text:
Why does the energy level of a particle in a box decrease with length?
I want to know the intuition of why L is in the denominator of the equation for the energy of a particle in a box. This helps explain why resonance is stabilizing without using molecular orbitals. Does it have to do with the Uncertainty Principle? As the length increases, uncertainty of position increases, so uncertainty of momentum decreases, consequently meaning that the kinetic energy is lower, lowering the overall energy of the orbital?

It has to do with quantum mechanics. The uncertainty principle is not an axiom, but a derived statement. As such, it can't be said to "cause" things, but can be called to explain them. With that in mind, your reasoning looks right to me.

In quantum mechanics, particles are waves, and as with any wave the longer the wavelength the less energy is in the wave.  You don't need the box to see this; infrared light has less energy in the photon than ultraviolet (the former makes you warm but the latter can damage your skin) and a slow moving free electron (lower energy) has a shorter associated wavelength than a fast-moving one (higher energy).  In the particle/box problem, the possible wavelengths are set by the dimensions of the box; a bigger box allows longer wavelengths -> lower energies.


text:
Confused on the peak intensities in this mass spectrometry problem
A problem from Introduction to Spectroscopy by Pavia et al. [1, p. 137]:

*4. An unknown hydrocarbon has a molecular ion peak at $m/z = 84,$ with a relative intensity of $31.3.$ The $\ce{M + 1}$ peak has a relative intensity of $2.06,$ and the $\ce{M + 2}$ peak has a relative intensity of $0.08.$ What is the molecular formula for this substance?

I normalized $31.3$ to $100\%,$ so the relative intensity $I$ was calculated as follows:
$$
\begin{array}{lr}
\hline
\text{Peak} & I/\% \\
\hline
\ce{M^+} & 100.00 \\
\ce{(M + 1)+} & 6.58 \\
\ce{(M + 2)+} & 0.25 \\
\hline
\end{array}
$$
The textbook is saying the molecular formula is $\ce{C6H12}.$ To me the answer doesn't make sense because if I calculate the $\ce{(M + 1)+}$ intensity using this formula from my textbook given for $\ce{C_xH_y}$:
$$[\ce{M + 1}] = 1.1x + 0.015y$$
Thus for $\ce{C6H12}$:
$$[\ce{M + 1}] = 1.1 × 6 + 0.015 × 12 = 6.78\%$$
However, $6.78\%$ is off from $6.58\%$ by about $0.2\%$. Am I missing something or is there some sort of acceptable tolerance for peak intensity differences?
Reference

Pavia, D. L.; Lampman, G. M.; Kriz, G. S.; Vyvyan, J. R. Introduction to Spectroscopy, 5th ed.; Cengage Learning: Stamford, CT, 2015. ISBN 978-1-285-46012-3.

Mass spectrometers can be slightly off due to calibration.

Here, the experimental errors are such that $6.78$ can be considered as equal to $6.58$.

Such problems in chemistry don't work out to infinite precision like pure math problems. You have to apply some chemical knowledge to the problem too.
The book formula is obviously an approximation to the exact solution. To find the exact values expected you'd need to use combinatorics.
Considering the chemistry:

7 carbons have a mass of 84, so a hydrocarbon molecule can't have that many carbons since there wouldn't be any hydrogen atoms.

5 carbons have a mass of 60, but that would mean at most 12 hydrogen atoms for an aliphatic molecule, so the mass could be at most 72.

Thus there must be 6 carbons in the molecule.

But then why did the problem add the peak intensity information if I all need to know is the mass of the molecular ion?

@Mohamed Why not ? In real cases, you need not to use all info that is available. In study cases,  extra info may be included, if it would come naturally, to train people to think what data to use.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Can single molecules of C and O2 react in isolation, and if so how will momentum be conserved?
I am trying to figure out how is it possible to reconcile the reality of exothermic reactions, which means that kinetic energy is transferred to (heats) the surrounding matter, with the principle of conservation of momentum, in particular when the product is a single molecule.
I have been exploring the question but found no sufficiently explanatory answer to the question of how the bond energy that is released may create movement (heat). The nearest to an answer that I can imagine is that the reactants enter some transitory combined state of vibration and that further contact with some other molecule results in the separation of each of them in opposite directions with equal but opposite momenta increments thus preserving the overall momentum, while at the same time reducing the vibration and stabilizing the product molecule.
If that were the correct answer (at least in some cases), I would like to know a bit more about the details of the process, which I suppose entail some description about how binding energy is transferred to that vibration.
In any case, it would be interesting to know whether is it actually possible for a single carbon atom and oxygen molecule to react (if they collide with the necessary energy) and produce carbon dioxyde or not, be it for the reason above or another one, since if they are in isolation, the transfer of vibration energy cannot be realized and thus the reaction could not be completed, and then I suppose that would eventually end with the spontaneous separation of the components.

First of all, you are asking several questions at the same time. We generally encourage focusing on one, which also helps to get quality answers. Also, what do oyu mean by " I would like to know a bit more about the details of the process, which I suppose entail some description about how binding energy is transferred to that vibration."? Do you you specific questions about the PES of this system?

I agree that formally there are several questions, but in my mind it's a single one, which may be summarized as: how can it be the proccess (in isolation or not) by which bond energy is converted into movement --macroscopically, heat-- in exothermic reactions while conserving momentum? And, sorry, I don't know what PES refers to...

It seems you would benefit from reading about astrochemistry. There are books and presumably websites where similar matters are thoroughly discussed.

There is no single molecule of C. That being said, the concern is legitimate and quite important, so I'd ask everyone to refrain from the close votes.

Thank you, Ivan. Of course I know C is not terminologically a molecule (but atom, as I write in the final paragraph), I just wanted to shorten the title, it's mostly irrelevant to the subject, I think. What I intended with it was to show a limit to the possible explanations of how bond energy is converted to motion, which is the deep question, as I try to explain since the first paragraph. While interesting, the astrochemical question is not my main concern here.

What I mean regarding astrochemistry is that there are books with a heavy physics/physical chemistry slant which discuss in detail the kinetics of reactions in space, where the distinctions between two-body reactions and three-body reactions are massively important. This includes a discussion of the conservation laws, vibrational and electronic excitation, collision parameters and cross-sections, etc. This background is strongly relevant to you, even if the specific reaction or environment is not. I would give you a reference, but I cannot remember the exact book I read 10 years ago.

Aha, I found it - The Physics and Chemistry of the Interstellar Medium. Chapter 4, Chemical Processes, contains a wealth of information.

I appreciate your effort, Nicolau, and I've got the book! Sect. 4.1.5 Radiative association reactions, shows that for some reactions during a vibrational transition a photon is emitted. As @fraximus explains in a comment to Neterin's answer, that becomes relevant in upper layers of the atmosphere, although as it's stated in Sect. 4.1.7 three-body reactions, where a third species is "thermalized", dominate exothermic reactions in laboratory settings.

$\ce{C + O2}$ is awfully complicated, so let's just pretend you've asked this:
In a single act of the reaction $\ce{H. + H .-> H2}$, how is momentum conserved?
That's a legitimate concern all right. After all, we are taught that this reaction does happen instantly, once given a chance, and that's in fact true. Also, we know that it releases a lot of heat. Now, heat is nothing but the motion of molecules; how does energy convert to the motion of one molecule as a result of one single reaction act?
It doesn't. The conservation of momentum forbids that, just as you reasoned. Chemical reactions are collective phenomena. Nobody cares about a single molecule.
Now what really happens to a single molecule which has just formed as a result of the mentioned reaction? That's really simple: the molecule is vibrating wildly, ready to break apart. Is has just enough energy to do so. It will do so half of the times, or maybe more often. But that doesn't matter. What matters is that sometimes the vibrating molecule will hit another molecule and sent both of them flying away in opposite directions, thus releasing a part of its energy and becoming more or less stable.
An emission of a photon is also an option, but that's another story.
So it goes.

Thank you very much again! I suppose that one can conclude that in isolation exothermic reactions are essentially unstable if not mostly impossible if the product is a single molecule. I'd also tend to agree that photon emission would be if only marginal, taking as reference black body radiation. Nevertheless, I am still intrigued about the details of the transitory, dynamic process by which the binding energy converts to that excess vibrational energy... maybe I could try some readable description?

@scifriend When the two atoms are colliding and forming a bond the potential energy of the system can be described as a function of the atomic distance, a curve, that has a minimum around bonding distance: https://demonstrations.wolfram.com/EnergyLevelsOfAMorseOscillator/
When the bond is formed the energy is coming from this potential energy difference between the minimum and the r -> infinity values. Imagine a sleigh that goes down to the minimum: it would go up again on the other side. Rolling back and force around the minimum is the vibration you are looking for.

Not sure exactly about C + O2, but in general the photon emission path is nowhere marginal. This is essentially what happens in upper layers of the atmosphere after the sunset. There are UV-broken molecules that want to recombine. There are few molecules around to "termalize" with. That's why the upper atmosphere slightly glows at night.

When two isolated atoms collide the total energy and momentum must remain with the two atoms so both are conserved overall. In fact in a reaction such as $\ce{H\cdot + H\cdot <=> H2}$ the hydrogen molecule only lasts for a few femtoseconds. This is because even though the bond is formed the atoms will still approach one another (total energy being constant, potential energy becomes more negative and kinetic energy more positive) and rebound as the atoms become very close. The 'molecule' only lasts for a single vibrational period, i.e. a few femtoseconds.
If, however a third body is present, say an inert molecule or atom and this collides with the nascent $\ce{H2}$ molecule then some energy can be taken away from the $\ce{H2}$ and it becomes stabilised. At this point it may radiate away some energy or suffer further collisions and so become thermalised. What happens depends on the relative rate constants for these processes.
In the atom-diatomic collision, e.g. $\ce{F + D2<=> D + DF}$, overall, total energy and momenta is again preserved if there are no other species involved. However, in this case the $\ce{D2}$ has translational, rotational and vibrational energy, this is then partitioned among the products depending on the nature of the potential energy surface describing the approach of the reactants and that of the products. Such 'reactive scattering' has been extensively studied in the gas phase under high vacuum conditions and in molecular beams. See Polanyi & Woodall, J. Chem. Phys. 57, 1574, (1972); Polanyi & Schreiber, Faraday Disc. Chem. Soc. 62, 267, (1977) and textbook by Steinfeld, Francisco & Hase, Chapter 9,' Chemical Dynamics & Dynamics'(Prentice Hall 1999);Levine & Bernstein 'Molecular Reaction Dynamics and Chemical Reactivity' (OUP 1987).

So does that mean H + H —> H2 won’t happen in the gas phase? As you’d need some kind of solid body to transfer energy to? Or is an excited H2 molecule colliding into another atom in the mixture enough to dissipate its energy?

