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Colour changing Lipstick
I have bought a colour changing lipstick with the colour Phloxine B.(Other ingredients, see below...)
Normally this colour changes to pink at pH 5.
So this means on skin it changes its colour because the human lips have a pH about pH 5-6.
I have made a test: I cut pieces of the lipstick and inserted them into a solution of HCl (no colour change), dest. water (pH7) (no colour change) and a solution of NaOH (colour change!)
My question is: Why doesn't it change its colour in water?
Ideas so far:

due to the solubility of the liphophile lipstick and the surface
tension of water, there is no real contact between water and the
lipstick possible and therefore also no reaction

the citric acid that is an ingredient of the lipstick solves in the water so the water gets acid and there is no rise of the pH
value

the reaction between water and the lipstick is not possible because water doesn't have enough OH- -Ions (but why does skin have
those ions?!)

On top of that i made a few more tests:
The lipstick changes colours when spreaded on: tables, paper, skin, plastic-lids
it doesn't change colours on: metall, glass, porcelain, plastic with a very plain surface.
Can you please help me out?

Ingredients: polyisobutene, caprylic/capric triglyceride, cera microcristallina (microcrystalline wax), euphorbia cerifera (candelilla) wax, prunus amygdalus dulcis (sweet almond) oil, cannabis sativa (hemp) seed oil, menthol, tocopheryl acetate, glycine soja (soybean) oil, tocopherol, caprylyl glycol, ethylhexyl palmitate, glycerin, caprylhydroxamic acid, beta-sitosterol, squalene, ricinus communis (castor) seed oil, ethoxydiglycol, citric acid, CI 19140 (yellow 5 lake), CI 45410 (red 27 lake), CI 61565 (green 6).

When you mix two samples with different pH, the resulting pH depends on the pH values of the two samples and their buffer capacity (a measure of "how much it takes" to changes their pH). In this case, pure water has a very low buffer capacity, so your idea #2 explains what you observed.
The skin has a higher buffer capacity, so it "wins" over the lipstick.
You mention the ingredient phloxine B as the one responsible for the color change. This substance has multiple names (red 27, red 28, acid red 92, phloxine B, phloxine O), perhaps based on what pH it was isolated in (as free acid or as salt). It undergoes fairly complicated changes with change in pH involving a lactone ring closure.

Source: https://www.mdpi.com/2673-8015/1/1/1
More common pH indicators like phenolphthalein also show this reversible formation of a lactone ring.

The first print of the colour index is from the 1920s, when colours and dyes were much more of interest for BASF, Hoechst, Lucius, Ciba, etc. than today (some of them no longer active as such).  At a time when chemical analysis relied on performing reactions to obtain derivatives, and combustion analysis; ahead of the advent of IR, MS, NMR; crystallography to the labs we (too often) take for granted.

The intentionally added colors are the ingredients behind the CI numbers where CI stands for the colour index, an inventory by the Society of Dyers and Colourists.  The numbers of five digits each are group dyes and pigments by the functional group deemed to be of principal influence for the color, though this does not correlate with the color perceived, nor exclude the presence of additional structure features which may influence the perceived color, or intensity.

CI 19140 (yellow 5 lake), CI 45410 (red 27 lake), CI 61565 (green 6).

While the printed catalogue is telephone book like heavy, Wikipedia has compiled a selection amended by the CAS registry number (an other index of chemicals, which is not constrained to dyes), and links to separate articles about the individual dyes, colorants, pigments.  Hence

CI 19140 / yellow 5 lake: tartrazine, CAS-RN 1934-21-0

CI 45410 / red 27 lake: phloxine B, CAS-RN 18472-87-2

CI 61565 / green 6:  quinizarine green SS, CAS-RN 128-80-3

Out of this set of three, phloxine is the red dye which eventually was discolored by the action of aqueous lye.  As mentioned in @Karsten's answer, $\ce{OH-}$ of the base opened the 5-membered lactone cycle.  This alters the electronic interaction of the remaining rings and the interaction of the molecule to visible light, i.e. what frequency of the spectrum is absorbed (and by consequence, the color perceived), and how probable such an absorption is).  The underlying principle relates to the one of phenolphthalein.
Image credit: structure formulae are extracted from property boxes of the corresponding Wikipedia pages.

@JeanClaudeDelaveine  Because I don't know your chemical education (there is at least one journal which includes this in its name), I refrained as much as possible from words too special to chemistry.

You can use special chemistry words, don't worry. I studied chemistry (haha who believes this now?!). Thank you, I will try to check this journal!


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Calculating simulatenous equilibrium concentrations given dissociation constants
Reactants A, B, and C form the complexes AB and BC according to:
$$ AB \rightleftharpoons A + B, \quad BC\rightleftharpoons B + C $$
Each reaction has a dissociation constant, $K_{dA}$ and $K_{dC}$, respectively, which are defined as:
$$ K_{dA} = \frac{[A][B]}{[AB]}, \quad K_{dC} = \frac{[C][B]}{[BC]} $$
Given starting concentrations $[A]_0, [B]_0, [C]_0$, how does one calculate the equilibrium concentrations of the five species in mixture?
I have tried a couple of different approaches to a solution. First off, I identified the three additional equations
$$ [A] = [A]_0- [AB] $$
$$ [B] = [B]_0 - [AB] - [BC] $$
$$ [C] = [C]_0- [BC] $$
Putting these additional expressions for [A], [B], and [C] into the equations for the dissociation constants ultimately yields two second-degree polynomials in [AB] and [BC] with a cross-term [AB][BC] which I have not figured out how to solve analytically.
$$ 0 = [AB]^2 - (K_{dA} + [A]_0 +[B]_0)[AB] + [A]_0[B]_0 + [AB][BC] - [A]_0[BC]$$
$$ 0 = [BC]^2 - (K_{dB} + [B]_0 +[C]_0)[BC] + [B]_0[C]_0 + [AB][BC] - [C]_0[AB]$$
The other attempt I made was to solve the system of five equations for $K_{dA}$, $K_{dC}$, [A], [B], and [C] numerically using root-finding algorithms. This works okay, but oftentimes the algorithm finds roots with negative concentrations which are obviously not desirable.
I have seen many solutions online for the classic case with one reaction where a so-called ICE-table is used, but I have found no extension of this for the case I have presented. Is there a way to solve this problem analytically?

It is often handy to make variable name substitutions by 1-letter symbols that are more convenient for algebraic manipulations. // Exact problem descriptions usually lead to a set of nonlinear equations, leading to a cubic or higher polynomial equation. It is painful or not possible to solve analytically. It can be solved numerically, or the problem can be often simplified, applying strong inequalities. E.g. famous pH=1/2.(pKa - log c) for weak acid solutions uses assumptions [H+] >> [OH-] and c >> [H+]

See also https://chemistry.stackexchange.com/a/171096/35806

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Is there experimental evidence to show boiling point increases with external electric field, and if so, by how much?
I was wondering whether anyone has demonstrated that a liquid will have a higher boiling point if it is placed in a very large electric field. I believe to be the case, since induced dipoles would be formed and dipoles would more heavily align, but I was wondering if there was any actual experimental evidence and how exactly boiling point grows with temperature.
For an order-of-magnitude check, I set up a model to predict the increase in boiling point.
We have the following formulae (considering induced dipoles only):
\begin{align}
U&=-k\frac{p^2}{r^3}N_A\\
p&=\alpha E\\
r&\approx\left(N_A\rho_m\right)^{-\frac{1}{3}}
\end{align}
Which gives
$$U=-k\alpha^2N_A^2E^2\rho_m$$
Given a change in molar density from $\rho_l$ to the ideal gas $\frac{P}{RT}$, we have
$$\Delta U=k\alpha^2N_A^2E^2\left(\rho_l-\frac{P}{RT}\right)=C_p\Delta T_{\text{boiling}}$$
In the case of benzene (initial boiling point $353.2\,\text{K}$, molar density $11385\,\text{mol m}^{-3}$, specific heat capacity $103\,\text{J mol}^{-1}\text{K}^{-1}$, dielectric strength $163\,\text{MV m}^{-1}$, polarizability $69.6\text{Å}^3$), my model would give a maximum $\Delta U$ of $59.0\,\text{J mol}^{-1}$ before dielectrib breakdown, or $\Delta T_{\text{boiling}}$ of $0.522\,\text{K}$, something that should be measurable. My model probably doesn't take into account a lot of things, but I'm sure that the actual maximum change in boiling point should only be half an order of magnitude away from my estimate hopefully.

related https://chemistry.stackexchange.com/questions/60289/in-practice-do-polar-molecules-actually-align-in-an-electric-field

https://chemistry.stackexchange.com/questions/108577/molecule-with-a-quadrupole-moment-in-an-electric-field

