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Why is octet extra stable?
In chemistry, it is agreed in general that the octet is extremely stable, through experiments except in some cases. But is there any specific theoretical reasoning for that?
According to me, if:

The $\ce{s}$ and $\ce{p}$ orbitals are full, they should repel in every orbital, causing destabilization.
There is no exchange of energy in fully-filled orbitals as it occurs in half filled orbitals.

So, shouldn't these two reasons make an octet filled atom unstable? Then, why is the octet stable and how does it counter the above arguments?

Consider adding an electron and removing an electron...

It isn't always so.  Try drawing a structure for a cyclopropenyl C3H3 ring with one double bond, putting the right charge on it to make every carbon atom have an octet.  But, the actually stable species has two electrons less than what you drew!  The link given above tells why.


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Is iron (III) oxide-hydroxide the same as iron (III) hydroxide?
Firstly, the Wikipedia article on iron(III) oxide-hydroxide says that $\ce{FeO(OH).H2O}$ is sometimes described as iron(III) hydroxide ($\ce{Fe(OH)3}$). If that is so, does iron(III) hydroxide not exist?
Secondly, the hydrated iron oxides section of the Wikipedia iron(III) oxide page says that on treatment of $\ce{Fe^3+}$ salts with alkali we get $\ce{Fe2O3·H2O}$ (also written as $\ce{Fe(O)OH}$) and not $\ce{Fe(OH)3}$. Is it the same as $\ce{FeO(OH).H2O}$ mentioned in the Iron(III) Oxide-Hydroxide page? If not, how are the two compounds different?
Thirdly, the iron(III) oxide-hydroxide page gives the following reaction:
$$\ce{FeCl3 + 3 NaOH → Fe(OH)3 + 3 NaCl}$$
Doesn't this directly contradict what's given in the iron(III) oxide page? And does $\ce{Fe(OH)3}$ here mean $\ce{FeO(OH).H2O}$.
Fourthly, the Wikipedia page on rust says that rust is composed of hydrated iron(III) oxides $\ce{Fe2O3·nH2O}$ and iron(III) oxide-hydroxide ($\ce{FeO(OH)}$, $\ce{Fe(OH)3}$). What are these hydrated iron oxides and how do they differ from hydrated iron(III) oxide-hydroxides $\ce{FeO(OH)·nH2O}$ described in the iron(III) oxide-hydroxide page?
Fifthly, the page on iron(III) oxide-hydroxide identifies $\ce{FeO(OH).H2O}$ as hydrated iron oxide, while the hydrated iron(III) oxide and rust page have me believing that $\ce{Fe2O3·nH2O}$ is hydrated iron(III) oxide. Are they wrong or is $\ce{FeO(OH).H2O}$ also a hydrated iron oxide?
I would also be interested in knowing the structure of these compounds, whether the water is just hydrogen bonded or if it takes part in some complex formation and where I can learn more about such unusual compounds.

Related for manganese: https://chemistry.stackexchange.com/questions/44159/what-is-the-structural-difference-between-mnooh-and-mnooh2

Start with my answer here: What is chemical formula of Bauxite? On oxides and hydroxides of aluminium - that's a quick explainer on notation.
Now, to your questions:

where I can learn more about such unusual compounds

Easy. Here: The Iron Oxides: Structure, Properties, Reactions, Occurrences, and Uses by R. M. Cornell and U. Schwertmann. ISBN: 3-527-30274-3. PDF link

The problem with iron hydroxides is that there are several polymorphs (same composition, different crystal structures), and they are often mixed together in one sample. Add to that that not all iron is fully oxidised, that these compounds are hygroscopic (they adsorb water moisture on their surfaces), and they can contain impurities of whatever was alloyed with the iron (nickel, carbon, chromium, etc etc) forming even more compounds.
A single polymorph rare exists by itself (but when they do, they are spectacular; examples: 1 2 3). Pure iron hydroxide is usually yellow, and pure iron oxide is bright red. Anything in between is going to be brown. When found in nature it's called "limonite", and when formed by rusting of iron, it's called "rust".

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Attack of ethoxide on epichlorohydrin

I eliminated the possibility of the first two options since they included an alcohol and the solvent has not been specified so we do not know whether it is polar protic or polar aprotic.
As for the next two options I drew both of their mechanisms are they correct?

The answer given is D. Is it because the carbon next to oxygen is more electrophilic?

C-Cl bonds are not that easy to break, the strained epoxide opens more readily. The subsequent ring closing is aided by the fact that it's intramolecular

Avnish: You may find the solution to problem #3 here of use. http://ursula.chem.yale.edu/~chem220/chem220js/PROBSETS/PS13spr/PS10-S13-ans/PS10-S13-ans.html

In general, you can expect an epichlorohydrin to be attacked nucleophilically only on the outer epoxide carbon. First, let me present the synthetic argument: In their total synthesis of RK-397, Burova and McDonald used epichlorohydrin as the starting point for the synthesis of their C10–C16 fragment. The fragment synthesis is shown in the scheme below which is taken directly from their publication.

They start with an enantiopure epichlorohydrin which they open using vinylmagnesiumbromide. The resulting alcohol is recyclised under elimination of chloride by distilling from $\ce{KOH}$. Starting from (R)-epichlorohydrin, their sequence exclusively generates (R)-33. Epoxidation gives 35 as a mixture of diastereomers which they separate using hydrolytic kinetic resolution with the Jacobsen cobalt catalyst.[1] This procedure acts only on the epoxide and selectively cleaves the (R)-epoxide to a diol. If there had been substantial amounts of (S)-33, 36 and 37 would each have been isolated as a mixture of diastereomers; however, both were isolated as single diastereomers.[2]
If there had been any direct attack of the vinyl Grignard on the $\ce{C-Cl}$ bond with direct displacement of chloride, the opposide enantiomer (S)-33 would have been formed (see scheme 1 below; 33 would have been inert to $\ce{KOH}$ and thus simply been co-distilled). This would have led to the aforementioned mixture of diastereomers further down the synthesis. Since no mixtures were observed, we can safely conclude that the attack only happens under cleavage of the $\ce{C-O}$ bond.

Scheme 1:  Hypothetical attack on $\ce{C-Cl}$ gives the opposite enantiomer; would lead to 36 and 37 being isolated as a mixture of diastereomers.
The synthetic argument already establishes that epoxide-cleavage must be orders of magnitude faster than chloride displacement. This does allow the question: why? One general argument is that the epoxide is highly strained, so a ring-opening attack will release the ring strain. Furthermore, $\ce{C-Cl}$ bonds are rather strong compared to other halogens (especially $\ce{C-I}$) and thus not as easily cleaved. Furthermore, we are likely looking at some hyperconjugation effects: the orbitals of the $\ce{C-Cl}$ bond and the $\ce{CHR2-O}$ bond can overlap as shown in scheme 2; this interaction leads to the chloride displacement attack being further disfavoured.

Scheme 2: Possible overlap of the $\ce{C-O}$ and $\ce{C-Cl}$ orbitals shown here as $\sigma_{\ce{C-O}}$ and $\sigma^*_{\ce{C-Cl}}$
Coming back to your question, since the outer epoxide carbon is marked and since that is where ethanolate will attack, only (D) is correct and your proposed mechanism checks out.

References:
[1]: S. E. Schaus, B. D. Brandes, J. F. Larrow, M. Tokunaga, K. B. Hansen, A. E. Gould, M. E. Furrow, E. N. Jacobsen, J. Am. Chem. Soc. 2002, 124, 1307–1315. DOI: 10.1021/ja016737l.
[2]: S. A. Burova, F. E. McDonald, J. Am. Chem. Soc. 2004, 126, 2495–2500. DOI: 10.1021/ja039618+.

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What are metal zip tags and belt buckles made of?

Metal zipper tags and belt buckles melt very easily and can be melted by heating it in the stove for 5 minutes. But other metals like aluminium refuse to melt even after hours.
If you leave it overnight, it dissolves pretty well in vinegar. I am sure they are made of metal, as I've melted it and cast some metal cookies. The metal is denser than aluminium, shiny when molten and cast, but surface tarnishes after 1-2 hours. I'm guessing it is some type of a zinc alloy.
The metal of the zip tag has the following characterisitics:

silvery
tarnishes into a dark gray
does not colour a flame in odd colours (like red , crimson , green etc.)
malleable (easy to bend a 5 mm thick square just by using leverage of hands and a surface to press down on)
brittle if bent a couple of times.
melts in a few minutes

So, what alloy or metal are they made of?

Possibly Woods Metal which is one of several low melting point alloys, see Wikipedia.

You were right, these are zinc alloys commonly used for making zips and stuff. It is called Zamak which contains:
Zinc (Zn): Around 90% by weight
Aluminum (Al): Around 4% by weight
Magnesium (Mg): Around 4% by weight
Copper (Cu): Around 2% by weight
It is used because it is resistant to corrosion, cost effective, as you've mentioned malleability (can be moulded into shapes for designs) and last but not the least it can be coated with other metals like nickel and chrome for shine, antimicrobial properties etc.
As you have mentioned Zamak whose melting point is  at around 385-410°C (725-770°F), which is relatively less for metal alloys.
Now coming to the tarnishing part, it is said to tarnish into a dark grey colour within a couple hours in contact with atmosphere due to the reaction between atmospheric oxygen and zinc present in the alloy.
2Zn + O2 → 2ZnO
[dark grey]
But based on the specific environmental conditions, there can be formation of zinc hydroxide (Zn(OH)2) and zinc carbonate (ZnCO3), which can contribute to the overall tarnishing of the alloy.


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In a two electrode system, is the applied current density referring to just the working electrode or to the auxiliary electrode as well?
I'm doing an experiment electrosynthesizing a polymer using a two electrode cell and it says to do it under 0.5 mA/cm^2. However, I am unsure if that means 0.5 mA per square centimeter of surface area on the working electrode or for both electrodes. If anyone could clarify this for me, it would be most appreciated.

It is normal in many laboratory electrochemistry experiments for the working electrode to have a much smaller surface area than the counter (auxiliary) electrode. This when the current density at the working electrode is 0.5 mA per square cm, the current density on the counter electrode is much smaller.
It is more common that someone will express the potential and the current density on the working electrode.
If for example we have a reference electrode which is at 0 volts (standard hydrogen electrode) and the potential of the working electrode is + 400 mV. We could have a flow of 1 mA out of the working electrode. As no current flows through the reference electrode all of this current flows through the counter electrode.
The electrical potential between the counter and the reference electrodes is known as the compliance voltage. This is the voltage required to create a flow of current which is just right to make the right amount of current flow through the working electrode to keep its potential at + 400 mV.
This is done using a potentistat, a range of different designs of such devices exist. I think that the most simple design and one of the best is the general design used by PAR (Princetown Applied Research) in the classic 173. It has been joked that at one point almost all electrochemists have used a 173, I rather like the 173 it is a nice machine. But we are not here to discuss classic electrochem machines.
Back to the design of a machine like a 173, If we set the drive at -400 mV then the top op amp will impose a current on the cell via the secondary electrode which will bring the potential of the reference electrode to -400 mV.

