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Compound Polarity Lookup
I know from the literature that the following compounds have decreasing polarity: Rosmarinic Acid -> TMF -> Sinensetin and I guess comparing compounds of similar structure it is manageable to compare bond types, groups etc etc by which to order their polarity, but what about compounds with very different structures, groups and bonds, especially when you want to rapidly order 10's-100's of compounds.
My questions:

Is there such a thing as an absolute measure of "polarity" for a compound by which other compounds can be compared/ordered?

Depending on 2) is there a resource whereby you can lookup your compound to obtain its polarity?

I am asking this question from a Chromatography perspective. I would like to know the order of separation for a group of compounds and how to know what "polarity" of mobile phase to start my experimentations with.

No. $\mathstrut$

In context of chromatography, It can be said in many cases that  solvent A is "more polar" than solvent B wrt analyte C, and at the same time, solvent B is more polar than solvent A wrt analyte D.

There are many measures, none is definite. For example dielectric constant, Donor number Partition coefficients.

@Poutnik - I am looking at this from a chromatography perspective. I am interested in estimating the elution order of a group of compounds and the choice of mobile phase that I need to make. I will update my question to give this information. Maybe you could answer based on this new information?

@tomdertech Always include eventual relevant circumstances in questions to prevent wrong assumptions and requests for clarifications.

The elution order cannot be easily theoretically predicted in general, except for big enough differences (which decrease with experience ) of analyte properties. It is near always combination of educated guess and experiment. For reverse HPLC on e.g. C18 columns, it can be interpolated from elution times of testing mobile phases MeOH+H2O, AN+H2O, THF+H2O, each mobile phase interacting with analytes in a different way.

Related chem.libretexts.org  - High Performance Liquid Chromatography

Polarity is an umbrella term which is popular among organic chemists. It is as vague as saying the weather is good today. "Good weather" may include, bright light, sunny, temperature, right humidity and so on.  Now read your own statement

Rosmarinic Acid -> TMF -> Sinensetin and I guess comparing compounds
of similar structure it manageable to compare bond types, groups etc
etc

Since you correctly use etc. etc. ...yes polarity includes et cetra et cetra. Its usage is discouraged by IUPAC.
The correct term is dipolarity which is a measure of the dipole moment (you can look them up in tables). Even that dipole moment is of no good because you have plenty of conformers and each conformer has a different dipole moment (at least in the gas phase).

It can be said polarity is multidimensional, as there are several phenomena that contribute in different proportions to the overall perceived "polarity". It is relative permittivity, dipole moment, polarisability, donor-acceptor properties, forming hydrogen bonds. It can be well illustrated by mobile phase choice in reverse HPLC, where 3 m. phases with the same elution time for main component (H2O + CH3OH/CH3CN/THF respectively ) have very different elution times for minor impurities, during determining the chromatographic purity )

Yes, another source of confusion from polarity is the chromatographic usage.

http://goldbook.iupac.org/terms/view/P04710


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How does it affect the equilibrium expression if the concentration of any component remains the same?
I am a high school student and I am very confused in Equilibrium expression, My confusion is that "Why we don't write concentration of solids and pure liquids in equilibrium expression?" Many people says that its because their concentration don't changes but how does it related to this? If concentration is same for them then what's the issue in it? I am not able to picturize how does it affect the equilibrium expression if their concentration remain the same?
please Help me regarding this using simple language that I can understand because I am at high school level. If anyone could help me to picturize this thing in my mind then it would be very helpful.

Well, it doesn't, so it can be ignored, as simple as that.

Well, you can either define an equilibrium constant with the formal constant concentration of a pure condensed phase in mol/dm3, or you can implicitly involve them in the equilibrium constant value, pretending their concentration is 1 mol/dm3. Why way is more convenient ? The same for the pressure variant of the equilibrium constant.

why we have to pretend that their concentration is  1mol/dm^3,.it can be of any value..

We do not have to. Neither we have to go along stairs forwards. We can go backwards as well. Multiplication or division by 1 is very convenient. You can omit it totally.

Suppose you want to study the equilibrium of the reaction of a metal like zinc $\ce{Zn}$ with water to produce $\ce{ZnO(s)}$ and $\ce{H2}$ :$$\ce{Zn(s) + H2O(g) <=> ZnO(s) + H2(g)}$$ At high temperature, the equilibrium constant may be written $$K_\mathrm{1}=\frac{[\ce{ZnO}]·p\ce{(H2)}}{[\ce{Zn}]·p\ce{(H2O)}}$$ In this expression three parameters are not changing : the constant $K_\mathrm{1}$, and the two concentrations of the solid stuffs $\ce{Zn}$ and $\ce{ZnO}$ in their own phases. The mass of these stuffs may change but not their concentrations, as they are in the solid phase. So it is no use keeping such a formula where three parameters are constant. It is better to group all constant parameters on the same side of the = sign, and define a new constant like $K_\mathrm{2}$ : $$K_\mathrm{2} = K_\mathrm{1}\frac{[\ce{Zn}]}{[\ce{ZnO}]}= \frac{p\ce{(H2)}}{p\ce{(H2O)}}$$ With this formula and a numerical value for $K_\mathrm{2}$ you may calculate immediately how $p\ce{(H2)}$ changes when $p\ce{(H2O)}$ is varied. No need to know the concentration of the solids. This may be useful if you want to optimize the yield in producing $\ce{H2}$ gas through this reaction.

so this new constant include all the constants in it, this also creates a problem  in my mind that the equilibrium constant tells us whether the concentration of product is high or low than the reactant at equilibrium, but if we are including the concentration of solid in this new constant then we would have get some higher value for Keq. which we MIGHT not get if the a reaction has some solid reactants in it, so it  creates a new problem ,

I don't see your point. The ratio of the concentrations of Zn in metallic zinc and ZnO in the solid ZnO is a constant, which should not be important, and probably between $0.1$ and $10$. I don't speak of the amounts, in mass or in moles. I just speak about the concentrations.

I am not only talking about this reaction, I am talking about those reactions in which those solids are involved which have a higher concentration.

Whatever the nature of the solids appearing in a chemical reaction, their concentrations stay constant

I know their concentration stay constant, but what I am saying is that, if we consider their concentration  to be the part of Keq. itself and do not count them in the expression then how can we predict that at equilibrium the concentration of product is high or reactant is high? we should consider them in the expression.

Only $K_2$ values are reported in Table of constants.

but in some reactions like esterification reaction in which water is formed as a product ,then we do consider its concentration also in in the expression, why is it so?

Whatever the reaction, when the solvent is also included in the reactants of a chemical reaction, the concentration of the solvant is considered as constant, and its expression is included in the equilibrium constant. It is not the cas in an esterification which produces water without being made in water. In this case, the amount of water is changing during the reaction. Here the concentration of water is not constant. So it remains among the variables that are combined to be equal to the equilibrium constant.

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trialkyl phosphine ligands in aqueous micellar Suzuki-Miyaura coupling
I have seen $\ce{P(tBu)3HBF4}$ salt used as a precursor to the phosphine ligands for aqueous Pd-catalyzed micellar Suzuki-Miyaura coupling, but why does it work - after $\ce{Et3N}$ has liberated the tributyl phosphine ligand why does it not deactivate in the presence of water?

And how you imagine "deactivation"?

Well there is a likelihood of free radical chain oxidation if there is oxygen present, which is why I thought it was odd that it was apparently stable for a reaction performed in water

It's not like it spontaneously combusts - it's not a problem if some of phosphine gets oxidised, isn't it added in surplus either way?

The phosphine ligand probably coordinates to palladium rather quickly, and the complex is air-stable (in the solid state at least). That may not render the reaction immune to oxygen, but likely does increase its resilience to it.

ok excellent cheers

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Weak acid titration curve
At the beginning of a weak acid titration curve (when titrated with a strong base), there is a sharp increase of pH before it levels out more due to the buffer solution that will occur. But why is that sharp increase there at the beginning? Is it because since the weak acid is weak, it is easier to increase it's pH when adding a strong base?

Why not to calculate pH as function of molar ratio base versus acid ? You will see then why.

Imagine adding your strong base to just water.

... but also listen to what Poutnik says and do the math!

@Karl - Thanks for the link, yes, it does answer my question. I will also try to do the math to get a deeper understanding myself.


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Can acetic or citric acid be too highly concentrated when used for cleaning copper?
I intend to clean a copper surface with acetic or alternatively citric acid. There are many different concentrations recommended by different sources on the internet. For example vinegar with 10% acid content diluted once more at a ratio of 1:9 with water.
I'm under the impression that neither acetic or citric acid actually reacts with pure copper, only reacting with the oxidisation products on the surface (creating verdigris and so on).
Does the dilution then serve a purpose other than saving on vinegar? What could happen if I were to use undiluted 25%-acid (or even higher) vinegar concentrate, i.e. a 25% aqueous acetic acid solution directly? Basically, what could go wrong if the acid is too highly concentrated?

Acid being solid may be a problem ;)

I suggest using commercial copper cleaner.

I remember cleaning old coins in suspended Na2H2-EDTA. ( Complexone/Chelaton-III ). Be aware acids are corrosive wrt copper if dissolved oxygen is present. It may easily form blue solution or cover of  copper acetate of copper-citrate complex.


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Is it safe to look at a mercury gas discharge tube?
As part of a chemistry course, I recently observed a demonstration in which the instructor electrified a mercury gas discharge tube (the tube looked something like this image). The glow from the tube appeared to be a light blue, but when I performed an internet search afterwards, I found that mercury gas discharge tubes apparently emit ultraviolet light, which can be harmful to the eyes and body. I was standing within one foot of the discharge tube (which looked something like this while lit) for several minutes. The tube was completely clear, which I am assuming means that it did not contain a phosphor to block the UV light (please correct me if I am wrong or if I am misunderstanding something).
The same demonstration was performed with a hydrogen gas discharge tube, which also was completely clear (see above).
Since I was looking at the tubes, could the higher frequencies of electromagnetic radiation (e.g. the ultraviolet light from the mercury) have damaged my eyes, skin, etc.?
After the demonstrations, my eyes seemed irritated, but I'm not sure if this was just a result of the placebo effect.

Most glasses absorb uv quite well. Ask your instructor.

"Placebo" not really. The correct term is Nocebo.

If you did not feel like sand in your eyes it was likely nocebo.

"Is it safe" is a question without a binary answer. It isn't safe if the intensity of the light is high and duration of exposure is long. But that is also true of bright sunlight.

Could it have caused irritation or injury? Possibly, but not likely, depending on a few factors:

The intensity of the emitted ultraviolet.
Proximity to the light source.
Length of exposure.