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1H-NMR: How does the oxygen in ethers act as a spin coupling barrier while the oxygen in hydroxy groups doesn't?
It is possible to observe coupling between a hydroxyl proton and other protons. Why does the oxygen atom in ethers prevent any further coupling and act as a sort of barrier between spin systems? Or is coupling technically possible, but the resulting $^4\mathrm{J}$ coupling constant is just too small to be observed?
[Edit]:
I initially asked this question because we learned in our spectroscopy lecture that proton couplings across R-O-R, R-N-R and R-S-R (with R ≠ H) aren't really possible.
It's nice to see that this is just a rule of thumb and that some coupling can occur.
Could you tell me if my reasoning for the coupling in the examples below is correct or not: Fermi contact interactions are usually the most important mechanism for spin-spin coupling. The carbon atoms in the furane derivate are $\ce{sp^2}$ hybridized, so there should be additional σ-π-spin-polarization interactions. The carbonyl-carbon of the formate ester also has a lower s-character. The large $^5\mathrm{J}$ coupling in the trioxaadamantane derivate is the most surprising. In contrary to what I initially thought, the coupling mechanism seems to be different from the stereospecific sigma-bond contributions that are responsible for "W-coupling" (so maybe some through-space mechanism?).
Here is the original adamantane paper for anyone whos interested: https://doi.org/10.1021/ja00958a011
Quote from the paper: "The coupling systems in I and VII differ from these in that rotation of 180 and 120° about the central bond has taken place. If one mechanism is responsible for all the five-bond couplings cited, it must be independent of the geometry of the central bond."
So let me ask a slightly better question: Is there any difference between $^4\mathrm{J}_{\ce{HCCCH}}$ and $^4\mathrm{J}_{\ce{HCOCH}}$ coupling in acyclic, unstrained,  $\ce{sp^3}$  hybridized molecules (i.e. do things like lone pairs play a role in spin-spin coupling and if so, why)?

Please edit your new post https://chemistry.stackexchange.com/questions/171293/1h-nmr-what-parameters-influence-spin-spin-coupling-between-protons-and-why to be more focused. It's ok that you posted a new question, given that you have evidently done research on this and your updated question did not get answered in this post. Also, before you state something like "why do oxygen atoms in unstrained, sp3 hybridized ethers usually prevent further coupling?" examine your assumptions. Is it really the oxygen? Maybe. But isn't the point of the question to find out if it is the oxygen?

Present your case more clearly. Look for couplings of the type $^4\mathrm{J}{\ce{HCCCH}}$ and compare these to $^4\mathrm{J}{\ce{HCOCH}}$.

Hans Reich's website remains a treasure trove of NMR information that you can consult for theory and values of J couplings. All of the following comes from that site.
First, on the origin of through-bond couplings:

The scalar coupling J is a through-bond interaction, in which the spin of one nucleus perturbs (polarizes) the spins of the intervening electrons, and the energy levels of neighboring magnetic nuclei are in turn perturbed by the polarized electrons.  [....] Because the effect is usually transmitted through the bonding electrons, the magnitude of J falls off rapidly as the number of intervening bonds increases.

On 4-bond proton J-couplings

Proton-proton couplings over more than three bonds are usually too small to detect easily (< 1 Hz). However, there are a number of important environments where such couplings are present, and can provide useful structural information. Coupling across π-systems are the most frequently encountered 4J couplings

As the following examples show, the oxygen nucleus in an ether or ester does not inherently interfere with a long-range coupling, and does not account for the sparsity of reported $^4\mathrm{J}_{\mathrm{HH}}$.

Thank you for anwering my post and showing me this interesting website. I will have to add my full comment to the post because of the character limit.


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"Cleaning" of HNO3 solution using distillation
In many industries, pickling baths are common. These are 10 cubic meter baths containing acid baths that metal components are dipped into prior to surface treatment. When the metals are dipped, some of their components are left in the acid bath and after some time he bath needs to be changed with new acid because the metals in the solution make the bath ineffective.
The old and metal contaminated acid is sent to destruction. But is it possible to remove the metals from the water and acid, to regenerate it?
Say we have a bath with 20% HNO3 in water and lots of metals. 20% HNO3 in water has a bp of 101°C which is slightly above water. If this was put through an evaporator, would both the HNO3 and the H2O evaporate and cooled down to a distillate that contains the water and acid? The metals won't evaporate of course and is left in the small amount of liquid that is left in the concentrate.
Is this possible? Also is it safe? I mean heating HNO3 creates nitrous gases, but if they're immediately cooled down and condensed, is it a problem?
Thanks in advance!

Interesting spelling of "distillation" with the letter 'e' which presumably is only used in Swedish language.

@andselisk Czech language uses "destilace" and I have always to self-check myself  not to write "destillation" too.

What is a piece of cake for $\pu{1 L}$ might become a nightmare for $\ce{10 000 L}$. So think twice…
$\ce{HNO3}$ forms an azeotrope with maximum of boiling point $\pu{120.5 ^\circ C}$ at $w(\ce{HNO3}) = 68\,\%$.

So the best would be to arrange the column distillation, with  first going just water at $\pu{100 ^\circ C}$ with just traces of $\ce{HNO3}$ and $\ce{NO_x}$.
Then the column head temperature skips to $\pu{120.5 ^\circ C}$ at $w(\ce{HNO3}) = 68\,\%$.
A simple fractional distillation may be more effective timely and energetically, but on the other hand with big loses of $\ce{HNO3}$.

Another option is the simple or fractional distillation:

Initial fractions having roughly 1/3 $\ce{HNO3}$ concentration, compared to original $20\ \% $ solution.
From the phase diagram $\ce{HNO3-H2O}$, it visually looks like vapor above boiling liquid have roughly these compositions:

$w(\ce{HNO3},\text{l})[\%]$
$w(\ce{HNO3},\text{g})[\%]$

20
7

40
20

50
30

68
68

All depends on what is the target concentration. If we start with $w(\ce{HNO3})=20 \%$, using a single fraction, we would end with less than $20 \%$, as the composition of the solution being distilled would converge to $w = 68\ \%$.
Unless we use more fractions and drop too diluted first fractions.
The third theoretical option would be to distill near everything, so we would reach near the same concentration, but with the ugly rest to be cleaned up.

Is it not just safer to leave it at 100 and let the HNO3 slowly trace its way over until there is nothing left (water nor HNO3)? The problem I see in a large scale would be not making a bomb...

The trick seems to be to take the water off first.  See reference in answer provided.  Keep in mind everything from the 1st distillation can be saved.

@Poutnik and what's left in the pot might go by a name worse than "bottoms".  But more water could be added in to make sure the pot doesn't gel.  Even 15% first pass might be workable.

The answer extended, including deleted comment.

I like the third option the best. The resulting concentration of HNO3 in the distilled fraction is not important, its easy to just add a little concentrated hno3 to up the conc or dilute it with water. Whats left sent to destruction, but it is much cheaper than to keep buying new acid and having to send the entire batch to destruction once it is used. Still worried about the NOx though... You think it is a problem at 20% or less to begin with? What could happen?

Try a lab scale experiment and you will see. I assume for concentrations under the azeotrope NOx should not be a problem, unless develop in original solution (like dissolving metals in HNO3 )

Yes, this would seem feasible.  Nitric acid forms an azeotrope with water and can be fractionally distilled to a concentration you desire after distillation from the "mother liquor".
Distillations in general are made safer when run under a hood.  Certainly wear eye protection and, as nitric acid is also an oxidizer, appropriate distillation apparatus should be used$^1$.  Inert gas such as N2 should used if there is any chance organics such as oils or grease may be in the pickling bath.
$^1$ Pyrex and Teflon sleeves

What it costs you to heat the mixture to boiling point may be more than a new charge of nitric acid.

True, but nitric acid is not inexpensive.  Could be a good way for the new chemists to earn their keep in the plant.  Run pilot scale first, and be aware of what is in the pickling bath beforehand.

The destillation in question would in a large scale be run by waste heat, and using a packing material that expands the surface area of the pickling bath, this lowers the temperature needed to evaporate it. It is only in theory as of now, needs a lot of work but might work.

@Evaporation123 show them it works lab scale first.  I don't see any major scaling issues here.  Vacuum distillation might help.  Control azo by adjusting vacuum.  Easier than heat.

Easier than heat but electricity to keep vacuum going on a large scale is a lot, with waste heat is available in endless amounts in my case, so heat consumption is not a problem.

@Evaporation123 but make sure there is any HNO3 left in the bath.  If anything lower than silver redox V is being pickled, such as copper or iron, nitric acid will be consumed.


text:
Why are some elements diatomic under normal conditions while others are not?
I see that the the standard form of many elements are diatomic molecules under normal conditions (eg oxygen, hydrogen and iodine are usually seen as diatomic molecules) and are all among the most electronegative, and I understand it has to do with energy and stability, but I just have no clue why other elements wouldn't be the same.
What explains why some appear as diatomic molecules and others do not?

https://chemistry.stackexchange.com/questions/32705/which-elements-can-be-diatomic

It's not really correct to say "diatomic elements" - it's implying certain universality, which just isn't there. Like every element can make diatomic molecules. And the ones being exceptions still can make diatomic "Van der Waals molecules".