The bottom op amp which is a I to V converter will hold the working electrode at zero volts (ground or GND) as far as the electronics is concerned. Thus for the chemist if they hold the view that the refernece electrode is the zero volts point then the working electrode will be at +400 mV.
If 1 mA flows into the working electrode and the value of resistor R is 1000 ohms then the output of the I to V converter (bottom op amp) will have to be -1 volt for the sum of the currents coming to the inverting (-) input of the op amp to be zero. We assume that the input resistance of an op amp is infinity. Thus if we attach a x-y plotter to the drive terminals (x to E or potential axis) and the (y axis to the current indication terminal) we will get the plotter to plot -400 mV, -1 volt.
A word of warning the Europeans and the Americans often follow different conventions on what sign to give a reducing current (cathodic) and what sign to give a reducing potential.
If I recall rightly the European convention is for bottom left to be reducing current and reducing potential.

Thanks for such a detailed explanation! Does this still hold when doing galvanostatic deposition rather than potentiostatic?

In a galvanstatic experiment a different circuit would be needed to keep the current constant. I am not so sure if the same need exists to make the counter elctrode bigger than the working electrode exists there. A lot will depend on what you are trying to do. Out of interest what is the polymer you are trying to make, is it by any chance polyaniline ?

Not polyaniline, but pretty close guess. I'm trying to make polypyrrole. The setup I'm using has a constant current source and ammeter with one working electrode which has a counter electrode parallel to it on each side. Everything else in the method I understand fine, it's just the current density that confuses me as it says to run the experiment under 0.5 mA/cm^2 using a two-electrode system without specifying any further.

Maybe you should put a coating of polyaniline on a Pt foil and then use a potentiostat to cycle it between the reduced and oxidized forms, The colour change is very nice to look at. I have no idea is polypyrrole is electrochromic or not. Maybe you will be able to tell us soon (get some photos please)

That sounds quite neat and I'll probably try make polyaniline at some point as well although for now I'm focusing on polypyrrole as, out of the conductive polymers, it exhibits the highest strain (I'm a physicist and am experimenting with them for use as artificial muscles).


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Different HCl concentrations and iron nails
Hey guys I am at high school and we have a project on designing our own lab.

My questions is will different concentrations such as 10%, 20%, 30% and 37% have an affect on the corrosion of an iron nail.
And will different concentrations(0%, 20%, 30% and 37%) have a different pH?
Is it possible to measure the corrosion on the iron nail for quantitative data and a graph?

We don't have to do the experiment, we just have to design one.
Sorry I am not good at chemistry, hope some one could take time to answer my question.

You must use similar nails for this experiments.

Different concentrations will corrode the nails at different rates. The rates won't increase beyond a certain concentration, no matter how much you increase the concentration. This must be a fun experiment to try
Depends on the medium you're going to use for diluting
Yes! You can calculate the time that each of it take to corrode the iron completely. A stopwatch will help

hey another question what do you mean by the medium?

The nature of the solvent that you're going to use in diluting. Well if you're dilute the acid HCL with relatively basic Water, the pH of the solution is going to change. The number of pH will increase

ohh kk thx a lot :)

By the way, how will I know the iron is corroded completely? Do I determine it with my naked eye? Sorry for the amount of questions :(


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Calculating the enthalpy of polymerisation of ethylene given the bond strengths

Given the average bond dissociation enthalpies of a $\ce{C-C}$ bond (say $x$) and a $\ce{C=C}$ bond (say $y$), find the enthalpy of the following polymerisation reaction (per mole of ethylene):
$$\ce{nCH_2=CH_2 -> [-CH2-CH2 -]_n}$$
where $n$ is a large integer and $x, y$ are in $\pu{kJ/mol}$.

According to me, this should simply be $y-x$ as for every double bond broken, a single bond is formed. But the source book of this problem states the answer to be $y-2x$ claiming that for every double bond broken, 2 single bonds are formed.
How is this possible?

The mistake I was making was not noticing that every time a double bond is broken, i.e.; completely broken, 2 single bonds are formed in its place.
One bond between the 2 carbons of the same molecule and another between two molecules, the former of which I forgot to consider.
Thus, the enthalpy is in fact $y-2x$

Then in that case, shouldn't there be 3 single bonds formed? 1 between the 2 Carbon atoms, then 2 more, formed at the free valencies; making the bond enthalpy $y-3x$


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Structure of the compound C9H7O2Cl?

A compound having formula $\ce{C9H7O2Cl}$   (O = oxygen) exists predominantly in enol form (A) and also in keto form (B). On oxidation with $\ce{KMnO4}$ it gives m-chlorobenzoic acid as one of the products. Identify (A) and (B).
(subjective question; no options were given)

I tried it and found the double bond equivalent as 6, 4 of which were utilized in benzene ring. Also, the benzylic carbon must have at least one hydrogen. Therefore I determined the following structure as (A):

But my book suggests (A) should be:

Please explain whether both are correct or only one product will exist.
Source: Joint Entrance Exam (JEE) 2003 India

The answer given by the book is correct because of the formation of 6 membered chelation ring. As well as extended conjugation provides additional stability.
As the structure drawn by you has only 5 memebered chelation ring as well as there is no active methylene group(acidic H) so it will not form.

Learn to draw structures such as [A] with proper bond angles.

The representation from the book is not the only possibility, but likely more stable than the one you drew and thus the more favored answer.

There are many constitutional isomers sharing the sum formula $\ce{C9H7ClO2}$.  Assuming a disubsituted benzene, one variation is to put the chloro- and the alkyl group in either ortho, meta, or para relationship to each other. (Side note: Hill formulae and sum formulae about organic compounds typically report the molecule's composition in the sequence about the number of atoms of carbon, then hydrogen, then by alphabetic order of the element symbol.)

For the sake of clarity, the structure formula depicted in the exam is redrawn below (left hand side):

The formula depicts the same molecule as the first of the three formulae depicting the tautomers which may coexist in the thermodynamic equilibrium with the $\beta$ dicarbonyl compound depicted on the right.
The use of the dashed line between oxygen of the carbonyl group and hydrogen of the enol highlights an intramolecular hydrogen bond in addition to the typical covalent bond between $\ce{O}$ and $\ce{H}$ of the alcohol. If you compare the two with each other, the covalent bond is stronger than the hydrogen bond. The attraction between the carbonyl $\ce{O}$ and the enol $\ce{H}$ is strong enough that a conformation similar to a six-membered ring is energetic favorable.  Here, in an analogy you may think of the hydrogen and oxygen atom attracting each other like opposite magnets:

(For comparison, the intramolecular $\ce{O-H}$ bond length in water is about $\pu{0.95 Å}$ (ref).)

The molecule you drew derives from a $\alpha$-keto aldehyde, a constitutional isomer of the above:

Even in a conformation most favorable for the opposition of hydrogen bond donor and hydrogen bond acceptor, the intramolecular distance is greater than in the earlier example of the $\beta$-keto enol:

Is the opposition of $\ce{H}$ of the enol and $\ce{O}$ of the carbonyl favorable (i.e., a lower energy conformation than other ones)?  Yes.  However given the hydrogen bond donor and hydrogen bond acceptor are farther away, the mutual attraction is less strong (remember Coulomb's law, electrostatic forces scale over distance $r$ by $1/r^2$) and the overall stabilization for the molecule is less important, than in the case of the $\beta$-keto enol.

The distances depicted in the images are obtained by running a quick and cheap MMFF86 force field optimization in Jmol.  More important than remembering these distances is to retain the overall trend (the closer the partners, the stronger the potential intramolecular hydrogen bond, the stronger the overall stabilization of the molecule).

Both the answers are correct but the most appropriate will be B as the chelation occurs because of hydrogen bonding which makes it more stable.
H-bonding is  possible in your structure but the ring is less stable
.

But there will also be H bonding i product 1

In your case the ring will be 5 membered which is less stable than 6 membered ring

Where is there a 5-membered ring?

What do you mean by 5 membered ring

There is no actual ring there

The 5 membered ring is the other ring than benzene which is formed due to a hydrogen bonding (on right) of image

It actually is [very weak ring ] but plays important role in stabilising the compound

Greater the association of molecules greater will be stability of compound

Is there any rule for stability of chelate ring?

Yep, greater the association of molecules greater will be the tendency to stay together which implies more stability.


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How does changing substituents change the electronegativity of the carbon atom attached to it?
I was wondering how electronegativity and charge of carbon atom is affected by switching the substituents.
For example, what will be the difference in electronegativity of carbon atom in $\ce{-CF3}$, $\ce{-CCl3}$, $\ce{-CBr3}$, and $\ce{-CH3}$.

@SourabhYelluru Electronegativity isn't conserved, but only transferable property. In every situation it's slightly different, so usually approximate/average values are used.


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Why is fluorine the best oxidising agent, but also the most basic?
(Please note that all comparisons I have made below are with respect to halide ions only.)
According to what I have learnt, the fluoride ion is the most basic ion, because it has the smallest size and thus the highest electron density. This implies that it has the highest tendency of sharing/giving away its electron density.
Why isn't the same logic applicable when reducing strength is compared? Fluorine should have the highest tendency to lose electrons, and hence should be the best reducing agent.

F2 molecules have quite weak and rather unusual bonds. F makes strong bonds this other elements, on the other hand.

You're looking at two different changes. The end state is different so it makes no sense to compare these two scenarios.
If fluoride is oxidized, it loses an single electron.
If fluoride acts as a base and donates a pair of electrons, it shares a pair of electrons.
In the first case, you now have fewer electrons around a very small charge. The effective nuclear charge on the valence electrons is quite high. The extra electron was attracted to this positive charge, and if you want the process to be favorable, you needed to stabilize that electron more than destabilization from removing it from the fluoride. But you're going to find something that's going to want an electron electron more than a fluorine atom...
In the second case, you have a high density of electrons fairly close to the nucleus. But you can stabilize the overall system by using some of those electrons to stabilize another positive charge. So, you're reducing some electron repulsion at the expense of attraction within the fluoride, but you're stabilizing another positive charge, which is overall favorable.

