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What is an enzymatic reaction for which a detailed tertiary structure showing all ligands is available on the PDB?
This is a follow-up to my earlier question here How do I obtain a 3D MOLFILE for the lactose synthase structure, especially the catalytic center? about the lactate synthase. I am using the lactate synthase and the paper I referenced only as an example. But sadly, it's not a great example because the authors have not submitted their (predicted) model of the exact ligand configuration during the critical catalytic step when actually the Galactose is added at beta-configuration to the Glucose.
I would like to find another example which of an enzyme, perhaps involving a metal ion too, for which we have pretty exact configurations submitted to PDB that show how exactly the reactants are aligned to create a specific reaction with a specific stereo-configuration, so we can see why exactly the reaction occurs on that atom and in that direction.
I thought the glycolysis pathway should be very well understood by now, but when I went to UniProt KB looking for the hexokinase, I do not see any tertiary structure model with any ligand. Besides, the glycolysis is a catabolic pathway, and I'd like to show an anabolic reaction, although if I remember correctly, many of the glycolysis steps are reversible.
Appreciate if you can point me to better examples that have lots of "snapshots" of the reaction publicly available.
EDIT comment:
This is what I am looking for: an anabolic enzymatic reaction, ideally with a metal ion involved in the reaction complex, and for which -- unlike in the case of the lactate synthase -- we have snapshots for the critical step of the reaction: the alignment of two ligands which get bonded into a single stereo-isomer. There is nothing dumb, nothing too broad, and nothing opinion-based about this question. The truth is, it's hard to find one by just trial and error fishing on PDB or UniProt.

How would you tell if something is best studied?

How do you define «best studied», what threshold(s) do you use to define «this is good enough»? Hunting for multiple X-ray snapshots of one reaction/type of reaction with enzymes reads like establishing a trajectory like Bürgi & Dunitz pioneered (the later recently embarked for his last way). So «Time-resolved structural studies of protein reaction dynamics: a smorgasbord of X-ray approaches» by Westenhoff et al. in Acta Cryst. (2010) A66, 207–219 (doi 10.1107/S0108767309054361, open access) may offer an entry about these bigger molecules.

@Buttonwood thanks for the paper reference, which sadly does not discuss any of the kinds of reactions I am looking for, just generally tells why it might not be easy but that it is the right thing. Not sure about all this negativity and going off on the word "best". But I added that comment to my original question.

Do people get crystal structures of the enzyme bound to its actual substrate? I was under the impression that it's easier to get the structure of a complex with a competitive inhibitor.

@GuntherSchadow - I have added a comment on your other question. The comments here are not intended to be negative, they are intended to get to the heart of the matter within the site's guidelines.

Following up on @orthocresol , it is very difficult to determine structures of wild-type enzymes with substrates except at cryo temperatures because the substrates react to products more quickly than any structural data collection method. More often you will see structures with inhibitors that mimic the proposed transition state or substrates bound to inactive variants of the enzyme. Some people generate computational models of the substrate in the wild-type active site and maybe even simulate the reaction, but those structures/simulations don't go to PDB usually.

@Andrew, I'd love to have even a predicted best guess model, but as you say, they don't usually go into the PDB. But where might they go? In the original lactose synthase paper I started with they have a model, but they didn't submit it to PDB and the authors are unreachable to try to get the model they had 20 years ago.

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@KarstenTheis I appreciate your support vote to re-open but I do not appreciate your edit and comment, they are just more useless formal Rechthaberei. There is everything very exact about a stereo-isomer selective synthesis. And you accused me of "yelling" elsewhere, I am not yelling, I am insisting on the truth. The close reason for this question is a lie.

The title needs to be edited as it is confusing and a possible cause for the closure. Ambiguous questions are often bound to be closed because the appropriate/best answer would not be clear. In this case the subject of the post is interesting but the question as written seems another, creating confusion. The question might be "are there enzyme crystal structures that show a reagent positioned to undergo a particular reaction mechanism (pathway)". I would suggest the OP at least edit the title to reflect the objective of the post...

Please note that comments critical of others are like screaming into the wind. You have little to nothing to gain and much to lose from them. Instead the best approach here is usually to try to address individuals using their handles (@name) to seek clarification or/and to edit your post in an attempt to address criticism.

Sometimes, the «click and mouse push» GUI of the databases are only a subset of possibilities to query a database. I recall this from trainings for Beilstein (when moderated by Beilstein institute and MDL) vs. access with keywords from the CLI (like STN/Télésystèms Questel) which requests a different (like in «complementary») training.  Perhaps the staff members of the databases in question are more likely to help you to identify a «good system» with «good coverage of experimental data».  There are some wizards in these techniques ...


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Why is the Gibbs free energy of a reversible reaction not always 0?
From my understanding, reversible proceses are those where the expression:
$$\Delta S_\mathrm{total} = \Delta S_\mathrm{system} + \Delta S_\mathrm{surroundings} = 0$$
is true for all for the entire duration fo the process. This would imply that the Gibbs energy must then also be 0 for all reversible processes, as:
$$\Delta G = T_\mathrm{system} S_\mathrm{total} $$
and $\Delta S_\mathrm{total} =0$ for reversible processes.
Where am I going wrong here?

Thermodynamical reversibility (as in reversible process) and reversibility of a reaction are very different concepts, dont mix them up.

What is the difference?

Why don't you write $\pu{\Delta G = \Delta H - T \Delta S}$ ? Why do you suppress $\pu{\Delta H}$ ?

@Benjaming Two unrelated concept using same / similar word.

EDIT: Read the comments by @theorist for why the answer below has problems, and check here to read about the opposing views associated with defining the entropy of the entire universe (literally, not as abbreviation for system plus surrounding).

$$ ΔG = T_\mathrm{system} \cdot ΔS_\mathrm{total} $$

This equation is incorrect, as the following counterexample will show: There are processes going from state A to state B which you can run either irreversibly or reversibly. As a state function, the Gibbs energy will change by the same amount. Also, the entropy change of the system will be the same. However, the entropy change of the surrounding will be different. An example is discharging a battery by shorting it (giving off heat) or by running a motor to do work. Thus, the equality you state can not be generally true.
The correct expression for the change of Gibbs energy is
$$ ΔG_\mathrm{system} = ΔH_\mathrm{system} - T_\mathrm{system} \cdot ΔS_\mathrm{system} $$
Notice that all of the quantities are for the system, not the surrounding or the universe. If you want to apply this to the entire universe, you get
$$ ΔG_\mathrm{universe} = - T_\mathrm{universe} \cdot ΔS_\mathrm{universe} $$
When the change of entropy in the entire universe is zero ("end of the world", "heat death of the universe"), the change in the Gibbs energy will indeed be zero. However, this is not helpful to describe chemical processes humanity will encounter.

Where am I going wrong here?

Just with that one claim. Otherwise, everything is fine.

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Why the count of the IR band of H2O and CO2 is different?
Why $\ce{H2O}$ shows three and $\ce{CO2}$ shows two fundamental bands in IR while they both are triatomic molecules?

Maybe something to do with symmetry and degenerate modes?

Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.

Related, see "Is carbon dioxide IR inactive?"

Molecules vibrate.  Some of the vibrations alter the polarizability of the electron cloud of a chemical bond, which renders them suitable for a characterization by RAMAN spectroscopy.  Some of the vibrations alter the dipole moment of the molecule; these are the vibrations characterized by infrared spectroscopy.  Both techniques are complementary to each other, and especially in larger molecules, there are vibrations which may be observed by either one.
For $\ce{CO2}$, a linear molecule, there is one vibration which does not alter the net dipole moment.  Two others coalesce upon observation since they match in energy. Chemistry LibreTexts contains an animation of all of them:

For water, already bent in the ground state, the vibrations IR spectroscopy may characterize are these:

Both animations are a verbatim copy from the entry on Chemistry LibreTexts about this topic.

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Owens - Wendt Model for Surface Energy of Solid - Liquid Interface
Owens - Wendt model is used for calculating surface energy on liquid - solid interface and it is given by following equation: $$ \gamma_{sl} = \gamma_s + \gamma_l -2(\sqrt {\gamma_l^d \gamma_s^d} + \sqrt {\gamma_l^p \gamma_s^p}) $$
So, if we use liquid and solid of known surface energy as well as their components (dispersive and polar contributions) we can calculate surface energy of the interface.
It is stated that model has 2 assumptions:

Total surface energy of any individual component (solid and liquid) is a sum of polar and dispersion contributions

Dispersion and polar interactions between solid and liquid on the interface contribute to decrease of surface energy of the interface as geometric mean of individual contributions

Given these assumptions:

How is the equation of this model derived (equation written in the question)?
Why is there number 2 multiplying geometric mean contributions for decreasing surface energy of the interface since formula for geometric mean doesn't include that number?

Related info for those not involved in the topic: nanoscience.com - surface-tension-surface-free-energy-and-wettability/


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Balancing redox reaction by removing spectator ion SO₄²¯
I aimed to balance the equation
$$ \ce{Cr2O^{2-}_7 + FeSO4 + H+ -> Cr^{3+} + Fe2(SO4)3} $$
My teacher explained me that since $\ce{SO^{2-}_4}$ is the spectator ion, I must proceed in this equation like this:
$$\ce{ Cr2O^{2-}_7  + Fe^{2+} + H+ -> Cr^{3+} + Fe^{3+} },$$
i.e by removing $\ce{SO^{2-}_4}$ ions.
After balancing,
$$\ce{6 Fe^{2+} + Cr2O^{2-}_7 + 14 H+ -> 6 Fe^{3+} + 2 Cr^{3+} + 7 H2O}$$
Now she asked me to add $\ce{7 SO^{2-}_4}$ to the left hand side of the equation; and $\ce{9 SO^{2-}_4}$, $\ce{3 SO^{2-}_4}$ and $\ce{1 SO^{2-}_4}$ to the right hand side.
Then the equation becomes:
$$\ce{6Fe^{2+} + 6 SO^{2-}_4 + Cr2O^{2-}_7 + 14 H+ +7 SO^{2-}_4 ->  6 Fe^{3+} + 9 SO^{2-}_4  + 2 Cr^{3+} + 7 H2O + SO^{2-}_4} $$
Which becomes:
$$\ce{6 FeSO4 + Cr2O^{2-}_7 +7 H2SO4 -> 3 Fe2(SO4)3 + Cr2(SO4)3 + 7 H2O + SO^{2-}_4}$$
Is it the correct procedure for balancing redox reactions?

There are many correct procedures. You may balance with $\ce{SO4^2-}$, or you may balance without it. In the end it is all the same.

