text:
Thermochemistry: On Units of Enthalpy Change and Implicit Given of Initial Temperature of Calorimeter
I would like to ask two separate questions. But first, so that we are on the same page, the experiment used a constant-pressure calorimeter, and it is a neutralization reaction of 1M of HCl and 1M of NaOH.
1. What is the difference between the units of ΔH: kJ and kJ mol^-1?
As I have read in Socratic at the end of the page, there are many units to represent the change in enthalpy. However, we are given a theoretical enthalpy of -2.79 kJ but my teacher gave an equation of Q_calorimeter/n_H2O thus my answer is -1.16 kJ mol^-1. How am I able to evaluate this by getting the percentage error? Should I multiply my answer to something with "mol^-1"? If so which mole would it be, the whole reaction or each species?
2. If you are not given data on the initial temperature of the calorimeter, is it correct to assume that its initial temperature is the initial temperature of the substance that is initially poured in (i.e. NaOH)?
In the other part of the activity, the initial temperature of the cold water is explicitly labeled also as the initial temperature of the calorimeter, and the cold water is initially poured into the calorimeter. Inferring from this, is my conclusion correct?

It is long time but the calorimeter equivalent makes sense as an absolute value to me. Perhaps I am wrong but it should be how much energy it takes. The calorimeter is that one calorimeter.

This is how I would proceed: 2) ask your teacher or schoolmates about such details. You are presumably allowed to take some liberties with respect to accuracy. 1) use standard propagation of errors (look this up if you are not familiar) to compute the uncertainty in the ratio heat/amount of water formed. As the reaction goes to completion the question is which reagent is in excess. You need to estimate uncertainties in the amount of the limiting reagent (either NaOH or HCl). I you don't know which is limiting assume they are equal and estimate the uncertainty in either one.


text:
Chemical potential and gibbs free energy
I would like to ask why the $\Delta_rG_1 = -RT\ln K + RT\ln Q$ is not equal to $Δ_rG_2$ numerically as shown in the figure where I derived $\Delta_rG_2 = RT\ln Q$ using a cyclic ring since $\Delta_rG$ is a state variable.

I am also aware that $\Delta_rG =  \mu_\text{product} -  \mu_\text{reactant}$, which should be the same at a particular state, and this is consistent with the constant gradient I have in the graph.
Hence, why is it that numerically, I cannot get $\Delta_rG_1 = \Delta_rG_2$? Thanks

I suspect you are attaching the wrong meaning to the different equations. The change in G at equilibrium is zero. The change in G when products and reactants are in standard states is $\Delta G^\circ$

Hi, thanks for the reply, I understood this, but my issue was that why ΔrG1 from a particular state to standard state not equals to ΔrG2 when that particular state moves to equilibrium. This sis because ΔrG is the gradient of the graph as shown in the main diagram. So, at any point, if you want to have a different end state (regardless of whether if it is standard state or equilibrium state), the gradient of the graph should not change, but numerically in my derivation, somehow it did. So, I am unsure where I went wrong.

I can't see what you did in your derivation, but, for what it's worth, here's my derivation.
My starting equations are $$G=\sum{n_i\mu_i}$$and, along your contour, at constant temperature and pressure,  $$\mathrm dG=\sum{\mu_i\mathrm dn_i}$$The changes in the number of moles of the various species are given by $$n_i=n_{i0}+c_in$$where $c_i$ is the stoichiometric coefficient for species i (a negative integer for a reactant and a positive integer for a product), $n_{i0}$ is the initial number of moles of species i, and $n$ is the molar progress of the reaction (number of times that the balanced reaction equation has been run through).
For an ideal gas mixture, $$\mu_i=\mu_i^{(0)}(T)+RT\ln{p_i}$$where $\mu_i^{(0)}(T)$ is the standard free energy of formation of species i at temperature T and $p_i$ is the partial pressure of species i:  $$p_i=P\frac{n_i}{\sum{n_j}}=\frac{n_{i0}+c_in}{\sum{n_{j0}}+(\sum{c_j})n}P$$
If we combine the above equations, we obtain:  $$\mathrm dG=\left[\sum{c_i\mu_i^{(0)}}+RT\sum{c_i\ln{p_i}}\right]\mathrm dn$$
But, $$\sum{c_i\mu_i^{(0)}}=\Delta G^{(0)}$$and $$\sum{c_i\ln{p_i}}=\ln{Q}$$Therefore, along the reaction contour, $$\mathrm dG=[\Delta G^{(0)}+RT\ln{Q}]\mathrm dn$$

Hi, thanks for your explanation, I learnt a lot from it. However when ΔG= dG/dn, the reaction from particular state to equilibrium, ΔG1=-RTlnK + RTlnQ and from standard state to equilibrium, ΔG(0)=-RTlnK. Since ΔG is a state variable, I derived ΔG2 from particular state to standard state as ΔG2 = ΔG1 - ΔG(0) = RTlnQ. From the graph of summation G against extent of reaction, the gradient = ΔG. Hence, at particular state at a certain moment of extent of reaction, the gradient is the same, so ΔG1 = ΔG2 should be the case. But he numerical calculation I have above, shows ΔG1 different from ΔG2.

You can't get from the standard state to any arbitrary state (and vice versa) unless the arbitrary state contains stoichiometric quantities of reactants and products to begin with and the total pressure in the arbitrary state is 1 bar.  In other words, the standard state does not lie along the reaction trajectory of an arbitrary starting state.

@ChetMiller In my answer, I started with the standard state, defined an arbitrary state along the reaction trajectory as Qt, and let it go to equilibrium. This way, you can ensure those three states are accessible in a single reaction mixture.

@KarstenTheis That's, of course, a valid starting state, but it is only at 1 bar.  Still, maybe that's closer to the path the OP envisioned.  I guess the more general path I treated includes this as a special case.

There are at least two fundamental issues you have to address. First, you have to distinguish between the Gibbs energy $G$ and the Gibbs energy of reaction $\Delta_r G$. In you diagram, one is the value on the y-axis (without defined zero point) and the other is the slope of the line, as labeled in your sketch. The second issue is that in the expressions on the left, the differences in Gibbs free energy of reaction should be distinguished from the values themselves by labeling the former $\Delta \Delta_r G$.
With those two considerations, there are two ways to look at the relationship between the three states. To define these three states in a way that one can turn into the other in a closed system (no transfer of substances into or out of the system, dn = 0), you could start with the standard state ($Q = 1$), choose a certain state while the reaction is running ($Q = Q_t$) and then let it go to equilibrium ($Q = K$).
Gibbs energy of reaction
Here are the expressions for the three states.
For an arbitrary $Q = Q_t$:
$$ \Delta_r G = - R T \ln K + R T \ln Q_t\tag{1}$$
For $Q = 1$ (standard state):
$$ \Delta_r G = - R T \ln K\tag{2}$$
For $Q = K$:
$$ \Delta_r G = 0\tag{3}$$
We can now write the differences between pairs:
$$ \Delta \Delta_r G_{1-2} = R T \ln Q$$
$$ \Delta \Delta_r G_{2-3} = - R T \ln K $$
$$ \Delta \Delta_r G_{1-3} = - R T \ln K + R T \ln Q$$
It all adds up. There is no special meaning to $ \Delta \Delta_r G_{1-2} $ though, and no reason that it should be the same as $ \Delta \Delta_r G_{1-3}$.
Gibbs energy
Specifying the Gibbs energy of the three states is impossible because of the issue with the zero point. It is possible, however, to say how one state is different from another, i.e. find relations for $\Delta G_{1\rightarrow 2}$, $\Delta G_{2\rightarrow 3}$, $\Delta G_{1\rightarrow 3}$. You would have to integrate over the extent of reaction $\xi$ in a way that $Q$ goes from 1 to $Q_t$ to $K$, using the relationships in the previous section. Because $G$ is a state function (and because the integral from state 1 to state 3 can be achieved by first going from state 1 to state 2 and then to state 3), it would add up correctly.

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Application of Markovnikov's rule with EWG's and Resonance
I want to ask a question concerning Markovnikov's rule.
Consider the following question:

We can tell that there will be two types of stabilisation upon formation of the carbocation:

Resonance via the phenyl group
sigma-hyperconjugation via the methyl group

When attempting this question, I thought the more stable carbocation would be the one where the positive charge lies to the left of the alkene double bond, and can be stabilised by the phenyl group, since resonance stabilisation is stronger than sigma hyperconjugation to yield the major isomer.
However, the answers state the opposite:

They state that the more stable carbocation is the one that can be stabilised by sigma-conjugation, yielding the alternate product, due to more carbon (groups) being attached to the carbon on the double bond. I thought it would be less stable as it is destabilised by the EWG aldehyde group and stabilised by the weaker sigma-hyperconjugation.
What is the correct interpretation of the answer?

When I tried modeling $\ce{[Mol\bond{...}X]+}$, both halogens ended up beta to the carbonyl group, so I'd conclude it really depends on the orientation with which the $\ce{ICl}$ reacts with.

@Martin-マーチン by "modelling $\ce{[Mol...X]+}$, both halogens ended up beta" do you mean the  halogen which forms the cyclic intermediate?

@Safdar In the model, there is no cyclic intermediate, or at least there are significantly different leg lengths. See for example image: Regioselectivity of bromination of alkenes, but it is much more pronounced.

Does Anchimeric Assistance have to do something with this?

To be honest, I don't understand and you don't make it clear how Markovnikov's rule would apply here. Could you clarify please.

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But what are anti-bonding pi-orbital? In search for an intuitive explanation
Imagine that you want to explain to an undergraduate why they have to to shade the pi-orbitals in a symmetrical way, i.e. dark on top (+), white on bottom (-) for two neighbouring pi-orbitals because they are bonding (or anti bonding), but then the student asks: why?!.. Can anyone  explain why this is without knowing about quantum mechanics and diving into molecular orbital theory? Intuitively, it makes no sense that an electron should 'care' why if they are 'on top' or 'on the bottom' of the nucleus.
I have a hard time both intuitively understanding, but also explaining why we care about bonding and anti-bonding orbitals. I know that I should draw them, and that one is stabilising while the other is destabilising. However, for a person with limited chemistry knowledge my reasons for why is simply:  'because it is so'. This is not very satisfying. I think this has something to do with the phase and wave equations (..etc) - but again,..this seems to go into heavy chemistry (..or?).
Question: Does someone have a nice analogy, or way of explaining why this is - and what anti-bonding orbitals are without any very high-level chemistry concepts?
I think there are others struggling with this concepts, as I found a few questions here on stack exchange that failed to give a satisfying answer. This might be because the concept is extremely complexes and not possible to understand without quantum mechanics... but thats why I want to ask the question here in hope of an intuitive explanation.
The linked questions:

Anti-bonding molecular orbitals
Why do higher orbitals have more energy?
Why is the antibonding pi bond higher in energy than the normal pi bond?