Zhe, could you add some examples for both cases with Fluorine, please?

@HolgerFiedler I don't understand the request...

„If fluoride is oxidizes, it loses an single electron.” - give an example of such a compound
“if fluoride acts as a base and donates a pair of electrons, it shares a pair of electrons.” - give an example of such a compound

I can do that, but I fail to see how that helps at all with understanding the OP's question...

A flourine atom is very electronegative, fluorine is the most electronegative element it has the highest effective nuclear charge.
If a F atom accepts an electron it then has the same electronic configeration as neon. This gives fluorine a great driving force to act as an oxidant.

Fluorine is the most electronegative element.
But due to its high effective nuclear charge it holds up its electron density very tightly, moreover its a non-metal so preferably accepts electron to form stable (F–) ion which has configuration of Ne.
F2 ------> F+ + e- is very favorable therefore it gets reduced very easily (reduced form F- is very stable than F2) and so it oxidises other elements and self gets reduced.

Then why is it that F- is the most basic halide ion? If it is stable, it should be less basic.

Let's take it in different sense, atom achieves stability by the octet rule so F- has a complete octet but there's a better chance of getting octet complete WITHOUT BEARING ANY CHARGE that's by forming a bond.

So by accepting a proton it forms HF which is more stable than F- as it HF is neutral. Also that's not with F2 because of high repulsion ( small size and high electron density) the molecule is unstable.

If you are referring to why the fluoride ion isn't the best reducing agent, it's because in order for anything to be a reducing agent it will be oxidized and reduces what ever it reacted with. Since Fluorine is the most electronegative atom out there, there is no way it can be oxidized and lose its electrons.

... fluoride ion is the most basic ion, because it has the smallest size and thus the highest electron density.

A look at the Periodic table and the ionization energies is always a good idea to see what’s going on.

The highest energies, required to remove an electron from the atom, are needed for the Noble gases. That means they are most aggressively reclaiming a removed electron back.

The less the number of electrons around the nucleus of Noble gases, the higher the aggressiveness to retrieval a lost electron.
The elements in the column before the Noble gases ($\ce {Fl}, {CL}, {Br}, ...$) are doing the same in an attenuated way. And more than that, in bounds these elements try to complete their electron shell to the Noble gas configuration.

Take the Noble gas compounds. Fluorine is able to remove electrons from Xenon:

The tendency to fulfill the electron configuration is higher for Fluorine when the tendency for Xenon to preserve its electron configuration. The Octet rule works better for elements with high ionization energy when for elements with less ionization energies. (Perhaps instead of ionization energy in this case we have to remember the electron affinity, but this seems to me to be more a scholastic difference in definition.) According the above image Fluorine has a ionization energy of roughly 17 eV, Xenon has about 13 eV. That is why the octet rule in this case “works” for Fluorine and against Xenon.

Noble gas, noble gas, noble gas. Question about halogens.


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How do pharmaceutical manufacturers ensure 100% purity in their medications?
What purification measures are taken to ensure no side products are present in the medication? Because hardly any organic reaction is perfectly efficient, I'm guessing that multiple recrystallizations are performed. Also how would recrystallizations even be performed on an industrial scale? How much product is then lost and can it be recovered or is not economically advantageous to do so?

The don't ensure 100% purity, they ensure that there are no harmful side effects beyond what was found in testing and is contained on that really big sheet of paper folded up really tightly in the box...

Most registered specifications will contain the main active pharmaceutical ingredient (API) at an assay of 98-102% to allow for hplc variation(the most common analytical technique for pharmaceuticals). The list of known impurities usually follows with their maximum limit ; this is usually derived from toxicology studies. But not always.
An exception is something like Sativex which has an assay of <90% for the main API, but the other ingredients are not detrimental but generally regarded as safe (GRAS); sterols, triglycerides.
The drug product then has the API in  specified range and can comprise most of the medication e.g. paracetamol tablets, or not very much e.g. acyclovir in ointment form or cisatracurium besylate in water for injection. Excipients are added which could be seen to reduce the "purity" of the medication, but they are present to aid bioavailability, taste etc.
Any technique you can think of can be used industrially... at a cost. So while liquid -liquid extraction is cheapest and easiest,  it rarely gives the desired purity. Recrystallisation is a cost effective technique and is frequently used industrially. Put solid in a reactor, pump in solvent, heat, cool. Extract, distil solvent,  solvent switch, heat, cool is also valid.
Yield loss from industrial scale recrystallisation is usually 5% or greater.  I have seen 30% for achiral material. I've seen 55% for a diastereomeric salt crystallisation (so at least 50% is to be expected).
Sometimes one crystallisation is enough; sometimes three are required. It depends on the chemical transformation efficiency and co-crystallising impurities.
As for recovery ; economic viability depends on the cost of the material. So at 20 dollars/kg, probably not. 15000 dollars/kg, probably yes.
Not all impurities are bad as they can be very similar to the API and simply have a different effect e.g. anti-histamine.  Some could genotoxic and are controlled at the ppm level.
Unknown impurities are tightly controlled to, typically, the total being no more than (NMT) 0.5%. Again, there are exceptions.


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Why is Frost's Circle an extension of Huckel's 4n+2 rule? What physics leads to the emergent property that the circle can predict relative MOs?
The Frost Circle is a useful mnemonic to determine the relative energies for the molecular orbitals of an aromatic ring. One vertex of the polygon of the cycle is drawn pointing downwards and the entire polygon is inscribed in a circle. At this point, the location of each vertex corresponds to the MO relative energies. What is the physical explanation as to why a Frost Circle can accurately be used to predict the relative energies of molecular orbitals in aromatic compounds? What quantum mechanical phenomena lead to this emergent property? I'm looking for a connection between an equation and the circle I suppose.

There is no property and no connection; it is just a mere coincidence.

@IvanNeretin in order to develop the visual representation in the first place, thought had to be taken in its development. I doubt anyone would intuitively happen upon the vertices of a polygon corresponding to MOs... surface level they don’t seem connected. From the below answer you can see that the original explanation did in fact stem from an equation and that equation was central to developing a more approachable visual model.

The MO energies are obtained from Hückel theory: see Pi molecular orbitals of polyenes (and the Internet in general, there are many excellent expositions) for more details.
Regarding the Frost circles, they do not actually have any meaning: they are simply visual representations of the energies obtained from Hückel theory for cyclic polyenes. The last line of porphyrin's answer is relevant here:

In the case of circular polyenes the solutions are $E_j=\alpha + 2\beta \cos(2\pi j/n)$.

where $n$ is the number of p-orbitals (e.g. for benzene $n = 6$) and $j$ is an integer satisfying $1 \leq j \leq n$. (The convention $0 \leq j \leq n-1$ is also sometimes seen, but it is mathematically equivalent.)
I am not a mathematician so some of the notation may be less than ideal, but the idea is as follows. Note that if you draw a circle with radius $r$, then the coordinates of any point on the circle $P$ may be described as $(r\sin\theta, -r\cos\theta)$ where $\theta$ is the (clockwise) angle that $\overrightarrow{OP}$ makes with the negative $y$-axis $\overrightarrow{OY}$. For example, here is a circle with $r = 2$. The angle $\theta$ here is equal to $\pi/4$, so the coordinates of the point $P$ are given by $(2\sin(\pi/4), -2\cos(\pi/4)) = (\sqrt{2}, -\sqrt{2})$.

Now we simply need to choose a circle with radius $r = -2\beta$ (note that $\beta < 0$, hence we need the negative sign) and select the six points on the circle where $\theta = 2\pi j/n$. The $y$-coordinates of these six points will then be the six MO energies. I have illustrated this for $n = 6$ and $\beta = 1$. We need to pick the points with
$$\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi$$
and these are $A$, $B$, $C$, $D$, $E$, and $F$ in order. Note that $\angle YOA = \pi/3$, $\angle YOB = 2\pi/3$, and so on. And note that whatever the value of $n$, there will always be the case where $j = n$ and hence $\theta = 2\pi$: that corresponds to the vertex at the bottom of the circle.


text:
Anti matter elements?
Quite a self-explanatory question I guess. I wonder whether there is any information or theory on this. I seem to like remember there is such an element with positron in center and electron rotating around but nothing else.
Can there be any use if we somehow get these? Don't think like immediate destruction, the electrostatic force between "normal" elements electron and the other types can still repel each other and prevent immediate annihilation, in my view.
Can there be any spot in universe with mostly antimatter?

Whether/where/how stabke antimatter is, is more of a physics questions. But assuming sufficient short time scales or stable antimatter, you can apply the very same quantum chemical methods to describe the chemistry of such elements. The one you are refering to is conveniently called positronium, and can be considered an excotic, super light weight hydrogen isotop

anti-hydrogen consists of an anti-proton with a positron in orbit. Not a positron and en electron. A positron is the anti-particle of the electron not the proton.

Antihydrogen has been made at CERN in small amounts.
This is made by the alpha experiment, from poistrons and antiprotons. So far they have not been able to do any chemistry experiments.
https://home.cern/about/experiments/alpha
https://home.cern/about/engineering/storing-antimatter

But also they have not observed any notable differences between the properties and the properties of normal matter. So it looks like chemistry would be the same.

There is no theory, nor is there a need for one. To put in in other words: yes, there is a theory, and that's the one you are already familiar with. Antimatter elements are precisely the same as ordinary matter elements, only the other way around. There "are" (read: "could be", though some were actually detected) antihydrogen, antihelium (all with the same energy levels as their real counterparts), also anticarbon, antiwater, anti-$\ce{NaCl}$, antiglucose, and pretty much everything anty-else.
To compose a substance of particles and antiparticles together is an entirely different question. Yes, there is a compound of electron and positron; it is called positronium and has the lifetime of a few microseconds or something. Electrostatic forces attract the opposite particles, rather than repel them (otherwise it would not be a compound, BTW), so it annihilates pretty quickly. There are other similar compounds, equally or more short-lived. Matter and antimatter do not sit well with each other; there is no going around it.
As for the spots of the Universe composed of antimatter, there was once a mainstream view that there must be many of those, about as many as those made of matter. Then the global symmetry broke, and things went... complicated. I'd rather leave it at that. Go ask at Phys.SE for a more comprehensive take.
So it goes.


text:
What is the smoke from KNO3 with sugar reaction?
So I was doing a reaction between $\pu{0.02mol}$ $\ce{KNO3}$ and $\pu{0.01mol}$ $\ce{C6H12O6}$ and there was a smoke. What is the chemical reaction, and what is the smoke?

Is it smoke or steam?

