text:
What additives can make an ethanol flame incandescent? (Historical setting)
How can I make the flame in a wicked lamp incandescent when using azeotropic ethanol (95% ethanol with 5% water) as the primary fuel in a 'hurricane lamp'? (Ethanol burns with an almost invisible flame. It is not suitable as a lamp-fuel for light.)
Would incandescence be improved by dissolving a little table salt in the azeotropic ethanol?
If yes, and if sodium ions in the flame are emitting radiation in the yellow part of the visible spectrum, what happens to the chlorine ions? Would that be an unsuitable solution from a chlorine toxicity viewpoint?
Or would incandescence be improved by dissolving a little tallow in the azeoptrope?
The problem is to be solved in mid-nineteenth-century setting in which ethanol is freely available and kerosene, paraffin and turpentine are not available.

I added a little sodium chloride to my homemade alcohol lamp and it looked like this: https://chemistry.stackexchange.com/a/164168/79678. The chlorine goes off in the exhaust or reacts with flame species. Interesting challenge.

In case of the ultimate goal is the light itself, there are camping lamps based on LPG flame and thorium dioxide(?)  fabric, that makes the shining. It could be used for ethanol too.

Thanks, Poutnik, but the setting is nineteenth century.

How about naphtha? The recipe of Greek fire is long lost, but naphtha  has been suggested as a plausible ingredient. Pine oil as well.

@Ed V, Thanks for pine oil idea. But I wonder whether it would be soluble in ethanol/water base-fuel? (Ethanol is feely available in the historical setting. Alcohols are the driver, except for the almost invisible flame if used as fuel without additives.)

I don’t know about the solubility of pine oil in the ethanol/water, but an emulsion might do the trick, kind of like modern day canned Sterno or napalm.

IHmmmmmmmm. I wonder whether an emulsion would suffer separation while moving up the wick. For the purpose of my nineteenth century setting, maybe that's of no consequence. Thank you.

Yes, a very little $\ce{Na+}$ goes a long way to make a bright yellow flame. Little would dissolve in EtOH with only 5% water, but that little should be more than sufficient.
Though the chlorine likely would react to a very small extent to form halogenated hydrocarbons, pre-1900, houses were quite drafty, and I don't believe much would accumulate, unless the lamps were used in closed-end tunnels or mines. Another issue with burning EtOH is that incomplete combustion produces acetaldehyde. Hot brass fixtures could act catalytically, creating larger amounts.
Consider "wood alcohol" (methanol) as a fuel, too. Due to tax laws, it is usually cheaper than ethanol, and was made by destructive distillation of wood waste. (Originally, I had misread the question and assumed that would be the fuel of choice.)
Also consider the limelight, which approached carbon-arc lamp brightness and intensity. "During the American Civil War in July and August 1863 calcium lights were used during the siege of Fort Wagner." Perhaps EtOH + forced air would be sufficient to get that lime glowing, though not as intensely as in oxy-hydrogen flame.

If the alcohol is 140-proof liquor (70%alcohol by volume) containing some fusel oil,  would that also burn in a Hurricane lamp? If yes, would that (with some NaCl) produce a brighter flame because of the presence of the higher alcohol?
Any comment on my tallow idea?
Thanks in advance.

Tallow might work -- you could easily test it at home. BTW, the wealthy, in the 19th century, used whale oil, or fish oil ("hooligan oil" among the Northwest Native American (First Nations, if you prefer). https://en.wikipedia.org/wiki/Whale_oil#Applications and https://www.fws.gov/story/salvation-fish See also "conspicuous consumption".

Thanks, DrMoishe. That adds to my body-of-knowledge.


text:
Integration of Artificial Intelligence in Chemistry
How is Artificial Intelligence impacting the design and discovery of new compounds and medications?
Given the rapid advancements in artificial intelligence and machine learning methodologies over the past decade, how are these computational tools being integrated into the realm of chemistry, specifically in the design and discovery of novel compounds and medications?
Moreover, how do these AI-driven techniques compare in terms of efficiency, accuracy, and innovation to traditional human-driven research methods? And, what potential challenges or ethical considerations arise when relying heavily on AI for such pivotal and impactful scientific endeavors?

In context of pre-synthesis computational screening of potential drug candidates, AI starts to compete with quantum chemistry computations.

Seems a bit broad in scope. You may want to find a more specific question that will not require an essay to answer. Questions regarding ethics are not generally for this site but you may look for a more specific question. Potential challenges sounds a bit broad and fuzzy, do you have any particular examples in mind?

@Poutnik I'm curious, do you have an example or news article for that? I know that ML approaches have been used and developed for pre-screening purposes in the last millennia. I wonder where the AI came in. As far as I know, this has never extended (practically) to the realm of QC computations as it is too costly. Many semi-empiric and dft-substituting approaches have been developed to tackle the inefficiencies. So I'm really curious where this is heading. Given the verbosity though, this is better discussed in [chat].

@Martin  Hi Martin, I posted the follow up post in Periodic table chat But no special new info about AI in O. chemistry.

Can you narrow the question?  Perhaps cheminformatics applying e.g. PCA for decades to recognize the most important contributions to a function (see the primer by Sidou and Borges, for example) is less visible to you than protein folding (keyword alphafold) which equally appeared in the press not specialized on chemistry.  There is at least one journal dedicated to the field, Journal of Cheminformatics. With Craig Plot by Ertl, for example, data sets can be analyzed for common substituents.  Perhaps your company wants to commercialize a product in the same niche as your competitor, but to circumnavigate their patent, the structure of your active molecule has to be sufficiently structurally different on paper to merit a separate IP protection while the ADMET properties are still similar enough (or better) to justify the investment.
AI can help in the identification of plausible routes of synthesis.  Formalized retrosynthesis e.g. by Corey has been a field of application for long; digitized databases of known reactions and publications crawled by OCR ease to build/improve tools like the open source AiZynthFinder (Reymond group, University of Berne), ASKCOS at MIT, or closed source IBM RXN for Chemistry.  (Too many to name them here, Wikipedia compiled a list.)  It equally extends to virtual catalyst screening, too (e.g., Yang et al.).  Similar to AI in ChatGPT, it can become a very helpful guide / source of inspiration to follow a route you are less familiar with, because your working group tackles a synthesis of motif A preferentially / always by a special method developed locally.  The potential danger to students (and users/workflows in general) is relying too much on these tools.  For one, what happens if you change your current group for one which does not have a subscription to this infrastructure [anymore]?  What are quality and coverage of raw data the AI accessed, and their digestion / model building (else AI is perhaps not about artificial intelligence, but an artificial idiot)?
Note, Artificial Intelligence Applications in Chemistry is one recognized keyword on the landing page for journals by The American Chemical Society.*  As by today, there are 7741 publications by this publisher alone.  The search by this link can be narrowed further e.g. by the type of publication (article, review, perspective), or/and topic (based on your question, biological chemistry, medicinal chemistry, organic chemistry possibly the more interesting sections for you).  With artificial intelligence alone (link), there are 10066 (and counting).  And ACS is only one (of the larger) publishers relevant to chemistry.
* one of the analogues about chemometrics.
(1) Sidou, L. F.; Borges, E. M. Teaching Principal Component Analysis Using a Free and Open Source Software Program and Exercises Applying PCA to Real-World Examples. J. Chem. Educ. 2020, 97, 1666–1676. doi 10.1021/acs.jchemed.9b00924.
(2) Ertl, P. Craig Plot 2.0: An Interactive Navigation in the Substituent Bioisosteric Space. J. Cheminform 2020, 12, 8. doi 10.1186/s13321-020-0412-1.
(3) Yang, W.; Fidelis, T. T.; Sun, W.-H. Machine Learning in Catalysis, From Proposal to Practicing. ACS Omega 2020, 5, 83–88. doi 10.1021/acsomega.9b03673 / entry on pubchem.

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Do the paired electrons of subshells' degenerate orbitals contribute to the total exchange energy?
For instance, in d7, there are 5 electrons with the same spin, and 2 electron with the opposite spin, meaning that there are only 3 unpaired electrons. So, which of these would be correct?

(i) Total number of exchanges taking place = 3? (only unpaired electrons)
or
(ii) Sum of individual exchanges of the same spin electrons, so 10 + 2 = 12?
Or is it something else entirely?
Also, if the first one is correct, then how do we explain the stability of d10 over d5?
ALTERNATIVELY:
Does the pairing of electron reduce the exchange energy?

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Quantum chemistry and covalent bonding
How can the formation of a covalent bond be described from a quantum perspective, and what implications does it have for traditional chemical bonding theory?
Specifically, in the context of many-electron systems where electron correlation becomes significant, how do modern quantum chemical methodologies, such as Hartree-Fock theory and Density Functional Theory, contribute to our refined understanding of covalent interactions? Furthermore, how does this quantum perspective challenge, extend, or complement classical notions of bonding, as represented by Lewis structures, valence bond theory, and molecular orbital theory?

https://chemistry.stackexchange.com/questions/710/fundamental-forces-behind-covalent-bonding

What is a covalent bond outside of a quantum perspective? How do you explain electronic structure outside of a QM perspective? You can't.

If there is anything like traditional chemical bonding theory, it is likely wrong and obsolete. Lewis structures are rooted in quantum theory. Valence bond theory and molecular orbital theory are approaches to solving for the wave function. Hartree-Fock is basically just an implementation of MO Theory. I don't understand this question.

