\ce{Cl-}]^2$$
What is the concentration that we use in the formula? Can we asume that the concentration/activity of the chloride in the left is the same as in the right?

The cell notation doesn't indicate, nor limit the shape of the electrochemical cell.  In the form used by you, it only describes a zinc electrode in a (presumed aqueous) solution of $\ce{ZnCl2}$ ($c = \pu{0.05 mol/L}$) on one side, and a silver/silver chloride electrode (which isn't the same as a silver electrode proper, because the activity of the $\ce{Ag+}$ now is noticeably tied to the solubility product of $\ce{AgCl}$) on the other.
In addition: even if presumed water indeed is the solvent, it doesn't yet consider temperature as parameter.

Wikipedia's article on the silver electrode has the following:

$$\ce {{Ag(s)}\ |\ {AgCl(s)}\ |\ KCl(aq)\ (3M)}$$

That makes sense because the chloride has to come from somewhere (or go somewhere), and you need to know its concentration to plug into the Nernst equation.
For the zinc electrode, it might be better to write:
$$\ce{Zn(s) | Zn^2+(aq) (0.05M)}$$
because here, the chloride ions are spectator ions (switching to different anions would not affect the potential, ideally).
UPDATE:
The net equation could be written like this:
$$\ce{Zn(s) + 2AgCl(s) -> ZnCl2(aq) + 2 Ag(s)}$$
However, this ignores the aqueous chloride ions in the half cell with AgCl that you need to make the reaction go. For the equilibrium constant expression, this would be more useful:
$$\ce{Zn(s) + 2AgCl(s) -> Zn^2+(aq1) + 2 Ag(s) + 2Cl- (aq2)}$$
with (aq1) and (aq2) showing the two separated aqueous compartments (with independent concentrations).
If you attempt to "cancel out" the chloride, you get this incorrect net equation:
$$\ce{Zn(s) + 2Ag+(s) -> Zn^2+(aq) + 2 Ag(s)}$$
The presence of an ion in the solid state should alert you to the problem. I think this is sufficiently tricky to ask the instructor about it, referencing the question and answers in this thread.

I added some information about the problem. We are asked to calculate the reaction quotient. Can we asume that the concentration of chloride in the cathode is fixed by the concentration of chloride in the anode?

This could be one reasoning. The other would be the assumption that you always use saturated KCl. Why don't you ask your instructor instead of here?

See my updates in the main answer, too.

The cell description is somewhat vague and leaves much to the imagination; it indicates a liquid junction to a solid-state reaction. The Ag-AgCl electrode is reversible to the chloride ion. Its typical use is as a voltage standard used at minimal, hopefully zero, current draw (AKA a good pH meter). This requires a definite or at least constant concentration of chloride ions. The half reaction is:  $\ce{AgCl +e- <=> Ag + Cl-}$. For the use as a voltage standard the electrode is immersed in a small container of chloride ion of definite concentration; the connection to the solution is usually a small restrictive flowing junction and the electrode system is immersed in a solution containing another electrode (for a pH meter a glass electrode) and the voltage measured.
Your cell suggests something else. Either a cell to oxidize zinc by reducing AgCl to Ag, or, at zero current flow, using the electrode as a chloride ion detector. If zinc metal were added to a solution of silver ions the silver ions would be reduced plating out silver metal. To make a cell there would have to be anode and cathode compartments connected by a salt bridge to keep the silver ions away from the zinc metal. The reaction is $\ce{2Ag+ + Zn <=> Ag + Zn++}$. The reaction quotient is $\ce{[Zn++]/[Ag+]^2}$. The pure silver and zinc both have activities defined as 1. (The reactions are in acid so there are plenty of negative spectator ions about.)
A second method to keep the silver ions away from the zinc is to enclose them in an almost insoluble salt such as AgCl (This makes the $\ce{Ag+}$ a poorer oxidant but it still works). A bare silver-silver chloride electrode is dipped into a solution containing $\ce{Zn++}$ and $\ce{Cl-}$ ion with a separated Zn metal electrode. [no need for the salt bridge] The reaction is: $\ce{Zn + 2AgCl <=> Ag + Zn++ + 2 Cl-}$. The reaction quotient is $\ce{[Zn++][Cl-]^2}$; the numerator is one because the activities of the pure Zn and AgCl are defined as 1. (This is an expensive way to make zinc ions.) When the bare electrode is paired with a standard enclosed Ag-AgCl (or calomel) electrode it becomes a chloride ion detector.


