text:
When I burn isopropyl alcohol (IPA), it burns orange. But when I burn ethyl alcohol, it burns totally blue. Why is this?
I burned them in a small aluminium tray. While IPA is burning orange, it produced the smell of soot, but while ethanol is burning blue there isn't any smell.
Also, ethanol made the tray really cold when I poured it on the tray before burning it, compared to isopropyl alcohol. How can this be explained?
Note: IPA concentration is 99%, Ethanol is 97%

Are you sure that your isopropanol is pure ? It does not look like so.

What percentage isopropyl alcohol? 99.5%? 70%? 91%? What grade (industrial, laboratory, etc.)?

The nice thing about very simple experiments, like this one, is that they can be repeated by others, if only to verify the reported observations. Unless this is done, explanations run the risk of getting too far in front of the skis.

@PeterMortensen it would certainly have to be ruled out before proposing something else more complicated! I've added a supplementary answer borrowing from Physics SE.

Interesting observation. The blue flame color of all hydrocarbon fuels is due to the emission small diatomic carbon species such $C_2$ or CH. There is nothing magical about IPA having a yellow flame. The yellow flame originates from incomplete combustion. There is more carbon per mole of IPA as compared to ethanol. Yellow flames are called reducing flames and blue flames are called oxidizing flames.
In older times when Bunsen burner was taught in detail, it was shown a blue flame of methane can be readily converted in to a yellow flame by altering the air supply valve. The yellow color, if you view through a spectroscope is a continuous spectrum (rainbow like), which shows that it is like a black body radiator. The black body radiator is nothing but glowing soot (carbon) particle, glowing chrcoal but a very small one. On the other hand, the blue flame shows band like structure. I once had a chance to view the blue acetylene flame with air with a diffraction grating. It was an amazing sight. The structure of colored bands was never seen before. They are called Swan bands. Unfortunately, I cannot find any color images in Google Images of Swan bands.
Here is one example from a 1857 paper by Plucker and Hittrof, "I. On the spectra of ignited gases and vapours, with especial regard to the different spectra of the same elementary gaseous substance". This more than 150 year old picture is not doing justice to what you see in reality of an extremely beautiful spectrum.

Swan bands are a topic in Flame Spectroscopy, Parts 1 and 2, by Radu Mavrodineanu and Henri Boiteux, Wiley, 1965. I do not have this classic book and it is extremely hard to get an original, though the reference above might be a later edition or reprint. Mavro used many exotic flame mixtures, including cyanogen and oxygen. I bet there are color plates in the book: they did a professional job in the old days! Maybe someone has the book and can check.

Thank you Prof. Ed, Mavrodineanu's edited book on Analytical Flame Spectroscopy is online on Internet Archive. Sadly no color plates there. I checked Swan's 1857 original paper, no figures. However, I was shocked to see the comment in the book "Draper in 1848 [26], on looking through a spectroscope at the cyanogen flame, said: "There was a spectrum so beautiful, that it is impossible to describe it by words or depict it in colors." I cannot agree more.

I saw the Swan bands from the blue portion of air acetylene flame of an atomic absorption spectrophotomer. Everyone can show it as AAS is common in every university.

Yellow: There is always the possibility of "contamination" with sodium - it doesn't take much to make a flame yellow. I am surprised that an alcohol with only one more carbon atom compared to ethanol and nearly the same boiling point (78 °C vs. 83 °C) would exhibit this behaviour (stearic acid has an 18-carbon chain). The azeotrope with water is 88 wt% (96 wt% for ethanol). Isopropyl alcohol vapor is denser than air - could that play a role?

Getting from the azeotrope to 99.5% may require some "chemical" means, potentially introducing small amounts of sodium. It seems one of the methods for breaking the azeotrope actually is adding NaCl (and destillation)...

If we are assuming IPA was contaminated for some reasons, why should we assume the OP had ultrapure ethanol. He can never have access to absolute ethanol which is dry and pure. The student clearly mentioned soot which implied that it was a reducing flame. Oil flames produce a lot of soot too, I am not sure if you have seen old oil lamps. They are still used in some some Indian temples during worshiping.

@M.Farooq Swan band sighting in this answer. I think that sodium would have to be ruled out before proposing anything else considering that there's already water in the alcohol. The ethanol may have less sodium than isopropyl just due to the differences in the way that they are distilled or otherwise manufactured.

@uhoh well, the "smell of soot", mentioned in the question, should indicate that at least partial reason for orangeness of the flame is continuous radiation from the soot. But yeah, it'd be best to do an experiment to check this. The simplest way to do this is the first option I did in my answer you linked to: using a diffraction grating. It would be quite easy in the conditions of this OP.

@Ruslan I got so excited remembering your answer there that I completely forgot about blackbody radiation from the soot! Where I live the alcohol available at the corner store is 70% ethanol believe it or not, I don't know where to get isopropyl here except a laboratory supply place. I'm going to go have some fun in the next few days, I've got a diffraction grating here somewhere...

IPA has a different carbon:hydrogen ratio than ethanol. There is more incomplete combustion occurring with IPA, hence the smoky orange flame and smell of soot. Ethanol combusts more completely, leading to a blue (soot-free) flame and no smell.
In response to your second question, ethanol likely has a lower latent heat of vaporisation than IPA, resulting in it evaporating rapidly. It takes a lot of heat energy away from the tray when doing this, leading to the tray cooling. A similar effect can be observed if you accidentally get some types of solvent on your hand and feel a sudden coldness as they vaporise, taking heat from your skin.

It has more to do with the length of the carbon chain and strong emission of C-C like radicals. Methylalkohol Burns by almost invisible flame. Erhylakohol with blue flame, sometimes with yellow in some flame parts. . IPA is expected to make more of yellow/orange color.

Yellow: There is always the possibility of "contamination" with sodium - it doesn't take much to make a flame yellow.

@Peter Mortensen It is true. But I suppose one can distinguish it from the BB-like radiation by colour and mainly the spatial distribution of colour across the flame. And there is the mentioned "smell of soot", leading to  C-C chains, what cannot be managed by sodium.

Expanding on @PeterMortensen's comments (1, 2) here is some other discussion of how a small contamination of sodium can lead to orange flames:
From Why does the humidifier make a stove's flame orange?:

From this answer to it:

OK, I've managed to measure some spectra using my Amadeus spectrometer with custom driver. I used 15 s integration time with the flame about 3-5 cm from the SMA905 connector on the spectrometer body.
Below the two spectra are superimposed, with the blue curve corresponding to the blue flame, and the orange one corresponds to the flame with some orange. I've filtered the data with 5-point moving average before plotting. The spectrometer has lower sensitivity near UV and IR, so disregard the noise there.
(Click the image for a larger version.)


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Mesomeric Effect and its Strength
While solving a question in organic chemistry, I needed to compare the strength of mesomeric effects of various groups.
I looked it up on Wikipedia, and it goes like this :
+M EFFECT ORDER :
–O− > –NH2 > –NHR > –OR > –NHCOR > –OCOR > –Ph > –F > –Cl > –Br > –I
-M EFFECT ORDER :
–NO2 > –CN > --S(=O)2−OH > –CHO > –C=O > –COOCOR > –COOR > –COOH > –CONH2 > –COO−
What is the way to understand such comparisons ? . Why does the -OH group show stronger mesomeric effect than the -NH2 group ? Also, why is an -OR group weaker than the -NH2 group ? Can somebody please explain ; I was unable to find a reasonable explanation through a google search.

Some background knowledge
You need to be aware of electronegativities, assignement of s and p character, relation of structure to the donating or accepting nature of the substituent, and most importantly, the overlap efficiency of the donor and acceptor positions (roughly estimated by overlap integral , although for the sake of this discussion, most of the mathematical rigor is not required)
I will be comparing most of the order for the +M effect, and provide some hints for the -M effect.
Group 1: from $\ce{O-}$ to $\ce{OCOR}$
This group of substituents consists of two kinds of donor atoms from the 2nd period, namely $\ce{O}$ and $\ce{N}$. As they are from the same period as $\ce{C}$ , therefore, overlap efficiency will be maximum in this case, for 2p-2p overlap.

Recall that p-p overlap> s-p overlap> s-s overlap as per overlap efficiency

Due to this reason, this group of substituents will have better +M effect as compared to the halogens, because as we move down the group 17, overlap efficiency for 2p-3p,2p-4p overlap and so on decreases.
However, a member of period 2 does exist among the halogens, namely $\ce{F}$ .But since it is more electronegative than both $\ce{O}$ and $\ce{N}$, it will again have lesser donating tendency as compared to this group.
Now, let's get down to the specific cases:
The group leader $\ce{O-}$ :
First,let us look at some enthalpy values:
$\ce{ O(g) + e^- -> O^-(g) 1st EA = -142 kJ /mol }$  (1)
$\ce{ O^-(g) + e^- -> O^2-(g) 2nd EA = +844 kJ /mol }$ (2)
As you can clearly see, adding an electron to $\ce{O-}$ ( or adding a second electron to $\ce{O}$) is an endothermic process, which means energy will have to be provided externally for pushing more electron density inside $\ce{O-}$.
That is pretty obvious as well, because you are forcing an electron into an already negative ion. It's not going to go in willingly! The second electron affinity of oxygen is particularly high because the electron is being forced into a small, very electron-dense space.[source]
Conversely, this means that the reverse reaction for (2) will be pretty favorable, in the sense that $\ce{O-}$ will quite readily give away electron density to a suitable acceptor atom,as is evident from the enthalpy values. Hence, $\ce{O-}$ leads the charge in terms of +M effect

It is to be kept in mind that the factor of compatibility of overlap still holds. Donation from $\ce{O-}$ to, say, $\ce{Si}$ will still not be as effective as that to $\ce{C}$

Subgroup 1:The curious case of $\ce{-NH2}$,$\ce{-NHR}$ and $\ce{-OR}$
First of all, since we have got neutral donor atoms in this subset, it stands to reason that simple electronegativity will be the major governing factor here to judge the overall donation tendency. Hence, both the $\ce{N}$ donor groups should be more donating than the $\ce{O}$ donor group.
The slightly peculiar judgement crops up when you are deciding between $\ce{-NH2}$ and $\ce{-NHR}$ . On a preliminary basis, you might say that since the $\ce{R}$ will most probably be exerting a strong +I effect ,and thereby the latter should have had more +M effect than the former.
But here's the catch: the electron density of the $\ce{N-R}$ bond will mostly shift towards N as compared to the $\ce{N-H}$ bond, but the donation to the benzene ring will ultimately take place by the $\ce{N}$ atom. And hence, it becomes important to look at what the donor nitrogens of these two groups have to say.
We are basically going to compare the bond angles of $\ce{H-N-H}$ and $\ce{H-N-R}$ .Due to the sterically bigger $\ce{R}$ group , the latter should be greater than the former. This means that the s-character will go up on the $\ce{N}$ atom of the latter, and this directly means that electronegativity for it will increase.