Some energy has to be taken away  so colliding with another molecule or atom can do this and make a molecule.  Collision with a surface will also remove energy. Most likely here though is that reaction occurs by a different mechanism which is atoms diffusing together when on the surface.

@porphyrin Thank you very much for your illustrative answer and references. However, I am still intrigued about how the chemical energy in, say, the O2 bonds can be released into vibrational energy in the CO2 product, even if the reaction is, as Ivan Neretin said, terribly complicated. Is that type of energy conversion a well described process? Could someone give a summary description or a reference?

Have a look for articles on LEPS potentials, and atom-diatomic reactions or the text books in my answer. The important point is that energy in translation vibration and rotation are not isolated from one another but are 'coupled' so that energy can flow between them.

@scifriend Carbon and Oxygen are complicated because there are so many effects that make a mess of any single, simple way to explain it. A simple model without explanatory power is not of any use to anyone. Carbon and Oxygen react through intermediary species, with rate determining steps (forbidden transitions), radicals and the product is rarely carbon dioxide, that is the final product of a series of unstable carboxides. You won't find a nice way to track what happens to momentum... I think...

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Molecular structure of iodine nonoxide
A question in an exam was as follows:

Iodine reacts with ozone gas to form a dark yellow solid $\ce{X}.$ Let the number of lone pair of electrons in un-ionised form of $\ce{X}$ be $m,$ number of lone pair of electrons in the anionic moiety of $\ce{X}$ be $n$ and the positive charge on the cationic moiety of $\ce{X}$ be $p$ units. Then what is the value of $\displaystyle\frac{m - p}{n}?$

This reaction of iodine with ozone is:
$$\ce{I2 + O3 -> I4O9 <=>I^3+(IO3^-)3}$$
Therefore the anionic moiety is $\ce{IO3-}$ and cationic moiety is $\ce{I^3+}$. However the first part of the question states un-ionised form. My assumption is that this is $\ce{I4O9}$ and they ask for the number of lone pairs on the molecular form.
The answer to this question takes into consideration that $\ce{I4O9}$ is an equimolar mixture of $\ce{I2O4}$ and $\ce{I2O5}$ which gives the answer to be $2.5$
However, from the abstract of J. Raman Spectrosc. 1985, 16 (6), 424–426:

The Raman spectrum of $\ce{I4O9},$ formed by the gas‐phase reaction of $\ce{I2}$ with $\ce{O3}$, has been measured. Freshly prepared samples of $\ce{I4O9}$ gave broad band spectra characteristic of an amorphous solid. Vibration bands at $780,$ $740,$ $620$ and $\pu{450 cm−1}$ were observed. It was established conclusively that $\ce{I4O9}$ is a distinct molecular species and not a mixture of $\ce{I2O5}$ and $\ce{I2O4}.$

If $\ce{I4O9}$ is a distinct molecular species, what is the molecular structure of $\ce{I4O9}?$

Regarding its ionic form, I don' think it is as simple as $\ce{I^3+(IO3^-)3}$. It is much more complex than that. From Ref.1:

This compound is likewise to be considered as an $\ce{I(III,V)}$ oxide and reacts with alkali hydroxide to give $\ce{I-}$ and $\ce{IO3-}$.
$$\ce{3I4O9 + 12OH- -> I- + 11IO3- + 6H2O}$$
Structurally, $\ce{I4O9}$ is possibly an iodate $\ce{I3O6+IO3-}$ (more
precisely $\ce{(I3O6+)_n.nIO3-}$) in which the isopolycation
$\ce{I3O6+}$ has a polymeric structure and is formulated as
$\ce{I_^{III}(I^{V}O3)2^+}$ consisting of twice as many pyramidal
$\ce{I^{V}O3}$ groups as square-planar $\ce{I^{III}O4}$ groups thus
$\ce{I4O9}$ would correspond to $\ce{I(IO3)2^+IO3-}$ which would
become $\ce{I(IO3)3}$.

Reference

Inorganic Chemistry By Egon Wiberg, A. F. Holleman, Nils Wiberg, Academic Press, 2001

The Wikipedia page states that

Tetraiodine nonoxide ($\ce{I4O9}$) has been prepared by the gas-phase reaction of $\ce{I2}$ with $\ce{O3}$ but has not been extensively studied.

However, looking at the reaction $\ce{I4O9 <=> I^3+(IO3^-)3}$ allows us to venture a good guess at the structure, which I have drawn below:

The central atom's orbitals are $\mathrm{sp^3d}$ hybridized and the molecule is expected to adopt a T-shaped structure:

Note that I don't have a reference to back this up: As already said by Wikipedia, no one seems to has done extensive research on this molecule, and I couldn't find any papers highlighting a study of this molecule (closest thing I could find was this paper, which merely states kinetics of oxidation of this molecule, rather than studying it's structure). If I do come across one in the near future, I will update the answer with the findings.


text:
Deriving an expression for reaction time in terms of the total concentration

Consider the dimerization of protein $\ce{A}:$
$$\ce{2A <=> A2}$$
with forward rate constant $k_1$ and backwards rate constant $k_{-1}.$ Derive the following expression for the reaction time in terms of the total concentration of protein
$$\frac{1}{\tau^2} = k^2_\mathrm b + 8k_\mathrm ak_\mathrm b[\ce{A}]_t,$$
where $[\ce{A}]_t = [\ce{A}] + 2[\ce{A2}].$

I assumed the 1st order in the forward direction and the 2nd order in the reverse direction. Rate law:
$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt} = k_1[\ce{A}]^2 - k_{-1}[\ce{A2}].\tag{1}$$
At equilibrium
$$k_1[\ce{A}]^2_\mathrm{eq} - k_{-1}[\ce{A2}]_\mathrm{eq} = 0.\tag{2}$$
I then substituted that into the rate law and factored out the rate constants with the aim of getting rid of all of the equilibrium concentrations. The equation indicates that
$$\frac{-1}{2}\frac{-\mathrm d[\ce{A}]}{\mathrm dt} = \frac{-\mathrm d[\ce{A_2}]}{\mathrm dt} \quad\implies\quad \frac{\mathrm d[\ce{A}]}{\mathrm dt} = -2\frac{\mathrm d[\ce{A2}]}{\mathrm dt}.\tag{3}$$
I then got stuck. I saw something where you define a displacement variable $x$ and then try to get the rate law in terms of $x,$ but I've tried that approach and keep running into dead ends.

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

In my opinion, there is no "reaction time" for this equilibrium. An equilibrium needs an infinite amount of time to be attained.

Just for fun, you can compute the time integral of (1) and substitute $\tau$ for the end time.Than you will see the final relation of reaction quotient Q and K.

What are $k_a$ and $k_b$?

Let the amount of $A_2=x$ and write down $dA_2/dt=dx/dt=2k_1[A]^2-k_{-1}[A_2]$ and find x the amount reacted in terms of $[A_0]$ and $[A]$ and substitute into the equation. Separate variables, ie, integrate in $x$ and $t$ separately. The integral in $x$ is just horrid but look for the constant square root term in front of the log terms.   This is equivalent to $ 1 / \tau ^2$. If $[A_0]$ is the initial A I found $1/\tau^2=4k_1k_{-1}[A_0]+k_{-1}^2$ which is related to your answer. (btw. should your $[A]_t=2[A]+[A_2]$ )

No I'm pretty sure the total [A] is correct. My professor literally copied and pasted it out of  our physical chemistry textbook

@Maurice the relaxation is first order (or pseudo first) and as such has a well defined lifetime or characteristic rate constant. The general solution is of the form $\ln( f(k,c))=-w(k)t$ where $f(k,c)$ is some (usually complicated) function of rate constants and concentrations and  $w(k)$ a function of rate constants and initial concentrations so has a constant value so that $1/\tau\equiv w(k)$.


text:
Binding energy vs. ionization energy
I am studying binding energy in the context of photoelectric spectroscopy, and I came across a graph of relative numbers of electrons vs binding energy (in units of megajoules per mole). The element in question is lithium. My chemistry resource (Khan academy, to be precise) noted that the binding energy for the furthest right peak corresponds to the binding energy for the electron in the 2s subshell, and that this is also equal to the first ionization energy. However, they go on to say that the the binding energy of the second peak is not equal to the second ionization energy, since the electrons are held more tightly by the nucleus once the outermost electron is removed. I actually have two questions about this. First, why are they more tightly held once the outermost electron is removed? Is the electric field of the protons somehow getting weakened by the presence of the outer electron? I don't think this would be the case, based on my understanding of electric fields. So what is the explanation?
My other question is why the binding energy is different from the ionization energy? Is it because the electrons are removed all at once versus them being removed one-by-one? Or is there some nuance to this situation I am not seeing?
I found a question on this website that seems to be asking a similar thing, but no one answered it. Here's the link to it anyway.
Chemistry: What's the difference between "binding energy" and "ionization energy"?
Thanks in advance for the help.

Isn't just one question? Or better, the question should be why PES can probe two energies "at once"? The rest is a duplicate.

First some basics....
Lithium has three electrons.

$\ce{Li}$ electronic configuration: $1s^22s^1$
$\ce{Li -> Li+ + e-\quad\quad\mathrm{Ionization\ Energy= 0.520\ MJ/mol}}$
$\ce{Li+}$ electronic configuration: $1s^2$
$\ce{Li+ -> Li^2+ + e-\quad\quad\mathrm{Ionization\ Energy= 7.298\ MJ/mol}}$
$\ce{Li^2+}$ electronic configuration: $1s^1$
$\ce{Li^2+ -> Li^3+ + e-\quad\quad\mathrm{Ionization\ Energy=11.815\ MJ/mol}}$
$\ce{Li^3+}$ electronic configuration: $1s^0$, i.e. no electrons
Ionization energies from Wikipedia

Thus the different ionization reactions remove the electrons from the outer most to the inner most electrons.
Below is an another image of the idealized spectra of lithium that has the binding energies stated. (from https://www.andrew.cmu.edu/course/09-105/S98Slides9.html)

$\mathrm{Binding\ Energy= 0.52\ MJ/mol}$ corresponds to $\ce{Li -> Li+ + e-}$
or $1s^2 2s^1\ce{->} 1s^2$, hence the first ionization energy of a Li atom.
However $\mathrm{Binding\ Energy= 6.26\ MJ/mol}$ corresponds to $\ce{Li -> Li+ + e-}$
or $1s^2 2s^1\ce{->} 1s^12s^1$, which is not the first or second ionization energy of a Li atom. Rather it is the binding energy of the 1s orbital in a neutral lithium atom.
Hence the photoelectron spectra is called the binding energy spectra because that is what it measures. Note that the sweep of the spectra would not have to cover the ionization energy. For example the spectra could have been taken from 5 MJ/mol to 100 MJ/mol.