text:
Could ammonium azanide exist?
In my first high school chemistry class, I misremembered nitrate as being $\ce{NH2-}$, rather than $\ce{NO3-}$, and wrote down a formula for "ammonium nitrate" that was $\ce{NH4NH2}$ (rather than $\ce{NH4NO3}$). Now I'm curious if that could actually exist.
I tried seeing if $\ce{NH2-}$ could exist as a polyatomic ion, and it seemed like it should, with the hydrogens being single bonds and the nitrogen having two lone pairs.I looked around for a while, and found azanide, which seems to look how I expected it to. I can find a few examples of it forming ionic bonds with alkali metals (where it's confusingly called "amide"), like lithium amide, and the article mentions silver(I) amide, but I still couldn't find anything by searching for "ammonium azanide" or "ammonium amide".
Is there some reason why $\ce{NH4NH2}$/ammonium azanide couldn't exist, or has it just not been looked for? Is there a way to predict what it would be like?
(I'm still in high school chemistry, so if I'm missing anything obvious that's probably why :p)

$\ce{NH2-}$ is very strong base, much stronger than $\ce{OH-}$, and would react with the weak acid $\ce{NH4+}$ forming $\ce{NH3}$.

@Poutnik This would make a great answer

Not so great. That's the obvious thing, if you knew their acid-base properties. Still, many strange things can happen, if you try hard enough. Water can be fully ionic, why not ammonia? But your thing is still considered ammonia, no matter how far equilibrium would be shifted - that's perhaps both a strength and a weakness of chemistry.

I have intentionally limited my comment to a simple, high school level. Answers to questions about existence of a specific compounds often depends on conditions. Many compounds generally assumed not to exist are realized to exist if conditions are extreme enough. Typical scenarios are cryogenic conditions in inert solid matrix, extreme pressures or interstellar space.

Technically, ordinary liquid ammonia may be regarded as a very dilute solution of ammonium azanide through its autoionization. (Water is similarly very dilute hydronium hydroxide.)
This technicality aside, we can indeed have ammonia convert to an ionic solid form $\ce{NH4^+NH2^-}$. Ninet et al1 calculated that under 150 GPa pressure at sufficiently low temperature solid ammonia switches from a molecular form to the ionic one referenced above, adopting two different ionic crystal lattice structures. Experimental evidence for this transformation was obtained from high-pressure IR spectroscopy showing the presence of $\ce{NH4^+}$ at 150 GPa. The figure below, taken originally from 1, shows the ionic structures and the temperature and pressure range for them on the lower right. Conditions suitable for forming such ionic phases (with ammonia alone or with ammonia and water together) may exist in the interiors of Uranus and Neptune.

Reference

Ninet, S., Datchi, F., Dumas, P., Mezouar, M., Garbarino, G., Mafety, A., Pickard, C. J., Needs, R.J., Saitta, A.M. "Experimental and theoretical evidence for an ionic crystal of ammonia at high pressure" Physical Review B., 89(17): art.n° 174103. (2014).

In standard conditions ammonium amide in pure form wouldn't be stable as it would turn into ammonia through proton transfer.
However, it is being taught that in liquid ammonia there is an equilibrium of self-ionisation present, similiar as in water, and that would result in extremally small amounts of ammonium amide in liquid ammonia.
Interestingly, there has been a theoretical study showing that solid ammonia under very high pressure could have a ionic (ammonium amide) structure: https://www.nature.com/articles/nmat2261