@Martin-マーチン Agree, not sure what is being asked. If I had votes I would vote to close for lack of focus

Quantum Chemistry and Covalent Bonding:
Quantum chemistry, a nexus between quantum mechanics and molecular science, furnishes an in-depth framework for comprehending and predicting the nuances of chemical bonding. By incorporating quantum principles, this domain offers a granular look into electron behaviors in atoms and molecules, surpassing the sometimes oversimplified interpretations of classical models.
Atomic and Molecular Orbitals:
Atoms are characterized by wave functions or atomic orbitals, which are mathematical representations of the probability density of an electron, indicating the likelihood of finding an electron at a given location. As two atoms approach each other, their atomic orbitals begin to overlap. This overlap can result in the formation of molecular orbitals — either bonding (when the overlap is constructive) or antibonding (when the overlap is destructive). It's crucial to note that this phenomenon isn't driven by "intent" but by the intrinsic properties and behaviors of electrons in the atomic orbitals.
Wave Function and Superposition:
At its core, quantum mechanics presents the notion of the wave function, which encapsulates the probabilistic nature of electron positions. Electrons in a bond don't reside in a fixed location; they exist in a state of superposition. This means they have a probability of being in multiple locations simultaneously. The Schrödinger equation provides a time-dependent wave function, representing the state of the system, encapsulating this electron behavior.
Electron Correlation and Many-Electron Systems:
A particularly challenging aspect in quantum chemistry is addressing electron correlation, especially in systems with multiple electrons. Electrons, being fermions, adhere to the Pauli exclusion principle and also repel each other due to their like charges. Capturing the behavior of one electron influenced by all others is a computational challenge. This is where methods like Hartree-Fock theory, which offers an approximation by treating electrons as moving independently in an average field created by others, and Density Functional Theory (DFT), which considers the electron density rather than the wave function, come into play.
Quantitative Approaches in Theoretical Chemistry:
Delving deeper into bonding characterizations, there are quantitative methodologies in theoretical chemistry. Atoms in Molecules (AIM) theory provides a method to study chemical systems using the electron density, allowing for the analysis of bond critical points and bond paths. Electron Localization Function (ELF) helps in visualizing areas where electrons are likely localized.Then, Natural Bond Orbitals (NBO) analysis provides insight into bonding and antibonding interactions, giving a clearer definition and quantification to the concept of covalent bonds.
Implications for Traditional Chemical Bonding Theory:
Quantum chemistry refines our understanding from traditional bonding theories. Lewis structures and valence bond theory provide intuitive visual models, but they don't capture the nuanced electron behaviors that quantum mechanics does. Molecular orbital theory, rooted in quantum principles, offers a more accurate depiction, especially for molecules exhibiting resonance or delocalized electrons.
In essence, while classical theories provide a foundational understanding, quantum chemistry and its myriad of methodologies grant us a more intricate, quantitative, and comprehensive insight into the fascinating world of molecular interactions and electron behaviors.

This answer now addresses the criticisms and integrates the suggested improvements.
I hope this provides a comprehensive dive into the topic!
For further information I recommend to read Ira.N.Levine "Quantum Chemistry".

"When two atoms come close together with the intent of forming a bond." QM does not ascribe electrons with "intent". They just are what they are.

I downvoted, for such a long answer its lacking quantitative content. This could be improved by mentioning quantitative approaches(AIM, ELF, ELI-D, NBO's, etc.) that exist in theoretical chemistry to characterize bonding and to give the conceptual idea of covalent bonds a clear definition.

I updated the answer, I always try to do my best....

This is informative, sure. I'm not sure it answers the question. But as I've stated above, I'm not even sure what the question is. | More to the topic though: I disagree that Lewis structures are traditional bonding theory. And given that VB and MO theory are congruent when used within the same framework, one cannot be more accurate then the other. I'm sure you mean the right thing, but it reads like there is resonance in MO theory.


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What is the difference between the VSEPR theory and its model?
I know that VSEPR was actually a model proposed by Sidgwick and Powell but then it was developed as a "theory" by Gillespie and Nyholm. Google says that a theory is more detailed and comprehensive as compared to a model. I just want to know what the orignal model stated and what further explanations were there in the theory


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Eigenfunctions of total angular momenta as linear combinations of Slater-determinants
In the Hartree scheme for many-electron atoms, the approximated Hamiltonian (in a.u.)
$$\hat{H} = \sum_{i=1}^Z \left(-\frac{1}{2} \nabla_i^2 -\frac{Z}{r_i} + V_\text{H}\left(r_i\right)\right)$$
is separable and commutes with the total angular momentum operators $\left\{\hat{L}^2,\hat{S}^2,\hat{L}_z,\hat{S}_z\right\}$ ($V_\text{H}$ denotes the radial mean-field potential).
In order to combine such commutation relations and the Pauli exclusion principle, one takes linear combinations of Slater-determinants that are eigenfunctions of the total angular momentum operators (i.e., a set of configuration state functions).
What guarantees that this procedure is always possible?

If two operators $\hat{A}$ and $\hat{B}$ commute then they have a common set of eigenfunctions. In your case, eigenfunctions of the four angular momentum operators are also the eigenfunctions of $\hat{H}$ and so is their linear combination.
You may want to refer to a related post for more details.


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Calculation of the retained mass after hydrolysis of an oligopeptide
I would like to ask about a problem that I am stuck on:

An oligopeptide is made up of glycine, alanine, and valine. Hydrolysis of X in $\pu{500 mL}$ of $\pu{1 M}$ $\ce{H2SO4}$ solution yields solution Y, condensation of solution Y yields mixture Z containing dipeptides, tripeptides, tetrapeptides, pentapeptides and corresponding amino acids. Burn half of mixture Z with a sufficient amount of air, absorb the combustion product into the excess $\ce{Ba(OH)2}$ solution, and it is seen that the weight of the tank increases by $74.225$ grams, the weight of the solution decreases by $161.19$ grams and at the same time $139.608$ liters (measured at STP) of inert gas has escaped. Allow solution Y to completely react with V liters of heated $\pu{2 M}$ $\ce{KOH}$ solution (used 20% excess compared to the required amount), evaporate the solution after reaction, the solid mass has a value of $m$. Find $m$. (Given $\ce{H} = 1$, $\ce{C} = 12$, $\ce{N} = 14$, $\ce{O} = 16$, $\ce{S} = 32$, $\ce{K} = 39$, $\ce{Ba} = 137$)

So far, I have only worked out that there are $\pu{1.195 mol}$ of $\ce{CO2}$ and $\pu{1.2025 mol}$ of $\ce{H2O}$ in the combustion reaction by utilizing the law of conservation.
The weight increase of the tank is the weight of $\pu{CO2}$ and $\pu{H2O}$ combined:
$$m_{\text{increase}} = m_{\ce{CO2}} + m_{\ce{H2O}} = \pu{74.225 g}$$
And, the weight decrease of the solution is due to the formation of $\ce{BaCO3}$
$$m_{\text{decrease}} = m_{\ce{BaCO3}} - (m_{\ce{CO2}} + m_{\ce{H2O}}) = \pu{161.19 g}$$
By substitution, I am able to find that $m_{\ce{BaCO3}} = 235.415$ which leads to the amount $$n_{\ce{BaCO3}} = \frac{235.415}{197} = 1.195 \Rightarrow n_{\ce{CO2}} = \pu{1.195 mol}$$ (conservation of the element $\ce{C}$)
After that, finding the amount of water is fairly simple.
However, after this step I am stuck and unable to progress through this question. I would like to receive some help to solve this question.

There is no solution Y left to react with KOH, as it was already condensed to make solution Z. so you have m = 112 * V g KOH.  :-P

This is a fairly simple peptide hydrolysis problem.
The inert gas is nitrogen:
$$n_{N_2}=\frac{139.608}{22.4}=6.2325 \text{ mol}$$
We can easily see that this nitrogen also contains the nitrogen that was already in the air.
Let's consider mixture Z, we will reduce its composition to the following "compounds": $\ce{C_2H_3ON, CH_2, H_2O}$
Let $n_{\ce{C2H3ON}}=2x, n_{\ce{CH2}}=2y, n_{\ce{H_2O}}=2z$
Half of mixture Z will have: $n_{\ce{C2H3ON}}=x, n_{\ce{CH2}}=y, n_{\ce{H_2O}}=z$
Then, the escaped nitrogen will be:
$$n_{\ce{N_2}}=n_{\ce{N_2}(air)}+n_{\ce{N_2}(rxn)}=n_{\ce{N_2}(air)}+0.5x=6.2325 \text{ mol}$$
Now, let's consider the combustion reaction, by looking at the elements, we can easily formulate the following system of equations:
$$
\left\{
\begin{array}{ll}
2x+y&=1.195 \text{ (C atoms)}\\
3x+2y+2z&=2\cdot1.2025 \text{ (H atoms)}\\
x+z+0.5(6.2325-0.5x)&=1.195\cdot2+1.2025 \text{ (O atoms)}
\end{array}
\right.
$$
Solving this simple system gives us $x=0.375, y= 0.445, z=0.195$
Then, the composition of mixture Z will be:
$$
\left\{
\begin{array}{}
\ce{C2H3ON}: 0.75 \text{ mol} \\
\ce{CH2}: 0.89  \text{ mol} \\
\ce{H2O}: 0.39 \text{ mol}
\end{array}
\right.
$$
Mixture Z is the important part in solution Y as it is the one that'll react with the potassium hydroxide solution. Let Z fully react with the $\ce{KOH}$ solution:
$$\ce{Z +KOH->Z' +H2O}$$
(Z' is the salt mixture)
By conservation of mass, we can easily formuate:
$$m_{\ce{Z'}}=m_{\ce{Z}}+m_\ce{KOH}-m_\ce{H2O}$$
We know that: $n_\ce{KOH}=n_\ce{Z}=n_\ce{C2H3ON}=0.75;\\ n_\ce{H2O}=0.39$
And: $m_\ce{Z}=m_\ce{C2H3ON}+m_\ce{CH2}+m_\ce{H2O}=57\cdot0.75+14\cdot0.89+18\cdot0.39=62.23 \text{ grams}$
Substituting all of them into our original mass equation, we have:
$$
m_\ce{Z'}=62.23+0.75\cdot56-0.39\cdot18=97.21 \text{ grams}
$$
Remember that when solution Y reacts with the $\ce{KOH}$ solution, not only does the Z part reacts, but the "acidic" part also reacts with it. What this means, is that the $\ce{KOH}$ here not only serves a part in creating the salt mixture, it also "neutralizes" the original acid. Therefore, we have to take the existence of the sulfuric acid into account:
$$
n_{\ce{KOH}(\text{total})}=n_{\ce{KOH}(\ce{Z})}+n_{\ce{KOH}(\text{acid})}
= 0.75 + 2n_{\ce{H2SO4}} =0.75+2\cdot(500\cdot10^{-3}\cdot1) =1.75 \text{ mol}
$$
This then leads us to the amount of excess $\ce{KOH}$: $n_{\ce{KOH}\text{(xs)}}=1.75\cdot0.2=0.35 \text{ mol}$
So, the solid mass contains Z', $\text{0.5 mol } \ce{K2SO4}, \text{0.35 mol }\ce{KOH} $
Hence, $m=97.21+0.5\cdot174+0.35\cdot56=203.81 \text{ grams}$.

Thank you saviour !!

To go forward, the amount of inert gas produced by the combustion must be taken into account. This inert gas is probably nitrogen $\ce{N2}$.
Burning these peptides produce $139.6$ L $\ce{N2}$, or $n$($\ce{N2}$) = $\frac{pV}{RT}$ = $\frac{101325·0.1396}{8.314·273}$ = $6.21$ mol $\ce{N2}$. This amount of $\ce{N2}$ was produced by one half of the hydrolyzed $\ce{Z}$ solution. The total amount of amino acid should have burned in producing twice as much, or $12.42$ moles $\ce{N2}$. As every amino acid has $1$ nitrogen atom, the total numbers of amino acids producing $12.42$ mole $\ce{N2}$ is the double, and is equal to $24.84$ mol amino-acid.
As a conséquence, the initial peptide contained $24.84$ mol amino-acid.
There should be something wrong here, because it is not possible to burn $24.84$ mol of any amino-acid, and getting only $1.195$ mole $\ce{CO2}$. Where is the mistake ?

