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What is a better laboratory set up for dehydration of ethanol to form ethene?
as shown in the question, I will have to design a lab on dehydrating ethanol to form ethene. The current method I am using is by heating up ethanol and sulfuric acid with a hot plate under a water bath and collecting ethene by water displacement method. But I wonder how I can improve this method, without changing the chemicals involved.
And I also want to find out how to identify the formation of double bond to test ethene is formed. I used potassium permanganate but because the ethene is a gas, I am not sure if there are other methods of testing it without having the gas escaping.
Thank you in advance.

Are you intending to collect the ethene?

Why improve a method working all right ? Is it even possible to improve such a method ? What do you mean by "improve" ? A good test for double bonds is to make your gas bubble through a diluted solution of bromine $\ce{Br2}$, which is brown due to the presence of bromine. In the presence of ethene (or of any double bond)  this solution gets discolored. And if you are not allowed to store bromine water, you may produce it just before use, by mixing bleach ($\ce{NaClO}$ solution) with diluted HCl and NaBr. And then destroying it by an excess $\ce{NaOH}$ after use.

@Waylander yes, the volume of ethene collected will be my dependent variable.

@Maurice may I ask for more details? How do you bubble bromine water to the ethene? Because the ethene collected by water displacement method, I am unsure how to add bromine water without having the ethene escaping once I remove the water from the tube which contains the ethene collected. Thank you.

@xxxx. Apparently your ethene is getting out of some plastic or rubber tubing bubbling through water. So get a fraction of the gas in a test tube under water. Close it with a stopper under water. Save the tube half filled with ethene. Prepare a 1 mL syringe filled with bromine water. Insert a needle between the stopper and the test tube. Inject this brown liquid through the needle into the test tube. Stir. You should see the brown color disappear. If you use more ethene, for example collected in a 100 mL glass balloon instead of a test tube, you should see small droplets of $\ce{C2H4Br2}$.


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Mechanism for Ce(IV) Oxidation of Naphthalene
S. Cotton's Lanthanide and Actinide Chemistry reports that naphthalene can be oxidized to naphthalene-1,4-dione using cerric ammonium sulfate (CAS), $\ce{MeCN, H2SO4},$ and $\ce{H2O}$.1 No mechanism was provided, and I was unable to find a complete mechanism for this transformation in the literature. From what I did find, Love et al.2 have reported the first step of this mechanism is the formation of the radical cation while the second step is the attack of a water molecule. After these two steps, the authors of the paper do not give any further information. I proposed a mechanism below, but I have a few concerns. To start, why is sulfuric acid used when protons seem to be lost during the transformation? I assume this is to prevent the cerium from reacting with the water to form hydroxides as it can be quite basic, but it is nonetheless unsettling.

References:
1Simon Cotton, In Lanthanide and Actinide Chemistry;  Inorganic Chemistry, A Wiley Textbook Series, John Wiley & Sons, Ltd: Chichester, England, 2006 (ISBN-13: 978-0-470-01005-1).
2Brian E. Love, Alexander L. Simmons, "Substituent Effects in the Oxidation of 2-Alkyl-1,4-dialkoxybenzenes with Ceric Ammonium Nitrate," Tetrahedron Lett. 2016, 57(50), 5712-5715 (DOI: https://doi.org/10.1016/j.tetlet.2016.11.042).

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Why doesn't relative humidity tend to 100%?
Relative humidity is defined as:
$$\frac{\text{pressure of $\ce{H2O}$  vapours}}{\text{vapour pressure of $\ce{H2O}$ at that temperature}} \times 100$$
Suppose the temperature is fixed and  I apply Le Chatalier's principle, then would relative humidity always try to be 100%?

Temperature and pressure are not fixed in the atmosphere. Infinite water is not usually available to try and saturate the air. Yes, water evaporates to try and hit high relative humidity, but the system is much more complex.

Sure, it always tries to be 100%, but rarely succeeds.

If atmospheric air masses had had constant temperature, the relative humidity would have converged to 100%.
But, because of dynamic nature of atmosphere behavior, there are processes of cooling down and warming up of air. The former eventually leads to water condensation and precipitation, while the latter leads to decreased relative humidity, compared to the initial conditions.
By other words, with all that raining and snowing, how one could expect atmosphere in the whole can ever get saturated ?

(The user served as an enlisted airfield meteorologist of the Czechoslovak army in 1989-90)

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Molecular Orbital Diagram
Chemists have revealed that $1 \lambda^{2}, 2 \lambda^{2}$ -disilene $\left(\mathrm{Si}_{2}\right)$ is diamagnetic while disulfur $\left(\mathrm{S}_{2}\right)$ is paramagnetic $(\mathrm{S}=1)$. Draw the MO diagram (valence shell only), determine the bond order for those two molecules and conjecture which element has higher $3 \mathrm{~s}-3 \mathrm{p}$ energy gap.
(Note that $\sigma_{\mathrm{g} 3 \mathrm{pz}}$ is lower-lying than $\pi_{\mathrm{u} 3 \mathrm{p}}$ in $\mathrm{S}_{2} .$ )
It seems that the paramagnetic type is not very necessary - however, I couldn't come up with the correct MO diagram in either case.

use the easy to find diagrams for C2 and O2 as guides

Welcome to Chemistry! This appears to be a homework question, please share your thoughts and attempts towards the solution. In short: It would be a good idea to post whatever you have come up with so far.

Related: Why is diboron (B2) paramagnetic?

@Andrew Would the diagrams of $C_2$ and $O_2$ be sufficient to draw the original molecular orbital diagrams? Could you please elaborate more about what you mentioned, still can not figure out the relation between it and the given problem.

@Martin-マーチン How diboron's being paramagnetic such related to the stated problem?! The original question asks to draw the MO diagrams with valence shell only. I would be pleased if you could provide more insights about the proposed problem specifically.

Can anyone suggest concrete ideas on how this question could be solved more rigorously in terms of formal diagrams?

The answer I have given in the question I have previously linked gives you generic MO diagrams for diatomic molecules with and without mixing. These should at least provide enough guidance to add the requirements to this question according to the homework policy. Please [edit] the question to include you thoughts towards the solution of the exercise. And while you are at it, please add a citation for the original source of the exercise.


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Is vapor-liquid equilibrium in contradiction with Pascal's law?
When in a closed container a small amount of liquid is introduced then at a fixed temperature $T$ we will have a vapor-liquid equilibrium and the pressure of the vapor at that temperature is called vapor pressure. In order to derive the relation between vapor pressure and temperature we must equate their chemical potentials i.e.:
$$μ_l (T,p)=μ_g (T,p)$$
What I don't understand is how is it possible for two phases to have the same pressure. For example when we have in a closed container liquid water with its vapor in equilibrium, the pressure of vapor can be approximated by ideal gas law. The liquid will also exert a pressure in the walls of the container (which can be approximated by another equation of state). Will the pressure of the vapor equal the pressure of the liquid? Initially I thought that it won't because the  pressure that the liquid exerts is:
$$p_l=p+p_g$$
where $p$ is the pressure from the corresponding equation of state for the liquid and $p_g$ the extra pressure that is transferred due to Pascal's law. But in a vapor-liquid equilibrium and in general when a system reaches equilibrium $T,p,μ$ must be same everywhere in the system. Where I got it wrong?
Edit
After reading the comments I think I should add the following. In order to find the coexisting curve that is the set of points $(T,p)$ where the phases coexist we must equate their chemical potentials. My problem is where the pressure ,when we write $μ(T, p)$, refers. Is the total pressure of the container (from both liquid and vapor) or the vapor pressure? I am thinking it must be the pressure of the container which must be the same throught both phases (equilibrium). But if is this pressure then I can't understand how for example Clausius-Clapeyron can be derived (where the $p$ refers to vapor pressure). Is the pressure of the liquid (neglecting gravity) so small that we can approximate the total pressure with that of vapor?

Hydrostatic pressure (your $p_g$) equals zero at the surface level (or in other words, at zero depth). And if a vessel has a hole below the surface, well, then the things won't be in equilibrium, because water will rush out.

If there is additional gas, than it is the total gas pressure for the liquid versus partial pressure for vapour.

The $p_g$ pressure refers to pressure due to gas. I have neglected gravity. I am asking if the pressure from the gas is transmitted to the liquid so the liquid exerts its pressure (due to the equation of state) plus the pressure from the gas.

It is quite trivial question from mechanics. Gas exert the same pressure on liquid as liquid on gas. If it was not so, the phase boundary would be moving until they would be the same.


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What do the phase space and slope fields of the iodine clock reaction look like?
The iodine clock reaction is a fun experiment shown to undergraduates to illustrate that chemical reactions can show periodic behaviour. Presumably, or hopefully, there is a system of differential equations for this system that can be studied qualitatively. Some qualitative approaches are visual, including slope fields and phase diagrams. An example of a qualitatively discoverable property of a system is the existence of attractors.
Has someone made visualizations of the phase space or slope fields for the differential equations governing the iodine clock reaction?


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Copper plating with electrolyte not working (Cu(OH)2 vs CuSO4)
I've been experimenting using electroplating, and I recently made a solution by placing two copper pieces connected to a voltage source, in water. I suspect that this created copper(II) hydroxide, as it matches the color of that specific solution (green-blue) and doesn't stay dissolved in the water (it sinks to the bottom after a day).
A lot of people use copper(II) sulfate for an electrolyte during copper electrolysis, so I assumed $\ce{Cu(OH)2}$ would work the same, but if doesn't seem to be. A coin was attached to the negative wire, copper was attached to the positive wire, but attempting to copper plate a coin in this solution has failed.
Why is that? Both copper(II) hydroxide and sulfate have $\ce{Cu^2+}$. Shouldn't that be attracted to the negative wire, and append the copper to the desired item?
Why does $\ce{Cu(OH)2}$ not behave like $\ce{CuSO4}$ during electrolysis?
I don't know too much about chemistry so if you explain it in simple terms, I would appreciate it!

If you are going to do anything differently than is normally done, you should understand what you are going to do and why it should work.(*)
The essential (but often not sufficient) conditions for successful electroplating by any metal are:

The used metal salts/compounds are soluble, so ions can migrate within the solution in applied electrostatic gradient. That is not the case of copper in copper hydroxide nor alkalic copper carbonates, which are insoluble.
The metal salt has sufficient concentration, so electrolytic current reacts primarily with this salt, at least on the cathode.
The current density on the plated electrode is low enough ( typically $\pu{10 mA/cm^2}$).

The solution composition for successful creation of shiny metal surface can be tricky, based on intensive empirical reasearch, involving solution modifiers like sugar or soluble polymers. The goal is mostly decreasing ion mobility and/or the rate of electrode reaction to support even metal deposition.

(*) It does not fully apply in the basic research, which uses often "what if.." approach, but this is not the case.