You'd need to calculate the dosage integrated over time.
That said, it is very unlikely any harm was done, because of the following:

The power supplies are limited to to only a few mA current. The total power is the arc voltage drop, ~30 to 100 VAC, times the current. For the supply abvove, that would be 100 VAC * 0.01 A = 1 W. One watt is low power for a mercury lamp.
Demonstration spectral lamps, such as this capillary mercury vapor lamp, usually have thick borosilicate glass capillaries. Borosilicate glass does not transmit shorter wavelengths well. Most of the UV would have been absorbed by the thick capillary wall.
The length of time was given as "several minutes", so the integrated intensity over time (i.e., total energy absorbed) was low.
That you state you were, "within one foot of the discharge tube," ~30 cm, is of concern, because some older discharge tube power supplies had exposed high voltage terminals. Though limited to 10 mA, considered below a dangerous current for electric shock, in most cases, one could still get burnt by the ~10,000 VAC arc, should one come in contact with exposed terminals, or could be startled and hurt by a muscular reflex action. Likely, though, the ends of the tube and associated wiring were safely enclosed.

So, all things considered, there might have been fumes in the lab or other cause of irritation, but I believe the UV light was not likely an issue.

Borosilicate glass does not transmit shorter wavelengths well. Most of the UV would have been absorbed by the thick capillary wall. This is not quite right. This type of tube emits a 365 nm line that is easily observable. That wavelength is not strongly absorbed in glass, and one can for example put it through a glass diffraction grating. However, that's a fairly long wavelength in the UVA band, and so not harmful at this intensity.

Without even knowing much about the tube itself, I can tell you the amount of UV it emits is a small fraction of what you're exposed to when you walk outside on a sunny day.
What matters for eye safety is actually power per unit area --- If you were to concentrate that light with a magnifying glass, and get it to focus on the back of your retina, then "maybe" you did some damage.
Note:  I'm not a chemist.  I just happened to spot this question on the list of 'open questions' in Stack Exchange.   I'm an electronics engineer, and one of my professional responsibilities is the application and understanding of the European laser/LED eye safety standards  (EN60825).   I actually use a mercury vapor lamp to calibrate our spectrometer.
If it really concerns you, next time wear a pair of sunglasses.  Even cheap ones do a very good job of blocking all types of UV.    You're not wrong to be concerned a little....  UV light from an arc welder can seriously damage one's eyes if you stare at it (this is why welders wear nearly black eye protection).   But the emissions from your little gas discharge tube is probably about a million times weaker than those from an arc welder.
It's like anything else -- Exposure time & exposure levels matter.   You can smoke a single cigarette and have no ill effects.  But smoke 20 a day for 20 years, much different story....

Even cheap [sunglasses] do a very good job of blocking all types of UV. One should be careful with cheap sunglasses that may not block UV but instead open your pupils up, allowing even more UV to enter the eye (see for instance https://healthcare.utah.edu/healthfeed/postings/2018/07/sunglasses.protect.php)

@WoJ    Yeah, I should have said that.   Funny thing, a few years ago I read in Consumer Reports this same topic, and they'd found even cheap sunglasses (nearly always) work fine.   I was curious, so I went and bought a dozen pairs from various dollar stores.   I found the same thing, every one blocked UV very well.   So there surely could be some out there that are just dark plastic, but you'd have to (per my limited experience) try really hard to find them.

as an ex-physicist I approve of your approach :) What you found out is good news.

Mercury discharge tubes such as germicidal lamps that are intended to emit short wave (UVC) radiation have to be made with an expensive quartz envelope rather than borosilicate glass.
Depending on the gas pressure they can be made to emit primarily UVC which is invisible and can be dangerous (at ~200nm) or UVA at ~400nm. Since you could see the discharge it was probably the latter, and probably any UVC was heavily absorbed by the envelope.
The 400nm emission line is actually longer wavelength than that used as "black light" in night clubs (~350 to 370nm) so less harmful.
The danger with strong UVC sources is that the light is completely invisible and does not trigger the blink reflex so you can get damage to the cornea (like snow blindness, usually temporary). Symptoms are described as follows:

Symptoms may not start for 6 to 12 hours after exposure to the
ultraviolet light. Symptoms include:
Eye pain.
A feeling that something is in the eyes (foreign body sensation).
Sensitivity to light (photophobia).
Tearing.

The 400nm emission line is actually longer wavelength than that used as "black light" in night clubs (~350 to 370nm) so less harmful. The 405 nm line is visible, not UV. Mercury tubes actually do emit a 365 nm line, which is UVA and not strongly absorbed by the glass.

The type of mercury discharge tube used for these classroom demonstrations has five strong lines in its emission spectrum. Four are visible, while one is  UV with a wavelength of 365 nm, which is in the UVA range. The wavelengths that are harmful to the human body are mainly the ones in the UVB and UVC bands, which have wavelengths shorter than 315 nm. This fact alone makes it very unlikely that you were harmed.
We can also think about the intensity. UVA is about 6% of the energy content of natural sunlight.
Ultraviolet is absorbed somewhat by glass, so some of the UV would have not have escaped from the tube. However, the absorption of the 365 nm UV line in a typical glass discharge tube of the types I've worked with is not very efficient. The UV line is clearly observable through fluorescence, and the intensity of the UV line is about the same as the intensities of the four visible wavelengths that it emits as well. There are shorter wavelengths emitted by some mercury discharge tubes, but in the classroom-demo tubes I've used, those wavelengths are not observable. This is probably due to the intentional design of the tube (electrodes, voltage, pressure, etc.), and the glass would probably also absorb these shorter wavelengths much more strongly.
So roughly speaking, there are five lines of equal intensity, and the UV emitted by the tube is probably on the order of 20% of the light it emits. (The exact proportion depends on the details of the tube, and can even vary between different parts of the same tube.) What this tells us is that the proportion of UVA to visible may be somewhat higher for the light from the mercury discharge tube than for sunlight (perhaps by a factor of 3 or 4), but not massively higher. Therefore the intensity of the light as sensed by your eye is not too bad a gauge of the intensity of the UVA from this source. From my experience doing this sort of demonstration in a classroom, I assume the light seemed much less bright to you than bright sunlight. The eye-brain system senses intensity sort of logarithmically, so I'm guessing that the brightness of the light was maybe $10^{-4}$ of the brightness of bright sunlight. The UVA exposure you experienced would therefore be many orders of magnitude less than what you would have experienced by being outside in bright sunlight for a similar amount of time.
Note that ultraviolet also has positive health effects. You need UVB so that your body can synthesize vitamin D.


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Why do iron objects leave black marks on laminate countertop
My over 30 years old laminate kitchen countertop had accumulated a lot of marks over the years that no cleaner would remove.  I finally gave up and decided to use chlorine bleach on it knowing that the bleach may damage it.  The bleach removed all the marks — looked like new.
There is only one unfortunate side effect: iron objects, e.g. cast iron, that come into contact with the countertop leave permanent black marks even a year after bleaching.  I suspect that the bleach damaged the seal on the countertop since I doubt any chemical residue from the bleaching would remain after a year.  On the other hand how could an over 30 year laminate countertop still have any kind of seal?
Any idea how I can prevent the black marks from iron objects?
Cheers and thanks.

There are a number of choices: a) accept it as patina, b) put a cloth over the countertop, c) rebleach after six months, d) try some wax, furniture polish, laquer, younameit to reseal the surface. There is no chemical, scientific answer to your question, how should anybody know what exactly your countertop surface is, now?

A photo might help, though.

I added a photo.

Never use a bleach on rust stains. Bleach can't remove rust it can only cause it become permanent. It will cause more problems. Rust stain will not dissolve by any typical home cleaner. There are commercial rust stain removers and they work like magic and in 5 minutes. Many of them should be available in a good grocery stores. The rust stain removers are either strong reducing agents or complexing agents and sometimes oxalic acid. Check with the manufacturer for safety as well.

Using a commonly available product that removes lime, calcium and rust, I was able to remove the stains.  Note that the manufacturer warns: "Vinyl or any other form of plastic can be easily damaged by the ingredient within the formulation." There was no damage to my countertop by the cleaner.


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Why does glycogen phosphorylase only act on the non-reducing end of the glycogen chain?
I read that glycogen phosphorylase only acts on the non-reducing end of the glycogen chain. This enzyme requires an inorganic phosphate molecule and PLP as a cofactor. The mechanism of the enzyme is as follows:

I attempted to workout the mechanism for the reducing end of the molecule and concluded that perhaps it results in a ring opening reaction.

Is this correct? If not what is the reason?

Glycogen phosphorylase does indeed phosphorylates the non-reducing end of a glycogen chain. It is unclear from the crystal structures how come this is an exo type enzyme. The prevailing catalytic mechanism is almost as you illustrated, involving the protonation of the $\pu{\alpha}$-1,4 link leading to the formation of a oxocarbenium ion and release of the long chain; nucleophilic attack by an inorganic phosphate to the anomeric carbon $\ce{C1}$ of the intermediate, leading to the final product. [1] Overall, the $\pu{\alpha}$ anomeric configuration is maintained from substrate to product, instead of a product with $\pu{\beta}$ configuration as shown on your drawing.
I think there are some unanswered questions about this mechanism:
a) How come there is little $\pu{\beta}$-D-glucose that is produced as a side-product? This is quite unexpected for such an unstable intermediate as an oxocarbenium ion.
b) How does the above mechanism account for the loss of 500-fold activity upon mutation of Glu672? [2,3] This is a huge decrease! I'm wondering whether this residue may act as a nucleophile forming a covalent intermediate with the residual saccharide?
Regarding the second question from the OP: I wouldn't think there is ring opening at the chain anomeric center as there is no evidence that this reducing end of the chain is presented in glycogen phosphorylase catalytic site. It is the non-reducing end that is presented in the catalytic site and the $\pu{\alpha}$-1,4 link that is presented at the catalytic center.
Sources

Duke, E.M., Hadfield, A., Martin, J.L., Clifton, I.J., Hajdu, J., Johnson, L.N., Reid, G.P., Trentham, D.R., Bruce, I. and Fleet, G.W. (1991) "Towards time-resolved diffraction studies with glycogen phosphorylase" Ciba Found. Symp. 161:75-86. DOI: 10.1098/rsta.1992.0064

Schinzel, R. and Palm, D. (1990) "Escherichia coli maltodextrin phosphorylase: contribution of active site residues Glutamate-637 and Tyrosine-538 to the phosphorolytic cleavage of $\pu{\alpha}$-glucans", Biochem. 29:9956-9962. DOI: 10.1021/bi00494a028

Schinzel, R. and Drueckes, P. (1991) "The phosphate recognition sites of Escherichia coli maltodextrin phosphorylase", FEBS Lett. 286:125-128. DOI: 10.1016/0014-5793(91)80956-4


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Chemical reaction of bronze tarnishing
Can anyone explain to me the chemical process of bronze tarnishing? According to Wikipedia:

Bronzes are typically ductile alloys, considerably less brittle than
cast iron. Typically bronze oxidizes only superficially; once a copper
oxide (eventually becoming copper carbonate) layer is formed, the
underlying metal is protected from further corrosion.

