text:
Can Bromophenol Blue Sodium Salt, and Pentane be safely mixed/in contact with each other?
I'm trying to determine if powdered Bromophenol blue is safe to mix with Pentane. I would only need a tiny amount of dye, just enough to tint the pentane.
I need to use pentane for it's specific freezing point but I may be able to mix the pentane with another color indicators (wont be using this for PH, just need color)
Pentane
https://www.sigmaaldrich.com/US/en/product/sial/236705
Bromophenol Blue Sodium Salt - https://www.fishersci.com/shop/products/bromophenol-blue-sodium-salt-fisher-bioreagents/BP11425

What make you think it's not safe to mix a dye with a hydrocarbon?

Safe yes. Immiscible - probably as the sodium salt is fairly polar I doubt you'll get much to go into the pentane

The SDS from the Fisher Scientific link you give lists only strong oxidizing agents as incompatible with your salt, so there is no safety issue with mixing the salt with the pentane apart from the flammability of said pentane.
Whether you get the solubility is a different matter. Salts usually don't go well with hydrocarbons, and the SDS does contain a line that this material is soluble in water (which makes solubility in the hydrocarbon even less likely).
A blue material that is more likely to work with pentane is azulene.


text:
What test should be conducted to test for the presence of Aromatic Functional Groups?
To give some background on the different tests I have went over: The Iron (III) Chlorine Test tests for the presence of phenols. However, phenols contain both the alcohol group and the aromatic group and not the aromatic group individually. The Bromine Water Test also tests for double bonds and single bonds, however, due to the presence of double and single bonds in the compound it will react to create another compound. If the test, when reacted with the unknown compound, is colorless then there is a presence of an alkene. However, if is no change in the color of the solution it would indicate the presence of alkanes. If there is both the presence of alkanes and alkenes within the unknown compound, it does not mean that there is the presence of aromatic compounds since the unknown compound could also be a conjugated chain.

Why are you doing this? I think you would be hard pressed to find a professional organic chemist today who is doing one of these tests to check for functional groups...

It was for my organic chemistry lab class.

OK. Not much to be done about that, but it does raise the question of why your laboratory class is teaching such antiquated methods...

If I understood your question correctly, you are asking how can you distinguish aromatic hydrocarbons from aliphatic hydrocarbons (alkane, alkene, alkyne) and you said that test for unsaturation (bromine water test) will fail because alkane and benzene will both show negative results. You are correct and in that case, you can use the ignition test. Place a small amount of compound on a spatula and apply flame. For highly unsaturated compounds like benzene, it will burn with a yellow, sooty flame. For a chemical test, you can use Friedel Craft reaction. Strong Lewis acid like aluminum chloride will react with benzene to give a bright orange-red complex. Alkenes and alkynes also reacts with $\ce{AlCl3}$ but it will give off a different color (usually more yellowish). Alkanes do not react with $\ce{AlCl3}$.
There is one more selective test for aromatic compounds - spot test using  2,4,7-Trinitrofluorenone but I don't think this is a readily available laboratory reagent.

The lowest aromatic is benzene so everything is up from there. Usually physical constants give a clue. With no heteroatoms and only aliphatic side chains aromatic rings sulfonate readily with concentrated sulfuric acid. They react with CHCl3 and anhydrous AlCl3 to give a wild mix of phenylmethanes that supposedly have bizarre colors. Simple alkyl benzenes don't react with Br2 in CCl4 or cold, neutral permanganate; more complicated ones the bets are off. The methods of choice are a combination of ultraviolet, infrared and nmr spectroscopy. As a test look up the physical constants of benzene and cyclohexane. Then figure a way or ways to definitely tell them apart.

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text:
Spontaneous reaction between chlorine and a hypothetical atom
Suppose, my hypothetical element is $\ce X$, whose first ionization energy is $\pu{200kJ/mol}$. An atom of this element will react with an atom of chlorine (assume that I've already dissociated a chlorine atom from a $\ce{Cl2}$ molecule). The first electron affinity of chlorine is approximately $\pu{349kJ/mol}$.
We take those two atoms inside a non-reactive box in a controlled room. Inside the box is vacuum. I intend to make the following reaction happen:
$$\ce{X(g) + Cl(g)-> XCl(s)}$$
Now, I'm curious as to the course of events.
Hypothesis 1:
One hypothesis is that the reaction will not proceed even if you get the atoms really close together. For the reaction to proceed, you must provide the required ionization energy to  $\ce X$ first. Then the removed electron from X will be acquired by Cl promptly. Energy will be released in this process. The energy released in this process will be greater than the ionization energy that was provided to $\ce X$ previously. In the final step, $\ce {X^+}$ and $\ce {Cl^-}$ will get stuck together due to the electrostatic force between them. In conclusion, the reaction will process in the following steps (in order):
$$\ce{X(g) -> X+ + e-}\tag{1}$$
$$\ce{Cl(g) + e- -> Cl^{-}(g)}\tag{2}$$
$$\ce{X+(g) +Cl^{-}(g) -> XCl(s)}\tag{3}$$
In this hypothesis, the reaction isn't fully spontaneous since heat/energy needs to be added in the beginning for the reaction to proceed.
Hypothesis 2:
Another hypothesis is that once you take the two atoms close together, electron transfer will take place i.e. the reaction will proceed even if you don't add any heat. In other words, in this hypothesis, heat/energy does not need to be added for the reaction to proceed: the reaction is fully spontaneous. Ionization energy will not need to be provided to $\ce X$ as the reaction is energetically favourable. The reaction will take place in the following steps:
$$\ce{X(g) -> X+ + e-}\tag{4}$$
$$\ce{Cl(g) + e- -> Cl^{-}(g)}\tag{5}$$
$$\ce{X+(g) +Cl^{-}(g) -> XCl(s)}\tag{6}$$
The commonality between hypothesis 1 and 2 is that there will be a net release of energy after $(1)$ & $(2)$ and $(4)$ & $(5)$. The net release of energy after $(1)$ & $(2)$ will be equal to the net release of energy after $(4)$ & $(5)$. $(3)$ and $(6)$ will be the same in both hypotheses. However, unlike hypothesis 1, $(4)$ and $(5)$ take place simultaneously while in hypothesis 1, $(1)$ takes place first and $(2)$ takes place second.
In this hypothesis, the reaction is fully spontaneous since heat/energy doesn not need to be added in the beginning for the reaction to proceed.
My question:

Which hypothesis is correct?

In context of this question, one should distinguish thermodynamical and kinetical spontaneity. It can be thermodynamically spontaneous even if there is needed such a high activation energy the process is never observed. Like some isotopes are thermodynamically radioactive with thermodynamically spontaneous decays, but kinetically stable.

Both hypotheses are conditionally correct, as each describes a different possible scenario. I assume hypothesis 2 would still have some activation energy, but much lower than X ionization energy.

Reaction's simply $$\ce{X(g) + Cl(g)-> XCl(g)}$$ That's it! No magical ionisation, no flying electrons, no ions.

@Mithoron "That's it! No magical ionisation, no flying electrons, no ions."- I'm not confident about the ionization and flying electrons part, but I'm positive about the formation of ions: in the formation of NaCl, Na+ and Cl- ions are definitely formed and they get stuck together.

Answer you linked is incorrect. https://chemistry.stackexchange.com/questions/14174/what-is-sodium-chloride-like-in-gas-state https://chemistry.stackexchange.com/questions/2511/will-gaseous-ionic-compounds-be-free-moving-ions

@Mithoron Sir, AmritanshSinghal used the Born-Haber cycle in his answer.

@Mithoron  It is correct I think (pic found in chemlibretexts). I'm having difficulty understanding why you think otherwise. Could you please explain it to me kindly?

Born-Haber cycle has hardly anything to do with mechanisms. Atoms of Cl and Na hit each other and NaCl molecule is created (and surplus energy needs to be released, but that's besides point). Bond it this molecule may have more ionic than covalent component, though.

@Mithoron, Sir, just to understand your position better, do you think this answer is incorrect as it wrongly presents the Born-Haber cycle as the reaction mechanism of $$\ce{Na(s) + Cl_2(g)->NaCl(s)}$$?

Well, it only says "following steps are involved in the formation" not exactly "mechanism", but that's still incorrect.

@Mithoron According to wikipedia, "In chemistry, a reaction mechanism is the step by step sequence of elementary reactions by which overall chemical change occurs." So, I think that AmritanshSinghal is describing the reaction mechanism as reaction mechanism just means a step by step sequence of elementary reactions, and AmritanshSinghal presented a step by step sequence of elementary reactions. Thoughts?

In this hypothetical case, formation of free ions is exothermic, in contrary to real endothermic cases. As consequence, X(g) +Cl(g) -> X+(g) + Cl-(g).

@Poutnik So, ionization energy will not need to be provided?

At contact of both atoms - no, it need not, as it is a different system than 2 separate atoms. Thete will be some activation energy, but lower than X ionization energy. Similarly as a planet could eventually escape a star easier, if some other star would incidentally come close to the first star.

@Poutnik What is the difference between when two atoms are in contact and when they are separate?

I agree with @Poutnik. It depends on the shape of the potential energy surface. If you get the atoms close together, you actually add energy to the system. Imagine you have to go from one valley to another one through a saddle point among two mountains and the two valleys. The ionisation energy moves you to the saddle and then you do not need energy to go downhill. If you put the atoms together, you just move closer to the saddle and the height you have to overcome is lower. This makes really high probability of quantum tunelling to the other side of the saddle and then downhill again.

@TomHardy What is the difference if there is another star near the Sun at the distance of the planet Neptun or not ? There is different electrostatic field for atom case and different classical gravitational field in the star case. // By other words, atoms at contact are not like billiard balls at contact. They are different and behave different.

When an electron moves from one atom to another -- at the contact -- to form a ion pair, it is not a free electron like in the IE context, but it is a bound electron all the time, with the energy lower than a free electron would have.

With respect to the Born-Haber cycle answer, the wonderful thing about state functions like enthalpy is that they are path independent, so we can use any steps we want in a calculation as long as the end points are the same. That is the fundamental logic behind the Born-Haber cycle - the calculation is not intended to represent the actual reaction steps but rather is a path comprised of steps whose energies are available.

Poutnik and Mithoron are disagreeing, so I'm not sure who is correct (point 1). Moreover, Mithoron disagrees with the answer I linked too (point 2). Light shed on both point 1 and point 2 would be much appreciated. (...)

(...) Moreover, I think that hypothesis 1 might be the correct one. This is because my book said, "The ionization energy necessary to remove one or two or three electrons from metal atoms is available in the environment, but to remove 4 electrons from atoms much more ionization energy is required, which is not readily available in the environment, so atoms of many elements fulfill their octet by sharing electrons. These bonds are called covalent bonds." My book seems to imply that ionization energy is non-negotiable.(...)

(...) For ionic bonds to form through electron transfer, one must provide the requisite ionization energy, without which the reaction will not proceed at all! Hypothesis 1 seems to be consistent with this idea, so probably hypothesis 1 is correct. I might be wrong.

@Poutnik If the ionization energy of X would've been greater than the electron affinity of Cl, would the provision of ionization energy then for the reaction to proceed be non-negotiable? [Assume that the two atoms are not in contact and are a measurable distance apart]

Should this be crossposted in Physics SE?