I think that the inert gas includes the nitrogen from the reaction and the nitrogen that was already in the air because mixture Z was burned in air which has roughly 80% nitrogen.


text:
What prevents an octahedral complex from always assuming the quadratic planar limiting case, due to the Jahn–Teller distortion?
When a given high-spin complex is Jahn–Teller distorted, this does not change the relative position of the total energy of the complex (assuming $\mathrm{d^{10}}$ configuration).
Indeed, the distortion only causes the degenerate triplet to split into a doublet and a singlet, and the degenerate doublet splits into two singlets.
In the case of a $\mathrm{d^4}$ high-spin complex, however, the total energy of the system is lowered, since the one ex-$\mathrm{e_g}$ electron is now in a singlet, which has a lower energy relative to the doublet.
What stops this process, which is lowering the total energy more and more? Why isn't every possible complex for a Jahn–Teller distortion automatically distorted in such a way, that we end up with a quadratic planar boundary case, as it can be observed e.g. in $\ce{[AuCl4]-}$?


text:
Convergence limit of emission spectra for atoms with multiple electrons
In high school chemistry, students are taught that the convergence limit in the emission spectrum of an element can be used to determine the first ionization energy.
However, what about the second, third or nth ionization energy? Wouldn't that be a higher energy and therefore appear as a higher energy / smaller wavelength spectral line? Shouldn't the convergence limit be the ionization energy for the 1s1 electron?

They have smaller wavelength - waaaay smaller. I guess they also have some "convergence limits", but what makes you think you'll even see them in the spectrum?

As you state, X-ray absorption spectrsocopy can determine ionization potential for inner electrons: https://en.wikipedia.org/wiki/X-ray_absorption_spectroscopy . It's elementary, just looking at things in a different light.

@Mithoron, The 2nd ionization energy of sodium is 4563 kJ/mol. This corresponds to a wavelength of 26 nm, which is UV. So it should be detectable with UV spectrophotometry.

Therefore, on a UV spectrum, I would think the line observed for the 2nd I.E. should be closer to the convergence limit than the line observed for the 3s electron. Wouldn't this make any calculations performed from the convergence limit inaccurate, because it's not necessarily corresponding to the 1st ionization energy, but for some n>1 I.E. ?

The "Convergence limit" is for a single excitation like in the Balmer series for a hydrogen atom. For the Bohr model, the Rydberg formula gives the wavelength of the emission line:
$$\frac{1}{\lambda} = Z^2 * R_\infty * (\frac{1}{n'^2} - \frac{1}{n^2})$$
where n is the principal atomic numb from which the electron  is decaying. So as $n -> \infty$  then $\frac{1}{n^2}$ converges to 0 and the wavelength converges to a limit.
So the same notion would apply to other excitations of electrons with multiple electrons, but of course there is no such "nice" mathematical solution for multi-electron atoms.
To get to the point, there is just no way to compare the excitation of a single electron with the excitation of multiple electrons.

There actually is a way to compare the excitation of a single-electron atom with that of a multi-electron atom. In quantum defect theory, the excited electron is considered to collide (at negative, i.e. bound energy) with the positively charged ion core. At long distance, the ion core acts like a point charge and hydrogen-like wave functions are expected and at short distance the scattering electron is disturbed by the field of the other electrons. Matching these regimes results in a phase shift of the wave function which manifests itself as a quantum defect ($n$ is no longer integer).

Note that the collision and therefore quantum defect depends on the orbital angular momentum $\ell$ and s, p, d, f electrons have different Rydberg series converging on the same ionization energy. Higher $\ell$ states have less core penetration, a smaller quantum defect and are more hydrogen-like.

@Paul - You correctly point out some of the nuances of multi-electron atoms. Hopefully it will suffice to say that the mathematics is very much more complicated than the simple Rydberg formula. I also should have also pointed out that multi-electron atoms require quantum mechanical calculations and the selection rules for electron transitions yield much more complicated spectra.


text:
Are Starch, Amylose and, Amylopectin reducing sugars?
Amylose and Amylopectin;
As per the information I've learnt, Amylose is considered a reducing sugar but Amylopectin is not because Amylose has a free "reducing end" which Amylopectin lacks, I fail to understand why we don't use a similar argument for Amylopectin as the Haworth structures for both are vastly similar, why is that that we consider a "reducing end" for Amylose but don't do the same for Amylopectin?
Starch;
Consists of both Amylose and Amylopectin, if Amylose is truly a reducing sugar, shouldn't the presence of Amylose in Starch be sufficient to classify Starch as a reducing sugar? or is it that we consider Starch non-reducing due to some other factors?

Are you sure that amylose is a reducing sugar, and that amylopectin is not ? Amylose contains more alcoholic groups than amylopectin. Is it sufficient to state that amylose is reducing and amylopectin not ?

In the institution I study at, our professor seemed to have taught us that but on clarification, he reiterated that Amylose is NOT a reducing sugar and neither is Amylopectin, he also mentioned something about Starch-Iodide reactions and that even in those reactions, starch wouldn't play much of a role; He concluded that Starch, Amylose and, Amylopectin are all non-reducing sugars

@Maurice The term "reducing" is used in a limited sense here to designate substances that have an aldehyde group or a group that reacts like an aldehyde.

[OP] As per the information I've learnt, Amylose is considered a reducing sugar but Amylopectin is not because Amylose has a free "reducing end" which Amylopectin lacks

It would be nice to know the source of this statement because it does not sound right.

[OP in comments] In the institution I study at, our professor seemed to have taught us that but on clarification, he reiterated that Amylose is NOT a reducing sugar and neither is Amylopectin, he also mentioned something about Starch-Iodide reactions and that even in those reactions, starch wouldn't play much of a role; He concluded that Starch, Amylose and, Amylopectin are all non-reducing sugars

Both amylose and amylopectin are made up of glucose building blocks. The aldehyde (or hemiacetal in the ring form) of glucose is called "reducing" because it reacts positive in Fehling's, Benedict's, or Tollen's test. When a glucose forms a glycosidic bond, it is no longer considered reducing because it does not react in these tests.
All the sugar linkages in amylose and amylopectin involve a hemiacetal group of one sugar, and a hydroxyl group of the other sugar. It turns out that both amylose and amylopectin, even though they contain thousands of glucose units, have a single reducing end each. Whether you can detect that single reducing end with one of the mentioned assays depends on concentration of these reducing ends; however, the signal will be weak if not absent.

[OP] Starch; Consists of both Amylose and Amylopectin, if Amylose is truly a reducing sugar, shouldn't the presence of Amylose in Starch be sufficient to classify Starch as a reducing sugar?

If one (or both) of the components have some property, then the mixture should show that property as well, unless there is some interaction or inhibition of the assays.  Technically, starch is not a sugar but a mixture of sugars (or more accurately, polysaccharides).

Amylose vs. Amylopectin – Reducing Ends:
Amylose is a linear polymer of glucose units linked by α(1→4) linkages. It has a reducing end due to its free anomeric carbon and a non-reducing end. In contrast, amylopectin is a branched polymer of glucose with α(1→4) linkages and branch points created by α(1→6) linkages. Despite having many non-reducing ends, it has only one reducing end. The difference between them is not based on the Haworth structures but on their linkage patterns and molecular architecture. Both are reducing sugars, but amylose, being simpler and linear, is often more associated with its reducing property.
Starch as a Reducing Sugar:
Starch, composed of amylose and amylopectin, possesses reducing properties. However, it doesn't elicit a pronounced positive reaction with Benedict’s test. This muted response is attributed to the limited presence of reducing sugar molecules, which are situated only at the ends of the carbohydrate chains. The term "reducing sugar" encompasses a spectrum: certain molecules demonstrate more potent reducing characteristics due to their abundance of free reducing ends. Both amylose (linear) and amylopectin (branched) possess reducing ends. So, while starch can technically be categorized as a reducing sugar, its reactivity is so minimal that it is almost indistinguishable from the background, leading many to treat starch as a non-reducing sugar for practical purposes.
References:
ScienceDirect. (n.d.). Amylopectin. Retrieved from https://www.sciencedirect.com/topics/medicine-and-dentistry/amylopectin
ScienceDirect. (n.d.). Amylose. Retrieved from https://www.sciencedirect.com/topics/agricultural-and-biological-sciences/amylose
Scottish Sensory Centre. (n.d.). Benedict’s solution. Retrieved from https://www.ssc.education.ed.ac.uk/BSL/chemistry/Benedictsolutiond.html#:~:text=(2.04)%20Benedict's%20solution%20turns%20orange,colour%3B%20it%20should%20remain%20blue.
Byju's. (n.d.). Limitations of Benedict’s Test. Retrieved from https://byjus.com/chemistry/limitations-of-benedicts-test/#:~:text=Why%20isn't%20starch%20detected,that%20has%20a%20negative%20effect.

@MetalStorm please raise a flag if you find something you suspect is amiss.


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Derivation of the Van 't Hoff equation
I was reading the paper Relaxation Kinetics of Ferric Thiocyanate (Goodall et. al, 1972) and I came across the passage

Reaction (1) is the simplest representation of the equilibrium between ferric and thiocyanate ions and their complex. The ratio $\frac{k_{12}}{k_{21}}$ is the equilibrium constant $K_c$, which has the value $\pu{1391 mol-1}$ at $\pu{298 ^\circ K}$ and ionic strength $\pu{0.5 mol kg-1}$.
$$\ce{[Fe(H2O)6]^3+ + SCN^-1 <=>[k_{21}][k_{12}] [Fe(H2O)5SCN]^2+ + H2O} \tag{1}$$
The enthalpy change in the reaction is $\pu{-5 kJ mol-1}$, since
$$\frac{\mathrm{d}\ln K_c}{\mathrm{d}T} = \frac{\Delta H}{RT^2}.$$

I am quite confused as to where the equation
$$\frac{\mathrm{d}\ln K_c}{\mathrm{d}T} = \frac{\Delta H}{RT^2}$$ came from, as after integrating it, I seem to get the equation $\ln K_c=-\frac{\Delta H}{RT}$, which then suggests that $\Delta H=-RT\ln K_c$. However, $\Delta G^\circ=-RT\ln K$. How can these two be the same?