Does this mean the tarnish that we see in the picture below is created by copper oxide turning into copper carbonate? If so, how do you write out that equation? What is being oxidized and what is being reduced?

I don't see much tarnish in the picture, if any at all. Well, yes, a little bit of CuO, maybe.

$\ce{2 CuO(black) + H2O + CO2 -> Cu2(CO3)(OH)2}$ ( found as green mineral malachite ). It can be seen on the old copper roof plating, once shiny metal, now weathered by green coating.
Similar compound
$\ce{2 CuO + H2O + 2 CO2 -> Cu2(CO3)2(OH)2}$  is  found as blue mineral azurite.

@Poutnik thank you for your reply! I’m a total beginner. What does $\ce mean?

it invokes mhchem extension for Mathjax. It probably stands for "chemical equation". See  MathJax

If you are seeing dollar signs, it means your phone does not know how to format the math and chemistry. Try to look at the question using a computer instead.


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What chemical reaction happens on worn out metal handles?
I randomly came across a picture of this. It's the comparison between right handed and left handed people. As more people use the door handle, the metal becomes more faded and turns gold color. What is this gold color underneath and why does friction between your fingers and the door handle cause this reaction?

Consider using more particular titles, this one is too general, being rather a click bait. If you was a writer, you would not probably name your book "What is this story about ?".

I have changed the title to be little more descriptive. Feel free to modify it if you udo not like it.

@Poutnik I probably wouldn't even use "chemical reaction" to describe grinding a metal surface. The question is pretty much the same as to why the chopped water melon being green from outside has red-colored cuts. Not much chemistry to begin with, really.

@andselisk I fully agree with you. But I have intentionally formulated it rather from POV of the OP, wondering about chemical reactions, than from my POV, backup up by some knowledge or experience.

It looks like a brass ( a class of alloys, containing mostly copper and zinc, being golden-like ), which surface was covered by protective plating , possibly by chrome or nickel(*). It was probably worn out mechanically and/or by corrosion due acidic sweat.
(*)Nickel plating could be worn out easier than hard and resistant chrome.

I understand nickel plate is more common on many decorative items today  because of lower "shine" compared to chrome.

@blacksmith37  Could be. I am no specialist here.  But I know Ni plating of Cu goes well by both galvanic and chemical way. I also remember watching Canadian documentary serie about airplane disasters, where in 1 episode the disaster cause was the worn out and snapped stainless steel string, because contained nickel caused faster mechanical wearing out than pilot expected. Routine controls require since than more frequent checks of stainless steel strings in small aircrafts. It was more corrosion resistent for maritime usage, but mechanically more vulnerable.

I watch most similar documentaries ; all produced by people who do not know the technology. Aircraft cables are cold drawn stainless ( 301, 302, 304 )  so an affect of nickel makes no sense,  The cables should be in a relatively dry corrosion free environment inside the aircraft.

@blacksmith37 That specific use was about a small pacific inter-island jumper, so corrosion was an issue, or at least that was reported as the reason why the usual material was not used. It was a naked string, not a cable, if cable means some isolation, which was moving along some guiders, what was mechanically wearing it out until the break.  301 has 6-8 % of Ni, 302 8-10%, 304  8-10.5% ( https://www.meadmetals.com/metal-products/stainless-steel/physical-chemical-properties). What do you mean by "So effect of nickel makes no sense ?"

Maybe a language difference ; steels are not made as "strings"  , steel comes a single wire, or multiple wires braded into cables.  Aircraft use cables so they have redundancy,  then even with a few individual wires broken the cable can still function. Nickel is homogeneously distributed in austenitic SS , so there can be no" failure of nickel". Cold worked 18-8 type stainless would be a poor choice of material for maritime corrosion.  As I said ,the people that produce these shows do not understand the technology.

OK, cables then, I am not native and may not know the jargon. Czech distinguish EN cable as CZ lano , while CZ "kabel" means cable or any similar encapsuled in mechanical/electrical isolation, usually plastic. Sure multiwire cable has redundancy, but as known from bicycle, it can rupture anyway.

It was meant steel alloy containing nickel has increased mechanical wearing out rate. It was not show, it was based on real disaster investigations.

BTW, I have noticed since I started use stainless steel cables for bicycle gearing and brakes, I need to change them more often, compared to non stainless steel.

Let us continue this discussion in chat.


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Why are alkenes considered nonpolar?
Alkenes have C–C sigma bonds, C=C pi bonds, and also C–H sigma bonds.
I know that C–C and C=C bonds will not give any polarity, but what about the C–H bond?
I think it should give some polarity because C is slightly more electronegative than H, so the electrons in the C–H bond will be slightly polarised towards C.
But alkenes are typically referred to as non-polar. Why?

It depends on the resultant of the electrical dipole moments, which anyways aren't so important. Of course you might find something if you look for small things.

"Alkene" is not a compound. You are overgeneralising. Look at specific examples, I recommend you start with the two simplest, ethylene and propylene.

"Non-polar" in chemistry means "not particularly polar", rather than "not polar at all".

Two reasons:

On the grand scale of things, the polarity of a $\ce{C-H}$ bond is very low. The electronegativity difference is less than $0.5$ which is generally the cutoff of a mostly unpolar bond.

Typically when looking at an alkene at large, the various (small) polarities of the different $\ce{C-H}$ bonds cancel each other out or come very close to doing so. In ethene, they cancel out completely for symmetry reasons. In propene, they should likewise cancel out to almost zero total dipole moment. The larger your alkene gets, the less exact this cancelling will be but the general trend remains the same.

A compound or a molecule being polar or non-polar is rather a fuzzy term.
Being non-polar does not mean no polarity of chemical bonds.
It means this polarity is low enough for atoms of such a molecule not to  interact with other molecules by relatively strong inter-molecular interactions. These include strong dipole-dipole interactions, hydrogen bonds or donor-acceptor bonds.


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Why burn methane to CO2 for isotopic analyis?
I am wondering why mathane is burned to CO$_2$ for isotopic analysis, e.g with a mass spec?
I have looked into several text books, they all describe how to do it but not why.
Is it because of a interference on mass 16? Analysing methane it it is would give you also information about the D/H ratio. So why?

There are different types of mass spectrometers with different modes of sample introduction, analyte ionization and ion selection/focusing. You cannot generalize "with a mass spec."
In many cases of isotope ratio mass spectrometry (IRMS), compound specific isotopic analysis (CSIA), or Gas Chromatography Combustion Isotope Ratio Mass Spectrometry organic compounds (not just methane) are burned and oxidized to carbon dioxide and water and analyzed as such to determine the total ratio of carbon-13/carbon-12 and hydrogen-2/hydrogen-1. However, combustion is not the only way to introduce compounds in the mass spectrometer.
Another secondary issue is that there if often more noise at low mass ranges. Methane is a small molecule (m/z =16) and carbon dioxide is relatively large (m/z=44).

Sorry for being not that precise. Let's consider ambient air. So methane is extracted and afterwards burned to CO2 because of noise. With noise you mean mass interferences like H2O?

Noise, means noise overall (from the instrument) not interferences.

You are using improper notations for isotopes. For example, "C-13" should be either carbon-13 or $\ce{^{13}C}.$

I have edited it. I know you like to have perfection in symbolism but one cannot be too orthodox or rigid in notations. Too much rigidity makes it easy to break. Do a Google Book search of "C-13" NMR and you will find book titles with C-13 NMR, and of course plenty of publications.

Well, they are either wrong, or use something like "… C 13" for radiopharmaceuticals. Look, it's not me who came up with the rules, but they make sense to me and allow for a hassle-free communication. Besides, the ACS Style Guide, which is taken exemplary in this community as well as by US publishers of chemical literature, also prescribes these notations. Also, we are still not quite there, the names of chemical elements should not be capitalized unless they are the first words of a sentence.

@andselisk - I think you are too rigid about formatting on this site. This site isn't an ACS publication. I certainly understood what C-13 meant. It sure is a lot easier to input than $\ce{^{13}C}$.

@maxw It doesn't matter whether it is easy or hard if it is wrong. However, please feel free to inform whether there is an applicable standard for writing "C-13". In the absence of one such notation should be treated similarly to a typo or a grammatical error, like "kemztri" in place of "chemistry". See, it is less symbols to type and is likely to be understood correctly in a context, but it is plain illiterate as it is simply not English language. Remember the analogy with the Tower of Babylon I use every time you raise the issue with the formatting guidelines being too complex.

@maxw It's part of my job as a mod to assure literate communication. See, the ACS Style Guide is not a hegemon. I cited ACS since M. Farooq is a resident of the US and these recommendations would be relevant for sure. Other authoritative sources such as IUPAC Red Book say the same, and this nomenclature is also used by ISO and in national standards such as DIN (Germany) or GOST (Russia).

@maxw Those who know the standardized notations around the world won't have to raise their eyebrows and wonder what was happening in OP's head at the moment of posting. That's the true way of making life easier, not accepting everything and being agreeable.

@andselisk, I respect ACS style guide and all other efforts by others to homogenize notations. My point in the comments was that the usage of "C-13" is not that uncommon in the literature even in respected Physics journals published by Americans. There are NMR books whose titles have "C dash thirteen" written on them  (American authors, of course they are field experts and the publisher is Springer). See for example springer.com/gp/book/9789400931619


text:
How do I prevent oxygen interfering when I prepare a FeCl2 solution from anhydrous FeCl2 powder?
I have a bottle of anhydrous $\ce{FeCl2}$ powder (from here; it is 98% pure and was kept in a glovebox after opening it for the first time to minimize reactions with oxygen) and want to make this into a solution. I work in a glovebox, but what I have done prior is that I prepared the reagents outside of the glovebox and then put them inside. I let nitrogen gas bubble through water and then add this water to a certain amount of the anhydrous $\ce{FeCl2}$ powder. After this, I put the solution into the glovebox.
However, I get nowhere near to the green color of the $\ce{FeCl2}$ solutions that I find online. My solution has a slight green hue, and after some time I find a little bit of orange precipitate at the bottom of my reagent. So I fear that there is some reaction between the $\ce{FeCl2}$ and $\ce{O2}$. What can I change about my method to prevent this?
The purpose of my research is to produce the mineral tochilinite ($\ce{6Fe_{0.9}S\cdot5[Mg,Fe](OH)_2}$). I mix the $\ce{FeCl2}$ reagent with $\ce{NaOH}$, $\ce{MgCl2}$ and $\ce{Na2S}$. I do not want any $\ce{Fe^{3+}}$ in my $\ce{FeCl2}$ reagent, because it is also not present in the mineral tochilinite. If it is present in the reagent, other minerals might start to form (such as magnetite).