(Think of the various hybridization schemes, namely $\ce{sp}$ , $\ce{sp^2}$ and $\ce{sp^3}$, the percentage s character in each, the bond angles for each, and the electronegativity trend of $\ce{sp}$>$\ce{sp^2}$>$\ce{sp^3}$ )

Hence, donation power of $\ce{-NHR}$ should be less than $\ce{-NH2}$

A similar argument will also explain why the +M effect of $\ce{-OR}$ is less than $\ce{-OH}$

Subgroup 2: $\ce{-NHCOR}$ and $\ce{-OCOR}$
Both these group are analogous structurally, and that's the key to their relative comparison for our purpose. On a preliminary basis, we see that the number of highly electronegative atoms( that is, $\ce{O}$, $\ce{N}$ and $\ce{F}$) has increased in both these groups as compared to the previous members of this group. $\ce{-NHCOR}$ has an $\ce{O}$ and $\ce{N}$, while $\ce{-OCOR}$ has two oxygens. This immediately tend to make them less donating than the previous members of the group.
Another reason to see why they would be less donating is because of the distribution of partial charges in the resonance hybrid. The donor atom's lone pairs in both the cases are conjugating between a $\ce{C=O}$ group and the benzene ring. Since negative charge is more stable on $\ce{O}$ than on $\ce{C}$, the oxygen of the $\ce{C=O}$ will end up possessing much of the negative charge which would otherwise be available for donation to the ring.
In other words, we can say that the charge will be dispersed over a competing electron sink, namely, the $\ce{C=O}$ .Hence, their mesomeric effects will decrease considerably as compared to other members of the group.
As for their individual comparisons, you can easily judge that by the electronegativities of their individual donor atoms, $\ce{N}$ and $\ce{O}$
The phenyl group
Wikipedia says:

It is generally considered an inductively withdrawing group (-I), because of the higher electronegativity of sp2 carbon atoms, and a resonance donating group (+M), due to the ability of its π system to donate electron density when conjugation is possible.

Drawing from this, one can say that since the phenyl group is donating by using bond pair of electrons(namely, it's $\ce{C=C}$ π bonds) rather than lone pairs(unlike all the previous groups which were using lone pairs of $\ce{N}$ or $\ce{O}$) therefore it will have lesser donation tendency than the preceding members as lone pairs are generally more "freely available" for donation, being attracted by a single nucleus rather than bond pair of electrons which are held by two nuclei.
Still, as it is a donation from a $\ce{C}$ donor atom, it has better +M effect as compared to the halogens.
Comparison between $\ce{-Ph}$ and $\ce{-F}$ is a tough case, as it may seem that fluorine which is using it's lone pairs should have been a better donor than $\ce{-Ph}$. To this I would say that maybe since fluorine is the most electronegative element, it's electronegativity would dominate the donating tendency by the lone pairs(in technical terms, HOMO-LUMO gap would be more in case of F's non-bonded electrons than the $\ce{C=C}$ pi-bonded electrons)
But again, this reasoning would seem unsatisfactory.  Anyway,the mesomeric effect of the $\ce{-Ph}$ group usually holds little chemical interest. It mainly serves as a control, and we are much more interested in the effect of hetero atoms on the ring as compared to the effect of one benzene ring on another(in case of mesomeric effect)
Group 2:The halogens
I have already given enough idea about them above, more details can be found here
So there you have it. A similar idea can be applied for deducing the logic of the -M ordering as well.

Hints for -M effect

$\ce{-NO2}$ has 3 highly electronegative atoms, and an electron sink in the form of $\ce{-N=O}$, so it should lead the pack.

Just like the previous ordering, we are again interested in the characteristics of the acceptor atom here. While for +M effect, it increased with decrease in s-character(and hence electronegativity) on the donor atom, and more freely available electron density

-M effect will increase with increase in s-character(and hence electronegativity) of the acceptor atom. Since in your list, the remaining substituents are $\ce{C}$ acceptors, then the hybridization of $\ce{C}$ needs to be considered.
Accordingly, $\ce{-CN}$ should fall next with an $\ce{sp}$ hybridized $\ce{C}$

Now most of the items I can see on the list are substituents with $\ce{sp^2}$ carbons, having different combinations of $\ce{O}$ and $\ce{N}$ attached to the acceptor atom. Your job would now be to checking the varying extents of cross conjugation happening at the electron sink directly attached to the acceptor atom(and thereby comparing the electron accepting ability of the same) by looking at the electron donating abilities of the attached heteroatoms

You can ask for further clarification if these hints should not prove to be sufficient.

Thanks Yusuf, I am able to understand most of the stuff now. I was taking the electronegativity into account, but was not thinking about the very overlap of orbitals required for the resonance. That should fill the gaps.

Incorrect.
Mesomeric effect follows the order:
NH2>OH>OR
Lone pair of electrons are more stable on O atom than N atom since O is more electronegative than N.
OH>OR due to Bent's Rule.

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Is this example considered liquifying a gas under pressure?
In my work place, we have a drum that has propene inside at a pressure of around 2,5 bar and at a temperature of around -15ºc and the boil-off of that drum is aspirate by a compressor and is compressed to a pressure around 11 bar to 15 bar and a temperature of around 65ºc to 75ºc. After the compressor the high pressure gas goes to a ventilator, and the gas is cooled down with the ouside air at ambient temperature and that makes the gas to cool down and to change to a liquid at high pressure. After this the liquid is stored at another drum.
My question: is this consider liquifying a gas under pressure?
Note: is not for school purposes.

Yes. It is a process to liquefy a gas under pressure.

@Maurice are there other ways to liquefy a gas by applying pressure?

As already noted by Maurice, yes you did liquify a gas under pressure.  In fact, the scope of application is so wide that this is equally known as Linde cycle.  It is highly possible that you have such an engine at home, either as fridge, or freezer to cool stuff, or as heat pump to warm a home.
Regarding propane: equally yes.  After banning Freon and other CFWs, ammonia (especially industry-scale, e.g. here), propane (more for domestic appliances like fridges) is one of the often seen heat carriers.
The limits regarding how much heat may be shuttled between the warm and the cold side depends on pressure and temperature accessible, formally called the Carnot cycle:

(source)

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What is meant by a 'hydrophilic' sphere?
So, I know that a hydrophilic material is defined as those with a special affinity for water, opposite to
hydrophobic, those that naturally repel water.
I am investigating the impact of spheres with water surfaces, and the splashes caused by them. However, one research paper focuses on the difference between the fluid dynamics of hydrophilic spheres and hydrophobic spheres, and the difference in the cavities they make when they sink underwater.
I understand what a hydrophobic sphere is, but what is a hydrophilic sphere? Like a rolled up ball of tissue paper? Or the water absorbing waterbeads/orbeez that get bigger(the ones children play with)?

Are you investigating something like this?
https://www.youtube.com/watch?v=mXaltOAVWL8

The answer to your question depends on the size level (microscopic or not). But it sounds like you are referring to spheres at a size that you can hold, or larger.
A hydrophilic sphere could simply be a sphere of some solid material (like plastic), coated with a layer of hydrophilic material/substance.
This page on hydrophilic coatings has some information on what this material may be.

Here is a educational reference: '19.2 COORDINATION CHEMISTRY OF TRANSITION METALS, to quote:

The coordination sphere consists of the central metal ion or atom plus its attached ligands. Brackets in a formula enclose the coordination sphere; species outside the brackets are not part of the coordination sphere. The coordination number of the central metal ion or atom is the number of donor atoms bonded to it. The coordination number for the silver ion in [Ag(NH3)2]+ is two (Figure 3). For the copper(II) ion in [CuCl4]2−, the coordination number is four, whereas for the cobalt(II) ion in [Co(H2O)6]2+ the coordination number is six. Each of these ligands is monodentate, from the Greek for “one toothed,” meaning that they connect with the central metal through only one atom. In this case, the number of ligands and the coordination number are equal.

So, in the presence of a water ligand, the adjective 'hydrophilic' is appropriate. The sphere, itself, is the central metal ion and associated ligand enclosed in brackets, where per my readings, the brackets are sometimes missing.

I suppose the OP has in his mind the size range micrometers to centimeters, and perhaps rather the latter than the former.


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Need an app that gives the name of organic compounds "step by step"
I need an app that gives the IUPAC names for any organic compound, and also gives the reasons of this nomenclature, step by step. Any help appreciated.

You need to learn how to name organic compounds. It´s not so hard, if you have a proper textbook.

This answer reflects a personal opinion and equally assumes you are student.
ACD's ChemSketch for Windows offers an Academic Version here which limits the «structure to name» algorithm to 50 heavy atoms (the most typically found in ochem) and three cycles.  Despite this numeric limitation, it equally covers heterocyclic chemistry, too:

After drawing and selecting the molecule, the GUI entry is found /via/ Tools -> Generate -> Name or the short-cut Ctrl + Shift + I.
Based on your «step by step», I speculate this could provide you an entry.  I recommend to use this program with its companion site, a searchable, dictionary-like documentation about the underlying rules (here); start with a simpler core motif which is known to you (e.g., furan), and which step-by-step is modified e.g., by a chain, by stereochemical informations as in the example of of the furanethanol.  The freely available Marvin sketch equally offers a plugin to convert structures to name.
Speaking of heterocyclic chemistry, the nomenclature here requires some additional training because of the side-by-side of «trivial» non-systematic names which may be used as a root, the a-nomenclature, and the systematic Hantzsch-Widman system.  As far I am aware, only Chemdoodle allows you to instruct the program which set of rules should be used; available only in the paid version.  For example, to name the silicon analogue of thiophene either $1H$-Silole, or Sila-2,4-cyclopentadiene.
Don't overly rely on these programs, though.  If you read the rules and literature, for example, you will find out that 1H-pyrrole is possible, but in cases without ambiguity called pyrrole only.


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Unique identifier for chemical substances?
One famous database for chemical substances is PubChem. If one reads the section about other identifiers than the PubChem ID for Ethanol one gets several CAS numbers (potentially deprecated ones as well), potentially several European Community (EC) Numbers and a lot of other identifiers which seem to be not really unique.
Is there a true unique identifier for chemical substances reasonable to be used in a database?

https://chemistry.stackexchange.com/questions/34563/pubchem-inchi-smiles-and-uniqueness is relevant, although not a direct answer. The short answer is : Yes, InChI but there are alternatives

This one is more relevant : https://chemistry.stackexchange.com/questions/110057/how-to-index-molecules-in-computer-programs

The point of CAS numbers is that they label mixtures and preparations, not just pure chemicals. If you want a 1:1 mapping from a chemical to a database identifier, CAS numbers aren't even the right tool to look at in your case.

@gilleain Seems like there is ununiqueness cause of "standard" and "non-standard" InChi as well with InChi.

In additon, polymorphs of the same compound may be filed under different CAS register numbers, too.  Example:  $\ce{TiO2}$, as anatase (1317-70-0), and as rutil (131-80-2) -- both tetragonal, but either $I4_1/amd$, packing fraction 70%; or $P4_2/mnm$, packing fraction 77%.  Source: Cardelli, Materials Handbook, 3rd edition 2018, Springer Cham/Switzerland, ISBN 978-3-319-39823-3, p. 910, table 10.8.  And since polymorphs may differ, e.g. in dissolution kinetics, this is relevant for pharmaceutical materials, too.


