text:
Why are the protons of ethene magnetically equivalent?
According to this website,

For two protons to be magnetically equivalent they not only have to have the same chemical shift, but they must also each have the same J coupling to other magnetic nuclei in the molecule.

(Emphasis in orginal text)
One example of this is 1,1-difluoroethylene, where the two protons couple to the fluorines differently (each one is cis- to one fluorine and trans- to another, so JHF is different than JH'F and JHF' is different than JH'F'). Why does this same logic not apply to ethene? I would think that each proton couples differently to the protons on the opposite side of the double bond,  leading to magnetic inequivalence. This answer says that the protons are equivalent through symmetry, but if the criterion for magnetic equivalence is having the same coupling constant with each magnetic nucleus, I think symmetric molecules could still have magnetically inequivalent protons. Are there cases where symmetry leads to magnetic equivalence, despite differences in coupling constants?
Thank you.

The same logic does apply. The protons in 1,1-difluoroethylene are equivalent, only they couple to the different fluorines differently. Same thing here, only without fluorines.

Magnetic equivalence is more rigorously defined in Levitt's Spin Dynamics (2nd ed., secction 17.5):

Magnetic equivalence is a strong form of chemical equivalence. A set of spins is magnetically equivalent IF

Condition 1: the spins have the same chemical shifts

AND

(Condition 2a: the spins have identical couplings to all other spins in the molecule

OR

Condition 2b: there are no other spins in the molecule.)

Condition 1 is the so-called 'chemical equivalence', and condition 2 is what differs between the cases of ethene and 1,1-difluoroethene.
In 1,1-difluoroethene, condition 2a is not fulfilled (cis and trans H–F couplings are different), and condition 2b is not fulfilled (when considering the equivalence of the two protons, there are obviously other spins, namely the fluorines).
In ethene, condition 2b is fulfilled: we can consider the four protons as a single group because they are all chemically equivalent, and there are no other spins present (ignoring the low-natural abundance 13C, etc.)
A more mathematical explanation of this is shown in Appendix A.9 of Levitt. Long story short, the H–H J-coupling interactions in ethene (of the form $\sum_{ij} 2\pi J_{ij} (\mathbf{I}_i \cdot \mathbf{I}_j)$) commute fully with the chemical shift interactions ($\sum_i \omega I_{iz}$) as well as the detection operator ($\sum_i I_{ix}$), so can be dropped from the Hamiltonian when calculating expectation values. nb: in the copy I have, eqn. (A.37) has a typo. It should read
$$\hat{B} = 2\pi J'(\hat{\mathbf{I}}_1\cdot\hat{\mathbf{I}}_2
 + \hat{\mathbf{I}}_2\cdot\hat{\mathbf{I}}_3
 + \hat{\mathbf{I}}_1\cdot\hat{\mathbf{I}}_3).$$

When we are determining whether protons are magnetically equivalent, do we only compare the couplings to chemically non-equivalent protons? Is this why we consider ethene’s four chemically equivalent protons to be one group? As another example, would the four protons in p-dinitrobenzene be magnetically equivalent because there are no chemically non-equivalent magnetic nuclei to compare couplings to, whereas in p-xylene, H2 and H6 would be magnetically equivalent and H3 and H5 would be magnetically equivalent, since we only compare couplings to the chemically non-equivalent methyl hydrogens?

@Akash First statement: Yes, that's right. Dinitrobenzene: Yes, that's right. The spectrum would consist of only one peak. Xylene: I'm not sure where you're going with this. The methyl hydrogens (all six of them) are chemically equivalent. In any case, it's probably not the best example, because the J-coupling is zero or so close to zero that it doesn't impact the spectrum. p-Difluorobenzene might be a better example; in that case I'd agree with your analysis. The spectrum has second-order effects. You can check all these out at https://sdbs.db.aist.go.jp

I meant that the methyl hydrogens and the hydrogens on the aromatic ring are chemically inequivalent, so you would need to examine the coupling between H3 and the methyl hydrogens and H2 and the methyl hydrogens, and since these couplings are different, H2 and H3 would be magnetically inequivalent. Was p-xylene not a good example because J coupling across 4 bonds is so small?

@Akash Oh, I see. Yes, you would be correct in principle. In practice, the coupling is probably too small to really see any second-order effects. (Perhaps that might be different if you got a high-resolution spectrum and looked really closely, but I don't have the ability to get a sample myself and record a spectrum.) It's worth pointing out that the quantity of interest is not 4J itself, but rather the difference in the two couplings of interest (one could say the 'extent of magnetic inequivalence'). 5J is basically zero, and 4J is so close to zero, though, so 5J - 4J is pretty negligible.


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Encoding molecule bond information in matrix form
I have the following problem: I have a pentacene molecule and would like to encode its bond information into a matrix form, to feed it later on to a Machine Learning (ML) algorithm (a Graph Neural Network, fyi).
The matrix has dimension $(36, 36)$ (since a pentacene molecule has $36$ atoms). Each entry $(i,j)$ represents the bond between atom $i$ with atom $j$.
My original idea was to encode a number depending on the bond type (e.g. single bond=1, double bond=2 and aromatic bond=1.5).
In my previous post though, I found out that the whole system is aromatic. Now my question would be: how could I encode this? Should I assign 1.5 to every bond?
I know this question might not have a definitive answer, but I am looking more at a way to justify my choices.
As a bonus: Do someone know of any ML paper where people encoded bond information?

Maybe this is not the right site to ask about this topic. However, I can suggest you two types of matrixes you can encode this information: the connectivity matrix (so just if atom i is linked to atom j); the distance matrix (so the arithmetic distance of atom i to atom j) which allow to bypass the bond-order problem.

Thanks for the suggestion :D I am however still curious if there is a way of using categorical information about the bond type.

@ixaixim, Have you checked out ECFP6 fingerprinting? If you have your molecule as a .mol2 or .pdb file, you would be able to generate this fingerprint using RDKit, which encodes the connectivity information in a 2048 bit vector. This fingerprint is theoretically not unique, but such collisions have been shown to be rare.

I only have the positions of the molecule atoms. Nothing more.

@ixaixim If you have the positions, and if you know the which atoms the positions correspond to, you can generate the .pdb file, and then continue.

As you have learned, in actual molecules, single and double bonds are not exclusive categories. Rather, bond order is a continuum for which one can use various metrics. How to best represent it likely depends heavily on what you are using the machine learning for.

There are, of course, multiple ways to encode the bonding graph into a matrix, for example:

Since you have pentacene, all bonds could be typed as aromatic
Since you have the 3D positions, you could encode the exact atom-atom distance / bond length
As some have mentioned in the comments, you can use atom environments (e.g., fingerprints) of the two atoms to encode information .. for example as (C.ar-C.ar).

All three methods have been used in the chemistry ML literature.
The first is more useful if you do not have accurate 3D coordinates, since it only requires the molecular structure (e.g., SMILES, 2D depiction, etc.).
It's hard to know which encoding would be the most useful -- in general one must try a few possible descriptors to find the best for your particular task.

source: Stack Exchange

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Integral Prescreening in calculation of Electron Repulsion Integrals?
I am trying to write a basic Hartree-Fock program that can calculate the total energy and molecular orbitals of any molecule at arbitrary basis sets largely as a self-taught effort.
I am able to calculate the required integrals, but the computational effort required to evaluate ERIs is prohibitively expensive even for benzene at a slightly larger basis set of 6-31g(d). I tried printing out the ERI matrix and saw that many of the integrals are very close to zero.
I understand that ERIs can be screened using the Schwartz inequality $(ab|cd) \le \sqrt(ab|ab)*\sqrt(cd|cd)$.
By my understanding, $(ab|ab)$ term is the ERI for the basis functions centered at $a$ and $b$. So, this should only save calculation costs of one ERI integral, right?

The exact number saved will depend on the system size and geometry, but apart from for very small systems it will be many more than 1. Consider the two methods for a system that contain $N$ basis functions
Firstly the naive method: Simply calculate $(ab|cd) \space \forall \space a,b,c,d$. This will clearly require $O(N^4)$ integrals to be evaluated.
Secondly this time with screening:

First calculate and store in memory $(ab|ab) \space \forall \space a,b$ This clearly is $O(N^2)$ integrals, which is insignificant with the $O(N^4)$ results we have to finally calculate
Next calculate the biggest value of $|(ab|ab)| \space \forall \space a,b$. Call it $I_{max}$ . This again requires an insignificant $O(N^2)$ operations
Now make a list of $a,b$ for which $|(ab|ab)| * I_{max} > tolerance$. These are the only integrals which you need to consider to be non-zero. Again this is at worst an insignificant $O(N^2)$ operations.
Now calculate the integrals $(ab|cd)$ where $a,b$ are taken from the list  just generated and $c,d$ are also taken from that list. All other integrals are (close to) zero
In the limit of a very large system this reduces the number of integrals from $O(N^4)$ to $O(N^2)$. This is a huge reduction, many, many orders of magnitude.

What physically is going on is the list you generate is the list of basis functions which have a significant overlap. Thus for a given $a$ the list of associated $b$ will be those basis functions in some sense close to $a$ in space. For a very large system this is independent of the total system size - each basis function only overlaps with a small subsection of the system, and increasing the system size does not increase the number of basis functions that $a$ interacts with. Thus for a given $a$ the number of overlaps in a large system is $O(1)$. There are $N$ $a$ hence the size of the list is $O(N)$. And you have a nested loop over the list, so the total work is $O(N^2)$
However note this is the limit for large systems, for small systems the savings will be smaller, and for very small systems there may be no measurable saving at all. This is simply because $a$ now overlaps with all the other basis functions, so no screening is possible.
Note also it is possible to reduce the limit to $O(N)$. But that requires more advanced methods.

Thanks for the explanation. |(ab|ab)| is the set of all (ab|ab) integrals right? I did not quite understand the reason for calculating the maximum of that set. Are you speaking of batch processing of ERI integrals?

Imax is useful as you can very quickly answer whether the a,b basis set pair is involved in any integrals - if |(ab|ab)|*Imax is less than your integral tolerance you can instantly ignore every integral that is of the form (ab|**) (and permutational symmetry related)

Note the above is just an outline of the ideas and one possible way of implementing the scheme. It's not the only way, but other ways will all be closely related conceptually  to the above.

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text:
IUPAC naming of lysergic acid diethylamide (LSD) heterocyclic core and its locants
Wikipedia article about LSD says its IUPAC name is
(6aR,9R)-N,N-diethyl-7-methyl-4,6,6a,7,8,9-hexahydroindolo[4,3-fg]quinoline-9-carboxamide
which is quite complex. Especially its didehydroergoline skeleton obscures it with fused heterocycle naming and lots of locants. How could it be broken down, so even a beginner would get it?
Skeletal formula for LSD structure looks like this:

Picture's in public domain, but lets credit Common's user NEUROtiker for it.
A good point to start is the parent compound. Here it's indolo[4,3-fg]quinoline.
"Indolo" is from indole, a heterocyclic aromatic compound, with this structure:

It's "that snowman on its side at the bottom right of the molecule must be the indole", citing OP here.
Quinoline, also called chinolin, is

It is "that inclined bigger snowman stacked on top of the first snowman". Still, it looks quite a bit different in LSD, doesn't it? That's because of this "hexahydro" bit - there's more hydrogens in there, even though two of them are substituted, which could lead to further confusion.
This [4,3-fg] bit says how the ring are connected, but how does it work? That's maybe most "advanced" bit of nomenclature-fu here.
Another question is how these locants for hydrogens - "4,6,6a,7,8,9" and chiral centers (6aR,9R) work - what's the relevant numeration here?

Comments are not for extended discussion; this conversation has been moved to chat.