In such a case there are 2 separate operations, with the first ionization independent on eventually following electron capture // Crossposting is frown upon on SE network.

This thought experiment needs more thinking.  Both scenarios are essentially Born-Haber cycles not possible mechanisms. Forming a positive ion always requires energy so that makes equations 4,5,6 not relevant to the second concerted mechanism. In the case of the concerted mechanism the X and Cl atoms must have sufficient KE to overcome electron repulsions as they approach and to continue to distort the atomic orbitals to form a molecular orbital. Ionic bonding requires a lattice energy to complete the cycle and two atoms will not suffice. Finally there must be a mechanism to remove the bond energy from the incipient bond. Radiation is a possibility.
The first proposal is not mechanistically reasonable. It requires external energy sources and the energy loss from the electron affinity is just lost and does not contribute to the overall reaction and the same applies to the final bond formation.   Considering an initiated reaction such as a free radical chlorination where final lattice heat provides energy to maintain the reaction is probable but such a mechanism would not necessarily involve free chloride ions.

There is a flaw in hypothesis 1 - getting the atoms "really close" will provide activation energy.
At sufficiently close distance the combined electron Pauli repulsion and nuclear core electrostatic repulsion will add significant energy.
For example, $2$ protons $7$ angstroms apart have about $200$ kJ mol$^{-1}$ of electrostatic repulsion between them.
Also, $200$ kJ mol$^{-1}$ ionization energy is the absolute maximum activation energy required.  As the two potential wells - the $\ce{X}$ nucleus and the $\ce{Cl}$ nucleus - approach, eventually the wells are close enough that an electron can tunnel from a $\ce{X}$ to a $\ce{X}$ to form $\ce{X+}$ and $\ce{Cl-}$ ions.
Alternatively, other bound electronic states across both atoms (essential a covalent bond) will form at much lower energies than $200$ kJ mol$^{-1}$ above the separated states.
Before that even, at much larger distances very weak dispersion interactions will draw neutral atoms towards one each over.  The long-range potential is attractive and the first minimum is likely close enough to transfer an electron.  The reaction is very likely to have no activation energy.
So diffusion will dominate to create dimers.  Depending on the temperature I would expect it to look like hypothesis 2.
These dimers will essentially be entirely ionic - pulled as close together as possible until the Pauli repulsion of the electrons stabilises the electrostatic attractions.
As the ions are more thermodynamically stable, the balance of the depth of the attractive potential against entropy will determine the size of aggregate species - from free ions (high temperature plasma) to dimers (NaCl gas) to larger oligomers or condensation/sublimation.

If the ionization energy of X would've been greater than the electron affinity of Cl, would the provision of ionization energy then for the reaction to proceed be non-negotiable? [Assume that the two atoms are not in contact and are a measurable distance apart]

If the atoms are very (infinitely) far apart and the ionization of X was greater (as is the case with all real elements), then the thermodynamically stable configuration is 2 neutral atoms.
They have to be close enough for tunneling, or for some activation energy to be added as I mention - then the situation is only stabilised by the electrostatic attraction between the two charged ions.

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text:
6-31G basis set of Silicon and the p-orbitals
Following is the 6-31G basis set of silicon. The numbers in the bracket denotes power raised to 10, e.g. $1.231(+3)=1.231 * 10^3$

Now silicon is $1s^22s^22p_x^22p_y^22p_z^23s^23p_x^13p_y^1$.
Now I am trying to understand the basis for 2SP and 3SP.
Lets consider 2SP. The exponent is identical for both $2s$ and $2p$ orbitals for ease of calculation. But I can construct only one 2p orbital from the information given in the table. There is no distinction between $2p_x,2p_y,2p_z$, atleast the p-coefficients are supposed to be different. Moreover each atomic orbital are going to contain two electrons at most. If there is only one $2p$ atomic orbital defined, in that case I can only insert two electrons in it. So then rest of the four electrons of $2p$ subshell go unaccounted for becuase I can not insert six electrons in a single atomic orbital, i.e. $2p_x$, lets say.
I am missing something. Can someone kindly help me out here?
EDIT:
Or is it that at $2p$ level, all $2p_x,2p_y,2p_z$ have the same functional form.
And then at $3p$ level, I have $3p,3p^{'}$ defined.
Then $3p$= $-0.0151774e^{1.65} + 0.0275139e^{5.4}+0.783e^{2.044}$
and $3p^{'}$= $e^{0.07238}$
Finally,
$3p_x=c_1*3p + c_2*3p^{'}$
$3p_y=c_3*3p + c_4*3p^{'}$
$3p_z=c_5*3p + c_6*3p^{'}$
Where all the coefficients $c_1, c_2,...,c_6$ are different.
This is the only explanation I can think of the above table. Can someone drop a pointer please?

In a hurry but briefly the above table is essentially coefficients for the radial dependency of the wavefunction. px, py and pz have the same radial dependency but different angular dependency - so they have all the same coefficients and exponents, but px is multiplied by x, py is multiplied by y and pz by z. When time  will write a proper answer talking about real solid harmonics


text:
Shielding vs electron-electron repulsion
Example of shielding:
 (source)
The last electron in the 6s subshell of $\ce {Cs}$ is shielded from the nucleus by the inner electrons.
Example of electron-electron repulsion:
The electron affinity of $\ce F$ is surprisingly lower than that of $\ce{Cl}$. This is because, in $\ce F$, there are 5 electrons in the small-sized 2p subshell, so the electron density is very high. Therefore, the incoming electron is heavily repelled by the 5 electrons in the 2p subshell through electron-electron repulsion. So, the electron affinity of  $\ce F$ is lower.
My comments:
Can the example of electron-electron repulsion of $\ce F$ be called shielding? I previously knew that electrons in the same shell don't contribute to shielding; however, I found a comment on the contrary.
My questions:

Do electrons in the same shell contribute to shielding?
Can the example of electron-electron repulsion in the case of $\ce F$ be called shielding?

As Q1: If you read wikipedia.org : Slater's rules - Rules, you would see they do. Q2: In a way, yes, but the shielding is rather considered in context of affecting the mean radial force, while repulsion involves substantial tangential force.

E.g. an electron in the lobes of the orbital 2p_x  partially tangentially repulses electrons in 2p_y and 2p_z, and also eventually its mate in the same orbital, especially when in the same lobe.


text:
Does the halogenation of an alkane also produce haloalkanes of a longer chain length?
Take a typical halogenation of an alkane reaction (the bromination of methane).
$$\ce{Br_2->Br· + Br· (in the presence of UV light)} $$
$$\ce{Br· + CH4->·CH3 + HBr} $$
One of the possible termination reactions is
$$\ce{·CH3 + ·CH3->C2H6} $$
Since there are still bromine radicals present, does this newly formed ethane then undergo this reaction?
$$\ce{Br· + C2H6->·C2H5 + HBr} $$
What's to stop this new ethyl radical from continuing this process again? Wouldn't the bromination of methane form secondary products of bromoethane and propane and so on?
How do you ensure, when synthesising a haloalkane through this method, that you only form the haloalkane of the specific chain length you need?
Am I missing something obvious?

Bromine radicals will be present in the majority as compared to alkyl radicals and thus making the alkyl chain elongation very less probable, in fact, less and less probable as you increase the number of carbons increased.

@NisargBhavsar When synthesising on an industrial scale, with thousands of mols of each, wouldn't there be at least a sizeable amount of unwanted bromoethane/propane/butane produced?

No, much more likely by-product is dibromomethane. Concentration of methyl radical is so small, their reaction with each other is negligible.

Also bromomethane isn't made like that in practice.


text:
Why does Ecell change with concentration but not with coefficients
I understand that the voltage difference is an intensive property, which does not depend on the number of moles. So the reduction potentials don't change when you multiply coefficients by a common factor. For example,
$$\begin{align}
\ce{1/2 O2 + 2H+ + 2e- &-> H2O} & E_\mathrm{red}^\circ &= \pu{1.23 V} \\
\ce{O2 +  4H+ + 4e- &-> 2 H2O} & E_\mathrm{red}^\circ &= \pu{1.23 V}
\end{align}$$
However, since the number of moles does not matter, why does the concentration change the value of $E^\circ$ vs $E$ (under non-standard conditions).
$$ E_\mathrm{cell} = E_\mathrm{cell}^\circ - \frac{RT}{zF}\ln Q$$

Concentration is an intensive property, isn't it? // BTW, as we usually do not say number of kilograms, we do not usually say number of moles either. We say mass and molar amount.

This part of my answer is a bit silly; the electrodes in an electrochemical cell "know" when you change the concentration, but they don't "know" when you change coefficients of your chemical equation on your sheet of paper.
More seriously, intensive properties do change when you change concentration (an intensive property). However, they don't change if you increase the size of the cell (AAA, AA, C, and D batteries all have the same voltage, but AAA drains faster than D).
As a general principle, measurements don't change just because you change your mental model or the chemical equation used to describe a situation.
To make things more complicated, the molar enthalpy is also an intensive property, but the value does change when you multiply all coefficients by the same factor. If you relate this to measurements, however, nothing changes. This is because the changed coefficients appear in the calculation of, say, the amount of heat a certain amount of reactants give off. So even thought the molar enthalpy is different by a certain factor, the calculated heat (which you can measure) is not because the arbitrary factor cancels out when calculating the heat given off.

Sorry this is another dumb follow-up question -- does this mean redox is concentration dependent (which I understand now after you explained, thanks!), and that the reverse reaction can happen depending on the concentration, even though the reverse reaction is not spontaneous?

@bread Spontaneity depends on concentration too. If you have a book which claims otherwise, it's wrong. So, everything you said is right, up until "even though the reverse reaction is not spontaneous". It is spontaneous; that's why it happens.


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How to compute the electronic energy reported by Gaussian from the AlphaOrbitalEnergies?
I followed the instructions on Wolfram Language & System Documentation Center — FCHK (.fchk) to analyze the *.fchk file.
I ran an energy calculation with Gaussian and converted the results to *.fchk files. The *.fchk files were imported to Mathematica. I extracted the AlphaOrbitalEnergies from MetaInformation provided in the *.fchk files. Is there a way to compute the electronic energy reported by Gaussian from the AlphaOrbitalEnergies?
For my example calculation of methane with HF/3-21G, the electronic energy is $\pu{-39.976406 Ha}.$
I tried adding up all the energies from AlphaOrbitalEnergies and I got $\pu{-2.18778 Ha}.$
Next, I tried to only add the energies of occupied orbitals and I got the value of $\pu{-13.7344 Ha}.$
How to obtain the electronic energy from the alpha orbital energy?