Just a thought: You are asking in the same place where you were unable to find a simple enough answer..

Would you accept the wave-particle duality? Probably I can explain it I guess

im not sure about that expression, but its sounds like something related to the fact that electrons can be described as waves? since waves have phases - it is Similar to the idea if positive and negative interference. In that case, why would this explain that we have anti-bonding orbitals? or at all two types of orbitals (bonding and antibonding).. in my simple understanding the electrons “bounce” all over the place and cant really be expected to be at one particular location at some time

Electrons do not exist as either particles or waves.. See Wave-particle duality. Attempting to answer something that has its basis in a different level of theory using outdated versions is dangerous and may lead to unneeded misconceptions..

The problem with analogies is that you have to draw upon experience, which is naturally rooted in the macroscopic world. However, this is of limited experience for quantum mechanical concepts. At best, with an analogy, your understanding is incomplete, and at worst, you develop misconceptions due to the baggage of constructing the analogy.

Can anyone explain why this is without knowing about quantum mechanics and diving into molecular orbital theory? I'm afraid that won't be possible. If you are talking about orbitals, you need a basic understanding of orbitals. The most basic ones will be atomic orbitals, but looking at molecules you will need molecular orbitals. Since this is an approximation to the wave function, you will need (at least some) understanding of quantum mechanics.

While intuitively a lot doesn’t make sense, if you think you understood quantum mechanics you haven’t understood quantum mechanics.

First, remember that the + and - on the lobes of an electronic wavefunction do not ever represent or suggest electronic charge. An electron is always negative; a proton (or nucleus) is always positive. In order to imbed this firmly in your mind without + or - signs, imagine one lobe of a p-orbital flashing red while the other flashes blue. In this picture, an s-orbital would flash either red OR blue.
Plus and minus signs are just the way mathematicians describe the orbitals: they don't use red and blue pens, they use + and -. Now take the orbitals and move them about. Wavefunctions have a wave nature; waves have a frequency; so two orbitals of the same kind on similar atoms will have the same frequency. Two hydrogen s-orbitals flashing red at the same frequency look like one object, that is, they seem to be bonded into a hydrogen molecule (if they are close enough). Their waveforms are constructive.
But if one flashes red while the other flashes blue, the picture is not so attractive - the individual atoms are not in sync, not bonded. If I now switch the metaphor to a musical (audio) frequency rather than a visual wavelength, the idea of wave addition or neutralization images better (see picture).

To clarify, the red and blue orbitals don't totally collapse each other and disappear - just where they might "touch". So each orbital, red and blue, still contains one (negative) electron, but there is no continuity of electron travel from and around one nucleus to the other, because there is a gap, a node, a location where there is no electron density, where they might come in contact. This represents two  atoms that have been dissociated from their bonded state as a molecule. They could have been bonded together using two red-flashing orbitals, but got excited by some energy input that brought one electron up to a blue energy which is now out of sync with the red orbital, and finds it repulsive. A $pi$ bond between two carbon atoms would show the two upper lobes of each atomic orbital to have the same color and the two lower lobes of each atomic orbital to have the other color: both upper and lower regions would be bonding. A $pi$ antibond would have a red and a blue in the upper region and a blue and a red in the lower region: repulsive!
So we usually show the direction of bonding or repulsion in bonding or antibonding molecular orbital by using + and - signs on the atomic (and molecular) orbitals, saving red and blue colors for PowerPoint presentations.


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Modifying a TLC mobile phase in order to increase the RF values
You are running a TLC using 50% ethyl acetate in hexane and find the resulting $R_f$ values are too low. How would you modify the TLC mobile phase in order to increase the $R_f$ values?
Would the hexane and ethyl acetate be considered the mobile phase/solvent?

Increase the proportion of ethyl acetate

50 vol% EA in hexanes already is pretty polar.  And yes, EA/hexanes is the mobile phase regardless how well the components of your analyte elute, or not -- this should be clear if engaging a TLC. Speculating you analyte is neither a low molecular weight carboxylic acid, phenol, primary amine (there is too few information provided by you), did you check the product is neutral (as in uncharged) and salt-free?

@Buttonwood really depends on what the substrate is. 1:1 Hexanes:EA is not to my mind particularly polar - that would be 7% MeOH in EA. It is worth running half a dozen trial mobile phases to see which one gives a decent separation.

@Waylander True, a high $R_f$ is not very useful if $\Delta{}R_f$ between the compounds of interest is little.

Question: Would the hexane and ethyl acetate be considered the mobile phase/solvent?

Answer is yes: Hexane/ethyl acetate systems ($0$ to $100\%$ ethyl acetate in hexane) are frequently used mobile phases for TLC technique. However, as Waylander pointed out in his comment:

The type of mobile phase really depends on what the substrate and product are. It is worth running half a dozen trial mobile phases to see which one gives a decent separation.

Also, it is  true, as Buttonwood pointed out, a high $R_f$ value is not very useful if $\Delta R_f$ between the two compounds (or more in the mixture) of interest is small unless it increase upon change of mobile phase. For example, TLC of the mixture of benzoic acid and benzocaine (ethyl 4-aminobenzoate) shows no separation in $50\%$ ethyl acetate in hexanes. However, when mobile phase is change to $5\%$ methanol in dichloromethane, it has shown $\gt 33\%$ separation in $R_f$. So it is a trial and error method, but worth trying.


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Is there a way to generate SMILES from ChemDraw files?
I have a set of ChemDraw .cdx files, and I would like to extract smiles strings from it.
Can Open Babel do that or maybe the rdkit in Python? Any other suggestions?

It appears OpenBabel can do this. I haven't looked in the menu for the specifics, but this web server hosts a version of OpenBabel and it is able to convert from cdx to SMILES.

ChemDraw can do it itself. Is there any special reason why you cannot use it in this case, so that we can better understand what you need?

If you're looking for a programming lib, here are docs on Indigo: https://lifescience.opensource.epam.com/indigo/api/index.html#reading-sdf-rdf-cml-multiline-smiles-files-cdx-binary

You can use our web tools: https://eln.chemotion.net/chemscanner. Click on "Scan Files" then "Scan for molecules", then select all the .cdx files that you want to read. After the results are finished, you can click on "Export" and choose "Excel"

Of course openbabel can do this (at least with release 3.1.0 from here).  To rewrite a file here and then, the GUI on top of the program on top may suffice for you.  However, you would skip a lot of the program's functionality if not trying to interact from the command line: for one, the conversion basically follows the pattern of
obabel -icdx input.cdx -osmi -O output.smi

which actually may be faster than clicking in an interface. For the other, that openbabel offers multiple interfaces to automate file transformations programmatically, i.e. to interact with programming languages like Python and C.  See the current documentation rather than the a bit outdated on the project's senior page.
Note 1: ChemDraw offers a structure to SMILES string conversion, too.  Be aware that a .cdx file may contain much more information, than a SMILES string may retain.  Seeking for a text-only representation of the molecular structure, axial chirality (P/M) is an example where a SMILES string is less suitable than, for example, an InChI string.
Note 2: Openbabel aims to interconvert chemical formats; on occasion with a visual output like .svg, or .png.  RDKit has a stronger emphasis on computing with the molecules (a.k.a. cheminformatics) as required while interacting e.g., with large sets of data and databases.  There is some overlap of the two programs, and «the trick» is rather to recognize where the two are complementary to each other.


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Why is potassium permanganate paramagnetic?
$\ce{KMnO4}$ is coloured due to charge transfer spectra. As Mn doesn't have any electrons, it shouldn't be magnetic. But, according to the magnetic susceptibility value given on Wikipedia, it comes out to be paramagnetic. How is it possible?

Welcome onnnthe ChemSE! Mn has a lot of electrons, about 30. I suggest to clarify this part :-)

@Cde124AB_ You may mean all Valence electrons are invokved  in bonding and there is none left to be unpaired. It may need MO point of view. Even O2 does not seem to be paramagnetic by simple valence electrons inventory.

Good question. Please see my comment at https://chemistry.stackexchange.com/questions/115841/magnetic-properties-of-potassium-dichromate#comment218490_115841 If I have time I will turn that into an answer. The reason is not trivial.


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Can the molecule POOF exist?
Obviously, it really would be $\ce{O=P-O-F}$.  Since $\ce{F-O-O-F}$ exists the $\ce{-O-F}$  part is obviously possible and the rest of it doesn't strike me as unreasonable, but whether it can actually hold together is way beyond my knowledge.  That $\ce{-O-F}$ says it would be some nasty stuff if it exists at all.

Almost certainly the proposed structure would be extremely unstable to rearrangement - there would be no O-F bond, and instead all atoms would be directly bound to phosphorus.

Being a native of Ohio, I was disappointed to learn that HIO2 apparently does not exist.  Like POOF, the structure of iodous acid would not be accurately rendered by OHIO, but still ... .

@OscarLanzi in a paper, it is formulated as HOIO and some research is also going on: here and here.

@NicolauSakerNeto I'm sure it's unstable if it exists at all, it would be quite prone to going poof (or more likely BANG!)

There exists related anions called monofluorophosphate ion ($\ce{PO3F^2-}$) and difluorophosphate ion, ($\ce{PO2F2-}$). The hypothetical neutral compound, $\ce{PO2F}$ is called phosphenic fluoride. It is also observed that irradiating potassium difluorophosphate with gamma rays create free radicals $\ce{PO2F^{.−}, PO3F^{.−} and PO2F2^.}$.
Moreover, there also exist a polymer $\ce{(PO2F)_n}$ which is a decomposition product of a white solid of composition $\ce{P7O10F15}$. It was apparently named polymeric phosphorus oxyfluoride species. Its structure has been proposed by Wannagat and Rademachers1:

It is suggested that the parent ions of Group I result from the simple
ionization of a linear polymer with the general formula, $\ce{(PO2F)_n}$,
where each of the terminal phosphorus atoms of the chain is bonded to
two oxygen atoms, one of which is a bridging atom, and to two fluorine
atoms.

Conclusion: There is no such compound called POOF, but there certainly is a radical ion and a polymer of exact formula.
Notes and References

High molecular weight phosphorus oxyfluorides by D.W.Muenow, O.M.Uy, J.L.Margrave, Journal of Inorganic and Nuclear Chemistry, Volume 31, Issue 11, November 1969, Pages 3411-3415, DOI: 10.1016/0022-1902(69)80324-6
Other decomposition products were $\ce{PF5, POF3}$ and $\ce{P2O3F4}$: Technique of Inorganic Chemistry, Hans Boegh Jonassen, Arnold Weissberger Volume 1
Interscience Publishers, 1963

What am I missing?  Why would PO2F be a radical ion and not neutral?