The answer to the questions as well as those in the linked post lie in the following:

the second peak is not equal to the second ionization energy, since the electrons are held more tightly by the nucleus once the outermost electron is removed

It is worth reviewing the basic principle of the PES experiment: you irradiate the  sample with high energy photons (with energy well above electron binding energies) and determine the count and kinetic energy of emitted electrons. From the kinetic energy and the energy of the incident radiation you compute the binding energy by simple arithmetic:
$$E_B=h\nu -E_K$$
This data you report as a spectrum of counts versus binding energy. The binding energy is the amount of energy required to eject a particular electron transforming it from a bound state in a specific orbital into a free (unbound) electron (in theory, with zero kinetic energy once it is infinitely removed from its source atom).
The question is, how might a measured binding energy differ from an ionization energy? Ionization energies are equal to or greater than binding energies. The binding energy of a valence electron is equal to the first ionization energy, which corresponds to the same event of ejecting the first valence electron. The problem is that second ionization energies are defined as the energies required to eject a second electron, after a first ejection has occurred, that is, from an ion. The second electron will occupy a lower energy orbital (it is more strongly bound). In the PES experiment, an electron in an inner shell (non-valence) can be observed directly, without a prior first-ionization step. The energy to remove the inner shell electron from the neutral atom (the so-called binding energy) is not the same as required to eject a second electron (the so-called second ionization energy).
The reason it takes more energy to remove a second or later electron is because electron-electron repulsion raises the energy of each and every electron in an atom. The usual convention is that (in general) bringing an electron closer to the positively charged nucleus lowers its energy, and bringing two electrons closer to each other increases their energy. Electrons repel each other, therefore if you have more electrons in an atom (the nucleus otherwise remaining the same) the ejected electron will be "pushed" away a little harder by the other electrons and it will take a lesser external energy input to expel the electron.

Don't you mean that the electron-electron repulsion decreases the energy of each and every electron? Unless I have it wrong still. The reason I say that it should decrease the energy is that it appears that the second or third ionization energy is greater than the binding energy for those same electrons. Sorry if I still have it wrong

@Aaron23 I edited my answer in an attempt to explain. It's a matter of convention. Adding electrons to an atom increases electron-electron repulsion (increase in energy), whereas removing an electron reduces the energy of remaining electrons.


text:
Ab initio molecular properties which can be experimentally measured and don't require derivatives/gradients
I'm attempting to teach myself HF and DFT by implementing various algorithms, but I've run into a snag: many molecular properties which are experimentally measurable appear to require taking gradients, which is considerable extra work (from a coding perspective). Are there any properties where I can start with a known optimized geometry, run my HF or DFT implementation, and then compare the results to experimental values? Pedagogically, I'd like to be able to motivate DFT by showing it clearly outperforms HF SCF in some experimental regime.

Many thermochemistry benchmarks, like the GMTKN55 (homepage listing most results, DOI), use fixed geometries and require only ground state total energies. Some of the subsets probably have back-corrected experimental references, but you would have to check each paper, I'm afraid. Note that for some properties, a dispersion correction is vital.


text:
How can I build the wyoming like montmorillonite unit cell in avogadro for a molecular dynamics simulation?
I am trying to build a unit cell for Wyoming-like montmorillonite, using Avogadro software. I am not sure how can I get access to the atom coordination and how to generate the .xyz file at the end? Is there any file that I can use as guidance during this process? Actually, I would appreciate any suggestions in this regard? please do not hesitate to share your knowledge with me.

you can just easily copy the Cartesian coordinates from Build->Cartesian Editor. And copy the text saving as .xyz (this is simple text file)

I would suggest you try and find the unit cell (.cif file) of the compound in question from the CCDC database, and then use Mercury (a free version is available at the CCDC website) to create the system.
You can of course use Avogadro if that is what you like, but in my experience the coordinate files it produces for super cells are quite messy.


text:
How to read a chromatography calibration curve?
I am a high school student who has been studying a bit of chromatography. I came across calibration curve during class and I trying to figure how to read it and use it? I have no clue how to read off the graph nor do I really now the use of it. Can someone please explain what information I can extract from this graph and possibly provide an example if that's fine?

I assume your teacher explained the HPLC separation. If injected a mixture of four ions you will get four peaks. Each peak has an area, which is proportional to the concentration of the substance.

Imagine you wanted to determine the concentration of chloride ions in your tap water. What you would do is that you will prepare several known concentrations of chloride ions using $\ce{NaCl}$ solutions, say $0$ to $\pu{350 ppm}$ (look at your calibration chart). Inject them one by one, and measure the peak area, let us say the second peak is that of chloride ion. The peak shape is Gaussian in real work but you can "estimate" them as triangles. Recall triangle's area is very easy to calculate.
First step: Plot the peak area for concentration. What you get is a calibration curve.
The next step is to find a mathematical equation that fits this data.
Fortunately it is linear in your case of the form $y=mC$ where $y= \text{peak area}$, $m=\text{slope}$, $C=\text{concentration}$ in $\pu{ppm}$.
Now you would inject your tap water in the HPLC, you do not know the concentration ($C_1$) but you do know the peak area.
The last step is called interpolation. You go from the known area, and read the corresponding concentration axis. Viola! Now you know the chloride concentration in your tap water!
Alternatively, you use the equation, $y=0.1312C$, so I know the area of the tap water chloride injection, say, it was $\pu{40.0 units}$, so $C_1 = 40/0.1312 = \pu{304.9 ppm}$.
In typical tap water it is around $\pu{100 ppm}$, I just made up those numbers.
Bonus: Rationalize the units of the peak area on the $y$-axis! Why it is conductivity times minutes?

Thank you so much for the help. I'm finally able to get a firmer grasp on this topic! Thank you again for the thorough explanation!

@Munchies, did you understand units of y-axis?

Yes, the y-axis indicates the area of peak, which is proportionate to the concentration of the substance right? Also, thanks for the help, it's really cleared things up for me!

Suppose it is a calibration curve of high performance liquid chromatography (HPLC) to determine concentration of chloride ion in unknown solutions. The detector is sensitive to charge ions, which gave the peaks, area of which is proportional to concentration of ions. I assume each peak area is highly specific for ion in interest. First you make a calibration curve using standard $\ce{Cl-}$ solutions with known concentrations:

The curve is a straight line, which goes through $(0,0)$ origin ($y = 0.1312x$) with an excellent correlation with $R^2 = 0.9995$ ($R^2 = 1.0$ is the best correlation). I insert an example of chromatography trace to explain how concentration of ion is related to area under the relevant peak. When running the standard solution, you'd find the correct retention time ($t_R$) for the chloride ion. Once you prepare your calibration curve, you can run your unknown sample (suppose it contains $\ce{Cl-}$ and $\ce{NO3-}$ ions) and find the area under the chloride peak (Nowadays, all peak areas of the chromatogram are automatically given by the software). Suppose it was $\pu{22.5 Absorbance/min}$ as indicated in the image. Since the equation of the best fitted straight line is given automatically ($y = 0.1312x$), you can submit the $y$-value to it and find the $x$-value, which is the concentration in $\pu{ppm}$:
$$22.5 = 0.1312x \ \Rightarrow \ x = \frac{22.5}{0.1312} \approx \pu{172 ppm}$$
Modern technalogy made these calculations easy by giving the equation for the calibration curve and the areas under the curve. In old-fashion, for example, you have to cut each peak from each printed chromatogram, and weigh them by analytical balance to correlate them for known concentration. Further, you plot the values in a graph paper and make the best fitted straight line. Once, you got the peak area value for your unknown solution, you mark that value in $y$-axis and draw a straight line parallel line to $x$-axis until it meets the curve. From that crossing point, you again draw a straight line parallel line to $y$-axis until it meets $x$-axis (see maroon lines in the image). The point second parallel line meet $x$-axis is the chloride ion concentration in your unknown sample.
Note that you cannot find the $\ce{NO3-}$ ion concentration in your unknown sample (see the insert) using the same calibration curve. The detector been used might show different response to the nitrate ion compared to chloride ions. Therefore, you must make different calibration curve for the nitrate ions using standard nitrate solutions.

Thank you for the really thorough explanation! I really appreciate the help!!!

The line goes according to : $y = 0.1312~x$ ; In other words the area of the peak is equal to $0.1312$ times the concentration of chloride in ppm. This result is extremely useful. If for example, in a later measurement you obtain a peak whose area is $y = 35$, it means that the concentration $x$ of chloride ion responsible of this peak is : $x = 35/0.1312 = 266.8$ ppm

If we take 1st and last points on your graph we'd see that:

At concentration $\pu{350ppm}$ your detector shows number $45$
At concentration around $0$ your detector shows number around $0$
So what if detector shows $175$ (which is $0.5*(350-0))$? You may conclude that the concentration is $\pu{22.5ppm}$ (which $0.5*(45-0)$)

Calibration curve gives you understanding of the relationship between the number that detector shows and the concentration. So that when you see some reading you could calculate the concentration.
Note that 2 points usually isn't enough because you don't necessarily have a straight line. So you need more measurements to see if it's true. If it's not true - you'll have to come up with more complicated equations than $y=0.1312x$ and your results will be less precise.

I was wondering why you divide by 2? Sorry i'm really new to this haha

@Munchies, I chose $175$ which is in the middle of the last and first point. Which means it's going to be in the middle of the concentration too, so need to multiply by $0.5$ or divide by $2$ - doesn't matter. I could've chosen another point - e.g. $300$ which is $0.85*(350-0)$, then to get the concentration: $(45-0) * 0.85$.


text:
Visualizing the e_g and t_2g orbitals in case of tetrahedral crystal field splitting
I am reading about the crystal field theory (CFT) for the case of tetrahedral coordination entities. I understand that the $\mathrm{t_{2g}}$ orbital is nearer to the ligand than the $\mathrm{e_g}$ orbital, from the data of the angles between the respective orbitals, central metal atom and the ligands. But I do have the following questions:

Deriving the angle for the $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ orbital mathematically.
The angle in case of $\mathrm{e_g}$ orbital is roughly $55^\circ,$ which seems to come from dividing the tetrahedral angle by $2.$ Also, the angle in case of $\mathrm{t_{2g}}$ orbital comes out as $35^{\circ}16',$ which I have no idea about.

Positional analysis.
Lee's Concise inorganic chemistry [1, p. 220] states that

Alternatively the $\mathrm{t_{2g}}$ orbitals are half the side of the cube away from the approach of the ligands, whilst the $\mathrm{e_g}$ orbitals are half of the diagonal of the cube away.

I believe the angles mentioned can be derived from the statement, but I'm not sure how the statement came about. It would be great if anyone could share some visual models of this part of CFT for tethrahedral coordination entities and suggest the right approach to understanding this.
Reference

Lee, J. D. Concise Inorganic Chemistry, 4th ed.; Chapman & Hall: London ; New York, 1991. ISBN 978-0-412-40290-6.