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Are there any tests which distinguish between triglycerides and polar lipids/impurities that can be conducted in a high school setting?
I am currently writing a research essay about using a mixed polar (methanol) and non-polar (heptane) solvent system to extract oil from mango seed kernels. One of the difficulties of using polar solvents is the fact that it also extracts polar lipids which can't undergo the transesterification process to produce biodiesel.
I need to find a method which can distinguish between non-polar lipids and polar lipid/impurities in the extracted substance. I've had a look online and read about the Sudan III test, but I'm not clear on whether it works only on non-polar lipids or just all fats in general.

What are the advantages of using methanol in combination in this situation, as opposed to heptane alone?

@Beerhunter Adding a polar solvent is supposedly able to increase the yield of triglycerides because it is able to extract (via hydrogen bonding) any remaining non-polar lipids held in polar lipid complexes.

Is this theoretical or empirical research?

@Beerhunter High school research with experimental data.


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What exactly is a 'chiral centre' in a carbon compound?
The typical progression (in books, YouTube videos) of explanation of chiral carbons in a compound is as follows:

Chiral compounds are non-superimposable by their mirror-image;
It is easily seen that only carbon atoms with all substituents different are chiral;
In a carbon compound with more than one atom, (2.) is exactly how you check for chirality;
It isn't necessary for a compound to be chiral despite having chiral carbons.

It seems to me that something between (2) and (3) is being eaten up. How does a chiral carbon in an chiral compound show the non-superimposability property? How does it satisfy the basic definition of chirality?

Well, you seem to be agreeing with (2): a chiral carbon is chiral, because it is not superimposable with its mirror image.

But if a compound is achiral, how does one check for the chirality of any one carbon in it?

I can only repeat what has already been said: you check the four substituents of a carbon, and if they all are different, then this atom is chiral. Whether the compound as a whole is chiral, that's another story.

I'm probably missing something very obvious. Just the same, I'll restate my question: Why does 'you check the four substituents of a carbon' work? Even if that's a, say, working rule, then what's the rationale?

You make two actual 3D models out of matchsticks and bubble gum, or maybe some other material. You rotate them this way and that, and try to superimpose them all day long, and you just can't. That's your rationale.

Your title doesn't exactly match the rest - if you have a carbon with 4 different substituents, this part of molecule is asymmetric. in theory it might end up still achiral if one substituent could flip to the opposite side of center. For carbon it does not happen, but various complexes show pseudorotation that does pretty much that.

Skipping axial and helical chirality altogether, no to point #2: there are examples of atom centred chirality without a carbon atom in the centre.  The careful oxidation of thioethers may yield chiral sulfoxides:

(Han et al., Chem. Soc. Rev. 2018, 47, 1307-1350, doi 10.1039/C6CS00703A).
Here, chiral requires just the two carbon substitutents to differ, because the third is oxygen, and the fourth the lone electron pair.  Thus, applying CIP rules as usual where the lone electron pair has the lowest priority, DMSO is not chiral, contrary to methyl phenylsulfoxide which is.
Just as with stereogenic centres around carbon, deprotonations on the adjacent prochiral carbon atom may proceed with preference for one hydrogen over the other:

(Solladié, Synthesis 1981, 185-196, doi 10.1055/s-1981-29378).

TLDR: They deal with similar, but different, notions of "chirality".

First, the definitions, from the IUPAC Gold Book:

asymmetric carbon atom
The traditional name (van't Hoff) for a carbon atom that is attached to four different entities (atoms or groups), e.g. Cabcd.
chirality centre
An atom holding a set of ligands in a spatial arrangement which is not superposable on its mirror image. A chirality centre is thus a generalized extension of the concept of the asymmetric carbon atom to central atoms of any element, for example N+abcd, Pabc as well as Cabcd.
chirality
The geometric property of a rigid object (or spatial arrangement of points or atoms) of being non-superposable on its mirror image [...]
chiral
Having the property of chirality. As applied to a molecule the term has been used differently by different workers. Some apply it exclusively to the whole molecule, whereas others apply it to parts of a molecule. For example, according to the latter view, a meso-compound is considered to be composed of two chiral parts of opposite chirality sense; this usage is to be discouraged. [...]

From the first and second, we see that the idea of a "chiral carbon" is unnecessarily restrictive, because other elements such as nitrogen or phosphorus can also have four different substituents (for elements in Period 3 and higher, this includes lone pairs). This isn't the point of your question, but is worth mentioning.
From the second and third, we see that the idea of a "chiral centre" is related to, but is not the same as, the idea of "chirality". The "chiral centre" refers to chirality in the local environment of the atom (it says that the atom is not superimposable on its own mirror image). On the other hand, the term "chirality" is almost exclusively used to refer to an object as a whole, i.e. the molecule and not just one particular atom. In fact, IUPAC discourages the usage of the word "chiral" in the local sense, as seen in the fourth definition.
The idea of "local chirality" (as defined by a "chiral centre") is useful, but one should not read too much into it. The presence of "local chirality" about a carbon atom (for example) can be a quick and easy guide to determining whether the molecule as a whole is chiral, and no doubt you will have seen examples of these. Why does this work? Very loosely speaking, if the "local chirality" isn't "cancelled out" by any other factor, then the molecule as a whole has some net chirality. More formally, if the molecule is to be achiral i.e. superimposable on its own mirror image, then the "local chirality" must be reflected onto something that is its own enantiomer (since it is locally chiral, it can't be reflected onto itself).
At the same time, a compound can have chiral centres but not be chiral: this is the case with meso compounds like meso-tartaric acid. Here there is local chirality at C-2 and C-3, but in a sense it "cancels out": C-2 reflects onto C-3 and vice versa. The opposite is true, too: a compound can be chiral but not have any chiral centres, like an allene.

To deal with just question 2 (which is not a complete picture as chiral centres may be non-carbon atoms and some molecules are chiral even without having an individual chiral centre).
A carbon atom with 4 different things attached to it is necessarily chiral because of its geometry. The bonds around the carbon will form an approximate tetrahedron and any tetrahedron with 4 different things on its corners will be chiral because its mirror-image will be different. This is a fundamental property of the geometry of a tetrahedron and can be tested on any scale by building such a tetrahedron and comparing it to its mirror image.
Try, for example, colouring the corners of a tetrahedral die with 4 different colours. Or build a tetrahedral structure  with a piece of plasticine and matchsticks, colouring the match-ends with 4 different colours. With any such structure you will find that its mirror image is different to the original.
To fully understand this you would need to understand the mathematics of symmetry and group theory. But those simple models should be enough to convince you that it is inevitably true.


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Which is more electronegative, Anions or Cations?

Electronegativity is the tendency of atoms in covalent bonds to attract electrons closer to themselves (I'll admit I realised ions do not form covalent bonds only after I finished writing).
Electronegativity depends on a variety of factors including the number of protons, the number of electrons in the outer shell and the ionic radii. In general it depends on the effective nuclear charge (or the pull)

Generally Atoms towards the left end of the Periodic Table are more electropositive (groups 1,2,3...) while those at the right end of the Periodic Table are more electronegative (group 17,16...). So $\ce{Cl}$ should be more electronegative than $\ce{Na}$.
Now coming to ions, it should be noted that in Chlorine the outermost shell is completely filled so its nearly impossible for Chlorine to continue attracting due to electron repulsion, so its EN must have reduced.
On the other hand, $\ce{Na+}$ is positively charged and so it should be attracting more electrons but isn't its octet already filled so should it be still attracting electron?

Is it incorrect to compare qualitatively the electronegativities of ions? I wasn't able to find much on the google about it.

Electronegativity is defined for elements, not for ions.

I would look at electron affinity and ionization energy instead, and read more on these subjects.

@BuckThorn and thanks, but could you check the comments I made on goarkz answer?

Electronegativity is defined for an atom in a molecule. It directly means that we talk about electronegativity of chlorine we are talking about its behaviour with respect to its molecules in general.
As I understand what you are trying to compare looks more like comparing Electron Affinity (or Electron Gain Enthalpy), which is different from Electronegativity.
For example:

E.N. of $C$ in $\ce{CH4}$ $<$ E.N. of $C$ in $\ce{CO2}$
E.N. of $Cl$ in $\ce{Cl2O7}$ $>$ E.N. of $C$ in $\ce{Cl2O5}$

Greater the oxidation state, greater is the electronegativity.

Thanks, but can you help me find Electron gain enthalpy. I found its value was -2.55 eV for Na+ and 3.7 eV for Cl-per atom at separate sources. Could you verify if possible? Also if it is true then is Na+ is much better at attracting electrons than Cl- as it is releasing energy when added and therefore more stable

nope those are not for ions, only for the parent atom

@AdilMohammed Firstly Electron gain enthalpy for Na+ is opp. of its Ionisation enthalpy, so for Na+ it is -496kJ/mol which is close to 5.1eV per atom. Secondly, if we blindly see them as +vely charged species and -vely charges species then yes they yes Na+ will "attract electrons" it means if there are sodium ions in water, lone pair of oxygen will get attracted to Na+ which is hydration of sodium ion. I hope you get the second point meaning, if not feel free to comment again


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Does all fire only emit light on its outermost shell?
This question is about light emission, which may overlap with physics, but I am most interested in combustion and types of flame (incandescence, petrochemical fuel flames, nuclear reactions).
Consider this article about candle flames. In it they describe how a candle flame only emits light where it has contact with oxygen (emphasis mine):

However the reaction can only occur where the air meets the wax vapour.  This cannot happen all the way though the flame, just around the outside.  So all flames are hollow.

How accurate is this? What parts of a flame emit light?
If flames have complex scattering and absorption properties distinct from the burning substance, does this only apply to the outside regions, with the interior being goverened by whatever fuel is burning?
I suppose "fire" is many different phenomena, so are the following flames also hollow?
A: Campfire
B: Rocket exhaust
C: Oil platform explosion
D: The Sun

The Sun is not hollow in any sense. Best not to extrapolate from a candle flame.

Emission does not need oxygen at all. It needs high temperature, that was achieved by previous oxidation by oxygen. In fact, it is the opposite - as the oxygen content decreases by reacting, emission increases, as the temperature is higher.

Does this mean that the article about candle flames is inaccurate? They demonstrate the effect.

Candle flames glow because they are sooty. Incomplete combustion leads to soot, which glows due to black body radiation from the heat. This can only happen in the outer part of the flame where there is oxygen because you need some combustion. Much of the inner part is vaporized tallow and does not have a high enough oxygen content to burn. This only applies to small-ish flames that are fed via atmospheric oxygen. Compare with a Bunsen burner, which does premixing, and the flame does not glow because combustion is nearly complete.

Flames are not all the same. The burning gases in a bunsen-burner flame (or a gas stove) are premixed with air enabling combustion throughout the flame. Candle flames don't do that and a re bad basis for judging other flames.

Thanks everyone. There is some really good and useful information in these comments. Is there any chance someone who is kowledgable could formulate a full answer? Especially the specific ABCD options and reasoning for them?