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Change of ionic packing upon crystals are pressurized
When rock salt crystal is pressurized, its FCC structure changes to BCC structure whereby coordination ratio changes from 6:6 to 8:8. At the same time packing efficiency of the crystal decreases from 74% of FCC to 68% of BCC. So will the crystal actually expand or contract during this process?

Your termination of BCC is incorrect about the $\ce{NaCl}$ crystal structure upon pressurization. It is actually simple cubic (SC as $\ce{CsCl}$) where eight $\ce{Cl-}$ ions ate the corners of unit cell and one $\ce{Na+}$ in the center or vise versa. Although either $\ce{Na+}$ or $\ce{Cl-}$ in the center, it is not BCC because ion possessing the center of unit cell is different from the ions at the corners of unit cell:

Thus, lattice constant $a$ can be expressed as:
$$a_\mathrm{SC} = \frac{2(r_\ce{Na+} + r_\ce{Cl-})}{\sqrt{3}} \tag1$$
For FCC $\ce{NaCl}$ crystal structure:
$$a_\mathrm{FCC} = 2(r_\ce{Na+} + r_\ce{Cl-}) \tag2$$
Since $a_\mathrm{SC} \lt a_\mathrm{FCC}$, we can conclude that upon pressurization, $\ce{NaCl}$ crystal structure would be  contracted.

Most information on the net says FCC changes here to BCC and not SC. This research data also points at this:http://jetp.ac.ru/cgi-bin/dn/e_016_04_0855.pdf

And most information on the net would be wrong.  The referenced paper smartly avoids BCC and SC in favor of just calling it CsCl-type.


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What is happening when KCl burns in a bunsen flame?
I did the experiment to qualitatively observe the different colours given off by different salts when placed in a flame.
But what is actually happening to the salt ?
Take $\ce {KCl}$ for example.
Does the $\ce{ KCl}$ react with $\ce {O2}$ ?
What does the reaction equation look like ?
Other salts include $\ce {LiCl}, \ce{BaCl2}$, $\ce{SrCl2}$, $\ce{CuCl2}$ and $\ce {NaCl}$.

I wanted to elaborate a little bit more so that the OP does not develop misconceptions. It is the potassium atoms not potassium ions that color the flame purple. While I was writing the answer, another poster emphasized the same point.
Caveat: In ordinary Bunsen burner type flames, emission from calcium, Sr, and Ba is from molecular compounds not atoms. The story of atomic emission is only true for alkali metals. The flame temperature is quite low in Bunsen burners. You need special flames to "see" the atomic emission of Ca, Ba, Sr.
Assume the processes when you introduce KCl in the flame:
a) The salt has to melt KCl (l)
b) It has to vaporize KCl (gas)
c) It has to decompose into constituents. Flames have free electrons, so potassium ions are converted to potassium atoms.
d) From the thermal energy of the flame, potassium atoms are excited. Once they de-excite, you see the violet/lilac color. It is an ultrafast process!
e) The flame temperature is so "low" on the scale of universal temperatures (stars, Sun) is that only 1-2% atoms get excited.
Now you may ask, what about the chlorine atoms, well, they emit in the deep ultraviolet and we cannot see them. The flame temperature is so low that it cannot excite chlorine atoms.
Keep in mind it is not a single wavelength. If you were lucky enough, you could use a pocket spectroscope and see red and violet lines (plus a persistent yellow line of sodium contamination).
Actually the strongest lines for potassium are 764 and 769 nm which are deep red. Many humans cannot see this deep red. So your flame would rather be deep red. Potassium atoms also emit less strongly at the other extreme end of the spectrum, corresponding to deep purple lines around 404 nm. The resulting "mix" of wavelengths appears to our eyes as lilac color!

"Flames have free electrons, so potassium ions are converted to potassium atoms." - Who has misconceptions here? There's no K cations in vapor in the first place, just molecular KCl https://chemistry.stackexchange.com/questions/14174/what-is-sodium-chloride-like-in-gas-state

Mr. Mithoron, Who said K cations, anywhere? Ever heard of flame ionization detector in gas chromatography? Flame is a conductor of electricity, it has a small amount of positive ions as well as free electrons. May be read this physics answer here to learn more about flames: https://physics.stackexchange.com/questions/45105/does-fire-conduct-electricity-why

@Mithoron, Have you heard of reducing or oxidizing flames in emission spectroscopy?

Flame isn't plasma. While ionisation equilibria give off some electrons, especially in case of atoms like potassium. K atoms in the flame are there because of thermal dissociation of KCl molecules ($\ce{KCl -> K + Cl}$), not reduction of cations with electrons

It is not such a simple picture as you are imagining. Have you used flame ionization buffers? Also, atomic population varies across the flame height for different elements. It is just not the temperature! Flame is such a complex system that people have just written complete books on flame emission spectroscopy.

I never realized that for some materials it's a molecular spectrum and for others its atomic, but then again I don't know anything about Chemistry :-) What about copper? In the recent Con-Ed transformer “fire”, what exactly produced the color of the huge blue glow over New York City? I have a "colorful" but somewhat speculative answer there; without a spectrum of the green glow it's hard to say for sure.

@uhoh, yes those are molecular copper compounds in the flame. Copper atoms resonance lines are in the ultraviolet.

Upon the heat of the flame, the crystals shatter and ...

The correct term is "the crystals vaporize"

is an ensemble of K+ and Cl− in regular proximity to each other. The
heat of the flame separates them, and excites them, independent of
each other, e.g., an outer electron of K+ is then lifted to a higher
level.

When ionic crystals vaporize, there are no longer
ions! The emission of spectral lines comes from
potassium or sodium etc atoms.
The temperature of Bunsen flames is not high enough
to make a plasma of ions.

Because solid KCl already consists of ions, I did not recognize that the flame of the Bunsen burner had to ionize the material.  Nor was I aware a monoatomic ion like $\ce{K+}$, thus in nobel gas configuration, would this easily be reduced back to K, which -- in condensed state -- tends to oxidize rapidly.  Even if there are the terms of «oxidizing zone» and «zone of reduction» while preparing the bead test or using the blow pipe.


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Effect of isotopes on H-Bonding
The following question was asked in one of the assignments my teacher has given.

Acetone ($\ce{Me2CO^16}$) on treatment with $\ce{H2O^18}$ gives a mixture of $\ce{Me2CO^16}$ and $\ce{Me2CO^18}$, the
latter being in slight excess. This may be explained by
(A) hydration of acetone is an equilibrium process
(B) $\ce{C – O^18}$ bond is slightly stronger than $\ce{C – O^16}$
(C) $\ce{Me2CO^18}$ forms stronger hydrogen bonds than $\ce{Me2CO^16}$ with water
(D)hydration of acetone is irreversible

I have doubt about option (C). According to me, (C) should be correct, but it is given incorrect. Why the $\ce{^18O...H}$ is weaker than $\ce{^16O...H}$ bonding?

Without thinking further about it, the question actually doesn't state what you have deduced. Just because the option is marked false, it doesn't mean that the reversion of the statement would be correct.

Is the reason given as B? Notice they ask for which option is the explanation and not the validity of the statements.

Can any conclusions really be drawn since the ratio of acetone to water isn't specified?

@MaxW I think OP's point is to just compare the strength of $\ce{^18O...H}$ bond vs $\ce{^16O...H}$ bond, which is a good question.

@AniruddhaDeb But $\ce{^18O...H}$ is not weaker than the $\ce{^16O...H}$ bond or atleast you can't conclude that from the fact that that is not the answer. Seems like more of a lack of understanding what the question asked for (which is ambiguous enough).

Related: https://chemistry.stackexchange.com/q/31234/ and https://www.nrcresearchpress.com/doi/pdf/10.1139/v66-393 and https://www.chemistryworld.com/news/oxygen-isotopes-help-scientists-probe-waters-structure/3003103.article

Hydration of acetone is an equilibrium process.
$$\ce{(CH3)2C^16O + H2^18O <=> (CH3)2C^18O + H2^16O }$$
The equilibrium constant of this reaction will depend on how the bond strengths of the $\ce{O-H}$ bond in water and the $\ce{O=C}$ bond in acetone is affected by isotope exchange. The isotope-dependent strength of hydrogen bonds might be a secondary effect, as well.
The ratio of heavy to light acetone will not only be dependent on the equilibrium constant, but also on the ratio of acetone to heavy water initially present.
So none of the statements are sufficient to explain the observation completely. Statements (a) and (b) are correct. Statement (c) is not directly related to the ratio of the species. You also have to consider the isotope effect on the strength of hydrogen bonds with water (as one acetone changes from light to heavy, one water changes from heavy to light).


text:
Interpreting a thermodynamics equation using a picture
Given
$$ \frac{\partial \Gamma}{\partial p}\Bigg|_{t} =\frac{\partial \Gamma}{\partial p}\Bigg|_{v}+\frac{\partial v}{\partial p}\Bigg|_{t}\frac{\partial \Gamma}{\partial v}\Bigg|_{p} \tag{1}$$
where
\begin{gather}
\Gamma(p,v) = vp^{2}\tag{2a}\\
pv=t\tag{2b}
\end{gather}
I'm trying to interpret $(1)$ using words or preferably a picture.
Note for notations: $p$ is pressure; $v$ is volume; and $t$ is temperature.
$(2 \text{b})$ is equivalent to the ideal gas law but with temperature in units of energy. I suspect the picture we're looking after is something along the sides of a $pv$ diagram involving an isotherm. Since $t$ is held constant in this setting, my conclusion is that $(1)$ represents the isothermal compression of an ideal gas, but frankly this conclusion is shaky and I'm not sure of my interpretation.

Do your $PVT$ in the text represent $pvt$ in the equations?

@MathewMahindaratne  yes I believe they do.

See my latest post on the other site.


text:
Which amino acids can be totally synthesized?
I can't find information regarding which amino acids (say the proteinogenic ones or in general) can be totally synthesized. Is there a reference?

Synthesized in the body? Synthesized chemically? Which one?

Also, amino acid is a general concept such as $\alpha,$ $\beta,$ etc. Thus question can be categorized as too broad.

@MathewMahindaratne proteinogenic amino acids are all $\alpha$

Glycine can be synthesized from acetic acid. First acetic acid can be chlorinated by reacting with chlorine gas. Then the chloracetic acid reacts with ammonium carbonate. This produces ammonium chloracetate, that can be transformed into amino acetic acid (glycine). See the details the procedure in typing Gycine Synthesis in Google.