Related: https://chemistry.stackexchange.com/q/53704/103369 and https://chemistry.stackexchange.com/q/37970/103369

The numbers and letters [4,3-fg] indicate the side where the fusion occurs and the relative orientation of the fused ring system. The direction of the numbers of the attached component (e.g. ‘3,4’ or ‘4,3’) corresponds to the direction of the lettering of the peripheral sides of the parent component (i.e. a, b, c, etc.).
The correspondig Subsection P-25.3.1.3 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), reads as follows:

(…)
Isomers are distinguished by lettering, continuously, each peripheral side of the parent component (including sides whose locants are distinguished by letters, for
example, 2a,3a) using the italic letters a, b, c, etc., beginning with a for the side numbered ‘1,2’, b for ‘2,3’ etc. To the letter as early in the alphabet as possible that denotes the side where the fusion occurs are prefixed, if necessary, the numbers of the positions of attachment of the other component. These numbers are chosen to be as low as is consistent with the numbering of the compound and their order conforms to the direction of lettering of the parent component. (…)

For the compound that is given in the question, the fusion occurs at the side of quinoline between the atoms 5 and 6, which corresponds to the peripheral side f, and also at the side between the atoms 6 and 7, which corresponds to the peripheral side g. Thus, the parent component is described using the name [fg]quinoline.
The attached component is indole. The fusion occurs between the atoms 3 and 3a and also between the atoms 3a and 4. Now there are two possibilities for the relative orientation of the fused ring system indolo[3,4-fg]quinoline as well as indolo[4,3-fg]quinoline.

The direction of the numbers of the attached indole (e.g. ‘3,4’ or ‘4,3’) corresponds to the direction of the lettering of the parent quinoline (i.e. a, b, c, d, e, f, g, etc.). For the compound that is given in the question, the direction is from 4 to 3. Therefore, the name of the fused ring system is indolo[4,3-fg]quinoline.
The fused ring system gets a new numbering of locants. Numbering is quite easy if you draw the structure correctly according to the IUPAC recommendations for orientation of fused ring systems (wich are beyond the scope of this question; see Subsection P-25.3.2.3 for details).

P-25.3.3.1 Numbering of peripheral skeletal atoms
P-25.3.3.1.1 The numbering of peripheral atoms in the preferred orientation starts from the uppermost ring. If there is more than one uppermost ring, the ring furthest to the right is chosen. Numbering starts from the nonfused atom most counterclockwise in the ring selected and proceeds in a clockwise direction around the system, including fusion heteroatoms but not fusion carbon atoms. Each fusion carbon atom is given the same number as the immediately preceding nonfusion skeletal atom, modified by a Roman letter ‘a’, ‘b’, ‘c’, ‘d’, etc.

When the numbering of a system is fixed like that, this numbering must be used. Suffixes and prefixes are added to the name, as required, using the new locants characterizing the fused system. In particular, the saturation of the compound that is given in the question is indicated using hydro prefixes as 4,6,6a,7,8,9-hexahydroindolo[4,3-fg]quinoline.

The principal characteristic group is expressed as suffix ‘carboxamide’.

The remaining substituent groups are added as prefixes (‘methyl’ and ‘ethyl’).

Finally, stereodescriptors are placed at the front of the complete name or name fragment to which they apply. So the complete name is (6aR,9R)-N,N-diethyl-7-methyl-4,6,6a,7,8,9-hexahydroindolo[4,3-fg]quinoline-9-carboxamide.


text:
When are frontier orbitals a better way to explain reactivity than the Hammond Postulate?
When explaining the ortho, para directing ability of electron donating groups in electrophilic aromatic substitution, one can use either frontier orbitals or the Hammond Postulate. The frontier orbital argument says that the HOMO has a higher coefficient on the ortho and para carbons, so the reaction primarily occurs on those carbons due to the better HOMO-LUMO interaction (I think you could evaluate the impact of this overlap quantitatively by looking at the effect of overlap on the resonance integral in the third term of the Klopman-Salem equation).
Using the Hammond Postulate, one can argue that the transition state of the addition of the electrophile to the aromatic ring resembles the intermediate because the reaction is highly endothermic and the energy of the transition state is similar to the energy of the intermediate. The intermediate (and therefore transition state) energy is lowest when the electrophile is added to an ortho or para position, because in these cases there is a resonance form with a positive charge on the carbon bonded to the donating substituent, which stabilizes the charge. This lower-energy transition state leads to a faster reaction.
Are both of these arguments valid? In general, are there certain situations where one approach is better, or in which each would lead to different predictions about reactivity? Do both of these methods account for the same factors? For example, how would frontier orbital theory and the Klopman-Salem equation account for the stabilization of positive charge by the electron-donating group?
One idea I have is that the relative importance of the frontier orbital argument may depend on how much the electron donating group affects the coefficients of the HOMO. For example, there is more electron density on the ortho and para carbons when the substituent is OMe or NH2 than when the substituent is CH3. Would this lead to an increased difference in HOMO-LUMO overlap between an electrophile attacking an ortho/para carbon and an electrophile attacking a meta carbon, which would mean that the frontier orbital considerations lead to more selectivity?


text:
How is it possible for the moon to contain any helium-3?
According to the following article, scientists have known the lunar surface contains deposits of helium-3 since the Apollo program. The Chinese Chang'e 5 mission appears to confirm it after returning from the moon with a new mineral containing the isotope. This raises a lot of questions. How is it possible to find any amount of helium-3 on the moons surface? It seemingly contracts what I've learned from science.
First, helium is a noble gas in all its isotopic forms, making it incapable of chemically bonding with other elements. Wouldn't any amount of it would exist as an isolated gas? Second, given the moon is about 1/6th of the earth's gravity due to its small mass. Wouldn't such a light gas as helium quickly dissipate out into outer space as would hydrogen on earth?
It could be argued that trace quantities of helium-3 could be trapped inside crystal or other molecular structures of the moon's regolith. helium-3, however, is a small atom. Wouldn't it easily escape the confines of a molecular lattice, especially when it's chemically unbounded to the structure?

A helium-3 atom is lighter, not smaller than helium-4 one. Finding He inside rocks with alpha decaying elements is common.

Here is a source (the source, perhaps): https://thehill.com/opinion/technology/3647216-china-has-returned-helium-3-from-the-moon-opening-door-to-future-technology/

"It is estimated there are 15 to 20 metric tons of helium-3 on Earth that can be exploited by humankind, but the reserve of this substance on the moon could be at least 1 million tons, experts have suggested." Source

@Kasten, That was the exact article I read.

@Poutnik, According to Karsten's article, the amount of Helium-3 is greater than here on earth, but I don't understand how the moon could have more of the alpha decaying elements than the earth. Being smaller, there's just less alpha decaying elements available. Moreover, it still doesn't explain how Helium can even accumulate in such quantities given the conditions on the moon.

Note that I said He (implicitly helium-4) , not helium-3. Part of trapped helium-3 may be formed by cosmic ray spallation (forced fission), similarly as most of Earth beryllium and boron, as these rays are more intense on the Moon surface due lack of atmosphere and magnetic field. // but main probably by the solar wind too as Nilay Ghosh says, as fast nuclei are better projectiles than bullets.

@Poutnik. OK. Makes sense now. HE-3 is formed via solar flares, but it doesn't explain how it could accumulate on the moon in any significant quantities since most of it would escape into outer space. I left a comment on Ghosh's answer addressing this.

It is not helium as free gas to escape.

Unlike Earth, which is protected by its magnetic field, the Moon is always bombarded with large quantities of Helium-3 by the solar wind (Helium-3 is emitted from the Sun and carried throughout the Solar System by solar winds). Since the Moon has a weak magnetic field and the atmosphere is extremely thin, Helium-3 is deposited in significant quantities, getting embedded in the upper layer of regolith (ref.2 on how helium-3 gets trapped in the regolith, thanks to OP). You can find more information in ref.1
References:

Estimated solar wind-implanted helium-3 distribution on the Moon by
Jeffrey R. Johnson,Timothy D. Swindle,Paul G. Lucey, Volume 26, Issue 3, 1999 DOI: 10.1029/1998GL900305
Taking advantage of glass: capturing and retaining the helium gas on the moon by Ao Li et al 2022 Mater. Futures 1 035101
DOI: 10.1088/2752-5724/ac74af


text:
Can limescale actually "etch" glass?
There are lots of articles on the internet about limescale "etching" glass due to the porous nature of glass. However, as I understand it, limescale is just calcium carbonate so can it actually cause permanent damage to the glass (or any other material)? Or is it just the case that it can be difficult to clean as the limescale is "embedded" rather than the glass actually being "etched"?

It may be that anything getting into the pores creates the same effect as etching, increasing opacity to light.  Calcium carbonate can be removed with vinegar.

Glass is hardly porous. I have a feeling you're reading wrong kind of articles.

Calcium carbonate, $\ce{CaCO3}$ is alkaline, and can, slowly, combine with the silica in glass. Consider that window (soda-lime) glass is made from lime, $\ce{CaO}$, derived from $\ce{CaCO3}$, as well as silica and other oxides, and that silica is also slightly soluble in water (cactus and maize contain silica phytoliths, derived from dissolved silica from the soil, for example).
Over a long period of time, the surface of the glass combines with $\ce{CaCO3}$, giving it a "frosted", translucent, appearance. Removing that surface coating is difficult, and, with tempered glass, perhaps impossible, because removing the thin surface of tempered glass can cause it to shatter. Acid cleaners can only remove the thinnest, newly-deposited, surface layer of lime.
Similarly, a glass bottle used for storing sodium hydroxide, $\ce{NaOH}$, becomes etched in a fairly short time. It's commonly found that $\ce{NaOH}$ reagent bottle have the ground-glass stopper permanently fused in place.
$\ce{NaOH}$ Bottle" />

Thank you, just the answer I was looking for! Out of interest, and I appreciate this is depedent on the circumstance, roughly how slowly is slowly?

From personal experience, a week's lime buildup on glass (and porcelain, which is glass-glazed) in our location, with fairly soft water, can be removed with gentle washing, and after a month, an acid cleaner is needed. Mileage may vary...


text:
Diffraction Derivation- relating the angle between deflected ray and original
This relates to diffraction from objects with a periodic structure.
I’m trying to relate the psi angle (diffracted angle) to the theta angle (angle between diffracted ray and the original non-deflected zeroth order ray. In this derivation but I just can’t get anywhere.
I’ve tried using the sin rule, basic trig, Pythagoras etc. I wondered if anyone could possibly spot the fundamental mistake I’m making here? An angle/geometry rule I’ve forgotten or am I going in the completely wrong direction with this?

It is a single beam (parallel or near-parallel because the distance to the source is large compared to the lattice spacing), so $d$ cannot be the spacing between "the two beams". You are missing a factor 2 (and the cited text got Bragg's law wrong and is using an unconventional definition of theta).

I’m trying to relate the psi angle (diffracted angle) to the theta angle (angle between diffracted ray and the original non-deflected zeroth order ray.

Braggs law has two steps to it, and the first one is often omitted. The terminology "reflection" is used because the angle of the incoming beam with the planes is equal to the angle of the diffracted beam with the planes. This ensures that the path traveled from source to detector is the same, no matter which atom on the single plane scatters the beam. So the angle of deflection (between incoming and diffracted beam) is twice the angle between the beams and the plane.
This is the same for reflections in a mirror, the angle of incidence and angle of reflection are equal, and the total deflection is twice this angle.
The usual terminology is to call the angle of incidence "theta", and the angle of deflection "2 theta". For example, building a goniometer, one angle that is changed (that between incoming beam, usually fixed, and detector, which is rotated) is named "2 theta stage".

I wondered if anyone could possibly spot the fundamental mistake I’m making here?