The answer is NO.
The article you quote makes a completely unwarranted generalisation that "all flames are hollow". This is true of some flames but only because the fuel that is burning is only able to burn when mixed with oxygen from the air. In those flames, the flame is "hollow" because only in regions where air can mix with the fuel can the fuel burn and the rate of mixing is dependent on the flow of fuel and the rate at which air can mix or diffuse into the fuel rich region. Bunsen burner flames are like this when the air is not premixed with the fuel gas (because the air intake on the bae is closed) and candle flames are like this because the wax has to evaporate to mix with air.
A bunsen flame when the air intake is open is not like that as the fuel gas and air are pre-mixed and can burn throughout most of the region of the flame.
Another factor worth considering is what causes the emission of light. In candles and many other poorly controlled flames, there is a lot of light from incandescence not from the underlying burning reaction. Incandescence involves black-body emissions from hot solid particles. Those particles are usually small particles of solid carbon produced from partial, incomplete burning of hydrocarbons. They emit light because they are hot. Candles are tuned to create a lot of these particles as this gives far more light than complete combustion. In contrast, a well-controlled bunsen flame emits very little light (the light is not black-body emission but comes from specific excited states in the reactants).
As to how this applies to your specific examples here are some analyses.
1. Campfires
Campfires are a mixture of various phenomena. They have flames  a little like candle flames. But they also have very hot solids which emit a lot of black-body radiation because they are hot. Some of the emitted gasses have little emission of light as they are like a bunsen flame, but the overall effect is dominated by the hot solids. And, of course, some camp fires have far better mixing of air and fuel than others so it is hard to generalise.
2. Rocket exhaust
Most rockets need to be as efficient as possible. So they tend to have very carefully designed mixing of fuel and oxidiser to achieve maximal energy output. They are not hollow and tend not to emit huge amounts of light (compared to the equivalent candle flame anyway).
3. Oil platform explosion
Impossible to generalise. But unintended explosions don't tend to be well controlled so will have some areas where mixing is good and others where it is not so some areas are like a candle and others like a bunsen. Incidentally, movie explosions emit a lot more light than real explosions because, well, hollywood likes visual spectacle not realism (see this youtube explanation and illustration of how movies deliberately add fuel to create poorly mixed candle like flames for more visually interesting explosions).
4. The sun
The sun does not emit light because of burning (at least not in the normal chemical sense). The sun emits light because the surface is very hot. And this happens because of nuclear fusion deep in its core (not what a chemist would call burning, though a nuclear physicist might, but what do they know?). The light we see is mostly black body emissions because the surface is hot. It is certainly not hollow either.

To add to this excellent answer, you might find it interesting that it takes photons originating from the core of the sun thousands of years to reach the sun's surface.

The light emission from rocket exhaust depends very much on what kind of fuel the rocket is burning. The old NASA Space Shuttle provided a very nice contrast of extremes: the main engines on the shuttle itself burned pure hydrogen and oxygen and had flames that were nearly invisible; the solid-fuel rocket boosters on the sides, meanwhile, burned a mixture of ammonium perchlorate, plastic binder and powdered aluminum (of all things!) and had exhaust plumes that were incandescent as heck. In pictures and videos of shuttle launches you can see both engines firing at the same time.

@IlmariKaronen Good point. I am also guilty of overgeneralisation!


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The safest and easiest way of working with solid LDA
I have a 25-gram bottle of solid LDA (lithium diisopropylamide), factory sealed. I'm going to setup an aldol reaction with it, but I don't know how to organize work properly.
I think of quickly weighing some amount into a flask flushed with argon and flushing the bottle with argon for storage. Then I'll dissolve the LDA in the flask in THF and use this solution. The only concern is stability of the rest of LDA in bottle.
If you have some ideas how to organize it better, please share.

The best way is to order solution of LDA in THF, which is available commercially in 1 M or 3 M concentrations. If you already have the solid, you may use it as suggested. We have used LAH in solid form (which is highly flammable) without trouble in that way (even using $\ce{N2}$ to flush it).

@MathewMahindaratne the solution is obviously better, but if I order it now it will arrive in couple months (we have delivery troubles)

Once you open it, it’s not going to last long without a glovebox. You’d be best quickly making up a few flasks of a solution and sealing them under N2. Much easier to dispense

I agree with @NotEvans. Dissolve the whole lot in THF under nitrogen and store it in the fridge. Titrate a sample every time you plan to use it


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Should I be worried about the waste from "sugar rockets" and other potassium nitrate fireworks?
"Sugar rockets" or sparklers made of metal fillings, potassium nitrate and some carbon compound are easily made, and there is a lot of tutorials and videos on the internet suggesting them as relatively safe and non toxic DIY fireworks.
However reading on the potassium nitrate reaction, it decomposes to potassium nitrite when heated. This compound is toxic, and is suspected to be mutagenic as well. It is also an oxidizer on its own.
So I have safety concerns and questions:

Do the charred remains of potassium nitrate fire experiments need to be handled as toxic waste?
If potassium nitrite is an oxidizer, what does it decompose to when burned along with a carbon compound?

Potassium nitrate is transformed into potassium nitrite only when heated alone. And it is not really toxic, as it is added in ham to prevent its too early oxidation in air.  When heated with charcoal, sulfur or any other oxydable substance, potassium nitrate is transformed into $\ce{K2CO3}$ or $\ce{K2S}$. For example, black powder is a mixture $\ce{KNO3 + C + S}$ that reacts violently according to ; $$\ce{2KNO3 + 3 C + S -> K2S + 3 CO2 + N2}$$ And if the explosion is made in contact with air, the hot $\ce{K2S}$ is quickly oxidized into sulfate according to $\ce{K2S + 2 O2 -> K2SO4}$
Sugar rockets are made of ordinary sugar $\ce{C_{12}H_{22}O_{11}}$ and $\ce{KNO3}$. This mixture reacts according to the two following equations : $$\ce{C_{12}H_{22}O_{11} -> 12 C + 11 H2O}$$ $$\ce{4 KNO3 + 5 C -> 2 K2CO3 + 2 N2 + 3 CO2}$$
So the ashes of any combustion reaction are mostly made of $\ce{K2CO3}$ which was the main constituant of the wood ashes. These ashes was used in the Middle Age for washing clothes by our grand-grand-mothers. They are not considered as toxic


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Why is the formation of a peptide bond not spontaneous?
Why is the formation of a peptide bond from a carboxylic acid and amine not spontaneous?
I'm not looking for the free energy calculation; I have it from several sources.  I'm just asking for an intuitive explanation.  Is it because the peptide bond is planar so there's a loss of free rotation and the delta S is unfavorable?  Is the resonance stabilization in the peptide bond less desirable because the other atom (N) is less electronegative?  Do two equally sized molecules somehow have more entropy than a big molecule and a small one (water)?

Maybe I am not understanding the question correctly, but peptide bond formation can be spontaneous. For example, you could react an amide with an acyl chloride, and the amide would form spontaneously. In biology, the bond is formed by reacting amine and carboxylic acid. That reaction would be difficult uncatalysed because the acid would stay as anion and amine in the protonated form, and they are not particularly reactive.

I should have contextualized this better, but I meant in biology.  All of my sources are uniform in saying that the biological reaction -- amine and carboxylic acid -- is not spontaneous and must be coupled to the hydrolysis of ATP.
e.g. https://pubs.acs.org/doi/10.1021/ed070p134

Oh I see. Sorry, I was thinking about the activation energy and the rate of the reaction, but you are asking about the thermodynamics and the overall equilibrium. I don't know the answer to that. It might not even be possible to explain it with chemistry. Sometimes the answer is just—"we did the QM calculation and it says this" or something like that.

Based on a quite old reference 1 (which I'm using because it's available free by open access), peptide bond formation at 25 C is unfavorable only because of a large enthalpy change, on the order of 1.5 kcal/mol (6.3 kJ/mol). The entropy change is actually favorable, with $T\Delta S$ being about 1 kcal/mol (4kJ/mol), so the net free energy change is ~ + 500 kcal/mol (approx. +2 kJ/mol).
The reason for both values being what they are is because the free acid and amine groups are ionized at neutral pH. Thus, the reaction can be represented as
$\ce{R-CO2- + ^+H3N-R -> R-CONH-R + H2O}$
This can be broken into two steps conceptually:

$\ce{R-CO2- + ^+H3N-R -> R-CO2H + H2N-R }$

$\ce{R-CO2H + H2N-R -> R-CONH-R + H2O}$

As we know, proton transfer from an acid to a base typically involves a large negative enthalpy change, so reversing that to neutralize both has a positive enthalpy change. Since the products remain neutral, we never get that energy back. And the bond energies of a neutral carboxylic acid and amine are not much different from an amide and water, so the neutralization dominates.
For the entropy change, we start with two molecules and end with two, and the degrees of freedom are similar. The authors of the cited paper propose that the positive entropy change is also related to the loss of ionization. They believe that the more structured hydration shell of the ionic species vs neutral is sufficient to make the free amino acids lower in entropy than the neutral products, making bond formation entropically favorable, but not enough to overcome the unfavorable enthalpy change.
Later paper have further shown that the amide formation reaction becomes favorable around 60 C because the favorable entropy change finally overcomes the enthalpy change at that temp.
1 Dobry, A., Fruton, J.S. and Sturtevant, J.M. (1952)  J. Biol. Chem. 195, 149–154.
(Available by open access here)


text:
What is meant by the term 'biosynthetic origin'?
I keep seeing the term biosynthetic origin used in medicinal related papers.
Does it just refer to the biological precursors to the molecule?
So, for example, the biosynthetic origin of Paclitaxel is terpenes?

Biosynthetic means : " has been synthesized by a vital process, and not in a lab". For example, sugar, chlorophyll, cellulose and proteins have a biosynthetic origin, because they have been produced in a living cell

Or produced in a lab by means of biosynthesis (typically using bacteria - the product of interest can even accumulate within the cells, from which is later separated).


text:
Notation for inclusion of water in a dissolution equation
Earlier I had a student come by wanting to know how to show the dissolution of $\ce{NH4OH}$ in water. I would think we would just write it: $$\ce{NH4OH(aq) <=> NH3(aq) + H2O(l)}$$ I say this as $$\ce{NH4OH(aq) + H2O(l) <=> NH3(aq) + H3O+(aq) + OH-(aq)}$$  I am struggling to understand when and why we put the $\ce{H2O}$ as a solvent above the equation arrows vs when we list it as a reactant. She seemed to believe the professor wanted the water to be expressed on the left as a reactant. Any assistance, please?
At first I believed perhaps this had something to do with whether the reactant is molecular or ionic and whether we are looking at ionization or disassociation but this does not appear to be the case as the book shows two examples of a molecular reactant in water differently.
$$\ce{HNO3(aq) \overset{H2O}{->} H^+(aq) + NO3-(aq)} $$ and
$$\ce{HCl(aq) + H2O(l) -> H3O^+(aq) + Cl-(aq)}$$

What is NH4OH or NH4? H+ always exists as H3O+ . (H2O)n. H+ = proton, but is used as a shortcut for the above. For eventual writing and formatting of chemical or mathematical formulas or equations, see how to use  MathJax

I made a typing error and fixed it. It was supposed to be $NH_4OH$. My apologies.