An example of a possible synthesis of methionine. Seems too broad

All of them can be accessed synthetically


text:
Why does Fe(II) have electrons in the 3d subshell?
The electron configuration for $\ce{Fe^2+}$ in my coursebook is given as:
$$ 1\mathrm s^2\, 2\mathrm s^2\, 2\mathrm p^6\, 3\mathrm s^2\, 3\mathrm p^6\,3\mathrm d^6$$
But shouldn't the $4\mathrm s$ subshell take precedent over the $3\mathrm d$ subshell according to Aufbaus principle?

For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details

Try writing the configuration of $\ce{Fe}$ and then remove two electrons from the outermost shell (higher n-value is farther)

Remember releasing of electrons is not the reverse order of their filling, as orbital energies change.

I would like to suggest this question as another dupe target, but if I vote to close it will be Golden-Mjölnir’d (dupehammered with a golden badge).

You might want to compare iron(II) not with neutral iron, but with neutral chromium.
Both iron(II) and neutral chromium have 24 electrons, but there the similarity ends:
$\ce{Fe^{2+}}:  [\text{Ar}]3d^6$
$\ce{Cr^{0}}:  [\text{Ar}]3d^54s^1$
Think of the Bohr model.  According to this model the energy level of electrons depends only on the shell number $n$, so we would expect 24 electrons to follow the $[\text{Ar}]3d^6$ configuration.  In real life that happens exactly only for single-electron atoms where there are only electron-nucleus interactions.  When there are electron-electron interactions they could fill the shells not in order, like the chromium atom described above.
But in a multielectron atom, if you add more nuclear charge you make the electron-nucleus interaction stronger, and the Bohr-model configuration becomes more favorable.  One might suppose that dozens of added protons might be needed to get chromium's $[\text{Ar}]3d^54s^1$ to the Bohr-predicted $[\text{Ar}]3d^6$.  Instead, the energies of the two configurations are so closely spaced that two extra protons, converting neutral chromium to ferrous iron, is enough.

Could someone also mention a concrete experimental way of determining this configuration perhaps spectroscopically? Electron configurations look like fairy tales now. How come these single electron wave functions were applied on complex ions of Fe and Co?

Oscar, That is a very hand waving answer by that person. He just scattered the answer with fancy names of X-ray techniques. Electron configurations "existed" well before XPS and other plethora of X-ray techniques were discovered.


text:
What does washing and drying a solution mean?
After combining two solutions, a direction in a 1970s British chemistry book says to wash the new solution with a mixture of baking soda, table salt, and water. Then it says to dry it with anhydrous magnesium sulfate.
What does washing and drying mean? Do the chemicals get physically mixed together? Doesn't that contaminate the solution?

This is standard for purifying substances.
To wash means to add your product solution to an aqueous solution (or just water, but frequently a saturated solution) to a separatory funnel. After shaking, you drain the lower layer (which is usually aqueous). This process removes water soluble impurities. This is frequently repeated.
Drying is accomplished by taking the organic solution of your product and mixing it with a desiccant (usually anhydrous sodium sulfate or anhydrous magnesium sulfate). This removes any water that was left from the washing in the above step. The desiccant may then be removed by filtration.

The solvent of this experiment should be non polar, and may be a hydrocarbon, or a chlorinated hydrocarbon. This organic phase may contain as impurities some acids and other polar substances soluble in water, that should be eliminated. This liquid is then mixed with an aqueous solution of soda $\ce{NaHCO3}$. Aqueous solutions and non polar substances are not miscible. But the soda reacts with the acidic impurities, producing a salt which is soluble into water. So, by stirring the two liquids, the impurities of the organic phase (salt plus other polar substances) will pass into water, and are eliminated from the organic phase. This organic phase is said to have been "washed" by the soda solution.
The organic phase has not the same density as the aqueous solution. So it is easy to separate the aqueous phase and keep the organic liquid. But the previous treatment has also added a few percent water which has entered the organic phase. This water must be eliminated in a second phase, called "drying". It is done by adding a powder to the organic phase which is insoluble in the organic phase but reacts with water. This powder is magnesium sulfate $\ce{MgSO4}$ which has a strong tendency to react and fix all water molecules in their vicinity. It produces a solid hydrate like $\ce{MgSO4·H2O}$ or even $\ce{MgSO4·7 H2O}$. After stirring the organic phase has lost all its water. The organic phase is called "dry".

The description you provided is essentially a shorthand for a standard workup procedure. After performing a reaction, one wants to isolate one’s product while removing the catalyst or other reagents and side products that one doesn’t need. For example, in many cross-coupling reactions an inorganic halide salt is an undesired side-product and a base such as sodium hydrogencarbonate may have been added to quench protons that form during the reaction. Likewise, a reaction may be catalysed by a small amount of acid (e.g. sulphuric acid or acetic acid).
The washing part of workup makes use of the fact that a lot of these side-products are hydrophilic while the organic products typically are not. By combining the organic phase with an aquaeous solution (they will not mix) in a separating funnel and shaking the funnel to ensure large contact areas, hydrophilic side-products can easily transfer into the water phase.
You mention that your method suggests using sodium hydrogencarbonate solution for washing; thus, I assume that your reaction was run in acidic conditions. For acidic conditions, I typically wash with sodiumhydrogencarbonate solution while for basic conditions I wash with ammonium chloride; both have the effect of neutralising the acid or base generating ionic compounds which are then more easily transferred into the water phase. Contrary to the other answer, I do not use saturated solutions for washing; rather, I add approximately equal parts of a saturated solution and deionised water because in my experience using saturated solutions can lead to a supersaturated aquaeous phase and subsequent precipitation of salts—not what I want.
Many a time the methods suggest extracting the aquaeous solution with the organic solvent. This is to ensure that the maximum amount of organic product is taken out of the aquaeous phase. For this, the organic phase of the first separation step is put aside, the aquaeous phase readded into the separation funnel and further organic solvent is added; rinse and repeat.
After the initial washing, I wash the combined organic phases with deionised water—this is to remove any remaining traces of acid or base that may have passed through the first step or ammonia that may have been generated by washing with ammonium chloride.
Now, I could immediately remove the organic solvent at the rotavap. However, the solution has been saturated with water; although water is only poorly soluble in most organic solvents poorly is distinct from insoluble; were I to remove the wet solvent at the rotavap most of the water would likely remain with the product. This is where drying comes in: the name essentially means removal of most of the water from the organic phase to ensure that only the organic product will remain in the flask at the rotavap.
Usually, drying is a two-step process. First the solution is washed once again with saturated (no dilution this time!) sodium chloride solution or brine. This highly concentrated solution serves to remove most of the dissolved water from the organic phase into the aquaeous phase. Second, a drying agent (usually anhydrous sodium sulphate or magnesium sulphate) is added to the organic solution. Neither of these salts will dissolve in the organic solvent but both are very hygroscopic and will draw moisture towards themselves. The resulting solution is considered ‘dry’ for workup purposes. The inorganic sulphate is filtrated off and the solution is ready for the rotavap. When performed correctly, no residues of water remain in your rotavap-dried product.


text:
If phosphorus trioxide is present in the atmosphere of Venus, could it not react with water to form (sufficient) phosphorous acid?
Sufficient phosphorous acid to be decomposed in the measured quantity of phosphine ?
From Phosphine gas in the cloud decks of Venus:

We also rule out the formation of phosphorous acid ($\ce{H3PO3}$). While phosphorous acid can disproportionate to $\ce{PH3}$ on heating, its formation under Venus temperatures and pressures would require quite unrealistic conditions, such as an atmosphere composed almost entirely of hydrogen.

But according to Vega mission results and chemical composition of Venusian clouds there is $\ce{P4O6}$ in the atmosphere which according to Wikipedia would react with water to form phosphorous acid.
However there is supplementary information in the chapter "Equilibrium thermodynamics in the atmosphere and surface" at the end of the article "Phosphine gas in the cloud decks of Venus":

As an example of our approach, we present a calculation for phosphorous acid ($\ce{H3PO3}$). This compound will spontaneously decompose on heating to form phosphoric acid  and phosphine; this is a standard laboratory method for making phosphine. Phosphorous acid is not stable in gas phase, but could in principle be formed in cloud droplets by reduction of phosphoric acid.

Edit for some additional informatiom.
Also from the supplementary information (page 13):

Reactions of $\ce{P4O6}$, $\ce{P4O10}$, $\ce{H3PO4}$ and $\ce{H3PO3}$ were considered (the last of these only in solution phase in the clouds),..

So not considered was the possible occurence of $\ce{H3PO3}$ below the clouds.
The Recent Evolution of Climate on Venus (page 23) states:

The evaporation of $\ce{H2SO4}$ occurs at about 48 km, the average cloud base.The vapor phase continues to exist down to 432 K (38 km), where it is thermally decomposed.

So below 38 km, and above 160 ⁰ C there would be water in the gas phase, free from sulfuric acid, that could react with $\ce{P4O6}$.
Could have that possibility been ruled out for some reason ?

You bring up an interesting point. Unfortunately this matter is heavily reliant on quantitative modelling of reaction kinetics in a very unusual environment, so I'm not sure any of us here have the capacity and resources to answer this satisfactorily. As Maurice mentioned, you'd pretty much have to ask the authors directly.

Even if there is $\ce{P4O6}$ in Venusian atmosphere, there is always no water vapor(only 20 ppm). This is not enough to make even a significant of phosphorous acid.

This is a follow up question based on my above comment.

Speaking of asking the authors directly, you can try this thread on Reddit's AskScience.

Maybe there isn't enough water? Wikipedia: "The water has probably photodissociated, and the free hydrogen has been swept into interplanetary space by the solar wind because of the lack of a planetary magnetic field.[23]"

@BuckThorn That is an intriguing question "Since there isn't enough water, how is phosphorous acid formation possible?"

The issue seems to be not the reaction between $\ce{P4O6}$ and water, but whether $\ce{P(III)}$ can form at all.  The OP's first reference says no whereas the second reference says yes.  Basically the two references used different thermodynamic assumptions.  The first one assumes an atmosphere in equilibrium while the second assumes only a partial equilibrium; in the latter case $\ce{CO}$ is allowed to form from nonequilibrium (radiation-driven) dissociation of $\ce{CO2}$, and minor species are then equilibrated with the $\ce{CO/CO2}$ couple.  In the case of phosphorous oxides this favors $\ce{P4O6}$ over $\ce{P4O_{10}}$, although the former could be oxidized by other species such as sulfuric acid.
The issue has become important because the recent discovery of $\ce{PH3}$ among the clouds of Venus has led to a hypothesis that this compound is generated by Venusian microbes, but if $\ce{P(III)}$ is available for a disproportionation reaction it could provide an alternative abiotic (non-life) source.  As noted above, the references disagree on this, and the question of whether this alternative to a biological source of $\ce{PH3}$ is viable awaits resolution.
Relevant to the question of whether $\ce{PH3}$ is a true biosignature is this question about organic compounds on Venus, which apparently spawned the current one.