@Waylander it is smoke. It looks like the smoke that comes from wood fires.

It isn't C6H12O2, it's C6H12O6.

It should be some mixture of N2, K2CO3, CO2, H2O, and some other things.


text:
Temperature of an aerosol can
Hey guys I just had a quick question regarding the temperature of a dust off can or other aerosol cans containing compressed gas. It is if I have a dust off can and I were to spray some gas from it, why would the temperature of the gas inside the can decrease. Is it because there is a reduction in both pressure and volume, or id it for some other reasons. I am probably gonna say that the reduction in pressure and volume cause a decrease in temperature as demonstrated by Gay-Lussac, but I am not sure. What do you guys think?

The air that remains in the can after some of the air has been expelled has just experienced something approximating an adiabatic reversible expansion.  It did work to push the expelled air out of the can, and, by the first law of thermodynamics, its internal energy (and thus its temperature) have decreased.


text:
Determining the chemical formula of a polyiodide Salt
I've got a compound with formula $\ce{N(CH3)4.I(I2)_x}$. By mass I've got $\pu{0.138 g}$ of $\ce{I2}$ present out of $\pu{0.189 g}$ of polyiodide added. How do I find $x$?
This is the exact part of the question that I got stuck. It's more of the concept that I am not getting instead an equation that I can just use.
We begin with using $\pu{0.25 g}$ of tetramethylammonium. Then used $\pu{0.825 g}$ of iodine, and finally collected $\pu{0.8 g}$ of polyiodide salt.
Then we titrated $\pu{5 mL}$ of the polyiodide solution with $\pu{0.2 M}$ thiosulphate solution. Final titre used is $\pu{5.40 mL}$.
Calculated below:

$\pu{1.09e-4 mol}$ of thiosulphate used to reach end point
$\pu{5.43e-5 mol}$ of iodine present in the $\pu{5 mL}$ polyiodide solution
$\pu{5.43e-4 mol}$ of iodine present in $\pu{50 mL}$ original sample
$\pu{0.138 g}$ of $\ce{I2}$ present in original sample
$\pu{0.189 g}$ is the full mass of original sample
Thus giving 73 percent iodine by mass

So for the final ratio i got $1:2.7$ which can be rounded to 1:3. The formula thus is $\ce{NMe4.I(I2)_x}$, where I think $x=3$.
But I can't seem to make sense of this. I mean when $\ce{I(I2)_x} = (\ce{I2$x$}+1)$ could it mean $2x+1=3$, therefore making $x=1$?


text:
In preparing sulphonilamide, why is aniline acylated if the acyl group has to be removed later?
Our professor talked about the six stages of the synthesis of sulfa-drugs, which are:

nitration of benzene
reduction of nitrobenzene
acylation of aniline
reacting acetanilide with chlorosulfonic acid
treating the product with a primary amine
subsequent hydrolysis of acetyl groups

If we skip step 3, and let aniline react with chlorosulfonic acid, and then go till step 5, could sulfonamide be synthesized? Why do we perform acylation of aniline if we just have to remove the acyl group by subsequent hydrolysis later?

Because aniline does not sulfonate in the ring unless you protect the NH2

The main problem encountered during electrophilic substitution
reactions of aromatic amines is that of their very high reactivity.
Substitution tends to occur at ortho- and para-positions. If we
have to prepare monosubstituted aniline derivative, how can
the activating effect of –NH2
 group be controlled ?

This can be done by protecting the -NH2 group by acetylation with acetic
anhydride, then carrying out the desired substitution followed
by hydrolysis of the substituted amide to the substituted amine

The lone pair of electrons on nitrogen of acetanilide interacts
with oxygen atom due to resonance as shown below:

Hence, the lone pair of electrons on nitrogen is less available for
donation to benzene ring by resonance. Therefore, activating
effect of –NHCOCH3
group is less than that of amino group.
Moreover, in the
strongly acidic medium, aniline is protonated to form the
anilinium ion which is meta directing.

However, by protecting the –NH2
 group by acetylation reaction
with acetic anhydride, the reaction can be controlled
and the p derivative can be obtained as the major product.

source: word to word from
NCERT(INDIA) CLASS 12 chemistry ;)
http://ncert.nic.in/NCERTS/l/lech204.pdf

Don't rely on NCERT. They are very misleading.


text:
Why does nitric acid have a different molecular structure from phosphoric and arsenic acids?
I remember that this really bugged me as an undergrad, and upon revisiting some inorganic chemistry notes I realized I still don't have a satisfactory answer for it. The structure of nitric acid, $\ce{HNO3}$, is quite different from the structure of phosphoric acid, $\ce{H3PO4}$ and arsenic acid, $\ce{H3AsO4}$:

Something tells me there should be an easy, intuitive argument explaining this, perhaps something about p-orbitals being more diffuse in phosphorus and arsenic? I find it kind of difficult to analyze this as a sort of "trend", since other groups don't really seem to have well defined oxyacids for the first two or three elements.
There is no oxyacid for water, but sulphuric and selenic acids are similar in structure; there is no oxyacid for fluorine either, but the oxyacids of chlorine and bromine are also analogous. In the carbon group something kind of strange also happens, with silicic acid (I believe) not really being a well-defined compound but a plethora of different things with different relative stabilities.

Phosphorus and arsenic are much bigger atoms with lower electronegativity so they can fit more atoms around them and oxygen can bond more easily.
Under extreme pressure it is possible to make the orthonitrate ion from nitrate and oxide. It rapidly decomposes in contact with water or carbon dioxide.


text:
What are the products of the reaction of methoxypropane with HBr?

The products of the reaction of methoxypropane ($\ce{CH3CH2CH2OCH3}$) with $\ce{HBr}$ are?

In my opinion, the products should be: bromopropane ($\ce{CH3CH2CH2Br}$) + methanol ($\ce{CH3OH}$). The first step is the cleavage of the ether. This can happen via the formation of a propyl carbonation or a methyl carbocation. Since the propyl carbocation is more stable, bromine will react with the same to form bromopropane (the other product thus formed is methanol)
But the products given in the solution were propanol ($\ce{CH3CH2CH2OH}$) + bromomethane ($\ce{CH3Br}$). So, what is wrong in my reasoning? And what is the correct reasoning?

Source: NCERT (India) 12th

The first step is protonation of the oxygen of the ether. This cationic species is then subject to nucleophilic attack by Br-. There are two potential sites of attack i.e. the carbons bonded to the protonated oxygen. Bromine is quite a bulky nucleophile, and of the two sites, the methyl group carbon is less hindered so this is attacked in preference, leading to bromomethane and propan-1-ol as your products.

so steric hindrance is the reason. Then, what would you say about this https://www.dropbox.com/s/dukb3mewv2vy8g5/SketchGuru_20180509133901.jpg?raw=1

Goes by a different mechanism. The protonated ether in that case fragments to give tBu cation, this is very stable and drives the reaction

With excess HBr propan-1-ol will also react to form 1-bromopropane.

In a real experiment yes, probably get measurable quantities of 2-bromopropane and propene too


text:
1,4-diphenyl-1,3-butadiene (E,E) and (E,Z) Isomer Newman Projections

I was wondering if it is possible to draw the Newman projections of the (E,E) and (E,Z) isomers of 1,4-diphenyl-1,3-butadiene in the following manner that I have? I wasn't quite sure about how to present the rings in such projections.

I don't know what tempts you to use the Newman projection for these compounds. Trying to draw them, you might have missed the existence of the actual conformers. (Not speaking of the confusing 120° angle representation of the planar system.)
Thanks to the totally conjugated system, the molecules tend to maintain a planar shape. The two conformations, ‘cisoid’ and ‘transoid’ (deprecated terms; recommended: s-cis, s-trans; discussed also here) resemble double bond configurations because of the stability of the planar conformation (sterical hindrance has to be taken into account as well, it can be significant for the s-cis E,Z).
For example, the s-trans and s-cis conformations of the E,E isomer can be represented as follows.

Trying to show the conformation at all the three single bonds at once wound be difficult, but could be done. And there's more missing info, so the two Newman projections would represent the (E,Z)'s conformers as well.
I am aware that excluding the hydrogens in the Newman projection is unorthodox. With or without H's, I don't think that the Newman projection is useful here, as for someone unaware of the planarity tendency, might consider e.g. torsional angle 90° much more favorable than the 0° one.


text:
Why do aluminum, magnesium, and calcium form white precipitate when mix with sodium hydroxide?
Why do aluminum, magnesium, and calcium form white precipitate when mix with sodium hydroxide? Shouldn't there be no reaction? Because sodium is more reactive than all of those element, it should be impossible for sodium to be displace by those three elements.
YouTube video - "NaOH + Al - sodium hydroxide and aluminum"

When adding aluminium, calcium or magnesium, you are most likely adding them as chlorides. They dissociate almost fully, similarly to sodium hydroxide.
Then you have a solution with these four elements:

Sodium cations
Calcium (for example) cations
Hydroxide anions
Chloride anions

The precipitate that forms depends on the solubility of the different combinations of anions and cations. Sodium chloride and hydroxide are soluble so they remain in solution. Metal chlorides are also soluble. However calcium, magnesium, and aluminium hydroxides are insoluble so they form the precipitate.
You’re basically neutralising the basic solution, formed a solution of sodium chloride and a “basic” solid precipitate.

Thanks for the answer! And why does the precipitation disappear when we add an excess amount of Sodium Hydroxide to solution containing Aluminum as the cation.


text:
Will we ever run out of gold, silver, copper and other important conductors?
Conductors are very important in technology. At the current rate of extraction how long until the world's reserves of copper are depleted?

I think you need to define "last". Last for what purposes, and in what quantities? Do you want to measure the time till their commercial use stops? Or till their total quantity on earth falls below 1kilogram? or when there is less than 1ypm trace amount left? Given your question, I believe you're looking for the first interpretation, but it's good to define your terms completely.

I would add that, unless for some reason we start shooting insane quantities of stuff in space, it's not like copper is going anywhere - it's all still here, in devices and objects subject to rapid obsolescence, and unlike - say - oil, it can be recovered. Although recent high-tech devices make separation more complex, reading around I see that currently recycled copper amounts to around 35% of the world annual copper usage.

35% Wow that's huge

@theenigma017: yep, and if reserves started to finish I'd expect we could do significantly better than that.

How long will these reserves last?