This is a common question in quantum chemistry.
(It's common enough that one colleague specifically mentions this every time he teaches a quantum chemistry class.)
I'm going to assume that when you say AlphaOrbitalEnergies that you have a singlet species, so the alpha and beta orbital energies are identical.
For a Hartree Fock calculation, the total SCF energy is:
$$
E=\sum_{i(\text { occupied })}\left\langle\phi_{i}|h| \phi_{i}\right\rangle+\sum_{i>j(\text { occupied })}\left[J_{i, j}-K_{i, j}\right]
$$
If I add up the occupied orbital energies (i.e., two electrons in each orbital), I get:
$$
\sum_{\mathrm{i}(\text { occupied })} \epsilon_{i}=\sum_{\mathrm{i}(\text { occupied })}\left\langle\phi_{i}|h| \phi_{i}\right\rangle+\sum_{\mathrm{i}, \mathrm{j}(\text { occupied })}\left[J_{i, j}-K_{i, j}\right]
$$
Notice that the second term is different. Notice that if I'm adding up the occupied orbital energies I'll be double-counting the term with Coulomb - exchange energies.
In principal, you could get the total energy from the orbital energies, but you would have to do work:
$$
E=\frac{1}{2}\left[\sum_{\mathrm{i}(\text { occupied })}\left\langle\phi_{i}|h| \phi_{i}\right\rangle+\sum_{\mathrm{i}(\text { occupied })} \epsilon_{i}\right]
$$
On the other hand, I believe the FCHK files include the total SCF energy already.


text:
Do both phosphoric acid and sulphuric acid prevent oxidation of ferrous ion in iron titration?
For the Determination of iron by Redox Titration experiment, I am finding the role of sulphuric acid and phosphoric acid. Then I found that other than $\ce{H3PO4}$ masks the yellow color of $\ce{Fe^3+}$ and makes the endpoint easier to see (Daniel C Harris, Quantitative Chemical Analysis 8ed. pg 350) by forming complex while $\ce{H2SO4}$ prevents hydrolysis and provide surplus $\ce{H+}$ ions in the solution to keep reaction proceeding; both can prevent air oxidation.($\ce{H2SO4}$, $\ce{H3PO4}$) Do both phosphoric acid and sulphuric acid prevent oxidation of ferrous ions in iron titration or is it just $\ce{H2SO4}$?
Titration against KMnO4

My understanding is that acidic solutions have a lower potential for the oxidation of ferrous ion by O2 so sulfuric will increase the oxidizing ability of permanganate and not accelerate side reactions. Phosphoric acid complexes ferric ion and has 3 effects. It increases the overall potential change at the equivalence point making endpoints easier to determine both potentiometrically and using a redox indicator and removes the ferric ion color. It should also hasten oxidation by O2. Phosphoric acid is more useful using dichromate oxidant than permanganate since permanganate is its own indicator. The best way to avoid reaction with O2 is to purge with N2 or add a pinch of sodium bicarbonate to the acidic solution and fill the flask with CO2, then add the phosphoric acid.


text:
Why are Slater-type orbitals used for atomic calculations instead of hydrogen-like orbitals?
Originally, Hartree–Fock atomic calculations were done by using numerical methods
to solve the Hartree–Fock equations, and the resulting orbitals were
given as tables of the radial functions for various values of r.
Since 1951, Roothaan proposed representing the Hartree–Fock orbitals as linear combinations of a complete set of known functions, called basis functions:
\begin{equation*}
\phi = \sum c_i \chi_i,
\end{equation*}
where the $\chi_i$ functions are some complete set of functions, and where the $c_i$'s are
expansion coefficients that are found by the SCF iterative procedure.
But in textbooks always says that "A commonly used set of basis functions for atomic Hartree–Fock calculations is the set of Slater-type orbitals (STOs)":
\begin{equation*}
    \chi_i = \frac{(2\zeta_i)^{n + 1/2}}{[(2n)!]^{1/2}} r^{n - 1}e^{-\zeta_i r} \cdot Y_{lm}(\theta, \phi).
\end{equation*}
As a rule, no further details follow. And this creates some confusion. It seems to me that it would be more logical to use hydrogen-like orbitals as a basis, since they are the exact solution of the Schrödinger equation.
All I could find was a sparing mention of Born and Hylleraas:

In 1928, it was already recognised by Born and Hylleraas that
the He atom could not be described by a CI expansion using the
H-like bound-state eigenfunctions.

Source: https://www.esqc.org/lectures/WK3.pdf
In this connection, I have a question: what are the main advantages/benefits of Slater-type orbitals compared to hydrogen-like ones?

what do you mean by nodeless?

@Grag May be you mean, the advantage of STO because it nodless? Then, yes. What is the advantage of nodeless functions?

@Greg "nodeless" - radial part of wave-function without nodes.

I don't know the answer but I will observe that it seems to me that you can construct hydrogenic functions as a linear combination of STOs. You can thus express an STO is a linear combination of hydrogenic functions. They are thus equivalent, and the simpler form of STO is probably easier to handle.

@IanBush One reason mentioned in Slater article Phys. Rev. 36, 57: "The nodes in the wave function are found [...] to be unimportant, [...] they come much nearer the nucleus than for hydrogen wave functions, and are therefore less important.  Consequently we neglect them entirely, taking as the radial part Of the wave function of one electron simply".

This may sound naive, but aren't Slater orbitals simply easier to calculate?

There are two criteria for a good set of basis functions:

The solution of the problem has to be in the basis, i.e. some combination of the basis functions has to approximate the real wave function reasonably well.
Using the set of basis functions has to be sufficiently computationally efficient that you can find the solution (or an approximation of it).

If it is easy to map the basis functions on something we are familiar with (like our mental model of the atomic wave functions 1s, 2s, 2p etc.), interpreting the results might be a bit easier.
The choice of basis set concerns the radial functions only. For the angle-dependent part, it seems like all calculations use the spherical harmonics.
From first principles, the Laguerre polynomials (so the solution to one-electron atoms) and the Slater orbitals individually model the wavefunction well at large distances from the nucleus, while a Gaussian function does not.
The Laguerre polynomials are the only ones that model the radial nodes in a single function.

[OP] It seems to me that it would be more logical to use hydrogen-like orbitals as a basis, since they are the exact solution of the Schrödinger equation.

They are not the exact solution of the Schrödinger equation for multi-electron atoms. In particular, the hydrogen-like orbitals are a bad model for the valence electrons in period 2 and higher elements, which are shielded quite a bit from the charge of the nucleus (in the one-electron model).

[OP] In this connection, I have a question: what are the main advantages/benefits of Slater-type orbitals compared to hydrogen-like ones?

In a paper on computational difficulties using Slater-type orbitals, the authors made the following statement:

Due to that fact, the hydrogen-like orbitals do not form a complete set (for finite n), they need orbitals of the continuum to be complete.

So according to this statement, hydrogen-like orbitals seem to fail criterion #1 (complete basis set) in a practical way. With current algorithms and hardware, Slater-type basis sets seem to fail criterion #2. Thus, the commonly used basis sets are built from Gaussian functions, with integrals that are easier to solve analytically.


text:
Acid hydrolysis of Birch-reduced Anisole
The following is a question from Advanced problems in organic chemistry by MS Chouhan:

Product $A$ is rather straightforward:

and on acidic hydrolysis, the following reaction takes place:

Here the book says that this product rearranges to (b), and they termed it the "3-carbon system tautomerism"
I found no reference for this interesting mechanism. How exactly does this rearrangement take place?

Not sure I've ever seen this being given a fancy name, but you tautomerise to form the conjugated enol, and then re-tautomerise back to the ketone, but protonating the gamma carbon instead of the alpha carbon...

This mechanism is wrong at the second step. Water attacks the cation at the carbonyl carbon to form the hemiacetal that collapses to the deconjugated ketone. Enolization of this ketone to form the conjugated dienol and gamma protonation leads to (b), the conjugated enone.

@user55119 it would really help if you could maybe sketch it out!

Lost Anisole: Before fulfilling your request for a diagram of the mechanism for the formation of cyclohex-2-en-1-one (9), a few remarks about the problem as presented are in order. Your product A is correct. Unfortunately, anisole (1) will not reduce with lithium in liquid ammonia in the absence of an alcohol such as ethanol. As to the loss of the methyl group in the mechanism presented, one cannot simultaneously have both protons and hydroxide available in acid medium. The only nucleophilic species in aqueous acid apart from the counterion is water.
Enol ether 2 is the Birch reduction product of anisole (1). Protonation of the more electron-rich double bond affords species 3, which suffers attack by water at carbonyl carbon not at the methyl group, giving rise to the hemiketal 4. This species is a typical intermediate in the formation of acetals and ketals in acid medium when alcohol is the solvent.  Here, with water in excess, the equilibrium is driven toward the unconjugated cyclohexenone 6 via protonated species 5.
Acid-catalyzed enolization of compound 6 will ultimately lead to the thermodynamically more stable dienol 8 as illustrated in 7. While dienol 8 may protonate at either the $\alpha$ or $\gamma$-positions, protonation at the $\alpha$-position leads back directly to cyclohexenone 6. This material will recycle with acid until $\gamma$-protonation occurs and the thermodynamic, conjugated cyclohexenone 9 is formed.

"As to the loss of the methyl group in ... the counterion is water." Can you please elaborate a bit on this? And yes, thanks for this excellent analysis!

If hydrochloric acid is the source of hydronium ion, then chloride is the counterion. If you have pH 1, then the hydroxide concentration is $\pu{[10^-15]}$. Not much hydroxide.


text:
Why 5-member rings prefer being cis-fused?
It always bothers me how whenever I am doing practice problems I never got a proper explanation as to why 5-member rings prefer(thermodynamically) being cis- rather than trans-fused( as opposed to decalins where the cis-fused rings are more stable due to gauche and 1,3-diaxial interactions). They seem to never give an explanation about the phenomenon besides "it is how it is".

I don't see an immediate reason why the cis fused would be stable. The cyclohexane would be happier having two equatorial substituents so it must have something to do with the 5-member ring.
Also on the topic, the equatorial and axial positions on a cyclopentane are also very confusing:

It seems that it was to equatorial positions on the same side of the ring and next to each other( the two right-most). How do you choose which substituents to put where and where does the bulk go- I cannot see a clear factor to decide?

Inner angle of a regular pentagon is $108°$. Tetrahedron angle is $109.47°$. These two values are not very different. At room temperature, the molecular vibrations may be sufficient to continuously pass from cis- to trans-cyclopentane

tldr; I think it's pretty close between the two forms - at least for unsubstituted hydrocarbons
Since we just worked on a project to understand ring puckering conformations, I was curious about this.
So I created the two hydrocarbons you indicated in your question and searched for the lowest-energy conformations using the GFN2 semi-empirical method. (It's quite good for ranking conformers in our recent benchmark.)
For the un-substituted framework, the calculations come to ~0.18 kcal/mol difference with trans slightly more stable .. and I'd guess the method has at least a ~1 kcal/mol error. So they're roughly the same at least with this theoretical method.
cis Isomer

trans Isomer

Both cyclohexanes seem free of ring strain, and all the torsions are close to ideal.
If I were to judge the various interactions, I'd say it's close. Perhaps the cis form has interactions between the 5-membered ring and the "bottom" axial hydrogens.