Is it possible your solid FeCl2 already contains some Fe(III) ? What is purpose of solution ? Can a reduction agens be applied, or Fe(III) hydroxide filtered out ?
Include in  the question all eventual relevant circumstances to prevent wrong assumptions and requests for clarifications.

I've added some information to the text. How would I go about filtering out Fe(III) hydroxides?

Try searching for filtering techniques, e.g. using columns with solid sorbents and consider also centrifuging. Be aware than depending on pH, there can be significant amount of dissolved $\ce{Fe^3+}$. Consider also using FeCl2 . 2 or 4 H2O instead. See https://en.wikipedia.org/wiki/Iron(II)_chloride

You could try mixing it with Fe(0) powder. It will reduce the Fe(III) ions. Also if it does not matter, which I am not sure you could add excess HCl to your solution, since it increases the stability of Fe(II). Best thing you can do if you don't want to add iron powder or HCl is to work in an inert atmosphere, still then there is a chance that there are Fe(III) ions present.

I would add some grains of metallic iron powder on the ground of the flask. Metallic iron reacts with ferric ions, according to $$\ce{Fe  +  2 Fe^{3+} ->  3 Fe^{2+}}$$

See if your group or an other (e.g., organic / organic catalysis) has a Schlenk line.  With some training, this allows you to reduce the interference of oxygen greatly, because you may evacuate your reaction flasks and to replace the air by an inert gas such as nitrogen or argon.  Nitrogen is relatively cheap because it equally is used for welding (as gas) or for cooling (think e.g., NMR spectrometers).  Argon gas, on the other hand has the advantage of a density greater than the one of air, thus you may cover and protect your substance under a blanket of argon.
Then, search for Schlenk flasks.  You identify these quickly for their connector to said Schlenk line. Besside round bottom flasks and tubes (a gallery, a second one), there equally are Schlenk flasks to perform a pressure filtration while working under argon enclosing a filter plate (example).
Then, degas your solvent.  In HPLC, bottled solvents for example are placed in the basket of the ultrasound bath while purging them with a steady, slow flow of argon to remove dissolved oxygen.  Do this for your deionized water.
Weigh in $\ce{FeCl2}$ in your dry Schlenk flask, and add the deionized water prepared under the inert gas.

My variation on the suggested synthesis is not the direct addition of, say, the metal iron (has already been suggested), but instead the partitioned introduction into the reaction chamber of an oxygen scavenger.
For example in one possible embodiment, in place of the direct intervention of Fe in the reactive solution mix, just place a small amount of freshly prepared fine powdered magnesium metal in the chamber, but not in direct solution contact.
My logic is that magnesium metal acting as an oxygen scavenger will anodically corrode (to MgO) with oxygen, thus mitigating the further possible action of oxygen on ferrous ions.
The procedure is then mixing of the FeCl2 reagent directly with NaOH, MgCl2 and Na2S, in the presence of a potential separated side surface reaction between Mg and any present oxygen.
If this fails, try a layer of Mg powder on the reactive mix. However, as in the case of suggested employment of Fe, the resulting chemistry may or may not favor the targeted formation of the mineral tochilinite ($\ce{6Fe_{0.9}S⋅5[Mg,Fe](OH)2}$).


text:
Why is the change in enthalpy computation for a change of state done like this?
Suppose that a system whose initial state is Pressure = $P_1$, Volume = $V_1$; and it is taken through a process after which final pressure and volume are $P_2$ and $V_2$ respectively. Now change in enthalpy is defined as $\Delta H= \Delta U +P\Delta V+V\Delta P$The thing which I am confused with is which pressure and which volume do we hold constant for their respective parts in the calculation? Till now I had dealt only with problems involving the $P\Delta V$ part, and it was calculated as $P_2(V_2-V_1)$, same as irreversible work. But when I came across a problem in which Pressure also varied with volume, the second part needed to be taken as $V_1(P_2-P_1)$.Why do we take $P_2$ as constant in one case and $V_1$ in the other?One reason which I thought of was that in an irreversible process, The pressure is first changed by a finite amount and then the volume varies. Thus the pressure throughout the process was $P_2$. This can be a rationale for the $P\Delta V$ part, but I am not sure if it is reasonable to say that the volume at start of the process was $V_1$, which resulted in this method of calculation. So.. what is the reason?

The expression you should use for a finite change is $\Delta H = \Delta U + \Delta (pV)$

$d(pV)=p.dV + V.dp + dp.dV$, $\Delta(pV) = p.\Delta V + V . \Delta p + \Delta p . \Delta V$. But dp.dV can be neglected with infinitely small error, so it is never explicitly mentioned nor used. $\Delta p . \Delta V$ cannot be neglected. Draw it as geometrical scenario of 1 rectangle consisting of 4 rectangles to see it.

@Pournik See my answer.

You're confused because the equation is incorrect.  It should read $$\Delta H=\Delta U+\Delta (PV)$$If you insist on writing it the way you did, then P should be $\frac{P_1+P_2}{2}$ and V should be $\frac{V_1+V_2}{2}$.  That will give the same result as the first equation I wrote.

Nice analysis. Not obvious on first thought, but if one imagines the geometry of 1 rectangle + 2 trapezoids, instead of 1+2+1 rectangles, it becomes obvious.

Actually, the process path doesn’t have to be straight lines.  The equations I gave are totally algebraic.

It is a state change, independent on path. You just divide rectangle deltap.deltaV and each triangle part add to p.deltaV or V deltap. In geometric picture it is very obvious. (Not really drawn it, just imaginating it )

THANKS.  I thought you were actually saying that they had to be straight lines.  Your assessment is correct, of course.


text:
Basic behavior of ammonia in aqueous medium
Thanks to the reversible reaction
$$\ce{NH3 + H2O <=> NH4+ + OH-},\tag{R1}$$
we know that ammonia acts as a weak base. However, I was wondering why it ends up forming ammonium $\ce{NH4+}$ and not azanide $\ce{NH2-}$ according to
$$\ce{NH3 + H2O <=> NH2- + H3O+}.\tag{R2}$$

Search the pKa of NH3/NH2-

@Waylander I mean at a molecular level. I know that its pKa explains this.

@daviddavid That's what you'd expect in liquid ammonia: $$\ce{2 NH3 ⇌ NH4+ + NH2-}$$

And why can't you expect this in an ammonia solution?

@andselisk Why does ammonia end up being NH4 instead of NH2?

It might, but the equilibrium lies far to the left. Hence, ammonia is not a very effective acid in water.

NH3 much easier accepts a proton than releases one.

As NH2 is a strong base and NH4 is a weak acid we then have a weak base (NH3)?

There are 2 separate conjugated pairs acid/base. What mystery is in that ? It is similar situation as H2PO4- Vs HPO4^2- Vs PO4^3-. . 2 pairs of acid/base as well.

The conjugate acid of a strong base (and NH2- is a strong base) is very weak. And a very weak acid like NH3 doesn't protonate water.

@Karl NH3 is a weak base ...

@daviddavid In your second equation above, it acts as an acid. That's what your question is about.

All what Karl and Poutnik are saying is right. But the real question is : Why ? Why is $\ce{NH3}$ more reluctant to loose one proton than $\ce{H2O}$, at the atomic level ? Why is $\ce{NH2^-}$ such a strong base ?
Let's compare this behaviour in the series $\ce{N-O-F}$. Starting with $\ce{F}$, we can observe that $\ce{F}$ holds one $\ce{H}$, only one $\ce{H}$, and it does not hesitate to give its unique $\ce{H}$ atom (as a proton) to a great variety of bases. The next in line, the oxygen atom $\ce{O}$ can hold two $\ce{H}$ atoms. This is one more than what does $\ce{F}$. Well, $\ce{O}$ in $\ce{H2O}$ is also more reluctant to loose one $\ce{H}$, and accepts to give it (as a proton) only to strong bases. The third in line is the nitrogen atom $\ce{N}$ which can hold three $\ce{H}$ atoms. It's still more than what does Oxygen. But $\ce{N}$ in $\ce{NH3}$ is still more reluctant to loose one of its three $\ce{H}$ atoms as a proton $\ce{H+}$. It even prefers stealing one $\ce{H+}$ ion from the neighborhood rather than giving it. Apparently the more $\ce{H}$ atoms the central atom has, the more it wants to increase this number. It looks like the human society : the more money you've got, the more you want to increase your fortune.
Edit :
Of course it would be interesting to see if this tendency $\ce{F-O-N}$ can be extrapolated to the carbon atom. If my reasoning is correct, the Carbon atom in $\ce{CH4}$ should be extremely reluctant to loose one $\ce{H+}$ ion for producing $\ce{CH3^-}$. Indeed I am not sure that the ion $\ce{CH3^-}$ does exist, and has a measured $p\mathrm{K_a}$ value. It should be extremely high. On the other hand, $\ce{CH4}$ should be eager to get one more Hydrogen. However the electronic situation is not similar to $\ce{F, O}$ and $\ce{N}$. In $\ce{HF, H2O, NH3, F^-, OH^-, NH2^-, H3O+, NH4^+}$, the central atom has always the electronic state of Neon. It is known that $\ce{CH4}$ does produce $\ce{CH5^+}$ ion in high pressure mass spectrometry. But the electronic structure of the carbon atom in $\ce{CH5^+}$ is unknown, to my knowledge. So I am not sure whether the tendency, valid for $\ce{F-O-N}$, can be extrapolated to the Carbon atom.

@KarstenTheis Methane is CH4 ...

You've only said that it is more reluctant to lose it but you haven't actually said why.

@David. The only reason is the result of experiment.

@Maurice I don't think you get it. Results have to be reasoned theoretically and checked practically (with experiments). I am asking for the theoretical explanation.


text:
Total Entropy of the Atkinson Cycle
I was researching for a question that asked me to find a ratio of temperatures, and I came across this, which would help me greatly:
Since, most of the temperature changes occur during the isochoric (2–3) and isobaric processes (4–1), a constant specific heat can be assumed during the isentropic processes (1–2 and 3–4).
The total entropy change in the cycle must be equal to zero, therefore
$T_2T_4^k=T_3T_1^k$
where k is the rerversible adiabatic index.
But, I don't understand why due tho these factors this equation is true. Is there any clarification as to why this is the case?