Procedures leading to balanced reaction equations are correct ones. The others are incorrect. So a better question is, if it is an optimal or effective procedure. // What  makes the task little weird is being a hybrid of a complete "molecular" approach and of the ionic approach, omitting all spectator ions/molecules. Generaly, the fully ionic approach is more effective, as there is less things to balance. When done, if needed, bystanders can be added rather mechanically.

Instead of the direct text formatting, you may consider as useful   Notation basics and How can I format math expressions or chem. equations on Chemistry SE. // 6Fe²﹢ + Cr₂O7²¯ +14H﹢ → 6Fe³﹢ + 2Cr³﹢ +7H₂O  versus  $\ce{6 Fe^2+ + Cr2O7^2-  + 14 H+ -> 6 Fe^3+ + 2 Cr^3+  + 7 H2O}$ : $\ce{6 Fe^2+ + Cr2O7^2-  + 14 H+ -> 6 Fe^3+ + 2 Cr^3+  + 7 H2O}$

@Poutnik I was wondering if $\ce{SO4^2-}$ is truly a spectator ion in this equation.

@MsBonniePython Nothing happens to $\ce{SO4^2}$ during the reaction, since the beginning till the very end. Remember that writing  $\ce{FeSO4}$ or $\ce{Fe2(SO4)3}$  in context of water solutions is very formal. As salts, strong acids and strong bases exist in solution as independent ions. There are $\ce{Fe^2+}$, $\ce{Fe^3+}$ and $\ce{SO4^2}$.

Take the time during one of the following days to familiarize yourself with the examples  mhchem provides.  It is helpful to use this extension for the body of questions, answers, and comments here on chemistry.se when formatting such equations.  Don't use it in the title of questions (because it is special and not understood by all browsers).

Balancing a reaction equation is a sequential process.  One approach is that

you identify the starting materials by their formulae, and the reaction products by theirs.

you balance the equation such the same number of atoms on the left hand side equates the number of atoms on the left hand side.
Depending on the reaction, there are some groups of atoms, or ions (e.g., $\ce{SO^{2-}_4}$) which possibly stay intact as such. If so, you may count sulfate ions, too.  But this is risky, as ions consisting of multiple atoms may be transformed, too.  E.g., the dichromate ion:
$$\ce{Cr2O^{2-}_7 + 14 H3O+ + 6e- -> 2 Cr^{3+} + 21 H2O  } $$
or the sulfate ion:
$$\ce{2 SO^{2-}_4 -> S2O^{2-}_8 + 2 e-} $$

you equally balance the total of charges on the left with the total of charges on the right hand side.  In the case of redox reactions, there are a few reactions where molecules of water are on the side of starting materials, or the one of reaction products.  These may help to balance charges and atoms, too.  Note, these should match with the reaction conditions, too; the above about the dichromate for example (many equivalents $\ce{H3O+}$) is in agreement that this reaction is more likely to happen in acidic media (high concentration  of hydronium ions), than in neuter or basic solution.

In some cases, it is necessary to repeat the second and third step a couple of times.
Re spectators: It may happen that you have ions/molecules both on the left, as well as on the right hand side of the arrow.  Say, for a hypothetical reaction you reach
$$\ce{5 H2O + \text{starting materials} -> 3 H2O + \text{products}}$$
which may be simplified to
$$\ce{2 H2O + \text{starting materials} -> \text{products}}$$
if the second equation still is balanced in the number of atoms on the left and right hand side, and equally is balanced for the total of charges on the left/on the right hand side.  The simplification, e.g.,
$$\ce{2 H2O + \text{starting materials} -> 2 H2O \text{products}}$$
into
$$\ce{\text{starting materials} -> \text{products}}$$
including the labeling «here, water molecules are mere bystanders of the reaction» requires that you know the reaction (with set conditions, is it possible that water/sulfate ions/iron ions react?) by observation, or (better: and) that you recognize in the balanced reaction equation that these neither are formed, nor consumed.  This however only is a necessary condition, not an always sufficient one about a spectator: overall, ideal catalysts of a reaction are neither formed, nor consumed either; yet in many instances a requirement to increase the rate of a reaction (e.g., the generation of ammonia from hydrogen and nitrogen by the Bosch process).


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Hydrated electrons produced by radiation vs electrons from electrostatic charge: are they equally reactive?
One of the species generated in water or aqueous solutions by the ionizing radiations are free electrons that quickly become "hydrated" or "aqueous electrons" (see for example this or this), and it is said that this is a chemically reactive species (reducing agent).
But the body can get charged with static electricity apparently with no harmful effect (for example here). In case of getting negatively charged, does the extra electrons also forms hydrated electrons? Are they equally reactive? If not, why not?

Hydrated electrons are extremely reactive. In contrary to liquid ammonia, one need extra fast camera with thousands frames per second to catch their blue color during reaction of alkali metals with water. // Electrostatically charged body  does not have the electrons hydrated, but they joined organic tissues, forming charged ionic structures.

Thank you! Could you elaborate a little or point me in the good direction to look for some info about those charged ionic structures? If most part of the cells is water, one may think that the electrons from electrostatic charge can be in this water too. Maybe they join the mollecules of the cell membrane?

There is no difference in electron reactivity between radiation-origin electrons and electrostatic electrons.
The difference is in the location of their landing zone and reactivity of their neighborhood. The former can land at any point within a bulk material in the reach of ionizating radiation. The latter land only on surface.
Electrons are too small with too high intensity of their electrostatic field to stay "naked" as they are. They bind to atoms, preferable to ones with high  electron affinity.
Electrons have  few major options:

Binding to some organic molecule, like  $\ce{R_1-A-R_2 + e- -> R_1-A^{-}-R_2}$ or $\ce{R_1-A^{+}-R_2 + e- -> R_1-A-R_2}$
Reducing some redox system like:  $\ce{\text{ox_form} + e- -> \text{red_form}}$
Reacting with active hydrogen, typically from $\ce{OH}$ group of water, alcohols or sugars, like $\ce{e-(aq) + H-OH -> H. + OH-(aq)}$, $\ce{e-(aq) + H3O+(aq) -> H. + H2O}$, $\ce{2 H -> H2}$


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Contact Angle Hysteresis
When solid surface is wetted and dewetted with some liquid, contact angle measured will not be the same. I am not sure what does process of wetting or dewetting have to do with contact angle or why should we get different values?
I understand that surface roughness, inhomogenity and solutes in liquid can affect contact angle, but what does it have to do with wetting or dewetting?

What is dewetting ? Is it drying ?

Yes, it is a process where contact area between solid and liquid is decreased while in wetting is increased.

And what is "In wetting" ?

Is "wetting" the advancing drop, while dewetting is retreating? Hysteresis is an observation that something goes differently one way from the way it goes back. Dewetting, i.e., retreating, is really retreating from a different surface (because it is totally wetted) than advancing over a totally dry surface.

Yes, wetting is increase of area of contact between liquid and solid by adding more liquid to liquid drop on solid surface while dewetting is reverse. What you said might be correct.

I believe recently wetted surface changes surface tension s-l and s-g, leading to different angle for gamma1 = gamma1 + gamma3 . cos alfa


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Is there any stereospecific enzyme in PDB that catalyzes an anabolic reaction and has an entry showing both reactant ligands?
I am desperately searching for an anabolic enzymatic reaction, ideally with a metal ion involved in the reaction complex, and for which -- unlike in the case of the lactose synthase -- we have snapshots for the critical step of the reaction in PDB that shows the alignment of the two ligands which get bonded into a single stereo-isomer. It's not easy by just fishing in UniProt because most PDB entries do not have ligands, and when they have, as with lactose synthase, then they don't have both ligands shown. I know this is very difficult to capture these precise moments in a reaction, which is why I am asking this question as there may be only a single or a hand full of such PDB entries in millions -- a literal needle in the haystack. I would take a best guess predicted model also, don't care if it is actually seen in x-ray imaging, as long as it has a PDB entry.

How about the active site of myosin with MgATP present?

@ToddMinehardt, that seems to be what the doi:10.1107/S0108767309054361 discusses as one example, but I want an anabolic reaction, some synthase, to show how one stereo-isomer is produced and not the other.

https://www.sciencedirect.com/science/article/pii/S0969212603001680

When an enzymatic reaction is studied using X-ray crystallography, there are multiple structures that provide information:

The apo-enzyme (empty active site)
Enzyme with inhibitor in the active site
Product complex
Substrate complex

Of those four types, (4) is often the most difficult to achieve. The first enzyme studyied by crystallography, lysozyme, still lacks a substrate complex structure as far as I know. The reason this is difficult is that crystallographic data collection takes time, and during that time the substrate will react. Strategies to still get information about the substrate complex are to use an active-site mutant (inactivated enzyme), to leave out substrates in multi-substrate reactions, or to work with caged substrates (e.g. caged ATP) and time-resolved crystallography.
To search for a stereospecific enzyme, you could go after reactions that have two possible ways of connecting two substrates (e.g. synthesis of lactose vs maltose, search for retention vs. inversion of configuration) or substrates that have a prochiral center (e.g. in the reaction of pyruvate to lactate). Often, textbooks will choose a system that is structurally well characterized to illustrate the stereospecific of enzymes, so looking in the table of contents of textbooks on enzyme mechanism might also be a search strategy.
Lactate dehydrogenase comes close to your list of requirements. In the reverse direction, it transforms pyruvate with its prochiral center stereospecifically to lactate.
The mechanism is described here and depicted here. Multiple structures with pyruvate exist, as listed here. If this is not a good example for the OPs needs, use the same kind of search strategies to find other examples.
It is likely that for a complete view of an enzymatic mechanism (which is just a hypothesis in any case) you would have to combine information from multiple structures.

Thank you for your explanations and links to the LDH. Indeed this is very close to what I was looking for! Thank you, thank you!


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example of endothermic luminescence?
Someone posed an interesting question to me the other day: is all production of light accompanied by the generation of heat? I have found that the question as posed invites vague answers, so I thought I would restrict the question to the case of chemiluminescence and sharpen the language to the following:
Is there an example of a spontaneously occuring endothermic chemical reaction which generates light as one of the (possibly many) products?
Assume for the sake of specificity that the reaction occurs at constant temperature and volume. Conceptually, I am envisioning the following energy diagram:

where the reaction begins with a system consisting purely of reactants $R$. These reactants are excited to a metastable product state $P^*$, with the energy for the excitation coming from the temperature bath. The meta stable state relaxes to the final product state $P$, emitting a photon $\gamma$ in the process. We then would have an instance of a process where a system simultaneously produces light and absorbs heat from its surroundings. Note that this is this case whether the energy of the products (neglecting the photon energy) is greater than the reactants (i.e. $P1$) or lower (i.e. $P2$), since in either case the combined energy of $Pn + \gamma n,\,n=1,2$ is greater than $R$.
So I would like to ask if there is a good example which comes to mind of such a reaction, either for final products $(P1, \gamma 1)$ or $(P2, \gamma 2)$?