Thank you for your comment. I guess that if substituents are added to the ring, the situation in terms of stability would be more clear cut.


text:
Gold cyanidation process
My chemistry book says the following:

In the metallurgy of silver and gold , the respective metal is leached with a dilute solution of $\ce{NaCN}$ in presence of  $\ce{O2}$ as a result  of  which the following reaction takes place:
$$\ce{4M + 8 NaCN + 2H2O + O2 -> 4Na[M(CN)2] + 4NaOH}$$

Here $\ce{M=Ag,Au}$. Now my question is since the complex anion $\ce{[Au(CN)2]-}$ is formed, the gold has been oxidized from its elemental state to $\ce{Au+}$ and apparently the $\ce{O2}$ has been reduced to $\ce{OH-}$. This is the place where I have some problem. In contrast to what we are taught from middle school that gold is highly resistant to oxidation by air (atmospheric oxygen), here it is clear that gold has been oxidised to $\ce{Au+}$ by air which is counter intuitive.
My Speculation
I think that when we say oxidation of gold by air does not occur it might mean that the process is very slow. I thought that the  reaction is a reversible one (as in case of oxidation of gold by $\ce{HNO3}$ where $\ce{Au(NO3)3}$ is formed but the reaction is so slow and the equilibrium constant for the reaction is so slow that practically the gold remains passive to $\ce{HNO3}$ but when we add $\ce{HCl}$ due to formation of chloroauric acid the equilibrium is driven forward):
$$\ce{Au +[O]<=>Au+}$$
Under normal condition this reaction is very slow and that's why we say gold is inert. But in this case as soon as $\ce{Au+}$ is formed the $\ce{CN-}$ ligand attacks it to form the complex $\ce{[Au(CN)2]-}$ which drives  the  equilibrium in the  forward direction by Le Chatelier's principle. But I couldn't find any such reliable resource for figuring out what exactly  is  going  on  here as any insight or if what I speculated was correct or not into the chemistry of the reaction will be gladly appreciated.

Note that for Ag, oxygen is not even needed and hydrogen is produced. The key is the stability of respective cyanocomplexes. making from both metals reduction agens, stronger for silver.

Standard redox potential for [Au(CN)2]-(aq) + e- <=> Au(s) + 2 CN- is -0.6 V.  ( https://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29 )

Macro- or microscopic amounts of material contain huge numbers of molecules. A very negative oxidation potential [calculate the potential for gold and O2in water] means that the reaction proceeds until the potential is zero and the equilibrium concentrations satisfy the equilibrium constant. For Au in water even with O2 present that is a very low concentration of Au+ [or Au+3 depending]. For homework do the math. Addition of cyanide, or other complexant, removes gold ions. Invoking LeChatelier's principle, Gold will dissolve until equilibrium is reached.
Equilibrium depends on the ratio of forward and reverse rates not on their absolutes. Gold is inert because its reactions with water are energetically not favorable not necessarily because the reaction is slow. Cyanide definitely slows the reverse reaction; whether it catalyzes the forward reaction requires research.
A hint: When writing projective equations write complete equations and balance them while thinking about the chemistry involved. Serious thought will help to answer questions.


text:
Partial reduction of anthraquinone to anthrone using stannous chloride?
I've heard of using $\ce{SnCl2}$ in the Stephen's reaction where you reduce a nitrile down to an aldehyde. I've come across a laboratory manual that discusses using tin in the partial reduction of anthraquinone1  so I guess it's possible but how would such a reaction work? Would the $\ce{SnCl2}$ act in a similar way to a dissolving metal reduction? That would create a secondary alcohol but I can't see how the alcohol can be further reduced to an alkyl.
Reference

Joaquín Isac-García, José A. Dobado, Francisco G. Calvo-Flores, Henar Martínez-García, Chapter 11 - Microscale Experiments, Experimental Organic Chemistry, 2016, Pages 371-408, DOI: 10.1016/B978-0-12-803893-2.50011-5

Luca: Your mechanism looks good except you haven't resolved the reductive elimination in the last step. Assuming that the carbonyl oxygen is protonated as in 1a, then two events must occur in some order: protonation of the hydroxyl oxygen and a 2-electron reduction of the protonated carbonyl (2). Elimination of water from 2 affords the "hydroxyanthracene" that tautomerizes to anthrone 4. Alternatively, the carbonyl is not protonated but the hydroxyl is. A one-electron transfer to the carbonyl provides ketyl 1c, which, by the addition of a second electron and protonation, affords 3. As in the first mechanism, the order of events is not certain.

According to this source Science Direct tin chloride will reduce anthraquinone to anthrone

Anthrone is a tricyclic aromatic ketone that can be used, for example for calorimetric determinations of carbohydrates in biological fluids. It can be synthesized by partial reduction of anthraquinone with several reagents, as sodium hydrogen sulfite, tin chloride, or tin.

Hello, I read the lab manual, what I'm curious about is how this reduction works mechanistically. Does it follow a dissolving metal style reduction, or does it proceed down a Stephen's reduction pathway, or something else.

I've added a mechanism of how I think the reaction proceeds. Thanks for the help!


text:
Why do Büchner funnels have no glaze on the top rim?
I have a question about Büchner funnels (just out of pure curiosity). The funnels I have seen have no glaze on the top rim. Actually, the question is why the top end is not glazed. Does it have any utilitarian meaning?
Or is it due to the peculiarities of production?
A quick googling yielded no results, but showed that the marketplaces, in the descriptions also often indicate the unglazed top edge.
Descriptions examples:
Funnels are glazed on inner and outer surfaces except for the rims.
Funnel is completely glazed porcelain except for the top edge of the funnel.
Buchner funnel with fixed perforated plate, glazed both inside and outside with the exception of the top rim.

Then why glaze the outside of the funnel?

The aspect of partial glazing of porcelain labware is not unique to Büchner funnels. This is typical of many porcelain items used in the lab. Other porcelain labware is also partly glazed. Check an evaporating dish. It is partially unglazed outside and completely glazed inside.
I do not know the exact reason for leaving the rim of this particular funnel unglazed but this has more to do with manufacturing rather than something fundamental to chemistry or filtration theory. As I recall damaged glazed porcelain funnels, it is possible that the rim is a high stress point. If the rim were also coated, cracks may start from the rim, they can propagate elsewhere, and this logic makes as to why one would leave the rim from glazing. It introduces a discontinuity. Porcelain does need glazing if it is to be used for reagents otherwise it is too porous and hence unwashable! Good for one use only.
Another reader (Snijderfrey) pointed out in the comments that during firing, the funnels have to sit somewhere and it is likely the reason that if the Büchner funnel is fired upside down, the rim does not get glazed.
Regardless, the rim not being glazed is a pretty old "tradition". One can see very old catalogues of the previous century with same specs. Today plastic Büchner, Pyrex funnels with sintered glass funnels are also very common, they do not need glazing and they are completely uniform.
I was able to access the first description of this funnel in the German chemical literature from 1888 by Eduard Büchner and an earlier description by Hirsch again in 1888. It is just a one paragraph description by Hirsch. His funnel is conical and Büchner simply made it flat. Nothing fundamental about porcelain or glazing is mentioned.

In the reports of the German Chem. Ges. 1886, p. 918, O. N. Witt
describes a filtration device consisting of a plate of glass, metal or
porcelain with multiple perforations, which is inserted into a funnel,
closed by a small filter disc and sealed against the funnel by a
larger one. This apparatus (Fig. 45), which is widely used in
thousands, especially in factory laboratories, often shows the defect
of not closing completely tightly, so that the filtrate is turbid,
especially at the beginning of the suction. To avoid this, I have had
porcelain funnels made in which the filter plate is firmly inserted.
Only one filter disc is then used for sealing, and the funnel can be
emptied and refilled without the plate moving and thus becoming leaky.
The apparatus is manufactured in the best quality by Max Kaehler
Martini, Berlin W, Wilhelmstr. 50. Huddersfield, March 3, 1888

Büchner simply changes the design...

Translated with www.DeepL.com/Translator (free version)

"This is typical of many porcelain items used in the lab." My porcelain plates, cups etc. at home are also not glazed at the bottom where they are supposed to stand on. I would think they also stand on those parts while firing.

@Snijderfrey, This could be another reason too. The items have sit somewhere.

@AChem first of all, thank you for such a detailed answer and references to original sources. Can you maybe mention the version of Snijderfrey in your answer as an update and I will accept it to close the question?

@Snijderfrey, thanks, looks like a very plausible reason.


text:
Why m-xylene is more basic than o- and p-xylene in tetrafluoroboric acid?

According to me, the more stable sigma complex should be ortho and para xylene as it stabilizes the positive charge through hyperconjugation but in the book the order is meta xylene is more stable than ortho para xylene. Why is hyperconjugation destabilizing in this case?
Source:
KS Verma Organic Chemistry for JEE Advanced

Draw resonance structures - these aren't benzyl cations.

For help, goo to Search on Chemistry at the top of the page and type m-xylene.

What stabilizes the carbocation is when the methyl groups are ortho/para to the protonation site. (The protonation site is where the proton actually adds to the hydrocarbon molecule.) Getting both methyl groups into these favored positions requires them to be meta to each other. So the most stable cation isomers are derived from meta-xylene.

But how do you know that the protonation site in meta xylene is not on the one of the -ch3 groups? That would make the meta isomer less stable than ortho/para.

It's not so much that we know, rather the molecule knows. Given a choice, meta-xylene will preferentially accept a proton where the methyl groups are favorably placed. In fact, trying to protonate elsewhere invites a hydride-shift rearrangement to get the carbocation into the more favored configuration.


text:
How are fission products formed?
I know this question may seem a bit simple, but I would like to know about how fission products of an element are formed. There seems to be no way for krypton-92 and barium-141, the immediate fission products of U235, to decay into a fission product like caesium-137 or technetium-99. Instead, I found they quickly decay to the stable nuclides zirconium-91, zirconium-92, and praseodymium-141, and the intermediate radioactive nuclides all have half-lives of within a few months.
So, my question is, how are these other fission products formed?

@Poutnik Then how are they formed?

@Poutnik But the fission of uranium-235 seems to only produce krypton-92 and barium-141.

It is false premise I have tried to negate since the beginning. See also Fission product yield

Comments moved to the answer.

i wonder why the popular texts as well as a lot of textbooks are concentrated exactly on the Ba / Kr example

@fraxinus Yes, that was why I had a misconception about nuclear fission.

There are many immediate primary fission products, with 2 probability peaks near 2/5 and 3/5 of original uranium-235 nucleon number. By other words, there are multiple ways how the nucleus can be broken, some more probable than others.

But the fission of uranium-235 seems to only produce krypton-92 and barium-141.

This is false premise. These are just the products of a single fission mode from many. In a classical analogy, there is multiple ways how to break a stone.
See also Nuclear fission products and Fission product yield


text:
Excitation of hydrogen atom
I studied the Bohr's model of atom that says electron can jump to higher energy level by absorbing photon, but the quantum model says we have more than that (quantum numbers).  Therefore, when the electron of hydrogen atom absorbs a photon, will it move according to the Aufbau principle (1s, 2s, 2p ...), or will it stay in the s subshell with an increasing value of n?

[...] will it stay in the s subshell with an increasing value of n?

Quite the opposite, in fact: it can't go from $\mathrm{1s} \to \mathrm{2s}$, $\mathrm{1s} \to \mathrm{3s}$, or any $\mathrm{1s} \to n\mathrm{s}$ for that matter. The selection rules for electronic transitions state that the quantum number $l$ must change by $\pm 1$, and starting from an s-orbital ($l = 0$) the only place you can go to is $l = 1$, i.e. a p-orbital.
However, there's no restriction on the principal quantum number $n$. So $\mathrm{1s} \to \mathrm{2p}$, $\mathrm{1s} \to \mathrm{3p}$, or any $\mathrm{1s} \to n\mathrm{p}$ are OK.
(Very loosely speaking, the reason for this is because a photon carries one unit of angular momentum, and essentially that must be reflected in the transition; the quantum number $l$ represents angular momentum, so it must change by one unit. Formally speaking, these are derived from angular momentum coupling rules, which is covered in typical graduate QM textbooks e.g. Sakurai. $n$ on the other hand has no relationship to angular momentum, so the only requirement for changes in $n$ is that the incident photon must have the correct energy.)
This is illustrated nicely in a Grotrian diagram, see https://en.wikipedia.org/wiki/Grotrian_diagram. The allowed transitions are depicted by arrows, and you'll see that they only correspond to $\Delta l = \pm 1$.