$\ce{NH4OH}$ does not exist. It is just a solution of $\ce{NH3}$ in water. If a solution contains simultaneously $\ce{NH4+}$ and $\ce{OH-}$ ions, these ions will immediately react by proton transfer producing $\ce{NH3}$ and $\ce{H2O}$. Compare the pKa to be convinced. Don't speak of a solution of $\ce{NH4OH}$

You still do not get it. There is no NH4OH in water. It is decades long taught nonsense. $$\ce{NH3(aq) + H2O <<=> NH4+(aq) + OH(aq)}$$

The professor must have made a typing error on the problem statement then. Thank you.

I would say rather conceptual than typing error.

see also Ammonium Hydroxide Name

Also Related https://chemistry.stackexchange.com/q/10230/4945

There are few conceptual and factical points:
There is no such molecule in water solutions like $\ce{NH4OH}$. Nitrogen atom is not able to form 5 covalent bonds. There is either hydrated ammonia $\ce{NH3(aq)}$, either hydrated ammonium and hydroxide ions, following equilibrium:
$$\ce{NH3(aq) + H2O <<=> NH4+(aq) + OH-(aq)}$$
respectively
$$\ce{NH3(aq) + H3O+(aq) <=>> NH4+(aq) + H2O}$$
$\ce{H2O}$ formula occurs in equations, if it explicitly takes part in the real or formal reaction and in the atom count enumeration. Otherwise, it occurs just implicitly, or explicitly as (aq) or formula above/below the arrow (see MathJax link in comments for coding it)
$$\ce{HSO4- <=> H+ + SO4^2-}$$
or
$$\ce{HSO4- <=>[H2O] H+ + SO4^2-}$$
or
$$\ce{HSO4-(aq)<=>H+(aq) + SO4^2-(aq)}$$
or
$$\ce{HSO4- + H2O<=>H3O+ + SO4^2-}$$
or
$$\ce{HSO4-(aq) + H2O<=>H3O+(aq) + SO4^2-(aq)}$$

Hydronium ion is often written as the shortcut $\ce{H+}$, or $\ce{H+(aq)}$, but this does not reflect reality. It would  be a "naked" proton or deuteron which immediately attach first available electron pair, usually of an oxygen atom in a water molecule or hydroxide ion. More correctly would it be $\ce{H3O+(aq)}$ or explicitly $\ce{H3O+. (H2O)_n}$, where n depends mostly on solution molar concentration.


text:
Can sodium borohydride reduce hemiacetals?
In the synthesis of Panacene by Canesi et al. [1], a step involves the installation of an $\ce{-OH}$ group on a double bond, using the oxymercuration-demercuration procedure. However, the product happens to be a hemiacetal, as shown in the picture. With a 78% yield, it seems that the hemiacetal is not reactive towards the added sodium borohydride. Is this true in general, that hemiacetals will not be reduced to diols by $\ce{NaBH_4}$? What about $\ce{LiAlH_4}$?

References

Cyrille Sabot, Didier Bérard, Sylvain Canesi, "Expeditious Total Syntheses of Natural Allenic Products via Aromatic Ring Umpolung," Organic Letters 2008, 10(20), 4629-4632 (DOI: 10.1021/ol801921d).

Hemiacetals generally reduce with NaBH4 because of the equilibrium with the open hydroxyaldehyde in base. Given some time, I think 15 will reduce to the diol. LAH should work but perhaps not for the demercuration. Notice in the paper the NaBH4 rxn was conducted at 0oC for 1 minute. Apparently, over eduction is a problem.

I agree. It makes one wonder what the other reaction products are. I strongly suspect that there will be a significant amount of the diol reduction product.

The $\pu{600 MHz}$ proton NMR of 15 is not clean, and a lot of unrelated peaks in it. I couldn't imagine how they predicted the product by just looking at it.

The major purpose of $\ce{NaBH4}$ in the second step of given reaction sequence is the demercuration, which stops at the hemiacetal as the applied condition is the limit (at $\pu{0 ^\circ C}$ for $\pu{1 min}$).
In general under usual conditions (at $\pu{25 ^\circ C}$ in ethanol for for an hour or two), $\ce{NaBH4}$ reacts with hemiacetals and reduce them to corresponding diols. For instance, $\ce{NaBH4}$ reacts with glucose, which is a hemiacetal in aqueous solutions, to give sorbitol, the corresponding diol (Ref.1). The reaction has been done in water and at room temperature ($\pu{20-25 ^\circ C}$) for $\pu{1-2 h}$. Thus, running the given reduction in $\pu{0 ^\circ C}$ for shorter time make sense.
Also, although lithium aluminum hydride ($\ce{LiAlH4}$) is too vigorous a reagent for reduction, under carefully controlled conditions, it can be used to convert lactone to lactol (cyclic hemiacetal such as monosaccaride) without reacting further to give the corresponding diol. However, excess $\ce{LiAlH4}$ would give the corresponding diol after reacting further with lactol (Ref.2,3):

References:

M. Abdek-Akher, J. K. Hamilton, F. Smith, "The Reduction of Sugars with Sodium Borohydride," J. Am. Chem. Soc. 1951, 73(10), 4691–4692 (DOI: https://doi.org/10.1021/ja01154a061).
2.Glen E. Arth, "Reduction of Lactones to Hydroxyaldehydes with Lithium Aluminum Hydride," J. Am. Chem. Soc. 1953, 75(10), 2413–2415 (DOI: https://doi.org/10.1021/ja01106a040).
Robert A. W. Johnstone, "Chapter 1.11: Reduction of Carboxylic Acids to Aldehydes by Metal Hydrides," In Comprehensive Organic Synthesis - Selectivity, Strategy and Efficiency in Modern Organic Chemistry, Volumes 8: Reduction; Ian Fleming, Volume Editor; Barry M. Trost, Ian Fleming, Editors-in-Chiefs; Pergamon Press: Oxford, United Kingdom, 1991, pp. 259-282 (ISBN: 978-0-08-040599-1).


text:
Does 50% hydrogen peroxide dissolve gold?
We have seen that the silver electrodes on a piezo element are being seemingly dissolved by exposure to our mixture of $50\,\%$ hydrogen peroxide. I was wondering if we were to switch to gold electrodes instead of silver could this mitigate the issue. I have not found much if gold is resilient to $\ce{H2O2}$ and was looking here to see if anyone has some expertise.

It doesn't dissolve gold.

..unless chlorides are present.

Not that chlorides would stand up over time in such a concentrated peroxide solution, either... .

Unless in strong acidic solution. In that case, gold can get dissolved.

It would dissolve gold if it also had concentrated nitric and hydrochloric acids.

I have no expertise, but Pourbaix does.  Let's take a look at some of his diagrams which offer a hint to what is going on, with both silver and gold.  Or more accurately, why gold is indeed less vulnerable to attack by hydrogen peroxide than silver.
Begin with just the diagram for hydrogen and oxygen in aqueous solution.  Ross[1] gives a diagram that includes hydrogen peroxide as well as the stable species hydrogen, oxygen and water (figure from this reference):

Pourbaix diagrams typically include a pair of parallel lines, here labeled (a) and (b) representing water reduction to hydrogen and oxidation to elemental oxygen respectively.  We see that hydrogen peroxide requires a higher oxidizing potential to form than elemental oxygen, by roughly half a volt depending on concentration.  This is no surprise, of course, since elemental oxygen plus water would be more stable.  Hydrogen peroxide is, of course, only metastable.
Now let's see the diagram for silver.  There is some variation from different sources, but Wei et al. [2] show the following, which is typical.

The boundary between elemental silver and various oxidized species (essentially, dissolved silver ions in acid and silver oxides in basic solution) actually lies below the line for water oxidation to oxygen, so in principle even dissolved air could attack silver given enough time, let alone the more powerful oxidant hydrogen peroxide.  In fact, at the potentials where hydrogen peroxide appears in the diagram from Ref. 1, one may even see the oxide of silver(II)!  We are lucky, actually, that the kinetics is slow enough so the silver electrodes have some lifetime.
Now it is time to go for the gold.  Huayhuas-Chipana et al. [3] adopted the diagram below from Ref. [4], showing that gold is not oxidized at all when water is oxidized to elemental oxygen (the higher of the two red lines) and that even hydrogen peroxide with its potential at half a volt higher is only on the borderline of beginning to oxidize gold.  More over, the soluble cation $\ce{Au^{3+}}$ appears only in fairly strong acidic solution, further hindering any dissolution of gold.  Thus gold would indeed be expected to hold up more strongly in your hydrogen peroxide solution.

References
1.
P. N. Ross Jr. (2010), "Oxygen reduction reaction on smooth single crystal electrodes", Handbook of Fuel Cells
Vol. 2.
2.
Weifeng Wei, Xuhui Mao, Luis A. Ortiz and  Donald R. Sadoway, "Oriented silver oxidenanostructures synthesized through a template-free electrochemical route", J. Mater. Chem., 2011, 21, 432-438.
3.
Bryan C. Huayhuas-Chipana, Juan Carlos Morales Gomero, Maria Del Pilar, Taboada Sotomayor, "Nanostructured Screen-Printed Electrodes Modified with Self-Assembled Monolayers for Determination of Metronidazole in Different Matrices", Journal of the Brazilian Chemical Society 25(9) (Sept. 2014).
4.
J. Van Muylder, M. Pourbaix, Atlas of Electrochemical
Equilibria in Aqueous Solutions, 2nd English ed.; Marcel
Pourbaix, National Association of Corrosion Engineers: New
York, 1974.


text:
Electrolysis of water using different kind of electrode and electrolyte safest as possible for splitting demonstration
I am planning to electrolyze water for an experiment. However, I have limited resources and want to do it as safely as possible

The resources that I have:

$\pu{12 V}$ DC battery
Graphite (in the form of a pencil lead)
Stainless Steel
Alligator lead clips
Sodium Hydroxide ($\ce{NaOH}$)
Sodium Sulfate ($\ce{Na2SO4}$)

In such a condition, what would be the best electrode and electrolyte for this experiment?

What kind of battery and pencil leads? A 12V car battery might burn through your skinny 0.5-0.7mm pencil leads (your "graphite electrodes") rather quickly if your salt solution is conductive enough and/or if the leads are sufficiently close to each other. Not to mention that $\ce{2H2O(\ell) \to 2H2(g) + O2(g)}$ doesn't require that much overpotential anyway...