Short answer: longer than you think and will outlast me, you, our children and our grandchildren. Possibly last forever.
Long answer: This is a fairly complex issue. There was a book published about this topic several years ago claiming that we are running out of metals. Turns out that the book had a major flaw in its reasoning, so the same publisher followed up with a second book refuting the first one. Both books are open access and freely available so I recommend having a look.
The problem lies in the definition of the word reserve.
What are reserves?
A reserve is an ore deposit (that is, a concentration of a certain metal) that we know exactly how much is in there to a good degree, and we have detailed plans on how to extract it. This means, we have plans to build a mine, we know the cost of running the mine, etc. In some countries (for example, Canada and Australia) you are not allowed to call something a reserve unless you have the plans. This is a legal definition.
What are resources?
On the other hand, if you know how much metal is in the deposit, but you haven't made the plans to extract it yet, it is a resource. You are not legally allowed to call it a reserve.

The work required to develop a resource into a reserve costs money. You have to employ mine engineers, you need to get environmental approval, you have to do a lot of lengthy and costly things. Therefore, you don't "declare" a reserve until you have to. If you are a mining company and you have reserves of 40 years, why bother converting more of your resources into reserves? And this is exactly the problem: alarmists and the media consider the reserves, without the resources. Of course we have limited reserves! Because it's pointless to turn the resources to the reserves if we still have reserves!
Quoting Antonio Arribas, an exploration geologist, from an enlightening article published in the SEG Newsletter:

One was published in issue 2605 of the international science magazine
  New Scientist in May 2007 (Fig. 5). In it, the author claims that,
  based on data by researchers from Yale University and the University
  of Augsburg, “if predicted new technologies appear and the population
  grows, some key resources will be exhausted” within what is
  graphically shown as an alarmingly short time frame (Cohen, 2007).
  Among the mineral resources covered, platinum, indium, and copper are
  particularly illustrative of the errors of interpretation contained in
  the article and the overall fallacy that this report represents.
According to the article, humankind should be running out of platinum
  within four years (15 years from 2007). However, despite production of
  2,000 t of platinum between 2007 and 2017 and a metal price drop of
  26%, the global reserves of platinum group metals have decreased only
  by 2.8%3 (data from USGS, https://minerals.usgs.gov/
  minerals/pubs/mcs/). Likewise, according to this report we should have
  run out of indium sometime between 2012 and 2017. Never mind that, as
  a key component of indium tin oxide (ITO), this metal has been present
  as a transparent thin film conductor in the touch screen displays of
  every smartphone and tablet manufactured since then—including the 1.2+
  billion iPhones sold by Apple alone since the product’s release in
  June 2007, barely a month after publication of the New Scientist
  article.
The second article and infographic selected were titled
  “Global resources stock check” and appeared in a June 18, 2012,
  article from BBC Future (www. bbc.com/future/story/
  20120618-global-resources-stock-check). In this report, the world
  supplies of indium, antimony, and copper, just to choose three, were
  projected to last through 2024, 2020, and 2044, respectively. I would
  argue that nothing but common sense is needed to believe that by the
  time of the Paris Olympiad in 2024, ITO-based transparent thin film
  conductors (or a new, and likely superior, substitute) will continue
  to be present in digital displays and thin-film photovoltaic cells,
  among many other applications. As for antimony, how do we reconcile
  the prediction that we should have only two years left of the metal
  now when, after production of
  1.1 Mt between 2011 and 2017, global reserves remain unchanged within the
  1.5- to 2.1-Mt range that has existed every year since 1997, when the U.S. Geological Survey started to record reserve estimates? In other
  words, for the time being and for the foreseeable future, the reserves
  of this metal are controlled by economics, technology, geopolitics,
  etc., and not by absolute limitations of availability in nature.

I may agree with you that we have endless resources, but can they also be converted into reserves with 100% efficiency? For example, resources that are located in more troublesome parts of the Earth (volcanic/flooded/stormy) would require more cost investment than profit output. So, from a commercial point of view, they won't really be useful, and hence remain untouched.

@GaurangTandon Everything depends on cost, which is driven by demand. New technologies emerge that allow economical extraction of metals that were previously in low abundance or inaccessible. You should also remember that recycling is getting more and more efficient with time. I recommend reading the second book I linked to in my answer, or at least parts of it.

One obvious and interesting example is that it's nearly economic to mine old landfill sites. Prices of some materials wouldn't have to go up much for that to be viable.  In fact they probably don't have to go up at all, as the costs will fall because the process is still in a protoype stage.

@GaurangTandon Which is exactly why the idea of "peak some resource" has always been a manipulation and doom-mongering. As prices increase and/or technologies improve, sources that weren't profitable in the past become profitable. There's probably more gold dissolved in the oceans than has been mined in human history, but it's not economical to extract it right now. But "being economical" is not a constant - it's a (complicated) product of supply and demand, technology, investment... If oil prices keep increasing, at some point bio-oil will become profitable, no intervention needed.

@ChrisH Today's junk is tomorrow's valuable resource :) This has happened so many times over in human history, it's crazy how it hasn't become ingrained in people's minds yet. They usually don't even have a clue of what kind of waste processing we're already doing. Heck, pretty much the whole petroleum industry (including plastics, medicine, fertilizers...) started as a way to do something useful with the byproducts of petroleum production (which was just burned for light and heat!). In some places, most of the profit from herding cows is from the sale of manure, rather than milk/meat.

@Luaan, my city is one of very few to use anaerobic digestion on sewage to produce methane - an obvious renewable resource that's easy to store compared to other renewable energy sources. And yet it's barely economic

@Luaan who's to say it hasn't become ingrained in people's minds yet? Many people give recycling a lot of thought, in particular environmental-activist folks (who by your tone I reckon you don't generally associate with). It may be that cost will eventually make such and such process economical, but that doesn't necessarily mean it's a good idea to let it come to this without any regulation. The necessary new methods may have more adverse environmental side-effects. And enforcing recycling before it's actually economically worthwhile may ease the process, avoiding dispersal/mixing etc..

@leftaroundabout I don't have a problem with people who care about the environment - I care myself. But "environmental-activists" aren't always people who do their research, and many of their proposals are harmful. Plenty of companies invest their own money in pre-empting processes before they are economical (and hoping they become worthwhile soon enough to make a profit on them), but I find it unethical to fund such ventures through tax money. You can never predict if it's a net loss or a net gain for the environment (or the taxpayers) either. Look at the costs too, not just the benefits.

"we know exactly how much is in there to a good degree" That sounds like you're saying we know exactly, approximately how much there is. So which is it? "Exactly" or "to a good degree"?

@Michael - This is true, but it obscures an important fact. If indium becomes so rare that it needs to be extracted from extremely low-concentration sources or recovered from old devices, and the price of my iPhone shoots up to $10,000 because it’s just that important, indium might not have disappeared, but nor are things just humming along. A product has changed from a relatively affordable tool to what will likely be a luxury. A plentiful resource has tremendous advantages over a scarce one.

@Obie2.0 but this will increase exploration projects and turning "resources" into "reserves' and prices will go down again.

@Michael - I’m not so certain of that. You seem to be suggesting that there wouldn’t be any long-term trends in the real price of resources no matter how the difficulty of extracting them changed, and I don’t think that’s true. Improved extraction  or production techniques can lower prices (just look at artificial diamonds) - why don’t you think increased extraction difficulty can increase them?

@Obie2.0 if prices increase too much, the products that are being used by them will become too expensive. This will drive technological development of alternative that do not use the metals. Look, this isn't black or white - and the economics of mineral exploration and extraction are much more complex that can be covered in a Stack Exchange answer. I highly recommend reading at least parts of the two freely available books I linked to in the answer.

@Michael - But what if the alternative simply doesn’t exist? We’re limited by physics in many respects. Certain phenomena that we take for granted are unique or nearly so to certain materials (water and carbon being two examples). It’s possible that something like an iPhone can be made without indium, if indium prices became too high  - but it might be inferior or different in various aspects.

Well, if we're going into "what ifs", then what if a new major deposit of indium will be found? The truth is that there are plenty of indium deposits that no one bothered developing to "resources" or "reserves" because there are enough reserves already. The metals are there - the prices are not high enough yet OR we still have our current reserves so we don't need to develop more. Read the quote in the end of my answer slowly and carefully, and you will understand that scaremongering of "we're running out" isn't not grounded in truth (pun intended).


text:
Why are there only less than 120 elements out there ?
Why stop at 118 , Why aren't any more stable elements possible ?

As the elements become heavier, they become progressively harder to synthesize. Have a look at this page:https://www.chemistryworld.com/news/beyond-element-118-the-next-row-of-the-periodic-table/9400.article

Or: https://chemistry.stackexchange.com/questions/38847/is-it-possible-that-atoms-with-120-protons-are-possible-but-that-atoms-with-119

The stability of elements is really a question about the stability of their nuclei. What happens with their electrons (and their chemistry) is irrelevant.
Some nuclei are stable and some are not. The unstable ones fall apart by a variety of radioactive mechanisms (alpha emission, beta emission, positron emission...) resulting in different nuclei (all these mechanisms alter the number of protons and neutrons in the nucleus and therefore give a different nucleus as a product). Some decays happen slowly (potassium 40 has a half-life of about a billion years but oganesson (element 118, the heaviest known) has a half life of about 1ms).
The reason why we only see ~118 elements is because we only see the ones stable enough to observe. Anything common in nature would need to have a half-life comparable to the age of the earth (or be produced as a decay product of something else that does).
The reasons why some nuclei are more stable than others is a complicated area of nuclear chemistry or physics. Some broad rules are known. Nuclei with even numbers of nucleons are more likely to be stable. Even numbers of both protons and neutrons are particularly favoured. Some "magic" numbers of nucleons seem to be particularly stable. And, broadly, the bigger the nucleus the higher the ratio of neutrons to protons needs to be to stabilise it. But, at some point, bigger nuclei just get less and less stable so we don't see them (we have only ever made a handful of oganesson atoms).
There might be some magic combinations protons and neutrons that make some super-heavy elements a little more stable, but we haven't found a way to make them yet.

'Nuclei with even numbers of nuclei ' that sounds complex. :|

@theenigma017 Well spotted. I meant nucleons. Now fixed.

@OscarLanzi Had I said "anything that exists in nature" you would be right. What I actually said was "anything common in nature". I think I'm still right for reasonable definitions of "common".

The decay products are not all that common.  They would need a "geological" or "cosmic" half-life to accumulate to "common" levels given the slow production rate from the long-lived progenitor.  If you don't believe this try solving for the concentration of B if A -> B has a half-life of 5 billion yrs and B -> C has a half-life of 5 million yrs.