No fundamental mistakes, just a little mistake: The total path difference is $2 q$, so for constructive interference, you need the condition $$2 q = n \lambda$$
If you plug that in, you arrive at Bragg's law. There is no need for any fancy trigonometry (like the tan(x) = sin(x) for small x), the derivation is complete once you fix that mistake. You should look up Bragg's law to make sure you know what the goal of the derivation is.


text:
Redox titration, complex formation effect on lg'K?
We have performed the following redox titration
(1): $\ce{MnO_4^- + 8H^+ + 5Fe^2+ <=> Mn^2+ + 4H_2O + 5Fe^3+}$
where we have used a iron(II)solution (the analyte) and titrated it with a permanganate solution (titrant). The iron(II) was added by measuring up $\ce{FeSO_4*7H_2O}$. There is also $\ce{H_2SO_4}$ and $\ce{NaNO_3}$ in the analyte solution. The permanganate solution was prepared by an instructor. When the equivalence point was reached we could calculate the original concentration of $\ce{Fe^2+}$ in the analyte solution, we have pe-values, E-values, the volume of titrant needed to reach the equivalence point and a lot more different values.
The equilibrium constant lgK for
(2): $\ce{Fe^3+ + e- <=> Fe^2+}$
is 13,02 (from a data collection). Our calculated value for lg'K based of the concentrations that was calculated in the titration in reaction (2) is 11,67.
We know that $\ce{Fe^3+}$ forms a complex with $\ce{SO4^2-}$
(3): $\ce{Fe^3+ + nSO_4^2- <=> Fe(SO_4)_n^{3-2n}}$
where n=1 or 2. This is supposed to be a reason for the difference in the theoretical value of the equilibrium constant and our calculated one(the difference between 13,02 and 11,67). Why does this complex formation make our calculated value lower than the theoretical one? How does this affect the equilibrium constant?
Another reason for this difference is the ionic strength in the solution, but in which way? What is influencing the ionic strength in this system?
(there have been no previous calculations with ionic strength or activities in this laboration and there is no other mentioning of ionic strength anywhere else in our instructions)
The calculation of our lg'K has been calculated as following:
$\ce{'K = \frac{[Fe^2+]}{[Fe^3+]*({e^-)}}}$
$\ce{<=>}$
$\ce{lg'K = lg[Fe^2+] - lg[Fe^3+] + pe}$
Where [] are the concentrations in Molar and () is the activity (we can not write $\ce{e^-}$ as a concentration but we have pe-values).

How are you calculating all these values and ultrafast reactions by titration (which technique)?

There are missing relevant details, context and background of the question. // What time passed between formulation of your questions and posting them ? What topic research have you done before asking?

I appreciation that you are thinking deeply. Okay, so you are using the Nernst equation. You have to check how the other author (person) got the value of 13.02? What is the citation and their experimenal condition? If those people employed different conditions, you cannot compare your values at all.

Also, curious about the meaning lg'K? What is the purpose of the prime over the log function?

The lgK value of 13,02 is a thermodynamic equilibrium constant, calculated at 25.0°C at 1 bar. It is calculated with the activities of the ions in solution in mind. Our lg'K of 11,67 is not calculated with the activities in mind. The temperature is therefore one reason for the deviation.The prime is only there to indicate that the activities is not included in the calculation, I guess you could also write "lgKc" to show that it is only calculated with the concentrations in mind.

@Tove, Yes, this is exactly what I am trying to ask: Who calculated the value of 13,02? Is there a reference? If yes, we have to see their conditions. Also remember that every measurement has an error. What was the standard deviation of the original reference and what is your standard deviation?

For numbering of equations, you use at the end with increasing numbers   \tag{1}.

I suspect the primary source for the error (≈10%) you are encountering between your calculated $K'$ value and the theoretical $K$ value is the deviation of the molar equilibrium concentrations of $\ce{Fe^2+}$ and $\ce{Fe^3+}$ from the activities of both ions.
The presence of many ions in a solution, especially when some possess multiple charges (like $\ce{Fe^3+}$ and $\ce{Fe^2+}$, not to mention the sulphate iron complexes noted) or are highly concentrated, contribute to a non-negligible ionic strength $I_c$. The reason for this is the following:
$I_c$ for a solution with $n$ charged species can be calculated with the formula:
$$I_c=0.5\sum_{i=1}^n c_i\;z_i^2$$
Where:
$z_i$ represents the charge of species $i$
$c_i$ represents the concentration of species $i$
When $I_c$ is relatively high ($I_c>0.01M$), the activity coefficient $\gamma$ of multi-charged ions deviate significantly from unity (are reduced significantly), and the ones with a higher charge tend to experience a higher reduction than those with a lower charge.
The activity of a species can be calculated using its activity coefficient and concentration in solution:
$$a_i=\gamma_i\;c_i$$
This means that the relationship between $K$ and $K'$ in our case is:
$$K=\frac{a_{\ce{Fe^2+}}}{a_{\ce{Fe^3+}}}=\frac{\gamma_{\ce{Fe^2+}}}{\gamma_{\ce{Fe^3+}}}\frac{c_{\ce{Fe^2+}}}{c_{\ce{Fe^3+}}}=K_\gamma\;K'$$
Or simply:
$$K=K_\gamma\;K'$$
We can observe that as both activity coefficients approach unity, $K$ approaches the value of $K'$:
$$\gamma_{\ce{Fe^3+}}≈\gamma_{\ce{Fe^2+}}≈1\implies K_\gamma ≈1\implies K≈K'$$
However, if $I_c$ is high enough to cause significant deviation, such deviation will be more impactful over $\gamma_{\ce{Fe^3+}}$ than it would over $\gamma_{\ce{Fe^2+}}$, since the former has a higher charge than the latter.
Just to give you an idea of the magnitude of error that arises when assuming ionic strength is negligible in cases when it shouldn't be, we will consider the theoretical activity coefficients for both ions at 25°C and a solution with an ionic strength of 0.01M:
$$\gamma_{\ce{Fe^3+}}=0.443$$
$$\gamma_{\ce{Fe^2+}}=0.676$$
$$K_\gamma=1.53$$
This means that the thermodynamic equilibrium constant $K$ would be more than 150% higher than the calculated equilibrium constant $K'$ with molar concentrations.
If you want to estimate the value of both activity coefficients in accordance with your experiment, you can use the Debye-Hückel equation.
After independently substituting all the parameters for $\ce{Fe^3+}$ and $\ce{Fe^2+}$, the resulting expressions are:
$$\pu{-log}\;\gamma_{\ce{Fe^3+}}=\frac{4.59\sqrt{I_c}}{1+2.97\sqrt{I_c}}$$
$$\pu{-log}\;\gamma_{\ce{Fe^2+}}=\frac{2.04\sqrt{I_c}}{1+1.98\sqrt{I_c}}$$
Or you can combine them and get a function of $K_\gamma$ in terms of $I_c$:
$$\pu{-log}\;K_\gamma=\frac{2.04\sqrt{I_c}}{1+1.98\sqrt{I_c}}-\frac{4.59\sqrt{I_c}}{1+2.97\sqrt{I_c}}$$
Plotting $K_\gamma$ vs $I_c$ would give:

I suspect the ionic strength of your particular solution at equilibrium has a value of approximately $4\times10^{-4}M$ since:
$$I_c=4\times10^{-4}M\implies K_\gamma=1.12=\frac{K}{K'}=\frac{13.02}{11.67}$$
However, if there were a significant difference between the experimental temperature and the theoretical temperature, that would also factor in the error, since both equilibrium constants $K$ and $K'$ are temperature-dependent. Measurement errors and impurities can also contribute to the error.

Funny is, I have never seen molality based I either, until in the link.

Me neither, apparently $I_m$ is more precise than $I_c$.

Perhaps, include the improved D-H equation (with W. link) in its symbolic form too, so the literal values do no look like magical numbers. A great answer, BTW.

This was a great answer, I will apply this to our data and see how this affects the equilibrium constant. And yes, I forgot to mention that the temperature in the laboratory was 23.0°C, which will have an impact.


text:
Why is only the (E,E)-isomer obtained in Claisen-Schmidt condensation?
I added 2 equiv of p-tolualdehyde to 1 equiv of acetone to make (1E, 4E)-1,5-di-p-tolylpenta-1,4-dien-3-one (NMR backed it up). Why is this the only isomer formed? Just sterics or something more?

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.

If your Claisen condensation was run with lots of heat (reflux?), as is typical, then you will obtain the thermodynamic product. For the alkenes, the (E)-isomer is more thermodynamically stable (because of sterics) so predominates under these conditions.


text:
Definition of Avogadro's constant
I've read this excerpt from Wikipedia countless times, but I'm still confused:

The Avogadro number is the approximate number of nucleons (protons or neutrons) in one gram of ordinary matter.

Shouldn't that be "in one mole"?

The quote you have used is there just behind the Avogadro constant([$\pu{mol-1}$])/Avogadro number(unitless) definition. Many true statements can be misinterpreted, if quoted or considered out of their context.

Notice the approximate. It is not the definition.

$\pu{1 mol}$ is by definition the amount of matter consisting of exactly $N_\mathrm{A} = \pu{6.02214076E23}$ particles(or generally any formal constituent objects).
The molar mass of nucleons ( protons or neutrons) is approximately $M_\text{nucleon} \approx \pu{1 g/mol}$.
Practically all the mass of ordinary matter is due its nucleons.
Therefore,  $\pu{1 g}$ of ordinary matter contains approximately $N_\mathrm{A}$ nucleons, as it contains approximately $\pu{1 mol}$ of them.

Imagine this:

I say that the current definition of kilogram is based on the fixed value of the Planck constant.
I then say $\pu{1 kg}$ is the approximate mass of $\pu{1 L}$ of water.
Does it mean I say the kilogram is defined as the mass of $\pu{1 L}$ of water? Of course it is not.

The subject and predicate should be reversed! One gram of matter contains approximately Avogadro's number of nucleons [protons + neutrons] [but not Avogadro's number of electrons unless the substance is protium or possibly the natural mix of H2]

@jimchmst Electrons were not talked about. :-)

Were not but they are pertinent.

@jimchmst Not really in this context.

maybe not but they are pertinent. What is wrong with getting one to think more about this. Also the Wikipedia statement is confusing to at least one person and seems foolish to me.

@jimchmst If I say 1 kg is the approximate mass of 1 dm3 of water, it does not meant 1 kg is defined as the mass of 1 dm3 of water.

The Wikipedia statement, if correctly quoted, gives a false implication about the definition of Avogadro's number and did confuse at least one person. Why defend it.  My parenthetical comment on electrons was to try to get a broader thought of what actually is in one gram of nuclides. As to the density of water no one seems to know what that is anymore.

@jimchmst It is there just behind the  definition. If he used a broader quote...  If it fooled 1 person, than it is still great. Many correct statements fool multiple persons misinterpreting that. Especially if quoted out of context.


text:
How can the heat equation be used to model temperature vs time in a solubility experiment?
Background:
I am doing experiments trying to map the practical solubility of a certain chemical species called an "activator" solution. This solution is essentially a sequential mixture of water, $NaOH$ pellets and aqueous sodium silicate. In short (without getting into too much background), my results will help me determine if I can use certain chemical species in some very environmentally friendly materials science applications.
One of the parameters to keep track of is the temperature of solution which changes the solubility of the activator. I need to model this as I've found identical solutions respond differently when the final sodium silicate ingredient is added to the aqueous $NaOH$ solution at different temperatures. Here is some experimental data from the lab showing T vs t:

Essentially, there are 2 main "periods" during the experiment:

The Initial Temperature Rise (i.e. from $T_i$ = 23.4 $^{\circ} C$ = ambient temperature $\rightarrow$ $T_{Max}$ = 97 $^{\circ} C$).
The Cooldown Period (i.e. from $T_{Max} \rightarrow T_i$)

Here is a picture of the experimental setup showing 2 x ~200mL of activator in a conical flask. Both activators were made the same initially but on the left, the sodium silicate was added at ambient temperatures (which caused precipitation) as opposed to the right, where the sodium silicate was added at around 80% of $T_{Max}$.