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Rate towards SN Ar reaction
In my organic chem notes,the following order for rate/ease of SN Ar reaction is given

The reason given is The -I effect which stabilizes the carbanion so produced in the reaction.
Turns out,my organic chem sir gave us 2 mechanisms for SN Ar
they are attached below

Since he introduced mechanism 1 first it seemed pretty intuitive
But if I consider mechanism 2,then the rates of reaction must be inverse(order of leaving ability)
My doubt primarily lies in the fact that 1)If the rate order is true,then why do we have two different mechanisms for two similar(more or less) reactions?
2)If one of the mechanism is incorrect,then which one is correct?
Sorry for my poor vocabulary and please explain the above(what seems to me to be a) fallacy.

Both the mechanisms you've mentioned exist, but they both have some kinetic difficulties. Your first mechanism (called $\ce{ArSN2}$) involves breaking the aromatic character of the ring, while your second mechanism (a form of $\ce{E1cB}$ leading to a benzyne intermediate) involves removal of a minimally acidic hydrogen.

So a good, strong base favours the second mechanism, while a deactivating group ortho or para to the leaving group (such as a nitro substituent) that would stabilise the carbanion intermediate, favours the first mechanism.
References: Nucleophilic aromatic substitution, wikipedia

Probably the SNAr reaction should have an electron-withdrawing group on it to make it chemically sensible. The given order F > Cl > Br > I is correct for SNAr reactions.

@orthocresol Yes, here I referred to an electron withdrawing group as a  'deactivating' group


text:
How does one calculate the electronic difference density in practice?
I am interested in calculating electronic difference density maps. These have been described for quite a long time by Bader and others. For example, below is the fourth of a series of papers which discusses these extensively.
Cade, P. E., Bader, R. F. W., Henneker, W. H., & Keaveny, I. (1969). Molecular Charge Distributions and Chemical Binding. IV. The Second‐Row Diatomic Hydrides AH. The Journal of Chemical Physics, 50(12), 5313-5333.
The idea is fairly simple. Use some electronic structure method (or in principle experimental data) to calculate the electron density of some atom or molecule by itself (but at the appropriate geometry) and then calculate the electron density of the molecule containing the atoms for which you have reference density distributions. Sum all the reference densities and take the difference from the reference densities. You then get a map with positive and negative regions that show areas of electron enrichment and depletion which hopefully provides some physical insight.
It is simple enough to calculate electron densities with most electronic structure packages and have the program print out the density in terms of expansion coefficients for the molecular orbitals which various other programs are able to read and visualize.
Now with that information in hand, is anyone aware of software packages which can do this operation I am describing? Or am I stuck with doing something like fitting the electron densities numerically and taking the differences in that way? If so, if anyone has done this before, some input in that regard would be useful.
This seems like such a simple thing to calculate, but I've been a bit confused. I know I've seen people make spin density difference plots for open-shell and excited state systems. This is such a similar calculation that it seems like one should be able to use the same tools, but I assume the latter calculations are buried in the electronic structure package somewhere and hence not easy for the user to adapt.
Any help is greatly appreciated.

How large are the molecules in question you refer to?  Smallish ones?  Larger proteïns?  Because Xray crystallography routinely uses calculated electron maps to refine structure models (example https://proteopedia.org/wiki/index.php/Electron_density_maps, or https://www.rcsb.org/pages/help/edmaps)  I speculate coot (https://en.wikipedia.org/wiki/Coot_(software), and pymol (e.g., http://www.drugdesign.gr/blog/how-to-display-an-electron-density-map-in-pymol) may at least adaptable (pymol with Python scriptable) for such a task, too.

@Buttonwood Quite small molecules. Ya my assumption is that I can kind of hack a number of programs to do this but if there is a standardized way I would rather do that so as to avoid making mistakes and being able to reproduce my results.

Possibly you have to align the coordinate systems of the «reference» and the «probe» molecule with each other in a first step, each carrying their electron density map like a knapsack.  If both are the same molecule, a Kabsch-like algorithm (e.g., https://github.com/charnley/rmsd) could be a start, which nicely runs with little manual intervention and offers output of the newly aligned $(x,y,z)$ about the atoms in a .xzy file.

Right. One definitely needs to be careful about this type of thing.

I know you can do this with Gaussian/Gaussview, though I realize that isn't very helpful if you don't have a license.

I actually have figured out an easy way to do it. I will answer the question when I get a chance.

If you have the electron density tensors then you should just be able to subtract them. The reason it's so easy in Gaussian is because they have a nice program that automatically reads cube files, however if you can read in these files the mathematics is trivial.


text:
How can I calculate the amount of HF in aquoues solution of NH4F and an aquoues solution of NH4HF2?
Dissociation of $\ce{NH4F}$ and $\ce{NH4HF2}$ in water is complete, I assume:
$$\ce{NH4F -> NH4+ + F-}$$
and
$$\ce{NH4HF2 -> NH4+ + HF2-}$$
Can I calculate the amount of $\ce{HF}$ in these solutions from this equilibrium?
$$\ce{HF + F- <=> HF2-}$$
$$\mathrm{K_2}=\dfrac{\ce{[HF^{2-}]}}{\ce{[HF][F^-]}}=4,7$$
If I know the concentrations of NH4F in water and NH4HF2 in water.

Is this a homework or book question? If so we'll need to see the exact question to help. Also we have a "homework" policy that requires you to show some work. // If this is some sort of other inquiry, please explain the context better so we can figure out how to help.

It's not homework. It's for my work. A reviewer asked me to calculate this for a paper I want to publish. But I'm so lost, I haven't done these calculations since the first year of Uni. Like I have some ideas on how to calculate this, but I'm not entirely sure. I prepared the solutions of NH4F and NH4HF2 in water and I need to calculate the amount of HF in these solutions. The concentrations of NH4F is 5 M and NH4HF2 is 0,18 M.

Ok, for a paper. In that case you'll want "reasonable" calculations not just some tripe 'theoretical" calculations. In other words, your solutions are so concentrated that you'll need activities, not the raw concentrations to do the calculations. Not sure where you'd find that data.

I think theoretical are enough, the reviewer asked to estimate the amount of HF.

$$\mathrm{K_2}=\dfrac{\ce{[HF2^-]}}{\ce{[HF][F^-]}}=4.7$$

WRONG SOLUTION: OP clarified that there are two separate solutions, not a mixture of the salts.
Assuming:
(1) that concentrations can be used instead of activities (bad assumption...)
(2) The concentration of the ions $\ce{H2F-}$ and $\ce{F-}$ is so large that the final concentrations will be the same as the initial concentrations.
(3) Since HF is a relatively weak acid, little $\ce{HF}$ will dissociate into $\ce{H+}$ and $\ce{F-}$.
Using equilibrium:
$$\mathrm{K_2}=\dfrac{\ce{[HF^{2-}]}}{\ce{[HF][F^-]}}=4.7$$
From 5 molar $\ce{NH4F}$, $\ce{[F-] = \pu{5 molar}}$
From 0.18 molar $\ce{NH4HF2}$, $\ce{HF_2^- = \pu{0.18 molar}}$
So plugging the number in:
$$\mathrm{K_2}=\dfrac{\ce{[HF^{2-}]}}{\ce{[HF][F^-]}} = \dfrac{0.18}{\ce{[HF]}\times 5}=4.7$$
$$\ce{[HF] = 0.008 molar}$$
For the final concentrations $\ce{[HF] \ll [H2F-]}$, so the assumption that the dissociation of $\ce{H2F-}$ can be ignored is valid. The $\ce{HF}$ will dissociate and acidify the solution, but still $\ce{[HF] + [H+] \ll [H2F-]}$

The NH4F and NH4HF2 are two separate solutions, c(NH4F)=5M, and c(NH4HF2)=0.18 M. For each of these solutions, that are not mixed I need to determine the concentration of HF. In the expression K2 for NH4F solution I only know the concentration of F- which is 5, but for conc. of HF2- and HF are unknown. This is where I get lost, because I don't know if I can assume this concentrations are maybe the same, or one is the same as F-

Let us calculate first the $\ce{NH4F}$ problem. The ion $\ce{F-}$ reacts partially with water according to
$$\ce{F- + H_2O <=> HF + OH-}$$
The equilibrium constant $K_\mathrm{b}$ of this equilibrium is defined by
$$K_\mathrm{b} = \frac{[\ce{HF}][\ce{OH-}]}{[\ce{F-}]}$$
The numerical value of $K_\mathrm{b}$ is needed, but it is not tabulated. However it can be obtained from the dissociation constant $K_\mathrm{a}$ of $\ce{HF}$ in water, which can be found in the tables : $K_\mathrm{a} = 10^{-3.17}$. So $K_\mathrm{b} = K_\mathrm{w}/K_\mathrm{a} =  10^{-14}/ 10^{-3.17} = 10^{-10.83}$. This numerical value is now introduced in the definition of $K_\mathrm{b}$
We can simplify by writing :  $[\ce{HF}]=[\ce{OH-}]= x$. If we admit that $x$ is small, $[\ce{F-}]$ =  $\pu{5 M}$. Solving this equation gives : $x^2 = 5·10^{-{10.83}}$ and $x = 3.84 \cdot 10^{-6}$. This is the expected result  : In $\ce{KF}$ $\pu{3 M}$, the concentration of $\ce{HF}$ is $\pu{3.84 \times 10^{-6} M}$.
Let's now calculate the problem of $\ce{NH4HF2}$ in a solution $\pu{0.18 M}$, where the following equilibrium is given :
$$\ce{F- + HF <=> HF2-}$$
In this solution, there was $\pu{0.18 M}$ $\ce{HF2-}$ in the beginning, from which $x$ is transformed into $\ce{HF}$ and $\ce{F-}$. So the final concentration of $\ce{HF2-}$ is $0.18 - x$. The equilibrium constant $K_2$ of this equilibrium is
$$K_2 = \frac{[\ce{HF2-}]}{[\ce{HF}][\ce{F-}]} = \frac{0.18-x}{x^2} = 4.7$$
This is an equation to solve, and the solution is $x = 0.116$. So the concentration of $\ce{HF}$ in $\pu{0.18 M}$ $\ce{NH4HF2-}$ is $[\ce{HF}] = \pu{0.116 M}$

Don't think that you can neglect $\ce{NH4+}$ as a weak acid.


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why are E2 and SN2 reactions not competing reactions?
I am quite confused about the competition of $\mathrm{E2}$ and $\mathrm{S_N2}$ reactions. My lecturers say $\mathrm{E2}$ and $\mathrm{S_N2}$ reactions are not competing reactions but most websites say they are. Please I would like to be clarified on whether $\mathrm{E2}$ and $\mathrm{S_N2}$ reactions are competing reactions, and if not, why?

If the question has been sufficiently answered, please accept it!