Although in multi-electron atoms a forbidden transition is not completely forbidden, just less likely.


text:
Melting and Freezing
Ok so I needed some help understanding one concept, for eg:- we are in an isolated room that has a room temperature of 0 degrees Celsius. And I bring in some water in the room, so eventually it will start freezing up at 0 degrees Celsius right?(doesn't matter the time), and when it's in an ice (solid) state its temperature is going to stay at 0 degree Celsius? if so then will it also not start melting if at 0 as the melting point of ice is also 0 degrees celsius, and as the room is 0 degrees celsius this will continue to loop?

As room is isolated at 0 °C, it cannot cause water freezing, as when it withdraws heat from water warmer than 0 °C it would warm itself  up above 0 °C. If water was 0 °C too, then there would be no heat exchange to freeze water.

thank you, but what if the room was big, big enough so that it wouldn't warm itself up entirely when cooling the water? so the water could use the cold temperature to cool itself? if possible, what would happen then?

If both room and water were 0 deg C then there would be no heat exchange, needed for phase transition, regardless of room size and heat capacity. If the room was colder than 0 deg C,then yes, water could freeze if room heat capacity was big enough.

what if water was let's assume 3 deg C, would it convert to ice? and as the melting point of ice is also 0 deg C would it then melt again? and it would continue to happen till the room was no longer 0 deg C?

Before asking, think, search and think again at least 1 hour.

alright will do

It'll never happen with you in the room.

"the room was big, big enough so that it wouldn't warm itself up" The function, for example, $1/x$ decays monotoneosly, yet never reaches or even crosses $0$. You cannot dilute the verdict "thermodynamically impossible" to the point where it suddenly becomes possible.

There's nothing wrong with asserting an infinite, isothermal heat bath.  We use this construction in thermodynamics all the time.  Clearly what the poster means by "big enough" to have its temperature be unaffected by the introduction of water is an infinite isothermal heat bath.  He just doesn't know quite how to express it.  Rather than telling him why the way he's phrased  it won't work, why not be helpful and suggest a construct that would, and then address his question?

But the scenario is isolated, not closed isothermal system. Whatever big 0 °C bath would not take away heat from 0 °C water to freeze it. We could even combine this bath (if water) and the brought water into a single unit. Would it partially free, if isolated? It would not.

That's just the OP using the wrong words because they don't sufficiently understand thermodynamics to ask the question properly. The OP's intent is clearly for the room to stay at 0C: "...and as the room is 0 degrees celsius this will continue to loop?" The OP probably used "isolated" to mean "insulated", thinking that would keep it at 0C. So someone just needed to explain to the OP that, if they want the room to stay at 0C, it needs to be an infinite isothermal heat bath...

....To use an analogy, suppose someone who didn't understand trucks went on an automotive site and asked how much gasoline a particular truck (that actually had a diesel engine) needs to go 100 km at 100 kph. What people should do is helpfully explain that this truck doesn't use gasoline, it uses diesel, and then give the fuel mileage. But if they acted like many of the answerers here, they would instead just unhelpfully say the question makes no sense because that model of truck uses diesel, not gasoline.

No, the water wouldn't cycle in an infinite loop between liquid and ice.
Let's assume you introduced three different containers of water at $T=\pu {0 ^{\circ}C}$ and $\pu{p = 1 atm}$ into an infinite, constant-temperature heat bath, also at $T=\pu {0 ^{\circ}C}$ and $\pu{p = 1 atm}$ . These contain: (a) liquid water; (b) a 50-50 ice-water mixture; and (c) ice. Let's further assume the containers are diathermal, i.e., they allow the flow of heat.
In order for the liquid water in container a to turn to ice, heat would need to flow from that container to the bath. In order for the ice in container c to turn to liquid, heat would need to flow from the bath to that container. And the same applies to container b: heat has to flow in to increase the proportion of water, and out to increase the proportion of ice.
According to classical thermodynamics, since they are all at $\pu {0 ^{\circ}C}$, there would be no heat flow between the water in the containers and the bath (no temperature difference means no heat flow). Thus they would all stay as they are.
However, statistically, I suppose it is possible you might have fluctuations that could cause changes. This is something I would need to think about further.

Sorry for asking the question with wrong words guys. I just hadn't reached the concepts of the thermodynamics stuff yet, so I didn't know about it. Absolutely sorry. But I figured theorist said that after the water was bought in the temperature of the water would go down to 0°C and then it would stop as they both have reached equilibrium. Thank you everyone for helping me out!

If ice and water are at equilibrium at zero Celsius the equilibrium is dynamic and the crystals and liquid and vapor will slowly redistribute about the enclosure. the relative amounts of each will not change unless energy is added to the mix.


text:
Affect of EtOH on Anionic Surfactant
I am not a chemist nor do I have a background in chemistry, so I know nothing beyond my observations.
Attempting to load SLS into a porous silica substrate by dissolving surfactant into 200 proof EtOH along with a lipid, held at 50c slow stirring on hot plate mag. Silica added once SLS and lipid are in solution. EtOH removed in Rotovap.
Question: does dissolving SLS in warm EtOH affect the surfactant's amphiphilic arrangement?

What do you mean by " surfactant's amphiphilic arrangement"? Do you mean CMC? Micelle size?


text:
Mass spectrum of N,N-dimethylaminoethane
I have been struggling with understanding how fragmentation works in mass spectrometry. An exercise my teacher gave me asked me to prove that a mass spectrum belongs to N,N-dimethylaminoethane, CH3CH2N(CH3)2. The mass spectrum is shown below:

I worked out that the peak at m/e = 58 is due to the loss of CH3 from the molecular ion (resulting in a fragment with an ethyl and methyl group bonded to an N+), but I can't figure out what the fragments are at the m/e = 44, 42 and 30 peaks. I've tried several ways to determine it, such as the cleavage of two bonds and rearrangement, but I still can't find it. How do you find out the fragments for those m/e peaks?
Also, in general, is there a clear-cut way to determine fragments at each m/e peak?

A candidate for m/z = 44 is $\ce{CH3N=CH2^+}$ or $\ce{CH3CH=NH2^+}$ . For m/z = 42, $\ce{CH2=N=CH2^+}$. For m/z = 30, $\ce{CH2=NH2^+}$

If you think you will have a lot of mass spec questions in the future, you may want to support this proposal for a whole site dedicated to mass spec.

"cleavage of two bonds and rearrangement"? There's hardly anything to rearrange here. Just cut some bonds... This is a quite simple example, and you complicate even the name of the compound,

A candidate for m/z = 44 is $\ce{N(CH3)2^+}$

@Mithoron Sorry for the confusion! The name of the molecule was given to me in the question––I do know that N,N-dimethylethanamine would be a better name to represent the molecule. I'm relatively new to mass spectrometry, so I'm not really familiar with 'simple' examples! Sorry for the trouble.

@user55119 Thanks for the answer! But wouldn't your suggestion for m/z = 44 have the m/z of 43 instead (the first candidate)?

I forgot an "H". The first m/z = 44 cation should be $\ce{CH3NH=CH2^+}$.


text:
NMR spectroscopy nomenclature: induced vs local vs effective magnetic field (B)
This is a simple question but I can't seem to find a place where they use the same nomenclature... I was wondering whether these are all the same thing: $B_\mathrm{induced}$, $B_\mathrm{local}$, $B_\mathrm{effective}$? I know $B_\mathrm{effective} = B_0 - B_\mathrm{shielding}$. But do all of them mean the same thing (as in the magnetic field felt by the nucleus in a particular environment)? Thank you!
Edit (including context)
Context for $B_\mathrm{effective}$:
$\mathrm{\omega}$ = $\mathrm{\gamma}$$B_\mathrm{eff}$ = $\mathrm{\gamma}$$(B_\mathrm{0}-B_\mathrm{shielding})$
I can understand this, but I wanted to know if this $B_\mathrm{eff}$ is the same as induced, and local.
Context for $B_\mathrm{local}$:

Context for $B_\mathrm{induced}$

Thank you!

+1 Welcome to Chemistry.SE! Please note the edits, we use MathJax to format equations and symbols: https://chemistry.meta.stackexchange.com/questions/86/faq-how-can-i-format-math-chemistry-expressions-on-chemistry-stack-exchange . I think it might be useful if you could also refer to the context in which $B_\mathrm{induced}$ and $B_\mathrm{local}$ are used.

The meaning of these words depends upon context. It may help if you could say what you are reading and quote passages where the words are used.

Agree with above posters - context is required.

In this context, i.e. the discussion of how the chemical shift arises, I'd assume that the induced magnetic field is the "internal" magnetic field, which is generated by electrons surrounding the nucleus, in response to the external magnetic field ($B_0$). This induced magnetic field generally serves to oppose the external magnetic field, meaning that the effective magnetic field is less than $B_0$. The equation you gave,
$$\omega = \gamma B_\mathrm{eff} = \gamma(B_0 - B_\mathrm{shielding})$$
suggests that $B_\mathrm{shielding}$ is to be identified with the induced magnetic field and that $B_\mathrm{eff}$ is what's left after removing that.
Note, however, that $B_\mathrm{eff}$ may mean something else in a different context (e.g. when discussing the vector model of NMR).
On the other hand, $B_\mathrm{local}$ is being used in a completely different scenario, namely that of nuclear spin relaxation. It refers to a magnetic field generated at the location of one nucleus due to other nuclei. (Think of every nucleus, or every NMR-active nucleus at least, being a bar magnet — $B_\mathrm{local}$ is the magnetic field generated by these bar magnets.)
This is in contrast to $B_\mathrm{shielding}$ above, which is generated by electrons around the nucleus of interest.
So none of them are the same.


text:
How to find out by what method the steel was made, having a microstructure?
I know that rolled steel will have grains elongated in the rolling direction and I don't see anything similar in the microstructures of these steels. If I'm not mistaken, then the irregular boundaries between the crystals means that they are randomly arranged.
I am also knowledgeable that extrinsic microstructural features such as entrapped refractory inclusions, aluminum oxide films, and porosity are not common in modern, commercially processed.
With that in mind, I think these microstructure patterns belong to forged rather than rolled steels. Do you agree with me? Do you see any other features that are characteristic of forged or rolled steel?
1.Ferrite,several oblong and globular slag inclusions can be seen.