On the other hand the elements you think diatomic can be also monoatomic, or even metallic.

@Mithoron. Can iron, aluminium, copper and lead make diatomic molecules ?

@Maurice Most likely. Exception I was thinking about is mercury https://chemistry.stackexchange.com/questions/108186/does-a-neutral-dimercury-molecule-exist and even this is arguable. Clear cut examples are likely only noble gases,

About lead, for example https://pubs.acs.org/doi/10.1021/jp810107c

Does this answer your question? Which elements can be diatomic?

Perhaps bigger issue is that this pretty much asks why all elements are as they are. Not only oxygen vs sulfur and nitrogen vs phosphorus, but all of them.


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Would the mechanism for the reaction be polar or radical?

I've disregarded the fact that the reaction could proceed via a concerted mechanism but am unsure how else it would go.
My thoughts are that it either goes via a polar mechanism or a radical mechanism. Where in the polar mechanism, the reaction would proceed via electron transfer between charged intermediates. In the radical mechanism, the formation of nitroso and phosphine free radicals would drive the reaction.
I'm leaning towards polar due to the reactivity of the nitroso and the strong polar nature of the system but unsure about the rate determining step and whether it would be addition/elimination (N=P formed first then N=O broken) or elimination/addition (N=O broken first, N=P formed after).

I think it's like addition-elimination, but I'm hardly sure.


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How is decarburisation achieved during Electric Arc Furnace steelmaking?
In the basic oxygen process, steel is made from pig iron (having a high carbon content) by oxidising the carbon with a high-pressure oxygen lance.
In an EAF, steel is made mostly from scrap, but it can also be made from raw materials such as Pig Iron, Direct-Reduced Iron, and Cementite (from the Iron Carbide Process).
Direct-reduced iron and Cementite still have relatively high carbon content compared to the desired amount in most steels.
In case steel is made primarily from these raw materials in the EAF, how is it decarburized? If it uses oxygen for this, how does this process essentially differ from Basic Oxygen steelmaking?


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Finding the molecular formula given combustion data

$\pu{150 cm^3}$ of oxygen were added to $\pu{20 cm^3}$ of a gaseous hydrocarbon D.
After explosion and cooling, the gaseous mixture occupied $\pu{130 cm^3}$
and, after absorption by potassium hydroxide, $\pu{90 cm^3}$ of oxygen remained.

$$\ce{C_xH_y + (x + \frac{y}{4}) O2 -> x CO2 + y/2 H2O}$$
$\ce{C_xH_y}$: $\pu{20 cm^3}$
Why is $(x+y/4) \ce{O2} = \pu{150 cm^3} - \pu{90 cm^3}$
and $\ce{x CO2} = \pu{130 cm^3} - \pu{90 cm^3}$ ?
I genuinely don't know why.why do we subtract 90 from both sides and not just the product? Why are we subtracting 90 from the $\ce {CO2}$

Convenient reference for text/formula formatting:
Notation basics  /  Formatting of math/chem expressions  /
upright vs italic  // For more:  Math SE MathJax tutorial.  // Not to be applied in CH SE titles.

Good practice is starting with symbolic algebraic expressions and keeping it this way until all is ready to plug in literal numbers. It helps in focusing on principles, mistakes are easier to spot, orientation is improved, Q/A is reusable and has bigger  permanent value.

Its not that hard to read off.

$\pu{20 cm^3}$ is the volume of the hydrocarbon.
$\pu{130 cm^3} - \pu{90 cm^3} = \pu{40  cm^3}$
is the volume of  $\ce{CO2}$ absorbed in the hydroxide, therefore $x=\frac{\pu{40 cm^3}}{\pu{20 cm^3}} = 2$.
$\pu{150 cm^3} - \pu{90 cm^3} = \pu{60  cm^3}$ is volume of spent $\ce{O2}$.
Therefore $x + \frac{y}{4} = \frac{\pu{60 cm^3}}{\pu{20 cm^3}} = 3 \implies y = 4 \implies \ce{C_xH_y} = \ce{C2H4}$.

I still don't understand why we are taking 90cm^3 from the co2

130 cm3 of final gas = 40 cm3 of CO2 + 90 cm3 of O2.

If starting from $\pu{150 mL}$ $\ce{O2}$, only a part of it is used to finally obtain a residue of $\pu{90 mL}$ $\ce{O2}$, it means that $150 - 90 = \pu{60 mL}$ $\ce{O2}$ has been consumed. Now added $\ce{KOH}$ after the combustion reacts with $\ce{CO2}$ and reduces the final gas volume from $\pu{130 mL}$ to $\pu{90 mL}$. This indicates that $130 - 90 = \pu{40 mL}$ is the volume of $\ce{CO2}$ produced. Let's rewrite these results:
$\ce{
  \pu{20 mL} C_xH_y + \pu{60 mL} O2 -> \pu{40 mL} CO2 + $a$ H2O
}$
The number of moles is proportional to the volumes. So that this equation can be also rewritten so, after division by $10$:
$\ce{2 C_xH_y + 6 O2 -> 4 CO2 + $a'$ H2O}$
To get $\ce{4 C}$ atoms at right, we need: $x = 2$ in the initial hydrocarbon. The equation can be rewritten:
$\ce{2 C2H_y + 6 O2 -> 4 CO2 + $a'$  H2O}$.
And now $4$ of the $\ce{6 O2}$ are required to make $\ce{4 CO2}$. The remaining $\ce{2 O2}$ (or $\ce{4 O}$) are required to burn the $2y$ $\ce{}$ atoms and form $\ce{4 H2O}$. So $a' = 4$ and $y = 4$. And the equation becomes:
$\ce{2 C2H4 + 6 O2 -> 4 CO2 + 4 H2O}$.
This can even be more simplified :
$\ce{C2H4 + 3 O2 -> 2 CO2 + 2 H2O}$.

@Maurice: I speculate, so far, the possibility to engage the \pu{} command (as in physical units) by mhchem (synopsis) escaped your eye.  Compared to your earlier approach, e.g. $60$ mL, the use of $\pu{60 mL}$ has the advantage that the space will not break if the construct approaches the end of the line. If you inspect the edit, you will see that it equally may be used within the $\ce{}$ of an chemical equation and can equally simplify the input / ease reading the post in case of a web browser with JavaScript disabled.

Why are we taking away 90 from the co2

Why does it mean that 90cm^3 co2 has been used?


text:
Sodium bicarbonate to sodium carbonate decomposition
Recently I ran the reaction:
$$\ce{2NaHCO3 ->[400°C] Na2CO3 + H2O + CO2}$$
I wanted an explanation why the sodium bicarbonate ($\ce{NaHCO3}$) decomposed into sodium carbonate ($\ce{Na2CO3}$), water ($\ce{H2O}$), and carbon dioxide ($\ce{CO2}$); and didn’t just turn to sodium bicarbonate in its gas form, since its boiling point is 333.6 °C.

You would not be considered less educated if you called H2O just water. Nobody calls it dihydrogen monoxide. BTW, names of compounds are not capitalized. // NaHCO3 starts to decompose at 50 °C before even melting. // Remember NaHCO3 is also called "baking soda" and it would not be used for baking if food would be carbonized first.

Sodium bicarbonate does not have a gas form at all.

So there are some compounds with only one state of matter?

I guess most of compounds decompose before melting. Simple inorganic or organic compounds able to be melted without damage are rather exceptions. // There is no bicarbonate in other than solid phase and for most of cations not even as solid // Have you even witnessed burnt food? Food is notorious for refusing to be melted or evaporated and then frozen and condensed. It prefers being decomposed, finally to some carbonized mess.

So that’s why is sodium carbonate?

The question is unclear. Sodium carbonate is stable enough to melt at 851 deg C, regardless of how stable or not is sodium bicarbonate.

One of the suppliers claims the boiling point is 851 degC but does say it decomposes at much lower temperature. I wonder where this claim comes from (it is pervasive).

Technical note: If you want a user, that is not the author of the commented question or answer, to be notified about a new comment, start the comment by @username. Otherwise, he may miss the comment.

@MatteoZervos  why did this happen? Because it is energetically favourable.

Does this mean it’s nucleus is excited or is this totally a different subject?

Even though it’s a bit dictatorial, thank you for editing my posts and composition them more correctly. I was on an iPad and didn’t know how to write “chemistry numbers”— the ones that are smaller and lowered down.

I wanted an explanation to the reason that the sodium bicarbonate ($\ce{NaHCO3}$) decomposed into sodium carbonate ($\ce{Na2CO3}$), water ($\ce{H2O}$), and carbon dioxide ($\ce{CO2}$); and didn’t just turn to sodium bicarbonate in its gas form, since its boiling point is $\pu{333.6 ^\circ C}$.