Your answer proves that all the comments of the "nobody here can answer this" variety under the question are "premature" in the Mark Twain sense :-)

"..this favors P4O6 over P4O10, although the former could be oxidized by other species such as sulfuric acid."  How does this correspond to the forming of phosphorous acid in cloud droplets by reduction of phosphoric acid ?

We have a three-way competition between reduction of the phosphorus from +5 to +3 (by CO), oxidation back to +5, and disproportionation of the +3 state to give phosphine and maybe other low oxidation state prodtcts.  We need more complex modeling and better experimental knowledge to sort this all out, hence the question "awaits resolution".

Made an edit to make it more clear that there is steam below the clouds , free from sulfuric acid that could react with P4O6.

I would like to note that the formation of CO by photochemical reaction of CO2 occurs near the cloud top. And because of the Hadley circulation on Venus near the equator SO2 is transported upwards from below the cloud deck that needs O for SO3 and could  favor P4O6 over P4O10.


text:
Methane Sensor electrochemical reaction
When using a methane sensor a voltage is applied and much like electrolysis if the voltage is above the decomposition potential a reaction happens. Flammable gasses like methane are easily oxidized and thus produce an anodic current. But what is the cathodic current reaction in a methane sensor in air? what is being reduced?

Is the CO2 reduced to CO carbon monoxide? Can the methane be oxidised to Methanol in the presence of water?

CH4 + 2O2 --> CO2 + 2 H2O is this the full redox eqn?

I’ve heard aqueous based (typically acidic) electrolytes produce hazardous substances. But what are they? This was what they used in the 60s iirc


text:
Maximum Local Value for the Electron Density
So the electron density is the function $\rho(\vec x)$ that associates a value of electrons per cubic angstrom, to each point $\vec x$ in the 3D space.
This information tells us how likely it is to find electrons at specific regions of space within a system. Integrating $\rho$ over a particular volume $\mathcal{V}$ gives us the number of electrons likely to be found, at any given moment, inside $\mathcal{V}$, denoted $n[\mathcal{V}]$.
$$
\tag{2}
\label{eq:rho-integrate-volume}
n[\mathcal{V}] = \int_\mathcal{V} \rho(\vec x) \mathrm{d}\vec x
$$
Integrate over the whole space $\Omega$ where the system is and you get $N$, the total number of electrons in the system.
$$
\tag{3}
\label{eq:rho-integrate-omega}
N = n[\Omega] = \int_\Omega \rho(\vec x) \mathrm{d}\vec x
$$
Electrons repel one another and are attracted to the atomic nuclei. I believe that particularly high values of $n[\mathcal V]$ are unlikely because as you pack more and more electrons in the same region, their repulsion would start to overpower their attraction to the nuclei, meaning the electron density would likely spread out more.
So my question is Is there a maximum value for $n[\mathcal{V}]$ above which the mutual repulsion of electrons would mean the density was unreal?
I understand that such a value would vary depending on promixity to the atomic nuclei (which would shield electrons from their own mutual interactions) and the size of the integrated volume $\mathcal{V}$. What I mean is if there is a known relationship between these factors that would give us the maximum amount of electrons allowed inside a particular region of space.

I can't really explain why, but this question irks me somehow. The electron density is a physical observable, it's not just a mathematical concept. It cannot be unreal. So what you measure is there. Repulsion on the other hand is not an observable. If your question is about a hypothetical case, I don't understand the parameters of the thought experiment. If your question is rather about the concepts of Density Functional Approximations, then that again would be a different story and much more involved. In any case, I don't really understand what you are asking.

I'm tempted to answer that n(V)<=N as rho(r) is positive definite, but don't as that's clearly not what is really being asked. But I can't formulate what mathematically IS being asked.

Short Answer:
There is a limit to the maximum value of electron density $\rho_e(\vec{r})$.
Long Answer:
It is true that electrons repel each other, so as the electron density increases, the interelectronic repulsion also increases. This means that you can find high electron density only if the nucleus-electron attraction is stronger than the e-e repulsion. You can get this in heavy atoms, as noted in Karstein Theis' answer.
The electron density near the nucleus of the heaviest elements we can synthesise would be the highest we achieve, at least on earth.
But there is a way to increase the electron density even more, and that is by pressure. This is exactly what happens in the cores of stars. The gravitational force for a star is large enough that it starts to push in on the material at the core (does not matter what it is, as long as it consists of nuclei and electrons).
Electrons are fermions, and have to obey the Pauli principle and Fermi-Dirac statistics. So, when you try to push electrons together in the same volume, they will push back. If there is a gas of electrons and we ignore the inter-electronic repulsion for a moment, the pressure exerted by it is — $$\ce{P_e=\frac{(3\pi^2)^{2/3}\hbar^2}{5m_e}\rho_e^{5/3}}$$
$\ce{m_e}$ is mass of electron, and $\ce{\rho_e}$ is the volume density of electrons.
This is from a non-relativistic treatment. If all electrons experience relativistic effects, (which you need to consider for high $\rho_e$ values) the pressure is $$\ce{P_{e,r}=\frac{hc}{4}(\frac{3}{8\pi})^{1/3}\rho_e^{4/3}}$$ This type of pressure is usually called the electron degeneracy pressure.
If the star is really massive, the core can experience huge amounts of pressure. Fortunately, as long as the star is "burning" there is an outward radiation pressure from the photons which holds everything together. What happens when the star consumes all its fuel (H and to a smaller extent, He) and stops burning? Well, there is nothing to hold the core against the crushing gravitation, and the electrons and the protons merge together to form neutrons, as the degeneracy pressure is not strong enough to hold back the gravity. This is a neutron star. (If heavier, it may form black hole).
In smaller stars, the electron degeneracy pressure wins, and the star can no longer contract. It stays as a white dwarf. The maximum mass where this can happen is called the Chandrashakhar limit, usually taken to be around $\ce{1.4\times solar mass}$. Note that this value is entirely derived by assuming that the electrons in stars behave as an electron gas (with relativistic corrections applied). You can include the electron-electron or electron-nucleus interactions by a more sophisticated treatment (see this paper, pdf link), but it only changes the value to 1.38 solar mass. So, the electron degeneracy pressure is so large here, that the other terms become negligible.
Then what is the value of the electron density at this mass limit? Well, the thing is, I can't find the value of $\ce{\rho}$ in any of the papers, but there must be a value for this. If I find it, I will update my answer.
Meanwhile, we can get a very rough estimate of the maximum electron density—
Assuming that the star is a sphere under hydrostatic equilibrium, the pressure at its centre must exceed $\ce{\frac{GM^2}{8\pi R^4}}$.(ref.) For main sequence stars heavier than sun, using the mass-radius relation ($\ce{R\propto M^{0.57}}$) gives a radius of about $\mathrm{0.8\times 10^9 m}$ for a star with 1.4 solar mass (by plugging in the mass and radius of sun, and taking ratio). The pressure at its centre must be greater than $\mathrm{5.01\times 10^{13} N/m^2}$. Considering completely relativistic electron gas gives a $\mathrm{\rho_e\approx 2\times 10^{36} m^{-3}}$. Obviously, this is a crude calculation, but you can see that there is a maximum value of electron density. Above the maximum value, the density becomes unfeasible not because of electron-electron repulsion, but because electrons and protons fuse in a nuclear reaction.

Single electron systems

I believe that particularly high values of n[V] are unlikely because as you pack more and more electrons in the same region, their repulsion would start to overpower their attraction to the nuclei, meaning the electron density would likely spread out more.

The main constraint on high electron density is not the repulsion between electrons, but the Heisenberg uncertainty principle. Even a single electron has a limited average electron density, given its average momentum (see e.g. here). For single electron systems (hydrogen-like ions), you get the highest electron density by increasing the nuclear charge. According to the virial theorem, as the potential energy of the electron increases (the magnitude, so tighter binding), the kinetic energy also increases. This will go along with a higher momentum, allowing the electron to occupy a smaller volume (i.e. have a higher density).
Multi-electron systems
For multiple electrons, theory gets complicated. X-ray diffraction gives experimental data related to electron density, but you need extremely high resolution (and extremely low static and dynamic disorder) to be able to map the electron density near the nucleus. In this study on LiF and NaF, the electron density at the lithium nucleus is estimated as 50 electrons per cubic angstrom using maximum entropy methods (panel b).

Fluorine, with a higher atomic number, will have higher electron density at the nucleus, but the density is cut off in the figure so you can't see the maximum value.

I understand that such a value would vary depending on promixity to the atomic nuclei (which would shield electrons from their own mutual interactions)

The innermost electrons would not be shielded, so the maximum electron density for a uranium nucleus with one electron would not be much different from one with 92 electrons. In fact, the binding energy of the inner electrons was used to figure out atomic numbers using Moseley's law.

Is there a maximum value for n[V] above which the mutual repulsion of electrons would mean the density was unreal?

Electron repulsion is not the main constraint on high electron density. To get high electron density, you need a high nuclear charge. At some point, relativistic effects become important, and for super-heavy nuclei, electrons get captured.
There was a comment from Martin about the use of the word "unreal". I agree with Martin that if you observe a certain high electron density, that is as real as it gets. You might run into a calculation claiming a ridiculously high electron density, and then might conclude that results are not plausible and warrant further research.
So where can I find high electron density?
That would be near the nucleus of the heaviest elements as long as they are stable at least for a while. Using the arguments made above using the uncertainty principle, the binding energy of inner electrons, and the associated momentum of the electron (or looking at the Bohr model), the radius $a_0$ should be proportional to the reciprocal of the atomic number, meaning the electron density increases with the cube of the atomic number.

The bounty system allows only incrementally increased rep rewards, so I wasn't able to assign the same amount to both answers. Sorry, Karsten, since you were first, you received initial smaller bounty. Both answers take different approach and are equally educational and are enjoyable to read.

@andselisk Thanks, your comment means more to me than the "imaginary points" (I think Martin coined the term). It was an interesting question, and I would have missed it if you hadn't tagged it.


text:
Do any of the geometry names (VSEPR) change if you use a double or triple bond instead of single bonds?
So I'm studying chemistry and wondering if any of the geometry names in VSEPR, like linear, trigonal planar, square planar, etc., change if you use a double or triple bond instead of single bonds? Are there any examples of such a thing happening? I tried it using a PhET simulation, but couldn't notice any change at all.

VSEPR theory classifies molecules based on the number of bonding/lone pairs surrounding the central atom to determine approximations in bonding angles. Because the repulsion forces between double and triple bonds are not significantly greater than single bonds, they are treated as single bonds.
Quoted from Wikipedia:

In VSEPR theory, a double bond or triple bond is treated as a single bonding group.