They haven't stopped at 118. The researchers at CERN and Dubna are looking into heavier elements. Such as element 120 and 122 etc. It's all about the "island of stability". https://en.wikipedia.org/wiki/Island_of_stability
Where the right amount of protons and neutrons increase the stability of the nucleus. It's going to be very difficult to get there due to the increased mass of the target nuclei and increased mass of colliding nuclei. So really the only limit to the elements scientists can create, is the tools they use

A different type of limit:  how far can we extend the Periodic Table?  This article, based on a joint study carried out at Massey University (New Zealand) and Michigan State University, suggests that relativistic coupling is seriously degrading the electronic shell structure by the time we get to oganesson (Z=118).  We may not see the shell structure survive the next period.  The Periodic Table, like the electron shells, would then end in a fog.


text:
As an atom get's bigger (on increasing atomic no.) , electrons have to move faster and faster around the nucleus, Why?
Does speed even make sense at such small scales ? Heisenbergs Uncertainty principle ?

It's not really the size and it's not the electrons in the outer shells that get faster. The charge of the nucleus is getting larger which means the inner electrons get closer and also faster.


text:
if difference of electronegativity between two atoms is 0.9 then which type bond will be formed an ionic or a covalent bond?
Bonds are formed by sharing of electrons .The difference of electronegativity determines which type of bond will be formed. If difference of electronegativity between two atoms is less than 0.9, then a non polar bond will be formed .If difference is between 0.9-1.7 then a polar bond will be formed with 50 percent ionic character .If difference is greater than 1.7 then it will be ionic  bond.But if difference is exactly" 0.9 ",Which type of bond will be formed?

Using electronegativity differences to determine bond character is just a rule anyway. It's not nearly as precise as you'd want it to be to properly answer this question.

This is as equally misunderstood as https://chemistry.stackexchange.com/questions/9222/why-are-bonds-ionic-when-the-electronegativity-difference-between-bonded-atoms-i?rq=1  and https://chemistry.stackexchange.com/questions/19568/what-happens-if-the-electronegativity-difference-is-exactly-2-1?rq=1

There isn't a clear limit between ionic and covalent bonds; it's more useful to think of covalent and ionic as two extreme models for bonding, with all real bonds falling somewhere in between in a continuum. We classify cases that fall close to one of those ends as "ionic" or "covalent", but in cases that fall somewhat in-between, it may not be easy to ascribe a label to them. Any limits (such as your values for electronegativity) are arbitrary and should be taken as a guideline and not as a strict definition.
In fact, as you advance in your understanding of chemistry, it makes more sense to talk about the ionic character of bonds than to talk about ionic and covalent bonds - with bonds close to the ionic ideal having a high percentage of ionic character and bonds close to the covalent ideal having a low percentage of ionic character. There are schemes to quantify ionic character from electron densities (either experimental or calculated); for instance, there's a classic expression proposed by Linus Pauling for diatomic molecules in the 1960s, as a function of electronegativities:
$$ f_{AB} = 1 -  \exp{[-0.25 (X_A - X_B)]}$$
which gives values for ionic character, $f_{AB}$, between 0 (purely covalent) to 1 (purely ionic). More modern expressions are far more complex and depend on other electronic properties, but they all give you fractions of ionic character in a continuum from pure ionic to pure covalent.
So I'd suggest to picture cases that happen right at an arbitrary boundary simply as "grey" cases, with a mixed character; in your case, at $\Delta X=0.9$, you'll have somewhat polar covalent bonds.


text:
Why is it so hard to turn plastics into something useful or decompose it?
Even after so many years, why haven't we yet come up with a solution to the plastic problem ?
Why are plastics so hard to decompose via chemical or physical processes into something harmless or useful ?

Its not hard, its expensive and the cost of new plastics has been so low that it was not worthwhile. Awareness of the 'plastic problem' is pretty recent and still has not reached some places.

I was looking at more like - you leave paper, cotton, organic stuff on the ground , it disappears in a few days-weeks , why does plastic stay there for years ?

Think about the chemical structure of the materials and what holds them together, mostly proteins and polysaccharides so esters, amides and acetal linkages. Thousands of enzymes evolved to cleave them. Plastic OTOH are all C-C bonds, fewer enzymes do this.

Oh i see.... :p

The (ironic? funny?) thing is, we have this great treasure of petroleum that we use to create many plastics precisely because nature once invented a chemical substance that could be built enzymatically, but, at that point in time, not be destroyed enzymatically. It was a great success, because organisms that built with it were built for eternity - you could gnaw them, but it would not benefit you because nobody could metabolize it. This giant wealth of indestructible organic matter accumulated and accumulated, and even existed for geological timeframes, getting submerged under sand and mud, slowly degenerating to become petroleum. Most lifeforms never caught up, which is why even today cows need special gut bacteria to crack the building blocks of the grass they eat. without them, they would benefit from the grass as much as we do - not at all. The grass would come out just as it went in, just more chewed. ---- So with the plastics that we produce, we have a similar conundrum, there are enzymes to break many of them down, but the creatures that posess them do not rely on them, and therefore do not produce a great deal of them. That's good because otherwise your Tupperware bowl might start being decomposed, and it's bad because all the plastic garbage (or its degenerated form) will be around for a very long time indeed (as were the plants that degenerated into petroleum - hundreds of millions of years, and that material is still around!).
Chemically and physically the problem is not as big.
The real problem is economics: To make a plastic bag, investment on the order of one tenth of a cent is necessary - so why not make one for every single piece of consumer goods? To store a discarded plastic bag on a trash-heap? A hundreth or thousandth of a cent. But to clean off all the muck, sort them, recycle them?  Two tenths of a cent. So a hundred percent more investment than to produce a new one, and ten to a hundred times more than to just stash it. This is where the government (could/should?) step in: Just tax the shit out of petroleum, and the industry will suddenly find it cheaper to recycle - and then they will, because already they can. (Why tax petroleum, not plastic-generation? Because what's needed for recycling is energy, and if it's just the plastic-generation that's taxed, the industry will burn petroleum to make energy to recycle plastic which would really defeat the purpose...)
You could also genengineer organisms to metabolize most plastics efficiently, but (apart from the (again: main point) economic problems) that would have serious consequences in regards to CO2 (metabolizing ultimately means reducing to CO2), technical safety (currently we rely on plastics ability to not be eaten in a plethora of use cases), and biological safety (who is to tell what might come from those super-metabolizers?) - so the safer way would be to go with the technical solution. Which ist just a matter of economics.

Most plastics are reusable . The problem is wastes are mixes of various different plastics and other materials . The cost to reliably separate them into specific materials is greater than the cost to make new plastic.


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How does Le Chatelier's principle apply to a gas phase reaction at constant volume?
This reaction is given in my book and the direction of shift of reaction on addition of $\ce{CO}$ and removal of $\ce{CO2}$ at constant volume is asked.
$$\ce{2CO(g) + O2(g) ⇌ 2CO2(g) + \Delta H}$$
If "at constant volume" was not specified, I understand that the forward reaction would occur since the reactant is added and the product is removed, but I do not know if this will vary with constant volume. What will happen in the case of constant volume? Does it matter that there is a change in the number of particles comparing reactants and products? Which direction will the equilibrium shift?

Since all your species are gases, for Le Chatêlier purposes, it's relevant whether you are working at constant pressure or at constant volume. At constant volume, total pressure will be proportional to the total number of gas molecules (of all species). You can figure out how will pressure change in your case; equilibrium will shift to counteract that change.

I understand that the forward reaction would occur since the reactant is added and the product is removed

You have to be careful about the physical state of reactants and products when making such a broad statement. In the reaction $$\ce{NaCl(s) <=> Na+(aq) + Cl-(aq)},$$
nothing will change if I remove some of the solid sodium chloride or add some - the equilibrium will not be disturbed. If I add some silver nitrate, though, silver ions and chloride ions will form a precipitate, the chloride concentration will drop, and the equilibrium will be disturbed.
In general, the concentration (or partial pressure) of at least one of the species has to be changed in order to disturb an equilibrium (unless of course you change K by changing the temperature).

What will happen in the case of constant volume?

This, I think, is supposed to make the question easier, not harder. If you add a substance at constant volume, you know that the partial pressure (the concentration) of that substance will increase while the other partial pressures remain the same.
At constant pressure, however, the other partial pressures change too. You could have a reaction where the small changes in partial pressure of the other species has a larger effect on the equilibrium, and a shallow application of Le Chatelier's principle will yield an incorrect answer:
$$\ce{A(g) + 2 B(g) <=> P(s)}$$
If this system is at equilibrium with equal partial pressure of A and B, and I add some A at constant pressure, the partial pressure of B would drop (I am "diluting" B by adding more A). Because B has the higher coefficient (higher exponent of the concentration of B in the equilibrium expression), there would be a reverse reaction toward equilibrium. This is not what you might expect.
So, the question avoids that complication by saying "at constant volume".

Let's see your formula (with the difference of enthalpy being negative):
$$\ce{2CO (g) + O2 (g) <=> 2CO2 (g)}\qquad(\Delta H^\circ<0)$$
At constant temperature, for Le Chatêlier purposes,

Adding $\ce{CO}$ produces that it react with $\ce{O2}$ to produce more $\ce{CO2}$ to counteract the effect of the applied change and a new equilibrium is established.
Removing $\ce{CO2}$ produce that $\ce{CO}$ react with $\ce{O2}$ to produce more $\ce{CO2}$.

When you are adding or removing a reagent or product, and all components are in a gaseous state, you are changing the pressure and you can study the reaction by $K_p=\dfrac{P^2_\ce{CO2}}{P^2_\ce{CO}P_\ce{O2}}$.
We also have $K_p=K_c$ $RT^{Δn}$:

$K_p$ is the equilibrium constant written with partial pressure of gases.
$K_c$ is the constant but being written by concentration of substances.
$R$ is universal gas constant.
$T$ is the absolute temperature (Kelvin degrees).
$\Delta n$ are moles $(\Delta n=n_{\text{products}} - n_{\text{reagents}})$.

If you change concentration of $\ce{CO}$ you are changing the pressure it, too.
But you can change also the pressure, by changing the volume. (For Boyle's Law: the absolute pressure exerted by a given mass of an ideal gas is inversely proportional to the volume it occupies if the temperature and amount of gas remain unchanged within a closed system).

For example, if you increase the pressure, volume will decrease and the system will shift towards it will be less molecules to counteract the change. In this case, to products. If you in decrease pressure, volume will increase and the system will shift towards reagents.
Other option is to change the pressure by adding or removing an inert gas, but since it don't intervenes in the reaction, the equilibrium is not affected.


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Is there an equivalent unit cell that can describe a bcc unit cell where all points are the same element?
Similar to how an NaCl-structure can be described as just a primitive cubic structure if all the lattice points were the same element or ion. I'm asking if there any such alternative for the CsCl-structure.