Problem:
I'm wanting to model the T vs t for the system (similar to the experimental data above) so I can see the temperature profile of different activators with different chemical ratios. It seems, fundamentally, to start with an energy balance with:
$$\frac{dQ}{dt} = Q_{in} - Q_{out} + Q_{rxn}$$
Assuming no heat input, a known value for $Q_{rxn}$ (from a previous question of mine) and $Q_{out} = Q_{Conduction,Plate} + Q_{Convection,surroundings}$ I'm wondering the following:

How should the system be modelled? is it good to model $Q_{out}$ with the heat equation and solve for T (as per pic below)? On that, how do I even solve for t? I have minimal knowledge of Fourier transforms.

Apologies, I have contextualised $Q_{out} = Q_{Conduction,Plate} + Q_{Convection,surroundings}$ in an edit to address your second comment. Do you mean separating each section into approximated fitted functions and then adding the functions together? I am not sure how long it would take to get to $T_{Max}$ despite my model for the value itself. I also want a predictive model for other solutions I could make, I don't believe fitting functions to one experimental dataset will be the same for the others.

I deleted my previous comments since I think they are not so helpful. I have just two questions: (1) you mix NaOH and silicate solution. The ensuing reaction is very fast on the timescale of measurement. That means heat production is like a pulse. Do you agree with this description? (2) Is the hot plate at the same T as the air?

If (2) is true, and you assume surroundings are at constant T, then I don't think you need to separate conduction and convection losses etc. Just use Newton's law and keep it simple: https://en.wikipedia.org/wiki/Newton%27s_law_of_cooling

(1) Sure for this situation, I wonder if a linear equation instead would be more fitting for solutions with larger volumes instead of a pulse though? On (2), that's great thank you! Although I'm having trouble validating the model using the equation as I can't seem to solve for the right r value properly (i.e with an example (x,y) = (t, T) = (t_max, T_Max) = (105, 97) with $T_{env} = 23.4$ and $T_{Max} = T(0) = 97$. When I try I get 0.0274 which makes it decay too quickly. Any help here?

Because, as a bonus I'd like to (at least somewhat) accurately predict when the system cools to ambient temperature again.


text:
Factor Group Analysis of Na2SO4 - Site Group Symmetry
I'm still trying to understand how to construct a factor group analysis for a given molecular point group and the resulting solid state group symmetry. For this I searched for some examples and found a good construction in a publication on $\ce{Na2SO4}$.
Sadly, the authors didn't give any information on the crystal structure and thus I had to test multiple versions. Their final group symmetry is given as $D_\mathrm{2h}$ which means it has to be a space group with the HM symbol of mmm. For $\ce{Na2SO4}$ this limits the selection to Cmcm ($\ce{Na2[CrO4]}$), Pbnm (olivine-structure) or Pbnn.
At this point I'm still unsure if the site-symmetry has some relation to the solid state group symmetry (any other than being a sub-group of it) or if it's related to the symmetry reduction of the molecular point-group by coordination.
An example of this is given here:

It's from K. Nakamoto, Infrared and Raman Spectra of Inorganic and Coordination Compounds and shows how the $T_\mathrm{d}$ symmetry of the free sulfate-anion is reduced by coordination to for example a metal. Now I always assumed that this is the first step you have to consider while constructing those factor group relationships.
So I took smaller and larger units from the crystal structure of all three given modifications of $\ce{Na2SO4}$ and made sure that they all have the sulfate sulfur in the center. Then I ran an automatic point group detection on them.

It seems like all three would return some symmetry elements and even using larger clusters resulted in the same symmetries (which you would probably expect from a periodic lattice).
The three site group (point groups) I could determine here were $C_\mathrm{s}$ and $C_\mathrm{2v}$. Besides that, comparing the non-trivial subgroups of $T_\mathrm{d}$ and $D_\mathrm{2h}$ would also offer $D_\mathrm{2}$ and $C_\mathrm{2}$.
When I compare this to the factor group analysis that Mabrouk et. Al. constructed in their Paper (DOI:0.1002/jrs.4374) it seems like they went for $D_\mathrm{2}$ only:

So either Mabrouk did a mistake while constructing them or used a crystal structure that is not in the ICSD or I'm wrong with my assumption that the site group symmetry is based on the symmetry reduction around the central ion of the molecular unit in question. Which is strange because other examples, even some starting from $T_\mathrm{d}$ arrive at the same conclusions as I when I use this method.
So if somebody has an idea where my mistake lies and how to correct it I'm open for any suggestions.
EDIT:
A minor editor here: I found another paper by R. L. Frost et. Al. about vanadates (DOI:10.1002/jrs.2906) and here the add more details to the site symmetry.

In vésigniéite,the two inequivalent Cu atoms and Ba atom  occupy C2 sites whereas V, O1, O2, O3, O4 and H1 atoms occupy Cs sites.

I understand this explanation as occupying sites and then Cs is given in the following table as using the site symmetry as given by the space group. So in case of $\ce{Na2SO4}$ in Cmcm the Sulfur resides on 4c which has an m2m symmetry that equals $C_\mathrm{2v}$. The only real example I found would be for this Pbnn structure, where S resides on 4d which means it has a 2.. symmetry. I don't fully understand how the dots work in crystallography but if it means that they could be two as well then 222 =  $D_\mathrm{2}$ which doesn't match my point group analysis though.

The orthorhombic polymorph of Na2SO4 crystallizes in the Cmcm space group type (n* 63) [Acta Crystallogr. B (1991) 47, 581-588]. Na atoms are located at two different sites: 4c and 4a. The site symmetry of 4c is m2m (m along x, 2 along y, m along z); the site symmetry of 4a is 2/m.. (2/m along x; no fixed symmetry along y, and z). BTW,  the site-symmetry is always related to the symmetry of the parent space group type.


text:
Is it possible to create Fischer projections (or similar) from SMILES expressions?
I've recently begun experimenting with drawing molecules from smiles expressions for use in an educational website. I am looking for a way to create simple structural representations of various molecules in metabolic pathways (Kreb's cycle, Gluconeogenesis, etc) in what I believe is Fisher projections (or similar?), as is often encountered in text books, for example the molecule Citrate:

Since I'm creating my website in Python I tried using rdkit but so far have only been able to create drawings with 110° angles.
The smiles formula I used here is courtesy of pubchem:

C(C(=O)[O-])C(CC(=O)[O-])(C(=O)[O-])O

Is there any open source tool (preferably in Python) that I could use to create images of the Fischer projections? Or is there a "trick" of how to do this with rdkit? I would be grateful for any help and please excuse if I have botched any of the terminology in question. Thanks!

I’m voting to close this question because it has been cross-posted on Matter Modelling and has already been answered there: https://mattermodeling.stackexchange.com/q/10002/724.


text:
Which Hydrogen in this structure is more deshielded/shielded?

Looking at the 2 hydrogens connected to the Carbon-carbon double bond, which hydrogen is more deshielded or shielded?

Presumebly the one in β.
Here I report the ChemDraw prediction.

To confirm that (that maybe is not necessary), you might run an HMBC experiment. You should be able to determine which hydrogen is closer to the carbonyl group. Probably the signal at 5.32 will couple with the carbon signal of the carbonyl group (around 170 ppm?)

HMBC experiments typically emphasise three-bond correlations - if anything I would expect the beta proton to show a crosspeak to the carbonyl carbon. Two-bond correlations are often absent or weak. Also, there's no peak at 5.32 ppm in the actual spectrum, it's only a crude prediction.


text:
Reaction constant contains one of the reactants
I need to provide the reaction rate functions of a system of reactions consisting of 5 reactions. One of them has the form: $$r_1 = \frac{k_a}{K_D + \frac{K_D}{K_I}X_4}\cdot X_2\cdot X_4.$$ I do not know how to handle the term of the reaction constant of this reaction rate function given that one of the reactants, $X_4,$ is inside the constant. What I ultimately need, is to write a differential equation concerning the reaction dynamics. How can I handle this problem ?

What's to handle? You know your concentrations: these are $X_1,,X_2$, etc. You know your rates. Well, the derivative of this is that, and so on.

If that's the rate expression, then it is. If $X_4$ is a constant value (i.e. the system has enormous amounts of it, then you can form one with a pseudo 'rate constant'

@user129173 sounds like you're dealing with a multiple reaction problem. What are the 5 reactions and individual rates given?

Reaction rates of complex multistep reactions in context of a system of linked reactions generally does not have form $\text{Rate} = k \cdot [\ce{A}]^a \cdot [\ce{B}]^b$, so $X_4$ is not inside the reaction rate constant. // See e.g. H2 + Br2 -> 2 HBr


text:
How do I determine the hybridization of terminal atoms?
I've read that terminal atoms do not undergo hybridization because there is no need to.
For example, CF4, C undergo sp hybridization, but fluorine do not hybridize.

How about CO2? Why is oxygen sp2 hybridized? Oxygen is a terminal atom.
 Oxygen, without hydridising, can already form a sigma bond and pi bond with Carbon. For example, Oxygen's px orbital bonding with Carbon's sp orbital forming sigma bond,
 oxygen's py orbital bonding with Carbon's p orbital to form pi bond.

Hybridization occurs in our heads.

Wait... what???

As chemists, we tend to have a restricted view of what bonds are and how they form. All orbitals of one atom interact with all orbitals of another. The only presence of the interaction is a bond, which can lead to a repulsive force or an attractive force and so to two atoms a part or two atoms bonded together, respectively.

@Bryan You heard what. Look at a world map with all them continents and oceans. They are real. Also, on the map there are meridians and parallels. They give the impression of huge walls that run from pole to pole. Are they real? No. They only exist in our heads. Same thing with hybridization.

Obligatory "All models are wrong, but some are useful."

The concept of hybridization of orbitals is a convenient way to the speak about the molecular orbitals in a molecule.
Consider the case of CH$_4$, where we would expect an sp$^3$ hybridized carbon interacting with four 1s orbitals of hydrogen. If you see the MO picture, you will see that there would be four orbitals of same symmetry and same energy at the frontier, with contributions from Carbon and Hydrogen atoms. Particularly, the MOs would have the form
$\psi_{MO}= a \psi_{1s}^{H1} + b \psi_{1s}^{H2} + c \psi_{1s}^{H3} + d \psi_{1s}^{H4} + e \psi_{1s}^{C} + f \psi_{2s}^{C} + g \psi_{2p_x}^{C} + h \psi_{2p_y}^{C} + I \psi_{2p_z}^{C}$.
Since the contribution from $1_s$ orbital of C would be very small, it would be easier to consider the four hydrogen orbitals as one group and the remaining carbon orbitals as a separate group of sp$^3$ character. If you do the same analysis for CO$_2$, you will see why oxygen is in sp$^2$ hybridization.
In other words, hybridization abstracts away all this information, while giving you the jist of how molecules form.

I would also add that obtaining the coefficients for that expansion often involves intensive efforts of numerical calculus, normally carried out by computers implementing quantum chemical softwares. Which is the entire reason why we have these qualitative heuristics, in order to do qualitative quantum chemistry in our heads.


text:
Why is cyanuric triazide environmental friendly and less toxic?
This is the first line on Wikipedia:

Cyanuric triazide ($\ce{C3N12}$ or $\ce{(NCN3)3}$) is described as an environmentally friendly, low toxicity, and organic primary explosive with a detonation velocity of about $\pu{7300 m/s}$, and ignition temperature at $\pu{205 °C}$

While it is written in the general WP article that "some" organic azides are classified as highly explosive and toxic (and citing the article in ref.), I want to know what determines its toxicity and why cyanuric triazide is less toxic?
Reference

How Dangerous Is Too Dangerous? A Perspective on Azide Chemistry, Daniel S. Treitler and Simon Leung, The Journal of Organic Chemistry 2022 87 (17), 11293-11295 DOI: 10.1021/acs.joc.2c01402

Umm, it's rather a relative thing. Well, for example KCN is much less toxic than botox ;>

Two differences from most organic azides: no hydrogen, and carbon is in effect fully oxidized (all four valence are to more electronegative atoms).