For readers - see Atkinson cycle

@Poutnik Apologies, was not aware, I have now edited the question

Let’s see your analysis so far.


text:
Do different eigenstates of total angular momentum have necessarily different energies?
Let $H$ be the Hamiltonian of a specific atom and $J$ the total angular momentum. Since $H$ and $J$ commute, they have common eigenstate.
So we can label the atomic states by their energy and total angular momentum $\phi_E^J$.
My question is, suppose we have a state $\phi_E^J$ and another state $\phi_{E'}^{J'}$,  do we have necessarily that $E\neq E'$, that is, do states with different total angular momentum have necessarily different energies?

You might have heard of hydrogen atom...

I  mean atoms different from hydrogen

Given a fixed momentum of inertia, the rotational kinetic energy of a system is monotonically related to the the magnitude of the angular momentum. Therefore, if the system has different angular momentum, it has different energy.

The magnitude of the angular momentum is  $J=\hbar\sqrt{j(j+1}$ and energy is momentum squared divided by twice the mass (or moment of inertia for rotation) then energy is $J^2/(2I) $ or $\hbar^2 j(j+1)/2I$ so depends on $J$, so different $J$ different energy. This is just the expression for particle on a sphere or rigid rotor.

well we could have accidentally symmetry

Accidental energy would only be the case if an external force acted (then it would only be approx the same, not identical), and symmetry is not accidental, because any symmetry would have to be taken into account in any proper calculation.


text:
IR absorption chart
I am taking a course in structural determination of organic compounds. Right now, we are giving the IR spectra.
I am looking for a table, as detailed as possible, of the absorptions. I hope you can help me.

Your best bet is the internet...

Google is your friend!   http://www2.ups.edu/faculty/hanson/Spectroscopy/IR/IRfrequencies.html

I recommend you stick to the table your prof gave you, because that is the one that will have to do when you have your exam. Otherwise Hesse, Meier, Zeeh surely has an English edition nowadays.

Thanks @Karl for bringing this book into my attention. It is highly rated on Amazon. Will try to buy it one day.

@Karl Of course there is (some of the entries here). I like to complement HMZ with Ernö Pretzsch's compilations from the other school in Zurich, too (Springer) and still have an edition with the CD.

The more detailed such a table is, the more time you will spend on it. Especially during a written exam, this is not good for you. In addition, a beginner's course intends to teach you concepts about the experiments, and plausible results. In an exam, it may be much more useful to extract from the IR spectrum «plausible hints about an aromatic system, a halogen and  nitrile group» and to complement these data by MS/NMR (e.g., o/m/p substitution) like in a typical synthesis lab. Despite a good IR spectrum may tell you substitution patterns (not so much the low resolution copies in a class).

A very nice comprehensive set of IR absorption group frequencies were compiled seventy decades ago in the Journal of Optical Society of America by Colthup, 1950, 40, 397-400.
It is freely available if you search "Spectra-structure correlations in the infra-red region" by Colthup in Google Scholar.
You should memorize some standard frequencies of very common functional groups.


text:
How is Potassium present in Potato (form) and where does it come from originally?
I was going through the concept of Boiling to reduce the potassium in Potatoes, which means $\ce{potato + water +heat -> potassium + X}$.  $\ce{X}$ is potato with reduced potassium. I am more interested in the composition of Potato, which I found to be close to starch $\ce{(C6H10O5)n}$. Then I started wondering how the Potassium is present in Potato (in which form) and from where it would have come from? (the soil itself?) But I researched and found that we use Potash nowadays, which has good Potassium.

Potassium is present in many living things. You have potassium in you, yet are not mainly potassium by a long stretch.

Potassium ions are natural part of electrolyte within animal and plant bodies.

Whoever told you that boiling a potato can reduce potassium has told you a fake story. You cannot evaporate potassium ions.
Potassium ion also comes from the natural soil minerals and it exists as potassium ion in water and in plants. I am sure you know that potato is an underground stem tuber. The sugars/starch come from the carbon dioxide in air. Many fertilizers contain potassium, so it is not surprising that K also comes from there.
Keep in mind that it is extremely hard to determine the total molecular composition of any natural stuff especially plants (thousands of compounds...)
Update: The OP presented a link from the Journal of Food Science (J Food Sci
. 2008 Jun;73(5):H80-5. doi: 10.1111/j.1750-3841.2008.00782.x.) with the title "The effects of boiling and leaching on the content of potassium and other minerals in potatoes"
The results are obvious, if you boil potato chunks (not the whole potato) in water, you will dissolve a fractional amount of potassium (ions) in water. You will have to throw away the water. The same should be true for sodium and other water soluble minerals. Potassium is not a surprise.

Maybe the OP wasn’t referring to evaporating potassium through boiling, rather, it was the removal of potassium through diffusion from the potato to the boiling water.

Probably, OP has clarify. One would rather worry about sodium rather K+, which is good.

Perhaps, as potassium is an active element in cardiology , OP may consider potassium daily intake.

@dval98 AFAIR ions are too large to undergo diffusion through cell walls. They are taken in by the plant via Active Transport

@AniruddhaDeb Absolutely, but the point I’m making is that’s what the heat is for. Boiling can break down the cell walls to allow ions to come out.

@M.Farooq Well, Interesting. Thank you for answer. Would it be possible to determine the molecular composition of just the part where Potassium exists? or would that be hard as well? (Also that boiling can reduce Potassium. which is proven through research and measurements - Example - https://pubmed.ncbi.nlm.nih.gov/18576999/ (there are plenty).

@dval98 So, Potassium is present as ions? (K+) inside Potato?

@sundar yea it exists in its +1 oxidation state. Potassium metal (0 oxidation state) reacts violently with water. It would not persist in a biological system.

@sundar Rather, potassium is present as a ion K+ everywhere in biological or geological material.


text:
Stereoselectivity of ring closure in intramolecular iodolactonisation

What is the product of the reaction above?

I am mainly confused between the following two options:

How do we decide the resulting stereochemistry of the $\ce{CH_2I}$ group?
The answer given is the one in which $\ce{CH_2-I}$ is coming out of the plane, i.e. (a).

The results of the iodolactonization example presented in this question were published in 1982 by Bartlett, et al.1 While @Aniruddha Deb has captured the essence of the reaction, there is more than one product formed in the reaction. A more detailed mechanism is offered, an issue not addressed by the original authors.
Carbonate 1 is a racemate. No products derived from reaction at the remote 7,8-double bond were detected. Understandably, the rate of closure is faster for a 6-membered ring than a 7-membered ring. A transition state for ring closure for a reversibly formed iodonium species will be favored having the carbon chain R equatorially disposed in the transition state. Transition state 2 is favored over that of 5 because the latter encounters a gauche butane interaction. In the iodonium cation the C-C bond is shorter than the C-I bond thereby having the former of greater steric consequence than the latter. SN2 attack on the iodonium cation is favored by the more polarized carbonyl oxygen than the oxygen attached to the t-butyl group. In the latter instance, establishing a positive charge on this oxygen is counterproductive because it is adjacent to a partially positive carbonyl carbon. This objection is in keeping with why amides are preferentially protonated on oxygen and not nitrogen. The cations 3 and 6 are now poised to lose a t-butyl cation, which, given the solvent acetonitrile, may lead to Ritter intermediates besides t-butyl iodide. In the event, racemic cyclic carbonates  4 and 7 are produced in 69% yield with the cis-substituted isomer 4 predominating over the trans-isomer 7.

P.A. Bartlett,  J. D. Meadows, E. G. Brown, A. Morimoto and K. K. Jernstedt, J. Org. Chem., 1982, 47, 4013. https://doi.org/10.1021/jo00142a002

Without further details, I would say that both products are possible.
The reaction that you have given here is an Iodolactonization reaction. You have correctly identified that the tert-butyl group would be removed and that the oxygen atom would prefer to form a 6-member ring. In terms of stereochemistry, however, it is tough to make a decision, as:

The stereochemistry of the ester group is not known
There are no more details on the reagents or reaction conditions.

Another note on the stereochemistry is that The iodine group would always be opposite to the attacking oxygen as this is an SN2-like reaction. Keeping these in mind, we can perform the reaction for the two different enantiomers of the substrate and see that both products are possible.

You are right that the oxygen must be anti to the iodine leaving group. But you haven't actually answered the question: what you have drawn are two enantiomers of the same diastereomer, whereas OP is asking about which of the two diastereomers are formed. You're not considering the alternative arrangement of the iodonium group, which leads to the other diastereomer. See: https://i.stack.imgur.com/ZReeA.png. The top is what you've drawn, the bottom is the other pathway which you need to consider. It's better to draw them as chair conformations...!


text:
What kind of bond exists between a metal and a non-metal with a difference in electronegativity of less than 1.7?
I'm being taught that the kind of bonds that exist between elements depends on the electronegativity difference between the elements. A difference less than 1.7 is covalent and a difference higher than 1.7 is ionic.
Separately, I've been taught that bonds between metals and non-metals are ionic. What happens when these two 'rules' disagree? When Magnesium and Phosporus bond for example.
I understand that there isn't always a definite border between covalent and ionic bonding, but what should my rule of thumb be? That bonds between metals and non-metals are always ionic, and that bonds between non-metals are covalent unless there's a difference of more than 1.7?

In principle, it should be possible to exhaustively list all (simple stoichiometric) binary compounds (about 20 thousand) and all the electronegativity differences between elements (about 10 thousand), then parse the results to at least qualitatively determine which rule of thumb is more accurate. Just for fun, I've set up a spreadsheet with Pauling electronegativity differences, and have coloured the combinations which result in combinations "on the fence" of the (somewhat arbitrary) value of 1.7. Some further processing might be enlightening.

Though I do think there is something to be learned from digging into the data, I must reiterate the fuzziness of what makes a compound "ionic" or "covalent", and that the rules of thumb are another layer of fuzziness on top of that. One rather quickly realises such simple descriptions are too superficial.

https://chemistry.stackexchange.com/questions/9222/why-are-bonds-ionic-when-the-electronegativity-difference-between-bonded-atoms-i


text:
Thermal properties of aluminium alloy 6063
I am looking for the operating temperature range both in a cold and hot environments (minimum and maximum operating temperatures) for aluminium alloy 6063.
I'm new to finding this kind of information efficiently, and I would appreciate a reference document as well.

Alloy properties are available in handbooks, such as ASM references. They will not give you an answer, since you need to analyze your design under the given data to determine if your design will perform within your requirements.

Or in otherwords: you need to hire a qualified engineer. ;)

Telling someone to hire a qualified engineer doesn't contribute anything to answering the question.

Well, the question behind looking for these data has not been formulated....