You have to calculate the finite integral between 2 T. See also van't Hoff equation on Wikipedia. For the infinite integral, you mustn't omit the integration constant.  Delta H =-RT(ln K_c  + C)

If the equation is written as it is in ($1$), the equilibrium constant $\pu{K_c}$ should not have any dimension. As this constant is given in $\pu{mol^{-1}}$, it means that there is no $\ce{H2O}$ at all in the equation, and that ($1$) should probably be replaced by : $\ce{Fe^{3+} + SCN^- -> [Fe(SCN)]^{2+}}$.

Unfortunately, the title of the question has no connexion with the question. The question is not related to "Relaxation Kinetics".

I have quoted the paper exactly exactly as it was written, and yes, I do apologise for the poor title choice, which I have changed.

Chemistry SE  (in contrary to some other SE sites) strongly recommends plain text question titles for index/search reasons and due possible displaying issues in question title lists. // Pay more attention  to prior search before asking, see https://en.wikipedia.org/wiki/Van_%27t_Hoff_equation

Is it this paper? https://doi.org/10.1021/ed049p675 Chemical Education isn't free (pun intended) so I can't check.


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Retinal 11-cis-retinal to all-trans induced by photons
In the reaction involving conversion of 11-cis-retinal to all-transform, why does the isomerization specifically occur at the 11th carbon in the molecule? When reforming the broken pi-bond, why wouldn't it happen at the cis-conformation again? Is this due to the trans-form being more stable in this case and that it has a lower enthalpy?

Yes, probably, all such changes are governed by that. From what I could gather, there's likely a steric issue there. It would help if you add an image of the reaction, including all the conditions.

@Martin-マーチン Ok, I just added an image!

related https://chemistry.stackexchange.com/questions/73289/isnt-11-trans-retinal-more-stable-than-11-cis-retinal https://chemistry.stackexchange.com/questions/64219/when-is-rotation-around-double-bonds-to-form-cis-trans-isomers-allowed https://chemistry.stackexchange.com/questions/107505/what-is-thermal-isomerization

There are more steps than the net reaction scheme posted by the OP. The trans-isomerization goes along with hydrolysis of a Schiff base, releasing the molecule from opsin. Several enzymes then recycle the 11-cis-retinal before it forms a Schiff base again with opsin, ready for the next light-induced isomerization to the trans form. Here is an overview from a 2010 review (doi: 10.1016/j.tibs.2010.01.005):

The isomerization from the all-trans to the 11-cis form (step iv) is catalyzed by the RPE65 enzyme containing an iron(II) center via a carbocation intermediate. The details are not yet clear (no well-resolved structures of the substrate or product complex), but the presence of an enzyme can help to isomerize a specific double bond rather than obtaining a product mixture. Also, by coupling unfavorable reactions with favorable ones, you can get a molecule that is kinetically trapped in a high-energy state, ready to undergo isomerization (step i) as a way to detect light.
The "why 11-cis" question is addressed in this computational paper: "The results show that the electrostatic interaction between retinal and opsin dominates the natural selection of 11-cis-retinal over other cis isomers in the dark state."


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Why is there difference in reactions in alcoholic and aqueous medium while both are polar protic solvents?
We use an alcoholic medium for reactions in which we want to prevent the characteristics of the anionic species and an aqueous medium for the reactions in which we want to control the activity of the anionic species (depends on how strong is our solvent).
The principle behind this is that alcohol solvates the cationic species only while water in an aqueous medium solvates both cationic as well as anionic species. The solvation results in neutralizing and reducing the mobility of the solvate and thus reduces the reactivity proportional to the degree of solvation.
Now back to my question Why does alcohol only solvate the cationic species despite being a polar protic solvent just like $\ce{H2O}$?

The permittivity epsilon is much greater for water (80) than for ethanol (24.3).

"alcohol solvates the cationic species only"? I wouldn't say it's true,

@Mithoron But then what's the reason for Why does alcoholic KOH prefer elimination whereas aqueous KOH prefers substitution?

@Maurice What does it interpret?

@D13G. Coulomb's law states that the force between two charges is inversely proportional to the factor epsilon. So in water it is easier than in ethanol to separate electric charges.

@Maurice so solutes should be less polarised in an alcoholic solution than in an aqueous solution but why do only the anionic part is not solvated?

I will come to an answer to this question by inductive reasoning.
If we compare acidity of $\ce{H2O}$ and $\ce{ROH}$:

$\ce{H2O ->[-H+] OH-}$ and $\ce{ROH ->[-H+] OR-}$
$\ce{OH- > RO-}$ [Stability]
$\implies$ $\ce{H2O}$ is more acidic than $\ce{ROH}$

As $\ce{pH}$ of $\ce{H2O}$ is 7 so it implies $\ce{ROH}$ is a base and as the $\ce{ö}$ has lone pairs so it is a lewis base and donating ability of lone pair is higher than that of $\ce{H}$, similar to polar aprotic solvents.
Therefore despite being a polar protic solvent, $\ce{ROH}$ acts as a polar aprotic solvent due to its Hydrogen atom is not acidic enough.


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Can someone explain why there is OH in the Standard Reduction Potentials Table
$$\ce{SO4^2− + H2O(l) + 2e− <=>  SO3^2− + 2OH−}$$
This is the half-reaction that I am having trouble understand.
I have read online that OH- is used to cancel out the protons(H+) in basic conditions, so does that mean that this half reaction only happens in basic environments? Or can this happen in acidic environments with the equation I balanced myself:
$$\ce{HSO3-(aq) + H2O(l) <=> HSO4-(aq) + 2H+(aq) +2e-}$$

Note that using photos/screenshots of text instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused in answers.  Specifically handwritten scripts can be difficult to decipher.  Consider  copy/pasting or rewriting of essential parts.
// Optionally, here are formatting guides for texts and formulas/equations/expressions.

I mean, such a simple equation is much better to write using mhchem and \ce{} MathJax syntax.

I will fix it, but do you know why there is OH? and if this reaction can happen in acidic conditions?

You can find the reaction version  for  acidic solutions too, with different standard reduction potential.

Most aqueous redox reactions take place only in acidic solutions or only in basic ones, because the predominant species are often very different between these environments. The reaction you pose, set up for basic solution, is part of the rule.
Even reactions that do not seem to depend on $\ce{H^+(aq)}$ or $\ce{OH^-(aq)}$ might require the proper pH to be realized. For instance, the reaction
$\ce{Zn^{2+} +2e^- <=> Zn^0}$
cannot be realized under standard state conditions except in acidic solution; neutralizing the acid results in the zinc precipitating as a hydroxide or oxide and making the solution basic leads to formation of anionic hydroxozincate or oxozincate complexes. On a Pourbaix diagram this shows up as the horizontal boundary between zinc ions and zinc metal terminating at the pH where precipitation takes over.

License

The standard reduction potential gives the potential that you would find under standard conditions. The standard conditions depend on how you write the half reaction. As written, you would have to measure the standard reduction potential at a hydroxide activity of one (i.e. pH of about 14). If you choose a different pH, you would measure a different potential (and you could extrapolate to standard conditions).
Whether the reduction would also happen at neutral or acidic pH depends on the half reaction you are looking at. It seems both basic and acidic work (I did not take the time to find a reputable source, though).
It is always possible to add $\ce{H+}$ on both sides to switch to acidic conditions. However, if you do that, the standard reduction potential of the new net reaction will be different. Also, without doing the experiment or looking it up, you would not know whether the reaction proceeds under these new conditions or not.
For example:
$$ \ce{IO3−(aq) + 3H2O(l) + 6e− <=>  I−(aq) + 6OH−(aq)}$$
turns into
$$\ce{IO3−(aq) + 6H+(aq) + 6e− <=> I−(aq) + 3H2O(l)}$$


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How to analyse emission spectra/absorbance spectra to show quenching
I have two samples with different chemical structures (but with the same core which is a fluorophore) and I need to prove that a structural change has resulted in the quenching of the fluorophore. Stern-volmer analysis would not be suitable as the molecular part which is believed to cause quenching is intramolecularly incorporated inside the molecule (not intermolecular). Sample A (without the quencher)'s max. emission intensity is much higher (sample A is almost two-fold higher), as compared to the Sample B (with the quencher).

(in case it's helpful, x and y has a difference of 1.04 nm.)
With the information above, can I conclude that the quenching is happening? Can I say, with the info above, that the absorbance baviours are similar but the emission behaviours are different?


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Acidic Strength comparison
I wanted to compare the acidic strength of aniline and water.
Water on losing hydrogen gives $OH^-$ and aniline on releasing H gives $\ce{Ph-NH-}$ and this -ve charge goes into resonance so the -ve charge is distributed, then which one is has more acidic strength??
If it is aniline then it should react with $\ce{NaOH}$, since water, which is comparatively weaker acid is formed as a product, then why does it not do so
Please tell me if i am wrong somewhere in my logic and which one is more acidic.
[NOTE: I am a class 12th student, so please try to keep the answer within the scope of a 12th grade student's knowledge]

You need two phenyl rings to get amine with acidity comparable to water (diphenylamine).

Usually what we call the Ka of aniline is the value of the anilinium $\ce{C6H5NH3^+}$ cation

The pKa of aniline is ~30 that of water is 14. pKa is logarhythmic scale so the aniline proton is approx 10^16 less acidic than water

Actually whether aniline will react with $\ce{NaOH}$ or not will depend on what medium we choose for the reaction.
If it is $\ce{dil. NaOH}$ then the $\ce{OH-}$ dissociated will be solvated so that it will no longer be a base however will be a neucleophile but as $\ce{-NH2}$ is attached at a $\ce{sp2}$ centre so can't do substitution reactions.
And if it is $\ce{conc. NaOH}$ or the medium is alcohol or polar aprotic then it can do acid-base reaction.

That's a good point. Even if OP's issue is more of "lets just get a hold on water pKa", it may be good to find its limitations early on.


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Rutherford's gold foil experiment thin sheets
I am struggling to understand why Rutherford needed to use extremely thin sheets in this experiment. I understand that gold was used, as it was relativley easy to achieve these thin sheets, but why do the sheets need to be thin?

What has this to do with chemistry? Isn't it a pure physics question?

@PeterMortensen, Is there a rigid boundary between modern chemistry and modern physics?

@Achem In a way, everything is related to chemistry, from astronomy to psychology. That does not mean chemists are always the best people to approach to get answers.