12V source would cause a violent rate of electrolysis, if the source is hard, or the source can be  damaged, if it is a soft voltage source without protection. Eventually both. Unless other measures are applied, consider serial resistor $R[\Omega] \approx \frac{10 [V]}{I [A]}$, where I is intended current, safe for both electrolysis and the power source.

Both sodium hydroxide and sulfate will yield same gases on the electrodes but sulfate solution will be neutral $\mathrm{p}H$ so you would want to stick with that. For the electrodes safest option is pure graphite because  stainless steel anode will probably produce hexavalent chromium which is not very good.


text:
How does temperature and other factors affect sigmatropic rearrangements?
The image below shows a [3,3] sigmatropic rearrangement. My question is why does the increased temperature favour the right side? The arrangement is technically the same except for the methyl groups on different carbons.

Here is another example:

Does this help for the Claisen rearrangement? https://chemistry.stackexchange.com/questions/107963/what-is-the-driving-force-for-claisen-rearrangements/108038#108038

The increased temperature is necessary to overcome the activation energy of the reaction. Many [3,3] sigmatropic rearrangements have very high activation energies. The products are favored at equilibrium because they are thermodynamically more stable. For example, in the first case, you are exchanging monosubstituted alkenes for disubstituted alkenes.


text:
What do we know about the structure of the amylopectin iodine complex?
I am studying a method of finding the ratio between amylose and amylopectin using the difference in color of the amylose-iodine complex(blue) and the amylopectin-iodine complex (red/brown).
Through the use of techniques like x-ray crystallography, the structure of the amylose-iodine complex seems to be fairly well understood, as iodine inside a helical chain of amylose.
However, I cannot find many papers that explicitly describe (report) the structure of the amylopectin-iodine complex. Perhaps the branched nature makes it more complex, or the variety in structure unlike amylose makes it impossible to find a general structure.
Are we aware of any certain structure of the amylopectin iodine complex? Any directions to relevant papers would be appreciated.

For clarification: What you search is a data set (e.g., as a .cif or .pdb file) to access the structure data like in the case of «Crystal structure of cellulose-iodine complex», Tashiro et al. 2019Polymer140, figures 7 (apparently based on diffraction data), 10/11 (apparently computation based)?

«Understanding Starch Structure: Recent Progress» by Bertoft appears to be an entry to consider, too.

For clarification, I am just looking for information, qualitative or quantitative, on the structure of the amylopectin-iodine complex. (or whether that can even be found, based on variety of amylopectin structure).
Tashiro et al. (2019) seems to cover both starch and cellulose, but only amylose and not amylopectin. For starch, it seems to be just the typical single helix structure.

Because amylopectin is less likely to form chanels than amylose, I think chances are lower to form a regular complex with iodine.  Because both are biopolymers, they are not a typical entry in CSD database, but as sugars not part of pdb.org either.  Maybe the author of this site used for the illustration of the «empty» structure of amylopectin X-ray based data and may hint where to find additional data (complexes with iodine). Nevertheless, continue literature search in scientific databases (web of science, scifinder).

A last resort may be to contact researchers active in the field (e.g., https://doi.org/10.1186/s40104-018-0303-8) since research in sugars has much tradition (example and its «cited by» below the publication).

Understood. Unfortunately I do not have access to resources such as web of science or scifinder, but I will continue searching with this in mind. Thank you very much for your advice.


text:
What is a cinammyl system?
I came across this term while searching for reference in March's Advanced Organic Chemistry.
It said:

LAH gives significant double reduction only in cinammyl systems, for example, with PhCH=CHCOOH.

This term is new for me and Google didn't have any  specific result for this.

Cinnamyl is the non-systematic name for the PhCH=CHCH2X system where X is a heteroatom, most frequently O or halogen

The structure you've provided is cinnamic acid.. now what do you think?


text:
Ester reacts with acyl chloride
I have worked with a somewhat long reaction sequence and got stuck on this last step. Im not sure if the entire sequence is even correct but this is what i came up with. Is this a possible reaction, if so, what is the product?
Thanks

Read about the formation of amides. You should use an equivalent of triethylamine as a "scavenger" base. Do you know why?  By the way, we are here to help, not do your homework. Let's see some effort.

The ester group is not involved in the reaction. Examine the role of the -NH2 group

Nucleophile  attacks electrophile, what can donate electrons in this scenario and what compound is electron deficient in this scenario..

Thank you for the answers.I will look into more about amides.


text:
Hydration in Chloral
Chloral consists of two OH bonds on the same carbon, which is usually unstable, and leads to the evolution of a water molecule, and converts into an aldehyde. However, this is not observed in chloral, and there must be a reason for this.
I have come up with 2 possibilities:

hydrogen bonding in chloral, although it is quite weak
high EN of the chlorines, which results in a huge partial positive charge on C

Which reasoning, if any, is correct, and why does it dominate over the other?

Usually hydrogen bonding is used to explain the stability of chloral hydrate.

As mentioned by you, the effect of two chlorine atoms is dominant

See where chloral is discussed in this earlier answer.


text:
Fluorescence Intensity and Fluorescence Quantum Yield
I plan to use ImageJ to analyze some fluorescence experiments where the pH changes. I will be using fluorescein as the indicator for the pH changes and it will be present in low concentration (like is typically the case when any indicator is used to measure a pH change).
As I understand it, I can get the relative fluorescence intensity from the grayscale intensity in ImageJ.  From the literature I have a chart of relative fluorescence quantum yield for fluorescein versus pH.  If I conduct two experiments, one at pH 12 and another at a lower pH (which is unknown), I think I can determine that lower pH simply from the grayscale intensities in ImageJ and the literature chart that I mentioned.  If the ratio of intensities is equal to the ratio of quantum yields in the literature chart, then I should be able to read the pH from the literature chart.
Assume all conditions for determining the grayscale intensity in ImageJ are identical, except that the pH of the liquid is different. So I'm going to assume that the molar absorptivity, path length, source power and concentration of fluorescein (all ionic and non-ionic forms) remain unchanged. For discussion purposes, let's say the Image J grayscale intensity at a pH of 12 is equal to 100 and at the unknown pH is equal to 60.  The ratio of these intensities is 0.6.  The question is can this ratio of 0.6 be assumed to be equal to the ratio of relative fluorescent quantum yields on the literature chart? So essentially, can a ratio of intensities from ImageJ be considered to be equal to the ratio of quantum yields, under the assumptions that I have outlined, so that I can read the pH from a chart that plots pH versus fluorescence quantum yield.

+1 Good question. You could search the literature or google to look if any experiment like this has been done before. I am not an expert on this field, but at first glance I think your reasoning is right.

Thanks for your reply, Shoubhik.  I have been searching the literature.  I haven't found a paper that shows a chart of relative intensity versus pH for fluorescein, but I have found a paper that shows charts for fluorophores typically used in biology.  For fluorescein, I've only been able to find a chart that plots quantum yield.

You must (a) check that the fluorescein absorption and fluorescence quantum yield remain the same as pH is changed and if not correct for this (b) correct for any absorption changes between solutions anyway, i.e. due to dilution etc. (c) Check that intensity scale of you detector is linear, you could use neutral density filters to do this. You reasoning about yields is correct.

Of course to get the proper intensity ratio you'd need a high pH such that all the fluorescein is unprotonated and a low pH so that all the fluorescein is protonated.

Fluorescence QY is a measure to be done carefully. Look for article explaining it step by step. After, your logic seems to be correct.

@porphyrin:  The chart from the literature shows how the fluorescence quantum yield changes with  pH.  Fluorescein can be used as a pH indicator.  Maybe I'm not understanding your point.

At MaxW:  The plan is to mix NaOH at pH 12 with a 5 or maybe 10% stoichiometric excess of HCL at pH 2 and monitor the fluorescence as mixing proceeds.  The reaction is instantaneous and I'm expecting diffusion controlled mixing. After complete mixing the pH will be of order 3 to 3.5.
I'll also be taking a photo of the base (containing fluorescein) mixing with water and a photo of  the acid (containing fluorescein) mixing with water as "control" images.

As the fluorescence yield changes with pH the absorption spectrum probably does too as may the emission spectrum (I would guess) so you will need to correct for this change. You also need to measure emission at at the same wavelength in all cases as the detection sensitivity depends on wavelength.


text:
Why is chloride measurment 0
In class we did an experiment to measure chloride in different water sources, including a lake.  We used Ion-selective electrode of Ag/AgCl technique.
We did the calibration curve and all steps and measurments in different sources of water are consistent, for example drinking water was a bit below 250ppm. But when measuring lake water we get a measurment of 0, why could this be happening?
I can't find a reason

Are you sure that the procedure was followed correctly?

Why can't this be true?

Assuming you have followed correctly the experimental procedure, the concentration in lake water may be lower then the sensibility of your electrode.

@IvanNeretin Unless they found a first lake of freshly distilled water in the world, that's impossible. They either did something wrong or there was interference from other stuff in water.

Measuring a concentration with an $\ce{Ag/AgCl}$ electrode cannot give zero, it can give a concentration "lower than $\pu{0.00xxx}$" but it cannot be zero, if a concentration is equal to zero, applying Nernst's law gives you a potential equal to ± infinite, which does not make sense.

Most likely this was an artificial "lake" sample and someone gave you distilled water instead. Making artificial water samples is not an unusual practice in teaching laboratories.
Natural water cannot have zero concentration of chloride ion. Sodium and chloride ions are very common contaminants even in pure reagents. There is no "zero" concentration concept in analytical chemistry at all. The proper practice is to write "Not detected" instead of saying 0 ppm. All it means that it was below the detection limit of the ion-selective electrode. Ion selective electrode usually measure samples in parts per million (mg/L) range. Typical limits of detections are in single digit ppm.


text:
Does the rate of effusion remain constant throughout the effusion?
Suppose a gas is effusing from a container through a small orifice at constant temperature and pressure. Does the rate of effusion (no of moles of gas effused per unit time) remain constant throughout the effusion or does it decrease with time as the no. of moles of the gas decrease in the container?

Welcome to Chemistry! What have you attempted to do with regards to the answer to this question? We need to make sure we aren't doing your homework for you.

It depends how you read the question. If p and T are constant inside the vessel then 'no' as the gas effuses more must be added to keep pressure constant, if pressure is constant outside then 'yes' as sooner or later there is no gas left to effuse :)

I am actually a student of Grade 12 preparing for a pre-medical entrance exam. So, I am clearing my doubts of chemistry here as I don't have any good chemistry teachers to whom I can ask these doubts.

For a container with a constant initial amount of gas at constant T, the rate of effusion decreases with time, since
$$\phi_N = \frac{\Delta P A N_A}{\sqrt{2\pi MRT}}$$
where A is the aperture, $\Delta P$ the pressure drop at the exit of the container, M is the molar mass of the gas, and T is the temperature. You can also consult a physical chemistry textbook to verify the equation.
Assume the pressure outside the container is zero (for simplicity) and assume the gas obeys the ideal gas law. Then the equation can be rewritten as
$$\phi_N = \rho_N A N_A\sqrt{\frac{ RT}{2\pi M}}  \\= \kappa \rho_N $$
where $\rho_N$ is the molar density of the gas in the container, and $\kappa=A N_A\sqrt{\frac{ RT}{2\pi M}}$ is a constant for a given container, gas and temperature.
Now since $\rho_N$ decreases as gas is allowed to escape, then so does the flow rate.