For example, sulfur trioxide & methanal are considered to be 'trigonal planar', and carbon dioxide, beryllium hydride and hydrogen cyanide HCN are considered to be 'linear'.

I see so the repulsion is not that different regardless of bond strength, so the molecular geometry remains the same, right?

@Apoqlite I may be a bit wrong in saying that their repulsions are 'not significantly greater' than single bonds (since their effects are noticeable), but the purpose of VSEPR theory is to provide simple and quick estimations to the bonding angles of a molecule, and treating multiple bonds as single bonds maintains that simplicity. In short, although it sacrifices accuracy, it is 'good enough' for the purpose of crude calculations.

For an explanation on why multiple bonds have greater repulsion forces than single bonds, read this as per one of the related questions.


text:
Mechanism of reaction of benzenediazonium chloride with ethanol to form benzene and acetaldehyde
In my textbook, all that was told was that ethanol is a mild reducing agent and hence reduces BDC while oxidising itself. Is there any mechanism or am I just supposed to remember this like a redox reaction?
I did search for a mechanism but did not find anything. I have no idea as to how this reaction works. Can someone please point me in the right direction?

If you use CH3CD2OH, I would imagine that you get benzene-d1, CH3CDO and N2. The rest is up to you.

In reality, benzene and acetaldehyde are the minor product of the reaction of benzenediazonium chloride with ethanol. The major product is ethoxybemzene (phenetole). Peter Griess, who had discovered diazonium salts in 1858, had reported benzenediazonium salt (nitrate or sulfate) with ethanol undergoes aforementioned redox reaction in 1864 (Ref.1). However, by 1901, it was confirmed that the products from the redox reaction is minor (about ~9%) when compared to major product, phenetole (about ~64%) when benzenediazonium chloride was used. By 1956, Kelley (Ref.2) found that the yield is even low as 3% and his student, Miller (Ref.1) later confirmed this result.
Now, it is believed that the reaction follows the scheme $\bf{A}$ and $\bf{B}$ (Ref.1):
$$\ce{Ar-^+N#NX- + RCH2-OH -> Ar-O-CH2R + N2 + HX} \tag{$\bf{A}$}$$
$$\ce{Ar-^+N#NX- + RCH2-OH <=> Ar-N=N-O-CH2R HX \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -> ArH + RCHO + N2 + HX} \tag{$\bf{B}$}$$
where process $\bf{A}$ proceeding through a $\mathrm{S_N2}$ displacement of the diazonium groups while process $\bf{B}$ involves a homolytic fission of the bonds to produce radicals which lead ultimately to aromatic hydrocarbon and aldehyde.
Proposed mechanism:
$$\ce{Ar-N=N-X -> Ar^. + N2 + X^.} $$
$$\ce{Ar^. + R-CH2-OH -> ArH + R-^.CH-OH} $$
$$\ce{AR-^.CH-OH + X^. -> R-CHO + HX} $$
It was found that when oxygen is added to the system rate of reaction has not changed significantly. The propagation steps with $\ce{O2}$ has been suggested as:
$$\ce{AR-^.CH-OH + O2 -> R-CHO + H-O-O^.} $$
$$\ce{H-O-O^. + AR-CH2-OH  -> AR-^.CH-OH + H-O-O-H} $$
Ref.1 and Ref.2 have reported the first order kinetics of this reaction with $k = 0.96 \times 10^{-4}$ at $\pu{25 ^\circ C}$ and $k = 2.03 \times 10^{-4}$ at $\pu{30 ^\circ C}$, which are consistent with other published data.
Note: More insight on mechanism, read Ref.3.
References:

Robert Warren Miller, "The stoichiometry of the reaction of benzenediazonium chloride with ethanol," MS Thesis 1957, University of Arizona, Arizona (PDF).
A. E. Kelley, PhD Dissertation 1956, Purdue University, Indiana.
DeLos F. DeTar, Takuo Kosuge, "Mechanisms of Diazonium Salt Reactions. VI. The Reactions of Diazonium Salts with Alcohols under Acidic Conditions; Evidence for Hydride Transfer," J. Am. Chem. Soc. 1958, 80(22), 6072–6077 (https://doi.org/10.1021/ja01555a044).


text:
How does Mars rover Curiosity's ChemCam's infrared laser produce a 100,000 K plasma plume (kT = 8.6 eV)?
Mars rover Curiosity's ChemCam uses a pulsed diode-pumped solid-state laser (DPSS) at 1064 nm to excite material on the surfaces of rocks up to 7 meters away, and a large aperture telescope to collect emitted light for analysis.
From NASA JPL's MSL Science Corer page Chemistry & Camera (ChemCam):

LIBS Instrument
The LIBS instrument uses powerful laser pulses, focused on a small spot on target rock and soil samples within 7 m of the rover, to ablate atoms and ions in electronically excited states from which they decay, producing light-emitting plasma. The power density needed for LIBS is > 10 MW/mm^2, which is produced on a spot in the range of 0.3 to 0.6 mm diameter using focused, ~14 mJ laser pulses of 5 nanoseconds duration. The plasma light is collected by a 110 mm diameter telescope and focused onto the end of a fiber optic cable. The fiber carries the light to three dispersive spectrometers which record the spectra over a range of 240 - 850 nm at resolutions from 0.09 to 0.30 nm in 6144 channels. The spectra consist of emission lines of elements present in the samples. Typical rock and soil analyses yield detectable quantities of Na, Mg, Al, Si, Ca, K, Ti, Mn, Fe, H, C, O, Li, Sr, and Ba. Other elements often seen in soils and rocks on Earth include S, N, P, Be, Ni, Zr, Zn, Cu, Rb, and Cs. It is anticipated that 50-75 laser pulses will be required achieve the desired 10% accuracy for major elements at 7 m distance.

Wikipedia's Laser-induced breakdown spectroscopy says:

LIBS operates by focusing the laser onto a small area at the surface of the specimen; when the laser is discharged it ablates a very small amount of material, in the range of nanograms to picograms, which generates a plasma plume with temperatures in excess of 100,000 K. During data collection, typically after local thermodynamic equilibrium is established, plasma temperatures range from 5,000–20,000 K. At the high temperatures during the early plasma, the ablated material dissociates (breaks down) into excited ionic and atomic species. During this time, the plasma emits a continuum of radiation which does not contain any useful information about the species present, but within a very small timeframe the plasma expands at supersonic velocities and cools. At this point the characteristic atomic emission lines of the elements can be observed. The delay between the emission of continuum radiation and characteristic radiation is in the order of 10 μs, which is why it is necessary to temporally gate the detector.

Question: What is the process by which infrared 1064 nm (1.17 eV) photons can produce an ionized plasma that results in the emission of UV photons as short as 240 nm (5.17 eV)? how exactly does a beam of 1.17 eV photons initially produce a "plasma plume with temperatures in excess of 100,000 K" where $k_B T$ = 8.6 eV.

Images from NASA.

Comments are not for extended discussion; this conversation has been moved to chat.

I try to convince OP that the critical passage he made is "Question: What is the process by which infrared 1064 nm (1.17 eV) photons can produce an ionized plasma that results in the emission of UV photons as short as 240 nm (5.17 eV)? how exactly does a beam of 1.17 eV photons initially produce a "plasma plume with temperatures in excess of 100,000 K" where  = 8.6 eV.". He seems to treat the laser as a black body. There is no upconversion of any photon. There is just heating of the sample through nir absorption. The heating might be difficult to model for shape/gradient , but it is heating.

@Alchimista No, I do not seem to do that at all! However your comment seems to suggest I don't understand the distinction. It's the exact opposite. The distinction is absolutely at the core of my question. Please double check my reply to your previous comment. The question sets up the difference between the two as part of the problem, and thermalization and implications of equipartition will be the nature of the answer.


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Any plastic for long term storage of Puriss. Ethanol?
I'm being offered to share some puriss Ethanol. Can I store it long term in any plastic container without the slightest contamination/reaction from the plastic? I have HDPE, PET, but I could buy another type if it's affordable (please don't answer PTFE).
If not, would an HCl > distilled water pre-rinse be sufficient to store it in a glass bottle?

Regarding trace ontamination, then generally, organic liquids are best stored in glass ( avoiding extraction ) and inorganic liquids in plastics ( avoiding trace metals from glass ). Surely there are exceptions as hydrofluoric acid or oxidating liquids.

I cannot say for certain so take my advice with a grain of salt, but if you cannot use PTFE, than HDPE is almost as good. Very few chemicals I have ever used have ever interacted with it and if you need something unreactive, HDPE is a pretty safe bet on top of being abundantly available.
PET, although very transparent, is very hit and miss when it comes to reactivity and I wouldn't recommend it.


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Brittleness of Ice?
I am curious as that I know that various materials can be subjected to fragility from ductile-brittle transition via temperature, where an item is more likely to fracture of shatter such as increase in strain.
I was wondering, what would be considered the equivalence for regular ice, I do not mean at 0 degrees Celsius where water becomes ice. I mean where it would likely cause ice to shatter more easily at lower temperatures?
Here are some links I found:

https://www.tms.org/pubs/journals/JOM/9902/Schulson-9902.html#ToC9
https://www.sciencedirect.com/science/article/pii/S0020768308004927

And things like https://www.sciencedirect.com/science/article/pii/0956715193903915 - there are a lot of articles on the ductile-brittle transition in ice. So, what are you asking for that the articles don't address?

@JonCuster guess I am trying to get a better understanding about at what point would ice be considered brittle that it would fracture or shatter easily with little effort.

I'm not a fracture mechanics person (although I know quite a few), but from glancing through the papers the answer is, well, complicated. This is not particularly surprising since brittle-to-ductile transitions and what they mean for a given mechanical loading are generally quite complicated as well.

@JonCuster I understand, it is quite tough to understand I thought that when temperatures get low it would mean the loss of energy, and without energy it cannot give cohesion with molecules and would fall apart.

"without energy it cannot give cohesion" is a fundamental misunderstanding. It takes energy to separate things, not keep them together. For example, we add energy to get from solid to gas. Interactions between molecules release energy, so lowest energy forms are solids

@Andrew, interesting, theoretically, would this be the same thing with Absolute Zero temperatures?

yes. The form of solid H2O at absolute 0 has even less internal energy than the crystals that form initially at 0 C.

One complication is that ice, unlike most solids, liquefies under pressure. At the grain boundaries, at higher temperature, ice can melt under strain and resolidify as the strain is relieved by the melting, leading to gradual deformation. So under gradual strain, ice is ductile. As mentioned in comments, this is hard to model.
BTW, ice was used to construct Pykrete boats during World War II, though they never reached the theater of war. It's safe to assume the properties were determined empirically.