You wouldn't have this question if you have spent some time to find out some scientific evidence of when and how $\ce{NaHCO3}$ starts to decompose instead of putting certain temperature, $\pu{400 ^\circ C}$, as the decomposition temperature. As Poutnik cleverly pointed out that in his comment, $\ce{NaHCO3}$ starts to decompose temperatures as low as $\pu{100 ^\circ F}$ (about $\pu{38 ^\circ C}$; Ref.1), thus, the temperature is even not closed to the boiling point of $\ce{NaHCO3}$ before its structure starts to change. Of course, the temperature at which decomposition occurs will depend on the extent of back pressures of the carbon dioxide and water vapors and particle size of the $\ce{NaHCO3}$ granules as well (Ref.1):

To be clear, the above figures (from Ref.1) are showing the sodium bicarbonate  thermal  decomposition data, fractional  conversion as a function of time, $t$, for three particle size ranges and six temperatures in the range $225$-$\pu{350 ^\circ F}$ ($107$-$\pu{177 ^\circ C}$), the temperatures of which are much lower than $\pu{333.6 ^\circ C}$!
The following conclusions are made based on Ref.1 experimental data:

The thermal decomposition of $\ce{NaHCO3}$, granules is chemical kinetically controlled in the temperature range of $225$-$\pu{350 ^\circ F}$ ($380$-$\pu{450 K}$). The shrinking  core  model provides a good fit to the data.
The rate coefficient, $k$, is of Arrhenius form,  with an activation energy of $\pu{20.5 kcal/mol}$ $(\pu{85.8 kJ/mol})$.
The decomposition rate of $\ce{NaHCO3}$, in the temperature range of $\pu{150 ^\circ C}$ ($\pu{302 ^\circ F}$) is not affected by $\ce{CO2}$ partial pressures as high as about $\pu{0.01 atm}$.

The authors have also provided an equation for the estimated decomposition times, $\tau$ in conclusion, which is not relevant here.
The following year of this work, another group has published extended work on the thermal decomposition of solid $\ce{NaHCO3}$ in the temperature range of $360$-$\pu{500 K}$ showing first order kinetic on decomposition at lower temperatures $(\lt \pu{440 K})$ in nitrogen atmosphere (Ref.2). According to the result, the mechanism for the decomposition process, which is essentially that of rearrangement of bicarbonate ions, is given as:
$$\ce{HCO3- ->[\Delta] OH- + CO2} \tag1$$
$$\ce{OH- + HCO3- ->CO3^2- +H2O} \tag2$$
The authors stated that:

Proton transfer, as indicated in reaction (2), will be facile in this case because of the chains of hydrogen-bonded bicarbonate ions present in the crystal structure; these can easily be visualized as reacting in pairs. Since the growth stage is unaffected by carbon dioxide or water vapor, it is likely that both reactions are essentially irreversible.

References:

Timothy C. Keener, George C. Frazier, and Wayne C. Davis, "Thermal Decomposition of Sodium Bicarbonate," Chemical Engineering Communications 1985, 33(1-4), 93-105 (ODI: https://doi.org/10.1080/00986448508911162).
Matthew C. Ball, Christine M. Snelling, Alec N. Strachan, and Rebecca M. Strachan, "Thermal Decomposition of Solid Sodium Bicarbonate," Journal of the Chemical Society, Faraday Transactions 1: Physical Chemistry in Condensed Phases 1986, 82(12), 3709-3715 (ODI: https://doi.org/10.1039/F19868203709).

O.K. Thank you for your answer. I also wanted to ask where you find all this information because it’s nowhere in surface Google.

Good question, Matteo ! How did Mathew Mahindaratne manage to find this reference ?

@Maurice: As in my profile, I teach at Texas A&M University, which has one of the best libraries in the country.

Also, been a member of ReserchGate help as well.

Thank you. I don’t have access to many other chemistry sources since pretty much I’m not learning any of these in school yet. I’ve been trying alone and I think I’ve been pretty good so far for 12. I’ll definitely view ReasearchGate and if im lucky some Texas A&M online documents!


text:
How to calculate number of all atoms in rings in a molecule?
Is there a way or a formula to find number of all atoms in all rings?
EDIT : I wanna explain it with pictures :
this is an anthracene which have 14 atoms in ring.
what thing it has that is 14?
2nd EDIT:
i have number of atoms bonds rings carbon atoms oxygen atoms and ... the connections.I want a formula with these things also i want it to work for no ring molecules like C2H4.

for example: anthracene have 14 members

Count them all one by one, that's the way. As for the formula, what good it would do if I say that it is "4n+2"?

im using it for programming project

Let's put it another way. By doing a programming project, you probably have some information about some molecule(s) encoded in your computer, in bits and bytes. What is that information? What is it that you know about your molecule(s)? Obviously not the number of carbons. Then what?

Try to be nice, folks. This person is a new member and is only looking for guidance.

That 'thing that has 14' is the Smallest Set of Smallest Rings, or SSSR. Calculating the SSSR is a perennial problem in cheminformatics. Treatises have been written about this subject.
Many ring perception algorithms have been devised, and they are too numerous to list here.
I would start with this paper:
Downs, G. M., Gillet, V. J., Holliday, J. D., & Lynch, M. F. (1989). Review of ring perception algorithms for chemical graphs. AUGUST. https://doi.org/10.1021/ci00063a007
which gives an introduction to the subject. As it's an old paper, you could then move onto:
May, J. W., & Steinbeck, C. (2014). Efficient ring perception for the Chemistry Development Kit. Journal of Cheminformatics, 6(1), 1–12. https://doi.org/10.1186/1758-2946-6-3
which then describes some more modern approaches to the problem.
EDIT
'i have number of atoms bonds rings carbon atoms oxygen atoms and ... the connections.I want a formula with these things also i want it to work for no ring molecules like C2H4.'
You're not asking much, are you?!
What you are talking about, in essence, is cheminformatics. Representing chemical structures using computers.
It's a well established and very significant discipline.
I'd start with this book:
https://www.amazon.com/Introduction-Chemoinformatics-Andrew-R-Leach/dp/1402013477
It's a really good introduction to the discipline.

so its a hard problem right?

No, not especially. It can be computationally expensive, but eh algorithms themselves aren't that difficult once you understand how to use them.
Our very own software, https://www.chem4word.co.uk/, uses the Figueras algorithm to calculate the SSSR. We use that because it's quick. But there are so many other algorithms you can just take your pick.
One of the originals  is detailed in
Use of Ring Assemblies in Ring Perception Algorithm
Journal of Chemical Information and Computer Sciences, (1975), 140-147, 15(3)
You got to do some library legwork. There is no way around that.

The easiest way to find the number of rings in a structure is to use Eulers formula: rings = bonds - atoms +1 . (Counting multiple bonds as single bonds). That will tell you how many rings you've got, but not what their members are.
For instance, for benzene, rings = 12-12+1 = 1. Napthalene = 19 - 18 +1 = 2. Etc


text:
What is the unit of molar density?
I'm reading Unit operations of Chemical Engineering, 5th edition by Warren McCabe.
They defined the "molar density $\rho_{M}$" as $\ce{\rho_{M}=\frac{n}{V}}$. The parameters $n$ and $V$ are the number of moles and volume as usual (they are indicated in the summary table of all the symbols at the end of the chapter). To my understanding, $\rho_{M}$ indicates how many moles of a molecule are there in a volume unit and its unit should be something like $\ce{mol/m^3}$ or $\ce{mol/cm^3}$.
$\rho_{M}$" />
However, the unit of $\rho_{M}$ in that table is $\ce{kg\cdot mol/m^3}$. Why is there the presence of mass?

kg mol is the way some engineering texts refer to kmol (1000 mol). g mol is the same as mol. lb mol is useful if you're converting from mol to lb, for example the molar mass of water is 18 g/mol or 18 lb/lbmol. It is defintely confusing in a way since you can read "kg times mol", which is nonsense, though it is less confusing if you write it altogether: kgmol instead of kg mol.

:) Thanks for your help. Now it makes a lot of sense. It was really confusing


text:
Do there exist any elastomers that do not contain carbon?
Elastomers (rubbery materials) that I know of either have a carbon backbone or a silicone backbone with carbon-containing side chains. Do there exist any elastomers that do not contain carbon atoms?
So far Google couldn't come up with anything, but I hope to get a more definitive answer here.

Yes, there are, e.g. poly(dichlorophosphazene)

Beat me by seconds... Polyphosphazenes in general are a good place to look. Why the downvote??

Are you thinking of "elastomer" in some specific sense other than just being "springy"? Or just being springy at normal conditions? Thin films of glass or many metals are pretty springy.


text:
Is there some other way to perform fractional distillation without using Vigreux Column?
So recently I was performing an esterification reaction of oils like peanut oil or any neutral oil where I got some products which are only possible to filter using fractional distillation. I already have a setup for simple distillation and I even have condensers like Liebig, Allihn and Graham. Still, unfortunately, prior to this reaction I haven't purchased a Vigreux column and since the cost of the Vigreux column is a bit out of budget. Is there any safe alternative for doing fractional distillation without the Vigreux column or is it worth purchasing Vigreux column for fractional distillation?