Can you add more context and some of your own findings?

I did do some preliminary searching, though, not having a chemistry background, it is difficult to come up with the right search queries to find examples. I figured it would be a good question to ask a community of chemistry enthusiasts, since they might be expected to be familiar with a wide variety of different chemical reactions.

As emission of photons must obey the momentum conservation law, it gives some momentum and related kinetic energy to the emitting atom. Even if this energy is very small, compared to mean thermal kinetic energy, it inevitably creates little heat.

Interesting question. My first thought was simply no, but it is certainly possible to do a non-chemical version using a heat source, a thermoelectric generator and a red LED. So the desired chemical system has to convert heat energy to light energy, with non-negligible efficiency.

@Poutnik the question is not whether a photon imparts a momentum kick to the atom upon leaving (though in fact it is also possible that the momentum kick acts to reduce the molecule's net momentum), but whether or not the light-producing reaction is accompanied by a net absorption of heat from its surroundings.

@EdV so you would claim that no chemical process exists possessing the energy diagram I describe above?

Is all production of light accompanied by the generation of heat? // Due nonlinearity, momentum decrease leads to lesser KE decrease than the vice versa.

No, but I do not know of any, for what that is worth. Maybe some chemiluminescence scheme, as you suggested, might do the trick. But the only ones I have heard about produce both light and heat.

I guess the supposed spontaneous endothermic process may have problems with the 2nd TD law. Delta H would by positive, and due conversion of heat to radiation, Delta S would be negative. Delta G would be positive and spontaneity is ruled out.

@Poutnik if we begin we a sample of pure reactants R, then we begin with very low entropy, so that the production of products P results in an increase of entropy which can compensate the increase from conversion of heat to light. Also note that the condition for spontaneity for a constant temperature, constant volume reaction is a decreasing Helmoltz free energy F = U - TS, not Gibbs free energy G.

I see, U have missed the constant volume condition. But I  am not sure, if TD would be happy with decreasing entropy of environment.

@Poutnik I did specify the constant volume condition in my question. Regarding your other point, whenever one has a system composed purely of reactants and no products, you will always at a finite temperature spontaneously produce products in some (possibly small) quantity, no matter the energy of the products. The entropy increase per mole of reaction diverges as the concentration of products approaches zero, which more than compensates any entropy decrease associated with the photon generation.

That is why I said I missed it,otherwise  I would say you have not mentioned it.

Unless your photon is produced with 100% yield from P* some heat has to be produced. Only a  few types of molecules such as 9-10 diphenyl anthracene appear to have almost 100% fluorescence yield.

After some digging around on Google Scholar, I came across the following article:
https://pubs.acs.org/doi/abs/10.1021/ja00499a052
which presents the following class of reactions:

All of these reactions are endothermic in that they have positive molar enthalpies, and also produce near-infrared light at around 1000 nm. Reaction 1 (i.e. $\ce{R_1} = \ce{H}$) in particular has the following energy diagram:

The relevant quantities here are:

$\Delta H_o$, which is the molar enthalpy for the production of the triplet, i.e. ground state, $\ce{O2}$ product, and
$\ce{E}_*(\ce{{}^1O2})$, which is the energy of the photon emitted in the singlet to triplet transition.

Interestingly, the authors appear to be assessing the endo/exothermicity of the reaction by $\Delta H_o$, not $\Delta H_o + \ce{E}_*(\ce{{}^1O2})$ as I argued using the diagram in my question. It seems to me that whether one uses the former or the latter to determine endo/exothermicity of the reaction depends on whether one assumes the emitted photon is absorbed by the temperature bath or if it is extracted and used to perform work. In the former case energy conservation would imply that the $\Delta H_o$ determines the heat absorbed by the bath, and $\Delta H_o + \ce{E}_*(\ce{{}^1O2})$ for the latter.


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What is "tail" in case of secondary structure of protein?
What is a "tail" in the case of the secondary structure of a protein?
How does it influence the "protein assignment" task?

Look at a model of a protein and see for yourself.

Please provide context, otherwise the answer is guesswork.

«Tail» probably is just one word to describe both N-terminus (amino/ammonium group) and C-terminus (carboxylic acid/carboxylate) of a sequence of amino acids/proteins.
The shape of a sequence of amino acids is conformationally constrained depending on the specific amino acids covalently bound together.  Empirically, this is described by $\varphi$ and $\psi$ in the Ramanchandran plot.
If the sequence growths, its interaction with e.g., cytoplasma becomes more and more important; thus curling up is one form of energetic optimization.  In the form of $\beta$-sheets, coils, etc., some of the conformations eventually lock-in because of the formation of intra- and intermolecular hydrogen bonds.  Whenever such ordering interactions are less many, the conformational flexibility of the protein chain increases again, i.e. multiple slightly different folding patterns differ little in energy and thus are «accessible».  If the structure of a large protein is resolved by NMR spectroscopy, the description of these regions then is more «more fuzzy». This may happen for the termini.  It equally may happen well within the sequence if the consecution of amino acid is not prone to the formation of e.g., $\beta$-sheets.


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Predict whether a reaction is likely to occur. If so, write a net ionic equation
a) $\ce{AgNO3(aq) + CuCl2(aq) -> }$
So the first step is to combine the inners and outers.
So I have $\ce{AgCl}$ and $\ce{CuNO3}$
I know $\ce{NO^-_3}$ has a -1 charge so then wouldn't $\ce{Cu}$ have a +1 charge?
But I see in the solutions that it says the resulting products would be $\ce{AgCl}$ and $\ce{Cu(NO3)2}$
Am I supposed to just know this? I don't know why I would assume $\ce{Cu}$ has a +2 charge?
I know that copper has a +1 and +2 charge. since $\ce{NO^-_3}$ has a -1 charge, I assumed that cooper would have a +1 charge. How do I know when copper has a +1 charge and a +2 charge?

You have copper chloride as a reactant: one copper ion with two chloride ions. Each chloride ion has -1 charge and copper chloride is neutral. So what is the copper ion’s charge?

ohhh okay that makes sense. thanks. so would my first step in a problem like this be to write down each element and their ions first before i combine them for the reactants?

Too broad. This reaction is simply a precipitation reaction: AgCl is insoluble.

+2 is the most stable oxidation state of copper and nothing else is stable in aqueous solution.

To add a little to the answer by @M. Farooq, we try to vault over that heap of hard acquired chemistry knowledge by providing beginning chemistry students with some guidelines, e.g, all sodium and nitrate salts are water soluble, carbonates plus acids result in evolution of carbon dioxide, etc. These are useful over-simplifications (white lies, if you will) and students who go on in chemistry learn that, sometimes a bit unhappily. But we have to start somewhere: chemistry is wonderfully complex and vast.

For future reference:  Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.

@josh copper(I) can be stabilized in water solution using an appropriate soft base ligand; thiourea is one such agent. But with direct water solvation it has to be mostly copper(II).

Your questions (or complains) are encompassing human knowledge acquired over a century and may be more! Science has several collections of known facts, huge body of information and knowledge. You have to know them just like in math you learned 2+2=4, and the area of the rectangle is length into breadth, and the ratio of circumference of a circle to diameter is always $\pi$.
Of course, you cannot read and find the history of someone did an elemental analysis and established the formulae of two solid compounds obtained from dissolving silver metal in nitric acid and copper oxide in hydrochloric acid respectively. Someone must have done an elemental analysis of the solids to find out the ratio of the elements in $\ce{AgNO3}$ and $\ce{CuCl2}$. You are right that you cannot find beforehand the charge on $\ce{Ag}$ in $\ce{AgNO3}$ if you had prepared the compound from scratch 150 years ago.
When chemists started studying the fact that some salts dissolved in water conduct electricity, with further refinements and experiments they were able to find the sign and the magnitude of the charge on the ions. Someone has experimentally determined it to be $\ce{Ag^+}$ and $\ce{Cu^{2+}}$ in these two salts.
You see there is a big body of factual knowledge in chemistry acquired by millions of experiments. You need a base to learn how chemical formulae are written. Once you collect information, you will start to see trends and patterns. Once you have “memorized” the basic charges and signs of common elements, patterns of precipitation, the question is answerable in a single line, as given by Prof. EdV. It is a precipitation reaction and silver forms insoluble salts when chloride ion is present.


text:
How to find polynomial factor of relative (M+n) peak when factoring more than 2 isotopes
I'd like to ask for an extension of Nicolau Saker Neto's answer here:
What is the relative size of the (M+2) peak?
I'm trying to calculate isotopic abundances of molecules with more than 2 isotopes. I've been experimenting calculating relative abundances involving $\ce{^{16}O}, \ce{^{17}O}$ , and $\ce{^{18}O} $ across $\ce{O_3}$ and $\ce{O_4}$. I've been googling and looking through books and haven't found any direct examples of someone doing this.
Laid out in the same format, and using $m_{16}$ to indicate the mass of $\ce{^{16}O}$, etc.:
$$\begin{align}
(0.9976m_{16} + 0.0004m_{17} + 0.002m_{18})^3 ={} & \ \ \ \ \ \binom {3} {0}(0.9976 m_{16})^3 \\[5pt]
      & + \binom {3} {1}(0.9976 m_{16})^2 (0.002 m_{18}) \\[5pt]
      & + \binom {3} {1}(0.9976 m_{16})^2 (0.0004 m_{17}) \\[5pt]
      & + \binom {3} {2}(0.9976 m_{16}) (0.002 m_{18})^2 \\[5pt]
      & + \binom {3} {2}(0.9976 m_{16}) (0.0004 m_{17})^2 \\[5pt]
      & + \binom {3} {2}(0.0004 m_{17}) (0.002 m_{18})^2 \\[5pt]
      & + \binom {3} {2}(0.002 m_{18}) (0.0004 m_{17})^2 \\[5pt]
... \\[5pt]
 & + \binom {?} {?}(0.9976 m_{16}) (0.0004 m_{17}) (0.002 m_{18}) \\
\end{align}$$
My real issue is with the factors involving the binomial for the triple-combination. Which I guess is a polynomial instead of a binomial. I know for sure the value multiplied by the product of % abundances should be 6. But I'm not understanding the logic behind why it's 6.
From what I understand, multiplying binomials is akin to to squaring them, but that would mean this would be 3 choose 1 squared, which is 9. Multiplying them could make sense if the second binomial had an $(n-1)$ for its upper value, this would give you 3*2. This makes sense as long as you approach the largest number first.
The above logic holds well until I test it on $\ce{O_4}$, where it seems to be too naive whenever all 3 isotopes are present.
I'm using a symbolic math solver to brute-force solve these to check my answers. The source of error is obviously coming from the binomial/polynomial term. Is there a proper way to calculate that for a molecule when $n$ different isotopes of the same element are present?