2. Ferrite only.
3. Pearlite abd ferrite.

4. Ferrite,pearlite,and slag.

Interesting micrographs of steel destroyed in high temperature service. It is a wrought steel (elongate inclusions) but not much other information available in the deteriorated condition. Chemistry will give some info, other than carbon which has been changed in service. Looks like exposure approximately in the range of 1100 F to get decarburization and creep/stress rupture. Some subsequent carburization around the rupture in Fig. 3. I have not seen microstructure like this since I retired long ago, memories.


text:
Determining iron(III) content in mixture of iron(II) and iron(III)
Problem

Outline a plan of an experiment to determine the percentage of iron present as iron(III) in a solution containing $\ce{Fe^3+(aq)}$ and $\ce{Fe^2+(aq)}$ ions. You are provided with zinc, a standard solution of potassium dichromate(VI) and dilute sulfuric acid. Zinc can reduce $\ce{Fe^3+(aq)}$ to $\ce{Fe^2+(aq)}.$

Answer

Titrate measured volume solution against $\ce{K2Cr2O7}.$
Reduce same volume solution with zinc.
Filter off excess zinc.
Titrate total $\ce{Fe^n+}$ using $\ce{K2Cr2O7}.$
$$x(\ce{Fe^3+}) =  \frac{c_2 - c_1}{c_2}\times 100\,\%$$

Questions
Firstly, why is the excess zinc filtered? $\ce{Fe^2+}$ has been oxidised to $\ce{Fe^3+}$ first by dichromate, and then we took a different aliquot and determined the amount of $\ce{Fe^3+}$ by reduction using zinc.
Secondly, the “titrate total $\ce{Fe^n+}$ using $\ce{K2Cr2O7}$” step confuses me. Would this be done in a volume that was the sum of the two aliquots taken previously? If so, does this mean I would have the amount of $\ce{Fe^2+}$ present in solution, yet the step using zinc would be made redundant?

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Notation basics ,  Formatting of math/chem expressions  and
upright vs italic // For more, see Math SE MathJax tutorial.

Suppose your solution contains $a$ mol $\ce{Fe^{2+}}$ and $b$ mol $\ce{Fe^{3+}}$. Titration with $\ce{K2Cr2O7}$ gives you $a$. To obtain $b$ you first add some zinc powder, before titration by $\ce{K2Cr2O7}$. Zinc will reduce all the $\ce{Fe^{3+}}$ ions according to $$\ce{2 Fe^{3+} + Zn ->  2 Fe^{2+} + Zn^{2+}}$$ So the solution contains now some $\ce{Zn^{2+}}$ ions which do not interfere later on. But it contains an amount of $\ce{Fe^{2+}}$ equal to $a + b$. So titration of this second solution requires more $\ce{K2Cr2O7}$ than previous titration. This second titration by dichromate ions gives you $a+b$.
To summarize, first titration gives $a$. Second titration gives $a+b$. Substraction gives you $b$. And the problem is solved.
But the second titration must be done without any residual zinc metal, because zinc may also be oxidized by $\ce{K2Cr2O7}$ according to : $$\ce{ 3 Zn + Cr2O7^{2-} + 14 H^+ -> 2 Cr^{3+} + 3 Zn^{2+} + 7 H2O}$$ If you titrate the second solution without removing the zinc metal, you will need a huge amount of dichromate, just for destroying all the excess of zinc metal added.


text:
Why does copper(II) sulfate solution turn black on addition of excess sodium borohydride?
In my laboratory experiment, I was supposed to reduce copper(II) sulfate to copper by means of $\ce{NaBH4}$, an indication of which was supposed to be a colour change from blue (copper(II) sulfate) to brown (denoting a suspension of copper nanoparticles). This much was true for a small dropwise addition of $\ce{NaBH4}$.
However, the solution turned black beyond a threshold point when I added sodium borohydride in excess. Is there a possible explanation for this?

Maybe it converted to copper(II) oxide.

Depending on size and concentration of particles, color can be deceiving, compared to the bulk or big particle
material. Could it be so intense brown looking like black? It seems to me improbable that by another addition of NaBH4 Cu gets oxidized to CuO. I was thinking about CuH, but it is not black either. ( https://en.wikipedia.org/wiki/Copper_hydride )

But if really added in excess and quickly, it could alkalize the solution and if hot enough, CuO could be formed.

Metallic copper can appear black if made of grains which are smaller than the wavelength of the light. Try to add a little diluted acidic solution. If the precipitate remains, it is metallic copper, If it gets dissolved, it is $\ce{CuO}$.

Why are you talking about $\ce{CuO}$ in the presence of strong reducing agent? Isn't sulfate reduced to sulfide and precipitate $\ce{CuS}$ ?


text:
Why is it that first order reactions never end?
I'm studying chemical kinetics in high school. We are studying Integrated Rate Equation of first order reactions - their derivations and graphs. Our teacher showed us a graph of:
Concentration of Reactant [R] at time t vs. Time (t)
In the graph showed to us, The graph line slowly becomes parallel to x-axis and stretches onto infinity which is the Time.
We were explained that first order reactions are never ending because as the concentration of reactant decreases, the Rate of reaction decreases at very large amount and hence the reaction keeps going on and never reachers zero concentration and thus doesn't end.
He gave us the example of radioactive decay, Chernobyl, how the nuclear recations there will never end as the radioactive decay is a first order reaction.
However I am not able to understand the explanation given to us:

first order reactions are never ending because as the concentration of reactant decreases, the Rate of reaction decreases at very large amount  and hence the reaction keeps going on and never reachers zero concentration and thus doesn't end.

It would be helpful if someone could help me understand this, maybe some mathematical correlation in the formula or something.
Any help is appreciated.

Think of the infinite series 1/2+1/4+1/8+1/16.... You will get closer and closer to 1, but only in the limit of an infinite number of steps will you reach it--because each step only goes half as far as the last one.  Of course, matter isn't infinitely divisible. If you have a reaction that only proceeds in the forward direction, like radioactive decay, eventually you will reach the point where you have 1 atom and then, when that decays you don't get 1/2 atom, then 1/4 atom , etc.  Rather, when that atom decays you are left with 0 radioactive atoms of that isotope—so it does reach 0.

This is simply an artefact of modeling the reaction with a perfect mathematical expression: exponential decay. In reality, matter is not infinitely divisible, so there will come a point where any first order reaction will run out of molecules.

Ever heard about Achilles and the turtle?

Also related: https://chemistry.stackexchange.com/questions/57075/is-there-no-actual-equilibrium

Note that basics of reaction kinetics were not created with the quantized nature of matter in mind. In this context, reactions of the 1st order do not end for the same reason why the function $\exp{(-x)}$ does not reach zero for any positive real $x$.
Due matter quantization, such reactions would end (unless there are other reactions producing the reactant) when the last molecule or a nucleus reacts or decays. But some time before that, it would not be a reaction of any particular order any more, as there would be strong quantum fluctuations for small counts of particles and the reaction rate would become more and more random.
For many scenarios it may be considered as if matter were not quantized and if the reaction does not end, as the eventual ending does not affect the kinetics at observed scope.
The differential equation of the reaction kinetics of the 1st order is
$$\frac{\mathrm{d}c}{\mathrm{d}t}=-Ac$$
and the function of the respective reaction kinetics is
$$c = c_\mathrm{t0} \exp{(-kt)}=c_\mathrm{t0} \exp{(-\ln{(2)}\frac{t}{T
_{1/2}})},$$
where $T_{1/2}$ is the half-time (or half-life) of the reaction or radioactive decay, with the meaning of the time the concentration or amount of radioactive isotope decreases to one half, regardless of the initial value.
BTW, only reactions of the zeroth order ends (in original continuous matter context), as their rate is linear. It applies also on the equilibrium state, where the forward nor backward reaction rates do not cease, they just become equal.

You wrote: "... as there would be strong quantum fluctuations for small counts of particles. For most scenarios it may be considered as it does not end."  How would quantum fluctuations prevent a sample of radioactive atoms from decaying completely?

@Theorist I have not said it prevents that. By the quote, I have tried to say that the reaction kinetic progressively stops following the simple exponential  function ( which was derived with statistics of  large numbers in mind ) and becomes more and more random. I have added some text.

I was referring specifically to your statement that "it does not end."  With radioactive decay, it does end -- eventually you reach the point where you have no more radioactive atoms.  And likewise, if there is no reverse reaction (or say the products are always swept away) eventually all the reactants will be used up, and that reaction would end as well.  I don't see how quantum fluctuations would prevent this.

No, I have said "For most scenarios it may be considered as it does not end." It implies many macroscopic measurements which end above the detection limit, or below it, implying it still goes on.

But for the scenarios you were describing, it does end.

"It does not end." and "for most scenarios it may be considered as it does not end." are different statements. Like the variable dependence is linear or can be considered as if it were linear ( but we know that it is not.) // The 1st order reaction DOES END for all systems, given enough time, if we do not consider backward reaction, but for many of the scenarios it can be considered as if it does not end. Like if we measure decay of isotope with the half-life of 30 years for 5 years, we can consider it does not end, as eventual ending does not  affect the measurement in any way.


text:
What is the paramagnetic susceptibility of sodium at around room temperature?
Proposed solution:
This question comes from Tanner's manual Introduction to the Physics of Electrons in Solids, at the chapter dedicated to the application of the Fermi gas model. The Fermi energy is $\mathsf{E}_F$.  The model used here is the free electron model, and the susceptibility is
$$ \mathcal{\chi} = \frac{\bf \mathcal{M}}{\bf B} $$
Assuming that the density of states at the Fermi level $D(\mathsf{E}_F)$ is approximated to: $$D(\mathsf{E}_F) = \frac{3N}{2} \mathsf{E}_F $$ where $N$ is the total number of electronic states. It is demonstrated that: $$ \mathcal{\chi}=\frac{3n\mu_B^2}{2\mathsf{E}_F}$$ where $n$ is the number of conduction electrons per unit volume given to be $n=2.5 \times 10^{28}\,m^{-3}$, $\mu_B$ is the Bohr magneton given to be $\mu_B = 9.3 \times 10^{-24}\,J\cdot T^{-1}$, the Fermi energy for $\ce{Na}$ is $\mathsf{E}_F = 5.2 \times 10^{-19}\,J$, and therefore the susceptibility is $\mathcal{\chi}=6.2\times 10^1\, MKS$, which can be converted in SI by multiplication of $4\pi \times 10^{-7}$, so the result is $\mathcal{\chi} = 7.8\times 10^{-5} \, SI$
Now my question:
This result is absolutely not consistent with the literature that states $$\mathcal{\chi} = 1.3\times 10^{-5} \, SI$$
What is the problem with Tanner or (most certainly) with my calculus ?

Bohr magneton value is not correct and formula for free-electron susceptibility should be multiplied by susceptibility of vacuum.

@10ppb - The susceptibility of a vacuum is zero, so what we're really talking about here is the relative susceptibility (I believe) - I'm not sure there needs to be a term by which this is multiplied. I also see a calculated value of 3.1 eV (McQuarrie, Statistical Mechanics) for sodium, which is slightly off from what the OP has given. I believe the error lies as you indicated with the value for the Bohr magneton; and the number density given/computed.

The formula given for chi is not a correct SI formula. It should be multiplied by mu_zero = 4pi10^-7 H/m, whatever you want to call it.

Many thanks @ToddMinehardt, I corrected the question according to your suggestions. There is still a factor of 4 between both results but this must be understandable from the model, ... and not from my calculation mistakes ...


text:
Nomenclature of tertiary amine derivative
I’m experiencing difficulties putting together a full name of the tertiary amine. One substituent is propyltrimethoxysilane $\ce{(OEt)3Si-CH2-CH2-CH2\bond{-}}$, and the two others are dimethylacetamide $\ce{(CH3)2N-CO-CH2\bond{-}}$ moieties:

Thus, could the molecule's name be something like 1,2-dimethylacetamide-3-propyltrimethoxysilane amine?