American Society for Metals has many publications. Temperatures below ambient are no problem for any aluminum alloys. They are commonly used for natural gas  separation liquefaction exchangers and piping down to −100 °C (guess at the temp.).
In the T6 condition (hardened) room temperature yield is 31000 psi; at 300 °F the short term yield is 20000 psi; 6500 psi at 400 °F, and falling fast as the temperature raises.
ASME Boiler Code will have allowable stresses which will be lower than these values. I looked in ASM Materials Handbook vol 1, 1961, it is probably not online. If this is long term, you will want creep or stress/rupture data which is not commonly available for aluminum; try in the aircraft industry.

You need to define what an "operating temperature" is, for what application.
An alloy that will work perfectly fine as a structural material for your kitchen oven will fall apart if used in aerospace applications.
Your best bet is to find alloys used in similar applications and argue, by analouge, that these alloys will also work with your application.
That said, experiment is king, and if there's any sort of safety risk, you need to investigate it before moving forward.


text:
Rate law for A + B → C → P
I need some clarification for the following assignment:

Derive the rate law for
$$\ce{A + B -> C -> P}$$
when $\ce{A + B -> C}$ is the slowest step and very slow.

My understanding is $\ce{A + B -> C}$ is the rate-determining step, so the rate law would just be
$$\mathrm{rate} = k[\ce{A}][\ce{B}].$$
I'm not sure if the question is asking about steady state approximation (SSA) or just a simple rate law. I know SSA happens if you assume the second step $(\ce{C -> P})$ is faster than $\ce{A + B -> C},$ meaning  $\ce{A + B -> C}$ is the slow step. I'm not so sure if my understanding of the question is correct.

Somewhat related: https://chemistry.stackexchange.com/q/83631/30836

The rate is determined by the slowest step, as you correctly assumed. The transition state of C->P is assumed to occur significantly faster than the first step in Organic Chemistry at least.
Unless you have an energy diagram for the reaction or any Le Chat, I think you have it right as rate = k[A][B].


text:
How does pH affect the spontaneity of biochemical processes?
A decrease in $\mathrm{pH}$ increases the hydrogen ion concentration, thereby decreasing $Q,$ and decreasing Gibbs free energy as mathematically expressed:
$$\mathrm dG =\mathrm dG^\circ + RT\ln Q$$
This means that the more acidic the environment is, the more spontaneous the reaction is. But don't biochemical processes operate at approximately neutral $\mathrm{pH} = 7,$ so enzymes function optimally (does not denature)? But why does the equation favor spontaneity at (for example) $\mathrm{pH} = 3$ than $\mathrm{pH} = 7?$
For specific case, such as the biological oxidation of ethanol? Does it mean that it will thrive at acidic environment based on the equation? But how can we explain for the specificity of biological reaction operating at their optimal pH ranges?
For example, in the stomach it requires an acidic environment so the equation holds true. But what about for other biochemical processes that operate at neutral pH (the equation cannot explain it)? Or is there actually another equation to utilize to explain how spontaneous biochemical reactions can be?

Provide more details and background, the best if on a particlar example.  Spend more time on clear and precise question formulation and by own searching for possible answers. Note that pH effect is reaction specific.

So this simply means that the effect of pH cannot be generalized, right? I was just looking for a way to explain spontaneity through the equation. But it seems it cannot account for the specificity of biological systems.

Remember that cells have organelles that can be held at different pHs. For example, the mitochondria creates a pH gradient that is used to make ATP. Also, the lysosome has a pH near 5, allowing it to better catalyze certain reactions.

Note that pH effect is reaction specific.
If a reaction is e.g. $\ce{A + B + H+ <=> C + D+}$ , or $\ce{A + B- <=> C + D + OH−}$,
then lowering pH supports the spontaneity.
If OTOH, the reaction is $\ce{A + B + OH-  <=> C + D−}$ , or $\ce{A + B+ <=> C + D + H+}$,
then lowering pH goes against the spontaneity.
It can be generalized by applying the Le Chatelier's principle, saying systems adapts to external changes by such equilibrium shifts that partially decreases the system chage caused by the external factor.
Lowering pH supports spontaneity of reactions consuming H+ and vice versa.
Increasing temperature prefers endothermic reactions consuming heat and vice versa.
Increasing pressure prefers reactions decreasing the total molar amount of gaseous components and vice versa.

Note that biological systems are much more complex than chemical systems.
Lowering pH may support thermodynamic spontaneity of some particular reaction. But an enzyme catalyzing its kinetic rate may start loosing its activity, get denatured, cells may stop producing the enzym, cells may stop overall life functions, or may die.


text:
Hydrolysis of Magnesium Carbide
I came across a question in which the following reaction was taking place:
$$
\ce{Mg4C3 + H2O -> CH3CH \bond{#} CH}
$$
Each $\ce{Mg}$ atom must be present in a +2 oxidation state, thus the total charge of the cation is +8. We thus have a $\ce{C3^{8-}}$ anion. The only structure of this $\ce{C3^{8-}}$ anion I could come up with was three carbons attached to each other in a linear fashion, with the terminal carbons each bearing a -3 charge and the central carbon bearing a -2 charge.
From this anion I would have assumed that the product formed upon hydrolysis is propane, but the product formed in the above reaction is propyne, just like when $\ce{Mg2C3}$ undergoes hydrolysis.
I'm guessing that the explanation for propyne being formed is the actual lattice structure of $\ce{Mg4C3}$, but I couldn't find any literature referencing this reaction or the structure of $\ce{Mg4C3}$.
The question is taken from:

V.K. Jaiswal, s-block elements, Problems in Inorganic Chemistry, Thirteenth Edition, Page 331, question 36

@Poutnik agreed, but if the 3 carbon atoms don't form a single anion, how will we get a product which consists of 3 interconnected carbon atoms?

The polyatomic ion C4−
3, sometimes called allylide, is found in Li4C3 and Mg2C3. The ion is linear and is isoelectronic with CO2. The C-C distance in Mg2C3 is 133.2 pm. Mg2C3 yields methylacetylene, CH3CCH, and propadiene, CH2CCH2, on hydrolysis, which was the first indication that it contains C4−
3.   https://en.wikipedia.org/wiki/Carbide

@Poutnik so are you claiming that $\ce{Mg4C3}$ consists of the $\ce{C3^{4-}}$ anion? Then how is the charge balanced, since Mg is in a +2 oxidation state, and thus the cation has a total charge of +8?

I have just quoted Wikipedia, I have not claimed anything. Note Mg2C3, not Mg4C3.

Typo? Should it be Mg2C3?

Yeah, seems a blatant mistake.

I don’t know if this is enough to prove that this is not a typo but this paper mentions Mg4C3

Probably they mistyped Mg2C3.  Lithium, with its diagonal relationship, does form a Li4C3 carbide with hydrolysis characteristics similar to Mg2C3.

Assuming the reagent is Mg2C3, one way to visualize the reaction can be as follows: Purely on oxidation state basis, each Mg will be charged +2, with 2 Mg atoms you get +4 positive charge. That means the anionic end has three carbon atoms required to bear a combined oxidation state of -4. Check it yourself that one possible way to have that is to make a cyclopropene-like moiety (C3H4). The two doubly-bonded carbons can bear a -1 charge, while the apex carbon can bear a -2 charge.

This cyclopropene-like moiety is extremely unstable due to the unbelievably high ring strain. Hence, in the presence of water, the ring will quickly open-up and hydrolyse by picking up protons from the water. This will straight-away give you propyne.


text:
Are Alloy Compositions Given in Weight Percent or Atom Percent
Often times I see alloys described like this,

Lockalloy is an alloy that consists of 62% beryllium and 38% aluminum

Are the precentages mentioned here in weight percent or atom percent?  Is there a standard somewhere that governs this notation?  I don't see this mentioned anywhere in the literature, how does everyone know when people are talking about weight or atom percentages?

Sometimes they will be nice and indicate ’wt.%’ or ‘at.%”, otherwise you have to look deeper at the document to see, or another source to compare. Metallurgy often uses weight percent, materials science atomic percent.

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwiGgbTP2f3sAhWaKs0KHWs5Ag0QFjACegQIAhAC&url=https%3A%2F%2Fmaterion.com%2F-%2Fmedia%2Ffiles%2Fberyllium%2Falbemet-materials%2Fmaab-006developmentofaluminumberylliumforstructuralapplications.pdf&usg=AOvVaw2njAxQKRimUzRiDZJelTYS would indicate it is by weight : "Aluminum-beryllium (AlBeMet® AM162) is a powder metallurgy composite of Beryllium 62% by weight and 1100 series pure aluminum, 38% by weight."

Engineers think weight per-cent . I can hardly remember  seeing atomic % , no doubt in some research paper.

Whenever it is left unspecified, it is weight percent. I don't know of an official standard, but it is a very common tradition.


text:
How do I find the pressure that would be present in carbon dioxide for this problem?
Problem:
For the reaction below, Keq = 0.0064 (at 825C).  If you place 12.011 g Carbon and 5 atm pressure of water vapour in a 2.0 L reaction flask, calculate the pressure of carbon dioxide gas that will be present once the reaction comes to equilibrium.
C(s)  + 2 H2O(g)  ↔  CO2(g)  +  2 H2(g)
Here is what I have done so far:

\begin{array}{l}K_{eq}=\frac{P\left(CO_2\right)\cdot P\left(H_2\right)}{P\left(H_2O\right)}=0.0064\ \ \left(at\ 825\ \text{°C}\right)\\\frac{\left(x\right)\left(2x\right)}{\left(5-2x\right)}=0.0064\\\frac{2x^2}{5-2x}=0.0064\\2x^2=0.0064\left(5-2x\right)\\2x^2=0.032-0.0128x\\2x^2+0.0128x-0.032=0\\x\ =0.1233\\∴\ \text{The pressure of carbon dioxide is 0.1233 atm}\end{array}
I have gotten the answer to be 0.1233 atm but the answer is supposed to 0.3 atm. So, did I do something wrong here? Could someone help me out here if possible?

Using photos/screenshots of ( even handwritten) text instead of typed text itself is strongly discouraged. It is impossible to index/search/reuse it in answers referring to it. In a  case  of handwritten text, it puts extra burden on responders to properly decipher it.
That all may lead to the question being ignored or even closed.
Consider copy/paste or retyping and using eventually  MathJax for expressions and formulas.

Similarly as in your prior question, you ignore power coefficients of pressures in the equilibrium expression.