Yes, E2 and SN2 are competing reactions.
Keep in mind, E2 reactions require a base whereas SN2 reactions require a nucleophile.
Nucleophilicity corresponds to the ability to donate a lone pair of electrons ‒ here, to a carbon bearing leaving group ‒ which is a kinetic property (time does matter).  Basicity corresponds to accepting a hydrogen, a thermodynamic property determined in the state of equilibrium (where time does not matter).
1.Both properties, nucleophilicity and basicity of a reagent may be in competition. For example, a sterically hindered compound which potentially could act as a nucleophile by the presence of a free electron pair will be considered a poor nucleophile and instead a good base.  Examples of sterically hindered bases are 2,6-lutidine (mild), potassium tert-butoxide, LDA (strong). The small methanolate, $\ce{CH3O-}$, on the contrary, may act simultaneously as nucleophile as well as a base.
2.Secondly, for SN2 reaction to take place, the substrate (the compound on which the nucleophile will react) should not be crowded, since it increases the energy in the transition state and the reaction becomes less favourable. In these cases, E2 can take place, provided the nucleophile here actually acts as a good base and the product formed is stable by hyperconjugation, resonance, etc.
3.Also, temperature is an important factor. If the temperature is high, E2 will be favoured owing to the thermodynamics of the reaction.


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Trying to duplicate a trumpet cleaning solution that isn’t available on the market
There is a product called “Herco Spitballs” which are small foam cylinders soaked in solution which are used to clean brass wind instruments like trumpets.  The “Spitball” is inserted into one end of the horn and then blown through the entire instrument.  The solution runs out long before the spitballs do, but the solution is not available.  I’ve written to the manufacturer and haven’t gotten a response.  The solution is .5% Amine Polyglycol Condensate, .5% Isopropyl Alcohol, and 99% inert ingredients (probably water).  I need to know what Amine Polyglycol Condensate is and where to get it.  I don’t need much.  8 ounces would last for years.  Can you help me to find some?

I think it is this stuff  https://www.sigmaaldrich.com/catalog/product/sial/07964?lang=en&region=GB

I´m sure you can just use distilled water.

Why not using a swap in sections accessible by remove of mouth piece and slides, prior / after the valves, as wood winds and e.g., French horns (around 3:09 min of https://www.youtube.com/watch?v=zslhbh5Exgg)?  While the disassembled b-trumpet on https://en.wikipedia.org/wiki/Trumpet shows not all corners are accessible by this, probably musicSE may be a better audience re alternatives for now since music dealers / conservatoiries (colleagues / teachers) probably are closed in your region as well.

I'm voting to close this question as off-topic because I think the question is more suitable for music.SE.

Brass cleaning compositions from home ingredients provided here https://www.thoughtco.com/make-copper-and-brass-cleaner-602200 .

Waylander is on the right track.  I’m just wondering whether it really is the same stuff.  Karl is right.  I can use distilled water, but that isn’t the question.  The question is, what is Amine Polyglycol Condensate, and where can it be obtained in small quantities?  Buttonwood is correct in thinking that the most thorough way to clean a trumpet is to take it apart, but one doesn’t do that every day.  Buttonwood’s vote to disallow the question is strange, because the question is where to get the chemical.  Ajkoer’s response is tangential—this is a search for a specific chemical.


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How much damage is done to ascorbic acid vitamin C content by sunlight and fluorescent light?
I left a large clear bag of ascorbic acid exposed to fluorescent light and sunlight for 48 hours.  The sunlight was direct for several hours in the afternoon.  How much has the effectiveness of vitamin C been damaged?  Thank you!

Ascorbic acid in solid crystallic form is much less sensitive to light and oxidation, compared to its solutions ( especially if in ascobate form ). Another lucky consequence is only the outer layer has been exposed to the light, the inner bag content is not disturbed.
It is hard to quantify the impact. Empirical evaluation could be if it changed the colour of the outer layer. But even if it did, the change of the overall acid content in the bag would be negligible.

I appreciate your responding to the question.  How much sunlight and/or fluorescent light would it take to degrade the vitamin C content of sealed bags of dry ascorbic acid powder?  What circumstances would be needed to reduce effectiveness of vitamin C by 5%?

Also, what is the ideal temperature for storing ascorbic acid powder?  Thanks again for your help.

Effectiveness would be the same, just amount would be 5% less. About 5% would have to be exposed to light, exposing it long enough.  For storage, room temperature, dark and dryness is enough.


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Electrolysis of copper without any copper initially in solution
Let's say we have two copper electrodes as shown in the image, and there is a solution with copper nitrate. In this case, I know that copper will deposit onto the cathode and copper from the anode will be oxidized and enter solution when a voltage is applied across the two electrodes. However, what if instead of copper nitrate, there was just NaCl (or any other electrolyte that wouldn't cause any problems like precipitation) in solution instead? Would current still be able to flow, and if so, would the electrodes still lose and gain mass as they did before? I am fairly certain that current does flow, but I am not really sure if copper will still deposit onto the cathode and vice versa.

It is a complex problem. Yes, NaCl solution is a conductor, so electricity will pass through the solution. Now in your situation both electrode are made of copper. Since electricity is passing through the solution, electrolysis must occur at both electrodes. You cannot have electrolysis only at one electrode.
Assuming there is sufficient salt in the solution, what will happen at the anode?
Chlorine gas will evolve. I will let you find relevant equations. Chlorine gas attacks copper very quickly and the anode starts to "degrade" quite fast, but will it pass into solution? Nope.
Let us see what is happening at the cathode?
Sodium ions cannot be reduced in water esp. on copper electrodes (on mercury they do), instead water will be reduced. What happens when water is reduced? It turns into hydrogen gas, which escapes, and leaves hydroxide ions behind. This process makes the solution basic.
Now coming back to the anode, since you have hydroxyl ions around, they precipitate copper ions. All you get is yellow orange junk in such a system. Cu(II) is blue so why an orange yellow mess? It is most likely due to Cu(I) compounds. Anode dissolves pretty fast but nothing happens to the cathode.

Thanks for the answer! I see that you stated that at the anode, chlorine will be oxidized. Do you have an explanation for why chloride will be oxidized but the copper on the electrode will not be?

Good point, look at the electrode potential tables for Cu(2+)/Cu and Cl (-)/Cl2 half cells. Which species will have a higher tendency to oxidize at the $anode$?

@M. Farooq Hm, I would suppose the copper oxidation would occur sooner than chloride oxidation, comparing the standard potentials being almost 1 V apart. Like Cu -> Cu2+ + 2 e-, or Cu + 4 Cl- -> [CuCl4]2- + 2e-
Unless current density is too high to be saturated by copper oxidation and with high enough voltage, chloride oxidation would start as well. I suppose oxidation to Cu(I) can happen as well.

@Poutnik, the oxidation potential of Cu to Cu(II) would be -0.34 V and that of chlorine would be -1.36 V with respect to SHE. Since there is no Cu(II) to begin with, it is hard to say. Under standard conditions, copper oxidation should occur.


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Too much leftover in zinc sulfate reaction?
I'm doing a reaction of zinc metal and sulfuric acid to get zinc sulfate. I've been trying to learn the chemistry lately so I tried to measure out correct molar amounts. My understanding from the equation is that it's equimolar:
Zn + H2SO4 = ZnSO4 + H2
Sulfuric acid has a molar mass of 98.079 g/mol and zinc metal is 65.38 g/mol. The density of sulfuric acid is 1.84 g/ml (yeah cubic centimeter but it's a ml), 98.079 / 1.84 = 53.3 ml/mol. I believe my acid to be about 96%-ish, so to give myself a fudge factor I measure out 60ml of acid, diluted with 120ml of water. I put in about 65 grams of zinc metal (over time). When the bubbling had mostly stopped I had a pretty much solid mangle of zinc metal at the bottom of the jar, and even some air bubbles trapped underneath where the liquid couldn't get to anymore. There was also a white precipitate.
I'm confused why my (mostly) equimolar amounts left SO much zinc after, far more than even 10 or so grams.
Edit: I filtered the liquid still left in the jar through coffee filters. Cheap ones, but I layered them and stuffed one into the stem. It took a long time to filter, and is black...I also added more distilled water and some 35% acid to dissolve the rest of the zinc in the jar. It's bubbling away happily now.

If you use about equimolar amounts of zinc and the acid, the reaction would about asymptotically converge to depletion of the substance with the lower molar amount. The white precipitate is a mark the solution is saturated with zinc sulphate heptahydrate. The solution becoming thickened by the suphate may further decrease the rate of zinc dissolution. You have said it "mostly stopped", that is what I am talking about. It got too slow in the end, as there is just little acid left.
Remember the activity coefficient of $\ce{H+}$ raises very fast with acid concentration, so e.g. 10 times diluted acid reacts much slower than 10 times.
I suggest to use bigger acid excess, and then let it partially evaporate, separate crystals and wash them to get rid of the acid. ( like icy water or methanol or ethanol, then evaporate the liquid ).

Excellent, thank you! Yeah there was still some very slow fizzing when I finally poured it off. I guess there was really no reason to use molar amounts, I could easily just have added zinc slowly until the fizzing stopped. Lessons for the future! Thank you for explaining it so well!


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Balancing redox equations with oxygens in multiple oxidation states
How do you balance this using the ion-electron method;
$\ce{CrO5 + H2SO4 -> Cr2(SO4)3 + H2O + O2}$
The oxygen has multiple oxidation states in $\ce{CrO5}$, and none of the sites I looked this up on dealt with that.

In my opinion, this question itself is questionable in regard to its possible advancement of academics (accept, to perhaps, illustrate the limitation of current scientific knowledge and the state of academic education).

This is a rather unusual case of what is discussed in answers like this one, where we circumvent problems with multiple atoms being oxidized or reduced by considering whole compounds as oxidizing or reducing agents.
Here, the whole-compound redox-active material is $\ce{CrO5}$, and as in peroxide disproportionations generally this is both an oxidizing agent and a reducing agent.  We thereby render
$\ce{CrO5 -> Cr^{3+} + (5/2) O2 + 3 e^-}$
$\ce{CrO5 + 10H^+ + 7 e^- -> Cr^{3+} + 5 H2O}$
We then apply the usual method of multiplying the first reaction by $7$ and the second one by $3$ to balance the electrons leading to
$\ce{10 CrO5 + 30 H^+ -> 10 Cr^{3+} + (35/2) O2 + 15 H2O}$
and reducing to "lowest (whole number) terms"
$\ce{4 CrO5 + 12 H^+ -> 4 Cr^{3+} + 7 O2 + 6 H2O}$
All that remains now is to add the spectator ions if desired, and we're done.

You just have to separate out the species that reacted and deal with them separately as usual after that, is that right?

Yes.  Here the "separation" is part of the CrO5 got oxidized to make oxygen and the rest got reduced.

But doesn't it go from +6 to +3 in both the cases?

The oxygen behaves differently, going up to 0 in one component but all of it down to -2 in the other.  We need to look at both, so the CrO5, not just the Cr, is being oxidized or reduced.