When you try to look up toxicity info for organic azides, they are often limited. Further, I read in one source that little data has been recorded for the toxicity of the majority of organic azides. The toxicity of sodium azide or hydrogen azide comes with the release of the anion which disrupts cellular processes.


text:
Can deuterium be separated from heavy water via electrolysis? It is an isotope of hydrogen, so I guess it should act in the same way?
I was wondering if it is possible to separate deuterium from heavy water via electrolysis.
Also, is this the way deuterium gas is obtained on an industrial scale, or are there better methods?

See https://www.iter.org/newsline/167/631 for example.

All hydrogen of heavy water is deuterium. It's electrolysis separates deuterium and oxygen.

Yes, Deuterium can be separated from heavy water  $\ce{D2O}$ by electrolysis. I don't know whether it is the best way of producing $\ce{D2}$.

Why would anyone produce deuterium gas on an industrial scale?

@Karl - because there is a demand for it for various things, from nuclear physics to bio-tagging molecules.

If by chance you are asking about separating deuterons and protons from water that has both, electrolysis can indeed be used for that, since the hydrogen gas produced is enriched in protium due to the isotope effects: https://www.sciencedirect.com/science/article/abs/pii/S0360319921029591

@JonCuster Sounds rather like lab-scale demands, not industrial.

@Karl - Canada exports roughly $30M of heavy water a year as the top producer. Not a commodity chemical, but not exactly lab scale either.

@JonCuster Yea, you need tons of it to run certain types of nuclear reactors, for starters. And to produce any kind of deuterated chemicals, which is pretty large market, etc. I was talking about deuterium gas. What is that good for? A few university labs and a very small number of speciality chemical companies buy a five-liter gas bottle per year, I expect.


text:
How to calculate the pH value of a Carbonate solution?
I read the similar questions suggested when submitting my question but they didn't help me How can one calculate the pH of a solution? How to calculate pH of the Na2CO3 solution given ambiguous Ka values

We are asked to calculate the pH value of a Carbonate Solution with a concentration of 0.005 mol per liter
They provide the numerical solution as 10.97, but not how to get it

Carbonate has the molecular form CO3^(2-)
I know how to calculate the pH of a strong base (which Carbonate seems to be)
pH + pOH = 14 so pH = 14 - pOH, and pOH = -log_10 (concentration = 0.005 mol per liter)
But I just don't get the same result as they give
I even assumed Carbonate is a weak base, and so used pKb = pOH^2/0.005 with pKb = 3.60 for Carbonate which I found online, and then solved for pOH and used pH = 14 - pOH, but even then I don't get their solution
I know that in water, Carbonate becomes Bicarbonate ( CO3^(2-) + H2O --> HCO3- + OH-), which then becomes Carbonic Acid (HCO3- + H2O --> H2CO3 + OH-)
All other exercises about pH they gave us, I always found the same result as they do. But for Carbonate, I just don't see what am I doing wrong. What am I missing ?
Thank you so much for your help

The equilibrium system of interest is:
$$\ce{CO3^{2-}(aq) +H2O(l)<=>HCO3^{-}(aq) +OH-(aq)}$$
Let:
A represent $\ce{CO3^{2-}}$
B represent $\ce{HCO3^{-}}$
C represent $\ce{OH^{-}}$
The $pK_b$ of $\ce{CO3^{2-}}$ at 25°C is approximately 3.67.
Initially, only $\ce{CO3^{2-}}$ and water are present. As an approximation, water is not included in the equilibrium constant expression, so at equilibrium we have:
$$C_A=C_{Ao}-x=0.005-x$$
$$C_B=C_{Bo}+x=x$$
$$C_C=C_{Co}+x=x$$
Which can be substituted into the equilibrium constant:
$$K_b=\frac{x^2}{0.005-x}=10^{-3.67}$$
Solving for $x$:
$$C_C=x=9.33\times 10^{-4}$$
The pOH of this solution would be:
$$\pu{pOH}=\pu{-log}\;C_C=\pu{-log}\;(9.33\times 10^{-4})=3.03$$
Finally, the pH of the solution is:
$$\pu{pH}=14-\pu{pOH}=14-3.03=10.97$$

Thanks a lot for your help ! It makes a lot more sense now. The methodology you used is for a weak base, I must have done something wrong somewhere, maybe I forgot to raise 10 to the -3.67 power to get Kb

The pH of a diluted solution of a weak base, like $\ce{CO3^{2-}}$ is given in the tables by the following formula $${p\mathrm{H} = \frac{1}{2} ( 14 + p\mathrm{K}_a + \mathrm{log} c_b)}$$ The acid $\ce{HCO3^{-}}$ has a $p\mathrm{K}_a$ value equal to $10.25$. Substituting this numerical value, plus $c_b = 0.004$M into preceding formula, gives $$p\mathrm{H} = 10.97$$


text:
How come nuclear waste is so radioactive when uranium is relatively stable with an extremely long half life?
Natural uranium has a half life of 4.5 billion years. Representing with the expression $m(0.5)^{\frac{x}{(4.5\times10^9 \times 364)}}$, where $x$ is in days and $m$ is the initial mass of uranium. In one day, only $4.2 \times 10^{-11} \%$ decays into thorium. It would take around $2.4 \times 10^{13}$ atoms for one alpha particle to be generated. About $\pu{4E-11 mol}$. Assuming $\pu{5 MeV}$ per alpha particle. With $\pu{1.6021773E-13 J}$ in a $\pu{MeV}$. It would take $\pu{2.5E2 mol}$ of uranium or $\pu{6 kg}$ of uranium directly aimed at a point with theoretically $0$ loss of energy to achieve just a single Sievert. Then how is it possible that it is leaking $\pu{10 Sv}$ an hour at Fukushima reactor 1, an approximate of $240$ per day. That would be an equivalent to $1.4$ tons of pure uranium with $0$ theoretical power loss all directly channeled through retroreflectors and a focusing array all with 100% efficiency, channeling all the alpha decay into a single point onto the detector. I just can't wrap my head around this.
TLDR: How come nuclear power plants release so much radiation when uranium is so relatively stable?

The whole point of fission reactors is that you extract energy by turning uranium into not-uranium. It's the not-uranium which can be many orders of magnitude more radioactive.

Also, nuclear power plants have very little to do with natural uranium.

@IvanNeretin  Unless they use heavy water moderator, but such designs like CANDU are not used much.

There is no need to use entriched uranium, there are reactor designs such as MAGNOX and CANDU which will run on natural uranium. It is interesting to note that much of the energy obtained from natural uranium or even low enriched uranium fuels is due to the fission of the plutonium which is generated by the neutron activation of the uranium-238.

How can a question about radioactivity be on-topic on a chemistry site? Shouldn't  it involve electrons?

@PeterMortensen Questions on radioactivity is very much on topic here.

As a pedantic note on your TLDR nuclear power plants work very hard to NOT release radiation. If they are releasing radiation, that's a very bad thing.

Radioactive isotopes (radionuclides to be more correct) are thermodynamically able to change into another nuclide, this decay has to be an exothermic event. For example if we consider beta decays for nuclides with mass numbers of 130. It is clear that Cadmium-130 atoms can be transformed by beta decay into lower energy atoms such as tellurium-130 by beta decay. But Xenon-130 is unable to convert itself into cesium-130.

The rate of the transformation (radioactive decay) is proportional to the number of the nuclei of the radionuclide in question. While with exceptionally high decay energies the decay rate is often higher, there not a clear relationship between decay energy and rate of decay. Things like odd vs even proton / neutron numbers and magic numbers are important.
The rate constant for the reaction is normally written as "lambda" the value of the constant varries greatly between very large (such as beryllium-8) and very small (such as thorium-232, bismuth-209 and plutonium-244).
The activity (A in Bq which is events per second) is given by the equation
A = lambda N
Where N is the number of atoms
You calculate lambda by dividing ln(2) by the half life in seconds
When fission of uranium or plutonium atoms occurs a vast range of different nuclides are created. The final decay products of these range from zinc through to the middle of the lanthanide series. You can imagine that a very wide range of different nuclides exist in the decay chains of the different initial fission products. If we consider if a zirconium-110 nucleus is formed by the fission process, all the radionuclides in the decay chain which leads to palladium-110 have short half lives (less than 15 seconds). So if a short pulse of fission such as a nuclear bomb detonation occurs then all the atoms with the mass of 110 will have decayed to the palladium in about two minutes.
If we were to make a solution of enriched uranium nitrate and deliver a strong neutron pulse to it and then five seconds later we were to separate the ruthenium from this mixture then the specific activity of the ruthenium in terms of Bq per mole would be exceptionally high. But given a minute or so then the radioactivity level of the ruthenium in the neutron bombarded uranium will be far lower.
Now if we were to leave uranium after a dose of neutrons and leave it it for millions of years then the vast majority of the fission products will decay away and then the radioactivity of the uranium will be larger than the fission products. There is a place in Africa where a series of natural nuclear reactors operated many millions of years ago. The vast majority of the fission products and the transuranium elements (such as plutonium) formed by these prehistoric reactors have decayed away to either stable nuclides (in the case of fission products) or uranium in the case of the plutonium which the reactor formed.
In the samples of the African fossil reactor materials clear signs of nuclear fission exist, while for a while (a long time ago) the fission product radioactivity was far greater than the either the uranium or plutonium radioactivity. The fission product radioactivity and then the plutonium radioactivity have no decayed away. Leaving the slowly decaying uranium still in the samples.

We have enough information on the rate at which that natural reactor worked to say that the fission product radioactivity was higher?

If anyone wants to look up this reactor, look for Oklo, Gabon.

The radioactivity associated with nuclear waste does not come from naturally occurring uranium, but from products associated with processing and using uranium. First off, uranium as it occurs in nature is not really nuclear fuel. It must be enriched in the more radioactive but more fissionable (and thus better energy source) uranium-235, leaving behind the more stable uranium-238, or uranium-238 is converted to more useful fuels such as plutonium-239.
Then after the fission process, what is left behind includes isotopes of strontium, caesium, and the aforementioned plutonium, which are the main contributors to the hazardous radioactivity of the waste.
From the National Resources Council:

During the fission process, two things happen to the uranium in the fuel. First, uranium atoms split, creating energy that is used to produce electricity. The fission creates radioactive isotopes of lighter elements such as cesium-137 and strontium-90. These isotopes, called "fission products," account for most of the heat and penetrating radiation in high-level waste. Second, some uranium atoms capture neutrons produced during fission. These atoms form heavier elements such as plutonium. These heavier-than-uranium, or "transuranic," elements do not produce nearly the amount of heat or penetrating radiation that fission products do, but they take much longer to decay. Transuranic wastes, sometimes called TRU, account for most of the radioactive hazard remaining in high-level waste after 1,000 years.
Radioactive isotopes eventually decay, or disintegrate, to harmless materials. Some isotopes decay in hours or even minutes, but others decay very slowly. Strontium-90 and cesium-137 have half-lives of about 30 years (half the radioactivity will decay in 30 years). Plutonium-239 has a half-life of 24,000 years.

I rolled back part of the previous edit. The text in the indented block is a direct quotation, and caesium was spelled without the "a" in the quoted material.