There are programs which solved this problem (e.g., this one) or some of the sketchers (e.g., gchemcalc, however without numeric access to the probabilities on the GUI; however it's source may be accessible).  Perhaps these may help you to compare your computed results quickly.  Good luck!

I think it is much clearer to frame this as a question in combinatorics, rather than chemistry.
In an $\ce{O3}$ molecule, the product of natural abundances gives you the probability that the molecule contains those isotopes. So the probability of having one $\ce{^16O}$, one $\ce{^17O}$, and one $\ce{^18O}$ is given by
$$p_{16}p_{17}p_{18}$$
where $p_n$ is the natural abundance of $\ce{^nO}$. However, there are many different ways in which you could accomplish this: the first oxygen could be $16$, the second $17$, and the third $18$; or the first could be $17$, the second $16$, and the third $18$; or so on. In general you should have $3! = 6$ different possible permutations of oxygen isotopes, and therefore this term needs to be weighted by a coefficient of $6$.
(If the above is unclear, consider the analogous maths problem of: how many different ways are there to draw one red, one green, and one blue marble from a box of marbles? Answer: $6$. The same is true of oxygen isotopes. Of course, they need to be further multiplied by the natural abundance, but that's what the term $p_{16}p_{17}p_{18}$ is for.)
Consider now the case of $16/17/17$, which is the fifth line in your above analysis. The probability is $p_{16}p_{17}^2$, but now the weighting factor is no longer $6$, because the $17$'s are not distinguishable. So our previous factor of $3! = 6$ has overcounted by a factor of $2! = 2$, and the real coefficient is $6/2 = 3$. Notice that this is equivalent to your $3 \choose 2$.
In the $16/17/18$ case, we haven't overcounted; or you could say that we have overcounted by a factor of $1!$ for each isotope. Obviously, $1! = 1$, so this has no influence on the result. But this shows us the way to generalise the problem at hand, which I assert:

The number of ways of choosing $n_1$ objects of type $1$, $n_2$ objects of type $2$, $\ldots$, and $n_g$ objects of type $g$ (where $g$ denotes the total number of types) is given by
$$\frac{n!}{(n_1!)(n_2!)\cdots(n_g!)}$$
where $n = n_1 + n_2 + \cdots + n_g$.

These numbers are not called polynomial coefficients, but rather multinomial coefficients; they are denoted by
$${n \choose {n_1 n_2 \cdots n_g}}.$$
Finally, in the case where $g = 2$ this reduces to the familiar binomial coefficients.


text:
PV=constant and V/T=constant then why PT is not constant?
We know that P1V1=P2V2 by Boyle's Law. Let's say this equation (i).
And we know that V1/T1=V2/T2 by Charle's Law. Let this equation be (ii).
Dividing (i)/(ii), we get P1T1=P2T2...
But it's completely wrong as Gay-Lussac's Law says
P1/T1=P2/T2
So where am I wrong ???

Boyle's law is valid when temperature is constant. Charles law is valid when pressure is constant. If you are using both laws to deduce an equation in P and T, you have to assume that both pressure and temperature are constant. If you assume that, both Gay-Lussac's Law and the equation $P_1T_1=P_2T_2$ hold good.

But if P and T are constant, no change can happen, as the number of moles has not changed.


text:
Why is the synthesis of graphene through alkyne trimerization not viable?
I have been looking for a rationale on this as massive polycyclic aromatic hydrocarbons have been synthesized this way, and with Graphene just being an indefinite expansion of aromatic rings, this has made me rather curious.

Alkyne trimerization gives substituted benzene, not graphene.

Yes, but the question is aimed at what prevents using this technique from forming Graphene? Would it no be possible to form a metal alkyne complex bonded to three diacteylene's, and hence, form graphene?

Trimerization of diacetylene will give you benzene substituted with some acetylene tails. Further trimerization of that will give you some linked system of benzene rings, but not graphene. See, it still has some hydrogens. Graphene has none.

Also: https://en.wikipedia.org/wiki/Alkyne_trimerisation?wprov=sfla1

PAHs with limited number of rings has been made from alkyne (e.g. here and here). But then making graphene from alkyne is still a long shot.


text:
Software for drawing biomolecules (with functional groups abbreviated)
I'm sorry if this is OT, but I find myself slowly going insane trying to draw common biomolecules such as Acetyl Coenzym A. The thing is, I just want to draw the acetyl part and then link that to a text that says "coenzym A" (or whatever), so something like this:

I have no need for the gigantic structure of actual Coenzyme A. I'm trying to visualize basic metabolic reactions such as the citric acid cycle etc. Having a 20-molecule-chain in every other step would just be distracting.
The software I've come across thus far don't support the typing part of the molecule as free text, they require me to pick an actual molecule from the periodic table. In the few cases where I can type text I can't have the molecule link to it.
Thus far I have tried chem4word 3.1 (only supports pre-defined functional groups), Pubchem sketcher 2.4, chemspace and Molview.

Resources for learning Chemistry — Software. ChemDraw and MarvinSketch both can do what you are asking about, both as label or as a functional group. chemfig allows to draw literally whatever you want in your $\mathrm\LaTeX$ document, Rmd file or exported PDF/image.

What is "OT" BTW?

Of Topic, other stackexchange sites tend to frown upon questions that don't have a specific answer (so what is the best software is considered a bad question since that's subjective)

Sounds related to the «HELM toolbar» Pierre Morieux/ChemDraw Wizard presented in his installment 4 about ChemDraw (around 9:17 mm:ss) here.  If done right, such a shortcut retains chemical meaning (i.e., offering an expansion from a mere string «CoA» to the corresponding structure if needed).

ACD labs ChemSketch is available in a limited freeware version that has this capability if you don’t want to pay for something more powerful


text:
Carbocation Stability question

Arrange, the following carbocation on increasing order of stability:

What have I though is that in second it will be most unstable, as $\ce{-OH}$ group will create most inductive effect (negative), further destabilizing carbocation, followed by (3) compound because, fluorine is at a distance, so inductive effect would be effectively weaker than (2).
So $(2)<(3)<(1)$
But in my book, answer is given:$(3)<(1)<(2)$, which I couldn't understand. Any, help is massively appreciated, and point out any mistakes in my explanation if any.

You have to consider resonance structures. Can you draw one for (1) where an electron pair from the oxygen atom is donated to make an extra bond and give all atoms a filled valence shell? What about (2), or (3) with the fluorine atom?

@Oscar Lanzi, resonance happens when lone pair and cation or anions are alternate/adjacent? I might be wrong so can you show resonance in proper answer?

Actually, you just need to know $1$ point that Mesomeric effect or M effect has a greater effect than Inductive effect or I effect. In this case, the -OH group shows +M (donating electron density) effect and the -F group shows the -I (withdrawing electron density) effect.
In compound $2$, -OH group shows the +M effect which stabilizes the positive charge.
Compound $1$ also has -OH group but due to a farther distance shows a weaker +M effect.
Compound $3$ has -F attached to it showing -I effect destabilizing the positive charge.
Therefore, final order becomes:  $(2) > (1) > (3)$

The fact that 1) is more stable than 3) can be explained by the concept of Neighboring Group Participation (NGP)
Oxygen being more polarisable and less electronegative than fluorine tends to delocalise its lone pairs to the C⁺ atom. (Nucleophilicity of oxygen atom is more due to greater polarisability which makes oh group capable of NGP).

[2] is a protonated ketone, nuff said. [1] is the second step in the pinacol rearrangement and is just about to rearrange to [2] so we really do not know how stable it is. The question to me is whether [3] will rearrange. F will attract electrons thru the sigma bond but can feed back thru a pi bond. Since I do not know how stable [1] is I cannot tell if [3] is more or less stable.


text:
Why is copper a better conductor of electricity than calcium?
How would we compare the electrical conductivity between copper and calcium? I understand that electrical conductivity of a metal is dependent on the number of delocalized electrons for that metal. However, in the case of copper which can lose variable number of electrons, how do we compare electrical conductivity with that of Calcium?

You don't. Conductivity can't be deduced that easily.

Copper has one conduction electron per atom, period.

https://chemistry.stackexchange.com/questions/117837/factors-that-influence-the-electrical-conductivity-of-metals https://chemistry.stackexchange.com/questions/58847/why-is-copper-a-better-conductor-than-iron

There is no simple explanation. Conductivity depends on the detailed crystal structure of the metal and how electrons are scattered when travelling though it. This is far from simple.

Comparing the conductivities must start with some data - then you start dealing with explanations. Otherwise, you are making predictions - no, not predictions - speculations. Get some data first, then compare, speculate, hypothesize, predict, check predictions, repeat.
As far as electrical conductivities, tables exist ( https://www.angstromsciences.com/elements-electrical-conductivity), and the screen shot below shows conductivities from nickel and zinc (bracketting copper) all the way thru more conductive elements, including calcium, and to silver, the most conductive element.
The sea of electrons is not equally smooth for all elements, nor equally deep. Semiconductors have interesting conductivity, and theories exist to explain such behavior (https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity), but it seems that comparing conductivities of very conductive elements is more difficult.
It is interesting that the conductivity of copper does not vary much whether it is fully annealed (very soft) or cold drawn (fully hard)(https://www.thoughtco.com/table-of-electrical-resistivity-conductivity-608499). Alloying copper has a much greater effect on conductivity.


text:
Chain transfer in polystyrene synthesis
while I was studying the many side reactions of radical polymerization, I stumbled across this source which illustrates the chain transfer side reaction in polystyrene synthesis:

Of course, there is a typo in this picture, it shouldn’t be -CH2-H- but -CH-H-, but you get the idea. A monomer radical can abstract a hydrogen atom from an already existing chain in the first step, so that it’s radical function is forwarded to the chain.
The chain then, in a second step attacks another styrene molecule from the back side of the double bond so that the more stable, secondary radical forms, so the molecule can also form branches.
Everything alright with me, but I don’t get the regiochemistry here. In the first step,a secondary carbon atom is attacked, which also forms a secondary radical intermediate; but wouldn’t an attack on the tertiary -CH group be more favorable thermodynamically as it is a tertiary radical intermediate which is also stabilized by the phenyl group? Or is it a kinetically controlled process, where more hydrogens can be subtracted on the bridging carbon atoms, so that after the chain transfer, the bridging carbons are tertiary as a result?
Any tips would really be appreciated!

Not sure but steric hindrance might play a role. Because if we attack on CH to form tertiary radical it will be having two phenyl group that might be unstable.

So bascially, if it comes to hindrance, that would indicate a kintetically controlled process, because it is more likely to attack on the bridge, and form sterically stable bridges, if I am not mistaken.