Without taking oxidation numbers into account, I have first balanced the Cr atoms, then the H atoms, and finally the Oxygen atoms. And I have found  :$$\ce{4CrO_5 + 12 H^+  -> 4 Cr^{3+} + 7 O_2 + 6 H_2O}$$
With the sulfate ions it gives :
$$\ce{6 H_2SO_4 + 4 CrO_5 -> 2 Cr_2(SO_4)_3 + 7 O_2 + 6 H_2O}$$

I found a 2019 source work on Chromium peroxide. To quote the key preparation equation:

Chromium peroxide (Cr(O2)2.H2O or brieﬂy CrO5) is an
  extremely potent oxidant. This compound is a product of the following reaction:
$\ce{(NH4)2Cr2O7 + 4 H2O2 + 2 H+ -> 2 Cr(O2)2.H2O + 4 H2O + ammonium salt}$  (1)

Here are associated corresponding comments from Wikipedia:

Chromate or dichromate reacts with hydrogen peroxide and an acid to give chromium peroxide and water.
$\ce{CrO4^{2−} + 2 H2O2 + 2 H+ → CrO5 + 3 H2O}$
After a few seconds, the chromium(VI) peroxide decomposes to turn green as chromium(III) compounds are formed.[1] To avoid this decomposition, it is possible to stabilize chromium(VI) oxide peroxide in a water-immiscible organic solvent such as diethyl ether, butan-1-ol or amyl acetate by adding a layer of the organic solvent above the chromate/dichromate solution and shaking during the addition of hydrogen peroxide. In this way, the chromium(VI) peroxide (unstable in the aqueous phase in which newly formed) is dissolved in the immiscible organic solvent. In this condition it can be observed over a much longer period.
$\ce{2 CrO5 + 7 H2O2 + 6 H+ → 2 Cr^{3+} + 10 H2O + 7 O2}$

So, in my opinion, due to instability, the reaction mechanics commences with chromium peroxide decomposition to $\ce{Cr^{3+}}$ and then the action of acid as follows:
$\ce{4 Cr(O2)2.H2O -> 2 Cr2O3 + 5 O2 + 4 H2O}$
$\ce{2 Cr^{3+} + 3 SO4^{2-} ⟶ Cr2(SO4)3}$
[EDIT] Some more comments per Wikipedia source above:

This compound contains one oxo ligand and two peroxo ligands, making a total of five oxygen atoms per chromium atom [...]
  Therefore, CrO5 is a good oxidizing agent (even better than chromates and dichromates due to the presence of two reactive peroxo ligands) but due to its low stability it is never used in organic syntheses.

So, one source claims one oxo ligand ($\ce{O^{2-}}$, but perhaps not reactive as the other 2019 source ignores it) and both agree on two reactive peroxo ligands, $\ce{O_2}{^{2−}}$. Based on the last statement on stability, and in particular, per both sources, stable in polar organic solvents (and not aqueous H2SO4), I suspect, the only valid path is likely multi-step, and further doubt that the reaction would ever be conducted in practice.

@ Ajkoer. How can you say that  Cr(O2)2.H2O is ... briefly CrO5 ? Two H atoms have disappeared.

I do not, that is the quote from the source paper, which implies, politely, that CrO5 is misleading shorthand, for the actual salt. Note the author use of this depiction of the salt in the key formation reaction which is likely more complex (radical intermediates, I suspect).

That being so, $\ce{CrO5}$ does exist.  As a neat compound it is too unstable to be of any use, but when stabilized by complexation with a suitable electron donor such as pyridine is is an effective oxidizing agent in organic synthesis.

Oscar: Do note per the Wikipedia comment on its stabilization with organics cites also "shaking during the addition of hydrogen peroxide", so perhaps Cr(O2)2⋅H2O2 shorthand for CrO5 seems speculatively more accurate. Also, the Wiki prep equation disagrees from the actual path, which apparently requires more H2O2.

"The structure of the pyridyl complex has been determined crystallographically.[3]"  Looks like a CrO5 complex to me.

Yes Oscar, and stable in polar organics, and not aqueous H2SO4.


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Does 1,3-diphenylprop-2-en-1-one give a positive Tollens test?
If we have 1,3-diphenylprop-2-en-1-one, will it give positive Tollens test? I think aldol reactions are reversible under basic conditions. Here there are mildly basic conditions, so will we get back the reactants of the aldol reation?
This question was part of JEE ADVANCED 2016 where they asked which will give Tollens reagent test. I feel it has mildly basic conditions for KOH, hence it should perform reverse aldol and thus this should be the Answer. But the official answer key says no.
Please help me out because I can't understand how to distinguish between the two types.

Reverse aldol on this sytem will not occur under mild condtions. The extended delocalisation of the system makes it pretty stable. I think you would need forcing condtions to get the reverse aldol

Ok then we must be using a strong base. oh- will not be sufficient here. But if it had not dehydrated then could we have done the retro aldol using weaker bases like oh-?

Yes, aldol products that still have the alcohol do not have the same level of stability,


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Use of Spiegeleisen in manufacture of steel
I recently read the following statements related to Manufacture of steel via Bessemer Process:

When no more $\ce{CO}$ is produced (i.e. blue flame disappears) a calculated amount of Spiegeleisen (an alloy of $\ce{Fe, Mn, C}$) is added and distributed throughout the mass by blowing air for some time.

Why do we need to add Spiegeleisen at the end? Is it to reduce the oxides of iron (that might have been formed)? Or due to some other reasons, like, say $\ce{Mn}$ to make steel harder and increase its tensile strength? But latter doesn't convince me, since we could have added $\ce{Mn}$ directly on order to get the same results. Also what's the meaning of calculated amount, how do we calculate the amount to be added, and why?

I didn't think Bessemer was used anymore. Spiegeleisen is more likely to be called ferro-manganese today. I expect it is added at the end to minimize losses of Mn to the slag. Mn does contribute to hardenability  but the primary reason for its use is to combine with residual sulfur to make MnS . MnS stringers can be harmful but are an improvement because otherwise the S will mostly be in grain boundaries reducing ductility and toughness. ( Bessemer process adds nitrogen to the steel reducing the toughness).  Calculated amount would be based on the total weight of Mn that must be added to get the desired content in the steel , such as 0.5 weight %.The added amount would need to allow for expected loss to slag, etc.


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Cation-anion radius ratio of manganese difluoride
How to determine the cation-anion radius ratio for $\ce{MnF2}?$

I tried to calculate the $r_\mathrm{c}/r_\mathrm{a}$ ratio using the relationships:
$$2r_\mathrm{c} = c$$
and
$$(2r_\mathrm{c} + 4r_\mathrm{a})^2 = 2a^2,$$
but that gives $r_\mathrm{c}/r_\mathrm{a} > 1$ with the given $a$ and $c$ parameters, which is impossible. $\ce{MnF2}$ has a rutile structure, so the predicted ratio should be between $0.414$ and $0.732.$

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. Please visit this page on how to make your future posts better:)

You assumed that the cations touch along the $c$ direction.  They don't.  In most ionic compounds you have only anion to cation contact and, therefore, any attempt to determine ionic radii from lattice parameters gives only the sum of the two radii, not the individual values.  There are exceptions such as lithium iodide, in which the lattice parameters enable determination of an ionic radius for iodide, but then they carry no information about the lithium ion.
In principle, you could use the data on lithium iodide, where there is iodide-iodide contact, to get a radius for just the iodide ion.  Then you turn to manganese(II) iodide, where anions contact cations, and get the sum of the manganese and iodide radii.  Then subtract the iodide radius to get radius for manganese ion.  Use that plus the lattice information for manganese(II) fluoride to get the fluoride radius.  It all works perfectly -- if the radius of a given ion is the same in all its ionic compounds.  Maybe it is to sufficient accuracy, maybe it isn't.


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Concentration when done wrt mass or volume
I have a question asked in jee main 2020. 10.30 mg of o2 is dissolved into a litre of sea water of density 1.03 g/ml. The concentration of o2 in ppm is?
Well I know that ppm is generally possibly mostly done as mg/L. Now in this question doing this we get 10.30 as the answer but the answer is 10 which suggests that it's done in some other mass related unit. How to understand when to use mass or volume? Please help me out.

Careful: ppm, like percent or radians, is not a unit. To get ppm, you must always divide two quantities with the same unit, so the unit cancels out, and then multiply by $1e6$.

@Karl Radian is supplementary fundamental unit...

The title and the body do not match. ( hardness versus oxygen )

Yes changed the title

@Zenix Have you got a source for that? Sounds like hairsplitting to me. ;)

@Karl check this page. But it's controversial :)

@Zenix Tnx! As suspected, a lot of hairsplitting. ;) In my opinion, sth. that is produced from a unitless quantity (i.e. a ratio of distances) via the application of a purely mathematical (i.e. trigonometric) operator cannot have a unit. Agreed, the case against ppm and  % is not the same as for "rad".

@Zenix In terms of dimensional analysis, the most toenail-curling equation ever imo is the one transforming radians into degrees. ;)

@Karl that amuses me too, ratio of 2 similar physical quantities (length) gives different numerical value in different "systems" (:

$1$ liter water weighs $1030$ g, and contains $10.3$ mg $O_2$. The mass concentration of $O_2$ in ppm is the ratio $0.0103 g/1030$ g = $10.0$ ppm.

Using significant figures, both the concentration of oxygen and the density have three. So the answer should be 10.0 ppm, not 10 ppm.

ppm is not necessarily defined for 2 quantities of the same unit.
There is ppm (w/w), ppm (V/V), ppm (w/V), ppm(n/n).
Sometimes these variants use an explicit affix, like ppm (V/V)=ppmv.
Due to the ambiguity of the default interpretation, it is highly recommended  not to use ppm / ppb / ppt anymore. If one has to use them and no explicit interpretation is provided, one has to stick to the most probable or common variant used in the given context.
I suppose in your case it can be both ppm (w/V) and ppm (w/w), probably the latter, but no guarantee for that.

Exactly that's why I got it wrong in jee main 2020

Next time, if I were you, I would comment the ambiguity and provided both possible answers, if applicable.

@Karl Thanks, partly because of not being native, partly "Android mistyping", partly my strange typos I am always shocked seeing.  :-)

@Karl ... but  "not to verb" versus "to not verb" is a moot point.
https://learnersdictionary.com/qa/Split-Infinitives
"Be aware that putting "not" or another adverb between "to" and its verb adds some emphasis to that adverb. For example, in the sentence "They decided not to stay another night" the phrase "they decided" is the most important information, but the sentence "They decided to not stay another night" tells us that maybe they decided to stay another night before, but now it is important that they will not stay."

@Poutnik I though you were not quite up to your usual standard! :)) And yea, the "to not" / "not to" can give a different emphasis, although here on second thought I guess it makes no real difference.


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Collagen digestion
As far as I know, pepsin is used very often in collagen solutions extraction. Since it only cleaves the telopeptides and leaves the triple helix chains of the collagen intact. I was wondering if there is any other enzyme that can do similar work to extract collagen?