"which are only possible to filter using fractional distillation" is a contradiction.  Either you filter, to separate solids from liquids, or you distill, which is a separation of liquids by different boiling points. Second, too little information about the product (perhaps it solidifies over time/can crystallize out) or scale (if the reaction is large enough, one may pack a column with beads of glass, though the dead volume is much larger than the one of a Vigreux column). Third fractional in fractional distillation is about the receiver, not the column on top of the round bottom flask.

So fractional is more about how you proceed in gently heating the mixture to separate in first place.  Depending on scale, and difference of the boiling points of the liquids to separate at a given pressure, a distillation may be performed well enough without a column at all, too.


text:
Why are some pH standard solutions 6.86 and 9.18?
I have calibrated pH sensors in the lab on several occasions and have used standards of 4.0, 7.0 and 10.0 usually. Recently I received a sensor and was tasked to calibrate it, but it requested standards of 6.86 and 9.18 in addition to the 4.0 — with no ability to change this. I see that you can buy these standards but they are kinda tricky to come by.
My question is: Why these specific numbers, and who uses these? (as I have never seen them before)

The pH 6.86 and 9.18 values come from NIST standard buffer solutions for pH calibration, as described in, for example, NBS special publication 260-53 (1988) (pdf available here).
It appears they chose those values because they are close to the desired values of 7 and 9, but are also easy to prepare. For example, the 6.86 results from combining equal concentrations of $\ce{KH2PO4}$ and $\ce{Na2HPO4}$ (see Table 1 of the above-referenced document), whereas the 9.18 buffer is a conveniently round number concentration (0.01 M) of borax.
The reason that you might be required to use these standards is to ensure that your calibration meets NIST standards.
(For those outside of the USA, NIST is the United States National Institute of Standards and Technology, formerly the National Bureau of Standards, responsible for national-level standardization of measurements.)

IIRC the formula isn't very sensitive either, you can mix it "by eye" and still get close to 3 digit precision.

Those strange $\text{pH}$ values belong to the primary $\mathrm{pH}$ buffer standard solutions, that have their $\ce{pH}$ value given by the definition of the practical $\mathrm{pH}$ scale. It means it is valid worldwide in context of metrology, not being limited to NIST. There is implied the standard temperature for electrochemical measurements $\pu{25 ^\circ C}$.
The round values belong to secondary standard $\mathrm{pH}$ buffers, that  are often tabelized for a suitable $\ce{pH}$ range. During their development, their $\mathrm{pH}$ values are adjusted using a pH-meter calibrated by primary standard buffers.
See e.g. Primary pH standards on iccontrols.com :

4.005  pH, 0.05 mol/kg potassium hydrogen phthalate
6.865 pH, 0.025 mol/kg disodium hydrogen phosphate + 0.025 mol/kg potassium dihydrogen phosphate
7.413 pH, 0.03043 mol/kg disodium hydrogen phosphate + 0.008695 mol/kg potassium dihydrogen phosphate
9.180 pH, 0.01 mol/kg disodium tetraborate
10.012 pH, 0.025 mol/kg sodium hydrogen carbonate + 0.025 mol/kg sodium carbonate

Note that disodium hydrogen phosphate comes as $\ce{Na2HPO4}$ (hygroscopic), $\ce{Na2HPO4 . 2 H2O}$, $\ce{Na2HPO4 . 7 H2O}$ and $\ce{Na2HPO4 . 12 H2O}$(the most usual, AFAIK). The crystal water has to be reflected in used masses of the phosphate and water, to keep the phosphate molar amount and the total water mass.


text:
how to improve self-constructed calorimeter (heat insulation)
Purpose:
My purpose of this project is to use some environmental friendly material to construct calorimeter instead of using styrofoam.
I tried:
I use two different size test tube as my calorimeter(as in picture).
And I tried two ways to improve heat insulation:
1.vacuum
2.silvered
But, these are not good enough to prevent heat loss.
Question:
1.Is there any suggestions to improve heat insulation.
2.Increase the space between two test tube would sufficiently prevent heat loss?

Use a vacuum-insulated coffee cup or Dewer vacuum flask?

1.Is there any suggestions to improve heat insulation.

There are two issues to address: the heat capacity of the calorimeter, and the insulation of the calorimeter. There are several papers in Journal of Chemical Education about replacing Styrofoam cups with something as cheap and convenient and safe while improving sustainability (search for "coffee cup calorimeter").
In a lab we recently did with students, we gave them bubble wrap, cotton balls, two-walled paper cups and a plastic cup (as insert, so everyone had the same chemical and thermal properties for the part touching the reaction mixture). The lab manual mentions some of the considerations.
In the photo below, you see the plastic cup wrapped in bubble wrap in a larger container, with the fitted foam lid in the background (which allows to insert the metal probe of a digital thermometer). I'm sure a couple of rounds of optimization wouldn't hurt for the design, but it is already a working model.

Looking at photos posted in the question, I would guess the biggest leak in the OP's design is at the top (convection of the air). An infrared thermometer with a small focal area makes it easy to investigate the points of heat loss. I would guess that the inner test tube has a fairly high heat capacity (roughly proportional to its mass). The Styrofoam efficiently creates small air compartments to insulate (little convection, little conduction) while having minimal heat capacity. You can correct for the effective heat capacity of the calorimeter if you characterize it carefully (either before the actual experiment, or by recording and analyzing a temperature trace), but if it is negligible, you save that complication. Many modern calorimeters work in an isothermal mode, where heat capacity and insulation become less critical.
A Dewar or any vacuum layer comes along with safety issues that should be considered carefully. The same goes for the flammability and chemical compatibility of the materials used for any design.

2.Increase the space between two test tube would sufficiently prevent heat loss?

If the space is too big and contains gas, you allow convection currents. The Styrofoam is closed-cell, so it does not allow for convection over large distances. I think the key for optimization is to measure the heat capacity and the heat flow in your current designs. Otherwise, you don't know what aspect to optimize.

For the determination of complex constants by calorimetry, I recall the inner vessel of the calorimeter (glass vessel with gently stirred solution A) being immersed in a much larger basin of well tempered water. After both the inner vessel as well as all reagent solution B were normalized to this bath temperature, they were mixed in the inner glass vessel of the calorimeter; temperature monitoring was a tiny Pt100 couple (4-wire connection to reduce the temperature error).

So thank you! These are the best suggestions I ever have. Im now trying different ways to minimize the heat capacity of the calorimeter.

One more question. After reading your post, I wonder if its possible for us to construct a calorimeters that works in isothermal mode. Thanks in advance.


text:
What is the charge of amino acid in alkaline conditions?
This problem refers to the protein buffer in a living system.
"If the blood pH becomes alkaline, there is a release of a proton from the
$\ce{NH3+}$ ion, which takes the $\ce{NH2}$ form."
I understand that the charge of $\ce{NH2}$ must be negative alone and neutral in an amino group. However, the subject of the protein buffer is very briefly touched upon in the text our lecturer gave us and it doesn't detail the reactions in different $\mathrm{pH}$.
I think that since $\ce{NH3+}$  loses a positive proton to balance the blood pH during alkaline conditions, this must mean it becomes neutral. The problem is: I'm not sure if $\ce{NH2}$  is actually in a group and I suspect that it might be alone, which necessitates that it carry a negative charge.
This is my current understanding of the problem.

See here: https://chemistry.stackexchange.com/questions/170512/how-to-determine-the-charge-of-amino-acid-at-certain-ph.

Protein buffer refers to the buffering action of amino acid side chains of proteins. The “$\ce{NH3+}$” is more accurately $\ce{R-NH3+}$, becoming $\ce{R-NH2}$ when losing a proton. Other side chains have $\ce{R-COOH}$ or $\ce{R-COO-}$ groups, also helping to buffer.
$\ce{NH2-}$ does not occur in aqueous (biological) systems, it is too strong of a base. However, it does occur in other solvents as part of chemistry in the synthetic lab.

@MuAlShammary No, $\ce{R-NH3+}$ has a positive charge, and it goes away with the proton, yielding $\ce{R-NH2}$. The nitrogen atom in the latter has a formal charge of zero, with two electrons in the lone pair, and three electrons involved in the three single bonds.


text:
Can water (liquid) be hotter than boiling point?
Can a liquid be heated passed its boiling point before converting to its gaseous form? I'm not referring to changing the pressure (suppose we fix the surrounding pressure).
Water boils at 100 deg C, but I guess I'm looking for clarification about how this happens.
Does the water inside (the parts of the liquid that are not at the border of surrounding gas) surpass 100 deg C? I guess I'm just looking for clarification since it's been some time since I've taken chemistry/thermodynamics and was curious.
Is it the water molecules at the edge of surface that will turn to gas once they hit 100 deg C? Are the water molecules in liquid form slightly higher "inside" the liquid?
I was looking at this article: https://en.wikipedia.org/wiki/Superheating
and was curious about the statement for "Cause" where it states: "For the bubble to expand, the temperature must be raised slightly above the boiling point to generate enough vapor pressure to overcome both surface tension and ambient pressure."
Thank you!