See if we talk about intermediate that intermediate will be more stable in ur case like removing H from tertiary carbon containing phenyl group but as final product it won't be much stable to have two phenyl group on single carbon. So intermediate is stable but product is not so I am not sure whether it will come under TCP or KCP.

The image shown in the question can be found with some context at https://www.open.edu/openlearn/science-maths-technology/science/chemistry/introduction-polymers/content-section-4.3.3
In that document, they say of chain transfer:

A similar mechanism accounts for the side branches in LDPE where it is a more important mode of termination than in polystyrene

This indicates that chain transfer is a rare event in a typical synthesis of polystyrene.

[...] but wouldn’t an attack on the tertiary -CH group be more favorable thermodynamically as it is a tertiary radical intermediate which is also stabilized by the phenyl group?

As this is a rare event, you don't expect any of the intermediates to be the most favorable ones.

Or is it a kinetically controlled process [...]

If the chain transfer is rare, you would expect faster processes to react with the radical intermediate faster than it forms (because these steps are less rare). This means it is kinetically controlled.
Moreover, all the steps with two radicals combining to form a bond would be expected to be kinetically controlled because once formed, the covalent bond would not fall apart into radicals (that is why you need initiator molecules in the first place).
References

Synthesis of Polystyrene and Molecular Weight Determination by 1H NMR End-Group Analysis, Jay Wm. Wackerly and James F. Dunn, J. Chem. Educ. 2017, 94, 11, 1790–1793
Preparation and Properties of Branched Polystyrene through Radical Suspension Polymerization, Wenyan Huang et al., Polymers (Basel). 2017 Jan; 9(1): 14
Self-Branching in the Polymerization of Styrene
J. C. Bevington, G. M. Guzman and H. W. Melville
Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences
Vol. 221, No. 1147 (Feb. 9, 1954),


text:
Which atom is the smallest atom?
Is hydrogen or helium the smallest atom? My teacher said that the smallest atom is the helium atom, but I think that the smallest atom is the hydrogen atom. It has a single electron and a single proton, whereas the helium atom has a pair of electrons, a pair of neutrons, and a pair of protons, which obviously occupy more space than the hydrogen atom.
My question is, which atom is the smallest atom of all?

There is generally applicable and mostly successful way called searching, usually providing bigger number of  more comprehensive answers in shorter time than interactive asking. The latter way is recommended only if the former one failed, as it spends resources of others by answering, instead of your own by searching.

SE and Q/A sites are very small part of internet. I do not discourage asking, I discourage asking without proper searching. // When you ask, it is expected you have thoroughly searched and thought about the topic, providing explicit summary of partial answers/ideas/thoughts you have got until then. It is quite common requirement across all StackExchange sites. // Resources are wasted, when answering what is already written on multiple places. // Atomic radius + Data page

Be aware the proper terms to search do not always match your question, see e.g. google: atomic radius data page

@Poutnik The data page has two answers depending whether you look at the empirical radius or the theoretical radius. I was surprised that hydrogen is not even the second smallest accourding to theory.

@huzaifaabedeen It is quite well possible, evaluating electron probability densities. OTOH, the Bohr model does not work for multielectron atoms.

The behaviour of electrons in atoms is not well describes by the classical points and orbits model of the atom. a more accurate modle for electrons is using quantum physics, in the is system the electron aren't at a specific spot, instead they are described by a quantity called the wave function which gives the probability of the electron being in a specific position.
from statistics you can then  work out the average position (the expectation value), this is the product of the position and how likely it is to be there (the probability)
if you do this for hydrogen and helium you get the figure below, where the solid lines are the probability distribution and the dashed lines are the average radius, which show that on average the electrons in helium are closer to the nucleus than in hydrogen.

hopefully that helps
Also if you want to break from the periodic table, there are atoms where the electrons are replaced by their subatomic cousins, muons. muons are heavier so are a lot closer to the nucleus so the atoms are a lot smaller, they are called Muonic atoms

I think something went wrong with the integration: the (average) theoretical radius of Helium should be larger than that of He+ (0.500).

The volumes of the nuclei are negligible compared to the atomic radii (like a "pea in a soccer/football/cricket stadium").
The higher the nuclear charge, the closer the electrons to the nucleus in a given shell.

Source: https://www.mdpi.com/1422-0067/3/2/87/htm

Helium has the smallest atomic radius.

@huzaifaabedeen No, electrons are considered to be point particles, with no internal structure we know of. Protons are made of quarks, etc., and are extremely tiny, but not point particles.

Maybe I should edit to say space visited by the electrons. @EdV

No problem: I was just trying to head off the OP’s notion that electrons were bigger than protons and neutrons. I upvoted, by the way!

Just the fact that a small bird is flying in larger area than an elegant is walking does not mean a bird is bigger than an elefant.

Please add the link to the source of the image.

@KarstenTheis I've initially upvoted your comment but "space visited by the electrons" is too orbit-oriented. Actually we shall speak of orbitals etc although it can go to far for OP. I would stick to the idea that the key is in the atomic radius. I also note that the question is someone surprising from a logical point of view.

@alchimista How would you say it, given the level of the question?

@huzaifaabedeen There is a quick-ish and exact answer compare hydrogen and the helium +1 cation using quantum mechanics, but the picture from the textbook suggests that the textbook is not up to quantum mechanics yet.

@KarstenTheis Good question.... But your answer now gives at least the facts.. Combining this answer with that of Nyra, the OP should get the picture.

Rutherford's gold foil experiment taught us that the protons and neutrons fill only a tiny portion of the atom, which we call the nucleus. The size of the atom is therefore only defined by the electron shells (orbitals), which in turn depend on the nuclear charge.
Your intuition about more electrons leading to a larger atom is correct, but more protons actually lead to a smaller atom. Which effect dominates depends on whether or not an additional electron finds a place in an existing electron shell.
To gain some intuition as to how this works, it may be helpful to do a Gedankenexperiment wherein we build atoms by adding protons and electrons one at a time, neglecting the neutrons which are not important here.

Imagine you start with one proton and one electron; this is the $\ce{H}$ atom. Let's say its radius is 1 (the Bohr radius).

Add one proton to the nucleus to obtain the $\ce{He+}$ ion. Because the central charge of the nucleus has increased by a factor of two, but the number of electrons hasn't changed, the radius of this ion is halved compared to $\ce{H}$, i.e. 0.500.

Now add an electron to this system to obtain neutral $\ce{He}$. This second electron joins the first one in the 1s shell, expanding the atom only a bit to a radius of 0.567. This is just a sliver larger than $\ce{He+}$ and considerably smaller than $\ce{H}$.

If you were to move on to $\ce{Li+}$, the additional proton would contract the radius to about 2/3 of $\ce{He}$, but when you add a third  electron to obtain $\ce{Li}$, it spawns a new 2s shell at a radius of 2.556, making the $\ce{Li}$ atom much larger than $\ce{H}.$

You can in principle continue this procedure, whereby each additional proton contracts the radius, and each additional electron expands it slightly. Whenever an electron shell is filled, adding another electron will cause a step-change in the atomic radius as can be seen in the jagged graph in one of the answers above.
Helium turns out to be the smallest atom on that graph, which is one of the reasons it was discovered in the sun before it was discovered on earth.


text:
Osmolality and osmolarity of infant formula
Actually, this question is about the following sentence, which I have seen (in various forms) in numerous papers (e.g. this one and this one):

The American Academy of Pediatrics advises that formulas for normal infants should have an osmolarity no greater than 400 mOsm/L (approximately 450 mOsm/kg).

I searched Google up and down but could not find the calculation or the conversion between osmolarity and osmolality of infant formulas. I might be missing something, but does anyone know or can find the relationship between those two values?

https://en.wikipedia.org/wiki/Osmotic_concentration the relation includes the plasma density and the dry content per volume.

From Wikipedia:
Osmolarity = osmolality * (ρsol − ca)
where:

ρsol is the density of the solution in g/ml, which is 1.025 g/ml for blood plasma.[5]
ca is the (anhydrous) solute concentration in g/ml – not to be confused with the density of dried plasma

p.s. I am surprised that the osmolarity and osmolality differ that much in numeric value. For other (non-medical) purposes I considered them almost interchangeable.


text:
Does the 1s orbital have two different phases?
How can the 1s orbital (in H atom for example) interfere both constructively and destructively (to give rise to bonding and antibonding molecular orbitals) with another 1s orbital, if the 1s orbital does not have nodes that make the distinction between phases? To interfere constructively/destructively the orbitals have to be in-phase/out of phase, but how is 'phase' defined in this case, as the H atom 1s orbital seems to lack nodes, and therefore two different phases...?

This isn't a new question, many people stumble over this conceptually. Some of the linked duplicates have a lot of maths, and the truth is that to truly understand it, you need to understand the QM that underpins this. But in the first one, I attempted to explain this with an imperfect (but easier to understand) analogy. Maybe that might help. If you know some linear algebra, then that will help, too.

To address the very last bit, the "in-phase" and "out of phase" is not within a 1s orbital, but rather a comparison between the relative phases of the two 1s orbitals. One may be (wholly) positive and the other (wholly) negative. (This is slightly simplified, because the only requirement is that they be equal and negative, so any pair of complex values $\pm \mathrm{e}^{\mathrm{i}\phi}$ works; the choice of $\pm 1$ is simply what happens when you have $\phi = 0$.  But that's going too far into theory.)


text:
How to predict colour of precipitate?
I am currently preparing for Junior science competitions and I am coming across many questions where colour of precipitate plays important role... Like

When compound $\ce{Hg2(NO3)2}$ reacts with $\ce{Na2CrO4}$ solution, the colour of the compound formed is--

The question is a bit long and the compound can be determined from a series of experiments given in question. the product is $\ce{Hg2CrO4}$ but guessing the colour is impossible. Is there any way to determine colour of compound i.e. precipitate.??

Actually the question is a a bit long and from the question the compound is determined. Let's assume the product can be determined easily. Is there any way to know it's colour??

May be the solution is in the details (I am looking for the hints there). Please post the entire question.

@Nilay Ghosh Looks perfect, Brilliant, Thanks

All chromates are usually yellow, orange or red. The mercury salts are a good example.

the mercury(I) chromate $\ce{Hg2CrO4}$  is brown at room temperature, and turns red at $100$°C. It changes crystallization system at high temperature.
the mercury(II) chromate $\ce{HgCrO4}$ is yellow at room temperature, and becomes red at $100$°C. This change is due to the formation of a basic salt.

With 1) the pH dependent equilibrium between chromate and dichromate, and 2) the more dark-red potassium dichromate than the more yellow potassium chromate: could there be dark mercury dichromate?  (I don't touch the question if it were Hg(I) or Hg(II).)