What is M in the equation?

What exactly is delta P in the equation? Is it equal to Pressure inside the container - Pressure outside the container? Is it equal to partial pressure of the gas? What is it?

@RIPANBARUAH Please read my answer again: M is the molar mass of the gas and $\Delta P$ the difference in pressure between the inside and outside (the drop in pressure). But for the second equation I assume that the outside pressure is zero, so that $\Delta P = P$.

Suppose the outside pressure is zero. Then can we say that delta P = Pressure inside the container = Partial Pressure of the gas?

@RIPANBARUAH Yes, that sounds correct.

Suppose we have a container in which we have filled a gas with some partial pressure $P_{gas}$. Suppose we have fitted a piston inside the container which is exerting pressure inside the container. Suppose the pressure exerted by the piston inside the container or on the gas is $P_{piston}$. Then is it safe to say that $P_{gas} = P_{piston}$?

@RIPANBARUAH Unless you happen to cross the boiling point at the T of your gas, yes.


text:
What makes up the conduction band?
Prologue/Rant about online classes
We have been taught the electron sea model to explain metallic bonding and conduction of electricity by metals. Then out of the blue, the band theory was introduced, without even giving the definition of what a band is. So I was lost in the recorded online class with no method of asking a doubt. So in an attempt to not wait a day for asking the doubts, I took it on myself to clear my doubts. So I surfed through the web and almost reached a conclusion.. before referring to a textbook.
The actual doubt
So as far as I know, the MOs (on a large scale, like 10 raise to 23 atoms scale) having similar energies appear as bands. The outermost band of occupied MOs is called the valence band. But what I am confused about is what makes up the conduction bands. I reached a conclusion from the chemistry libre texts that conduction bands are made from the combination of the unoccupied atomic orbitals. But when I referred to my reference book, it stated that they are the anti bonding MOs of the valence band.
So what actually makes up the conduction band? Is it one of the two I mentioned above? Or is it a theoretical set of energy levels where an electron can move freely to conduct electricity?

When you do H+H -> H2, you combine two partially occupied AO's to get both the bonding and antibonding MO's. The unoccupied atomic orbitals are not a part of the picture at all. Same thing here.

Ok, but.. how does it answer my question tho?

In essence in a molecule all the molecular orbitals can be expressed as linear combinations of the atomic orbitals, and you occupy the lowest energy ones with the appropriate number of electrons. Note you don't refer back to the original occupations at all, except to get the total number of electrons. A solid is just a very, very big molecule. The valence band is the highest  energy occupied linear combinations, and the conductions band is the lowest energy unoccupied linear combinations.  [yes, basis set completeness, translational symmetry, etc. let's keep it simple]

The point is that the bands can overlap / electrons have the energy to move within the conduction band.

@Alchimista Strictly electrons don't move in a conduction band any more than they move in any molecular orbital - it is a stationary state solution of the Schrödinger equation and the probability density does not change with time. The point is that in a metal an infinitesimally small electric field will mix the conduction band with the valence band due to the zero band gap (and assuming the relevant MOs overlap in some sense), and it is that leads to a nett movement of charge.

@IanBush what is the net movement of charge you refer to if not motion of electrons? Of course motion refer to a contacted piece of metal, I doubt your picture as any current in it.

@IanBush but I agree that OP asked for bands formation.

@IanBush strictly speaking "a solid is just a very big molecule" might confuse many. Obviously you referred to covalent or metallic solids.

@IsnBush - no, the valence and conductance bands do NOT ‘mix’, at least not in proper solid state physics descriptions of band structures.


text:
Ln2 Dewar Filling
For cooling of a measurement device (HPGE) i need around 30 Liters of LN2 (Liquid Nitrogen). I'm a hobbyist but, familiar with the basic rules of Ln2 (Eyeshield,Gloves, Ventilation, Pressure Danger) and also have a professional O2 sensor by hand.
My current problem is, that i own a special 30L dewar for the detector, in the region where i live, suppliers of LN2 will not fill my dewar because firstly they want to sell(rent) their storage units, and for liablity issues because my dewar is old ;)
Even if they could fill it, the dewar including Ln2 would weight over 35kg, which is kinda heavy to lift into the third floor by staircase.
My question is now, can i fill the dewar from smaller dewars? The suppliers rent out 10l Dewars. Can i "pour" the LN2 via a funnel in my dewar? Or is there any other solution that does not result in expensive devices?
Or is is possible to fill via a hose and gravity? (by placing the source dewar higher than the target dewar and let the Ln2 flow by gravity, so only one pump-action is needed to start the process)

Doing it by pouring seems OK. Though I have no idea of what might be special with your old dewar. Safety P issue arise whit more complicated despenser. If you get all in open dewar and your one is open as well, I don't see trouble. Just do it slowly at beginning otherwise you'll loose quite a lot.

I don't know as well, but all commercial suppliers said they can only fill their own, or some standard dewars, they declined to fill customer dewars with no labels etc.

that's understandable. But you can certainly pour LN, as a gesture in itself. If your dewar can be filled like this, just use the gloves and do it. Go slowly so that the bottom can cool first otherwise it take more than 35 l :)

This will definitely work, but make sure your old dewar's isolation is not broken (this will show by condensation forming on the outside of the dewar) and use a plastic funnel as suggested by Buttonwood.

Mechanically yes, considering the weight, pouring from a smaller Dewar is easier than from a larger one.  It takes some practice to aim, of course.  Using an additional plastic funnel consumes some of LN2 because the funnel obviously has to be cooled before LN2 passes the bottom of the chute.
Not knowing your experimental setup: would be an other mean of cooling the detector than LN2 acceptable for you?  Some of the ICCD cameras for example cool their sensor with a Peltier element, which in turn is cooled with water of a liquid thermostate.  This reliably allows temperatures around $\pu{-20 ^\circ{}C}$ of the detector.  Or, as a less expensive approach, why not backing your sensor with a trough made of metal, and fill the trough with a mixture of of dry ice and isopropanol?
Re pumping LN2: The larger mobile Dewars I've seen deliver LN2 by their own pressure like a spray can:

In the simplest form, beside a pressure indicator (I) and a safety valve (S), they have a metal tubing (H) reaching to the inner bottom of the vessel.  H is removed to fill the tank, but in idle state, H is shut and there is a small, occasional hissing pressure release from the safety valve S because still some the LN2 evaporates.  To obtain the LN2, you place your small Dewar below the hose H and open the valve of H.  Thus, opening the valve at H allows the liquid to pass the hose toward the Dewar.  There may be additional valves to eventually increase the pressure above LN2, and additional indicators to report the content of LN2, etc.; all omitted in the drawing for clarity:

(modified illustration, based on one here).

HPGE's are built for LN2, they have a long dipstick that goes into a Dewar, they need to cool down for ~6hrs before you can use them, if they go "warmer"  you have to fully warm them up, and cool them down again. Some modern HPGE's bring electric cooling aggregates, but those are terrible expensive and produce a lot of heat, noise and electrical consumption. The minimum operation temperature is usually ~100K the lower, the better the resolution
About your illustration, i think i will try this by applying  a rubber foil on the and then push air in with hand pump.

As witnessed by a colleague working with a squid magnetometer during a humid summer day, forgetting to close the entry knobs for LN2 and He, trapped water may become a mechanical problem.  But if you keep the setup all well and dry, at small scale, away from air's $\ce{CO2}$ then there is, as stated, only oxygen to have an eye on.


text:
Reaction of Copper and concentrated sulphuric acid
We know that dilute $\ce{ H2SO4}$ doesn't react with copper because copper has higher reduction potential ($0.33$) as compared to hydrogen ($0.00$).
However, it reacts with conc. $\ce{H2SO4}$ and gives $\ce{CuSO4}$ with $\ce{SO2}$ and $\ce{H2O}$.
Why does this reaction take place?

Use $\ce{<insert-compound>}$ to format chemical compounds over <sub> (insert compound)</sub>, It's easier and looks better

You can imagine concentrated sulfuric acid as an oxidising agent which supplies [O]. The copper takes the [O] and become CuO while sulfuric acid become sulfur dioxide. Then CuO reacts with remaining sulfuric acid to form $\ce{Cu2SO4}$. This is a grossly simplified version of what probably happens, but it gives you a sense of why conc. $\ce{H2SO4}$ can react with Cu.

You can a frame the nersnst equation , and see that the term coreesponding to concentration of SO4 2- comes in denominator so increasing concentration might make E cell >0 .

@RajatGolechha, Nernst equation should never be applied on concentrated solutions.

https://chemistry.stackexchange.com/questions/145740/reaction-with-sulfuric-acid

Concentrated sulfuric acid contains the molecule $\ce{H2SO4}$ that makes a redox reaction with copper metal, Diluted solution of sulfuric acid do not contains any $\ce{H2SO4}$ molecule. They are all transformed into $\ce{H3O+}$ and $\ce{SO4^{2-}}$ ions, which are both unable to make a redox reaction with copper.


text:
Why are Dispersion Interactions Stronger in Molecules with Bigger Molecular Mass?
Usually there are two explanations:

When dipoles are induced due to movement of electrons in the molecule, they are bigger in molecules with more mass since they have/include more atoms (has sense to me)
Bigger molecules have bigger contact or "surface area"

What does "surface area" between molecules have to do with how strong dipoles are induced?

Possibly repeating some other Q/A like this one: https://chemistry.stackexchange.com/q/104537/81. certainly much overlap if not a duplicate.

Thank you. Here is a quote from the answer: "More surface, more opportunity for the fluctuations to interact with other molecules." What does that mean? What I would say regarding how strong dispersion forces are is that the bigger dipole the stronger the interaction. Dipoles are induced by fluctuations. Dipole will be bigger if more charges are separated and if they are separed by greater distance so that centers of positive and negative charge are separated by greater distance. What does that have to do with the surface area?

The fluctuations are random. Their individual strength does not change with surface area. But the number of random fluctuations does change with surface area: more area, more opportunity for fluctuations. Hence the overall sum of the forces between two molecules varies but not the individual scale of the dipoles.

Fluctutations do cause shift of centers of positive and negative charge in a molecule. If molecules are non polar this creates dipole. What does number of fluctuations have to do with how strong dipoles are created?

The fluctuations are driven by quantum uncertainty in the position of electron clouds and so appear randomly. The strength of an individual random varying dipole has nothing to do with the size of the molecule. But the number that can occur depends on the number of electrons. The electron clouds capable of interacting with another molecule depends on the surface area.