Yes it can.  It is one of the reasons why, performing a distillation, you gently stir the liquid in the lower round bottom flask.
If you distill under reduced pressure, superheating of liquids is even more prominent so that you may add boiling chips to the (still cold) stirred liquid, or that you rotate the flask containing the mixture from which you want to remove e.g., the solvent of reaction with a rotary evaporator

(image credit: Wikipedia).
In the later case, the distillation no longer occurs from the bulk of the solvent, but from the thin film of the liquid on the inner wall of the rotating round bottom flask.

That's certainly a device, thanks for the response. Follow up question: In a standard beaker, if heat is being applied at the bottom, when the water hit's a full boil, is there any part of the water in the center of the fluid's body that is hotter than 100? Or the water molecules nearer to the flame, would they ever go over 100 and remain in liquid form in the beaker. I know the heat would distribute, and the water at the surface would begin to evaporate at 100 (again, 1atm pressure). Does water being surrounded by other water keep it in liquid form despite being hotter than boiling pt?

The device is an older one some (teaching) labs not exchange for more modern ones e.g., this one, or this one because of the price tag. For temperature, you need multiple molecules (Boltzmann distribution of energy) because a single molecule only has its own energy. Boiling includes energy corresponding to a temperature plus surpassing latent enthalpy of vaporization.

You can not form a liquid with only one, or two, or three molecules; a liquid is an ensemble (at leas to form a droplet). Within a beaker, a molecule close to the heating plate at the bottom surely can accumulate more energy than molecules close to the upper surface, but to enter the gas phase, it has to travel up (against hydrostatic pressure) and beyond the surface (surface tension). Bouncing with other molecules, energy is exchanged with them (i.e., the energy can drop below the critical threshold, it can "cool down"). A bit like vents.


text:
Thiol vs carboxylic acid reactivity to acyl chlorides?
I want to form a thioester between some acyl chlorides and NBoc-cysteine. Is the free thiol so much more reactive than the carboxylic acid in cysteine that the side product formation of carboxylic anhydrides is negligible?

I think so that's the case according to Ref.1. For example:

Your $\ce{R^1}$-group can be one you wish to have. A typical procedure to prepare S-Acylation of N-protected-cysteines is given in the reference as follows:

To a solution of N-protected-L-cysteine ($\bf{4a-c}$; $\pu{1 mmol}$) in THF ($\pu{5 mL}$) was added a solution of acyl 1-benzotriazolyl $\bf{5}$ ($\pu{1 mmol}$) in THF ($\pu{5 mL}$) in the presence of $\ce{KHCO3}$ ($\pu{1 mmol}$). The heterogeneous mixture was then stirred at room temperature for $\pu{12 h}$. THF was evaporated, and the solution was acidified with $\ce{HCl}$, and the solid formed was taken in ethyl acetate and washed with $\ce{HCl}$ ($\pu{1N}$, 3×) and brine (3×) and then dried over sodium sulfate. Evaporation of ethyl acetate under reduced pressure afforded N-protected S-acyl cysteine.

References:

Tarek S. Ibrahim, Srinivasa R. Tala, Said A. El-Feky, Zakaria K. Abdel-Samii, and Alan R. Katritzky, "Cysteinoyl- and Cysteine-containing Dipeptidoylbenzotriazoles with Free Sulfhydryl Groups: Easy Access to N-terminal and Internal Cysteine Peptides," Chemical Biology & Drug Design 2012, 80(2), 194-202 (DOI: https://doi.org/10.1111/j.1747-0285.2011.01303.x).


text:
Will ethyl 2-amino-1H-imidazole-5-carboxylate undergo ester aminlysos with the primary amines of bPEI efficiently in methanol?
I was about to set up the following reaction, but before I do, I wanted to see if anyone has some suggestions or thinks this will not work:

24mmol Nboc protected ethyl 2-amino-1H-imidazole-5-carboxylate
2mmol 1,800Da bPEI (should have ~12 amines available for reaction according to this https://www.sciencedirect.com/science/article/abs/pii/S0014305720317316)
10mL methanol
5 days at room temp
Ive seen reactions like this performed under nitrogen, is this absolutely necessary and should I also bubble nitrogen through my reagents for a while before letting it run?

Also, should I add something to catalyze the reaction? Is there anything less exotic than these ionic liquids that I might have on hand that would help but will not cleave my Boc? https://chemistry-europe.onlinelibrary.wiley.com/doi/abs/10.1002/cctc.201100221

The nitrogen is to keep the reaction mixture dry as MeOH will absorb water from the atmosphere if left open, it also reduceds evaporative loss of MeOH.

PEI stands for Poly(ethyleneimine). There would be variety of PEI solutions (e.g., PEI1800), each of which have different numbers of free primary amine groups. Therefore, make sure the solution you are using have the correct amount of $\ce{NH2}$ groups available for the reaction. In Ref.1 (the reference you have given as example), the authors have capped different number of $\ce{NH2}$ groups in PEI1800 using different amounts of capping agents (reaction with at different molar ratios) as needed (capping agent examples: Imidazolyl and guanidyl). Although the procedures are different for each and different capping reagent, the authors have let each capping reaction stirring at $\pu{75 ^\circ C}$ for $\pu{8 h}$ with nitrogen.
The important fact you have forgot to mention when using PEI is that you may need to lyophilize the material in order to remove all water in the solution before subjected to undergo further reactions like capping $\ce{NH2}$ groups. According to the Ref.1, PEI1800 was lyophilized before use and the product was also lyophilized after, before they use them in the intended research. Also, in the text of the Ref.1 says that free $\ce{NH2}$ groups in PEI1800 is 16 (not 12 as you mentioned). Thus, if you want to cap all of free $\ce{NH2}$ groups, you may need to uses excess capping reagents.
As Waylander suggested and the authors of the Ref.1 have done, keeping the reaction under nitrogen would be better. However, a 5-day reaction time is too long and asking for trouble. Thus, I'd like to suggest to use a catalyst to shorten the reaction time. The one good, cheaper catalyst would be cyanide ions $(\ce{NaCN})$, which would dissolve in methanol (Ref.2). If you like to use an organic catalyst instead, 1,5,7-triazabicyclo[4.4.0]dec-5-ene (TBD) would be a good addition (Ref.3) in aminolysis of ester. Regardless of what you have used, your protective group would be safe as long as you have avoided acidic conditions.

References:

Jie Song, Dongli Wang, Jing Wang, Qing Shen, Cao Xie, Weiyue Lu, Ruifeng Wang, and Min Liu, "Low molecular weight polyethyleneimine modified by 2-aminoimidazole achieving excellent gene transfection efficiency," European Polymer Journal 2020, 140, 110017 (11 pages) (DOI: https://doi.org/10.1016/j.eurpolymj.2020.110017).
Thomas Hoegberg, Peter Stroem, Michael Ebner, and Sten Raemsby, "Cyanide as an efficient and mild catalyst in the aminolysis of esters," J. Org. Chem. 1987, 52(10), 2033–2036 (DOI: https://doi.org/10.1021/jo00386a025).
Cyrille Sabot, Kanduluru Ananda Kumar, Stéphane Meunier, and Charles Mioskowski, "A convenient aminolysis of esters catalyzed by 1,5,7-triazabicyclo[4.4.0]dec-5-ene (TBD) under solvent-free conditions," Tetrahedron Letters 2007, 48(22), 3863-3866 (DOI: https://doi.org/10.1016/j.tetlet.2007.03.146).

I dissolved my ester and bPEI in methanol, rotovapd it dry, then redissolved into methanol that i freshly dried with magnesium sulfate, added half a handful of molecular sieves, suba sealed the flask and flushed with nitrogen. You think it might turn out okay or should I just reset it up once the catalyst comes in next week? Didnt have any cyanide on hand

Did your PEI in powdered form to bigin with? If so, your set up would be good enough for me. If PEI is in a solution, lyophilization (freeze-driing) is important.

I didnt lyophilize it before hand, im using a 99% pure viscous liquid thats been open and in use for a couple years in our lab. i suppose it might still have had some moisture even after I rotovapd it since it was still bubbling slightly when I stopped the rotovap, at the time I thought that was either the ester or the PEI starting sublime/evaporate. Instead of lyophilizing can I also just dissolve it in methanol and dry that with MgSO4 before mixing with my ester and then do the rest of my process?

Do the reaction in small scale to see it would work. If it didn't give a good yield, you should lyophilize the substrate.

Yeah fair enough, thanks very much for bearing with me and answering all that!


text:
Dependence of the permittivity on solvent mixing ratio
I am doing DFT calculations on an organometalloid compound to assess whether it dissociates into an ion pair in different solvents. I suppose that the more polar the solvent is the more stable this compound should be in its dissociated form. To achieve this I want to successively increase the permittivity of the solvent mixture using PCM. It was stated that this compound is  well soluble in a 1:1 mixture of water and acetonitrile and now I am wondering if I can just calculate the mean permittivity value of both solvents and imply it in my PCM calculation to simulate this solvent environment. So my question is if the permittivity is directly proportional to the mixing ratio of two solvents.