As said in the comments, the whole point of fission reactors is that you extract energy by turning uranium into something else and that "something else" is much more radioactive than uranium itself. The nuclear waste can be categorized according to their radioactivity:

low-level waste (LLW)
intermediate-level waste (ILW)
high-level waste (HLW)

The vast majority of the waste (90% of total volume) is considered LLW which contributes to 1% of the total radioactivity. On the other hand, HLW accounts for 3% of the total volume of waste, but contains 95% of the total radioactivity.

The HLW is primarily both uranium fuel that has been used in a nuclear power reactor and is "spent" or no longer efficient in producing electricity (Spent fuel is thermally hot as well as highly radioactive and requires remote handling and shielding) and the fission products (why the fission product is radioactive is mentioned in Oscar's answer).
When the nucleus splits, most of the energy is released immediately and carried off by coolant to do useful work. This energy continues to be released for thousands of years after the atom splits. This delayed emissions mean that nuclear waste is still highly radioactive. Even the afterglow heat that is produced is what makes the waste hazardous.

The amount of HLW worldwide is currently increasing by about 12,000 tonnes every year. A 1000-megawatt nuclear power plant produces about 27 t of spent nuclear fuel (unreprocessed) every year.
References:

https://en.wikipedia.org/wiki/Radioactive_waste
https://world-nuclear.org/nuclear-essentials/what-is-nuclear-waste-and-what-do-we-do-with-it.aspx
https://whatisnuclear.com/waste.html
https://world-nuclear.org/information-library/nuclear-fuel-cycle/nuclear-wastes/radioactive-wastes-myths-and-realities.aspx

A point worth making is that we don't have any choice about the products of uranium fission being much more radioactive than the starting fuel. The laws of physics decide that.


text:
Calculating energy difference between two 1-4 dimethylcyclohexane conformers
The following question was asked in a quiz at my college

Draw the most stable and least stable chair conformer for 1,4-dimethylcyclohexane and calculate the
energy difference between these two. Given that the value for CH3-H diaxial interaction = 3.8 kJ/mol and
CH3-CH3 gauche interaction = 3.8 kJ/mol. Ignore the gauche interaction with the ring C atoms

I drew the Newman projection but I am unable to understand how to proceed further as to which CH3-H pairs to consider and which not to, also in the Newman projection will the two methyl even interact?

Look here.

A second thought. Conformational structures and/or molecular models are better than Newman projections for what you are trying to accomplish.


text:
Could a beach have green or blue sand due to a nearby copper deposit? How poisonous would such a beach be?
I am in the midst of a world building project and am wondering about green sand from copper oxide/patina material. The material could either be washed into the water from a nearby land deposit or be coming from an exposed deposit on the lakebed or seafloor. Would it be possible to have vibrant blue or green, or would the particles break down onto something of more boring color? How poisonous would such a beach be with a high copper concentration? How would the salinity of the water affect the situation?
Any answers would be much appreciated. I haven't been able to find any real-world examples of this happening.

Green sand would be harmless breakdown products of mafic/ultramafic rocks, such as those found in volcanically-active areas. These beaches in Hawaii are green, read on about the details. Forget about copper.

Had you asked if it was possible for a beach to be entirely covered with deep red garnets (not gem quality), I could have verified that it was possible: here is a photo I just took of some of the garnets I scooped up when I visited that beach about 30 years ago. The whole beach looked like that, with very little sand. So maybe a peridot beach would suit your world building project!

I have heard that some soils in the former Belgian Congo are blue /green in some areas. Some copper deposits are exceptionally rich with a few per-cent copper.

BTW, podzol soils may be initially a blue-green, and change to yellow-brown after exposure to air for some minutes. This is not due to copper, but ferrous iron oxidizing to ferric. It's an interesting effect noted on digging into soils in pine forests and taiga (krummholz).

About blue sands: It contains Ilmenite, which is dark blue in color.  Some part of Sri Lankan beach contain ilmenite and is a export item for the country.


text:
Computing molecular integrales in STO-3G basis with the Obara Saika recursion formulas
I'm trying to implement the Hartree-Fock algorithm on a water molecule using the STO-3G basis. This far I have only computed the overlap integrals with the Obara-Saika formulas but I'm running into problems with the ones involving p-orbitals. My question is,on formulas like this one,
$$
(p_i||s) = (P_i-A_i)(s||s)
$$
What function is exactly the $(s|$ on the right hind side? My guess would be that it is the same as $(p_i|$ (same exponent and coefficient) but with $l_i = 0$ instead of 1. In case this is indeed the correct interpretation I'll leave the code I have written, and hopefully my mistake is really obvious.
import sys
import math
import numpy as np
from scipy import special
from scipy import linalg

def d_fact(n):
    product = 1.0
    for i in range(3,n+1,2):
        product *= i
    return product

class primitive_gaussian():

    def __init__(self, alpha, coeff, coordinates, l1, l2, l3):

        self.alpha = alpha
        self.coeff = coeff
        self.coordinates = np.array(coordinates)
        self.l = [l1,l2,l3]
        self.N = (2.0*alpha/math.pi)**(3/4)*(4*alpha)**((l1+l2+l3)/2)*(d_fact(2*l1-1)*d_fact(2*l2-1)*d_fact(2*l3-1))**(-1/2)

def overlap(i,j):

    N = i.N*j.N
    cicj = i.coeff*j.coeff

    zeta = i.alpha+j.alpha
    xi = i.alpha*j.alpha/zeta
    AB = i.coordinates-j.coordinates
    AB2 = np.dot(AB,AB)

    P = (i.alpha*i.coordinates+j.alpha*j.coordinates)/zeta

    if((np.count_nonzero(i.l)==0)and(np.count_nonzero(j.l)==0)):
        return N*cicj*math.exp(-xi*AB2)*(math.pi/zeta)**(3/2)
    elif((np.count_nonzero(i.l)==1)and(np.count_nonzero(j.l)==0)):
        i_m = np.nonzero(i.l)
        si = primitive_gaussian(i.alpha,i.coeff,i.coordinates,0,0,0)
        return (P[i_m]-i.coordinates[i_m])*overlap(si,j)
    elif((np.count_nonzero(i.l)==0)and(np.count_nonzero(j.l)==1)):
        j_m = np.nonzero(j.l)
        sj = primitive_gaussian(j.alpha,j.coeff,j.coordinates,0,0,0)
        return (P[j_m]-j.coordinates[j_m])*overlap(i,sj)
    elif((np.count_nonzero(i.l)==1)and(np.count_nonzero(j.l)==1)):
        i_m = np.nonzero(i.l)
        j_m = np.nonzero(j.l)
        si = primitive_gaussian(i.alpha,i.coeff,i.coordinates,0,0,0)
        sj = primitive_gaussian(j.alpha,j.coeff,j.coordinates,0,0,0)
        if(i_m==j_m):
            return (P[j_m]-j.coordinates[j_m])*overlap(i,sj)+overlap(si,sj)/(2*zeta)
        else:
            return (P[j_m]-j.coordinates[j_m])*overlap(i,sj)

def kinetic(i,j):

    zeta = i.alpha+j.alpha
    xi = i.alpha*j.alpha/zeta
    AB = i.coordinates-j.coordinates
    AB2 = np.dot(AB,AB)
    P = (i.alpha*i.coordinates+j.alpha*j.coordinates)/zeta

    if((np.count_nonzero(i.l)==0)and(np.count_nonzero(j.l)==0)):
        return xi*(3-2*xi*AB2)*overlap(i,j)
    elif((np.count_nonzero(i.l)==1)and(np.count_nonzero(j.l)==0)):
        i_m = np.nonzero(i.l)
        si = primitive_gaussian(i.alpha,i.coeff,i.coordinates,0,0,0)
        return (P[i_m]-i.coordinates[i_m])*kinetic(si,j)+2*xi*overlap(i,j)
    elif((np.count_nonzero(i.l)==0)and(np.count_nonzero(j.l)==1)):
        j_m = np.nonzero(j.l)
        sj = primitive_gaussian(j.alpha,j.coeff,j.coordinates,0,0,0)
        return (P[j_m]-j.coordinates[j_m])*kinetic(i,sj)+2*xi*overlap(i,j)
    elif((np.count_nonzero(i.l)==1)and(np.count_nonzero(j.l)==1)):
        i_m = np.nonzero(i.l)
        j_m = np.nonzero(j.l)
        si = primitive_gaussian(i.alpha,i.coeff,i.coordinates,0,0,0)
        sj = primitive_gaussian(j.alpha,j.coeff,j.coordinates,0,0,0)
        if(i_m==j_m):
            return (P[j_m]-j.coordinates[j_m])*kinetic(i,sj)+kinetic(si,sj)/(2*zeta)+2*xi*overlap(i,j)
        else:
            return (P[j_m]-j.coordinates[j_m])*kinetic(i,sj)+2*xi*overlap(i,j)

def make_matrixes(molecule):

    nbasis = len(molecule)

    S = np.zeros([nbasis, nbasis])
    T = np.zeros([nbasis, nbasis])

    for i in range(nbasis):
        for j in range(nbasis):

            nprimitives_i = len(molecule[i])
            nprimitives_j = len(molecule[j])

            for k in range(nprimitives_i):
                for l in range(nprimitives_j):

                    S[i,j] += overlap(molecule[i][k],molecule[j][l])
                    T[i,j] += kinetic(molecule[i][k],molecule[j][l])

    return S , T

molecule_coordinate = [[0.0,0.0,0.116321],[0.0,0.751155,-0.465285],[0.0,-0.751155,-0.465285]]

# create H2O molecule - sto-3g basis
O_pg1sa  = primitive_gaussian(0.1307093214E+03, 0.1543289673E+00, molecule_coordinate[0], 0, 0, 0)
O_pg1sb  = primitive_gaussian(0.2380886605E+02, 0.5353281423E+00, molecule_coordinate[0], 0, 0, 0)
O_pg1sc  = primitive_gaussian(0.6443608313E+01, 0.4446345422E+00, molecule_coordinate[0], 0, 0, 0)

O_pg2sa  = primitive_gaussian(0.5033151319E+01,-0.9996722919E-01, molecule_coordinate[0], 0, 0, 0)
O_pg2sb  = primitive_gaussian(0.1169596125E+01, 0.3995128261E+00, molecule_coordinate[0], 0, 0, 0)
O_pg2sc  = primitive_gaussian(0.3803889600E+00, 0.7001154689E+00, molecule_coordinate[0], 0, 0, 0)

O_pg2pxa = primitive_gaussian(0.5033151319E+01, 0.1559162750E+00, molecule_coordinate[0], 1, 0, 0)
O_pg2pxb = primitive_gaussian(0.1169596125E+01, 0.6076837186E+00, molecule_coordinate[0], 1, 0, 0)
O_pg2pxc = primitive_gaussian(0.3803889600E+00, 0.3919573931E+00, molecule_coordinate[0], 1, 0, 0)

O_pg2pya = primitive_gaussian(0.5033151319E+01, 0.1559162750E+00, molecule_coordinate[0], 0, 1, 0)
O_pg2pyb = primitive_gaussian(0.1169596125E+01, 0.6076837186E+00, molecule_coordinate[0], 0, 1, 0)
O_pg2pyc = primitive_gaussian(0.3803889600E+00, 0.3919573931E+00, molecule_coordinate[0], 0, 1, 0)

O_pg2pza = primitive_gaussian(0.5033151319E+01, 0.1559162750E+00, molecule_coordinate[0], 0, 0, 1)
O_pg2pzb = primitive_gaussian(0.1169596125E+01, 0.6076837186E+00, molecule_coordinate[0], 0, 0, 1)
O_pg2pzc = primitive_gaussian(0.3803889600E+00, 0.3919573931E+00, molecule_coordinate[0], 0, 0, 1)