The dichromate ion $\ce{Cr2O7^{2-}}$ does not give a precipitate with $\ce{HgCl2}$, but is gives a brownish yellow precipitate with $\ce{Hg(NO3)2}$, according to Treadwell's Qualitative Analysis.


text:
Stability of Lithium hydride over Sodium hydride
Why is LiH more stable to heat compared to NaH?
The lithium cation is harder than the sodium cation. The hydride ion is soft. Hence, I would assume that LiH would be more stable compared to NaH, owing to the better interaction that results between a less hard and soft species (NaH) over the interaction between two species of more extreme hardness (LiH).

Softness and hardness are not good indicators of thermal stability.

https://chemistry.stackexchange.com/questions/71598/confusion-in-comparing-melting-point

What could be some attributes that hardness and softness are precise indicators of? @ToddMinehardt

Resistance to abrasion and deformation (here). Google "Mohs scale" and go from there.

Wait. I was referring to hardness and softness in regard with the HSAB theory.

Ah. I go to hardness/softness (1) mechanical properties; then (2) in reference to calcium and magnesium concentration in water. Hence the confusion on my part -


text:
Is the heat (or work) of the system always equal and opposite to that of the surroundings?
I know energy must always be conserved by the first law of thermodynamics, but is there ever a scenario in which a system say releases some amount of energy as work, and the surroundings gains equal and opposite energy to that as heat (as opposed to work). As in, must heat/work released be absorbed stritcly as heat/work?
Let me know if I'm misinterpretting something or if this question doesn't make sense.

The heat supplied to the system comes across the boundary with the surroundings, and comes from the surroundings.  The work done by the system on the surroundings is equal and opposite to the work done by the surroundings on the system.

Other thing is, the system or surrounding can transform the provided work to heat. The opposite is also possible, but with restrictions of the 2nd TD law.


text:
Is a tetrahedral complex with two symmetric bidentate ligands optically active?
If the shape of of a complex $\ce{M(AA)2}$ (with $\ce{AA}$, a bidentate ligand) is tetrahedral, then is it optically active?
Currently I know that it is optically inactive, but I want to know the reason for this. About which plane is this symmetrical?

The plane which passes through two A and between the other two A, for example.

@Ivan Neretin please elaborate your reasoning

To OP: you should use $\ce{...}$ specifically for chemical formulae. Plain dollar signs $...$ are to be used for mathematical expressions. What's the difference? Well, for one, the former is upright and the latter italics. Upright $\ce{I}$ $\ce{I}$ is the symbol for iodine, but italics $I$ $I$ represents something else entirely, like inertia, nuclear spin, or the identity matrix. Please read: FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange? for more information.

Get in contact with other (more senior) students/TA's/your instructor and get access to a molecule kit (an example, no affiliation) to build a model of these complexes.  If you can't, use small cubes cut of a potatoes (as atoms) and toothpicks (as bonds).  Then build these complexes.
An idealized representation of a tetrahedral complex around a central ion with two bidentate ligands is shown in the animation below: carbon atoms in black, nitrogen in blue, the transition metal in ocher:

As mentioned by @Ivan, it is possible to define a plane across two atoms of nitrogen and the central ion to bisect the molecule in hemispheres like a mirror.  A second one is just $90^\circ{}$ to the one shown, too.  Whenever you identify at least one of such a mirror plane across the whole molecule, the complex/the molecule/the object is achiral.  And if a molecule is achiral, it is not optical active.  In your study of chemistry, you are going to encounter this later on again with meso compounds in organic chemistry, e.g., tartaric acid.
For the sake of clarity of the representation, I omitted to display the hydrogen atoms.  Second, independent of the former, was to a assume the five membered rings as flat.  While such a conformation is not much likely,
it may be -- depending on what you aim to highlight in molecular symmetry -- a permissible simplification.
(A nice approach to train oneself recognizing symmetry in molecules is, beside using molecule kits, the set of pages around Symmetry@Otterbein.)

Your animation was was very helpful...


text:
pKa of the alpha proton of an alpha amino ester
Is there a theoretical pKa value for the proton on the alpha carbon of a methylated amino acid, or just a general alpha amino methyl ester?

It will depend on what is attached to the N. In the O'Donnell alkylation procedure where the N is protected as a benzophene imine, the pKa of the glycine proton is estimated at 18.7 and the alanine proton 22.8  source here


text:
Prochirality and major/minor products
I got the following exercise question on configurations, which the textbook does not have an explanation that matches this question, so could someone please tell me whether my attempt is correct or not:

My attempt:
A) both the (3R, 4R) and (3S, 4R) products are chiral since for each of them, there are chiral centers and they are not meso (no plane of symmetry observed).
B) since the alkyne anion preferentially attack carbonyl at re face i.e. attack from the bottom viewing from the figure of the question (top view) since the assigned rotation is anticlockwise, I get something like this after attacking (my thinking when drawing this: attack from 'bottom', then rotate the entire molecule such that the carbon linked to the bulky group is on the same plane as the alkyne group):

the configuration of this (major product) is S. Moreover, the mixture is optically active since S:R = >50:<50 (not racemic mixture).
May I ask whether this approach is correct or not? And for B), is the reason that re face is attacked preferentially is because of the steric hindrance by CH3, which is out of the plane i.e. on the si face?

I don't think this is a real reaction. The pKa of the benzylic proton is around 15, it will just quench the acetylide anion (pKa 25), forming the enolate and racemising the chiral centre

You need to consider the actual conformation of the aldehyde that reacts. You've assumed that the reacting conformation is the same as the one that is drawn in the textbook, but that's not necessarily true. Look up the Felkin–Anh model, for example.

@Waylander Oh you mean the acetylide anion is going to just grab the H+ bonded to the carbon which also bonds to CH3? (Is it called alpha-H?) And after this base-catalyzed racemization (I just known), and the reaction still take place by forming (3S, 4S) (3R, 4R) (3S, 4R) (3R, 4S) products, or any acetylene added will just grab the H+ again?

Any acetylide will grab the H+ and form acetylene so you will have no nucleophilic attack on the aldehyde. When the reaction mixture is worked up the chiral centre will have racemised

@Waylander's point is that racemic aldehyde will be the product should deprotonation prevail. No acetylene anion can add to the resultant enolate. Should addition occur as proposed, then follow orthocresol's lead.


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How to balance this half-equations
HNO$_3 \rightarrow $ N$_2$O(g)
I'm looking at the solution and it says that the above reaction can be modified as
NO$_3^-$(aq) $\rightarrow$ N$_2$O (g)
I don't understand why. Can someone please explain?
I understand the steps afterwards but I don't understand this initial step.

Have you heard about strong acids?

yes. so because hno3 is a strong acid, that's why?

You do not really need to answer that, do you?

Essentially, it's because nitric acid is a strong acid. At concentrations where we generally perform aqueous chemistry, it is essentially completely dissociated to give $\ce{H^+(aq)}$ and $\ce{NO3^-}$. So the nitrogen-bearing species that does the oxidizing is primarily the dissociated ion $\ce{NO3^-}$. As you balance the reaction you will also see the $\ce{H^+(aq)}$ as part of the balancing process for charge and hydrogen.
You may want to compare this methodology with that described in this case, where the acid is only a weak one (hypochlorous acid). You therefore have to design your half-reactions to simultaneously balance protons and electrons so that the acid does not have to dissociate strongly.


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Why does magnesium hydroxide solution solidify after passage of time?
I have been trying to make a suspension of magnesium hydroxide in water. Around 40% magnesium hydroxide w/w. Initially it is fluid, but as time passes, it solidifies. What is causing this?

Gravity, time and cohesion. It is common process for all suspensions and happens in large spatial and time scale in nature.

Is your container open to the air?

@Poutnik, I can imagine it settling down, which I have seen in many suspensions, but never it creating a massive solid mass with no separate water layer

@Loong: yes. When I put it in a bottle, it still solidfies, but takes longer. Will a hydroxide get oxidised? What would be the oxidative product? Is this some sort of cross-linkage? I’d be grateful if you (or anyone) can point me to a book/reference explaining this.

@Maurice, thanks! That sounds logical. Will test the solidified mass for carbonate anion and see…

You could simply have lost enough water from evaporation to exceed the saturation index - note that $\ce{Mg(OH)_2}$ is sparingly soluble in water to begin with...

The solidification of a $40$ % suspension of $\ce{Mg(OH)2}$ in water is probably due to the action of $\ce{CO2}$ from the air, which reacts with $\ce{Mg(OH)2}$ producing $\ce{MgCO3}$ which is insoluble in water. Remember that $58~ g$ $\ce{Mg(OH)2}$ produces $84~ g$ $\ce{MgCO3}$ and $174~ g$ lansfordite $\ce{MgCO3⋅5H2O}$, as described in the Handbook of Chemistry and Physics. This is enough to absorb all the water from your suspension. Remember also that $\ce{Mg(OH)2}$ is practically insoluble in water : $0.9~ mg/L$.
Ref.: 1. Charles D. Hodgman, et al., Handbook of Chemistry and Physics, The Chemical Rubber Publishing Co. Cleveland, $1961$, p. $523$ and $600$.

Wikipedia. Lansfordite is a hydrated magnesium carbonate mineral with composition: $\ce{MgCO3·5H2O}$. Landsfordite was discovered in $1888$ in a coal mine in Lansford, Pennsylvania. It crystallizes in the monoclinic system (space group P21/c) and typically occurs as colorless to white prismatic crystals and stalactitic masses. It is a soft mineral, Mohs hardness of $2.5$, with a low specific gravity of $1.7$

Milk of magnesia (~8% Mg(OH)2 + additives) can be shaken to make a homogeneous dispersion that settles slowly - even without additives - over a period of hours. As the Mg(OH)2 settles and becomes more concentrated, it settles more slowly until after many hours or days, the dispersion settles to a somewhat stiff paste which can be tilted without flowing, but not strong enough to be called solid. A more concentrated dispersion made by stirring powder into water will settle in the same way, most probably into a thick paste with some supernatant water. More concentrated suspensions (or slurries), with additives, are available commercially up to 60% Ref 1). Just a little more than 60% Mg(OH)2 might not even be possible to be stirred into a homogeneous fluid.
Magnesium hydroxide has a layered crystal structure (like clays)

(green is Mg, red-gray is OH) in which hydroxyl groups are available for hydrogen bonding to water.
A more complete description of the experimental conditions would be helpful. For instance, how strong is the solid that forms? How much clear liquid appears above the solid phase (so that the actual concentration of Mg(OH)2 in the solid phase can be estimated)? While the formation of an apparent solid could be due to atmospheric CO2, this path seems unlikely in the closed bottle. Forty grams of Mg(OH)2 plus sixty grams of H2O would require thirty grams (15.3 L) to completely convert the Mg(OH)2 into Lansfordite (or maybe this happens only on the surface - but the closed bottle still poses a problem - where does the CO2 come from?).
Intense homogenization separates many layers, but they eventually come together again by gravitational settling, perhaps even reforming crystals. Magnesium hydroxide is prepared from seawater by precipitation with NaOH, over a period of settling for many hours; some new crystallization can be observed by x-ray diffraction (Ref 2).
I think the problem will go away if you restir the solid, breaking down any crystallization that might have helped stiffen the suspension. Or shake the (closed) bottle vigorously. These things happen.
Ref 1. https://www.thioguard.com/treatmentPlants_slurry.php
Ref 2. Marian Turek and Witold Gnot, Ind. Eng. Chem.Res. 1995, 34, 244-250


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Reaction of aqueous KOH and alcoholic KOH with alkyl halide
CONTEXT:
Here's reaction of alkyl halide with aqueous $\ce{KOH}$

and here's the reaction for dehydrohalogenation by alcoholic $\ce{KOH}$

Question
Initially $\ce{KOH}$ is aqueous. On reacting with alkyl halide, it produces alcohol product. Does the alcohol not make the initially aqueous $\ce{KOH}$ alcoholic as a result producing alkene in the later part of the reaction? If so, wouldn't the reaction initially produce alcohol and produce alkene from the remaining reactants?