I have some trouble understanding your question. Apparently you think orbitals can be reduced to a point ?

@Maurice I understand that orbitals cannot be reduced to a point, but for the angles we just need the directions which the lobes are pointing in alongwith the direction of approach of the ligands. I think that does clarify things up a bit. But still there's some confusion about the statement from my textbook quoted in the OP.

Due reasons of symmetry, there are $\textrm{t}{\textrm{2g}}$ and $\textrm{e}{\textrm{g}}$ sets of d orbitals in octahedral complexes. However, in tetrahedral complexes, these sets are reversed in energy, obviously, and now called $\textrm{t}_{\textrm{2}}$ and $\textrm{e}$, note the missing g.

draw a cube so that the x,y,z axes mark the centers of the faces. Place the 5 d orbital on the axes and diagonals as they are normally aligned. The four ligands in a tetrahedron are in corners of the cube, positioned so that they (obviously) form a tetrahedron. Then use the geometry to find the angles easily.

As you may want to calculate angles the d orbital functions in cartesians ( and without normalisation constants) are $Y^0_0=(3z^2-r^2)/r^2,; Y^2_{-2}=(x-iy)^2/r^2,;Y^2_{-1}=z(x-iy)/r^2,;Y^2_2=(x+iy)^2/r^2,;Y^2_1=-z(x+iy)/r^2$ where $ r^2$ is the distance to any point.


text:
Differences in using molten NaCl compared to aq. NaCl in a galvanic cell having copper anode and cathode
Since we know that copper anode will release $\ce{Cu^2+}$ ion, so I was wondering whether the cathode will thicken as copper has a lower position on the electrochemical series.
Based on my research, I found that some said that cathode will only thicken if the solution is stirred, some said that hydrogen gas will form.
Hence, are there any difference if electrolyte used is in a molten state compared to in an aqueous state?

At the cathode, metallic sodium will be created. At the anode, Cu will get dissolved.

Missing water in molten salt is quite a difference.

Be aware of distinguishing galvanic and electrolytic cells.

A galvanic cell with the same electrodes ( copper ones in this case ) is usable as a galvanic cell only if cathodic and anodic electrolyte differs in the concentration of electroactive components.
E.g. if we can use 2 different concentrations of $\ce{CuSO4}$ solution, separating them by a diaphragm or a salt bridge. The electrode with more concentrated solution around would become cathode with the more positive potential and vice versa.
Putting 2 copper electrodes into homogeneous solution of $\ce{NaCl}$ ( or molten one ) does not form a galvanic cell, unless some minor undefined differences create some residual voltage.
In case you did mean an electrolytic cell with moltan salt instead, anodic copper would be dissolving.
At the cathode, only metallic sodium would depose initially. Since some copper ions reach the cathode, copper would start priority deposition. Sodium deposition would depend on the applied electrode potential.


text:
Connecting a computer to Genesys 10S Spectrophotometer
Has anybody ever connected a computer to a Thermo Scientific Genesys 10S Spectrophotometer without their (costly!) software?  I'd like to pull absorbance data off of it to perform analyses.  There are several ways I could think of to do this:

A USB flash drive could be plugged into the front of the instrument, but none of my drives are recognized by the spectrophotometer.
The PC could be connected as an external printer, but I haven't found a way to receive the data without ThermoFisher's software.
The PC could be connected as an external printer, but set the instrument to believe it is a PCL format printer.  Could the data be intercepted and converted to a usable form?

I appreciate your input.

I recommend reading the manual. From the manual: " Only text (no graphics) can be output via the USB connection to a computer." The text is whatever is shown on the instrument panel. Not sure this includes the actual spectral data.

Your first idea cannot work, unless you write a firmware addon for the instrument. A daunting task. ;-)

I have no experience with Genesys 10S Spectrophotometer, but when it comes to transferring data with a specialized hardware on a lower level it often helps to use a USB 2.0 drive of < 4 Gb capacity formatted to FAT/FAT32.

user andselisk brings up an interesting point. You could try calling Thermo Scientific and ask about a USB data stick for the spectrometer. They sell them for other spectrometers.

ThermoFisher were the first people I called.  Their only suggestions were buying their software (~$625) or taking data on a flash drive.  They suggested Sandisk drives, but I hadn't thought about reformatting the drive I already have.  I'll try that...

This is a long comment....
The webpage specification says that there are three USB connections.

USB Type A port for USB memory device (front panel)
USB Type B port for optional computer connectivity (rear panel)
USB Type A port for external printer (rear panel)

The printer uses the HP PCL 3.0 specification via a USB cable. You should be able to get the HP specification and emulate that to get the data. The spectrometer doesn't know what it is communicating with...
The type A port on the front is supposedly for a USB stick for data storage. I'd guess that the Thermo Scientific software formats the stick in some screwy manner. You'll need the Thermo Scientific  specification to duplicate that. You might be able to get someone with the software to format some USB sticks for you. No idea what format the data format would be. I'd also guess that the Thermo Scientific software manual doesn't detail the format of USB data.
The USB Type B port on the back must have some bidirectional communication protocol. Unknown if Thermo Scientific would release details of the communications specification even if you bought the program. If you know someone that has such a computer instrument setup then you could install a sniffer in the USB communications line and figure it out. A lot of work.

I thought about emulating a printer today, but I don't know if it would send quantitative absorbance data.  Do you have any tips for printer emulation?

The HP specification would only enable the data transfer. I have no knowledge of what information would be printed. I'm sure that Thermo Scientific  could tell you what is printed, probably even given you a few printouts from a demo.

I'll go to them and see what I can find, then.


text:
How can reversible reactions be exothermic or endothermic?
So this may be a dumb question, but because the forward reaction of a reversible reaction releases the sane amount of energy as the backwards reaction absorbs (because the bonds that are formed from the forwards reaction are broken by the backwards reaction and vice versa) and at equilibrium both reactions occur at the same rate, surely in a reversible reaction the rate of energy release to the environment is equal to the energy absorbed by the environment. So how can there be any net loss/gain of energy to the environment like what is needed for an exothermic/endothermic reaction? I don't understand.

A reversible reaction starts with a chemical composition which is out of equilibrium. So the system releases some energy to the environment or absorbs energy from the environment.The chemical composition of the system slowly changes, and evolves towards the final state of equilibrium. At equilibrium, no heat is exchanged with the environment, and both reactions occur with the same rate.

Okay so its what happens before equilibrium is reached that releases/absorbs the energy. Got it. Thanks!

The gist? Reactions rarely start at equilibrium.

See $\ce{NO2}$ - $\ce{N2O4}$ demo, https://chemdemos.uoregon.edu/demos/NO2N2O4-Equilibrium-Demonstration# , to see how temperature affects equilibrium.


text:
Does 2‐indanone form a stable hydrate?
Following up the question Formation of a stable hydrate.
I understand that for cyclopropanone the change in hybridization is the reason for stabilization of hydrate since it relieves the angle strain.
So, can this be further be given as a reason to state the hydrate of 2‐indanone (structure below) will be stable? I suppose upon hydration the angle will change from 120° to 109.8°, bringing the angle closer to that of a pentagon that is 108°.

I am confused if this is appropriate or if it's true. If it's wrong, please explain.
P.S. After going through the comments I understood by considering keto-enol tautomerisation that hydration might not at all be favourable.

The cyclopentane ring is not perfectly flat.

Need to remember that this compound will be more extensively enolised


text:
Nomenclature of unusual nitrogen functional group
I am trying to create a proflavine derivative containing a 1,2,4-oxadiazole-5-one functional group, the first step produces an interesting functional group that I have no idea how to name:
$$\ce{[Pro]-NH-CH=N-OH}$$
I am trying to find literature on the above functional group so I can analyze proton NMR to see if I have actual created the above molecule.

Does the $\ce{-NH-}$ group in $\ce{[Pro]-NH-CH=N-OH}$ correspond to one of the $\ce{-NH2}$ groups of proflavine?

The HC=N-OH part is usually called an aldoxime, so I might say a "formamide aldoxime" if I were describing it to someone.

IUPAC aside, such compounds are usually called amidoximes.

According to Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name of the compound $\ce{H2N-CH=NH}$ is methanimidamide. The name formimidamide may be used in general nomenclature. The name formamidine is no longer recommended.
The letter locants ‘N’ and ‘N'’ are used to differentiate among the different nitrogen atoms. Thus, the compound $\ce{R-NH-CH=N-OH}$ would be named as N'-hydroxy-N-…methanimidamide.


text:
How can an electron being a wave have such property as spin?
Here's what I know about electrons.
Electrons have wave-like properties and the number of wavelengths in the $n^\text{th}$ shell is equal to $n(\lambda).$ Also, I read in my book that they have another property called spin.
But I couldn't imagine how a wave can have a property of spin. If this property is just a mathematical model, then is this property compatible for a wave i.e.  Is it appropriate to use this property of spin with waves ?

Note that this topic fits better to Physics SE site.

I will probably never understand why it's often suggested that the questions about electrons are better suited for Physics.SE and at the same time the questions about dimensional analysis for some reason are perfectly fine for Chemistry.SE.

In fact, spin is probably easier to define for a wave than for a particle. https://en.wikipedia.org/wiki/Angular_momentum_operator

@andselisk I would it narrow down to some questions about electrons. Behaviour of electrons in context of atomic and molecular  configuration is fine with me at CH SE. OTOH, nature of particle spin ? It is not so much about chemistry. It does not mean chemists cannot explain it. Some are surely familiar with spinors and Dirac equation ( not me ). About dimensional analysis, both physics and chemistry use many of shared and specific quantities, so it kind of belongs to both.

@orthocresol As Dirac equation is a wave equation, I would think so.

I guess some hard core physicists may claim chemists do not understand what the spin really is.

Assigning a classical physical meaning to spin is actually a big mistake that (unfortunately) many textbook makes. Spin doesn’t mean that the electron is a spinning ball.

@Greg I didn't mean to say spinning electron but a spinning wave.

@Ankit neither one

related https://chemistry.stackexchange.com/questions/139348/what-does-an-electrons-spin-of-0-5-and-minus-0-5-signify https://chemistry.stackexchange.com/questions/69805/schr%c3%b6dingers-equation-and-spin-quantum-number https://chemistry.stackexchange.com/questions/58020/do-electrons-really-spin

An electron is not a wave nor a particle. An electron is usually described as a quantum object with some wave-like properties and some particle-like properties.
Some of them, like particularly a spin, do not have direct counterpart in our familiar macro world. It is a kind of a mysterious, specifically quantum property of all elementary particles and  atomic kernels, that interact with angular momentum and interchange with it its values.
Non-relativistic quantum mechanics has no theoretical idea/model about the observed spin phenomena, that would come e.g. from resolving the Schrödinger wave equation. It was  added just as an ad hoc extension. OTOH, the relativistic Dirac wave equation brings spin as the natural part of the quantum mechanical model.
Note that the classical idea about a spin as a rotation related angular momentum of a fast spinning small ball is wrong. Considering the electron mass, energy and the limit of the maximum electron size, an electron simply cannot rotate fast enough to have such amount of classical angular momentum, related to its eventual spinning.
Both orbital and spin contributions to the total angular momentum have their counterparts as orbital and spin magnetic momentum, if quantum objects have nonzero electric charge or charge nonhomogenity (like neutrons ).


text:
How to synthesize 4‐[(4,4‐dimethylpentyl)oxy]‐2,2,8,8‐tetramethylnonane starting with 4,4‐dimethylpent‐1‐ene?