What power coefficients are you implying? @Poutnik

Yes, Sir, I know the general formula for the equilibrium constant as evident from my answer, but the idea that I want to know is how the answer is 0.3 atm and where did I make a mistake. I am sure it is a very simple mistake that I made, so if you could point that out to me or anyone else, that would be great. Likewise, knowing what I did wrong here, I can have a better understanding of how to do questions like this in the future. I hope you understand. @Poutnik

No, you do not. Think and search again.

Sir, the formula for the equilibrium constant is evident at the first step of my solution. If there is anything wrong with that, let me know. @Poutnik

You should stop asking and start searching.  https://en.wikipedia.org/wiki/Equilibrium_constant

Okay, Sir, I will start searching... But really I just want to know that little mistake that I made.

Or here https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/The_Equilibrium_Constant

If you find your mistake yourself, it gives you more, than if you are just told. BTW, the site prefers teaching fishing to giving fish.

Yes, I understand, Sir, as you said, "I should stop asking and start searching". That was what I was doing some time ago, but I greatly appreciate you giving me a head start. Thanks.

@JabanBaroose the term $P_{\ce{H2O}}$ in the denominator of the equation should have a power of 2, since the coefficient of $\ce{H2O}$ in balanced chemical equation is 2.

@Eyyboss Thank you for your reply. I have already got the answer that I was looking for as it was a very simple error that I made. Since Poutnik here couldn't help me out, I asked an online friend of mine who really helped me clarify the mistake that I made. But I greatly appreciate you for giving me an answer, even if it is too late.

I'm not sure if I'm late to answering your question but I had the same problem with my calculations and I figured out what I did wrong. You forgot in your Keq equation the exponents: (x)(2x)^2/(5-2x)^2 = 0.0064 and then you square root both sides and go from there until you get a quadratic equation and like you did before just use the quadratic formula to solve for x. Hope this helps


text:
Aluminum Monoacetate Solubility in Aqueous Medium?
Does anyone have access to the aqueous solubility value of  aluminum monoacetate?
I would expect it to be slightly soluble but I've searched the internet and haven't found any solubility values. : /

I re-edited the link and it seems to work for me. From wikipedia: "Aluminium monoacetate, also known as dibasic aluminium acetate, and formally named dihydroxy aluminium acetate, is a salt of aluminium with acetic acid. It has the formula Al(OH)2(CH3COO)"

Good. Now the compound is legitimate, after all. Well, I also expect it to be slightly soluble.


text:
Why is boron inert to acids and bases?
I was reading my school text book and in the segment related to the reactivity of group 13 elements with acids and alkalis it was written:
Boron does not react with acids and alkalis even at moderate temperatures
Earlier it said that it is weakly acidic so I attributed that to it's non reactivity with acids but I can't seem to find any reason why it doesn't react with bases.

I was more interested in the reason but thanks

That's not the way life works. Negative assertions don't require explanations; positive do. Why didn't Donald Trump climb Mount Everest? Er, well...

Ok ok i got your point XD, but i can't say that if a question is asked about it

Crystalline boron is quite similar to diamond. I doubt you'd ask the same about it?


text:
"Bromobenzene" vs "bromophenyl"
What is the difference between the keywords Bromophenyl and Bromobenzene used in organic chemistry.

Phenyl is an alkyl from benzene, the same as ethyl from ethan. I admit, more logical alkyl name of benzene would be benzyl, but it is already historically reserved for the phenyl-CH2-  alkyl. Bromo- is just a prefix for bromine substituent.

@Poutnik  do you mean we use "phenyl" when the ring joined as a alkyl group to the parent chain

Anywhere where you would use ethyl instead of ethan, you use phenyl instead of benzene. As benzene is $\ce{C6H6}$, while phynyl is $\ce{C6H5}-$. By other words, phenyl is benzene with 1 H removed, making it a single bond substituent.

Bromobenzene is a complete compound, bromophenyl is a description of a unit in a larger compound
Chemists abbreviate some groups to simplify the names of more complex compounds. So while bromobenzene is a complete name for a compound, bromophenyl signals that the unit occurs as part of a larger compound.
In this case–to abuse strict chemical terminology a little–there is a bromobenzene unit attached to to 2 carbon on an ethanol unit and attached at the 2-carbon on the bromobenzene unit so we say 2-bromophenyl not the more complicated "bromobenzene with the hydrogen at the two position replaced by an ethanol unit with the connection being made at the 2-position on the ethanol."


text:
Looking for a non-toxic, environmentally safe highly colored or intense dye
I need a dye that is "highly colored" or intense.  I want to put it in water and flow the water in the form of a liquid sheet that is around 100 microns thick.  A number of spray nozzles use this principle to produce droplets.  The liquid is first formed into a sheet that breaks down after a few milliseconds into droplets.
The issue is that at such thin thicknesses, colors often disappear even if they look very intense in a tank.  I'll be working with around 0.2 gallon per minute to 0.5 gallon per minute flowrates, so something that hopefully can be added in relatively small quantities.  But more important is that the dye is non-toxic and environmentally safe.
Thanks for any suggestions.

There is many env-safe-nontoxic dyes, but If by env-safe is meant easily degradable, it gives limited options. Hot candidates are plant dyes, like anthocyanins, but it may be difficult to separate them from plant matrix.  It may be possible to bye purified ones.

I can't address the environmental issues, but would a fluorescent dye work better? You could illuminate your device with UV light, and the emitted light should show up better than a non-fluorescent dye.

Go to the next supermarket, and buy some of these dye solutions specially made for food coloring.

@user137  I've considered fluorescent dyes, but I'm going to be shooting high speed video (10000 fps or more) and a lot of light is required.  Are you aware of any fluorescent dyes that fluoresce under LED light?

You might try fluorescein, a commonly used tracer fluorescent dye. It will fluoresce very intensely when excited by UV LEDs, such as the ones I mentioned here. Just use a UV blocking filter (or long pass filter) so you see the green fluorescence and reject the UV excitation light.

@Ed V.  I need massive amounts of light because of the high-speed video and I might be using macro lenses as well. I'm not sure that UV LED, which I presume requires a dark ambient will work.  Isn't fluorescein used as a tracer in bodies of water?  Doesn't it provide some degree of fluorescence in natural sunlight?

I did some checking and you can get UV LED flashlights with 100 LEDs. Ebay has them cheap. Typically, these are 365 nm or 395 nm. These are serious light sources. And, yes, sunlight works fine and you can use a simple Fresnel lens to concentrate the sunlight. Or use a light pipe: a glass rod drawn down somewhat to act as a light funnel. If I remember correctly, some commercial cleaning products have fluorescein in them. Best of success with your project! (I think I will buy a couple of those UV LED flashlights and some fluorescein.)

Many organic dyes which are fluorescent can be carcinogenic like the famous Rhodamine dyes. You can try a couple of tricks. One of best inks, at least for demonstrations, is the yellow highlighter's ink which fluoresces bright green in water (see my photo below). If you can shine a bright white LED or a bright blue LED, it will fluoresce very strongly. You can use an optical filter in front of the camera that can block blue light and let the green and longer wavelengths in. With that you can increase the sensitivity.
Note that fluorescence intensity is directly proportional to the incident light intensity in dilute solutions.

@ M Farooq:  Thank you for your response.  So it sounds like fluorescein should provide a strong fluorescence under a bright white LED light. I'll be using a 300 watt equivalent LED light bulb.

Yes. Fluorescein is best in alkaline media. Better to use optical filters, shine blue light over the liquid and block blue light on the camera. You need cut off filters.

This is a good time of year, a few weeks past Halloween, to buy heavily discounted “glow in the dark” chemiluminescent light sticks. Carefully cut open, they have the various fluorescent dyes used to produce the light stick colors. The old ones had the dye in the glass ampoule: I still have three of them. Newer light sticks may have the oxidizer (hydrogen peroxide) in the ampoule. Either way, the dyes are available for testing with whatever set up is ultimately used. For fluorescein, I would use the disodium salt. And the filters are essential.


text:
What happens when unsaturated primary alcohol is reacted with acidic potassium dichromate?
It is clear that the hydroxy group in the substrate gets oxidised to carboxylic acid. But what happens to the double/triple bond in the unsaturated primary alcohol when it reacts with acidic potassium dichromate?


text:
What is the optimal voltage for storing a 3.7 V lithium polymer battery?
According to this website, storing a lithium polymer (LiPo) battery/cell at around 40% capacity would be more beneficial for the lifespan of the LiPo cell comparing storing the battery at 100% capacity. The author also stated that "it is best to store LiPo batteries at 3.6 V to 3.8 V. This is applicable for standard LiPo batteries that hold 4.2 V per cell when fully charged."
However, the instruction of an RC helicopter (with a LiPo battery (3.7 V (4.2 V when fully charged), 180 mAh capacity)) I bought stated that if I am not using the model, I need to make sure that its voltage is at least 3.9 V at storage.
The manufacturer, upon inquiry, told me that I should store the battery fully charged and only recharge it after the voltage drops below 3.9 V. I.e., my correspondent's answer and the instruction of the helicopter is quite different from the website mentioned before.
Who is the right one? The website or the instruction and the manufacturer, according to the working principle of LiPo battery?
Some other questions relevant for the maintenance of the battery (question added that were not asked originally):
Would the charging frequency be a determinant for battery lifespan? Usually how fast would the battery lose voltage by self-discharge (is it dependent on the voltage the battery has)? Also, when charging, does the charge in the LiPo cell build up linearly (so that I will know when is 40% charge reached)? Would it be better if I do not charge the battery fully even right before using the battery in the RC helicopter?
There is a undervoltage cut-off for my RC helicopter. What is usually the percentage of the charge at which the undervoltage cut-off is reached?
A/two question(s) out of curiosity:
Would the time needed for charging a completely new and empty cell to full for the very first time be different from the time needed for charging an already-used empty cell (maybe not completely empty, since the RC helicopter has undervoltage protection) to full? What is the difference between these two charging process?

The $\pu{3.6-3.8 V}$ range is a good general choice, but it may be battery-specific. The particular voltage for 40% charge may differ for different cell technologies, e.g. various deviations of electrode materials and due to cell aging.
The optimal storage conditions, as you mentioned, are more often expressed as charge/capacity % ratio. Usually, the optimal storage conditions are said to have about 40% of the maximal charge. For longer term storage, it is advised to be kept in a fridge, but not to get below
$\pu{0 ^{\circ}C}$. Note that by the semi-empirical van 't Hoff's rule(*) chemical reactions (= battery deterioration) speed up typically 2-4 times per temperature increase by $\pu{10 ^{\circ}C}$.
Lithium-ion/lithium polymer (Li-ion/LiPo) cells reportedly lose typically 20% of capacity per year when fully charged, but just 4% at 40% charged state.
For practical purposes, sometimes a trade off is made between battery lifetime and battery readiness.