It depends on the conditions. You'll note that the 2nd reaction is heated. In practice all these reactions produce mixtures of alkene and alcohol - the proportions of the mixture depend on temperature, solvent, reactant concentration, phase of the moon etc

@Waylander what do you mean by phase of the moon?

I mean that you can run this reaction several times under what you think are identical conditions and get differing results. It is also difficult to replicate the reported results of other researchers. It is a reaction that is much taught but seldom used in the real world as there are better procedures that give more reliable results

https://chemistry.stackexchange.com/a/15737/30929


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Are terminals alkyne reduced by Na/liq NH3?
By the mechanism, it appears that any alkyne should be reduced, but I couldn't find any source supporting or disapproving it. One possibility is that first acid-base rxn takes place but still the reduction should happen.
Please help.

According to this source  here

The Na /NH3 reduction of alkynes does not work well on 1-alkynes unless certain modifications are made in the reaction conditions.

edit: Thanks to @Buttonwood for providing details on the modifications required. Addition of ammonium sulfate to the reaction mixture gives quantitative yields of alkene according to this edition of Modern Methods of Organic Synthesis here

can you share problem 14.39 also of the above-mentioned source?

@Govind It possibly is a typo since problem 14.39 is about subsequent complications to have both ethynylmagnesium bromide and EtMgBr in one flask which may lead to double deprotonation and metallation (thus, «simply» a story of $pK_a$). And $\ce{NaNH2}$ equally is a base strong enough to deprotonate a terminal alkyne. Problem 14.40 of the same source however states the hassle of $\ce{3 RC#CH -> RCH=CH2 + 2RC#C^- + 2Na^+}$ with $\ce{Na/NH3}$ may be prevented by the addition of $\ce{(NH4)2SO4}$ to the reaction mixture.


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Does solvate isomers exist for coordination compounds with ammonia solvent as ligand?
I could find many sources confirming the hydrate isomerism(with water as solvent) but I couldn't find any information about any other solvent. In all the sources "hydrate/solvate isomerism" is written but no example with any other solvent.
I came across this question:

The total number of possible isomers for $\ce{[Pt(NH3)4Cl2]Br2}$ are?

If we consider $\ce{[Pt(NH3)3Cl2Br]Br.NH3}$ and $\ce{[Pt(NH3)3ClBr2]Cl.NH3}$ also, then the answer should be 12. Without them, the answer would be 6.
Any help will be appreciated.
PS: I have seen this. But I don't think the reasoning is correct because hydrate isomers do exist.

I can produce no example nor completely refute their existence. However, they should be significantly rarer than hydrates, because ammonia is both a much better ligand than water and is much more volatile, both factors which should destabilise ammonia solvates.

Also, $\ce{Cu[PtCl6]·18NH3}$

Also I would like to ask, does all the 6 ionisation isomers do exist in real, or are they only theoretical?

Some links: https://www.researchgate.net/figure/Experimentally-observed-decomposition-temperatures-Tdec-for-selected-metal-borohydrides_fig4_261564719 ... https://www.researchgate.net/figure/Conditions-for-the-ammonothermal-crystal-growth-of-metal-halide-ammoniates-and-of_tbl2_281290526

if you consider hydrate isomerism, shouldn't you (well theoretically) consider ammoniated isomers too?

@napstablook what do you mean by ammoniated isomers? I think it makes sense to consider isomers with NH3 outside the coordination sphere as the OP suggests, are you referring to such isomers as "ammoniated isomers"?

@BoxBoxBoxBox yes you are thinking right.

@NilayGhosh but the compound you have mentioned does not have the same molecular formula so it cannot be an isomer.

I am considering complexes which have ammonia outside of coordination sphere. It's surprising that we have hydrates but rarely do we encounter ammoniates like these. @Ritil


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Should we consider reactant as the product of a no reaction?
If a reaction doesn't take place, should we call the reactant as the product? What is the convention? Let me illustrate the question with an example from JEE Advanced 2021, paper 1:

The reaction of Q with $\ce{PhSNa}$ yields an organic compound (major product) that gives positive Carius test on treatment with $\ce{Na2O2}$ followed by addition of $\ce{BaCl2}.$ The correct option(s) for Q is(are)

In option C, the usual SNAr won't take place, but if we consider the substrate for the next step then it would give a positive test.
Also, can SN2 take place on carbon of $\ce{R-S-Me}$ and form $\ce{Ph-SMe}$ and $\ce{R-S-}$ anion? By thermodynamics, the equilibrium should lie in forward direction, but I don't know whether kinetics will support it or not.

You're overthinking it, to be honest, and that's probably why you realise it sounds a bit silly. If it doesn't react, there is no reaction, and no product. Ideally, you shouldn't have to resort to [philosophy / twisted logic / wordplay] (delete as you deem appropriate) to answer a question.

It would be helpful to say what the Carius test is for

Without the reactions conditions being specified this question is difficult to answer

What about the 2nd possibility of Sn2 that I mentioned?

I do not think that a likely reaction

@Govind I agree with your reopening reason, but please refrain from adding comments to your question framing your own question as "silly" or requesting mods to rush-review your post. If anything, your question (again, not a bad one, but a bit strange, I think) was closed by the community users, not mods, and the community users are as close as one gets to touch the aforementioned "spirit" of SE.

For those who don't know the Carius test with addition of barium chloride is for organic sulfur.
Without the exact reaction conditions it is a bit difficult to be definitive however, it is clear that A & D will react with sodium thiophenoxide under mild conditions. I think this is the answer the question-setter expects. I think it possible with the right conditions, possibly using copper catalysis, that B could be persuaded to react. C will not react.


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Effect of substituents on aromatic character
Does the presence of substituents (electron withdrawing or electron donating) on an aromatic system have any effect on the aromatic character?
In other words, would it be correct to say that benzene has "greater" aromatic character than aniline? (Because in aniline, the lone pair on nitrogen can interact with the aromatic ring through delocalisation.)

No, it's not correct.

One should look at the hydrogenation of the compounds you have in mind. You can start with toluene and do the classical analysis always presented for benzene. This would give an answer at least if aromaticity is treated as a thermochenical property. Another possibility is to compare H-NMR spectra using the idea of ring current. And so on.

It is also a matter of usage of terms. Electron withdrawing substituents slow the rates of electrophilic substitution and the effect is usually expressed as increasing the energy of the intermediate-activated complex and not on changing the energy of the ground state. This makes some sense and it is easy to see the effect.  However, electron rich compounds such as hydroquinone or aminophenols are very easily oxidized suggesting that the resonance orbitals are of higher energy. I have been unable to find heats of hydrogenation for various substituted benzenes, they might give a clue. At the moment without some actual data it might be fine to  suggest some research either in the literature or in the lab into the subject. My feeling is that there is an effect and the data is already out there.


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Finding the order of decreasing rate of solvolysis of the following molecules with aqueous ethanol

In this question, I need to find the order of decreasing rate of solvolysis of the above molecules with aqueous ethanol.
I think this can be done by finding the species that can form the most stable carbocation intermediate (and consequently, solvolysis will easily take place).
Now, the third molecule is obviously going to produce the least stable carbocation (since it can only form a secondary carbocation and no hydride shift is possible).
That leaves with us the first two molecules.
In the first molecule, the carbocation can be stablised by resonance. I believe that hyperconjugation from the methyl group will also help.
In the second structure, the carbocation formed will be tertiary and it too is relatively stable.
I had this notion that resonance in the first structure would make the carbocation formed by it the most stable.
So, my answer was 1>2>3.
But it seems that I am wrong. The correct answer is 2>3>1 (ie. C).
Could someone tell me why this is the correct answer (and also clarify the mistakes in my assumptions)?
Thank you for the help.

Vinyl halides really don't undergo solvolysis so c is the only possible answer

The reaction can be safely assumed to be an SN1 reaction since, the solvent (aqueous C2H5OH, which is the nucleophile as well) is a polar species.
(2) shall yield a tertiary carbocation and (3) shall yield a secondary carbocation by the cleavage of the carbon-halogen (C-Br) bond.
However, in case of (1) cleavage of the said bond to yield a carbocation would not be that easy. Acquiring of double bond character is responsible for the strengthening of the bond and hence, difficult cleavage.
Further, the more electronegative a species is, the worse it is at handling positive charge. Hence, even if the carbocation does manage to form, it would be unstable. That's because the positive charge would reside on an sp2 hybrid carbon atom and as the 's' character of hybridisation in an atom increases, so does its electronegativity.
The stability of carbocations so formed would be: (2)>(3)>(1). And this order shall also mirror the decreasing rate of solvolysis of the given molecules.


text:
Is there any magnetic resonance spectroscopy that based on orbital angular momentum?
The different momenta of electron and nucleus are described by different quantum numbers.
(a) l and m$_l$ describe the orbital angular momentum and quantization on its component along an arbitrary direction, respectively.
(b) s and m$_s$ describe the electron spin angular momentum and quantization on its component along an arbitrary direction, respectively.
(c) I and m$_I$ describe the nuclear spin angular momentum and quantization on its component along an arbitrary direction, respectively.
Those space quantizations can be exploited for spectroscopy. For example, m$_I$ $\implies$ the direction of nuclear spin angular momentum w.r.t. external magnetic field is quantized $\implies$ when magnetic field applied, because of the multiple possible directions w.r.t. external field, the energy of initially degenerate states will split according to the direction of angular momentum (i.e. according to the value of $m_I$), and the gaps between the now non-degenerate states can be probed by spectroscopy, and that gives structural information.
magnetic resonance spectroscopy based on $m_I$ = NMR
magnetic resonance spectroscopy based on $m_s$ = EPR
Is there any kind of similar magnetic resonance spectroscopy based on $m_l$? When an external magnetic field is applied, won't the energy of electron also split according to $m_l$? But I do not recall any spectroscopic method based on $m_l$.