My attempt:

I don’t know how to add another chain. I think I probably need to react with another A molecule, but I don’t know the mechanism. Can anyone help me with this?

Divide your alkene starting material in two portions - one twice the amount of the other.
Take the larger portion and do an anti-Markovnikov addition of HBr to it using peroxide or light reference to give the terminal bromide (1-Bromo-4,4-dimethylpentane).
Take the smaller portion and ozonolyse it, working up with Dimethyl Sulfide to give the aldehyde (3,3-dimethyl butanal).
Take half the bromide material and form the Grignard with Mg metal. Add the Grignard to the aldehyde. Work up and isolate the resulting secondary alcohol. Dissolve the secondary alcohol in DMF at 0c and form the alkoxide with NaH, add the second portion of the bromide to the alkoxide solution. Work up, isolate and purify to obtain the desired product.


text:
Type of chirality in polycyclic compounds
Ignoring whether or not this molecule could exist, is there a name for this type of chirality? This molecule's mirror image cannot be superimposed on itself but there are no chiral centres.

Even though the bottom boron only has three substituents, the fact that they are all joined at the top gives it similar geometrical properties as if it were tetrahedral.
The InchI is InChI=1S/C4H8B2OS/c1-2-6-7-3-5(1)4-8-6/h1-4H2

Could it be an example of Axial Chirality? Where the axis is through the boron atoms?

how would boron exist in bridge head position? Is there a point in discussing chirality for something that may not even exist?

@safdar I'm sure you could substitute the atoms for something that could exist instead. I chose boron instead of carbon because trigonal planar atoms are inherently not stereocenters. The geometry is what matters.

Something like this? https://chemistry.stackexchange.com/questions/136695/r-s-configuration-of-bridging-carbon-in-bicyclic-system

This atom's environment wouldn't be trigonal planar - pyramidalisation with ring strain. Still it's planar chirality.

https://chemistry.stackexchange.com/questions/35470/why-is-trans-cyclooctene-chiral

You don't need atoms or bonds to define chirality. Just the colored balls will suffice. I think it is an example closer to planar chirality: P or Rp.  (C-C-B-S). http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/planar.html

About "but there are no chiral centres" : not clear if you are aware, but they are not needed for a molecule to be chiral.


text:
Effective core potential and Gaussian basis set
Is using the effective core potential (ECP) and Gaussian basis set simultaneously in ab initio calculation a valid approach? For example, is it correct to use ECP for one of the atoms and, instead of a basis set which is optimized for ECP, to take a normal Gaussian basis set (say, 6-31G) when we perform calculations for a molecule with a Hartree–Fock method?
P.S. I know technically I can do it in the quantum chemistry packages (for example, I have checked it in MOLPRO), but I don't know whether it is valid physically.

Using a basis set that is not designed for the effective core potential is not a good idea. The problem is that an ECP is designed to produce the effective potential of core electrons, and an all-electron basis set will include basis functions meant to model the core electrons. At best, those core basis functions are poorly optimized for your problem. At worst, you will run into SCF convergence problems.
I always recommend to use ECPs with basis sets designed specifically for them.

Thanks for your useful answer. You mentioned, the corresponding basis set is optimized for core electrons. So, If I add some diffusion function on those basis set (to improve the description of the continuum in my system), is it valid physically or not?

A normal all-electron basis set will include the minimum number of basis functions for the core electrons, but multiple basis functions for each outer valence orbital. Also possibly higher angular momentum functions (polarization) and low exponent functions (diffuse). If your all-electron basis set does not have diffuse functions it might be quite helpful to add them, as they often improve accuracy. Beware that diffuse functions often increase computational cost and cause SCF problems more than other functions, especially on condensed systems. That's why they are often not included by default.

Sure, it can be valid physically. You can use the full-electron basis set, like 6-31G*, and the ECP with associated basis set, like Lanl2dz, for instance, in calculations of transition metal compounds or compounds with heavier non-metals (Sb, Te, etc.) as an example.
Important here would be to compare results obtained with available experimental data (such as bond distances/angles, reaction energies, UV-vis spectra, etc.) or results of similar calculations in order to see if the simultaneous use of ECP and full-electron basis sets is valid and would not give you some weird results.


text:
Why does the titanium go to the tungsten filament in the crystal bar process?
I've tried to read a report by the US atomic energy commission [1], but couldn't really find the answer, or maybe I missed it.
Why is the gaseous titanium deposited on the tungsten wire/filament during the van Arkel–de Boer process?
Reference

Petersen, A. W. Preparation of Metallic Titanium by Film Boiling (Thesis); UCRL-2523, 4393936; United States, 1954. DOI: 10.2172/4393936.

$\ce{Ti + 2I2 <=> TiI4}$ is an endothermic reaction.  In the chamber, temperatures are high enough that there is $\ce{TiI4}$, but the tungsten filament is so hot that $\ce{TiI4}$ decomposes again.  Or, to cite the Dutch Wikipedia: forward reaction at $\pu{600 ^\circ{}C}$, backward reaction at $\pu{1200 ^\circ{}C}$ for a $\Delta{}H = \pu{−427 kJ/mol}$.  A bit like a halogen lamps, if you still have at your bicycle.

By "gaseous titanium", do you mean "titanium tetraiodide ($\ce{TiI4}$) vapor"?
Van-Arkel De-Boer process is a purification process of titanium and zirconium. Basically, what happens is that the impure metal, let's say titanium is heated in iodine environment at a temperature of $\ce{250 ^\circ C}$ to form volatile titanium tetraiodide ($\ce{TiI4}$) vapor. The impurities are left behind, as they do not react with iodine. This vapor is then  passed over a hot tungsten filament at $\ce{1400 ^\circ C}$ for which the vapor gets decomposed and pure titanium is deposited on the filament and is removed. The iodine is reused. The overall reaction is:
$$\ce{Ti _{(impure)} + 2I2 ->[250 ^\circ C] TiI4 ->[1400 ^\circ C][W filament] Ti_{(pure)} + 2I2}$$
The thermodynamics of this reaction is discussed in Buttonwood's comment.

Except that Buttonwood stated it upside down (although the point is clear, that the dissociation is endothermic)


text:
Determining the mass of a substance using calibration curve produced by HPLC
I hope everyone is doing well and safe! Yesterday, I was fortunate enough to be taught about how to read a chromatography calibration curve through this stackexchange! Special thanks to those who explained to me patiently and meticulously. Anyways, I've tried to solve some questions regarding this field but ended up no-where with this one. Could someone run me through the procedure of how to solve this question? Huge thanks in advance!!! <3 <3

Info: A $\pu{1.0mL}$ aliquot of a drink was diluted to $\pu{50.0mL}$ with deionized water. A sample of the drink was run through the HPLC to produce the calibration curve below. In this experiment, the mass of caffeine in the drink was determined by HPLC. The peak area obtained for this diluted sample was $2800$ arbitrary units.

Question: What is the mass of caffeine in grams, in a $\pu{330mL}$ can of this drink?

Answer = $\pu{825mg}$
P.S.: Could someone explain to me the significance of the peak area ($2800$ arbitrary units) that is obtained? And thank you for the constant support and help. Much love to this community!!!

The arbitrary peak area allows you to use the calibration curve. Since the calibration curve doesn't go up to 825 mg you must take the dilution into account.

Read my answer once again where I talk about running a sample in your previous question.

To calculate the peak area, you must be in front of the curve defining the peak. Usually it is a curve $y(x)$ starting at $y$ = 0 on the left, then suddenly growing up like up the hill in the middle of the page, where $y$ > 0, and going down to $y$ = 0 at the right part of the page. Now you look for the two inflexion points, one before and one after the maximum value of $y(x)$. You draw the tangent to the curve through these points. These two curves define a triangle with the base line $y$ = 0. You calculate the surface of this triangle, in $cm^2$ or in $mm^2$. That is the peak area.

Thanks you for the explanations. I really appreciate the help!!

Warning: This is also a homework question. However, this is more like mathematics (geometry) than chemistry, I decided to help you. Nonetheless, please keep in mind that we won't help you in all future perspectives, if you do not show your effort to solve the problem, regardless more mathematics involved.
As  M. Farooq pointed out that this problem can be solved using the answers to your previous question, specially my old-fashioned approach. Here is how you could do it. Analyze your graph thoroughly to find two different points of graph where it perfectly crossed the cross-lines:

For me, these two points are: $(35,2000)$ and $(60,3500)$. Therefore the tangent of the slope ($\alpha$) of line is:
$$\tan \alpha = \frac{3500-2000}{60-35} = \frac{1500}{25}=60$$
Now, take another arbitrary point in the graph. Say it is $(x,y)$. Then:
$$\tan \alpha = \frac{y-2000}{x-35} = 60 \ \Rightarrow \ y = 60x - 100 \tag1$$
The equation $(1)$ is the straight-line equation of the graph.
When $y = 2800$, $x= \frac{2800+100}{60} = \pu{48.33 mg/L}$. This is for the diluted sample. I think it is fair to leave you some work to do. So you can back calculate to get your answer. When I did, it came close to $\approx \pu{798 mg/can}$.
Note: The error ($3.27\%$) is due to the finding of the slope. Also there are few flaws in the question. The calibration curve is made with HPLC traces of known standard caffeine solutions (not using diluted samples of the drink). The peak area of $2800$ arbitrary units was obtained by the injection of diluted sample of the drink ($1$ to $\pu{50 mL}$ dilution).

Hello, thanks so much for taking your time to help me out. I really appreciate the help and it has definitely solidified my knowledge in chemistry. Again, thank you! Stay safe!

I was trying to connect the ties between the 48.33mg/L of the diluted sample to the 798 mg/can but I'm not ending up with your answers and i'm getting 1292.5 mg/can. :( Could you point out the steps that you would take. Sorry about the inconvenience.

I redid my calculations and it was very close to the actual answer. Could you tell me if my steps are correct or if there is another way of solving this? The dilution factor is 1:50 since a 1mL aliquot was diluted to 50mL with water. So I multiplied the 48.33mg/L by 50 = 2416.6 mg/L. Since the can contains 330mL, 2416.6 x 0.33 which leaves me with 797.445 mg/can?