H1_pg1sa = primitive_gaussian(0.3425250914E+01, 0.1543289673E+00, molecule_coordinate[1], 0, 0, 0)
H1_pg1sb = primitive_gaussian(0.6239137298E+00, 0.5353281423E+00, molecule_coordinate[1], 0, 0, 0)
H1_pg1sc = primitive_gaussian(0.1688554040E+00, 0.4446345422E+00, molecule_coordinate[1], 0, 0, 0)

H2_pg1sa = primitive_gaussian(0.3425250914E+01, 0.1543289673E+00, molecule_coordinate[2], 0, 0, 0)
H2_pg1sb = primitive_gaussian(0.6239137298E+00, 0.5353281423E+00, molecule_coordinate[2], 0, 0, 0)
H2_pg1sc = primitive_gaussian(0.1688554040E+00, 0.4446345422E+00, molecule_coordinate[2], 0, 0, 0)

np.array(molecule_coordinate[2])]
O_1s  = [O_pg1sa, O_pg1sb, O_pg1sc]
O_2s  = [O_pg2sa, O_pg2sb, O_pg2sc]
O_2px = [O_pg2pxa, O_pg2pxb, O_pg2pxc]
O_2py = [O_pg2pya, O_pg2pyb, O_pg2pyc]
O_2pz = [O_pg2pza, O_pg2pzb, O_pg2pzc]
H1_1s = [H1_pg1sa, H1_pg1sb, H1_pg1sc]
H2_1s = [H2_pg1sa, H2_pg1sb, H2_pg1sc]
molecule = [O_1s, O_2s, O_2px, O_2py, O_2pz, H1_1s, H2_1s]

S , T = make_matrixes(molecule)
for i in range(7):
    for j in range(i+1):
        print("{:.6f}".format(S[i][j]),end="\t")
    print("\n")

for i in range(7):
    for j in range(i+1):
        print("{:.6f}".format(T[i][j]),end="\t")
    print("\n")

Right after I posted the question I found the problem... I'll leave the question just in case someone runs into a similar issue. The thing is when i made this s orbitals from the p orbitals y inadvertently recalculated the normalization constant, which I believe is wrong. Now i copy the object and only change its momentum to zero. There was a second problem, more mundane, the coordinates I put where expressed in angstroms rather than atomic units. After changing this two things I got the expected results.

For future reference, questions like these probably find a bigger and better suited audience at [mattermodeling.se].


text:
Why is the imine (C=N) bond so weak?
Nitrogen is similar to carbon in terms of electronegativity and size. MO theory suggests that the more electronegative N would result in a lowering of the of the π-bonding energy, E(πCN), below the E(πCC) of ethylene. As I understand it, the MO energy trend should be:

Eπ(CO) > Eπ(CN) > Eπ(CC) > Eπ*(CO) > Eπ*(CN) > Eπ*(CC)

Meaning the trend in bond strength should be:

formaldehyde (H2C=O) > azomethine (H2C=NH) > ethylene (H2C=CH2)

However, bond dissociation energy implies that the C=N bond is the weakest compared to C=C and C=O. https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociationenergy.pdf

formaldehyde (H2C=O) = 732 kJ/mol
ethylene (H2C=CH2) = 682 kJ/mol
azomethine (H2C=NH) = 644 kJ/mol

https://chemistry.stackexchange.com/questions/164270/why-does-a-carbon-double-bond-break-readily-but-this-is-not-so-for-other-eleme

To me the argument you make appears to be only taking the pi bond into consideration, which cannot be explained by dissociation energies (or enthalpies). For your argument you might want to focus on something like hydration energies to come to an understanding about the pi bond. For examples ethene to ethane or formaldehyde to methanol. When you look at bonding enthalpies you need to consider the stability of the fragments, too. (FWIW you can use MathJax in combination with mhchem to make you post look beautiful.)

Thanks for the comments, in reading more on the subject I can see why using bond energies as a measure of bond strength can be somewhat misleading. "Carbon–Nitrogen Double Bonds (1970) by C. Sandorfy pg 6-7" has a great discussion of this exact question. Essentially, bond energies are highly dependent on molecular composition and structure and "transferability to bond strength can only be approximate". In fact the bond energies referenced match my expected values of C=O > C=N > C=C.
Edit: Using an alternative bond strength observable like bond length or IR frequency stretch is preferable.


text:
Why might a prepared 1% solution of glucose take 2 hours to give maximum, stable reading on a glucometer?
I put 1.000g of glucose powder into a 1 L volumetric flask and filled the flask to the 1L line with room temperature distilled water.
I then inverted several times for 10 minutes to mix until there was no visible glucose crystals left and tested the solution using an oxidase glucose strip.  Even though the glucose was all clearly dissolved, the reading I got after 10 minutes were only around 40 mg/dl.  I waited an hour and the readings were now about 65 mg/dl.  After 2 hours the readings were averaging about 125 mg/dl, a value that remained stable after 4 hours, and after 12 hours.
First off, the actual concentration was 100 mg/dl, but the test strip and glucometer estimate blood glucose, and real blood has other components that will effect that reading, so the reading of about 125 was not unexpected, but WHY might it take 2 full hours to achieve a peak and stable glucose reading from the solution?
The only ideas I had were that 1) the "dextrose" powder (lab grade) that I used MIGHT have been in polysaccharide form, maybe some monomers combining to make di- or tri-saccharides in the container which were able to hydrolyze in distilled water at room temp.  I don't think this can happen without enzymes though.  Perhaps some yeast enzymes in the air could have gotten in and done the job.  2) The glucose molecules had some degree of oxidation which made them un-affective to the test strip, but that the oxidation reversed in solution over time.  This also doesn't make sense since I don't expect oxygen to spontaneously bond to a glucose molecule nor reverse without a catalyst.  3) maybe the glucose molecules bond loosely to something like carbon dioxide or nitrogen that comes off gradually in the solution.
I have repeated several times with the same results and time frame.  The stable/peak glucose reading did seem to occur faster with warm water but reached the same peak value.  I can't imagine that the powdered glucose is not completely dissolved short of 2 hours.

Crystalline glucose is the alpha anomer, but the enzyme in the glucometer is specific for the beta form. They interconvert slowly in solution. See https://www.jofem.org/index.php/jofem/article/view/555/284284366

"Crystalline glucose is the alpha anomer". Hmmnnn, yes, I concur.


text:
Why would a white film form on cutlery but not glassware in a dishwasher?
An odd effect I notice is that I when wash dishes in a commercial dishwasher there is a white film left behind on steel cutlery but not on the glassware, which appears to have no residue. Why would this occur?
I know that the local tap water has high salt content, but if the residue is indeed the salt, why would it preferentially coat the cutlery but not the glasses and dishes?
The cleansers being used are standard commercial liquid detergents and a rinse aid.

At a guess, the white film is not corrosion of the metal cutlery, but salt adhering to it. Experimentally increase the amount of rinse aid used, to see if that helps. As to why more rinse aid might be needed for metal than glass, glass is more easily wetted by water than metals such as stainless steel. If it works, please let me know and I'll make my educated guess an "answer" :-)

I suspect this is coming from oils and fats in the steel cutlery (as a result of hard water forming a film precipitated soaps). Certainly, glasses will not have oil or oily products. As a control test, put washed and clean steel dishes and see if a white film is still there. These second reason could be the fact that water jets in the dishwasher are not reaching the steel cutlery properly with full force. You could try changing the way you put dishes, there could be physical blockages.

@AChem Are you suggesting there is oil inside the steel itself?

@ShakaBoom, No, oils/fats from food. Why don't you post a picture? A picture is worth a 1000 words :-)


text:
Is there a concise way to calculate pH of a solution after being diluted with water?
I do not have a chemistry degree or even much knowledge in the subject at all. I'm actually an auto detailer and this is why I have an interest in pH levels. All of our chemicals obviously have pH levels associated with them. I am trying to figure out a way to calculate the new pH of a solution when diluted with water. I have tried to do some research online about it but it brings in things like molarity, of which I know nothing. I'll give an example of what I need:
This chemical has a pH of "12-13" according to the SDS sheet. So let's say 13 for simplicity.
We use this chemical in two different dilutions: 10 parts water to 1 part chemical, and 4 parts water to 1 part chemical.
How would I calculate the pH level of the new solution?

Simplest way is to just measure it. Unless you know what compounds are in there, you could only make a guess (probably wrong one). If everything about compounds was simple there wouldn't be much need for chemistry as a separate science discipline.

It would help alot if could list the chemicals in your pH 13 solution.  But, just for example, diluting a pH 13 solution of caustic soda (lye) 1 to 10 will still be a very strongly basic pH 12.  This is because pH numbers are logarithmic.

The calculation will depend on whether the chemical of interest predominantly behaves as an acid or a base, whether it's strong or weak, and whether it's monoprotic or polyprotic.
Concentrated pure aqueous solutions of a chemical with a pH below 7 are typically acids, while concentrated pure aqueous solutions of a chemical with a pH above 7 are typically bases.
Assuming the chemical "B" is a strong monoprotic base and water autoionization is negligible, the moles of hydroxide ions $\ce{OH-}$ of a solution with volume $V_B$, before dilution are:
$$n_{\ce{OH-}}=10^{\pu{-pOH_o}}\;V_B$$
After diluting by adding a certain volume $V_w$ of pure water:
$$C_{\ce{OH-}}=10^{\pu{-pOH_o}}\frac{V_B}{V_B+V_w}$$
Taking the negative base 10 logarithm on both sides:
$$\pu{-log}\;C_{\ce{OH-}}=\pu{-log}\;(10^{\pu{-pOH_o}})-\pu{log}\;\left(\frac{V_B}{V_B+V_w}\right)$$
Which is equivalent to:
$$\pu{pOH}=\pu{pOH_o}-\pu{log}\;\left(\frac{V_B}{V_B+V_w}\right)$$
Or in terms of pH:
$$14-\pu{pH}=14-\pu{pH_o}-\pu{log}\;\left(\frac{V_B}{V_B+V_w}\right)$$
Solving for pH:
$$\pmb{\pu{pH}=\pu{pH_o}+\pu{log}\;\left(\frac{V_B}{V_B+V_w}\right)\quad\text{(for SMB)}}$$
For a strong monoprotic acid, the derivation is analogous, except no conversion between $\pu{pH}$ and $\pu{pOH}$ is necessary, so the resulting expression is:
$$\pmb{\pu{pH}=\pu{pH_o}+\pu{log}\;\left(\frac{V_A+V_w}{V_A}\right)\quad\text{(for SMA)}}$$
Where in both cases, pH represents the final value (after dilution), while $\pu{pH_o}$ represents the initial value (before dilution).
For either case, adding a very high volume of water will make the pH of the solution asymptotically approach the value of 7, but it will never reach it.
Some examples of monoprotic strong bases are sodium hydroxide ($\ce{NaOH}$) and potassium hydroxide ($\ce{KOH}$).
Some examples of monoprotic strong acids are hydrochloric acid ($\ce{HCl}$) and hydrobromic acid ($\ce{HBr}$).
For a weak monoprotic acid, the calculation is as follows:
The initial concentration of the acid solution can be set in terms of $x$:
$$C_{Ao}=\frac{x^2}{K_a}+x=x\left(\frac{x}{K_a}+1\right)$$
Which means the total moles of acid before dilution is:
$$n_{Ao}=C_{Ao}V_A=x\left(\frac{x}{K_a}+1\right)V_A$$
At equilibrium, before dilution, the equilibrium constant expression in terms of moles is:
$$K_{n_o}=K_a\;V_A=\frac{y^2}{n_{Ao}-y}$$
At equilibrium, after dilution, the equilibrium constant expression in terms of moles is:
$$K_{n}=K_a\;(V_A+V_w)=\frac{(y+z)^2}{n_{Ao}-(y+z)}$$
Substituting $n_{Ao}$:
$$K_a\;(V_A+V_w)=\frac{(y+z)^2}{x\left(\frac{x}{K_a}+1\right)V_A-(y+z)}$$
Where:
$$x=10^{-\pu{pH_o}}$$
$$y+z=10^{-\pu{pH}}\;(V_A+V_w)$$
Substituting above and simplifying:
$$K_a=\frac{10^{-\pu{pH}}}{10^{\pu{pH-pH_o}}\left(\frac{10^{-\pu{pH_o}}}{K_a}+1\right)\frac{V_A}{V_A+V_w}-1}$$
Solving for pH:
$$\pu{pH}=\pu{log}\left(\frac{K_a+\sqrt{K_a^2+4\times10^{-\pu{pH_o}}\;(10^{-\pu{pH_o}}+K_a)\;\frac{V_A}{V_A+V_w}}}{2\times10^{-\pu{pH_o}}\;(10^{-\pu{pH_o}}+K_a)\;\frac{V_A}{V_A+V_w}}\right)$$
Conversely, for a weak monoprotic base:
$$K_b=\frac{10^{\pu{pH}-14}}{10^{\pu{pH_o-pH}}\left(\frac{10^{\pu{pH_o}-14}}{K_b}+1\right)\frac{V_B}{V_B+V_w}-1}$$
Solving for pH:
$$\pu{pH}=14-\pu{log}\left(\frac{K_b+\sqrt{K_b^2+4\times10^{\pu{pH_o}-14}\;(10^{\pu{pH_o}-14}+K_b)\;\frac{V_B}{V_B+V_w}}}{2\times10^{\pu{pH_o}-14}\;(10^{\pu{pH_o}-14}+K_b)\;\frac{V_B}{V_B+V_w}}\right)$$
For simplicity, we can define:
$$\beta_A=2\times10^{-\pu{pH_o}}\;(10^{-\pu{pH_o}}+K_a)\;\frac{V_A}{V_A+V_w}$$
$$\beta_B=2\times10^{\pu{pH_o}-14}\;(10^{\pu{pH_o}-14}+K_b)\;\frac{V_B}{V_B+V_w}$$
So the final expressions become:
$$\pmb{\pu{pH}=\pu{log}\left(\frac{K_a+\sqrt{K_a^2+2\beta_A}}{\beta_A}\right)\quad\text{(for WMA)}}$$
$$\pmb{\pu{pH}=14-\pu{log}\left(\frac{K_b+\sqrt{K_b^2+2\beta_B}}{\beta_B}\right)\quad\text{(for WMB)}}$$
An example of a weak monoprotic acid is acetic acid ($\ce{CH3COOH}$).
An example of a weak monoprotic base is ammonia ($\ce{NH3}$).
The calculation for polyprotic acids and bases is significantly more complex.
As a final note, all volumes need to be in liters (L), and the dissociation constants $K_a$ and $K_b$ are dimensionless.