It didn't really answer my question what does number of fluctuations have to do with how strong dipoles are created in a molecule? Number is bigger in bigger molecules, yes. What ultimately determines the strength of dispersion forces between molecules is how strong dipoles are able to be created in a molecule, not number of fluctuations itself.

Precisely wrong. The size of the fluctuating dipoles does not vary (they are just probabilistic quantum fluctuations that don't relate to surface area at al). Therefore, the overall force between molecules (all the dipoles in each molecule interacting) depends on the number not the strength of the dipoles (which doesn't vary).

Hmm, I wouldn't agree. There aren't dipoles in a molecule there is only one dipole. If centers of positive and negative charge aren't at the same point in space dipole is created and dipole moment of a molecule becomes different from zero. Fluctuations of electrons create dipoles in non polar molecules. How can there be more than one dipole on level of a molecule? We can maybe talk about dipoles of particular groups within a molecule if within specific group (which is non polar) there is separation of centers of positive and negative charge and that can justify what you said.

Also, how can strength of dipoles not vary if they are created by random motion of electrons? How can random motion create unchanging dipole?

You are not getting the key distinction between dispersion forces and other intermolecular forces driven by permanent dipole interactions. these are totally different forces and the interactions from permanent dipoles are usually far stronger. But not in non-polar molecules where the dominant interactions are dispersion forces.

This is the same answer someone already posted here, but thank you.

Can you tell me what the important distinction is?


text:
Auto-ionisation and Dilution
When you dilute an acidic solution, pH increases towards 7 and pOH decreases. However, the decrease of pOH indicates that the concentration of hydroxide ions increases. This seems to contradict my understanding of how equilibrium systems respond to external disturbances, as it suggests the auto-ionisation equilibrium of water shifts to favour the ions (according to Le Chatelier's principle) to such an extent that the hydroxide ion concentration increases from prior to the dilution. However, I was under the impression that equilibrium systems can never fully undo the change due to the bi-directional nature of equilibria, meaning the reaction is using up ions as they are replaced. Please help to clear up my confusion! Thank you.

Recall that $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$
Where $\ce{[OH-]}$ is the concentration of hydroxide ions.

Suppose a solution has $10^{-7}$ moles of $\ce{[OH-]}$ ions. Then this implies $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$ and $\mathrm p\ce{OH} = -\log_{10}\space [10^{-7}]$ which is equal to $7$.

Suppose a solution has $10^{-4}$ moles of $\ce{[OH-]}$ ions. Then this implies $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$ and $\mathrm p\ce{OH} = -\log_{10}\space [10^{-4}]$ which is equal to $4$.

So as you can see, lower the $\mathrm p \ce{OH}$, the higher the concentration of $\ce{OH-}$ ions. Le Chatelier's principle is not violated.
(How I like to remember this is - the strongest acids have the lowest $\mathrm pK_\mathrm a$. So that extends to $\mathrm p\ce{OH}$ too. This is all due to the fact that $\mathrm p\ce{OH}$ is the negative logarithm of the $\ce{[OH-]}$ concentration in solution.)

Note that the chemistry typography expects chemical formulas and p operator to be upright, in SE site context implemented via using \ce{}  and \mathrm{} MathJax formatting.

@Poutnik Hmm..I'm still new to the formatting. However, is it correct now?

$\mathrm p K_\mathrm a$ is denoted as $\mathrm p K_\mathrm a$ and it's $\ce{[OH-]}$ written as $\ce{[OH-]}$. Also don't leave spaces since it creates unnecessry white space

Very well. Almost correct. But note the bracket [] are used to express molar concetration of compound inside. For literal numbers or variable , it better to use implicit or explicit paretheses, like $\log{\pu{e-4}}$ $\log{\pu{e-4}}$  or   $\log{(\pu{e-4})}$ $\log{(\pu{e-4})}$. $10^{-4}$ can be used as well, but \pu{} has advantage to be used together with units )

I'm still a little confused as to how this doesn't contradict Le Chatelier's principle. When an acidic solution is diluted, the concentration of hydroxide ions initially decreases before being partially replaced. Therefore why is it that there is a net increase in the final concentration of hydroxide ions, leading to a decreased pOH? Thank you

You can find useful  Notation basics and How can I format math/chemistry expressions on Chemistry SE.

@latin333 Dilution causes [OH-][H+] < Kw.  LCHP pushes it back to Kw.

Yes, however when I studied diagrams of concentration vs time in relation to equilibrium shifts, the concentrations of species after disturbances was always partially offset in the direction of the disturbance as the system is never able to fully undo the change. For example if there was an increase in concentration of X, even though the system shfited to decrease the concentration of X, there would still be a net increase in concentration of X. Hence my question remains as to why there isn't a net decrease in hydroxide concentration due to the original dilution.

@latin333 - the water that is added to perform the dilution has 10^-7 M HO- because of water autoionization. If the solution being diluted is acidic, initial [HO-] is less than 10^-7, so "dilution" actually results in an initial increase of [HO-] (to the weighted average of the initial and 10^-7), and the concentration decreases back towards initial during equilibration.

The autoionization equilibrium of water is satisfied irrespectively of $\mathrm p\ce{H}$. That's why $\mathrm p\ce{H}$ and $\mathrm p\ce{OH}$ add up to a constant:
$$K_\mathrm{w}=a_{\ce{H3O+}}a_{\ce{OH-}}$$
$$\rightarrow \mathrm p K_\mathrm{w}=\mathrm p\ce{H} + \mathrm p\ce{OH}$$
This is a bit different from the typical scenario in which Le Chaterlier's principle is invoked because $\ce{H3O+}$ and $\ce{OH-}$ are on the same side in the reaction equilibrium, that is they are both products of the autoionization of water.


text:
Why does the carboxylate ion show positive inductive effect?
While solving some problems on general organic chemistry, I came across this question that asked for the nature of inductive effects of $\ce{-CH_3}, \ce{-CO_2^-}, \ce{-Br}, \ce{-NH_3^+}$.
And in the solution to that problem, it says $\ce{-CO_2^-}$ shows $\ce{+I}$ effect.
How is it possible that the carboxylate ion with its two electronegative oxygen atoms shows a positive inductive effect instead of attracting a negative charge towards itself?
Please provide clarification on this.

https://chemistry.stackexchange.com/questions/63181/does-the-carboxylate-group-show-a-positive-or-a-negative-inductive-effect?rq=1
Comment on this answer might be a answer to your question

Why would a negatively charged species attract a negative charge?


text:
How is Electrode Potential Developed on Inert Electrode?
How does redox couple develop its potential if electrode isn't the part of the redox system like platinum in SHE or Cl2/Cl- in water solution for example? Usually it is discussed how electrode develops its potential when it isn't inert like Cu/Cu2+ system when copper electrode itself is important for developing potential. What about inert electrodes?

What is mysterious in providing an electron from the electrode to reduce a molecule/ion, or accepting it when oxidizing  them?

Who said it was? So, electrode acts as electron donor or acceptor we can say for equilibrium to develop?

@Dario Then why don't you ask similar questions for the connecting wires?

So, let's take a solution of copper sulfate in a beaker and insert an active electrode: a copper wire. OK so far? We connect this wire to a voltmeter.
Then lets connect another wire (we are quite well off, even rich, so we will use a shiny platinum wire) to the other connection of the voltmeter. Platinum is nice, inert, always shiny. How shall we connect to the solution? Just stick the inert wire into the solution! The wire is inert chemically in that it does not replace copper from copper sulfate or hydrogen from water, but it is not an insulator - it does do something: it carries electrons when they are pushed or dragged.
If we had inserted an inert electrode (Pt) into the solution first so we had two inert Pt electrodes, well, nothing would happen. Nothing that we could measure. Those little ions and molecules are hopping all over the place, but to no net effect.
So the copper atoms near the solution can donate electrons which then float thru the voltmeter to the platinum which can accept those electrons, handing them off to water, forming OH$^-$ and H$^.$ (two of which will form H$_2$). This occurs more easily than the other way around. (It's even a bit more complicated: copper doesn't do this so easily, but if O$_2$ is available, it will consume the hydrogen atoms produced at the Pt wire and drive the equilibrium toward copper dissolution. I just didn't want to start off the discussion with an active metal like zinc, and zinc wires, etc., because it's more difficult to paint a picture with unfamiliar objects. I probably could have used aluminum wires...)
The philosophical question then is: Is the inert electrode really inert? At the copper electrode, a copper ion just plops down on the metal (yeah, some activity), while at the platinum electrode, electrons come in, go out into H$_2$O, split it, form OH$^-$ and H$^.$, help that form H$_2$, then bubbles - it's a crazy mess compared to the copper electrode.

The kinetic rate constant of oxidation reactions ( electrons taken away by an electrode ) climbs exponentially with increasing electrode potential.
The kinetic rate constant of reduction reactions ( electrons provided by an electrode ) climbs exponentially with decreasing electrode potential.

The electrode potential converges to the value where both reaction rates match each other. In case of multiple redox systems, then the potential converges to the value when the total oxidation rate equals to the total reduction rate.


text:
Does RMS velocity depend on number of gas molecules in a gas sample?
The expression of RMS velocity clearly has square root of number of gas molecules in the denominator and square root of squares of velocities of the gas molecules in the numerator. Then why RMS velocity does not depend on number of gas molecules? By looking at the expression, it seems as if it does depend on number of gas molecules.

Define "depend on." Does any average depend on the number of items going into the average?

RMS velocity does not make much sense, as gas does not move anywhere.
RMS speed converges for n->infty to sqrt(3RT/M). So, it depends on n as the given formula value is the limit value for a large molecule set, similarly as temperature.

No, it does not depend directly on the number of molecules. The expression is
$$c_{rms}=\sqrt{\frac{3RT}{M}}$$
where $T$ is the temperature in K, $M$ is the molar mass in kg/mol and $R$ is of course the gas constant in J mol-1 K-1.
However, it is possible to have a situation where the number of molecule affects the temperature of the gas, which would change the r.m.s. velocity. For example, in Joule-Thompson effect, increasing the pressure causes the gas to heat up. It might be possible to do this by pumping in more gas into an insulated constant volume vessel. (although it sounds a bit contrived to be honest!)