Some background about Li-Ion/LiPo cell chemistry:
The negative anode is formed by lithium graphite intercalate $\ce{Li_\mathrm{x}C6}$, x<=1
The positive cathode is formed most commonly by $\ce{Li_xCoO2}$
The electrolyte is most commonly formed by $\ce{LiPF6}$ dissolved in mixture of dialkylcarbonates $\ce{R1-O-CO-O-R2}$ (**)
About electrode reactions from Li-Ion wikipedia article:

The positive electrode (cathode) half-reaction in the lithium-doped cobalt oxide substrate is
$$\ce {CoO2 + Li+ + e- <=> LiCoO2}$$
The negative electrode (anode) half-reaction for the graphite is
$$\ce{LiC6 <=> C6 + Li+ + e^-}$$
The full reaction (left to right: discharging, right to left: charging) being
$$\ce {LiC6 + CoO2 <=> C6 + LiCoO2}$$
The overall reaction has its limits. Overdischarging supersaturates lithium cobalt oxide, leading to the production of lithium oxide, possibly by the following irreversible reaction:
$$\ce {Li+ + e^- + LiCoO2 -> Li2O + CoO}$$
Overcharging up to 5.2 volts leads to the synthesis of cobalt(IV) oxide, as evidenced by x-ray diffraction:
$$\ce {LiCoO2 -> Li+ + CoO2 + e^-}$$
In a lithium-ion battery, the lithium ions are transported to and from the positive or negative electrodes by oxidizing the transition metal, cobalt ($\ce{Co}$), in $\ce{Li_{1-x}CoO2}$ from $\ce{Co^{III}}$ to $\ce{Co^{IV}}$ during charge, and reducing from from $\ce{Co^{IV}}$ to $\ce{Co^{III}}$  during discharge. The cobalt electrode reaction is only reversible for $x < 0.5$ (x in mole units), limiting the depth of discharge allowable. This chemistry was used in the Li-ion cells developed by Sony in 1990.

In addition to the above, there is another deteriorating process, related to keeping Li-Ion/LiPo cell long time at fully charged state ( typical for laptops permanently connected to AC, unless special coutermeasures are involved ). There is a side tendency of lithium ions in the highly charged cell state to form metallic lithium at surface of graphite-lithium intercalate. Normally, this is reversible. But there is ongoing slow side reaction with the dialkyl carbonate solvent:

The Reaction of Lithium with Dimethyl Carbonate and Diethyl Carbonate in Ultrahigh Vacuum Studied by X-ray Photoemission Spectroscopy
Abstract: The reaction of dimethyl carbonate (DMC) and diethyl carbonate (DEC) with clean metallic lithium in ultrahigh vacuum was studied by the use of X-ray photoelectron spectroscopy with the temperature-programmed reaction methodology. Both molecules are of interest as solvents in ambient-temperature lithium batteries. The solvent molecules were condensed onto the surface of an evaporated lithium film at $\pu{120 K}$, and spectra were collected as the sample was warmed in ca. $\pu{25 K}$ to $\pu{30 K}$ increments. The reaction of either DMC or DEC with lithium was initiated at $\pu{180 K}$, a temperature much lower than their bulk melting temperatures, producing lithium methyl carbonate, methyllithium and lithium ethyl carbonate, and ethyllithium, respectively. At temperatures greater than $\ce{270−300 K}$, the lithium alkyl carbonates start to decompose with $\ce{Li2O}$, elemental carbon, and alkyllithium as products on the surface. Both DMC and DEC are more reactive toward metallic Li than another carbonate solvent, propylene carbonate, which we have studied by the same methodology. Because methyl and ethyllithium are highly soluble in the parent solvent, electrodeposited $\ce{Li}$ is predicted to have poor stability in an electrolyte composed of either DMC or DEC.

(*) Not to be confused with the van't Hoff's equation - the van't Hoff's rule is about temperature dependency of kinetic rate, the van't Hoff's equation is about temperature dependency of equilibrium constant. But both are related.
(**)( dimethyl carbonate, diethyl carbonate, methylethyl carbonate, ethylen carbonate, propylene carbonate )
---- Feedback to comments
Being fully charged helps to prevent deep discharges, what is not the same as saying it is optimal storage condition. Note that storing cells fully discharge is worse than fully charged.
About half charge is optimal for lifetime and full charge is optimal for the time before needed recharging. But self discharging is not dramatic. LiPo have reportedly self-discharge rate approximately 5% per month.
This discharging is probably not linear, and its rate would be decreasing. Also, consider it as illustrative number only. It may differ for different cell technologies and age. Discharge cutoff cuts the connection before a real 0% charge. So the effective 0% charge > real 0% charge. Generally self discharging rate may increase with battery age, but it can be an individual number.
The usual mode of Li-ion/LiPo charging is by a constant current, until the threshold voltage is reached. Since then, charging switches to the constant voltage charging, until the charging current drops below a threshold value. The point of mode switching is typically between 70 (old cell)-80(new cell)% of the full actual capacity.
All that means the charging current is constant until the charge mode switching, and then it is asymptotically decreasing toward zero, until the current threshold cutoff stops the charging.
For lifetime, more shallow charging cycles is better than fewer deep charging cycles. The former has bigger total provided energy value for the cell lifetime. E.g., just illustratively, if a cell lifetime is 300 0-100% cycles, it has lifetime e.g. 800 30-80% cycles, giving 4/3 times bigger total charge. And it goes further for even more shallow cycles. But, unfortunately, shallow charging strategy is seldom practical.
So, if you know you are going to use the battery shortly after charging, it is the best to fully charge it, or at least to 90%. If you do know it will not be used for some time, it is better to charge the about discharged battery just partially, e.g. just 30-40% of the typical full charging time, depending on the residual charge.
The charging time depends on the battery state. A new one may have bigger capacity and generally different remaining charge. Note that new batteries are shipped partially charged, never empty. So bigger free capacity = longer charging time. The same free capacity, but older battery = longer charging time, because the second stage kicks in sooner due higher internal battery resistance.
The discharging cut-off voltage is vendor and technology dependent. It is usually somewhere between 3.0-3.5 V. Note that voltage as a function of charge is strongly nonlinear. More discharged = steeper voltage fall. The same vendor and technology dependency is for the cutoff charging voltage. The percentage itself is hard to tell, but estimated from the voltage curve just few percents of the capacity, perhaps 2-3%.

Thank you for your answer. The issue is, the manufacturer told me to store the battery fully charged not because of battery readiness, but because of the prevention of deep discharge by having too low voltage.

I think the manufacturer means that if the battery is charged to full before storage, one can wait for longer to recharge it. Would charging frequency be a determinant for battery lifespan? Usually how fast would the battery lose voltage by self-discharge (is it dependent on the voltage the battery has)?
Also when charging, does the charge in the LiPo cell build up linearly (so that I will know when is 40% charge reached)?

Thank you for your answer. Another question to make sure: Am I correct that after the threshold of mode switching of charging is reached, the battery will be charged faster (since the voltage no longer increase)?
Also would it be better if I do not charge the battery fully even right before using the battery in the rc helicopter?

Thank you for your answer. Another question out of curiosity: would the time needed for charging a completely new and empty cell to full for the very first time be different from the time needed for charging an already-used empty cell (maybe not completely empty, since the rc helicopter has undervoltage protection) to full? What is the difference between these two charging process?

I see, thank you. Would that also mean that new batteries are also subjected to the 5% self-discharging rate per month? Am I correct that 0% charge is reached when the undervoltage cut-off is reached when using the rc helicopter?

Thank you for your hint. Done. (I did not include all questions because some of the questions are prompted by your answer.)

I just updated my question for another question: What is usually the real percentage of charge at which the undervoltage cut-off is reached?

Comment about Q update is better to place under your question, not under my answer.  If targeted to a particular user, write @username  .  Your comment is replied on the bottom of my answer.

Thank you for your answer. I updated the question directly in the question.

I mean the comment about the update, not the update itself. Always ask if it is related to your questions or to answers. about comments in general, they should be used for exchanging inportant information, that belongs to answers and questions. Main comment purpose is Q or A clarification, or tips for improvements.

This is mostly from the outside, but what are the chemical (and physics/mechanical) explanations? For instance, why isn't there a memory effect?

@PeterMortensen Because Li based cells do not use nickel hydroxides. The answer is not going to cover all aspects of maintenance of rechargables of various chemistry.


text:
How can I mask the Chromium (III) and Chromium (VI) ions in my solution?
I want to conduct a quantitative analysis on my acid etching solution for my Aluminum pre-treatment process. The easiest way for me to conduct my quantitative analysis of Al is through complexometric titration with EDTA and back-titration with standardized Zinc solution.
My problem is that there's also Chromium ions in my solution and it can bind with the EDTA which reduces the volume of Zn solution I utilize in my back-titration. [I have confirmed my dilemma because the solution I am using was analyzed through Atomic Absorption Spectroscopy (AAS).]
How can I mask the chromium ions in my solution so I can proceed with the complexometric titration?

If I summarize, you want to make an EDTA titration of $\ce{Al^{3+}}$ in a solution containing also $\ce{Cr^{3+}}$ and $\ce{Cr^{6+}}$. The $\ce{Cr^{6+}}$ ions are usually in the form $\ce{CrO4^{2-}}$ and they don't interfere with $\ce{Al^{3+}}$ in EDTA titrations. But they can be eliminated by precipitation. The best thing to do is to oxidize $\ce{Cr^{3+}}$ into $\ce{CrO4^{2-}}$. In the old Treadwell's book, they propose to take $5$ mL $\ce{Cr^{3+}}$ solution, add $0.5$ mL of a saturated solution of $\ce{Br_2}$ + $0.5$ mL $\ce{KOH}$ $4$ mol/L. The reaction is finished at room temperature in $5$ minutes, and $\ce{Al^{3+}}$ ions do not interfere. $$\ce{2 Cr^{3+} + 3 Br2 + 16 OH^- -> 2 CrO4^{2-} + 6 Br^- + 8 H2O}$$ At the end of the reaction, a solution of barium chloride $\ce{BaCl2}$ or baryum hydroxide $\ce{Ba(OH)2}$ is slowly added to get rid of the chromate ions, by the reaction $$\ce{Ba^{2+} + 2 CrO4^{2-} -> BaCrO4(s)}$$ After filtration of the insoluble $\ce{BaCrO4}$, the obtained solution does not contain any more chromium, it can be acidified and the $\ce{Al^{3+}}$ ions can be titrated by EDTA.

So, I need to oxidize and precipitate the Cr3+ because I can't proceed with the quantitation of Al3+ with just masking it? That's way more difficult to do in the perspective of cost and accessibility. Thank you for the answer.