Oh, so something like this? https://www.facebook.com/NiceRink/posts/ice-cracking-in-extreme-cold-why-ice-cracks-a-quick-note-on-ice-crackingit-will-/10152472389555221/


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Why is the first ionization enthalpy of molecular O2 less than atomic O?
A cousin of mine asked for my help to answer one of his homework questions. It stated-

Explain why- The first ionization energy of molecular oxygen $\ce{O2}$ ($\pu{1175 kJ mol-1}$), is lesser than the first ionization energy of $\ce{O}$ ($\pu{1314 kJ mol-1}$)?

It took me some time trying to find out perfect reasons for it, because generally, bonded electrons are harder to remove than valence electrons in atoms. But indeed, not just with $\ce{O2}$, but $\ce{F2}$ ($\pu{1515 kJ mol-1}$) also has a first ionization enthalpy lesser than atomic $\ce{F}$ which is $\pu{1681 kJ mol-1}$. (I wonder why there aren't any more talked of!). So I ended up taking a "class" on the basics of MOT, trying to reason that "the energy required to remove an HOMO electron from an ABMO in $\ce{O2}$ is lesser than removing an electron from a bonding orbital. That could probably lead to this anomaly."
But this wasn't convincing enough for me. I could sense a lack of conviction, a whole list of factors that could go wrong, also with the possibility of a much simpler explanation. But what made me post it up here, despite its dubious look, is the lack of convincing information addressing this issue even on the internet. Hence, a proper answer would be helpful for the community at large.

A similar explanation (kinda supporting mine), can be found here. Better answers are still awaited.

Did you check the values for $\ce{N2}$, or other such atoms?

Also, the fact that linear combination of atomic orbitals leads to a lower energy state(bonding orbitals) and a higher energy state (ABMO) seems to be pretty well defined since this would mean that you need more energy to reach zero potential for a bonding orbital w.r.t atomic orbital compared to the anti-bonding orbital where the actual energy of electron is higher than the atomic orbitals. In terms of throwaway numbers. If AO had an energy of -12 eV, then BMO would have a lower value(E$<$ 12 eV) and ABMO would be closer to the zero value than the AO..

The highest-energy electrons in O2 are antibonding. Think about what antibonding means: it doesn't just mean higher in energy than a bonding orbital. It is antibonding with respect to the constituent AOs.

"generally, bonded electrons are harder to remove than valence electrons in atoms.".  Maybe.  But in O2 are the highest electrons actually bonding?  Look again at the molecular orbital structure.

@orthocresol Yeah, I missed that. So you feel there aren't any other factors to be accounted for, and that my thoughts were in the right direction?

@Oscar Lanzi You can check out the post again, I had no problem dealing with the MOT. Just that it didn't seem to make it conducive enough for such large changes in enthalpy. I felt there could lie quite complex factors behind plain sight. Hence, I wanted a well-documented answer. Though, now it seems I wasn't very much wronged...

A cousin of mine asked me to share her answer here:

In the molecular orbital diagrams shown in textbooks, the highest occupied molecular orbitals of $\ce{O2}$ and of $\ce{F2}$ are shown as antibonding. If the molecules and their cations had the same bond length, and the electrons in a species didn't "talk to each other", and the orbital energies of an isolated atom were not influenced by the approach of another atom, you could predict that the ionization energy of the molecular species is lower than that of the atomic species. The argument would be that these HOMOs have higher energy than the corresponding atomic orbitals (see comments to question).
As these assumptions are not quite true, it is more complicated, but luckily the quick and dirty analysis yields the correct answer.
According to the MO diagrams, the bond order of $\ce{O2}$ is 2 and that of the cation $\ce{O2+}$ is 2.5 (after all, we are removing an antibonding electron). Indeed, the bond length of the cation is shown as smaller in this source. I did not find a bond length for the $\ce{F2+}$ species. This shows that you have to at least adjust the MO diagram for the change in bond length.
To get a rationale for the ionization energies in a quantitative manner, you could calculate the energies of the four relevant species ($\ce{O, O+, O2, O2+}$) and look at the differences.

So to summarize, you do concord with my views as well as fears. Well maybe then there really isn't any significant impact with those electronic forces and changes in bond length, as I thought. As these assumptions are not quite true, it is more complicated, but luckily the quick and dirty analysis yields the correct answer- that is probably the whole essence of it. I also came up with an interesting comparison, the IE1 of $\ce{Xe}$ is actually less than atomic $\ce{O}$. Never thought of that one before...


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Three stacked arrows, what does it signify?
I was checking the mechanism for the Lossen Rearrangement, when I came across the chemical reaction denoted as follows:

I couldn't find any resource where this notation was talked about.
What do the three arrows stacked as follows mean? A possible example where this is applied in literature would be appreciated.

Multiple reactions (I came across this in some IChO papers, so not a canonical reference)

It indicates there are several reactions in between with different reaction conditions before you achieve the final product. @Aniruddha Deb says it simply (I thought you know better Safdar! :-)).

There is an explicit answer in section 3. Other Horizontal Arrows of prof. Santiago Alvarez' published work [1, p. 594]:

[…]
Finally, multiple arrows (7) are used as an ellipsis for several intermediate steps in a chemical reaction.

Analogously, Nature Research Chemical Structures Guide (PDF) suggests using stacked multiple arrows as the multistep synthesis/reaction arrows:

Please use each type of arrow below for the indicated purpose.

Reference

Alvarez, S. Chemistry: A Panoply of Arrows. Angew. Chem. Int. Ed. 2012, 51 (3), 590–600. DOI: 10.1002/anie.201101767.
Nature Research. Style Guide for Chemical Structures; 2021. Accessed 2021-06-03. (PDF)

It means that after multiple (unspecified) steps, the substrate on the left is transformed into the product on the right. It can be mechanistic steps or synthesis steps (to infer from the context).
Here is an example from the literature where the authors have added "steps" to make it perfectly clear. (Nicolaou, Kyriacos Costa, et al. "The Diels–Alder reaction in total synthesis." Angewandte Chemie International Edition 41.10 (2002): 1668-1698.)


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How does liquid gallium interact with a gold leaf
I would like to know what happens when gold leaf comes into contact with liquid gallium. Will the gold dissolve into (alloy with) the gallium, wet the gallium or have no reaction with gallium?
The reason I am asking this question is to figure out if I could gild some gold leaf onto a base metal (that is compatible with gallium) and use the gallium as an 'adhesive' between the gold and base metal. I am planning on using this as a gold plated cathode in electrolysis experiments.

I’m pretty sure the gold leaf will simply dissolve in the gallium. If I find anything definitive, I will post it as a comment.

The wikipedia article on “Colored gold” says blue gold alloy is an alloy of gold with either gallium or indium. No idea how it is made.

@Ed V: It has to be adding gallium to melt gold. Melting point of gallium is just above room temperature, thus, I can't imagine how gold dissolve in liquid gallium unless it is above melting point of gold (boiling point of gallium: $\pu{2500 ^\circ C}$)..

@MathewMahindaratne I think this is simply analogous to amalgamation: gold amalgamates very readily, so the melting point difference between Au and Hg does not factor in. In any event, a piece of gold leaf is inexpensive and presumably already available and a few drops of gallium are also inexpensive, so the experiment could be done easily. I think it is just a matter of dropping the gold leaf onto the gallium and seeing what happens. Alternatively, Ga wets glass, so wet a microscope slide, drop the Au leaf on it and see what happens.

https://www.sciencedirect.com/science/article/pii/S0364591610000830 covers the Au-Ga phase diagram. There are 6 intermetallic phases, 4 of which are stable at and below room temperature. None of those could be considered an amalgam. The kinetics of phase formation are not covered.

@EdV - from my perspective, Ga is a pretty boring and normal component of many binary phase diagrams. Perhaps the weirdest one is the strong positive heat of mixing in the liquid phase with Hg, leading to a liquid miscibility gap. But otherwise? Nope, pretty normal.

Broadly speaking, looking at the Au-Ga phase diagram one gets:

(Now, this is based on a paper from 2011, a more recent 2017 paper is open access).
Several things to notice (or get from the paper):

The solubility of gold in solid gallium can be considered to be zero.

The solubility of gallium in gold is not zero, but is on the order of a few atomic percent.

Slightly above the melting point of gallium, the solubility of gold in the liquid is also very small.

There are multiple intermetallic phases across the diagram, 4 of which are stable at (and below) room temperature.

Formation of an amalgam can be ruled out.

So, it is a pretty normal binary phase diagram. Now the phase diagram does not provide kinetics information, so the time needed to actually form any intermetallic compounds (and if those formed near room temperature would even be the equilibrium phases) is not known.
Now, if we look at how gold leaf is used, there is no glue used - the leaf is place, smoothed and hammered into place, and it just sticks. So try just sticking it.

Jon Custer - you really are correct on about 98% of what you put there...but you really need to look into how gold leaf is used. Yes there are glues involved....it doesn't just simply stick into place.


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Graphene in Lithium-ions batteries: why?
I have recently been reading about the amazing discovery of a super-material: Graphene. This one carbon-layer chemical is not only one of the most resistant materials ever known, but also one of the best heat and electric conductors.
At some point, the article mentioned that the implementation of graphene in batteries - such as the lithium ion battery - would revolutionize the industry. I then found in a closet the dusty chemistry book we used in school, and I took a look at the electrochemistry section (I still remember I had enjoyed it back then).
Now that I understand - or rather believe to understand - again how a Lithium-ion battery works, I can't figure out how graphene should contribute to its improvement. Apparently, both electrodes are made of materials that can incorporate lithium ions into their crystal structure and release them easily. One would, therefore, typically employ a lithium metal oxide such as $\text{LiCoO}_2$ as cathode and graphite as anode. (By the way, what could you use as electrolyte?)
Anyhow, as I read in this paper, graphene is usually somehow incorporated to the cathode... Notwithstanding, I don't understand how this should improve the conventional lithum batteries. What are its key properties in this process?
Please consider that I am by no means a professional chemist, but rather chemistry enthusiast who has the knowledge of high-school. Thanks in advance.

As electrolyte is usually used Li[PF6] dissolved in mixture of dialkylcarbonates. ( esters of carbonic acid). Using graphene can improve availability of the carbon layers for Li+ ions, compared to graphite.

The shape and electronic structure in graphene also contributes to its high conductivity which means that basically electrons can flow through it more easily then other materials

Lithium has a standard redox potential equal to -$3.04$ V. Coupled with any compound at the cathode, it could hardly produce cells yielding more than $4$ Volts, because part of the available energy is needed and lost to extract the external electron of the Lithium atom. The performance of a cell built with a Lithium anode could be more important if the Lithium was already ionized. That's what is done with graphene. When introducing Lithium atom in a graphene structure, its electron is not so strongly attracted by the Lithium nucleus. Lithium inserted between graphene layers behaves as if it was already partly ionized. That is why Lithium in graphene produces an anode which can deliver about $5$ Volts if coupled with a convenient cathode made of Lithium and Cobalt(III) oxides. Graphene plus Lithium may be considered as a mixture of Lithium ions $\ce{Li+}$ and "nearly free" electrons delocalized in the graphene structure.