Thank You! Very interesting, but I am confused
even more as before. So, N,N-dimethylacetamide could be named as 2,2-azanediylbis?  What about silane moiety?  Maybe you could, please, suggest a literature where I could study this nomenclature a bit more...

@Iryna The literature is pretty much IUPAC Blue Book. You might also want to check out References about Nomenclature, List of Loong's nomenclature answers and How to ask and answer nomenclature questions?

The suggested name “1,2-dimethylacetamide-3-propyltrimethoxysilane amine” is not correct for multiple reasons. What are the locants 1, 2, and 3 referring to? There is only one “dimethylacetamide” in the name but there are two in the structure. There is a “trimethoxysilane” in the name but it's a triethoxysilane in the structure. The order of substituents in “3-propyltrimethoxysilane amine” does not correspond to the structure, which is 3-(triethoxysilyl)propan-1-amine.
(Also the name “propyltrimethoxysilane” in the question and the label “3-methoxypropylsilane” are wrong.)
Concerning characteristic groups, amides are senior to amines. Thus the smallest parent structure that contains both functional groups would be 2-amino-N,N-dimethylacetamide and not “2-(dimethylamino)-2-oxoethan-1-amine”.
In this case, however, another rule is relevant. Since there are two identical parent structures (the two N,N-dimethylacetamide units), which are not alkanes, multiplicative nomenclature is preferred to substitutive nomenclature.

P-51.3.1 Preferred IUPAC multiplicative names
For a multiplicative name to be categorized as an IUPAC preferred name, certain restrictive conditions must be met. Multiplicative nomenclature is preferred to substitutive nomenclature for generating preferred IUPAC names to express multiple occurrences of identical parent structures, other than alkanes when
(1) the linking bonds (single or multiple) between the central substituent group of the multiplicative group and all subsequent structural units are identical and
(2) the multiplicative groups, other than the central multiplicative group, are symmetrically substituted; and
(3) the locants of all substituent groups on the identical parent structures, including suffix groups, are identical.

Therefore, the identical parent structures are expressed by the multiplicative name 2,2'-…bis(N,N-dimethylacetamide). The preselected prefix azanediyl derived from the preselected parent hydride name “azane” is now used for the multiplicative substituent group $\ce{{}-NH-{}}$. So the complete multiplicative name is 2,2'-azanediylbis(N,N-dimethylacetamide).
The rest of the name is formed using the usual substitutive nomenclature. Thus, the complete name of the structure that is given in the question is 2,2'-{[3-(triethoxysilyl)propyl]azanediyl}bis(N,N-dimethylacetamide)


text:
Why do non-metals not form Ionic bonds(Ti-C, Sc-P) and instead form covalent compounds?
Is there something like, "sharing or gaining electrons gets harder", or is there something that lays hidden from common sight?
I am looking for a High-School Level Answer, pls something provide me with this answer..

If you already know that, I don't see why you'd need some "lies for students".

See: https://chemistry.stackexchange.com/questions/43966/can-carbon-and-titanium-form-an-ionic-bond

Ionic bonds are created when there is big enough difference between attraction of valence electrons by respective atoms. The measure of this atraction ability is electronegativity.
In reality, there is no purely ionic nor purely covalent bond, with the exception for the same atoms, like $\ce{O2}$, $\ce{N2}$, $\ce{F2}$ etc.
There is always some charge shift, but OTOH this charge is never fully transferred to either of atoms, while still bound by a mostly ionic bond.
Quoted from the Ionic bond link:

Purely ionic bonding cannot exist, as the proximity of the entities involved in the bonding allows some degree of sharing electron density between them. Therefore, all ionic bonding has some covalent character. Thus, bonding is considered ionic where the ionic character is greater than the covalent character. The larger the difference in electronegativity between the two types of atoms involved in the bonding, the more ionic (polar) it is. Bonds with partially ionic and partially covalent character are called polar covalent bonds. For example, Na–Cl and Mg–O interactions have a few percent covalency, while Si–O bonds are usually ~50% ionic and ~50% covalent. Pauling estimated that an electronegativity difference of 1.7 (on the Pauling scale) corresponds to 50% ionic character, so that a difference greater than 1.7 corresponds to a bond which is predominantly ionic.

The estimated electronegativity difference for 50% of ionic bond character is 1.7. For $\ce{Ti-C}$ is 0.92, for $\ce{Sc-P}$ it is 0.83, therefore the covalent bond character is dominant.

Element
Pauling electronegativity
Electronegativity difference

Ti
1.63
0.92

C
2.55

Sc
1.36
0.83

P
2.19

As Poutnik describes in his answer, low-electronegativity nonmetallic such as boron form bonds with low ionic character because in general, their atoms are not strongly electronegative enough to fully withdraw electrons even from highly electropositive metals (which still do require some ionization energy).
Nonetheless, incorporating such atoms into structures whose molecular orbitals are more strongly electron-accepting than the individual atomic orbitals can lead to compounds having more ionic character. For instance, in magnesium diboride ($\ce{MgB2}$) the boron atoms are arranged in sheets isostructural with graphite layers, but the boron atoms offer only three electrons apiece whereas the carbon in normal graphite layers offer four. The resulting electron deficiency means the boron is easier to ionize negatively, as indicated in this answer:

Nishibori et al. [1] found that at room temperature, magnesium is essentially fully ionized to Mg2+ while the boron remains neutral; the negative charge is associated with the interstitial regions as if to constitute metallic bonds. This still represents two-thirds of the theoretical charge separation for an ionic model and in that sense, the bonding between magnesium and boron may be deemed predominantly ionic. At 15 K the electrons become more localized so the boron now has a significant negative charge and the percentage of theoretical charge separation exceeds 80%.

Cited Reference

Eiji Nishibori, Masaki Takata, Makoto Sakata, Hiroshi Tanaka, Takahiro Muranaka and Jun Akimitsu, "Bonding Nature in MgB2", Journal of the Physical Society of Japan 70:8 (2001), 2252-2254, https://doi.org/10.1143/JPSJ.70.2252.

Thanks For This detailed answer, so does it mean that, for a ionic bond to take place, the electronegativity diff needs to be high enough, which is not, in case of Ti-C, or Sc-P?

Ionic bonding is possible with a low electronegativity difference if the electron transfer is favored by molecular-orbital structure. In magnesium diboride this is provided by the boron atoms assembling into sheets; with TiC and ScP there is no such structure with either element and then the low electronegativity difference gives more covalent character.


text:
Is haloform reaction considered to be disproportionation?
A question in an exam listed a number of compounds and the task was to find the number of compounds which will undergo disproportionation in conc. $\ce{NaOH}$. One of the compounds was $\ce{CCl3CHO}$, which will undergo haloform to give $\ce{CHCl3}$ and $\ce{HCOO-}$. Since one of the carbons is getting reduced and the other is getting oxidized, I considered it to be undergoing disproportionation. According to this answer, as long as the same element gets oxidized/reduced, it is disproportionation; over here, the same element, carbon is getting oxidized/reduced, hence I think it is disproportionation, but the answer says otherwise.
It is impossible for the same atom to get oxidized and reduced at the same time, like for example $\ce{P4}$ in $\ce{NaOH}$, obviously not the exact same atom is getting oxidized and reduced; but the difference between $\ce{P4}$ and $\ce{CCl3CHO}$ is that the phosphorous atoms which are getting reduced are indistinguishable from the ones which are getting oxidized, which is not the case in $\ce{CCl3CHO}$.
So is $\ce{CCl3CHO}$ in conc. $\ce{NaOH}$ considered to be disproportionation or not?

Canizzaro is disproportionation, this isn't. To some extent, you even answered why. BTW I'm unimpressed by your exam question, concept of disproportionation is largely useless, especially in organic chemistry.

And https://chemistry.stackexchange.com/questions/27368/cannizarro-reaction-of-chloral/27379


text:
Can I expect the same (or very close) peak areas from herbal extracts prepared with the same plant/solvent ratio?
I prepared an extract by using 100 mg of plant and 2 mL of ethanol, filtered and injected 20 μl into the HPLC system equipped with Waters® XSelect (150 mm × 4.6 mm, 5 μm) column. If I prepare an extract of 50 mg of plant and 1 mL of ethanol (therefore the same plant/solvent ratio) and also inject 20 μl into the HPLC system, can I expect the same peak (or very close areas), right?
Is the most important thing to maintain the solid/solvent ratio and to prevent the process from being “scaled-up”, e.g. 100× to other factors being relevant? Or am I wrong?

Comments are not for extended discussion; this conversation has been moved to chat. @OAS Please append necessary details regarding HPLC method and further findings from the chat discussion to your question (you can [edit] the question at your own pace).

I would summarize the extensive chatting to this answer:

Putting away the constant HPLC part, focusing solely on the sample preparation.
Assuming the dry powdered plant leafs samples are properly and representatively prepared.
The same extraction procedure, applied on variable mixture volume, can have slightly different extraction efficiency results.
E.g. using the same mixing frequency, bigger volume may be in average mixed less vigorously, or may reach different temperature due conversion of mixing energy to heat.
Considering all steps contributing to the method repeatability error, the extraction scaling error ( extraction efficiency, evaporation of volatile solvents) may be negligible.
Justification may need verification, e.g. by analysis of variance of the least square method for substance content as function of extraction scaling. There are standard formulas to test the statistical significance of the nonzero slope value.
If extraction scaling does not have significant impact, than it is trivially obvious the same sample/solvent ratio should lead to the same result. BUT...
While it should lead to the same mean result value, the repeatability error may still be affected by the sample size. 50 mg is not much and the sample granularity and particle non-homogenity may manifest more strongly.

If the result is well reproduceable at 100 mg / 2 ml, 50 mg / 1 ml should give very much the same outcome.
Changing the batch size by an order of magnitude could give a significant change. You never know until you tried. ;) Repeatedly!

So are you saying that I can probably calm down and breathe because the change do not involve an order of magnitude?

Yea, probably. ;) Some things scale very badly. For example if you compare a process in a half-filled to a fully filled vessel.


text:
According to Maxwell-Boltzmann distribution, what is probability distribution function proportional to?
If, according to the Maxwell-Boltzmann distribution
$${f(v)}\propto\exp\left(-\frac{\varepsilon}{kT}\right),\tag{1}$$
which is the equation from which the whole final equation is derived, then why is the final equation
$$f(v) = 4\pi v^2\left(\frac{m}{2{\pi}kT}\right)^{3/2}\exp\left(-\frac{mv^2}{2kT}\right)?\tag{2}$$
Clearly, here
$${f(v)}\propto v^2\exp\left(-\frac{mv^2}{2kT}\right).\tag{3}$$
Where have I made a mistake in understanding this?

The link you shared goes to a not-so-informative website. What's $\epsilon$? Let's say that $\epsilon = mv^2$. The proportionality statement still holds.