@AChem presumably to Rutherford chemistry was equivalent to either stamp collecting or a subset of physics ;)

The sheets were thin to maximize the amount of forward scattered signal. As AChem explains, even air would provide a barrier to subatomic particles (thanks @MattBlack for pointing out the glaring oversight), so it was necessary in addition to work in vacuo. The thinner the sheet, the greater the number of alpha particles that would scatter through the material and be detected. In addition, a thin sheet would reduce the probability of interaction of a alpha particle with multiple atoms, simplifying interpretation of the scattering pattern.
In any case, it is easy a posteriori to justify the design of an experiment and assume it arose as Athena did from Zeus' head, but in practice it is usually an iterative process involving trial and a lot of error. The final design is a result of necessity, based on available tools and materials, and of history and accident as more is learned about the system in question and features are tweaked. There is some analogy in the process of natural selection, except the iterative filter employs human reason.  The word serendipity is also commonly used, although this refers to the plain luck in which one observes the unexpected. That would also have played a part. For instance, some alpha particles back-scattered. That was unexpected according to some atomic models of the time (such as the plum pudding model) and would have been simpler to interpret for a thin sheet (altogether eliminating the possibility of multiple scattering).
Finally, don't assume that something was not attempted simply because a textbook does not mention it. Certainly experiments were performed with thicker foils, and they should have been consistent with the principles gathered from the neater thin foil tests.
The following is an excerpt from a nice brief summary of the story of this discovery on the APS website:

In 1907 Rutherford returned to England, to the University of Manchester. In 1909, he and his colleague Hans Geiger were looking for a research project for a student, Ernest Marsden. Rutherford had already been studying the scattering of alpha particles off a gold target, carefully measuring the small forward angles through which most of the particles scattered. Rutherford, who didn’t want to neglect any angle of an experiment, no matter how unpromising, suggested Marsden look to see if any alpha particles actually scattered backwards.
Marsden was not expected to find anything, but nonetheless he dutifully and carefully carried out the experiment. He later wrote that he felt it was a sort of test of his experimental skills. The experiment involved firing alpha particles from a radioactive source at a thin gold foil. Any scattered particles would hit a screen coated with zinc sulfide, which scintillates when hit with charged particles. Marsden was to sit in the darkened room, wait for his eyes to adjust to the darkness, and then patiently stare at the screen, expecting to see nothing at all

The main reason is that Rutherford and Geiger were both interested in exploring the interactions of alpha particles with thin sheets of matter. Mainly, they were rather fascinated by the causes of their deflection. It was known that alpha particles could be stopped by air in relatively short distance. Studying thin sheets of different metals (not only gold) and their interactions with radiation was a hot topic in early 1900s. Those physicists were developing calculations for scattering.
Almost all papers of Rutherford are available freely to the public. Use Google Scholar to read the original works.
Rutherford, Ernest. "LXXIX. The scattering of α and β particles by matter and the structure of the atom." The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science 21.125 (1911): 669-688.

The scattering data for the β rays, as well as for the α rays indicate
that the central charge in an atom is approximately proportional to
its atomic weight. This falls in with the experimental deductions of
Schmidt. In his theory of absorption of β rays, he supposed that in
traversing a thin sheet of matter, a small fraction α of the particles
are stopped, and a small fraction β are reflected or scattered back in
the direction of incidence. From comparison of the absorption curves
of different elements, he deduced that the value of the constant β for
different elements is proportional to nA$^2$ where n is the number of
atoms per unit volume and A the atomic weight of the element. This is
exactly the relation to be expected on the theory of single scattering
if the central charge on an atom is proportional to its atomic weight.

Note that in older times, atomic mass was called atomic weight.

"Older times", wow.  The term "atomic weight" may have fallen out of favor, but it's the one that was used in all the textbooks I studied from.  I'm not that old.

@JohnBollinger, I am comfortable with atomic weights but some like to split hairs.


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Pure MEA vs Aqueous MEA Solution for CO2 Scrubbing
I was wondering if using pure MEA (Monoethanolamine) or a MEA solution using water would be a better solution for scrubbing $\ce{CO_2}$.

Based on this image, I would think the key reaction would be able to occur with both pure MEA and an MEA solution. However, in this reaction, it takes 2 amines to react with one molecule of $\ce{CO_2}$. The other reaction involves water, so I don't think it would be able to take place with pure MEA. It seems to only use 1 amine to react with one molecule of $\ce{CO_2}$, so theoretically it would would increase the $\ce{CO_2}$:MEA ratio, but I don't know the magnitude of that reaction compared to the main one.
So, it seems that having an aqueous MEA solution might improve efficiency. But the issue with an aqueous solution is that when the MEA is heated to release the CO2, it is above the boiling point of water, so the water would have to be replenished. Therefore, I would like to use pure MEA.
Ultimately, my quesion is:
Is it reasonable to use MEA as a pure liquid rather than in an aqueous solution?

If pure MEA would work OK, it would be used. It's less polar solvent without water and, perhaps counter-intuitively, less basic, afaict.


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Anthoncyanins and its maximum absorbance value
Why does maximum absorbance occur for Anthocyanins at the 500 ~ 600nm range. Additionally how can one explain the high absorbance at lower pH for anthocyanins compared to higher pH values ?

Review the guides Asking and How to ask to prevent misunderstandings,  need of clarification, objections, downvoting or closure.

This was heavily downvoted, presumably because you don't provide more background, for instance have you researched the question yourself?

I've actually worked with Anthocyanins before, very fun to observe the color change when the pH of the solution is adjusted.
The color of anthocyanins is a consequence of their molecular structure and their interaction with light in the visible spectrum, specifically in the 500-600 nm range (blue-green to orange-red light).
Refer to this paper for all the specifics:
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5613902/#:~:text=The%20color%20of%20anthocyanins%20is,in%20an%20increasing%20pH%20condition.

Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference.

As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.


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Are there any compounds with pentagonal pyramidal molecular geometry?
As a high school student, I am curious whether there are any compounds with pentagonal pyramidal molecular geometry. I know that most compounds having hexacoordinated central atom and having $\sigma = 6$ and $\mathrm{lp} = 1$ $(\ce{XeF6},$ $\ce{IF6-})$ have either capped octahedral or distorted octahedral shape, or they have perfect octahedral shape $(\ce{ICl6-},$ $\ce{SeF6^2-}),$ where the lone pair goes to the stereochemically inactive orbital.
Are there any compounds hexacoordinated central atom and $\sigma = 6$ and $\mathrm{lp} = 1$ which have pentagonal pyramidal geometry?

Tellurium (https://doi.org/10.1107/s0108270195005154; https://doi.org/10.1107/S0108270195004902; https://doi.org/10.1107/S0108270195004914; https://doi.org/10.1107/10.1021/ja00228a026; https://doi.org/10.3891/acta.chem.scand.44-0139; https://doi.org/10.1107/S010827019600234X; https://doi.org/10.1016/j.poly.2007.11.007; https://doi.org/10.1107/10.3891/acta.chem.scand.49-0834) and bismuth (https://doi.org/10.1039/c1dt11207d; https://doi.org/10.1016/S0022-328X(03)00435-2; https://doi.org/10.1002/anie.201509629) have quite several examples.

@andselisk You should convert that into an answer.


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Is it possible for a surplus of electrons to be generated during a chemical reaction?
I was wondering whether a chemical reaction can produce excess electrons after the completion of the reaction, in addition to generating by-products (chemical compounds) and heat.
In other words, can there be instances where the reactants share the required electrons among themselves and release surplus electrons to achieve a stable configuration?

If a chemical reaction produces electrons, it is not a reaction. It is a half-reaction. One example is the oxidation of aluminium Al to Al(III), according to :$$\ce{Al -> Al^{3+} + 3 e-}$$

Well, you may dissolve sodium in liquid ammonia and get something which kinda can be described as "free electrons", or more correctly, "solvated electrons". This, I believe, is the closest that you can get.

It is certainly possible to generate a local excess of electrons during a reaction but not to leave an excess at the end. The electrostatic forces of any excess are large and will cause something else to happen. For example, the temporary excess of electrons generated when lumps of sodium are thrown into water is the primary driver of the resulting explosion (a coulomb explosion).

What " surplus electrons"? You don't magically summon electrons there's only what was earlier.

The thing remotely similar is the spatially separated releasing and capturing electrons in galvanic or electrolytic cells. But the local net charge is controlled  by enormous eventual potential change in case of charge disbalance.

Sure. It happens all the time when an electropositive metal is dissolved into an appropriate solvent. Ammonia is the best known solvent capable of holding such solvated electrons, but other amine solvents such as ethylenediamine can do so as well. In synthetic applications the electrons are consumed again when another material is added to set off an additional reaction.
Water can in principle have solvated electrons too, but in that solvent the electrons react so quickly that they are difficult to catch experimentally. So if you are studying aqueous redox chemistry, you find that the electrons are treated as virtual or transient species and therefore must be balanced out of any observed chemical reaction.


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What kind of adhesive could I use as a binding agent for calcium chloride?
I want to create a paper-thin layer of calcium chloride. This sheet should still be capable of absorbing water and allowing gas to permeate downward when it becomes saturated with water and is exposed to sunlight on the top surface. I've tried a sandwich of a regular plastic sheet (top surface) and a gas-permeable membrane PDMS (bottom surface), with a layer of salt bound with commercial white glue between them. Unfortunately, the glue prevents any gas from reaching the bottom surface.
I wonder what I should use as a binding agent for my salty sheet while maintaining its water absorption and gas permeability capability.

Sorry, but this is unlikely to work if exposed to ambient air, except perhaps atop an arid mountain. $\ce{CaCl2}$ is sufficiently hygroscopic that it dissolves in the water absorbed, i.e., deliquesces. Even if the adhesive were porous but yet able to hold onto the resultant solution, the droplets would plug the pores of a membrane.
Try another desiccant (water absorbent) that does not deliquesce, such as anhydrous calcium sulfate, $\ce{CaSO4}$, sold with water indicator as Drierite.
Some SWAGs:

For adhesive, you might try gelatin, as used in photographic emulsions, which allows water to permeate.
Cellulose membrane (sold as dialysis tubing) is water-permeable and might be used to retain the desiccant.


text:
Why is pH the most commonly known acidity scale when there are more intuitive ones?
Consider the switch from $$\operatorname{pH} = -\log_{10}\left(\left[\ce{H3O+}\right]\right)$$ to $$\operatorname{acidity} = +\log_{10}\left(\frac{\left[\ce{H3O+}\right]}{\left[\ce{OH-}\right]}\right)$$ in water.
I'm calling it "acidity" just so that I can refer to it below, don't worry about the name too much.

The neutral point is 0, when $\left[\ce{H3O+}\right] = \left[\ce{OH-}\right]$.

Acid solutions have a positive "acidity" instead of < 7.

Basic solutions have a negative "acidity" instead of > 7.

Both increase in strength as their value increases in absolute value.

The scale is open on both sides, which makes sense because there is no reason to restrict pH to [0, 14]. I know it's not restricted in reality, but for some reason I was taught that at some point.

In water, jumping from one scale to the other is as easy as $$\operatorname{acidity} = 2 \times (7 - \operatorname{pH}).$$

So why is that metric not used? I'm not a chemist so admittedly I have not seen  enough chemistry works and maybe this scale exists, but in any case it's not mainstream while it seems like it would be easier to understand for the common folk.

pH as an acidity parameter is just its secondary usage.