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Propene with Hydrogen. Name the catalyst used
Which industrial process uses a similar reaction?

If you are talking about formation of propane(alkene to alkane) , then the catalyst commonly used is Nickel along with heat. Also, Platinum or Palladium can be used instead of the above combination.
The process or reaction is Hydrogenation

More precisely Catalytic hydrogenation


text:
If ions are floating around in a solution, how do we know they combine to form a compound in a solution?
I apologize if my question sounds confusing I am not sure how to properly convey it here's an example:
For the reaction between lead nitrate and ammonium phosphate you get the reaction:
$$\ce{3Pb(NO3)2 (aq) + 2(NH4)3PO4(aq) -> Pb3(PO4)2(s) + 6NH4NO3(aq)}$$
but the ionic equation is
$$\ce{3Pb2+(aq) + 6NO3-(aq) +6NH4+(aq) + 2PO4^3-(aq) -> Pb3(PO3)2(s) + 6NH4+(aq) + 6NO3-(aq)}$$
so how do we know the compound $\ce{6NH4NO3}$ is formed if its just $\ce{6NH4+}$ and $\ce{6NO3-}$? Do the ions combine and then immidiately disassociate again? I'm very highly confused I would appreciate an explanation, and i apologize if this makes no sense.

In a beginning chemistry course, we just say all ammonium salts are soluble. So the solution in your example has ammonium and nitrate ions just bouncing around. Filtering out the lead phosphate and evaporating the water would give solid ammonium nitrate, with some lead phosphate impurity.

ahhhhh I see so what it's saying is that ammonium nitrate can be formed not that it is actually being formed?

Exactly! We also say, in the beginning chemistry course, that all nitrates are soluble, so ammonium nitrate is soluble on both counts. As the water evaporates, you get to the solubility limit of ammonium nitrate. Then, with further evaporation, solid ammonium nitrate is formed, unless the solution is becomes supersaturated. Even then, solid ammonium nitrate will form once the supersaturated solution gets nucleated by whatever, i.e., particles, scratches on the inside of the vessel, beard droppings, etc.

oh I see, thank you i was having trouble with this a lot, one other quick question that's kind of related to this that i was having some trouble with is that when we make a net ionic equation of the equation above, I don't understand how it's possible that some ions don't react at all or are unchanged?

Spectator ions are necessary for charge and mass balance, but they simply do not react under the circumstances. Really, it is as simple as that: there is no free energy favorable reaction for them to do.

In the second equation $\ce{6NH4+(s)}$ should be $\ce{6NH4+(aq)}$

Oh thank you, and oh I did not notice that I'll fix that right away

User EdV comments above get to the heart of the matter. To summarize:
Your first equation is typically how a chemist would write the reaction.
$$\ce{3Pb(NO3)2 (aq) + 2(NH4)3PO4(aq) -> Pb3(PO4)2(s) + 6NH4NO3(aq)}$$
The equation is a "shorthand" explanation. What a chemist reads is:

On the left hand side is that that two solutions are mixed together.

Since $\ce{Pb(NO3)2}$ has an "(aq)" modifier it will be in solution, but since it is an ionic compound it will not be in solution as a molecule, but rather as the $\ce{Pb^{2+}}$ and $\ce{NO3-}$ ions.
Since $\ce{(NH4)3PO4}$ has an "(aq)" modifier it will be in solution, but since it is an ionic compound it will not be in solution as a molecule, but rather as the $\ce{NH4+}$ and $\ce{PO4^{3-}}$ ions.
Since the RHS shows a reaction the two salts can't be in one solution simultaneously. Therefore the salts must have been dissolved in separate solutions and the solutions were mixed.

On the right hand side $\ce{Pb3(PO4)2}$ precipitates and $\ce{6NH4NO3}$ remains in solution.

Also since $\ce{Pb3(PO4)2}$ has a tiny but finite solubility, a very small amount of $\ce{Pb3(PO4)2}$ will remain in solution.
Since $\ce{NH4NO3}$ has an "(aq)" modifier it will be in solution, but since it is an ionic compound it will not be in solution as a molecule, but rather as the $\ce{NH4+}$ and $\ce{NO3-}$ ions.

The "ionic equation" is harder to understand because of its verbosity. Did the $\ce{NO3-}$ ions come into the solution with the $\ce{Pb^{2+}}$ ions or the $\ce{NH4+}$ ions? Being chemists we can puzzle it out, but the equation is less succinct.
$$\ce{3Pb2+(aq) + 6NO3-(aq) +6NH4+(aq) + 2PO4^3-(aq) -> Pb3(PO3)2(s) + 6NH4+(aq) + 6NO3-(aq)}$$

All this has already be already mentioned in EdV answers.

@Maurice Thanks, but no problem from my point of view: anyone is welcome (and encouraged) to post a good (or great) answer that benefits both the OP and the site.


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Periodate oxidation of sugars

When Methyl D-glucopyranoside is treated with $\ce{HIO4}$, what is the amount of $\ce{HIO4}$ (in moles) consumed per mole of the sugar?

I have studied that $\ce{HIO4}$ oxidises only those vicinal diols which are cis to each other. But here only one pair is cis, so shouldn't the answer be 1?

The cis/trans issue has been addressed: https://chemistry.stackexchange.com/questions/270/effect-of-periodic-acid-on-cyclohexane-derivatives/114836#114836

The vicinal-glycols can be cleaved to corresponding aldehydes and/or ketones in high yield by the action of periodic acid ($\ce{HIO4}$) or lead tetraacetate ($\ce{Pb(OAc)4}$). This oxidative cleavage of a carbon-carbon single bond provides a two-step involving cyclic intermediate reaction mechanism with high-yield. A generally accepted equation for these oxidations (A) is shown below:

The reaction using $\ce{HIO4}$ is called Malaprade Reaction (Ref.1), while that involving $\ce{Pb(OAc)4}$ is called Criegee Reaction. Malaprade Reaction is described as:

The oxidation of adjacent diols with periodic acid or its salt in aqueous solution is generally known as the Malaprade reaction. Several solvents have been used to increase the solubility of organic substrates and the reaction proceeds faster under acidic conditions. This reaction has been further extended to the cleavage of $\alpha$‐hydroxy carbonyl compounds, 1,2‐dicarbonyl compounds, $\alpha$‐amino alcohols, $\alpha$‐amino acids, and polyhydroxy alcohols, and successfully applied for structural analysis.

Both cis- and trans-glycols (e.g., cis- and trans-cyclohexane-1,2-diols) undergo the oxidation with $\ce{HIO4}$ or its salts. As a rule, cis-glycols react more rapidly than trans-glycols and there is evidence for the presence of heterocyclic intermediates as shown in the diagram (B). For example, the publication by Buist, et al. (ref.2) revealed that cis-cyclohexane-1,2-diol cleaves about 20-times faster than trans-cyclohexane-1,2-diol. However, the equilibrium between the diol and the cyclic periodate intermediate (faster first step) slightly favors the trans-isomer (the ratio of cis/trans equilibrium constants is $400:1000$; Ref.2) while the rates of cleavage of cis-cyclohexane-1,2-diol to products is much faster compared to its trans-isomer (the ratio of cis/trans rate constants is $3300:165$; Ref.2). Thus, theoretically your believe of only cis-vicinal-diols would get oxidized with $\ce{HIO4}$ is completely wrong.
Therefore, theoretically, each available $\ce{-C(OH)-C(OH) -}$ function would get oxidized ($\alpha$-methyl-D-glucopyranoside has 4 such functions), costing one $\ce{HIO4}$ molecule per function (Ref.3). However, rate of oxidation is depend on conditions, steric effects, $\mathrm{pH}$, etc. For example, the oxidation of $\alpha$-methyl-D-glucopyranoside by a solution of $\ce{HIO4}$ in pyridine has cleaved only the $\ce{C}$3$-\ce{C}$4 bond while the reaction of 6-O-trityl-$\alpha$-methyl-D-glucopyranoside is less selective under same condition and is directed to a greater degree to the $\ce{C}$2$-\ce{C}$3 bond (Ref.4).
References:

Zerong Wang, "Malaprade Reaction (Malaprade Oxidation)," In Comprehensive Organic Name Reactions and Reagents; John Wiley & Sons, Inc.: New York, NY, 2010 (https://doi.org/10.1002/9780470638859.conrr406). ISBN: 9780471704508.
G. J. Buist, C. A. Bunton, J. H. Miles, “149. The mechanism of oxidation of α-glycols by periodic acid. Part V. Cyclohexane-1 : 2-diols,” J. Chem. Soc. 1959, 743-748 (https://doi.org/10.1039/JR9590000743).
Glen Dryhurst, In Periodate Oxidation of Diol and Other Functional Groups: Analytical and Structural Applications; First Edition, Pergamon Press: Oxford, United Kingdom, 1970. ISBN-10: 1483128385.
R. G. Krylova, S. N. Ryadovskaya, L. I. Kostelian, A. I. Usov, “Periodate oxidation of α-methyl-D-glucopyranoside, 6-O-trityl-α-methyl-D-glucopyranoside and 6-O-tritylcellulose in pyridine,” Bulletin of the Academy of Sciences of the USSR, Division of chemical science 1972, 21, 2005-2009 (https://doi.org/10.1007/BF00854627).

But i have been given answer as 2 to this problem.

As I said, it's depend on a lot of factors. See, same compound needed only one periodic acid in Ref.4. You make the choice based on mechanism.


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What happens with the boiling point of a liquid in a vacuum
This may be a silly question, but when I was reading my chemistry textbook, there was this following statement:

"A liquid boils when its vapor pressure equals the external pressure acting on the liquid
  surface. At this point, bubbles of vapor form within the liquid. The temperature at
  which a given liquid boils increases with increasing external pressure."

But what would happen if that liquid was sealed in a vacuum, or at least submitted to very low pressure? Would the boiling point be a lot lower?

Yes, with lower pressure the boiling point lowers too. // Think about it. Water boils at a lower temperature in the mountains than it does at sea level because there is less air pressure in the mountains.

"Sealed in a vacuum" is an oxymoron, a contradiction in terms, much like "frozen with fire". If you seal a liquid in a flask containing nothing else but vacuum, then a part of the liquid will quickly evaporate and fill the flask with vapor, so it would no longer be a vacuum.
The said vapor will exert some pressure, depending on the temperature and the nature of the liquid. For example, with water at room temperature, it will be a lot lower than the normal atmospheric pressure. Indeed, the boiling point of water at that pressure will be a lot lower than the usual $100\;^\circ\rm C$. Guess what? It will be the same as ambient temperature. That's how equilibrium works.
Now what if we just keep the liquid at vacuum by pumping out the vapor as soon as it forms? Well, the simplified† answer is: the liquid will evaporate gradually, bit by bit, until nothing is left. There will be no more liquid, and consequently no more boiling point. That's what happens with the boiling point in a vacuum.

† In fact, it is more complicated. As the liquid evaporates, the rest of it gets progressively colder. At some point, it will get cold enough to freeze, and then you'll be dealing with ice. The molecules will still be leaving its surface, but slower and slower, and the complete sublimation might well take forever, or even longer. That's why the space rocks are still there after all these years.