The value $\pu{48.33 mg}$ is in $\pu{1000 mL}$ of diluted sample. Thus, find how much in $\pu{50.0 mL}$. That value is in $\pu{1.0 mL}$  of original solution. Then find how much in $\pu{330 mL}$. What you have got is correct, but calculation operation is wrong.

Ahhh i see. Thanks so much for the help!! Really appreciate it. :)


text:
Do both iodine and potassium iodide turn dark in the presence of starch?
I will do a titration to standardise a solution of sodium thiosulfate with potassium permanganate solution and potassium iodide solution.
My book says that when the potassium permanganate solution begins to lighten, add a few drops of starch indicator and titrate until the solution turns clear.
This seems to suggest to me that starch turns $\ce{I_2}$ dark but not $\ce{KI}$ as some of the $\ce{KI}$ solution would be still in the flask. Is this the case?

Yes, that's right.

@Ivan Great, so just to confirm, starch indicator does not turn KI solution dark ?

Yes, starch is an indicator for I2 and not for KI.

Starch is a long ribbon or a long filament made of a great number of glucose units attached to one another like the wagons in a train. But this filament is wound in a helicoidal way. And the inner part of this helix is a sort of long hole. And the dimension of this hole is just big enough to allow the iodine molecule $\ce{I2}$ to enter and stay there, maintained by van der Wals forces. But when the iodine molecule is inserted in the starch helix, it produces a "starch-iodine" complex which has an intense blue color.

Cool , I didn't know that. But what about KI ? Does that turn dark with starch ?

@Maurice Much as I respect your expertise and commitment, I can't help but notice that you've chosen an excessively long and convoluted way to say "yes, that's right".

@ Kantura. Starch and KI do not react with one another, and do not produce any color.

@ Ivan. You may be right. But I cant't prevent me from thinking that Kantura did not know how to measure the surface of a peak. And now he knows.

Thanks all , my question was answered and I welcome and appreciate the deeper understanding

At the molecular level, it is $\ce{I3-}$ and higher aggregates

I want OP to understand that iodine ($\ce{I2}$) is not the one giving the starch solution deep blue color. It is rather triiodide ion ($\ce{I3-}$), which would form in presence of excess iodide ion ($\ce{I-}$):
$$ \ce{I2 + I- <=> I3-} \tag1$$
This is an important aspect in iodometry titrations, because $\ce{I3-}$ is very water soluble compared to partially or not particularly water soluble $\ce{I2}$, which is one of the two major disadvantages of using $\ce{I2}$ as a titrant. The second disadvantage of using $\ce{I2}$ as a titrant is $\ce{I2}$ is somewhat volatile. Consequently, there is a significant amounts of dissolved iodine from the solution by evaporation (sublimation?). Therefore, adding enough iodide ($\ce{I-}$) to iodine ($\ce{I2}$) solution would make it to overcome both of these disadvantages by the reaction in equation $(1)$ (i.e. in the presence of $\ce{I-}$, $\ce{I2}$ reacts to form $\ce{I3-}$, which is highly soluble, and most importantly, not volatile). In addition, $\ce{I3-}$ ions give the needed deep blue color with the indicator, the starch solution.
For example, the color changes in one perticular iodometric titration is depicted in following image:

Keep in mind that when you have known concentration of potassium permanganate solution and excess potassium iodide solution together in acidic medium, it is actually become a $\ce{KI3/KI}$ solution. The major chemical species present in the solution is $\ce{I3-}$:
$$ \ce{MnO4- + 8 H+ + 5 e- <=> Mn^2+ + 4 H2O } \tag2$$
$$ \ce{2I-  <=> I2 + 2e-} \tag3$$
Sum of $2 \times (2)$ and $5 \times (3)$ gives:
$$ \ce{2 MnO4- + 16 H+ + 10I- -> 2Mn^2+ + 5I2 + 8 H2O } \tag4$$
The excess of $\ce{I-}$ react with freshly formed $\ce{I2}$ to give:
$$ \ce{I2 + I- <=> I3-} \tag1$$
Dilute triiodide solutions are yellow (as shown in $(b)$ in the image), more concentrated solutions are brown, and even more concentrated solutions are violet (as shown in $(a)$ in the image). If you add starch solution at the beginning, excess $\ce{I3-}$ would destroy the starch structure. That's why you need to titrate dark color to yellow color with thiosulfate solution first before the addition of starch. That time, $\ce{I3-}$ concentration is dilute enough, yet give a dark blue color by making the $\ce{I3-}$-starch complex (see the insert in the image at right hand side). The end point would be dark blue to very pale pink becace of the presence of $\ce{Mn^2+}$ ions (not colorless as shown in the image).


text:
Does davis safety lamp detect explosive gases in ship compartments?
I feel managers at Bangladesh ship breaking yard are not properly caring for safety of workers. Many workers die in explosions while salvaging metal.  documentary available  on YouTube by National Geographic Channel.
My question is does Davis safety lamp work in compartments  of a ship since it detects explosive gases in coal mines.

A safety lamp does not necessarily detect explosive gases... its fine screen just prevents propagation of the flame from inside to outside. There are cheap VOC detectors, e.g. https://www.amazon.com/methane-detector/s?k=methane+detector .

It deflects rather than detects. Also yes, it should work in ship compartments. But wait... is that the primary reason for the explosions? Do the guys in Chittagong use open fire for lighting? I highly doubt that.

Sure, and you can't put them behind the Davy's safety mesh. So the said device is hardly of any help.

@Dr.Moishe " miners learnt to show the presence of firedamp by looking for changes in the shape and colour of their candle flames. For example, on approaching danger, the tip changed from yellow to greenish-blue". Secondly, $30 is a huge amount in bangladesh. They can't send every gas cutter with a voc

Early 19th century safety technology is a bad substitute for early 20th century workplace safety regulations. And yes, those companies can afford proper ventilation and gas detectors. It's just a matter of establishing and enforcing regulations.


text:
Is is valid to compare reaction quotient to $K_p$ as well as $K_c$?
Can I use the reaction quotient compared to $K_p$ to answer the following question?

The solid $\ce{XY}$ decomposes into gaseous $\ce{X}$ and $\ce{Y}$: $\ce{XY(s) <=> X(g) + Y(g)}$ with a $K_p = 4.1$ at $\pu{0 ^\circ C}$. If the reaction is carried out in a $\pu{22.4 L}$ container which initial amounts of $\ce{X}$ and $\ce{Y}$ will result in the formation of solid $\ce{XY}$?

a. $\pu{5.0 mol}$ of $\ce{X}$, $\pu{0.5 mol}$ of $\ce{Y}$
b. $\pu{2.0 mol}$ of $\ce{X}$, $\pu{2.0 mol}$ of $\ce{Y}$
c. $\pu{1 mol}$ of $\ce{X}$, $\pu{1 mol}$ of $\ce{Y}$
d. none of the above
If I had been given $K_c$, I could put the different answer choices into the equilibrium expression and obtain $Q$. If $Q$ turned out to be smaller than $K_c$, then that would suggest that products would be favored. Contrariwise, if $Q$ was larger than $K_c$, then reactant(s) would be favored. So, I would simply see if any of the choices resulted in a $Q$ larger than $K_c$.
But, I was not given $K_c$. Must I convert $K_p$ to $K_c$ before making the comparison with $Q$?
I suspect the answer is "yes" (i.e. I must do the conversion) because (1) the respective values are very different and (2) I get different answers to the question depending on which value I use. This leads me to believe that I cannot compare $Q$ with $K_p$. Any additional insight into why this might be the case would be appreciated.

$K_p$ has a unit. It may be in bar, or in atm, or any other pressure unit. What is it ? $K_c$ has also a unit. Which one ?

You do not need the sought conversion. This is not hard as you think it is. Follow the concept.

$K_p$  should not have units, its log could never be taken if it did. In old texts units are used but this habit was changed years ago by dividing as necessary by 1 unit of concentration/pressure etc . (In practice this means just using numerical value)

Since you are given the volume, temperature and initial amount of each gas, you can compute initial partial pressures as $n_iRT/V$ and from these the product $Q_p=p_X\cdot p_Y$ and compare this to $K_p$. If $Q_p>K_p$ then solid will form:
a. $\pu{5.0 mol}$ of $\ce{X}$, $\pu{0.5 mol}$ of $\ce{Y}$
$Q_p=2.50 \rightarrow$ no solid is formed
b. $\pu{2.0 mol}$ of $\ce{X}$, $\pu{2.0 mol}$ of $\ce{Y}$
$Q_p=4.01 \rightarrow$ no solid is formed
c. $\pu{1 mol}$ of $\ce{X}$, $\pu{1 mol}$ of $\ce{Y}$
$Q_p=1.00 \rightarrow$ no solid is formed
I assume that the pressure unit is atmospheres.

I thought this might be the case when I first did the problem. But then I had second thoughts since the "test" values given were in moles and not pressures (and therefore I thought I needed to use $K_c$). This confusion is why I asked the question. It seems, from your answer, that I CAN use $K_p$ and compare it with Q. Thank you. Also, thank you M. Mahindaratne, because your answer shows that, in this case, the pressures are equal to the number of moles.

The reaction of interest is:
$$\ce{XY(s) <=> X(g) + Y(g)} \tag1$$
Thus, $K_p = P_\ce{X}\cdot P_\ce{Y} = 4.1$ since $P_\ce{XY} = 1$ in given condition because it is a solid. If we assume $\ce{X}$ and $\ce{X}$ are real gases, the given conditions are such that $P_\ce{X} = \frac{n_\ce{X} RT}{V} = n_\ce{X}$ and $P_\ce{Y} = \frac{n_\ce{Y} RT}{V} = n_\ce{Y}$, no matter what units are used for $P$.For example:
$$P_\ce{X} = \frac{n_\ce{X} \ \pu{mol} \times  \pu{0.082 L atm K-1 mol-1} \times  \pu{273.15 K}}{\pu{22.4 L}} = n_\ce{X} \ \pu{atm}$$
Suppose, initial amounts of $\ce{X}$ and $\ce{X}$ are $x$ and $y$ and amount of $\alpha$ reacted to produce $\alpha$ amount of  $\ce{XY}$ at the equilibrium (to be realistic, $\alpha \gt 0$). Thus, at the equilibrium: $P_\ce{X} = x- \alpha$ and  $P_\ce{Y} = y- \alpha$. Therefore:
$$K_p = P_\ce{X}\cdot P_\ce{Y} = (x- \alpha)(y- \alpha) = \alpha^2 - (x + y) \alpha + xy = 4.1 \tag2$$

If $x = 5$ and $y = 0.5$: From the equation $(2)$, $$ \alpha^2 - 5.5 \alpha + 2.5 = 4.1 \ \Rightarrow \ \alpha^2 - 5.5 \alpha - 1.6 = 0 \\ \therefore \ \alpha = \frac{5.5 \pm \sqrt{5.5^2 + 4 \times 1.6}}{2} = \frac{5.5 \pm 6.05}{2} $$
Thus, $\alpha = 5.78$ (not realistic) or $\alpha \ne -0.28$ (because $\alpha \gt 0$). Therefore no reaction.
If $x = 2.0$ and $y = 2.0$: Again, from the equation $(2)$, $$ \alpha^2 - 4 \alpha + 4 = 4.1 \ \Rightarrow \ \alpha^2 - 4 \alpha - 0.1 = 0 \\ \therefore \ \alpha = \frac{4 \pm \sqrt{4^2 + 4 \times 0.1}}{2} = \frac{4 \pm 4.05}{2} $$
Thus, $\alpha = 4.025$ (not realistic) or $\alpha \ne -0.025$ (because $\alpha \gt 0$). Therefore no reaction.
If $x = 1.0$ and $y = 1.0$: Similarly, you can prove that there is no reaction.