(comment made into answer)

You certainly can construct what is called a primitive cell containing one atom for a bcc lattice.  There could be more than one way, but here is my method.
Start with the usual cubic representation.  Pick any face and connect the midpoints of its edges to make a smaller square.  Do the same with the opposite face.  This gives a pair of squares which are then the bases of an elongated square prism, whose lateral edges connect the two opposing faces you "contracted".  This cell, which contains just the central atom of the original bcc structure, is a primitive cell.  In this cell the lateral edges, which we did not "contract", are about 1.41 (actually, the square root of 2) times as long as the basal edges.
How does this replicate to make the lattice? If you just scoot the primitive cell over to "catch" a neighboring atom on either side of the lateral faces, you find that you must move the adjacent cell "up" or "down", in the direction perpendicular to the bases, to get the second atom centered properly.  The "up"/"down" move is half the height of the prism.  But once you see this, you can repeat this process and get all the atoms involved in the entire cubic representation.
Last question:  I said the primitive cell is an elongated square prism.  It no longer has cubic symmetry.  Where did the cubic symmetry go?  The answer is you can't make a primitive cell in the bcc lattice with the full cubic symmetry.  Instead, note that when you modified the cube you had the choice of three pairs of opposite faces to "contract".  Thus there is really a combination of three primitive cells with different orientations.  The cubic symmetry emerges from the combination rather than any one primitive cell.

Yes there is. In your particular case, it is also the primitive cubic lattice.
A hint is in the name - BCC = Body Centered Cubic Any arrangement that is BCC is also primitive cubic, for the isolated constituent building blocks.

I guess I just don't see that though, where can I see the primitive cubic lattice in a BCC system?

I can't see a primitive cubic lattice either.  I end up with a tetragonal one instead.

@OscarLanzi I have perhaps misunderstood the question. If you take the lattice of element A in an AB bcc cell, the elements A and elements B form primitive cubic lattices with each other. Not elongated prisms and not tetragonal - however if the question supposes you replace A with B I might understand your comments. (?)


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Calculating the Radius of a Molecule
Assuming the molecule is a sphere, would the radius of amylopectin be $9.17*10^{-22}~cm^3$? I arrived at this answer by deriving the molar volume by dividing the molar mass ($828.72~g/mol$) of amylopectin by its density ($1.5~g/cm^3$). Then I divided the quotient by Avogadro's constant to arrive at my aforementioned answer. However, I don't know if this is correct so please correct me if I am not.

The value you quote is a volume not a radius, so you need to take the cube root. Proteins are no more than a few nm ($10^{-9} $m) in radius.

As @porphyrin pointed out, you got the volume of the particle, but should assume it as a cube instead of a sphere to avoid void volume trap among spheres on packing (see answer of  TIME RUB below). Thus, the cube root of your answer gives you the length of a side of that cube.

When you divide the mass of the solid by its density then you get the volume of the object which means not only the volume of the molecules also the volume of voids. Now the volume you get is = volume of the particles + volume of voids. So the radius you get is = radius of void + radius of molecule.  So in order to solve the problem you must which kind of packing is this. And then you should know it's packing efficiency. So now the volume you got should be multiplied by the packing efficiency to get the actual volume of molecules. If you say why do they have voids. Then I'll  say you can't fill a particular shaped object with small sphere's so packing efficiency can't be 100%. So all you need to know is the kind of lattice formed by the molecules of amelopectin. And now what you got in your question is the volume of single particle + the volume of void. Well if you wanted to get the radius from it then it would be error as I said before. But in that case also you had to put $V=\frac{4}{3}\pi(r^3)$= the volume of the single molecule. And in this way you get the radius as 6.02*10^-8cm. And then get the radius. But actually taking the molecules as cube gives you a better answer though this would not give the proper answer if amilopectin molecules aren't cubic. But certainly you can fill a shape with cubes more efficiently (without having much voids) so in that case $a^3=volume you mentioned which is 9.17*10^-22 cubic cm. That gives a as 9.7*10^-8cm as the side length.

@brooklyn I have edited my answer. Please state if you have doubts.


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Structure of trichloroborazine
I came across a question that asked the correct structure of trichloroborazine. The two options provided were :
a) the chlorine atoms attached to the three N atoms, and
b) the chlorine atoms attached to the 3 B atoms
My idea was that, since, in borazine, the N atoms have a formal +1 charge, the Cl would more readily bond to it, making (a) the correct answer.
But, turns out, it's actually (b) that's the correct structure.
Why is it so, and what is wrong with my reasoning?
Edit: If both (and other) structures are at least theoretically possible, then which one is the most stable and/or most common one, and why so?

The big issue I see in your reasoning is that formal charge is a formality, not a real charge. Don't use it where real charges matter.

@zhe I get that, yes, and I only employed that line of reasoning because I couldn't think of any other. It was a case of "better this than nothing". I'd love to know the actual/correct logic...

I'm going to put it out there that your question, as phrased, doesn't make sense. If the question was "what is the correct structure of water", then fine, H–O–H is the correct structure. However, the name "trichloroborazine" is ambiguous and can refer to either of the two structures you have chosen, or potentially even more as there is no stipulation as to where the chlorines are attached to. So, there is no single correct structure corresponding to the name "trichloroborazine".

Granted, (b) is probably a more stable isomer than (a), or any of the other possible isomers. But then you need to rephrase the question appropriately.

@orthocresol I understand that there are other possible structures; however, the question was given as I have asked it. Also, a quick Google search for trichloroborazine does provide results only for 2,4,6 trichloroborazine (with cl attached to B atoms). So, are there any reasons for this structure being the most common/stable one?

@orthocresol Also, I have edited my question to hopefully clarify what I'm looking for

Yup, that's much better!


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Why does 1 mol. of any gas at STP occupy only 22.4L Volume when it should occupy the volume of entire container in which it is kept?
Let us say we have a container of 100L capacity and a gas inside it is kept at STP. According to the law the gas should occupy only 22.4L of volume but shouldn't the gas occupy the entire volume of the container in which it is kept(100L in this case) as gases expand without limit?

Simply, the amount of your gas in 100-mL container at STP is more than $\pu{1 mol}$.  Actually, it is a little less than $\pu{4.5 mol}$. As you said, yes, gas expand or compress, but other two factors (P & T) will change according to the amount of gas in the container, according to $\mathrm{PV=nRT}$.

Not 100 mL but 100 L.  Stuff happens.

Let us say we have a container of 100L capacity and a gas inside it is kept at STP.

ok, let us think about this and let us assume ideal gas for this whole topic.
We got a container with 100 L volume at 0°C and at 1bar. This means there has to be 100/22.4 mole of gas in there, so roughly 4.5 mole of gas, not just one.

According to the law the gas should occupy only 22.4L of volume but shouldn't the gas occupy the entire volume of the container in which it is kept(100L in this case) as gases expand without limit?

If we put one mole of gas into a 100 L container it will expand to the whole volme but that means that temperature and/or pressure have to change and then it is not STP any more! If we put 1 mole of gas into a 100 L container and we keep the temperatur at 0°C the pressure would go down to 0.224 bar.


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Avogadro's gas Law doesn't make sense? (it does actually)
It says equal volumes of all gasses under same conditions of temp. and pressure contains the same number of atoms/molecules.
Now suppose you have 1L of Chlorine in a box and 1L Hydrogen in another box , now the comparitively huge size of the Cl atom means less of it will fit in the box , whereas lots of hygrogen will. ( breaking the law)
But, well this should have an effect in pressure , how does all that boil down to avogadro being right ?

Molecules are very small indeed so what you say about not fitting in is not true; there is plenty of space.  If you don't believe me work out what fraction of the volume of 1 litre Avogadro's number of molecules of each of your gasses would occupy. You can easily find the bond lengths and atomic radii needed, perhaps even molecular volume.

The gas laws assume ideal gases which thus ignore atom/molecule size and atom/molecule interactions.

I dint get it But if you scale that up many many times you should actually "see" the difference @porphyrin

@porphyrin we can fit more chlorine but that will lead to a higher pressure in the chlorine box !

Oh it definitely makes sense since it allows us to describe a lot of systems with very simple formulas.
The problem is that this law assumes "ideal gases" (just like the ideal gas law) where atoms/molecules are "points" (no volume) and they don't interact with each other. This works well at rather low temperatures, low pressure and for gases without strong intermolecular interactions.
But you are correct, there are problems and the law doesn't work in a lot of cases. For example if you go to high pressures the volume of the indiviual atoms/molecules does play a role, since if they are bigger the available space will be smaller and the pressure will be higher than predicted by the ideal gas law.

Exactly this was what i was looking for it had to have that point molecule assumption, I've been occasionally thinking about this for like a year !

our teacher never mentioned that !

You're right that it should have an effect on pressure-and it does. It's just that it's practically a negligible amount.
One of the assumptions of the Ideal Gas Model is that the volume of the molecules is negligibly small compared to the volume occupied by the gas. This assumption allows us to simplify the complexity in the derivation. Note how the only mention of the atom's volume is in the assumption and left completely out of the mathematics.

Exactly this was what i was looking for it had to have that point molecule assumption, I've been occasionally thinking about this for like a year !


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Why do carboxylic acid derivatives show a more upfield shift than carbonyls in 13C-NMR?
The prototypical 13C NMR shifts of carbonyl carbons is in the range 190–200 ppm, however, carboxylic acid derivatives have a 13C NMR shift in the range 160–180 ppm instead. What structural factors cause this shift upfield? Shouldn't the carbon in the carboxylic acid be more deshielded than the analogous carbonyl, since there are two oxygens instead of one?

No time for a full answer now but this mostly has to do with the $\Delta E$ term in the paramagnetic shift, cf. Gunther NMR Spectroscopy 3ed p410 and https://pubs.acs.org/doi/pdf/10.1021/ed077p905

Shouldn't the carbon in the carboxylic acid be more deshielded than the analogous carbonyl, since there are two oxygens instead of one?

Well, no. I love this misconception and I'm always having fun when doing oral exams about reduction reactions and grignards, which is basically the same problem.
If you look at reactivity of acids/esters and ketones/aldehydes against nucleophilic attack (grignard reagents or hydride reagents) you will see that ketones and aldehydes are much more reactive towards nucleophiles than acids/esters are. So why are ketones and aldeyhdes more electron deficient on the carbonyl carbon?
Yes, in case of acid/ester you got two oxygens both with electron withdrawing -I properties, but the additional oxygen also got lone pairs so a +M effect is possible. This actually pushes more electron density towards the carbon than the -I is pulling away leaving the carbonyl carbon in acids and esters overall more electron rich than in aldehydes and ketones.