I doubt it's so simple.

You can consult reference (1) and others cited therein. Nikitin and Lyubartsev explain:

Experimental data 26 [Venables and Schmuttenmaer] show that the dielectric permittivity of water–acetonitrile mixture is about linear in terms of the volume ratio.

My interpretation of volume ratio (or more accurately, volume fraction, which is used in Fig. 5) is that it is computed from the equivalent volumes of pure components that would be combined, so it can be translated linearly into a molar fraction. The alternative is that it implies application of a way of computing the partial molar volume of the components, since the density of the mixtures is not linear in the mole fraction (see Fig. 4).
References
(1) Nikitin, A. M.; Lyubartsev, A. P. New Six-Site Acetonitrile Model for Simulations of Liquid Acetonitrile and Its Aqueous Mixtures. J. Comput. Chem. 2007, 28 (12), 2020–2026. https://doi.org/10.1002/jcc.20721.
(2) Venables, D. S.; Schmuttenmaer, C. A. Far-Infrared Spectra and Associated Dynamics in Acetonitrile–Water Mixtures Measured with Femtosecond THz Pulse Spectroscopy. The Journal of Chemical Physics 1998, 108 (12), 4935–4944. https://doi.org/10.1063/1.475902.

Hmm, the pic seems rather to be in vol fraction, as it seems to me very  suspicious 1% AN would have so low permitivity. OTOH, why to simulate permitivity of the pure AN?

Yes, the axis label shows vol% but I assume that is an error and fraction is meant. Also, in the text they speak of vol ratio which is inconsistent. Not the tidiest reference but it links to the experimental source which the OP can consult.

This helps a lot.


text:
What are the reactions between Zinc acetate dihydrate and Methoxyethanol?
I am synthesizing a sol-gel solution with Zinc acetate dihydrate + Methoxyethanol and ethanolamine as stabilizer, and I wonder what are the reactions that are taking place in the process. Could you help me please?
More details:
I had zinc atetate dihydrate (sigma aldrich 99.999% trace metal basis) and I added 0.913 g to 20 mL of Methoxyethanol. It was stirred at 60 °C during 30 minutes. Then, 1 mL of ethanolamine (stabilizer) was added, and the solution was stirred during 90 minutes at 60 °C.
After that, it was kept aging at room temperature during 24 hours, and then stirred at 60 °C during 3 hours (were the solvents vaporized and the sol gel was formed).

For the lack of preparation, the presentation of this question is a bad one. "Somewhere", "someone" had the idea to mix these chemicals in "some ratio"/"somehow".  You possibly pick up the mode opératoire from a literature reference, so cite it. Chances are that the author(s) of the original publication in turn reference to prior work, too.

Edited, sorry for the lack of information

No, @Poutnik I created my account 3 hours ago!

@Chemistry I see. I apologize, coincidences may happen anytime. I have posted the above because such deletions/repostings occur frequently.

Zinc acetate ($\ce{Zn(CH3COO)2}$) is known to form the complex $\ce{Zn4O(CH3COO)6}$ with loss of acetic anhydride on heating.  2-methoxyethanol and ethanolamine will also evaporate, leaving the Zinc complex as your sol gel.

Ethanolamine will react with the acetic anhydride. It also has a boiling point of 175C and2-methoxyethanol has a boiling point of ~125C so I don't see them evaporating off very readily.

@Waylander at 60C they will both have a vapor pressure.  Ethanolamine will react with the anhydride, but equilibria over days has most of the volatiles leaving.

What is the reaction between ethanolamine and the acetic anhydride? Does methoxyethanol do not react?

Also, how is the complex formed? I'm chemical engineer and I don't really know about mechanisms and I'm quite lost... If you could put the reactions please, it would be marvellous.

@Chemistry the amine is "stronger" than the free alcohol group in reacting with the carbonyl group of the anhydride.  This is undergraduate organic chemistry.  Team up with an organic and analytical chemist.


text:
Is delta H = nC_p dT only valid for ideal gas for an isothermal process?
The related question is:

$10\ \mathrm g$ of argon gas is compressed isothermally and reversibly at a temperature of $27\ \mathrm{^\circ C}$ from $10\ \mathrm l$ to $5\ \mathrm l$. Calculate enthalpy change ($\Delta H$) for this process $R=2.0\ \mathrm{cal\ K^{-1}\ mol^{-1}}$.

I know that
$$\mathrm dU=\left(\frac{\partial U}{\partial T}\right)_V\,\mathrm dT+\left(\frac{\partial U}{\partial V}\right)_T\,\mathrm dV$$
In case of ideal gas it will be equal to $nC\,\mathrm dT$ but as it is not given it is ideal gas. I have no idea on how to proceed.

$dH=nC_pdT$ is valid only for an ideal gas.

Look up in a thermo book how to determine the effect of pressure on enthalpy for a real gas.

@ChetMiller I know that but how to solve it then.

What is the equation for dH in terms of dT and dP?  Do you know that equation?

Well, you could get a fairly good first iteration by using ideal gas law, then plug into your known equations to refine. OR you could of course locate an expression for dH as @ChetMiller suggest.

If you treat it as an ideal gas, the enthalpy change is zero.

@ChetMiller In all the ways suggested, It seems at some point I need to consider it as an ideal gas, yes I know for ideal gas enthalpy change and internal energy change both will be 0.

$$\left(\frac{\partial H}{\partial P}\right)_T=\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$

I think I don't know how to use it. It seems above my current level.

To apply this, you need to know the equation of state f argon, or an approximation to the equation of state (such as the law of corresponding states).

This might not be the expected textbook answer. For most practical purposes, however, you can treat the gas in this example as ideal gas. The example is about a monoatomic gas, well above its critical point temperature $(27\ \mathrm{^\circ C}\gg-122.46\ \mathrm{^\circ C})$, well below its critical point pressure $(0.62\ \mathrm{bar}\ll48.63\ \mathrm{bar})$, and well below its critical point density $(1.0\ \mathrm{g\ l^{-1}}\ll535.6\ \mathrm{g\ l^{-1}})$; therefore, ideal gas behaviour may be safely assumed.
In case the ideal gas approximation is not good enough, you would usually not try to find a better equation for a real gas in the real world; you would use tabulated numerical values instead. For state functions like internal energy $U$ and enthalpy $H$, this approach works so well because they depend only on the current equilibrium thermodynamic state of the system and not on the path that the system has taken to reach that state.
You need a set of quantities that describe the equilibrium states of the system. Usually temperature $T$, pressure $p$, and volume $V$ are used to describe a state point of a gas. However, you can as well use temperature $T$, mass $m$, and volume $V$ from the question. You can directly calculate the density $\rho$ from the given values and then use the values for temperature $T$ and density $\rho$ to find the corresponding tabulated values for the specific enthalpy $h$ of argon. For the values in the following table, I used NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9 with an arbitrary number of five significant digits. On this occasion, I also looked up the values for pressure $p$ and specific internal energy $u$; however, those are not needed for this example. Using the values for specific enthalpy $h$ and the given value for mass $m$, the enthalpy $H$ of the sample can be directly calculated.
$$
\begin{array}{lllll}
\hline
\text{Quantity} & \text{Quantity} & \text{Unit} & \text{Value} \\
\text{name} & \text{symbol} &  & \text{State 1} & \text{State 2} \\
\hline
\text{mass} & m & \mathrm g & 10.000 & 10.000 \\
\text{temperature} & T & \mathrm{^\circ C} & 27.000 & 27.000 \\
\text{volume} & V & \mathrm l & 10.000 & 5.0000 \\
\text{density} & \rho & \mathrm{g\ l^{-1}} & 1.0000 & 2.0000 \\
\hline
\text{pressure} & p & \mathrm{bar} & 0.62448 & 1.2485 \\
\text{specific internal energy} & u & \mathrm{J\ g^{-1}} & 93.602 & 93.508 \\
\text{specific enthalpy} & h & \mathrm{J\ g^{-1}} & 156.05 & 155.93 \\
\hline
\text{enthalpy} & H & \mathrm{J} & 1560.5 & 1559.3 \\
\hline
\end{array}$$
As expected for a gas with approximately ideal behaviour, the resulting enthalpy change is very small: $\Delta H=1559.3\ \mathrm J-1560.5\ \mathrm J=-1.2\ \mathrm J$