Hence, answer is "$(d)$. none of the above."


text:
How does a shock, such as the snap of the disk in a reusable hand warmer, trigger the re-crystalization of a supersaturated solution?
Reusable handwarmers are made of something dissolved nearly to its saturation point in hot water (Sodium Acetate in the case of HotSnapz) at a high temperature, and then cooled to a temperature at which some would normally be insoluble, but remains dissolved due to a lack of a site to grow a crystal from. When the crystal finally grows, triggered by popping a little disk, the process is exothermic and it releases heat.
What I'm wondering is: how does popping the little metal disk inside the hand warmer allow nucleation to begin?
EDIT: I got a bit more information, but nothing definitive. What I’ve heard is that the metal disk may either be initiating nucleation by releasing tiny metal particles when it snaps or by snapping quickly enough to cause cavitation. The first seems unlikely, because if bits of metal were released and functioned as nucleation sites, the warmer wouldn’t be reusable - the crystals could just reform on those metal bits. Cavitation seems more plausible, though I still don’t understand how that would allow crystallization to begin.
EDIT 2: I did some experimenting, and a whack also starts nucleation, not just the disc snapping. So, any shock can cause the sodium an acetate ions to come out of supersaturated solution. How?

I am sure the thermodynamics of supersaturated solution must be complex. In older times in order to initial crystallization in a beaker, one would slightly rub the beaker with a glass rod. A single scratch or a microscopic (invisible) glass particle would initiate crystallization. Similarly, I think, bending the metal disk provides new nucleation sites on its surface.

Recall superheated water in a microwave which is not willing to boiling, the moment you add a spoon or something solid it boils violently. It must be a similar phenomenon.

Related: https://www.pnas.org/content/113/48/13618

I don't have first hand experience with super-saturated solutions, but could it be that they precipitation would occur on stirring? The snap could be considered a vigorous quick stir.

Edit: Contribution of enthalpy and bond formation
Edit 2: Changed the description of entropy
After thinking for sometime, I realized that I neglected to mention three things:

The Driving force of precipitation reactions depends on the charge and size of the ions which corresponds directly to the magnitude and order of the hydration shells formed upon dissolution of the salt.

The enthalpy as a driving force of the precipitation of the salt. (Remember the Gibbs Free energy equation has two terms, entropy and enthalpy)

Additionally, the formation of ionic bonds can be lower in energy (more stable) than the electrostatic interactions present in solution. The release of energy during salt formation is transferred to the surroundings as heat. This would be a negative enthalpy (heat flowing out of the system) and if large enough, would overcome any decreases in entropy stemming from the formation of highly order crystalline structures and drive the precipitation.

In this case it is probably the enthalpy term that is driving the reaction, since you are putting a ton of heat into the system to dissolve the salt, once the solution cools, there is a lot of potential energy that is stored in the position of the atoms in the system. This is then released as heat upon the formation of the crystal structures.
Entropy changes in precipitation reactions:
In order to make a supersaturated solution, you have to add energy to form hydration shells around the ions (entropy in the system is decreasing). The thing to remember is that these hydration shells decrease the dispersal of the ions relative to that in the crystalline structure for some salts. When you heat and mix the solution with excess solute, you are providing sufficient energy to form hydration shells. When the solution cools (Slowly), the hydration shells remain intact, however, the system is unstable and applying enough energy to the solution will disrupt the hydration shells and crystallization of the salt will occur. The large amount of heat that is released is coming from the increase in the micro-states available for the ions (gain in entropy) in removing the hydration shells around the ions. The hydration shells are less dispersed than the crystalline structures of the salt and that is why entropy increases during some crystallization reactions, making this process exothermic.
Here is why the disc snapping is important:
Even though this process is very exothermic, it still requires activation energy. The mechanical force you are referring to (i.e. snapping the disc or whacking the heating pad) is doing just that by simply transferring energy to the molecules, through vibrations, which provides the activation energy for crystallization.
Just as walking can cause energy transfers from the mechanical forces of your legs to the molecules of the floor (i.e. through friction) and increase the temperature of the surface, some of the force applied on the disc transfers to the solution and if it is enough, will overcome the activation energy barrier of the process and crystallization will begin.

Interesting that the supersaturated solution - a liquid - is more ordered than NaOAc.3H2O crystals plus some saturated solution.

@JamesGaidis I thought about this specific case and realized that it is probably the enthalpy that is the driving force since sodium is a non-acidic cation with a relatively small Z^2/r value. However, I would refer you to this link if there is any doubt about entropy as the driving force for other precipitation reactions. https://people.wou.edu/~courtna/ch412/ppt2.htm

Interesting link! But entropy doesn't drive, it is an explanation, just like enthalpy is a result. What two forces, or rather what force plus what blockage are operating here? NaOAc .3H2O crystal structure has Na+ surrounded by 6 oxygens in a distorted octahedron (https://onlinelibrary.wiley.com/doi/abs/10.1107/S0567740876002367). But hydrated Na+ is all over the place, from 4 to 8 waters (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3250073/). Cooling a solution must enable a slide into one metastable (yet fluid) structure rather than the more stable crystalline form.

@JamesGaidis indeed, entropy isn’t a literal force. I agree, the term driving force is misleading. I also agree with what you said about slow cooling to a meta stable structure. Just to double check though, by blockage, are you referring to the actual forces that are preventing this process from having no barrier to proceed? (I.e. 0 activation energy)

Ah, I hadn't thought about the hydration shells. That makes sense - a shock would be enough to disturb the water molecules surrounding the ions. Thanks for the explanation!

How are entropy and order related? There are several articles in the Journal of Chemical Education which discuss that entropy and order have nothing to do with chemical processes. This order related entropy is information  entropy coined by Shannon. Unfortunately, all the gen. chem. books repeat this garbage.

@dval98: Sure. In the metastable state, electrostatic forces hold the ions in place. If you apply much smaller forces to the system, it returns to its equilibrium. If you apply a much larger force to rearrange a few ions, they may form the more stable solid, causing a gap or strain, which causes more rearrangement to ultimately form the stable solid. It is fascinating to watch the supersaturated solution crystallize apparently slowly, leisurely, but imagine the trillions of ions playing musical chairs under the direction of electrostatic force!

@M.Farooq I apologize for the inaccuracy of my answer and will be corrected tomorrow morning. I take responsibility for propagating such “garbage”. From my current understanding, Boltzmann would describe it as an increase in micro-state’s of the system. Even though some processes may lead to more disordered systems, there isn’t really a way to quantify it and it is not necessary for disorder to increase to have a positive entropy?

@M.Farooq As for SMI, the increase in entropy is an increase in the information needed to define the positions of the particles? Please correct me if I’m wrong or maybe send a link to one of those articles.

@dval98, Thanks for your message. Yes, please talk in terms of microstates. My understanding on entropy is very little but all I know that all what is taught from high school to university is hand waving. Neither the teachers understands it nor the student.

See https://pubs.acs.org/doi/abs/10.1021/ed076p1385, for example. Frank Lambert passed away but he was a big critic and many authors correct this "messy room", shuffled cards" stories fed to students.

@M.Farooq thank you. I really appreciate this. Part of me trying to answer others questions is also an attempt to expand my knowledge.

@dval98, Entropy is indeed a complex topic and very challenging. I tried to read Clausius original work (translated from German) but it is so engrossed in hardcore heat engines and mechanics that I gave up. One needs a very good grasp of thermodynamics and mathematics in which chemists are generally weak. It is certainly is not what is taught today. Part of me is now discovering how much junk was fed to us. This is why you will see me questioning the very basics once again-so that I can correct myself.

The problem with crystallization of a supersaturated solution is that all molecules are equal. No particular molecule is "entitled" to be the first to start the first crystal. An external event is needed to "nominate" the place of the first crystallization. This external event can be an impurity, a sudden shock, an irregularity at the surface of the container. But once the first face of the first microcrystal is created, all other molecules "know" where and how to develop the crystallization process.

What physically happens to the molecule that allows it to become the first mircocrystal?

In the aqueous solution, the ions are surrounded by adsorbed water molecules. These molecules must be removed for making the first arrangement of ions like in the crystal. A sudden shock may be enough to expel for a while these adsorbed water molecules around a given ion. Once the first line of ions is done, the rest is following.


text:
Chemical difference between hand sanitizer vs hand wash?
Is there any major difference between hand sanitizer and hand wash?
Is it safe to use hand wash only alone, without use of water?

Major difference is sanitisers need not water as they need not to be washed away.

The major practical difference is that sanitizers need not water as they need not to be washed away.
Hand sanitizers are usually ethanol or isopropanol, diluted by water and enriched eventually by a minor amount of active disinfectant and/or thickeners. When applied, part of it evaporates, rest is absorbed by skin.
Hand wash is similar to shampoo, shower gel or liquid dish washer. All of them are basically concentrated solutions of detergents, anionic ones ( typical example sodium laureth sulphate ) , cationic ones ( like cocamidopropyl betaine ) and non ionic ones. Different product classes differ in the detergent concetration, their mutual ratio and in minor and auxiliary components.
Long skin exposure to the undiluted hand wash is not desired. It is not some dark top secret of cosmetical companies. Prolonging any good effect may eventually lead to turning it into a bad effect. Long skin exposure to water is not desired either.
Practically and short term, who wants sticky fingers ? From the chemical POV and longer term, skin maceration by concentrated detergent causes extreme skin fat extraction and various related and/or consequent effects that I am not going to comment as a chemist.
All is also a question of dilution. More diluted the hand wash is, more the hand wash properties converge toward the sanitiser in context of stickyness and fat agressivity. OTOH, the sanitizing effect of hand wash would decrease with dilution. Chemists do not dare to define the boundary of yet sufficient sanitation effect.
The aggressiveness of the detergent effect usually decreases in the order

liquid cloth wash
liquid dish wash
hand wash
shampoos and shower gels
shower gels with fatty components, having therefore a body lotion feature.