Generally speaking, liquid boils at the temperature, at which its saturated vapour pressure is equal the external pressure.
With external pressure going down, boiling temperature goes down as well.
When external pressure goes down too much, the boiling point may meat the freezing point of the liquid and the liquid freezes. There is no boiling any more, but sublimation. (That is the way of "dry ice" production from liquified carbon dioxide.)
For very low vapour and external pressures, there may be needed little liquid overheating, as the vapour pressure must deal with external pressure AND surface tension, acting against bubble forming.


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Does anyone know a database that shows the effect of electrolyte concentration on the production of hydrogen gas during electrolysis of water?
I'm currently studying in the International Baccalaureate model. This high school model asks their students to do an individual work called the Internal Assessment. One of the subjects I study is chemistry, and my Chemistry Internal Assessment is based on the production of hydrogen (a source of energy) in the water electrolysis as a way to save electrical energy. I was going to do an experiment to measure the production of hydrogen in the electrolysis of water with various concentrations of four different solutes (NaCl, NaOH, sulphuric acid and sodium bicarbonate, but any electrolyte will do). However, with the Covid-19 outbreak, I cannot go to the laboratory, so in order to get the data I need, I searched on google databases with the information I wanted, but I found barely nothing. These links (https://www.scirp.org/pdf/NR_2013032814044640.pdf) (https://pubs.rsc.org/en/content/articlehtml/2017/se/c7se00334j) are the only useful pdfs I found, and I wonder if anyone could help me giving me some databases with the data I need.

I am afraid the amount of H2 produced  by electrolysis does not depend on the concentration or of the formula of the chosen salt or hydroxyde, provided it is a soluble alcaline compound. It only depends on the intensity of the currant.

it should, as it enhances electrical conductivity. moreover, there is a point where the solute concentration is so high that the electrolysis of water will stop, and the electrolysis of solute begins. that is the point of my investigation.

The problem of performing an electrolysis in a solution containing nearly no water is a tricky problem. it may happen with concentrated sulfuric acid and concentrated solutions of NaOH. By memory, it seems that electrolysis of sulfuric acid produces peroxydisulfuric acid. I am not a specialist in this domain, but it deserved to be studied.

Of course is a problem, I want to see what is the optimal concentration of solute (the maximum velocity at the production of hydrogen and oxygen), where the solute does not react with either the anode and cathode. In the laboratory I was only going to use sodium bicarbonate for safety reasons, but now that I redirected my study and I use databases, I don't have to worry about that, that is why now I use more solutes.

Sodium bicarbonate is the least candidate to start such a study. it is not enough soluble in water.

I know, but that is not the problem. I want to compare these 4 solutes (NaCl, NaOH, sodium bicarbonate and sulphuric acid) to see the best one and the best concentration. I´m just 17, and the international Baccalaureate don´t ask to do an extensive study, that is why they say (the international baccalaureate) that is very important to mention the limitations of our studies. Even though is barely soluble, the sodium bicarbonate is save, but again, that is not the issue, because I´m not doing any experiment, the only thing I want is a database.

Let us continue this discussion in chat.

Alejandro, you are right, the electrolytes will make a difference. The difference is not in terms of hydrogen production per se, as Maurice mentions in the comments, it will be in terms of energy saving. Electrolytic reactions are "exact" in the sense that if 100 electrons reach the electrode from a battery, 100 hydrogen ions will be reduced. In other words,
1 mole of electrons will reduce exactly 1 mole of a monovalent ion, 1/2 mole of a divalent, 1/3 mole of a trivalent ion and so on.
You would probably like to have an electrolyte which has the least electrical resistance. Now you may recall that the proton in water (= hydrogen ion) has the least electrical resistance known today. What comes next? Hydroxide ions.
All you have to search and study is the conductivity of electrolytes at various concentrations.

that is very helpful. Thank you, but even though hydrogen has the least electrical resistance, oxygen has not. This is very important, because we could storage hydrogen and oxygen produced at the electrolysis to then, in a galvanic cell, combine these two elements to produce energy. If the oxygen does not electrolyse, we would lose an important element.

@Alejandro. What do you want to say, when stating that "oxygen does not electrolyse". As written, this sentence does not seem to have any meaning

Sorry, I wanted to say oxidize

I mean hydrogen ion has the highest conductivity not hydrogen. Be careful in terminology. Whenever there is a reduction, there is an oxidation.

You need to look for the data in a different way. You can use electrical conductivities to get the current flow the solution. From the imagined current flow you calculate how much hydrogen would be generated.

I did find a work, 'Electrolyte Engineering toward Efficient Hydrogen Production Electrocatalysis with Oxygen-crossover Regulation under Densely Buffered Near-neutral pH Conditions', with comments of interest.
I start with some background from the introduction:

In recent decades, drastic progress in solar fuel production has occurred: photovoltaic cells can generate electricity from the sunlight,[1−4] whereas photoelectrochemical [5] and photocatalytic [6−11] water splitting can directly produce hydrogen and oxygen by harvesting the sunlight. An electrolyzer can electrochemically split water molecules,[12,13] which can be conjugated with the photovoltaic cells.[14,15] Importantly, during the photoelectrochemical and photocatalytic water splitting, what takes place on the surface is electrocatalysis, in association with the photonic processes at the bulk of photon absorber and the interface between the photon absorber and electrocatalyst.[11] Electrocatalysis will thus certainly play a crucial role in the solar fuel production process in the future.

And further concentration-related statement taken from the abstract, where HER refers to the hydrogen evolution reaction:

The choice of electrolyte in terms of its identity and activity drastically altered the HER performance. Electrolyte properties (activity coefficient, kinematic viscosity and diffusion coefficient) accurately described the mass-transport contribution, which was easily isolated when a highly active Pt catalyst was used. The HER rate on the Pt was maximized by tuning the solute concentration (typically 1.5 – 2.0 M). Moreover, the kinematic viscosity and oxygen solubility under such densely buffered conditions governed the oxygen mass-transport flux in the electrolyte, which in turn tuned the cross-over flux. At near-neutral pH, as high as 90% selectivity toward the HER was achieved even under an oxygen saturated condition, where only a 40 mV overpotential was needed to achieve 10 mA cm−2 for the HER. This information can be regarded as an important milestone for achieving a highly efficient water splitting system at near-neutral pH.

Of potential interest are associated graphs, starting with Figure 1, on Page 31, to Figure 9.


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Rocket Candy Fuel - Common Alternatives to KNO3?
I'm sort of a model rocket enthusiast, but the most common fuel for model rockets, Rocket Candy, requires Potassium Nitrate as an oxidizer to oxidize Powdered Sugar. Potassium Nitrate is not necessarily expensive, but it's not very common, and the only process I found to make it at home took almost a year to produce product. I'm wondering if anyone has found another suitable oxidizer for the sugar, preferably not a nitrate unless it is one that is common, as I am not allowed to store nitric acid in my apartment to synthesize the nitrate. As a side note, I happen to have a significant amount of ammonium sulfate, and I am also wondering if this might be a viable substitute. I know that sulfates aren't always as good of oxidizers as nitrates, but I also know that ammonium salts tend to be better oxidizers. If anyone has some idea as to a common household compound that would be a suitable replacement for Potassium Nitrate as a fuel oxidizer, or if Ammonium Sulfate is good at oxidation, I ask you to help me out here.
Thanks for your consideration!
 ~ mpprogram6771

As an update, further investigation reveals that Ammonium Sulfate increases the production of tar and residue, which is not something that I want.

until you do that successfully your real name is going to show, and your account won't properly be configured.

Yes I know it's crazy! The registration process has so many bugs. You just have to re-commit to Materials Modeling now.

Your new account is not committed to anything: https://area51.stackexchange.com/users/208834/mpprogram6771

I am afraid, potassium nitrate is the most common and probable the least expensive oxidizer. Sulfates are definitely not usable in rocketry. When I was a high school student, we used for simple rockets the filtration paper soaked in solution of sodium chlorate, then sold in technical quality as a total herbicide, probably banned now. It is more reactive than nitrate or perchlorate, so potentially more dangerous., It was little hydroscopic, compared to potassium chlorate, that we used with sugar and metallic salts for "Bengal fires" for TV shows.( as young helpers of the TV chemistry consultant

@mpprogram6771: If you click on your Area51 account: https://area51.stackexchange.com/users/208834/mpprogram6771, you will see that you are not committed to anything, can you please try to commit to Materials Modeling again?

oh, shoot! Sorry :/

Here is one substitute, per Wikipedia:

Ammonium perchlorate composite propellant (APCP) is a modern fuel used in solid-propellant rocket vehicles. It differs from many traditional solid rocket propellants such as black powder or zinc-sulfur, not only in chemical composition and overall performance, but also by the nature of how it is processed. APCP is cast into shape, as opposed to powder pressing as with black powder. This provides manufacturing regularity and repeatability, which are necessary requirements for use in the aerospace industry.

Here is a source: 'Perchlorate: Sources, Uses, and Occurrences in the Environment', detailing its possible creation:

A number of processes have been patented for the production of perchlorates. In general, sodium perchlorate is manufactured electrolytically using sodium chlorate as the feedstock. Potassium perchlorate and ammonium perchlorate are produced in a second step by reacting sodium chlorate as a water solution with other chemicals to create concentrated solutions of either potassium perchlorate or ammonium perchlorate, depending on the added chemical(s). Perchlorate crystals are then precipitated from the solution and are dried to produce a homogeneous dry granular product that is shipped in sacks and drums for sale to manufacturers of various perchlorate-containing end products.

I like where you're going with this, and your answer was very helpful, but do you know of a common household source of any of the chemicals listed above, either the compound itself or it's precursors? I'm kind of going for something accessible.
However, If you truly could find a household source for Ammonium Perchlorate, I've read about how great it is for rocket fuels, but I couldn't find a good source or convenient way to produce it. Even if you can't, your answer is greatly appreciated and I've learned a lot from it. Thanks!


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Symmetries and degeneracies of the s,p, and d orbitals
Consider the molecular AB_8 (complex with central atom A and 8 B ligands (coordination number 8), for which I determined the point group D_4d) and determine the symmetries and degeneracies of the s, p, and d orbitals of A.
So I believe the general procedure here is to determine a reducible representation for each s, p, and d orbital of the central atom, but how do I do that? Is the s orbital just A1? Is p just B2 and E1? How do I computer these representations?

The s orbital is always $A_1$. The others transform the same as their corresponding Cartesian function (e.g. $p_x$ transforms the same as $x$). You can determine how the function transforms under the operations of the group, though this usually given in the the third and fourth columns of the character table.

I was just starting to figure that out! It was much simpler than I was imagining. Thank you for your response!

There is some more information in this answer https://chemistry.stackexchange.com/questions/58609/understanding-group-theory-easily-and-quickly/58613#58613


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What are the fundamental principles that are found throughout chemistry?
In the same way that biology has the fundamental idea that all organisms want to reproduce and pass on their genetic information, and that evolution accounts for the diversity of life, what are some of the ideas that tie chemistry together? Apart from all the specifics, what is chemistry really about at its heart?
Apologies if this sounds like a really strange question.