Btw, pH = - log ([H3O+]) is just approximate equality for low enough ionic power. More exactly, the thermodynamic activity should be used.

This is not the worst thing we do because of traditions.

What is the use of this new scale ?

@Maurice Same as pH but with intuitive signs and neutral point.

You need to involve pKw=f(T). acidity = log ( (a(H+))^2/Kw(T) ) = pKw(T) - 2. pH

This was suggested on April 1st 2019 but did not gain traction. Sometimes, the historical definitions prevail even when there are (with hindsight) better definitions. For instance, learning chemistry might be easier if the electron had a positive charge and the proton a negative charge.


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Is the order of 1st ionization energies and order of the work function between some metals the same?
I understand that the ionization energy and  the work function are two different terms both used in different contexts. But as a highschooler, I have learnt more about comparing 1st ioniztion energies of elements than their work functions. Can those methods of comparison be applied to the work function as well. Currently, I don't see a reason why they should not work. Also, looking at some data about work functions it also seems to follow the same order as 1st ioniztion energies. For example, the ionization energies as well as the work funcitons of Li, Na, K are in the order - Li > Na > K. Will this be neccessarily the same and if no, how good of a guess can one make for determining the order of work functions using this assumption?

No, it's not true in general. Also work function isn't a constant.

The 1st ionization energies of elements is measuring the energy needed to remove an electron from an "isolated" atom. The work function is measuring the energy needed to remove an electron from a bulk solid. In bulk metals the orbitals combine to create a "conduction band" of very loosely bound electrons. So you don't have "isolated" atoms anymore. Conversely in an bulk ionic solid like sodium chloride you have essentially sodium cations and chloride anions, not sodium and chlorine atoms. So don't expect excellent correlation between the first ionization potential and the work function.


text:
Help with replicating Fionn Ferreira's microplastic experiment
Hi I am an eighth grade student doing my first science fair project on microplastics. The first step of my experiment is replicating F. Ferriera's microplastic experiment where he used magnetite to remove microplastics from water. The problem is that I have been unsuccessful in doing this. I have used plastic glitter, microbeads and epoxy resin. None of these were attracted to the iron filings in water nor were they removed when I stirred them together and inserted a magnetic rod. What am I doing wrong? I don't think it's the plastics, is it the filings? I tried inserting a magnetite rock to see is that did anything, no luck. Help!

Apparently he uses or used ferrofluids, which not magnetite only. It is a suspension of magnetite in some fluid. The size of microplastics should be very small, ideally not visible to your eyes.

Add more details about your experiment. Did you use a ferrofluid or just magnetite particles. Magnetite particles alone will not work.

Thank you so much for responding quickly. I just used iron filings. I looked on Amazon and found a ferrofluid for purchase. I am going to add 1/4 teaspoon of blue epoxy resin pigment to 1 cup water and stir. Then I will add 1/4 teaspoon of the ferrofluid to the beaker and stir. I then place my Magnetic chip collector (long magnet) in the beaker and pray that it collects everything (ferrofluid and all the blue epoxy resin).  My 'real' experiment was going to be designing a way to efficiently remove the plastic from the ferro-solution, like a sustainable filter. Thank you again for your help!

I am not sure about your proposed experiment in the comment (never did this). The key point is that the plastic has to be really small. Your original plastic "particles" were pretty large. I am sure your teacher must be advising you correctly. Also iron filings are not magnetite.

Cool science project!
What are ferrofluids?
These are magnetic liquids. Imagine a liquid that a magnet can pull. They're made by mixing oil and magnetite (a kind of magnetic rock). Ferreira found out that these magnetic liquids love to stick to microplastics when they are in water.
How did he do it?
He added the oil and magnetite (which together make the ferrofluid) to water. Then, he put in a bunch of microplastics. When he stirred it all up, the microplastics stuck to the ferrofluid.
Next, he used a magnet to pull out the ferrofluid. Since the microplastics were stuck to the ferrofluid, they got pulled out too! And voila, the water was a lot cleaner.
Did it work well?
Yes! Ferreira tested this over 950 times and managed to remove 88% of the microplastics. That's even better than he thought he'd do!
Lab Technique: Always prioritize safety. When I was your age I got a fair number of burns, cuts, and scares and destroyed plenty of good equipment. Wear safety gear always, and remember that gloves themselves will react with some chemicals. Never underestimate the power of taking a few deep breaths and really thinking things out.
Ask for Help: If you're stuck or unsure about something, ask your teacher or an adult. Adult supervision is highly recommended.
Sources: http://www.oceansplasticleanup.com/Projects_Plastics_Oceans_Cleaning/Fionn_Ferreira_Google_Science_Fair_Education_Prize_Ocean_Microplastics_Cleaning.htm
https://www.businessinsider.com/microplastics-water-pollution-solution-from-google-2019-8
https://www.fionnferreira.com/

This doesn't address the key question at all. It merely restates the experiment in a very verbose and meaningless way.


text:
Equilibrium constant of the formation of ferric thiocyanate
I was reading the paper Relaxation Kinetics of Ferric Thiocyanate (Goodall et. al, 1972) and I came across a passage which read

$\space$ Reaction (1) is the simplest representation of the equilibrium between ferric and thiocyanate ions and their
complex. The ratio $k_{12}/k_{21}$ is the equilibrium constant
$K_c$, which has the value $1391{\text {mole}^{-1}}$ at 298°K and ionic
strength 0.5 mole $\text{kg}^{-1}$ ($\textit 8$)
$$\ce{[Fe(H_2O)_6]^{3+}}+\ce{SCN^{-}}\underset{k_{21}}{\stackrel{k_{12}}{\rightleftharpoons}} \ce{[Fe(H2O)5SCN]}^{2+}+\ce{H2O} \space\space\space\space\space\space(1)$$

However, the paper later makes clear that reaction (1) cannot be single step

Reaction (1) is insufficient to explain the mechanism
of the equilibration, since the relaxation rate decreases
with increase in the concentration of hydrogen ion in
the solution. This is explained by introducing eqns.
(2)-(4). Most of the reaction proceeds through the
hydroxo complex, which is present in amount determined by the pH and the acid dissociation constant of
the hexaquo complex, $\ce{K_{OH}} = 1.89 \times 10^{-3} \text{mole}^{-1}$ at
298 $^\circ$K and ionic strength 0.5 mole $\text{kg}^{-1}$ ($\textit{10}$).
$$\ce{[Fe(H_2O)_6]^{3+}} \xrightleftharpoons{\text{fast}} \ce{[Fe(H_2O)_5OH]^{2+}}+\ce{H^+}\space\space\space\space\space\space(2)$$
$$\ce{[Fe(H_2O)_5OH]^{2+}}+\ce{SCN^-}\underset{k_{43}}{\stackrel{k_{34}}{\rightleftharpoons}}\ce{[Fe(H_2O)_4(OH)SCN]^{+}}\space\space\space\space\space\space(3)$$
$$\ce{H^+}+\ce{[Fe(H_2O)_4(OH)SCN]^{+}\xrightleftharpoons{\text{fast}}\ce{[Fe(H2O)5SCN]}^{2+}}\space\space\space\space\space\space(4)$$

How then can it be possible that $K_c = k_{12}/k_{21}$ for reaction (1), since it is not elementary?

Measuring $K_c$ for eqn 1 is ok, as it is $k_{12}/k_{21}$ even if these individual values are unknown.  However, further experiments shows that protons play a role and so the scheme 2-4,. The product of rate constant ratios should produce $K_c$ so $k_{12}$ and $k_{21}$ are not simple rate constants but comprise other terms, according to the expanded scheme.


text:
What is the relation between cross-section and spectral line intensity?
I was told by my supervisor that there is a relation between these two and he talked about a coefficient connecting the two of them. I am currently brushing up a lot of chemistry as I'm a physics student currently working on an internship mainly working with physical chemistry.

Review the guide How to ask and Asking FAQs to prevent need of clarification, objections, down-voting or closure.

A plane wave moving along $z$ has intensity $I$ that decreases as
$$dI=-\alpha Idz$$
which yields Beer's law
$$I=I_oe^{-\alpha z}$$
where $\alpha$ is the absorption coefficient in cm$^{-1}$, at frequency $\omega$, for a transition from state $i$ to $f$, and which depends on the population difference and on the optical cross section $\sigma_{if}$ (cm$^2$) as
$$\alpha_{i,f}=\sigma_{i,f}(N_i-N_f)$$
and, as is almost always the case, $N_f\ll N_i$ then  $\alpha_{i,f}=\sigma_{i,f}N_i$.  As $N$ is the number density it is common to replace this with concentration (mol/dm$^3$) and then Beer's law becomes
$$I=I_oe^{-\epsilon [C] z}$$
where $\epsilon$ is the extinction coefficient in dm$^3$/mol/cm at a given wavelength.


text:
Reaction of ammonium sulfate with an acid?
Consider the following

Which statements are properties of an acid? Statement 1: React with ammonium sulfate to form ammonia. Statement 2: Red litmus turns blue.

Obviously stataement 2 is false. However, I don't seem to understand statement 1. From my understanding, ammonium sulfate is a salt, meaning there shouldn't even be a reaction between an acid and ammonium sulfate. However, ammonium sulfate does react with a base. Ammonium sulfate is a neutral salt, so why does it even react with a base?
Since ammonium sulfate does react with a base, it is making me suspect that there is a reaction between ammonium sulfate and an acid. So what is this reaction mechanism? Why does it occur?
Please clarfiy the respective reactions between ammonium sulfate with a base or acid.

$\ce{(NH4)2SO4 + H2SO4 -> 2 NH4HSO4}$ resp. $\ce{SO4^2- + H+ -> HSO4-}$.

What does resp mean?

Resp. is the abbreviation for respectively.