Can you explain what -I and +M means?


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How does calcium chloride act as a sequestrant (food science)?
This Wikipedia page lists common sequestrants (with respect to food science). As stated on the page:

A sequestrant forms chelate complexes with polyvalent metal ions, especially copper, iron and nickel, which can prevent the oxidation of the fats in the food.

To me, this behavior makes sense for every example listed except calcium chloride. I can certainly see chloride complexing these metal ions, but even then, I typically don't expect the chloride ion to reduce catalytic capabilities much.
I assume the calcium cation has an effect on these processes somehow, too. Otherwise, I would expect regular table salt to have the same effect, but, while discouraging bacterial growth, I don't think sodium chloride reduces the rate of oxidative rancidification.
Could it be as simple as the desiccant effect of calcium chloride keeping the polyvalent metal ions from dissolving? It would seem like a massive amount of calcium chloride would be needed to have the desired effect if that was the preservation mechanism involved.

Sodium chloride actually increases the rate of rancidification as per this paper: https://onlinelibrary.wiley.com/doi/pdf/10.1111/j.1365-2621.1950.tb16478.x. Astonishingly, calcium chloride increases rancidification even more than sodium chloride.

@ShoubhikRajMaiti So, that would suggest that the Wikipedia page is inaccurate? Or that the using the term "sequestrant" for calcium chloride is inappropriate? The Wikipedia page for calcium chloride also says that the EU considers it a sequestrant.

If think that if the poster of the question wants to know how calcium chloride acts as a sequestant, then they should make sure it truely acts as a squestering agent. And also please tell us what substance the calcium chloride sequesters. I know that calcium chloride can sequester ammonia, but unless we know what is being sequestered we can not be able to explain how the sequestering occurs.

@NuclearChemist well, seems pretty obvious that the "poster of the question" put the quote from the Wikipedia page, which specifically mentions chelating polyvalent metal ions (specifically copper, iron, and nickel) directly in the text. Since a sequestrant in food science refers to a compound that chelates polyvalent metal ions, I'd guess those polyvalent metal ions were what the calcium chloride is supposed to be sequestering.

No I do not think it is clear that the original poster is looking for a thing to sequester metal ions, the problem is that words are not always used in the same way by people from different academic backgrounds. You are right that "sequester" is often used in relation to metal ions but it is not the only use of the word.

@NuclearChemist I mean, how much clearer could I get than "A sequestrant forms chelate complexes with polyvalent metal ions, especially copper, iron and nickel, which can prevent the oxidation of the fats in the food"?

You are getting into the problem of "some vs all", we can agree that some reagents such as EDTA and potassium cyanide can mask or sequester transition metal ions. But this does not mean that all sequestering agents are designed to sequester metal ions. If we consider for a moment cholestyramine this is a medcine which is able to sequester bile acids (it is a ion exchange resin with strong basic groups). This clearly is not targeting transition metal ions.

@NuclearChemist Have you looked at the Wikipedia page? Maybe it should be called "Sequestrant (food science)" or something, but maybe I should have used that qualifier here. I'm referring to "sequestrant" here as the term is used in food chemistry/food science.

I do not regard wikipedia as a source worth quoting, it is a interesting social experiment but sadly it is a poor source. Please come back with a better and more trustworthy source.

@NuclearChemist As unnecessary as I find this, see "Food Additives Data Book," chapter 12, page 820.  https://doi.org/10.1002/9781444397741.ch12

When I get the time, I will read the source and give my opinion

@SendersReagent I don't really know. This is confusing.

@ShoubhikRajMaiti The more I look into it, the more I think that calcium chloride is called a sequestrant in food science even though it doesn't meet the food science definition of a sequestrant. I think it's more of a general preservative, but I'm not completely (or even mostly) sure what the deal is here.

Food Scientist here. Calcium chloride is typically used in high concentrations in pickles to keep them firm (give that desired bite) and also add a slight salty flavor without increasing sodium content. It is also used in potato chips to reduce formation of acrylamide. Calcium chloride is highly soluble in water and will dissociate into Ca2+ and 2Cl-. It is used to sequester metal cations. The metal ion complexes with the free Cl-. It has also been found that calcium ions may reduce iron absorption, and is a divalent cation that can form complexes of its own. Calcium chloride can stabilize colors, flavors, antioxidants, and also improve oxidative stability. Fats and oils readily oxidize in the presence of metal ions, so a sequestrant such as calcium chloride is used to form a complex with the metal. Additionally CaCl2 is used to stabilize gels such as those formed with alginate. Another popular sequestrant, or chelating agent is EDTA.

As I mentioned in the question, though chloride can complex metal ions, I wouldn't expect it to do much in that case. Chloride is pretty weakly coordinating, so I wouldn't expect it have much effect on the metals' catalytic abilities.

Is it possible the calcium excess can cause competitive inhibition of metal ions taking part in redox reactions ? So it could achieve the similar effect by different, non chelating way.


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Why isn't pure ortho hydrogen possible to obtain?
In school, I've learned that we can only get an ortho hydrogen content upto 75%. Why is that we can obtain a pure sample of para hydrogen, but not ortho hydrogen? What prevents us from isolating ortho hydrogen?

Just how are you going to isolate it?

@IvanNeretin I couldn't imagine a method myself, but it seems weird to me that we are able to get 100% para hydrogen but not 100% ortho, when the only difference is in nuclear spin. Also the fact that the highest possible concentration we can obtain is 75%, why couldn't we go further than that?

Should seem more weird that some separation can be done, when the only difference is in nuclear spin, which is awfully close to "no difference at all". In general, "why can't we" is almost never a question worth asking. Why can't we shoot an arrow into the Sun, really?

@IvanNeretin In general, 'why can't we?' is a question worth asking. These tend to be the most useful and fruitful for advancing one's intuition. (That's my humble opinion at least.) Your arrow-to-the-sun is an interesting mechanics question, quite advanced even once the whole picture is taken into account; I would probably upvote such a query if it showed some effort.

Ortho and para hydrogen differ in the alignment of their nuclear spins. The two hydrogen atoms each have a nuclear spin of 1/2 which results in a total nuclear spin of 0 or 1. The states with zero spin are called para and those with 1 ortho. The combination of the nuclear spins also results in the formation of a new spin wave function. There are 3 possibilities to get a nuclear spin of one with an even wave function and there is one possibilitie to get a spin of zero with an odd wave function. The overall wave function of the hydrogen molecule is given by the product of the electronic, vibrational, rotational and spin wave functions and should we antisymmetric with respect to exchange of the fermionic hydrogen nuclei. Because the electronic ground state of molecular hydrogen is even and the vibrational wave function of a diatomic molecule is always even, the wave function of para hydrogen has to be combined with the even rotational states and ortho with the odd rotational states. The absolute ground state of molecular hydrogen is thus the rotational state with zero rotation and is therefore para. The next state, with one quantum of rotation, is ortho.
You can only convert ortho and para hydrogen into each other using collisions or chemical reactions. To make pure para hydrogen, people use a catalyst at very low temperature. Because of the low temperature, all produced hydrogen will eventually end up in the absolute ground state, that is, para hydrogen. You cannot make ortho hydrogen this way because it does not represent the absolute ground state of the system. The 75% comes from the degeneracy of the Boltzmann factor: at low temperature (with no reaction catalyst) there are three times more ortho molecules with a spin of 1 than para molecules of zero spin.
Although you cannot make pure ortho hydrogen, you can study it by using narrow lasers that only excite odd rotational states.


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Is Biotin water-soluble or fat soluble?
Vitamin $\ce{B_7}$ or Biotin is a member of the vitamin B complex. Members of the vitamin B complex are usually known to be water soluble. However, this webpage mentions that biotin is water soluble. Again, according to another webpage, biotin is neither fat soluble, nor water soluble.
So, is biotin soluble in fat or water? As different sources are contradictory, please provide a reference.

Water solubility
The CRC Handbook of Chemistry and Physics, 97th ed., Physical Constants of Organic Compounds, p. 3-48, describes biotin solubility: soluble in water, ethanol; slightly soluble in diethyl ether and chloroform.
Descriptive terms for solubility, p. 1-38 (solubility – parts of solvent required for 1 part of solute (recalculated to mass%)):

very soluble – less than 1 (> 50%)
freely soluble – 1 to 10 (50% to 9.1%)
soluble – 10 to 30 (9.1% to 3.2%)
sparingly soluble – 30 to 100 (3.2% to 0.99%)
slightly soluble – 100 to 1000 (0.99% to 0.1%)
very slightly soluble – 1000 to 10000 (0.1% to 0.01%)
practically insoluble, or insoluble – >= 10000 (< 0.01%)

Aqueous Solubility and Henry’s Law Constants of Organic Compounds, p. 5-141, states that biotin solubility is 0.35 g kg−1 H2O (0.035 mass%) at 25 °C.
This solubility value would be described as very slightly soluble.
I don't know how to explain this inconsistency.
Octanol/water partition coefficient
.. However, low water solubility does not necessarily mean high fat solubility. A standard measure of lipophilicity is octanol/water partition coefficient (or its decadic logarithm, log P). Positive means more soluble in octanol (fats), negative means more soluble in water.
Experimental log P values for fat-soluble vitamins:lit

vitamin A +5.68
vitamin D3 +10.24
vitamin E +12.18
vitamin K0 +2.20

Typical water soluble vitamin:

vitamin C −1.85

Some B vitamins:

vitamin B1 −3.93
vitamin B2 −1.46
vitamin B4 −0.09
vitamin B7 +0.39
vitamin B8 −1.68
vitamin B13 −0.38

Biotin (vitamin B7) is interestingly slightly lipophilic, or little more fat-soluble than water-soluble.
Biological systems
Even though biotin solubility in water is quite low, it is still classified as water soluble. Originally, biotin (named vitamin H, originally) was discovered in extracts from egg yolk or liver, which are rather fatty tissues, in methyl ester form, practically insoluble in water.
On the other hand, biotin was originally not found in egg white, because it contains avidin protein, which binds biotin so strongly, that it's one of the strongest non-covalent interactions known.
Biotin is often bound in dietary sources to proteins as amide on lysine. By digestion, it's metabolised to biocytin, then to free biotin.
So the fat or water solubility view may be somewhat simplistic.

Solubility in water 22mg/100ml at rt according to this source: 1  and this one 2  which also notes that it is recrystallized from hot water
20mg/100ml according to the Sigma Aldrich catalogue 1