There are atoms. They bind to each other... and that's pretty much this for general stuff. If you have a feeling you need to apologise while writing a question, then consulting it in chat before asking may be a good idea.

Atomism - All (regular) matter is made of atoms.

Chemistry has so many fields, and the question could have so many answers. I would say organization: no chemistry, no organized matter in the universe, no life, no minerals, no drugs, no polymers, no quantum mechanics...

Your idea about biology is not quite right. Species survive if they generate offspring. Which is a bit too trivial to count as a fundamental principle.

I meant in the sense that species do need to pass on genetic information otherwise they will die out, regardless I think it gets the point across. I was thinking about this for a bit and came to the conclusion that the fundamental idea that ties chemistry together is that electrons and atoms want to lower their energies. This is why they form bonds so that they are more stable. Idk I do feel like I'm missing something massive I just can't figure out what

Is a virus an organism?

It's in between like right on the line of living and non living, which is why scientists are so confused about viruses, since they pass on genetic information but do not perform any of the other activities of life, my example wasn't that good but I guess it gets the point across

There is absolutely nothing confusing about viruses. They are free floating DNA or RNA, which, if it finds its way into a cell, gets replicated along with the regular DNA or RNA of the cell. If the cell has no regulatory mechanism (because the genetic information is foreign, and has no builtin regulation), it produces too much of it, until it dies.  In that case, the virus is a pathogen for the given organism.

The fundamental principle of chemistry is probably the conservation laws : energy and matter. Maybe also the universal attraction between positive and negative charges and the repulsion between similar charges. These laws have no exceptions, as far as I know.

One might argue that those are in fact the fundamental principles of physics.

@Ivan. Of course. The fundamental principles are the same in physics and chemistry.

One might ask what is the reason for the very existence of chemistry as a science, then.

@Ivan. That is a good question. In France, chemistry is a chapter of physics, at school, like optics or electricity.

Of course it is, because the alternative, quoting Baron Rutherford, would to sort it under "stamp collecting".

@IvanNeretin Mandatory xkcd.

This is an inherently subjective question. It's very difficult to come up with an answer that isn't just a fundamental principle of physics. My personal opinion is the following:

Chemistry is the study of interacting electrons.

Now, obviously the electrons are only near enough to interact because the positive charges of nuclei don't let the electrons fly away. Fundamentally, however, all of chemistry can be described in terms of electron densities, which is the basis of density functional theory.
This is not just a theoretical construct either. In every field, people are talking about different types of bonds and different electronegativities. If you think about organic chemistry, reaction mechanisms are described in terms of electron-pushing, which is some highly abstract way of talking about the dynamics one might find if you could solve the time-dependent Schrodinger equation for arbitrary lengths of time.
Even things that seem to have nothing to do with electrons, such as nuclear magnetic resonance, are often useful because they key in information about the where electrons are localized via chemical shifts.
Also, this definition manages to distinguish chemists from physicists because very often physicists don't care about solving for the energy of something to the sixth decimal place. Whereas, as a computational chemist, I sometimes worry that I might actually need to get that electronic energy out to the ninth decimal place...
So, my opinion is that chemistry is the study of interacting electrons, and chemists care very deeply about the details of those interactions.

It is perfectly right. But is it really an answer to the question ?

I think so. I'm just saying there's really one fundamental principle. There are definitely lots of other ideas and quantities which show up repeatedly, but their relevance to chemistry is in some way connected to the interactions of electrons.

You seem to have a strong bias against nuclei. I would say the electrons play the game but the nuclei create the field.

@Buck Thorn Haha fair enough. I've got nothing against nuclei. The chemistry wouldn't happen without them. It's just that much of the variety in chemistry seems to come from how the electronic structure of a material responds to it's environment. This response is due to a change in the electrons. Obviously they are changing from some base configuration which was only adopted because the nuclei were there in the first place. Your point is well taken though.

I'd argue that nuclear interactions are no less essential, it's just that the computational task is focused on the wavefunction of the electrons, and that has caused chemists to forget the importance of the nuclei. The difference between two neighboring elements in the periodic table is primarily due to a difference in the nuclei, which alters the relative energy of electrons that interact with the nucleus. The fermionic nature and electrostatic repulsion between electrons is obviously important, but would you say nuclear-nuclear repulsions are not equally important?

In other words, the diffuseness of the electronic wavefunctions does not make the compactness of the nuclear wavefunctions any less important to chemistry.

It seems like you are arguing that all chemistry can be described within the Born-Oppenheimer approximation which is not the case. It is often a valid approximation, but there are countless examples of places where 'nuclear quantum effects' are needed to capture the basic properties.

@PJ R Nothing about what I said implies that is the case. There's no doubt that nuclei are very important in chemistry as there are no elements without. This is an inherently subjective and the discussion we're having is very semantic. Basically, I am taking it as a given that it is possible for electrons to interact because the nuclei shuttle them around. Even when nonadiabatic effects are important, the primary concern is still in properly describing the electronic interactions, but then this must be done on multiple PESs.

I only answered the question because I had actually thought about this before and view electrons as being the prime object in chemistry. I'm not saying nothing else matters. Only that very often, the electrons are the most important part of the problem. On the other hand, vibrational spectroscopists tend to think about the nuclei in the field of the electrons, so I'm sure many vibrational spectroscopists would disagree with this view. Ultimately, whether you look at nuclei or electrons are two sides of the same coin, as the BO approximation tells us.

Chemistry is different from physics in that its focus has historically been on understanding what gives rise to the variation of matter, rather than asking "what is matter", which is one of the more profound questions physics asks. By extension chemistry also concerns itself with how matter can be transformed within this manifold of possible forms.
Chemistry is an extremely broad field that encompasses just about anything you can think of, as reflected in the great variety of subfields with hyphenated or compound names: geochemistry, astrochemistry, biochemistry, physical chemistry (not to be confused with chemical physics), nuclear chemistry, analytical chemistry, inorganic chemistry, organometallic chemistry, organic chemistry, chemical engineering, petrochemistry, agrochemistry, food chemistry, quantum chemistry, electrochemistry, among many others and including only those that include "chemistry" in their name (unlike say molecular biology, protein science, materials science and many other fields which are arguably also chemical sciences, or contain a strong chemical component).
This is why chemists proudly state that chemistry is the central science.
For more information on the great scope of chemistry you may want to check what branches exist within a professional society such as the ACS (American Chemical Society).
Addendum: Given the question "what are the central principles found throughout chemistry", emphasizing the role of the electrons makes sense, since describing their behavior is a key concern for understanding the structure and reactivity of compounds. But I'd say the fundamental principles are "quantum mechanics" and the "principle of charge" (the latter given as an answer elsewhere) and not restrict myself to the electron.

I did not include computational chemistry in the list because it is not about the chemistry of computers, it is theoretical.

Chemistry seeks to describe, among other things, chemical changes in the world. All spontaneous changes obey a law, and that is that they increase the entropy of the universe. Hence, a fundamental principle of chemistry could arguably be:
dS > 0

As for the conservation laws, the increase of entropy is not specific to chemistry. It is also valid for physics.

@Maurice Of course, but as you said, the fundamental principles are the same in physics and chemistry.


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How to transfer Gel into liquid
We’ll know there are few easy method to make liquid into gel form
But whats the method for reverse it
To transfer gel into liquid form

I suppose to transfer is a wrong verb, implying displacement instead of transformation.

There are many forms of gels and there is no universal method.
The method with the highest probability of success is heating the gel up, turning it into sol,  but it is temporary and reversible.
Generally, forming gel may involve irreversible 3D linkage, that cannot easily be undone without destruction of gel structure and content.
If chemical structure of gel is known, there can be chosen a reaction leading to  break the polymer structure, if applicable with the overall gel content.


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Active masses of solids and liquids
In the calculation for equilibrium constant, we take the active mass of solids and liquids as unity. Intuition tells that more amount of a solid reactant should result in more product being formed, but according to the formula for the equilibrium constant it has no role in determining that.
Is it unity in all possible cases? How can we assume it to be one and have it not affect the equilibrium calculation?

Chemists don't call it the active mass, but rather the activity. The activity of a solid is taken as unity if the reaction is happening in a liquid or gas. The activity of a liquid is taken as unity if the reaction is happening in a gas. // Consider putting some calcium carbonate in pure water. A tiny amount will dissolve. But once the solution is saturated it doesn't matter if there is one microgram or 1 kilogram of calcium carbonate excess, no more calcium carbonate will dissolve.

You might view the relevant principle as follows: components whose activity changes during a reaction tend to put a stop to a reaction before they are completely consumed. Components with constant activity (independent of the amount of other reactants), on the other hand, can be completely consumed (exhausted): they can be limiting reagents, provided they are less abundant (stoichiometrically) than the other reagents. If a reagent with constant activity is exhausted before equilibrium is reached the reaction stops - it is then a limiting reagent. Reagents whose activities change during a reaction (because their concentration changes, for instance) are in principle not exhausted this way because their activity increases too much for the reaction to proceed any further. Reactions involving only reagents with variable activity tend to go to an equilibrium in which none of the species is completely exhausted. If reagents with constant activity are present in excess (such that their initial amount exceeds that at equilibrium), then they will not limit the extent of reaction. Note also that if the increase in activity with dilution of a reagent with variable activity is negligible, its concentration at equilibrium may be so low that it can be considered exhausted (and limiting).
A good example of the relevant principle is a mixture of two reactive solids or two immiscible liquids of constant activity that react at the interface between them. In either case, if the reaction is favorable ($Q\gt K_{eq}$) then it will proceed until one or both of the substances has been completely consumed (both being completely consumed if present in stoichiometric proportion). On the other hand, when a solid dissolves into a solvent, the activity of the solid remains constant and it will continue to dissolve provided the activity of the dissolved species does not increase above that of the solid. If it doesn't, then the solid is completely dissolved (exhausted).
If a reaction is already at equilibrium, adding more of a solid won't make a difference (provided there are no surface effects). But if you are not at equilibrium, it will if the amount present is limiting. A good example again is solubilization of a solid in a solvent. Added solid will continue dissolving until you reach the saturation point. Past this point the concentration of the compound in solution does not change any further. In addition, with saturation the activity of the compound in the solid and solution phases becomes equal.

In case of solid being the limiting reagent, If we add more solid reactant, the reaction should proceed further. But according to le chatelier's principle, which depends on the equilibrium equation, it has no effect. Am I understanding something wrong here?

@AnanyaPateriya If the reaction is already at equilibrium, adding more of a solid won't make a difference. But if you are not at equilibrium, it will if the amount present is limiting.

A good example again is solubilization of a solid in a solvent. Added solid will continue dissolving until you reach the saturation point. Past this point the concentration of the compound dissolved in solution does not change any further. In addition, with saturation the activity of the solid in the solid and solution phases becomes equal.