There is no rule that salts don't react with acids or bases. This may be taught in some introductory classes, but if so, it is a huge oversimplification. In fact, it is a big enough oversimplification that we should probably just call it a lie. It is definitely something you should unlearn if you learned it.
Ammonia is a base. When you attach a proton, you get an acid (ammonium). So a salt like ammonium sulfate is not just a salt, but also an acid (though a very weak one). Therefore, if you later add a strong base to an ammonium salt, the base can steal that extra proton on $\ce{NH4+}$ and give back ammonia. This reaction tends to be obvious because ammonia is a gas, so it bubbles away. Ammonium is not a base, though, so it won't react with acids -- statement #1 from the question is false.
Salts can react with acids too. Let's take sodium carbonate as an example, formed from the reaction of sodium hydroxide and carbonic acid. The carbonic acid loses its protons and becomes carbonate, a base. It's a weak base, but it is a base. So if you add a strong acid to sodium carbonate, carbonate will accept two protons from that acid and you get carbonic acid back. Carbonic acid decomposes into carbon dioxide and water, so once again you get a gas (carbon dioxide) that bubbles away, and a nice visible reaction.
In fact, you can have salts that react with both acids and bases. Based on the previous two examples, you might guess that ammonium carbonate is one, and so it is. If you add a strong base you will get ammonia. If you add a strong acid you will get carbon dioxide.
So where does the original lie come from? Well, in many salts, the ions involved are so weakly acidic and/or basic that it's approximately true that they don't react with acids and/or bases. Sulfate is an example. It is a very weak base, though as a commenter noted, it can accept one proton to form bisulfate. So there will not be a notable reaction with an acid, certainly not one that releases ammonia. As long as you stick to using strong acids and bases to make your salts, you'll end up with salts where the ions are very weak acids and bases, and therefore are essentially unreactive to acids and bases.[1]
[1] Even this is a lie. Not all strong acids are equally strong, and other factors can cause reactions to occur. Sodium chloride reacts with concentrated sulfuric acid to produce HCl gas. The driving force in this reaction is the evaporation of the gas.


text:
Neopentane melting point
Given that neopentane and n-pentane are both hydrocarbons derived from pentane with five carbon atoms each, but differ in their structural arrangements — with neopentane being more branched and n-pentane being a straight-chain isomer — why does neopentane exhibit a higher melting point than n-pentane? How do their molecular structures and the inherent intermolecular forces associated with them influence this difference?
I probably asume that it’s due to its geometry and symmetry, so that it’s easier to pack in solid state, but any more detailed explanation?

It is boiling that can be explained easily. Melting points are another story altogether. Crystal structures are tricky.


text:
Conformational analysis
In organic chemistry, much emphasis is given to the study of the conformers of alkanes and the broader topic of conformational analysis. Can you elucidate why the spatial arrangements and rotations around single bonds in alkanes are so crucial that they merit such focused attention? How does understanding these basic structural variations provide a foundation for grasping more complex concepts in advanced organic chemistry? Moreover, why do introductory organic chemistry courses prioritize this subject, and how does it set the stage for deeper exploration in subsequent coursework and research?

Conformers of alkanes aren't important at all. Conformers of proteins are, but good luck starting education with them - you need simple examples and that's what you get.

Barton writes in his 1969 Nobel lecture:

Every chemical reaction has a transition state. Many transition states have a
well-defined preferred geometrical requirement.

So the conformation (or the distribution of conformations) is one driver of chemical reactivity. Conformation also determines the shape of a molecule, which in turn is important for understanding whether and how molecules attach to the binding sites in receptors and enzymes.

[OP] Can you elucidate why the spatial arrangements and rotations around single bonds in alkanes are so crucial that they merit such focused attention?

The reactions of alkanes are not as diverse as those of other families of compounds, but the principles of conformational analysis (rotation around single bonds, steric hinderance, conformations of rings) are the same.


text:
Uncertainty when beaker goes by 50s
I am helping my daughter do her chemistry homework and am stumped by this one. I have a beaker that seems to count by 50s.
That makes uncertainty calculations awkward. What is the “next unit down”?
How should one calculate uncertainty for this?

Vaccano, No need to feel bad! Chemistry teachers should stop asking these questions, especially the uncertainty by arbitrary rules. It is pure non-sense. The reason is that there is no true answer to such questions. What you can do is that you can mentally draw 5 divisions or 10 divisions between the two marks (corresponding to 50 mL). Estimate the volume. Since there are 50 mL graduations, the so-called uncertainty would be 50 mL/5= +/-10 mL. Some would suggest to divide 50 mL/10 =+/-5 mL uncertainty. The volume is more than 125 mL but not 140 mL. Estimate the value mentally or on paper.

This video might be helpful https://www.youtube.com/watch?app=desktop&v=BkBZn794kSA

There are three independent but related things: Resolution, precision and  accuracy.

Related: Using significant figures when measuring with an instrument with marks other than powers of ten

A real beaker says "approximate volume".

So the volume is approximately between 100 mL and 150 mL. So maybe 125 mL with an uncertainty of 25 mL. Your daughter would know best because she could apply exactly the rules she learned in class.
There is no universally accepted answer because the beaker is not made to measure volume accurately.


text:
What is the diffusivity of colloidal Formazine in water?
I have synthesized Foramzine in an aqueous solution (turbidity = 4000 NTU), and will be further diluting it. I would like to know the diffusion coefficient (diffusivity/mass diffusivity) of colloidal Formazine in water. Is there existing literature where I could find the information?

A lot of the basic work on reactions and diffusion was done using colloids so there should be lots in the literature. In the meantime you could use the Stokes-Einstein eqn. if you know the radius $\displaystyle D=k_BT/(6\pi\eta r)$ where $K_B$ is the Boltzmann constant, $\eta$ the solvent viscosity (Pa s).


text:
Make ethanol with sourdough
I have seen a lot of videos on internet and to make ethanol they ferment sugar with yeast but is it possible to ferment sugar using sourdough ?

Seems like Seasoned Advice SE would be the right place for this?

Yes. BTW, a local bakery was forced to add filtering to remove alcohol vapor. Though I support smog reduction, it was regrettable to me that the luscious aroma of baking sourdough bread was removed, as well.  ψ (heartfelt sigh)

Essentially yes, according to this sourdough site

When yeast breaks down glucose, it transforms it into carbon dioxide and ethanol, both by-products are formed in equal parts. So, for every glucose molecule, two molecules of carbon dioxide and two molecules of ethanol are formed.

OP is asking about sourdough culture which is a complex mixture including a lot of bacteria as well as multiple yeasts, so it would be good to address the effects of that diversity relative to a pure yeast culture as is often used for ethanol production

If you have ever smelled an overripe sourdough starter you will have no doubt that it contains quite a lot of ethanol.

@Andrew  You are correct, if you wish to contribute a second answer exploring this then be my guest


text:
Why is glass a mixture and not a solution?
While studying mixtures and solutions, I read that glass is a mixture and not a solution.
Then I searched over the Internet about it, but I got the same answer that it is a homogenous mixture.
But nowhere did I find any meaningful answer about what the difference is in a homogeneous mixture and solution.
People talked about particle size. But if we zoom enough even salt in water is not completely homogeneous.
So, I got some questions here:

Is there any difference in a homogenous mixture and solution?

Is there any clear cut boundary of what size particle are considered a solution or mixture?

Why is brass or any other alloy considered a solution while glass is not?

Last, but not least - Why is air not a solution despite oxygen, nitrogen, etc. molecules being so small?

Much of this is semantics. Certainly in a glass (or the liquid it is formed from) there are free energy effects from enthalpies and entropies of mixing that will influence whether you end up with a one-phase or two-phase (or more) chunk of glass.

Solution is a homogeneous mixture.

I wouldn't think it odd for someone to call air a solution. Maybe a bit nonstandard as people don't usual, but it certainly does fit the definition.

@Hearth - indeed, air is pretty much an ideal solution, with only an entropy of mixing term to deal with.

[OP] Is there any difference in homogeneous mixture and solution?

According to an intro-college textbook, there is no difference: "A homogeneous mixture, also called a solution, exhibits a uniform composition and appears visually the same throughout."
If you look at the definition from IUPAC (the professional organization of chemists), the definition of solution is a bit vague:

[IUPAC] A liquid or solid phase containing more than one substance, when for convenience one (or more) substance, which is called the solvent, is treated differently from the other substances, which are called solutes. [...]

So it does not mention the requirement that the phase be homogeneous.

[OP] Why air is not a solution despite oxygen, nitrogen etc molecules being so small.

Gas mixtures fit the definition of a solution as homogeneous mixture. However, gases are always completely miscible with each other, so at equilibrium they would always be a homogeneous mixture. Maybe the term solution is not used for the gas phase because there is no reason to distinguish, or because there would be no solute-solvent interactions.

[OP] Why glass is mixture and not solution?

Without a source for the statement, and a specification of what types of glasses it applies to, there is no good answer. If someone claimed that glass is a heterogeneous mixture instead of a homogeneous mixture, there would be ways to test that.
It is possible that glass is so different from other materials that it makes sense to classify it with its own term. Here is one source to study the weirdness further.

I just want to thank you for showing clearly where the quotes are from. I wish this was more widespread in SE!

I feel like there's more to a "solution" than the convenience of the idea. For example, would volume changes matter when referring to a solution vs a mixture? If we mix sand with water, the volume is the sum of the two separately. When we dissolve salt in water the volume is less than the sum of the two. I don't know about glass too much, but it seems to me like the only volumes lost are gasses while heating.

But, when considering a solution of water and alcohol, I can see the convenience argument when deciding which is the solvent of the other.


text:
Drying salts with HCl solution
I'm following a protocol from some old literature (1962), and I want to make sure I'm being safe. It asks you to dissolve some salts in 150 mL water + 25 mL concentrated HCl, then evaporate to dryness at 100 °C. So the question is whether the acid is dilute enough that this is safe to do in a benchtop drying oven (i.e., not inside a fume hood). I imagine back in 1962, you might have done this drying on steam bath inside a fume hood, but I haven't seen a steam bath around for ages. If this is not safe, please suggest how it could be done safely (I do have a fume hood, but the oven won't fit inside).
Telling me to check with my PI or safety officer isn't going to be helpful in this case.

It depends on  the nature of the salt.There are a huge amount of different salts. Some support  a temperature much higher than 100°C. Some are decomposed already at 100°C or a little bit higher. Without knowing the nature of your salt, nothing can be proposed.

@Maurice, I believe it's the HCl fumes that are of concern, not the solid.

What is the salt to isolate?

Personally, I'd go for it - back then, you'd have a scotch in one hand and a cigarette in the other. Just do it outside or in your garage.

The $\ce{HCl}$ vapor given off certainly could be an issue. It's not only a health hazard, but would likely corrode the oven. Many stainless steels and nichrome are attacked at elevated temperatures by moist $\ce{HCl}$ vapor.
Since you've got less than 200 mL, you might dry the container, e.g., a beaker, inside a desiccator that can take that mild heat. Add a safe alkali (e.g., $\ce{CaCO3}$ or  or $\ce{MgOH}$) to the desiccant mix, enough to neutralize 25 mL of concentrated $\ce{HCl}$. If there is an exhaust from the oven, you might pipe that through some marble or limestone chips.

The application of vacuum of a membrane pump (down to $\approx \pu{20 mbar}$, they should have a mark stating the permitted pressure next to the valve) on desiccators is easier done, than gentle (and evenly distributed) heating of them.  To place their lower part in a water bath is clumsy and impractical.