We normally assume that the molecules are in contact with a heat bath at a fixed temperature $T$ and then the Maxwell distibution of speeds in range $v\to v+dv$ is
$$ F(v)dv = 4\pi n\left(\frac{m}{2\pi k_BT}\right)^{3/2}v^2e^{-mv^2/2k_BT}dv$$
It rises due to the $v^2$ term and falls due to the exponential giving a curve with a gentle maximum. The average speed is found from the integral over all possible speeds or
$$\langle v\rangle =\frac{1}{n}\int_0^\infty vF(v)dv=\cdots= \sqrt{\frac{8k_BT}{\pi m}}$$
where $n$ is the number of particles of mass $m$ and $k_B$ is the Boltzmann constant. You can see that $n$ cancels out. So it does not matter if there is one particle measured many, many times or many just measured once, the result is the same.
The rms is found by a similar argument
$$\langle v^2\rangle =\frac{1}{n}\int_0^\infty v^2F(v)dv=\cdots= \sqrt{\frac{3k_BT}{m}}$$
Notes. The average of a normal function $f(x)$ is $\displaystyle \langle x\rangle=\frac{\int xf(x)dx}{\int f(x)dx}$  and in this distribution $\int F(v)dv =n$


text:
Why is the equilibrium constant increasing in this exothermic reaction?
A problem asks, if you have a positive $\Delta S$ (positive change in entropy) for an exothermic (meaning a negative $\Delta H$, or negative change in enthalpy) reaction and temperature increased, what would happen to $K_{\mathrm{eq}}$ and $K_{\mathrm{eq}}/Q$ (where $K_\mathrm{eq}$ is the equilibrium constant and $Q$ is the reaction quotient)?
The answer states that $K_{\mathrm{eq}}/Q$ and $K_{\mathrm{eq}}$ would increase. This is because if you use the thermodynamic equation, $\Delta G = \Delta H - T \Delta S$ (where $\Delta G$ is the change in Gibbs free energy and $T$ is for temperature), as you increase temperature you get a more negative number (since $\Delta S$ is positive and $T\Delta S$ is being subtracted). This would lead to a more negative $\Delta G$. Then, using another thermodynamic equation, $\Delta G = RT\ln(Q/K_{\mathrm{eq}})$ (where $R$ is the ideal gas constant), if $\Delta G$ is decreasing (becoming more negative), then the term $(Q/K_{\mathrm{eq}})$ must also be going down. This means that $K_{\mathrm{eq}}/Q$ must be going up and therefore $K_{\mathrm{eq}}$ increases, as temperature increases for this exothermic reaction.
So my issue is, I thought for an exothermic reaction, where heat can be represented as a product, I thought increasing the temperature would shift the reaction to the left and decrease $K_{\mathrm{eq}}$. Could someone clarify this for me? I tried looking this up with Chemistry Libretexts, but it didn't seem to clarify anything. The table seemed to suggest that we do use heat as a component of the reaction in endothermic or exothermic reactions to predict the $K_\mathrm{eq}$ change.

Heat is not a product.

@Zhe  Some intro chemistry texts suggest that students think of thermal energy as if it were a product or reactant to help understand Le Chatelier's Principle.  It's not an unreasonable anzats.

To get better understanding think of it as, "The reaction wants to release  heat but if the ambient temperature is too high than it would be difficult for the reaction to do so, whereas at lower ambient temperature it will be easy for the reaction to liberate heat. " This is not accurate but just a way in which you can think intuitively about such conditions.

@Jay The problem is wrong.  The temperature-dependence of $K_{eq}$ is determined solely by the sign of $\Delta H$. To understand this, note that $\Delta G = -RT \ln K_{eq}= \Delta H - T \Delta S => \ln K_{eq} = \frac{\Delta S}{R}-\frac{\Delta H}{RT} $.  I.e., you were looking at the T-dependence of $T \ln K_{eq}$ instead of $K_{eq}$.  When you look at the latter, you see the T-dependence switches from the entropy term to the enthalpy term.

@theorist I was taught that myself. It may give the correct answer, but the reasoning is wrong (obviously, since your comment nicely highlights the correct relationship). So I oppose it on those grounds.

An exothermic reaction has a reduced equilibrium constant at higher $T$ because while the contribution of the change in the entropy of the system is a fixed quantity (for a small $T$ change), the effect of transferring heat to the surroundings is reduced at higher $T$ (because it causes a smaller change in the entropy of the surroundings). Mathematically this is described by the van't Hoff equation:
$$\left(\frac{\partial \ln K_\textrm{eq}}{\partial T}\right)_p = \frac{\Delta H^\circ}{RT^2}$$
On the other hand the change in $\Delta G$ with temperature is given by $\Delta S$:
$$\left(\frac{\partial\Delta G}{\partial T}\right)_p = -\Delta S$$
Since $\Delta G =\Delta G^\circ +RT\ln Q = -RT\ln (K/Q)$ one can write
$$\left(\frac{\partial\Delta G}{\partial T}\right)_p = \left(\frac{\partial\Delta G^\circ}{\partial T}\right)_p +R\ln Q\\ = -\Delta S^\circ+R\ln Q$$
Therefore if you want to see how $ K_\textrm{eq}$ depends on $T$ you should focus on $\Delta H^\circ$. If you want to see how $\Delta G$ depends on $T$ you should inspect $-\Delta S^\circ+R\ln Q$.
It is somehow unfortunate that the temperature dependence of $\Delta G$ and the equilibrium constant are different, because it leads to endless mix-ups, but it reminds you that they describe different things.

Thank you for that response! So, are you saying for an exothermic reaction at higher temperature, that Keq decreases, while Keq/Q increases, because of the reduced effect of transferring heat at higher temperatures (meaning we disregard T*delta S). It's hard to deduce that from delta G = delta H - T * delta S . Would that mean that for Keq/Q to increase, while Keq decreases that Q decreases by a greater amount? Do you have a good educational website/webpage that would describe this scenario in great detail? This is very confusing.

If you like the answer, please upvote and/or accept the answer.

I still had two additional questions.

@Buck Thorn: I'm just curious that van't Hoff equation is about standard enthalpy change, isn't it?

@MathewMahindaratne You're right, thank you.

@Jay The effect of changing T is shifting $K$ with respect to Q (which we consider fixed in this analysis). Assume the reaction is exothermic. Increasing T will make $K$ smaller making $K/Q$ smaller therefore $\Delta G=-RT\ln (K/Q)$ more positive.

@Jay Q reflects the current concentrations of the sample, so it can be anything really. It does not necessarily change with T. I would look through additional questions on chem SE, for instance this: https://chemistry.stackexchange.com/questions/49342/does-exothermic-solvation-mean-solute-is-more-soluble-at-low-temp?rq=1

You have a right to be confused. The first enthalpy is a product of a reaction and it is unjustifiably ignored because it is usually removed and discarded.
At equilibrium,
$$\Delta G = \Delta H -T\Delta S =0$$
For an exothermic reaction, $\Delta H$ and $\Delta S$ are both negative. This means that raising the temperature will decrease the equilibrium constant. This is exactly what le Chatelier's principle predicts.
The confusion generated by the different signs is just from the definition that negative energy is more favorable and positive $K_\mathrm{eq}$ is more favorable.  Also adding to the confusion is that standard free energy an enthalpy changes are calculated with the reactants and products in their standard states. Both sets of concentrations are changed to the equilibrium conditions here.

You can use MathJax to format here.. A full list of what you can do

In an exothermic reaction, $\Delta H$ is negative.  By contrast, whether a reaction is exothermic places no constraint on $\Delta S$.  It can be positive, negative, or zero.

@NisargBhavsar  What do you mean "for equilibrium, Δ and Δ both must be negative"?  Remember that when we say that Δ is negative for an exothermic rxn, we are referring to the standard enthalpy change:  "exothermic reaction:  A reaction for which the overall standard enthalpy change ΔH⚬ is negative." https://goldbook.iupac.org/terms/view/E02269

@NisargBhavsar   The temperature-dependence is determined by the standard enthalpy change, $\Delta H^\circ$, which occurs in the expression $-RT \ln K_{eq} = \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$.  When you write $\Delta G = \Delta H - T \Delta S=0$ at equilibrium, that is the actual reaction $\Delta G$, not the standard $\Delta G^\circ$.  If a reaction is exothermic, then $\Delta H^\circ < 0$.  However, $\Delta H^\circ < 0$ doesn't necessarily mean $\Delta H< 0$ at equilibrium.


text:
Carbon atoms types in amino-acids
I'm looking for a table or another kind of schema where I can find which type of carbon atoms (sp3, sp2 etc.) are present in each amino-acid.
In my biochemistry books these informations are not available.
I looked also on pubchem (e.g. glycine), and I found some information about alanine, aspartic acid, methionine.

For those who downvote my question, please specify why in order to improve it

Can you please provide a list of amino acids about which you want the information.

Hello @NisargBhavsar I'm interested in the 20 amino acids present in the proteins of our body.

I didn't downvote you but it is beneficial if you put how much work you have been done so far. Also, please clarify your info (e.g., see Nisarg Bhavsar's comment elsewhere).

For instance, carboxylic carbon  and carbonyl oxygen in glycine is $\mathrm{sp^2}$. The rest of the atoms in the molecule is $\mathrm{sp^3}$ ($\ce{C,N,O}$. It is more complicated than that but for your level of knowledge, this is good enough

Also remember, the oxygens on carbonyl groups in both structures (alanine and aspartic acid) you have just put are $\mathrm{sp^2}$.

It is not tabulated because you should be able to figure it out for yourself if you ever need to. If you pick up a good general chemistry textbook somewhere it will explain hybridisation. And there are lots of resources on the Internet to learn from, too. Mathew's comments are pretty much it, actually.

Hi @orthocresol ok maybe the informations I'm looking for are not so important to write them in a table. I'll try to retrieve them by myself.

Sr. No.
Name
Structure
Hybridization and Number of Carbon

1
Serine

$$sp=0\\sp^2=1\\sp^3=2$$

2
Threonine

$$sp=0\\sp^2=1\\sp^3=3$$

3
Tyrosine

$$sp=0\\sp^2=7\\sp^3=2$$

4
Asparagine

$$sp=0\\sp^2=2\\sp^3=2$$

5
Glutamine

$$sp=0\\sp^2=2\\sp^3=3$$

6
Glycine

$$sp=0\\sp^2=1\\sp^3=1$$

7
Alanine

$$sp=0\\sp^2=1\\sp^3=2$$

8
Cysteine

$$sp=0\\sp^2=1\\sp^3=2$$

9
Valine

$$sp=0\\sp^2=1\\sp^3=4$$

10
Leucine

$$sp=0\\sp^2=1\\sp^3=5$$

11
Isoleucine

$$sp=0\\sp^2=1\\sp^3=5$$

12
Proline

$$sp=0\\sp^2=1\\sp^3=4$$

13
Methionine

$$sp=0\\sp^2=1\\sp^3=4$$

14
Phenylalanine

$$sp=0\\sp^2=7\\sp^3=2$$

15
Tryptophan

$$sp=0\\sp^2=9\\sp^3=2$$

16
Aspartic acid

$$sp=0\\sp^2=2\\sp^3=2$$

17
Glutamic acid

$$sp=0\\sp^2=2\\sp^3=3$$

18
Lysine

$$sp=0\\sp^2=1\\sp^3=5$$

19
Arginine

$$sp=0\\sp^2=2\\sp^3=4$$

20
Histidine

$$sp=0\\sp^2=4\\sp^3=2$$

Fantastic table @NisargBhavsar

Which app you used to draw the structures?

@Eclipse239 I used the online interface at py-chemist.com