How would I calculate the pH level of the new solution?

There is no general formula to calculate this. It depends on what components are present in the solution.

I am trying to figure out a way to calculate the new pH of a solution when diluted with water. [...] We use this chemical in two different dilutions: 10 parts water to 1 part chemical, and 4 parts water to 1 part chemical.

If the solution is buffered, the pH changes little when diluting by these amounts. When the solution is not buffered, the pH moves a bit in the direction of neutral (i.e. pH 13 is now pH 12, and pH 2 is now pH 3).
The easiest estimate is to assume the pH did not change much, and have the same precautions as for the undiluted stuff ("chemical").

From this reference we find:

pH + pOH = 14

Log of [OH] = -1.  [OH] is 10$^-1$ or 0.10 M.
Assuming the pH of the diluent is around 7, very little of the hydroxide will be neutralized.  The solution is only diluted.
A pH 13 solution of NaOH diluted 1 to 10 will be 0.010 M.  Log of [OH] = -2. Working the above formula, pH still be a very strongly basic pH 12.
Similarly, a pH 13 solution that is diluted 1 to 4 will be 0.025 M. Log of [OH] = -1.6, yielding a pH of 12.4.
From a safety standpoint, both diluted solutions remain highly basic.


text:
Average molecular distance between hydrogen gas
The whole question goes like this:
Calculate the de Broglie wavelength of hydrogen molecule at 300K (Take the translational energy to be 1.5kT) Compare this to the average intermolecular spacing for hydrogen gas at 300K and 1 bar.
I know how to do the first part easily. I don't know how to approach the second part. Is it supposed to be a known fact or is there some way to calculate it?

From the ideal gas state equation and Avogadro constant, you get molecular density. Then, you can calculate statistical count of molecules as function of radius. Or, you may assume regular cubic placement.

The volume containing in average 1 molecule of H2  : $$V=\frac{RT}{pN_\mathrm{A}}$$

I am sorry but could you please elaborate? I didn't quite get your idea. We find volume containing one molecule of h2? But I don't know the total volume so?

Use the handle symbol @ + username to send a message.

If you assume the molecules are uniformly distributed in a grid and you know the volume of one cell in the grid you can compute the distance between adjacent grid centers.

@Sj2704 It is independent on the total volume.

From the ideal gas state equation, applied to hydrogen, the molar volume is:
$$V_\mathrm{m} = \frac{RT}{p}$$
Using the Avogadro constant $N_\mathrm{A}$ as number of molecules in $\pu{1 mol}$, we get the volume, statistically containing in average $\text{1 molecule}$ of $\ce{H2}$:
$$V=\frac{RT}{pN_\mathrm{A}}$$
The reciprocal value $\frac{pN_\mathrm{A}}{RT}$, used below, is the molecular density in $\pu{m-3}$.
The cube root then gives the length of the edge of the respective cube:
$$L=\sqrt[3] {\frac{RT}{pN_\mathrm{A}}}$$
that gives the mean distance of centers in the tightest case of 2 facing cubes.
For $R=\pu{300 K}$ and $p = \pu{e5 Pa}$, $\sqrt[3]{\frac{RT}{pN_\mathrm{A} }} \approx \pu{3.46 nm}$
For the most distant case of 2 cubes touching by the corner, it is
$$L=\sqrt{3} \cdot \sqrt[3] {\frac{RT}{pN_\mathrm{A}}} \approx 1.73 \cdot \sqrt[3] {\frac{RT}{pN_\mathrm{A}}}$$
So the true mean will lay between these two extreme values.
More exact value would be integration over randomly placed molecules.

Another approach is to use the molecular density to calculate radius where are 2 molecules.
$$2 = \frac{pN_\mathrm{A}}{RT} \cdot \frac 43 \pi r^3 $$
$$6RT = pN_\mathrm{A} \cdot 4 \pi r^3 $$
$$ r = \sqrt[3]{\frac{3RT}{2 \pi pN_\mathrm{A} }}= \sqrt[3]{\frac{3}{2 \pi}}  \sqrt[3]{\frac{RT}{pN_\mathrm{A} }} \approx 0.782 \cdot  \sqrt[3]{\frac{RT}{pN_\mathrm{A} }}$$
and then
$$ D \approx 1.564 \cdot  \sqrt[3]{\frac{RT}{pN_\mathrm{A} }}$$
This is the maximal value, the more proper one would be to compute the mean distance of two molecules randomly place in a sphere of a given diameter. But this step is already beyond my skills.
But, by stochastic Monte Carlo analysis, it looks like the mean distance of two randomly placed objects in a sphere of radius $r$ is $$L_\mathrm{mean} = \frac{4}{3}r = \frac{4}{3} \sqrt[3]{\frac{3}{2 \pi}}  \sqrt[3]{\frac{RT}{pN_\mathrm{A} }}=  \sqrt[3]{\frac{32}{9 \pi}}  \sqrt[3]{\frac{RT}{pN_\mathrm{A} }} \approx 1.042 \cdot \sqrt[3]{\frac{RT}{pN_\mathrm{A} }} = 1.042 \cdot \pu{3.46 nm} \approx  \pu{3.61 nm}$$


text:
Reaction of ester and and sodium ethoxide
I tried using C2H5ONa as the base, and then removing the alpha hydrogen, so that I can carry out SNGP, but can't proceed after this
please help :)

What is SNGP? Please provide more detail for what you have tried.

Look here to find your answer.

So the anion you created will attack the ester group of another molecule of diethyl succinate to initially give a beta-keto-ester. The reaction regenerates EtO- so this beta-keto-ester is deprotonated and reacts with an internal ester group in a classic Dieckmann Condensation to give initially 2,5-Dicarbethoxy-1,4-cyclohexanedione. The second step of hot aqueous acid hydrolyses the esters to acids which then decarboxylate giving cyclohexane-1,4-dione (product 1).
There is an Organic Syntheses procedure for this exact reaction from which I have borrowed the reaction scheme.


text:
how does enolate affect an elimination reaction?
From my class:

But why wouldn't the reaction proceed through this way, which makes more sense? Since the reaction above generates an enolate bearing a negative charge first, making the molecule less stable, whereas the one I proposed is just a straightforward E2 reaction.

Read this article to find some explanation.

Both can occur, but the top drawing allows for easier formation of a "free" electron pair.
In general breaking C-C, C-N, etc. bonds has a rather high activation energy barrier, so the direct E2 is going to be pretty slow. The top diagram basically splits the activation energy barrier into a deprotonation step and an elimination step, making the overall transformation easier.
The negative charge will also be somewhat balanced by the positive charge of the nitrogen.  At five bonds apart they aren't going to interact that well, but there's definitely a bit of stabilization there too.


text:
what's the difference between gas and vapour?
In the lecture, our teacher said that the significant difference between gas and vapour is we can't liquefy gas just by compressing it  but it's not the same in the case of vapour."
So, I'm confused about the possible reason behind it, Is it because vapour contains both the liquid and gaseous phase?

Generally, if you are confused by stuff, taught in school, the first thing you should do is reviewing available offline and online sources. Asking here should be rather the last than the first step how to get the answer.

Wikipedia: Vapor

In physics, a vapor(US) or vapour(UK, CA) is a substance in the gas phase at a temperature lower than its critical temperature, which means that the vapor can be condensed to a liquid by increasing the pressure on it without reducing the temperature.

Note that for some reasons, I have not found vapor definition in IUPAC Gold Book, the authoritative source of definitions of terms used in chemistry.
Gases(1) = vapors(1) + (permanent) gases(2)
Every vapor(1) is a gas(1), but a gas(2) is not a vapor(1).
Gases(2) that are not vapors(1) are historically called permanent gases (or just gases). That does not mean they cannot be liquefied.

In everyday life of chemists, a vapor(2) is often used in a narrower sense as a gaseous phase of substance with boiling point above room temperature. It is eventually mixed with other gases like air, if the total gas pressure is greater that saturated vapor pressure.
Water forms both vapor(1) and vapor(2), while gaseous chlorine is vapor(1), but not vapor(2). Chlorine is often consider as a gas, but it can be liquefied at room temperature by pressure about $\pu{8 atm}$. Similarly for propane, isobutane or butane, components of $\text{LPG}$.

This is an interpretation on general use of the terms. A vapor is a gas in possible equilibrium with its liquid [or solid] at a temperature below its critical T and either in contact with the liquid or at the equilibrium vapor pressure. If not at a sufficient pressure to condense and not in contact with the liquid the vapor behaves as a gas [nonideal] and the term becomes colloquial. For example, water in the atmosphere is "water vapor", CO2 in the atmosphere is a gas or simply "carbon".

Out of space. [water gas! has an entirely different, archaic, meaning] This is in American English; I have no idea what is going on across the pond [we buy "gas" they buy "petrol", hopefully after fossil fuels are gone that problem will vanish] or in other languages. Language can cause minor problems that can be alleviated by concentrating on the actual chemical and physical properties of gases and the equilibria among solids, liquids and "vapors" [or "gases"].