Thanks @Maurice for the answer! There is one question I have left: you claim that "When introducing Lithium atom in a graphene structure, its electron is not so strongly attracted by the Lithium nucleus" ... why does this happen?

@ Dr. Mathva. Generally speaking, an electron gets more stable if the limits of its domain of existence are increasing. So when the lithium atom touches the huge $\pi$ system made of the infinite number of $sp^2$ Carbon atoms present in the graphene assembly, this lithium atom has tendency to offer its electron to the graphene $\pi$ layer


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How to remove CH3 labels from a molecule's image? [OpenBabel in Python]
I would like to remove CH3 labels from svg images generated in openbabel in Python.
I am thinking I should use obConversion.SetOptions("w",openbabel.OBConversion.OUTOPTIONS) and replace "w" with another character. This does not work. Do you have any ideas? The image is below, for reference.

obConversion.AddOption("C",openbabel.OBConversion.OUTOPTIONS)

Referenced openbabel documentation. Setting it to "C" will remove terminal carbons.


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Specific Heat Ratio for a perfect gas mixture
I am reading Rocket Propulsion Elements by George P. Sutton & Oscar Biblarz, 9th Edition. In the fifth chapter, I was introduce to the specific heat ratio k for the perfect gas mixture, Eq. 5-7. Without explanation the Equation was stated as follows:
(5-7) $k_{\text{mix}} = \frac{(C_{p})_{\text{mix}}}{(C_{p})_{\text{mix}}-R'}$
From Thermodynamics I know,
(a) $k_{j} = \frac{(C_{p})_{\text{j}}}{(C_{v})_{\text{j}}}$ ; Species specific heat ratio
(b) $R_{j} = \frac{R'}{\mathfrak{M}_{j}}; $ Species Gas constant equal to Univ. Gas constant over Molar Mass of species $\text{&}$
(c) $R_{j} = (C_{p})_{\text{j}} - (C_{v})_{\text{j}}$ ;Mayer's Formula for species
So, it seems to me that Eq. (5-7) should instead include the mixture gas constant $R_{\text{mix}}$ in the place of the Universal gas constant, $R'$.
I tried to derive Eq. (5-7) from Eqs. a, b, and c without success. Can someone clarify the derivation of Eq. (5-7) to me?

It seems to me, Eqn. 5-7 is referring to molar quantities.

In general for a perfect or ideal gas,
$$C_p=C_V + R'$$
(using your notation) where the heat capacities are molar quantities.
It follows that for a perfect gas mixture $(C_p)_\text{mix}=(C_V)_\text{mix} + R'$.

I agree. I believe my issue was due to a misunderstanding. $C_{p}$ is the molar specific heat, and $c_{p}$ is the mass-basis specific heat. Thank you!


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Strange Distillation Behavior
I was attempting a fractional distillation with 1mL water and 1mL propylene glycol, and I’ve been encountering strange behavior with the heating of the mixture. The heating curve shows strange “spikes” in temperature rather than a continuous heating curve. I have ran a standard water and ethanol distillation to check for defects in my laboratory equipment, but the curve for that looks completely normal, and it seems to me like the equipment isn’t the issue. I would love any possible help in explaining this abnormality.

Can you describe the equipment and what item is having its temperature monitored and what kind of thermometer? Are you using vacuum? Vigreaux column? Spinning band?

The issue might be that distilling one liter and one milliliter is indeed different at least in controlling the T.

Can we know more about the placement of the thermometer?


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Copper anode dissolves in electrolysis of copper sulfate solution
I did an experiment with copper sulfate where you place two copper rods in an aqueous solution of copper sulfate. You apply an electric current and copper metal builds up on the cathode. I noticed that the copper anode dissolved. Does the dissolved copper from the anode just turn into more copper sulfate? If not, what does it produce?
This is NOT a homework question. It’s an experiment that I performed out of pure curiosity.

What happened to the cathode?

In my question, I said that copper built up on it. I probably should have added that it didn’t dissolve.

You have replicated electro-rafination of copper. Copper ions from the being dissolved anode are replacing the copper ions being deposited on the cathode.
Therefore overall electroneutrality of the solution is kept. The local neutrality is managed by electromigration of both ions $\ce{Cu^2+}$ and $\ce{SO4^2-}$.


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Performing a Redox Titration on a Vitamin C Sample Solution Containing Cellulase enzymes
I was wanting to performing an Iodometric Redox Titration on a Vitamin C sample solution created from allowing diced citrus fruits to sit in a water + cellulase solution. Because vitamin c is water soluble, the water +enzyme solution should possess a fair Ascorbic acid concentration, to which I planned a redox titration for its analysis.

This method determines the vitamin C concentration in a solution by a redox titration using iodine. Vitamin C, more properly called ascorbic acid, is an essential antioxidant needed by the human body (see additional notes). As the iodine is added during the titration, the ascorbic acid is oxidised to dehydroascorbic acid, while the iodine is reduced to iodide ions.
$$\ce{ascorbic acid + I2 -> 2I− + dehydroascorbic acid}$$
Due to this reaction, the iodine formed is immediately reduced to iodide as long as there is any ascorbic acid present. Once all the ascorbic acid has been oxidised, the excess iodine is free to react with the starch indicator, forming the blue-black starch-iodine complex. This is the endpoint of the titration.

Explanation taken from https://www.canterbury.ac.nz/media/documents/science-outreach/vitaminc_iodine.pdf.
I was wondering it the water + cellulase solution I’d be titrating on was viable for this way of determining Vitamin C. Will cellulase interfere with the reaction which governs this titration?

Why you want to add a cellulase solution to the reaction mixture?

@MathewMahindaratne To catalyze the hydrolysis of the cellulose cell wall of the citrus fruit placed inside the water solution, and see how that effects Ascorbic acid release

In that case, I wouldn't think cellulase affects iodination titration, but think it may hydrolyze starch indicator so that end point might not be visual.

@MathewMahindaratne Cellulase only catalyzes the breakdown of cellulose, but starch is made up of amylose and amylopectin polysaccharide chains. In this case, cellulase wouldn’t interact with the starch indicator, correct? However, if this were the case, should I use a higher concentration starch indicator solution?

I've searched for that. But it wasn't clear whether cellulase catalyzes the breakdown of only $\beta$-D-linkages or not because some literature uses starch linkages as substrate. Easy way to find this is run a sample titration with known ascorbic acid concentration (you need to scandalize your iodine solution anyway). Use extra sample with and without cellulase and compare the results.


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What lab equipment did Marie Curie use to isolate radium?
My experience in growing crystals for condensed matter physics has been sealing grams of material in ampoules which get heated in laboratory furnaces, so I don't have the experience to understand how Marie Curie managed to go through tons of pitchblende to isolate radium, especially given her limited resources. How were the samples heated, and how would she have kept her samples from getting oxidized? It seems like making lots and lots of ampoules would be impractical.

Why is making lots of ampoules impractical? That is what chemists did.

Marie Curie's work was detailed in the French Journal L'Actualité chimique, $352$, p. V, May $2011$. It is called "Dossier Marie Curie" and was published just $100$ years after Marie Curie's Nobel Prize Conference in $1911$. She used beakers, Bunsen burners, filter paper and funnel. That is all.
It could be summarized this way. Marie Curie extracted first all the heavy elements from $30$ kg of the mineral pechblende, using successive and selective precipitation techniques. At the end she obtained $40$ g pure $\ce{BaCl2}$, with Radium as only impurity ($10$ mg $\ce{RaCl2}$ ; $0.025$ ppm).
Now starts the real purification process. She dissolved this $40$ g $\ce{BaCl2}$ in the minimum amount of boiling water ($68$ g), let it cool down. She obtained a cristalline deposit $\ce{D1}$ containing $16$ g $\ce{BaCl2}$ and $8,3$ mg $\ce{RaCl2}$ ($0.05$ ppm) : the proportion of Radium has increased in the recrystallized solid. She redissolved these $16$ g $\ce{BaCl2}$ in the minimum amount of hot water ($27$ g), let it cool down, to obtain a recrystallized deposit $\ce{D2}$ containing $6.4$ g $\ce{BaCl2}$ and $6.9$ mg $\ce{RaCl2}$. Once again the new deposit $\ce{D2}$ contains more radium ($1$ ppm). Repeating this procedure twice leads to deposits $\ce{D3}$ ($2.6$ g $\ce{BaCl2}$ and $5.8$ mg $\ce{RaCl2}$, $2$ ppm)} and $\ce{D4}$ ($1.0$ g $\ce{BaCl2}$ and $4.8$ mg Ra, $4$ ppm) ($4.0$ ppm). The proportion of radium increased steadily, but the mass of enriched $\ce{BaCl2}$ decreased.
Waste waters from $\ce{D1}$,$\ce{D2}$,$\ce{D3}$,$\ce{D4}$ are all saved. mixed and evaporated so as to save the rest of $\ce{BaCl2}$ ($39$ g) and the rest of radium ($5.2$ mg}. The whole series of recrystallization is repeated, and repeated over and over again. See the detailed report in l'Actualité chimique, if you are interested.
After repeating this tedious series of similar recrystallizations may be more than one thousand times, and starting again and repeating it with $40$ times the original amounts of pechblende (30 kg), she was able to get a final sample made of $0.17$ g $\ce{RaCl2}$ containing only $0.01$ g $\ce{BaCl2}$. This is $\ce{RaCl2}$ at  $94$% purity. Enough to get the Nobel Prize in Chemistry.

After searching Google Books, I found autobiographical notes from Marie Curie. Apparently, it was radium chloride that was isolated first, not elemental radium. Salts can often be precipitated from solution at much lower temperatures than those needed to melt many inorganic materials, and you typically wouldn't need to shield a solution from air, so ampoules wouldn't be needed. This would be the case for fractional crystallization, which she mentions in her autobiographical notes. (She also mentions spending whole days "mixing a boiling mass with a heavy iron rod". I'm not sure if she's referring to mixing molten material or a solution in water or acid.) She also says they eventually used a factory; as time passed and their work became "more and more known", she was able to do the "crude processes" in a factory, allowing herself "more time to the delicate finishing treatment".
source: the book Pierre Curie: with Autobiographical Notes by Marie Curie

Great information, but could you provide a link to the source?

Most of the work of elemental separation was done by dissolving the minerals in various acids and them using chemistry to separate the salts. She certainly didn't melt the minerals.