The 'additional' terms are obtained when transforming the probability distribution from 3-D distribution into magnitude of the velocity vector (averaged over direction) using spherical coordinates. Referenced from https://ocw.mit.edu/courses/5-62-physical-chemistry-ii-spring-2008/resources/28_562ln08/

Closely related, nearly  duplicate // You may confuse Boltzmann and Maxwell-Boltzmann distributions, with the latter based on the former, adding the v^2 factor from dV=4pi.r^2 dr, being integrated for the radius spherical coordinate.

The funny thing is both posts share the same OP.

@Poutnik I thought it was familiar...

@ToddMinehardt yes, the link is not the actual source that the expression was taken from. As mentioned in my comment on orthocresol's answer, it was taken from Peter Atkins' Physical Chemistry. Also Poutnik, sorry for the bother, but could you please go more into what the difference between the Boltzmann and the Maxwell-Boltzmann distribution is? Is it that, as orthocresol mentioned in his answer, the Maxwell-Boltzmann distribution is a 'vector form' of the Boltzmann one?

@theawesomenerd I have thought you got it in the linked Q/A. The BD is very general distribution, about probability of a state with given energy. The MBD  is specific BD application on translational kinetic energy. As all velocity vectors with given speed share this energy and the probability is therefore further weighted by the 4piv^2 factor, what is surface of the sphere in velocity vector space.

Oh I understood now @Poutnik. Sorry for the bother, again. So the BD is like the quadratic formula while the MBD is analogous to a formula for deriving the square root of a given number, right?

Not sure if this analogy fits. MBD is like "how many ways can the molecule have the given E_kin (proportional to  4.pi.v^2)"  times "what is probability it could have such E_kin (proportional to exp(-E_kin/(kT))"

@Poutnik I like to think of it as the "degeneracy factor" $g$ in the Boltzmann formula $p \propto g\exp(-E/kT)$, although I'm not sure whether that's 100% technically correct.

@orthocresol  Yes, I have also thought to use the term degeneration. I think it is correct but I was not sure if it would clarify or confuse things for the OP.

@Poutnik I did not literally go for the same relation as between BD and MBD. I meant in my analogy that the latter can be derived from the former. Either way, all the things are clear now.

You could see how MBD could be derived from BD in your prior Q/A.

The $v$ in your first expression (that $f(v) \propto \exp(-\varepsilon/kT)$) most likely refers to the true velocity which is a vector
$$\vec{v} = (v_x, v_y, v_z),$$
whereas the $v$ in the Maxwell–Boltzmann distribution refers to the magnitude of the velocity
$$v = |\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}.$$
The latter is more useful to a chemist because we don't really care which direction the particle is moving in, only its speed. However, note that there are many possible combinations of $(v_x, v_y, v_z)$ which yield the same magnitude $v$; so the formula needs to be adjusted for this. The factor ends up being $4\pi v^2$, which is the surface area of a sphere with radius $v$ (you can think of the surface of this sphere as representing all possible velocities with magnitude $v$).
My suggestion would be to look for a derivation in a good physical chemistry textbook, which should explain this more thoroughly.

Actually, I did read the first expression that I gave, in Physical Chemistry by Peter Atkins, but I really couldn't share the link because I have a hard copy, and I couldn't find any similar derivation on the internet. So, I had to  link that website. But the book did in fact give $f(v)\propto e^{\frac{-\epsilon}{kT}}$

Also, I found that the derivation in the same mentioned book is wrong in many places such as equating $f(v)$ with $f(v_x).f(v_y).f(v_z)$.

I think I understand now. We were taking $\vec{v}$ and therefore only one combination of $v_x$, $v_y$ and $v_z$ is present(due to only one direction). Thus we need not add the $4πv^2$ part to accommodate for the total volume in the velocity space. Is this what you meant @orthocresol?

I think that's pretty much what I said, yeah.


text:
Is LCP working well in real experiments? or is it more of an oversimplified principle?
I have a question for people who have applied Le Chatelier's principle (LCP) to their experiment or have knowledge about people who have done that to increase the yield.
Essentially, I am curious
"Is LCP working well in real experiments? or is it more of an oversimplified principle?"
I have a chemical reaction. A+B <-> C+D <-> E+F (liquid phase reaction)
This reaction starts with only A and B and C+D are wanted outcomes and E+F are unwanted outcomes. Since the amount of E+F is relatively small compared to C+D, people are happy most of the time but I am curious whether I can reduce E+F more by applying Le Chatelier's principle (LCP).
I am thinking of adding a large amount of F at the beginning of the chemical reaction so that the relative amount of E is minimized compared to C+D. F can be removed by a simple procedure.
My idea is theoretical and I have zero experience in applying LCP to real experiments. I want to hear from you before I devote my time to an experiment.

Le Chatelier's principle is completely provable using theory, so it can't be "oversimplified" - it's not a simplification of some other, more complicated theory. It is very much used in practice, one example being azeotropic removal of water to drive condensation reactions. That said, it's not going to be easy to predict how well it will work in your specific use case; you likely need some quantitative data like equilibrium constants and concentrations.

What you want is the https://en.wikipedia.org/wiki/Law_of_mass_action . Like the name suggests, it's a solid law. Le Chatelier is useful for multiphase reactions at variable pressure and temperature. And there are complicated reactions where Le Chatelier, if taken too literally, fails. You will find some if you search on this site.


text:
Can you use more concentrated hydrogen peroxide solution for obtaining iodine from PVP-I?
In NileRed —
Making iodine video Nigel Braun extracts iodine from the medical povidone-iodine. He says that the quantities of hydrogen peroxide needed is $\pu{500 ml}$ of $3\,\%$ hydrogen peroxide.
I did the calculation, and found out that it contains $\pu{15 ml}$ of hydrogen peroxide diluted with water. Can I use $\ce{167 ml}$ of $9\,\%$ hydrogen peroxide since $\pu{167 ml}\times 0.09 = \pu{15.0 ml},$ e.g. the same amount of $\ce{H2O2}$?

I remember near identical question posted recently here. You can, if you use also 333 mL of water. Otherwise there may or may not play role the different concentration and total volume.

It may provide better iodine yield with lower volume of more concentrated peroxide. But generally, it is always good to try to reproduce the procedure exactly, unless you do know very well why you intentionally deviate.

So do I need to use the same amount of water and the same amount of h2o2 or just the same amount of the 9%? Does the amount of water matter?

What is your interpretation of reproduce the procedure exactly, unless you do know very well why you intentionally deviate.  ? // BTW, I remember seeing near exactly same question recently here.

It's just I have 9% hydrogen peroxide and I want to use it instead of diluting it

Then try it. // Note that SE questions are supposed to be self-contained, giving sense even without link content, which should provide just additional information or refer to info source. CH SE site users are not big fans of watching youtube links to understand question. You should recreate the procedure by words here.

Yeah I know, I just mentioned the source. Because SE site users like to see resources too

Yes, they do, together with explicitly written essential content of the linked material, in this case the discussed procedure of iodine production, not to be forced to watch the video to realize, if the missing water would or would not matter.

Let us continue this discussion in chat.


text:
Amides from carboxylic acid and ammonia?
Would it be possible to synthesize amide by simply reacting carboxylic acid with ammonia and then driving off water at higher temperatures? Or would ammonia be also driven off in the process?
Google says that this should be possible, but I'm aware that some water based inks work by having ammonia neutralized acid polymer (polymer with COOH groups), then on drying the ammonia will leave thus leaving resin that will not solubilize in water again. For example styrene-co-acrylic acid and ammonia.
Thanks

It is not a very good process as you tend to just form the ammonium salt of the carboxylic acid. To get decent yields you have to activate the acid first.

@Waylander: I agree with Waylander. Best example is an amino acid, which exists as a Zwitterion in aqueous solutions.

Yes, phthalimide is prepared in this way.


text:
derivation of activity coefficient from Margules laws?
basics
In Margules activity model, the excess Gibbs free energy is expressed as:
$$
\frac{G^{ex}}{RT}=X_1 X_2 (A_{21} X_1 +A_{12} X_2) + X_1^2 X_2^2 (B_{21}X_1+ B_{12} X_2) + ...
$$
meanwhile, we know the relationship between activity coefficients of components and the excess Gibbs is:
$$
G^{ex} = RT(X_1 \ln \gamma_1 + X_2 \ln \gamma_2)
$$
and based on the above, I see many conclusions give:
$$
 \ln\ \gamma_1=[A_{12}+2(A_{21}-A_{12})x_1]x^2_2
\\ \ln\ \gamma_2=[A_{21}+2(A_{12}-A_{21})x_2]x^2_1
$$
which would be the result of partial differential of equation1 with respect to $X_1$，$X_2$ separately.
But how?
take $X_2$ for example, and only consider the first item.
shouldn't it be:
$$
\begin{align*}
\frac{\partial (G^{ex} / RT)}{\partial X_2} = \ln \gamma_2 &= \frac{\partial \{X_1 X_2 (A_{21} X_1 +A_{12} X_2) \}}{\partial X_2} \\
&= A_{21}X_1^2 + 2 A_{12} X_1X_2
\end{align*}
$$
???
can anyone explain it to me?

The approach is fine, but you are partially differentiating incorrectly, watch the definition for the natural logarithm of the activity coefficient
\begin{align}
  \ln(\gamma_1)
  & = \bigg[\dfrac{\partial(nG^E)}{\partial n_1}\bigg]_{P,T,n_2} \\
  & = \dfrac{\partial}{\partial n_1} \bigg\{n\bigg[\frac{n_1n_2}{n^2}\bigg(A_{21}\frac{n_1}{n} + A_{12}\frac{n_2}{n}\bigg)\bigg]\bigg\} \\
  & = \dfrac{\partial}{\partial n_1} \bigg[\frac{n_1n_2}{n^2}
(A_{21}n_1 + A_{12}n_2)\bigg] \\
  & = \dfrac{n_2n^2 - 2nn_1n_2}{n^4}(A_{21}n_1 + A_{12}n_2) + A_{21}\frac{n_1n_2}{n^2} \\
  & = \dfrac{n_2n - 2n_1n_2}{n^3}(A_{21}n_1 + A_{12}n_2) + A_{21}x_1x_2 \\
  & = \dfrac{n_2 ( n - 2n_1)}{n^2}\bigg(\frac{A_{21}n_1 + A_{12}n_2}{n}\bigg) + A_{21}x_1x_2 \\
  & = x_2(1 - 2x_1)(A_{21}x_1 + A_{12} x_2) + A_{21}x_1x_2 \\
  & = x_2 [(1 - 2x_1)(A_{21}x_1 + A_{12} x_2) + A_{21}x_1] \\
  & = x_2 [2A_{21}x_1 + A_{12}x_2 - 2A_{21}x_1^2 - 2A_{12}x_1x_2 ] \\
  & = x_2 [2A_{21}x_1 - 2A_{21}x_1^2 + A_{12}x_2 - 2A_{12}x_1x_2 ] \\
  & = x_2 [2A_{21}x_1(1 - x_1) + A_{12}x_2 - 2A_{12}x_1x_2 ] \\
  & = x_2 (2A_{21}x_1x_2 + A_{12}x_2 - 2A_{12}x_1x_2) \\
  & = x_2^2 (2A_{21}x_1 + A_{12} - 2A_{12}x_1) \rightarrow
 \boxed{\ln(\gamma_1) = x_2^2[A_{12} + 2(A_{21} - A_{12})x_1]}
\end{align}