You might look into the Zeeman and Paschen-Back effects https://en.wikipedia.org/wiki/Zeeman_effect

I think Zeeman effect cannot be considered as "pure" $m_l$, because Zeeman effect led to the concept of electron spin.

@M.Farooq Indeed! These were just the first things that came to mind, and I hoped they might nudge someone towards what might be an answer

@Ian Bush, What about Stark effect? Basically, I think the OP wanted to see how $m_l$ is experimentally detected. Interesting question.

@M.Farooq I almost mentioned the Stark effect, but didn't because of the OP's focus on magnetism. But it's certainly very closely related.

@IanBush, Right, I think there is no magnetic spectroscopy analogue for detecting $m_l$.

@M.Farooq Is there any known reason why is that so? Certainly, $m_l$ does express some kind of space quantization, and so seems like that could be exploited for spectroscopy.

@TheLearner, I may be wrrong but there is no magnetic moment associated with orbital motion. The $m_l$ values become observable in Stark experiments. I will try to modify your question properly and ask it in SE physics.

@M.Farooq: there is actually a magnetic moment associated with orbital angular momentum. Classically this is easy to picture as a moving charge generates a magnetic field that can interact with the external field. This interaction is however roughly a factor of two smaller than the interaction of the electron spin with the external magnetic field (compare their respective $g$ factors). In a strong magnetic field, orbital and electron spin angular momentum decouple from each other and recouple to the external field axis. This should be observable with EPR-like techniques.


text:
How do I quantify a mix of SiO2 (amorphous), silicon (amorphous, polycrystal), graphite (dust) in a solid sample?
I collected solid samples from a plugged tube-side of a heat exchanger. These deposits include compounds such as hexachlorodisilane, hexachlorodisiloxane, pentachlorodisilane, graphite dust and amorphous/polycrystalline silicon dust. In the process of collection, the sample has been exposed to moisture, and hence the disilanes should have hydrolyzed to silica/HCl, siloxanes, etc. and the graphite and silicon should remain unaltered.
How do I go about quantifying each component? I am expecting this will be a multi-step analysis with separation/extraction followed by a combination of analytical techniques.

It is always beneficial to state what instruments you have access to.

I would recommend sending a sample to a company or lab that can run a Raman spectrum, an ATR infrared spectrum and an SEM-EDAX. The companies will probably run samples for free if you can convince them you want to buy an instrument[or are very nice]. Careful preparation of standards should give reasonable quantitative data from at least one of the methods for each component. I do recommend that much more care be exercised in sample collection and preservation.


text:
Does Fischer esterification occur when mixing alcoholic drinks with lemon juice?
I would assume it does happen because the ethanol in the drink would be in an acidic environment because of the lemon juice and would react with the citric acid in it. But the thing is, how come we don't notice the formation of this ester when we drink, let's say, vodka mixed with lemon? Is it because the concentration is just too low for actually tasting it?
I have already searched some properties of the product of the reaction between ethanol and citric acid and found out it is odorless and colorless, but didn't find anything about its taste.

Beside kinetics considerations... Odour is, by the way, a very important part of what commonly we refer to as "taste". For the rest looks at the concentration of the reactants.

A mix of alcohol and acid with some acidity given time will reach an equilibrium of alcohol, ester, water and acid. At the pH levels and concentrations of a vodka martini with a twist it might take a long time but it will happen.  The question is how long do you wait before you drink it.  This is an important factor in wine production and why wine changes with age.  http://www.diwinetaste.com/dwt/en2007066.php
If water is removed esterification is a reasonably fast reaction taking minutes to hours for completion, but if water is in excess or acid is low I can imagine it requiring months to years for equilibrium.


text:
At constant temperature, can the solid phase be more stable than the gas if pressure decreases?
In a phase diagram, the triple point is the point at which the three phases can coexist. If we consider, for example, CO2

If we go below the triple point by decreasing pressure and keeping temperature constant, we get to the gas phase.
My question is: Is there any substance for which, from the triple point, if the pressure is decreased by keeping the temperature constant, the solid phase is reached instead of the gaseous one?
As shown in the sketch below:

In order for that to happen, the solid phase would need to have a larger volume (lower density) than the gas phase (and of course the liquid phase as well).  While there are of course cases where the solid is less dense than the liquid (water), I can't think of a single case where the solid is less dense than the gas.

Related https://physics.stackexchange.com/questions/669506/material-which-becomes-less-dense-as-pressure-is-applied

That's not really possible. To see why simply express the Gibbs' free energy for each of the phases of the homogeneous substance in differential form:
$$dG = Vdp - SdT$$
We assume constant composition.
We can identify the volume $V$ as the partial derivative of the free energy with respect to pressure at constant temperature:
$$V = \left( \frac {\partial G}{\partial p}\right)_T$$
Since volumes are always positive quantities, the general response of a pure substance at constant temperature to a decrease in pressure is a decrease in free energy ($(dG)_T = Vdp$). The only way we can have a solid with a lower free energy than the corresponding gas as we lower pressure is therefore if there is a discontinuity (break from monotonous behavior) marked by a sudden increase in the free energy at the transition point from solid to gas. But since $V$ is, by definition, greater in a gas than in a condensed phase such as a solid, the free energy change at the transition point from gas to solid is $(d \Delta G)_T = \Delta Vdp >0$ if we decrease the pressure ($dp<0$), so that the solid is identified as the less stable phase.

It may be difficult or impossible with a pure substance, but a solid may precipitate from a supercritical fluid when the pressure is decreased and the fluid becomes an ordinary gas.
For instance, in principle a solution of sodium chloride in water could be heated and compressed unto the supercritical range of the water. Under supercritical pressure the water would remain dense enough to solvate the ions of the salt, which therefore remains dissolved; but if the solution is decompressed and the water becomes steam, the sodium chloride loses its solubility.

There are many ways to change the solubility, changing the ratio of solvent, evaporating the solvent, or adding common ions to name a few. What I was thinking was if there is a process where the total entropy will increase when the same substance goes from liquid/gas to solid. This may seem like a contradiction, but it is not. In the process you mention a solid is produced and the entropy of the universe increases as a net result. But... is it possible to decompress a liquid/gas and turn it into a solid? I think not, since there will be nothing else to increase the entropy.


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why is $f_{ES}=\frac{[S]}{[S]+K_M}$
I'm working my way through the chapter on enzymes in "Biochemistry" with Stryer et. al, 9th edition. In the current paragraph it's claimed that the fraction of active sites bound to substrate in an enzyme can be approximated by:
$$f_{ES}=\frac{V}{V_\mathrm{max}}$$
And fair enough, that sounds reasonable. It also claims, however that:
$$f_{ES}=\frac{[S]}{[S]+K_M}$$
This i can't quite grasp. Why would this be the case?
I tried to arrive at this expression by rewriting $V_0=V_\mathrm{max}\frac{[S]}{[S]+K_M}$ in terms of $V_\mathrm{max}$ and then dividing the expression for $V_0$ with the expression for $V_\mathrm{max}$. So far, no such luck.

Chemistry SE site strongly recommends plain text titles for index/search reasons and due possible displaying issues in question lists.

If you already know that
$$V_0=V_\mathrm{max}\frac{[S]}{[S]+K_M}$$
you divide by $V_\mathrm{max}$ to get
$$\frac{V_0}{V_\mathrm{max}} = \frac{[S]}{[S]+K_M}$$
Both sides of the equation tell you what fraction of active sites is occupied by substrate.


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How does the boiling point of water inside a pressure cooker change with temperature?
My question is why a pressure cooker would affect the boiling point at all.
To my understanding, the boiling point of a substance is defined by 'The temperature at which the vapor pressure of the substance equals the surrounding pressure' and according to the Wikipedia article regarding pressure cooking, 'In a sealed pressure cooker, the boiling point of water increases with increasing pressure'.
Now say we fill a pressure cooker with water and turn on the heat for a while. Eventually, it'll start vaporizing and all the pressure accumulated inside the pressure cooker due to vaporization of water can be considered 'vapor pressure', but wouldn't that only mean that the vapor pressure has merely risen? Why should that affect the boiling point at all, when the 'surrounding pressure' itself is still the same? Does a change in vapor pressure also affect the boiling point in any way?
After a bit of thinking, I also started wondering if the expansion of the air inside the cooker alone was enough to increase the pressure sufficiently, but I am not really too sure about it and I couldn't find much clarity on why this pressure increase occurs otherwise.

So when the pressure cooker hits 100C all the water should turn to vapor. Now what? Given the volume difference of liquid:vapor is roughly 1:1000, unless you put very little water in the pot it simply can't fit all the vapor.

Partly related: Minimum amount of water in pressure cooker

The surrounding pressure in the boiling point context is the pressure acting on the liquid, what includes also partial pressures of eventual vapors.
There is always some air in a (fully)  sealed cooker and the pressure is always higher than vapor partial pressure, so boiling point is always higher than water temperature.
$$T_\mathrm{boil}=f(p_\mathrm{air}+p_\mathrm{water}(T))$$
That is obviously not true for a pressure cooker with a pressure relieve vent. With the vent, air gradually escapes when the vent relieve pressure is reached. The boiling point raises only to the temperature, at which the saturated vapor pressure equals to the vent relieve pressure.

Okay so we can say the now vaporized water exerts a certain pressure on the liquid, and that in itself causes the boiling point to go up.

Sure. Pressure of vapor is not different to pressure of other gases.


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Rate of Decomposition of Sodium Thiosulphate
I want to calculate the activation energy of the decomposition of sodium thiosulphate, as a part of my school research project. I am aware that I must use the Arrhenius equation where the rate of reaction is measured at different temperatures. I will be using HCl as the acidified solution.
The problem which I am facing is in measuring the rate of reaction. The standard method is to use measure the time taken for the sulphur precipitate to cover a 'mark' which is placed at a specific position on the conical flask. This method only provides the time, which can be converted to 1/time. Is this suitable for the Arrhenius equation? If I need the rate in terms of mol/s*dm^3, what method can I use to measure the change in concentration?

See this video https://www.youtube.com/watch?v=FdBqD4ylOdw

Thanks for your comment. The video essentially neglects the amount of sulphur formed which is not what I'm looking for. I need a method to measure or approximate the concentration of the sulphur precipitate formed.