25 mL  of 32% HCl will weigh 28.975 g at 20ºC. This is about as much as you need to clean a toilet (which then gets flushed down the drain, to no great environmental damage). Ref 1.
If you do the preparation outdoors with a moderate wind, you could assume safety. Otherwise, you could trap the HCl or divert it. Trapping the HCl would involve a lot of glassware and preparation.
Diverting it could be done simply with an aspirator (Ref 2)(purchased for less than $5).

Attach the aspirator to a water source and lead a tube from the aspirator to the vessel to be heated.

Ref 3 . Place this tube at the opening, as in the image, or loosely, slightly inside. If placed loosely, the aspirator will draw air from around the vessel opening - and from within the heated vessel - along with the HCl.
For greater safety, you could consider reducing the size of your experiment. In addition, depending on the salts involved, you might be able to attain complete solubility with less water - especially since the mixture will be heated to 100ºC.
Ref 1. https://www.handymath.com/cgi-bin/hcltble3.cgi?submit=Entry
Ref 2. https://www.nascoeducation.com/chapman-aspirator-filter-pump-sb53131.html?gad=1&gclid=EAIaIQobChMItIPa8baXgQMVIQizAB2iMQE2EAQYByABEgK-d_D_BwE
Ref 3. Setting up a water aspirator pump for vacuum distillation

On the back of an envelope: you start with $\pu{25 mL}$ of concentrated $\ce{HCl}$.  Often seen among the suppliers, this is about $\pu{37.5 mass\%}$, or $\pu{12 mol/L}$ (reference), clearly corrosive.  A dilution in $\pu{150 mL}$ of water yields a total volume of about $\pu{175 mL}$.  With a dilution by a factor of 7 (you recall $n = c \cdot V$), the concentration now equates to $\pu{\approx 1.7 mol/L}$.
Your question does not detail out what salt you are going to isolate.  An example could be light weight (organic) amines eventually isolated (thanks to differences of p$K_a$ values) as ammonium hydrochlorides.  Here, the addition of the acid  is just the necessary quantity to obtain the corresponding protonation, with practical no remaining corrosive acidity of $\ce{HCl}$ left. (Check with a droplet of the diluted solution on a stripe of pH paper.)
What scale we talk about?
$$n = c \cdot V$$
$$n = \pu{12 mol/L} \cdot \pu{25E-3 L} = \pu{0.3 mol}$$
For anilinium chloride obtained from aniline, for example, this corresponds to about $\pu{39 g}$ of product (with a reported density of $\pu{1.68 g/cm^3}$, less than $\pu{25 mL}$ occupied volume if it were a compact powder).  So pick your heating plate from the shelf, a wide beaker, and  a magnetic stirring bar to concentrate the diluted solution with gentle heating while stirring.  This is a small setup to fit into a hood.  Add the diluted solution in portions; when nearly dry, remove the magnetic stirring bar and stir with glass rod instead.  As for OSHA, keep the sash down, protect your eyes, skin, clothing.  (It doesn't leave a good impression about the chemistry and work ahead if you can't ask a TA / colleague [of your, or of an other group of the department], nor supervising PI for a safety check...)

I'm the PI, and the safety officer doesn't have enough practical knowledge to be of any help. I appreciate you for taking the time to give this some thought

@ike9898 Multiple days after addressing the question, you still do not indicate the chemistry intended which would narrow, and possibly yield a fitting answer.  That is to say: HCl used to purify an inorganic salt, to digest a compound (salt metathesis), to precipitate an amine (the example above) as a couple of options.  This is, not only because chemistry.se has such a category to vote for closure of a question, but more general, an instance «lack of detail, too little information shared».

Some steam baths require a steam source, either a connection to piped stream ("house steam") or a local steam generator. There are, however, items that look very much like the old steam baths (concentric rings which you remove to match the size of your vessel), but instead of being connected to a steam line, they just sit on an ordinary hot plate (inside a fume hood). Presumably this is slower than using house steam but that's acceptable for infrequent use.

As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.

And it does not answer, "please suggest how it could be done safely." Your proposal vents 25 mL HCl to lab!

What is the purpose of the HCl. If it doesn't react you are evaporating 25 mL of concentrated HCl. A steambath is a temperature control at 100C aka boiling water. Direct heating of such a solution will concentrate HCl to the high boiling azeotrope about 20% HCl Why not simply distil off the water and acid and collect the distillate?

I'm new here. I've rewritten my answer to be more generally useful. I'd appreciate it if you could remove the down votes


text:
What is the reason why galvanic series is in it's specific order?
I'm trying to understand why galvanic series molecular entities are in the order they are in.
Why does $\ce{Cu(s)}$ reduce $\ce{Ag^{2+}}$ ions but not $\ce{Pb^{2+}}$ ions?
there seems to be some kind of small correlation with electronegativity but there's a lot of exceptions for it to be true.
ie:
the electronegativities of Li and K are 0.98 and 0.82 respectively and K is more noble, but K and Ca electronegativities are 0.82 and 1 and Ca is more noble in this case?
If you go down the galvanic series the nobility of metals sometimes increase steadily for a long way but then start decreasing with no seeming connection to electronegativity or octets or half octets.
ie from Mg to Sn the electronegativity increases step by step. (going through the nobility order)(Mg 1.31, Al 1.61, Zn 1.65, Fe 1.83, Ni 1.91, Sn 1.96, Pb 2.33 but then decrease, H 2.2, Cu 1.9 and then increases again Ag 1.93, Au 2.54)
So is there any connection between electronegativity and the nobility of a metal? and if so why is there so much exceptions?
If not then what determines the nobility of a metal?
does ionization of a metal notably change the molecular entitie's electronegativity?
Here's a picture to illustrate with the order of nobility and arrows to clarify the direction.

What do you mean by "galvanic series molecules"? A molecule is an electrically neutral object, consisting of 2 or more atoms, connected by covalent chemical bonds. Then there is a molecular entity, covering all kinds of neutral or charged smallest particles of matter.

Review the guide How to ask and Asking FAQs to prevent need of clarification, objections, down-voting or closure.

You may want to review search results for the targeted search site:stackexchange.com OR site:libretexts.org OR site:hyperphysics.phy-astr.gsu.edu OR site:wikipedia.org  series electrochemical Gibbs energy eventually using alternative keywords.

@Väinö For future reference: for the body of questions, answers, and comments, chemistry.se offers to use mhchem as a comfortable method to add chemical equations.  Here, it was applied only in the one line where it is most suitable.

Thanks for the advice! Seems I have a lot to learn about the culture here and I will start with the sources you provided.

Chem+Math Expression formatting reference:
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There is relation between metal electronegativity and the position in the electrochemical / galvanic metal series. But the metal electronegativity is just one of the factors determining the position.
The position is determined by the value of the standard Gibbs free energy of the reaction.
$$\ce{M^{n+}(aq) + $\frac n2$ H2(g) -> M(s) + n H+(aq)} \quad \quad ΔG^{\circ}_\text{r}=ΔH^{\circ}_\text{r} -TΔS^{\circ}_\text{r} \tag{1}$$
and the related standard reduction potential of the formal "half-reaction":
$$\ce{M^{n+}(aq) + n e- -> M(s)}\quad \quad E^{\circ}_{\ce{M/M^{n+}}} = - \frac {ΔG^{\circ}_\text{r}}{nF} \tag{2}$$
where $F \approx \pu{96485 C mol-1}$ is the Faraday constant.
The metal position on the galvanic series is determined by the value of this potential, the most negative (the least inclined to oxidize anything) for alkali metal ions, the most positive (the most inclined to oxidize anything) for precise/inert metal ions.
The sign of the reaction Gibbs energy change determines the reaction spontaneity. Spontaneous reactions have this sign negative.
The relation of the potential and the Gibbs energy may seem on the first look suspicious, until one reminds himself the equation of electrostatics $\text{d}E = U . \text{d}q$, where $E,U,q$ are potential energy, potential and electric charge, which defines the favourite unit of energy $\pu{1 eV}$.


text:
Dissolution of CO2 over time in a high pH solution
I asked this question in the biology stack-exchange but thought it might be more appropriate here.
I am running a batch experiment using penicillin bottles where I have bacteria growing for 21 days in a highly alkaline media (pH 10). I am measuring the CO2 and O2 content in the headspace over time, but I am specially interested in the CO2 production. Usually I can simply use the CO2 content measured with a gas-chromatographer (GC) but because of the pH of my media, the CO2 will be dissolved into my media. This I have seen in the sharp drop of pH over time as well.
I want then, to calculate the amount of CO2 that has dissolved into my media. I know I require Dalton's Law of partial pressures and Henry's constant/Law. The thing is that, I am a bit confused because as I measured the pressure inside my closed bottles, the pressure in these bottles became negative. So my calculations are returning a negative value for my concentration of gas in the solution. Furthermore, I do not know if the formulas I used (Dalton  and Henry), take into consideration the pH of my aqueous media.

Note that more than 99,9% of the dissolved carbon dioxide will be present at pH 10 as bicarbonate and carbonate anions, so the effective gas solubility would be more than 1000 times bigger than in pure water. // See Carbonic acid and there embedded chart Carbonate_system_of_seawater

pH 10 solutions would absorb practically all CO2 that comes in contact to them, possibly causing suction/underpressure. // By other words, you do not deal just with gas dissolution equilibrium, but also with three chemical equilibrii, two of them pH dependent as acid-base chemical reactions.

Wouldn't a "sharp drop of pH over time" have an effect on bacteria growth?


text:
What methods are there for drying oils?
I am trying to generate thin films of sample for TXRF testing, but my samples contain a significant amount of oil in them. My slides are analyzed vertically, so the oil simply drips off.
I have access to a muffle furnace and can ash my sample, but I am hoping to check for mercury content and that will drive it away into the air.
What are some things I could do to turn these samples into a stable film?

TXRF?

You can dry wet samples, but not oily samples.

Drying oil is a thing, though.

@Karsten yes, that method! It's a great tool for measuring PPM/PPB levels of metals in things, however the sample needs to be a thin film on a disc in order to work. Perhaps there's a way to convert the oils into something more stable?

I believe you mean that you want to firmly fix the oil so that it does not drip- not remove water from it. The only way is perhaps to freeze it. For a chemist, drying only means removing water.

@AChem, there are "drying oils" such as linseed oil, that oxidize in air to form a "dry" film.

Could you coat the slide with an oleophilic substance (e.g., melamine: https://pubs.acs.org/doi/10.1021/acsami.7b13626)? Or chill it in liquid N2 until it congeals on the slide (though you'd have to get it into the analyzer before it warms)?

@DrMoishePippik, After adding oils to another oil "sample", the concentration values will be wrong from XRF. The sample has been diluted in a non-reproducible way.