Probably nobody would say using undiluted dish washer as hand sanitiser substitute is a wise thing to do. Hand wash is significantly less harmful, but still...

Long skin exposure to the undiluted hand wash is not desired - is it? I never came across such a warning. Could you explain more about this / state a reference?

I do not have any handy reference, but it is not a dark top secret. Long skin exposure to water is not desired either. Anybody can do simple experiment on his/her own hands. Note that initially, there were no warnings against putting cats into microwave oven to dry them either. But some less wise Americans forced many vendors of many devices to create multipage list of what you should not do.

From UCLA Health1:

Right now, the CDC recommendation is to wash hands whenever possible.
Hand sanitizer is more precise at killing bacteria and most viruses,
but hand washing can effectively remove all dirt, microbes and
chemicals on the hands. All elements of hand-washing are key—friction
for removing debris, soap to emulsify dirt, chemicals and microbes and
running water to remove the debris.

And Deb Blog2:

When considering the effectiveness of hand washing and sanitizer use,
we need to recognize that human pathogens and opportunistic pathogens,
exist in multidimensional continuum in terms of survival, chemical
resistance to antimicrobial compounds, and efficacy of hand washing
etc.  Because each microorganism exists in their own multifaceted
conditions for existence and survival, there can be no “one size fits
all” approach to control and prevention of disease in respect to hand
hygiene.
At one end of the hand washing efficacy continuum are viruses that are
harder to remove from hands than bacteria and the even larger parasite
cysts and eggs.  This is because size matters.  The smaller the
pathogen, the more easily trapped in the cracks and creases of our
epidermis and our stratum corneum.  Here we need to realize that
aggressive or “good quality hand washing” is more critical in going
after viruses as opposed to our usually larger prey, bacteria.
Multiple studies indicate that in addition to surfactant action,
friction (good old fashioned elbow grease) at the rinsing stage is
perhaps the most critical period of hand washing in its ability to
dislodge viruses from skin surfaces. That said, you only have so many
highly aggressive hand washings (hand washing events) a day to remove
viruses before hands begin to dry out and become sore, even when using
the gentlest of soaps.
Hand washing men and women. In another continuum of antimicrobial
action we have organisms at one end like C. difficile whose spores are
entirely resistant to alcohol hand sanitizers (as discussed in the C.
diff blog).  And indeed, norovirus is not far from C. diff in terms of
chemical resistance. To protect hands from these compounds we must
rely on use of the proper gloves where appropriate. For norovirus
generally considered fairly resistant to chemical sanitizers there is
accumulating evidence that efficacy of high tech sanitizers is as good
as an average or substandard hand wash.  This should be kept in mind
when trying to balance skin health with risk reduction and frequent
aggressive hand washing has begun to degrade skin barrier function.
Fortunately at the opposite end of the continuum of antimicrobial
action are a very large number of microorganisms, bacteria, viruses
and fungi that are easily killed by hand sanitizers. Here hand washing
while still necessary, is less important when no visible soil is
present and hand sanitizers can be relied upon not only to kill these
bugs but also provided much needed emollients back to the skin
surface. Skin actually becomes healthier the more the latest
generation of hand sanitizers are used.
Remember, use of sanitizer in healthcare and food venues was
implemented to protect hands from the potentially damaging effects of
too frequent hand washing (and drying).  We are not always going to
know which microbial hazard we are up against.

References:

https://connect.uclahealth.org/2020/03/10/handwashing-vs-hand-sanitizer-whats-the-difference/
https://info.debgroup.com/blog/bid/270757/hand-washing-or-hand-sanitizer-which-is-better-and-why

For a detailed analysis and description, check out this:
Andreas F. Widmer, Replace Hand Washing with Use of a Waterless Alcohol Hand Rub?, Clinical Infectious Diseases, Volume 31, Issue 1, July 2000, Pages 136–143, DOI: 10.1086/313888


text:
Aluminum powder from aluminum sulfate?
I recently did the experiment where you drop a piece of iron in an aqueous solution of copper sulfate. After a while, the piece of iron gets coated with a fluffy coating of copper that can be dried and mashed into a powder. Apparently this happens because iron is more reactive than copper, according to the reactivity series.
Can I do the same thing by making an aqueous solution of aluminum sulfate, then dropping a piece of magnesium metal in? It should work because magnesium is above aluminum on the reactivity series, right?

No, you can't. $,$

Metallic aluminum cannot be obtained from a solution of its salts. It is because pure aluminum reacts with water, a little bit like sodium. Of course a piece or a sheet of aluminum does not usually react with water, because the metal is covered by a thin, colorless and waterproof layer of aluminum oxide, which prevents the contact between aluminum and air. And if you remove this layer with a knife, the oxide layer will immediately be remade by air.
The only way of showing that aluminum reacts with water is to dip the aluminum piece in a solution of Mercury chloride. There will be a surface reaction like $$\ce{2 Al + 3 HgCl2 ->  3 Hg + 2 AlCl3}$$ Then the mercury Hg will make an alloy with aluminum. And this alloy cannot be covered by the alumina shield. So such a piece of amalgamated aluminum will be quickly oxidized in air, and will produce long white filaments of $\ce{Al2O3}$ or $\ce{Al(OH)3}$ if abandoned in air for a couple of minutes due to one of the following reactions $$\ce{4Al + 3 O2  -> 2 Al2O3}$$ $$\ce{4 Al + 3 O2 + 6 H2O ->  4 Al(OH)3}$$ This production of alumina filaments is very interessant to observe. Of course it is not allowed any more today, because of the toxicity of mercury.

I did another experiment with copper sulfate where are you stick two copper rods in an aqueous solution of copper sulfate. You apply an electric current and copper metal builds up on the cathode. I noticed that the copper anode dissolved. Does the dissolved copper from the anode just turn into more copper sulfate?

@ Electronic Noob. What you have done is exactly the industrial method for purify impure copper. This impure bloc is at the anode, and pure copper gets deposited at the cathode. The other metallic impurities fall on the ground of the flask, where they form a deposit that can be eliminated later on.


text:
How to experimentally determine the chemical potential of salt water?
I have NaCl dissolved in water. How do I find the chem potential?

It depends of the amount of oxygen gas dissolved in the water.

@Maurice what do u mean? how do i measure this?

@Maurice i thought i could get chemical potential by dividing enthalpy by number of moles

Sorry. I was thinking of the redox potential, when I spoke of dissolved oxygen.

The chemical potential in what regard? Freezing point of solution, vapor pressure of water, conductivity, redox potential of $\ce{Cl-}$. and so on. Are you considering oxygen and carbon dioxide in the water? "Chemical potential" needs some context.

@MaxW sorry, i refer to freezing point of solution. the chemical potential of salt water.

A simple aqueous salt solution contains water and salt. One does not generally discuss the "chemical potential of a solution" but rather that of the individual components. Chemical potential is defined by the following equation
$$\mu _i = \left(\frac{\partial G}{\partial n_i}\right)_{p,T}$$
where G is the Gibbs free energy and $n_i$ is the amount of substance i in the system (the solution).
One may want to discuss the Gibbs free energy of the solution, which can be related to the composition and chemical potential of individual components as follows:
$$G = \sum_i \mu_i n_i $$
The chemical potential requires a reference scale, so is usually reported as a difference relative to that of pure water, and in lieu of potential one can report the activity since there is a simple relation between the two ($\mu_i \propto RT\ln(a_i)$).
Water activity can be derived from the partial pressure of water vapor in equilibrium with a sample. The activity is equal to the ratio of the partial pressure for the sample to the partial pressure of pure water at the same temperature. There are a variety of methods that can be employed to determine the partial pressure of water, see for instance the Wikipedia.

i wanna find the chemical potential of salt water

@RazeXIX I congratulate you on this ambitious goal, but it might help to understand a little better exactly what you are trying to accomplish. Measuring the chemical potential is not exactly like counting beans.

what other information do i provide oof

@RazeXIX What are you going to do with the "chemical potential" values? If you can explain this it would go some way to explaining what it is you actually want to know. My guess is you are looking for either the chemical potential of the water in a salt solution, OR the chemical potential of the ions.

I am looking at freezing point depression, dissolving salt should reduce the chemical potential of the solution

@RazeXIX When you look at colligative properties such as freezing point depression it is the chemical potential of the solvent that is of interest (or rather, for an ideal solution it is what you measure). So if you are measuring freezing point depression in water, then you would be interested in the chemical potential of water.


text:
What do you call different formations of polycyclic molecules, such as the 'bent' phenanthrene vs. the 'linear' anthracene?
Are there terms, IUPAC or not, that describe multiple carbon rings fused in different shapes, like the aforementioned anthracene and phenanthrene?
Also, I thought I read about a term describing carbon rings that only attach at one carbon, as opposed to two....

Anthracene and phenanthrene are isomers. I am not aware of a more specifiс term, though it might  exist. Rings with one common atom are called spiro compounds (as in "spiral"; that's not somebody's name).

You could say which type of isomer they are, so constitution isomer would be somewhat more specific. But I‘m not aware of something more specific than that, either.

Naphthalene, Phenanthrene, Anthracene all are "fused"-Bicyclo/Tricyclo rings

The answer to the final line is: spiro.


text:
What does A. R. in A. R. sodium nitrite mean?
I'm reading about a reaction from a 1970s organic chemistry book. It references A. R. sodium nitrite. What does the A. R. mean?
I see from online that "Ar" means aromatic ring. Is that the same thing?
But I've also seen "Sodium Nitrite, Granular, AR®" with a registered trademark, implying it might be a specific brand of Sodium Nitrite, but that didn't make sense. https://www.chemical.net/sodium-nitrite-acs
Are there different types of sodium nitrite?

Can it stand for Analytical Reagent, an equivalent to P.A. ( Per Analysis ), as the substance purity level ?

A.R. probably Analytic Reagent as a mark of purity. Maybe later on it became a trade mark.

This. AR = Analytical Reagent

I don't know if I have completely answered, but here's my go:
AR : (ANALYTICAL REAGENT) : Reagents essential for analytical purpose and research work having high purity. If reagent meets the requirements of the American Chemical Society Committee on Analytical Reagents, it will be denoted as an AR (ACS) reagent.
Also see:

https://www.researchlab.in/?product-details/grade-of-purity/
https://www.avantorsciences.com/pages/en/grade-definitions#:~:text=AR%E2%80%94The%20standard%20Macron%20Fine,an%20AR%20(ACS)%20reagent.

EDIT:And note that it isn't a Analytical ring

Thank you very much for setting me on the right foot. I was previously leaning towards the ring thing, but didn't think it really made sense.

@GFL Welcome..Anytime!