text:
Why is C a matrix and not a vector in Roothaan Equations?
I'm trying to figure out why $C$ is a matrix and not a vector in the Roothaan Equations. Below, I have a derivation which I have taken from this PDF in which, to me, $C$ appears to be a vector. But when I look at other derivations (for example this derivation) $C$ is a matrix, although I can't figure out why $C$ is a matrix, I feel it is poorly explained.
So this is my derivation where $C$ is a vector:
Here is the basis function for 1 electron:
$$
\Psi_a = \sum_{i=1}^n c_{ia}\phi_i
$$
This is the Fock operator for this electron $a$:
$$
f_a = h_a + \sum_{J=1}^{N/2}(2J_j(a)-K_j(a))
$$
Above, I'm using $a$ to represent a particular electron, so $J(a)$ and $K(a)$ are, with a loose notation, Coulomb and exchange operators for electron $a$.
So we now have an equation:
$$
f_a \Psi_a = \epsilon_a \Psi_a \\
f_a \sum_{i=1}^n c_{ia}\phi_i = \epsilon_a \sum_{i=1}^n c_{ia}\phi_i \\
$$
This is $1$ equation for $n$ unknown coefficients, so to get $n$ equations we can left multiply each side by each of the $n$ $\phi_i$ and integrate over all space with respect to electron $a$'s location. This would give us $n$ equations:
$$
\int \phi_j f_a \sum_{i=1}^n c_{ia}\phi_i = \int \phi_j\epsilon_a \sum_{i=1}^n c_{ia}\phi_i \\
\sum_{i=1}^n c_{ia} \int \phi_j f_a \phi_i = \sum_{i=1}^n c_{ia} \int \phi_j\epsilon_a \phi_i \\
\sum_{i=1}^n c_{ia} F_{ji} = \epsilon_a \sum_{i=1}^n c_{ia} S_{ji}
$$
Where $F$ and $S$ form two $nxn$ matrices, and the matrix equation becomes $FC=SC \epsilon_a$
$$
F = \begin{bmatrix}
    F_{11} & F_{12} & \dots \\
    \vdots & \ddots & \\
    F_{n1} &        & F_{nn}
    \end{bmatrix}
\qquad
S = \begin{bmatrix}
    S_{11} & \dots  & S_{1n}\\
    \vdots & \ddots & \vdots\\
    S_{n1} & \dots  & S_{nn}
    \end{bmatrix}
C = \begin{bmatrix}
    C_{1a}\\
    \vdots\\
    C_{na}
    \end{bmatrix}
$$
In this case, $C$ is a vector, and I believe we are solving for only 1 electron $a$ and it's energy $\epsilon_a$. Keep in mind I believe this is wrong. Above, $\epsilon_a$ would be a constant. The left side has dimensions $(n\times n) \times (n \times 1) = (n \times 1)$ and the right side of the equation has dimension $(n\times n) \times (n \times 1) \times (1 \times 1) = (n \times 1)$
So if we wanted to solve for two electrons, $a$ and $b$, I believe we would just solve one set of these matrix equations for each electron. The Fock matrix and coefficient matrix would be functions of the electron: $F(a)C(a) = S(a)C(a)\epsilon_a$ and $F(b)C(b) = S(b)C(b)\epsilon_b$
Edit: Shown below, I tried to form one matrix equation for the case of two electrons by concatenating the $C$ vector and Fock matrix but I can't make the dimensions work:
$$
F =
\left(\begin{array}{@{}c|c@{}}
  \begin{matrix}
  F_{11a} & \dots & F_{12a} \\
  F_{n1a} & \dots & F_{nna}
  \end{matrix}
  & 0 \\
\hline
  0 &
  \begin{matrix}
  F_{11b} & \dots & F_{12b} \\
  F_{n1b} & \dots & F_{nnb}
  \end{matrix}
\end{array}\right)
C = \begin{bmatrix}
    C_{1a}\\
    \vdots\\
    C_{na}\\
    C_{1b}\\
    \vdots\\
    C_{nb}
    \end{bmatrix}
$$
$$FC=SC$$
So in this scenario $C$ is still a vector but I could not figure out how to work in the energy while keeping the dimensions correct because now the energy is a $2 \times 1$ vector $[\epsilon_a, \epsilon_b]$.
Anyway I clearly have a major assumption wrong. Could someone please show me how to properly form these equations with $C$ as a matrix?
EDIT
Responding to PAEP's answer, while I do know the correct form, I just don't get why that is the form. Thus the need for a derivation.
For example if I multiply out the correct form which you supplied:
$$
FC = \begin{bmatrix}
    (F_{1,\bullet} \times C_{\bullet, 1}) & (F_{1,\bullet} \times C_{\bullet, 2}) & \dots \\
    \vdots & \ddots & \\
    (F_{n,\bullet} \times C_{\bullet, n}) &        & (F_{n,\bullet} \times C_{\bullet, n})
    \end{bmatrix}
$$
Where $(F_{i,\bullet} \times C_{\bullet, i})$ is the dot product of row $i$ of $F$ times column $i$ of $C$
Similarly,
$$
SC = \begin{bmatrix}
    (S_{1,\bullet} \times C_{\bullet, 1}) & (S_{1,\bullet} \times C_{\bullet, 2}) & \dots \\
    \vdots & \ddots & \\
    (S_{N,\bullet} \times C_{\bullet, n}) &        & (S_{n,\bullet} \times C_{\bullet, n})
    \end{bmatrix}
$$
then,
$$
SC\epsilon = \begin{bmatrix}
    (S_{1,\bullet} \times C_{\bullet, 1})\epsilon_{11} & & & & \\
    & (S_{2,\bullet} \times C_{\bullet, 2})\epsilon_{22} & & & \\
    & \ddots &  & \\
    & & & (S_{n,\bullet} \times C_{\bullet, n})\epsilon_{nn} \\
\end{bmatrix}
$$
Can you derive for me $F_{i, j}$ in this case? If you look at my original derivation of $F$ it was built from 1 electron only. But this $F$ looks to be built from all electrons yet it has no additional terms.

I don't think it's more complicated than that $c_{ia}$ defines a matrix that transforms from the basis ${\phi_i}$ to ${\Psi_a}$. Since the latter are vectors $c_{ia}$ has to be a matrix.

This seems appropriate for the theoretical chemistry sister site.

Matter modelling is not theoretical chemistry. It is matter modelling. Why on earth can't we have questions on basic theoretical chemistry techniques like Hartree-Fock on the chemistry site?

@user1271772 This is definitely more appropriate here, than it is on MM SE. We've had questions like this way before there was the other site and it fits right in here with our canon of existing ones.

@IanBush Who "on earth" said that we can't have such questions on both sites? Martin, I totally disagree with you that this is "more" appropriate here than on MMSE. Just because Chem.SE has "had questions like this way before" MMSE existed, doesn't mean that Chem.SE is more appropriate. I'm happy that this question got an answer here. But that was only 2 hours ago, and if the OP isn't satisfied, I'm just letting them know that there's another place that's focused almost completely on questions like these.

@user1271772 This is neither the place nor the time to have a discussion about what is on topic where. The question was posted here, it is on topic here, there is no reason to confuse a new contributor with a comment like the one you have posted. I have asked you this before: If you advertise the Matter Modeling site, which is fine, then please do it in a way that it is unambiguous about question still being on topic here, that cross-posting is as much avoided as possible, and that makes your own  intentions clear.

I think the problem lies within the Fock "Operator", it is dependent on its own solutions, which is the reason the HF-Equations have to be solved iterative. As such it is not a one-electron operator. But I'm not sure how to answer the question as I am currently not really understanding the whole argument. I have to brain about this. It would probably help if you defined all things you used there, $J$ and $K$ and $h$ and the lot…

@Frank, I would try to give you a more detailed answer. This answer is inspired in Lewars book which I strongly recommend you consult.
It is important that you take into account than when you applied the Hartre-Fock self consistent method you are optimizing a single electron spin-orbital but you are considering the rest of the electrons of the atom or molecule you are studying. And as @Martin mentioned, the Fock operators $\hat{F}$ are pseudooperators in the sense that in order to obtain the optimized spin-orbitals you have to modify the orbitals that appear in the operators as part of the optimization process.
Lets start with the Hartree-Fock equations
$$\widehat {\rm{F}}{\kern 1pt} {\psi _i} = {\varepsilon _{\rm{i}}}{\kern 1pt} {\psi _i} $$
Expressing $\psi _i$ as a linear combination of a set of orbitals
$${\psi _i} = \sum\limits_{\nu  = 1}^K {{c_{\nu i}}{\phi _\nu }}$$
you would obtain
$$\begin{array}{l}\widehat {\rm{F}}{\kern 1pt} \sum\limits_{\nu  = 1}^K {{c_{\nu i}}{\phi _\nu }}  = {\varepsilon _{\rm{i}}}{\kern 1pt} \sum\limits_{\nu  = 1}^K {{c_{\nu i}}{\phi _\nu }} \\
\sum\limits_{\nu  = 1}^K {{c_{\nu i}}} \widehat {\mathop{\rm F}\nolimits} (1){\phi _\nu } = {\varepsilon _i}\sum\limits_{\nu  = 1}^K {{c_{\nu i}}} {\kern 1pt} {\kern 1pt} {\phi _\nu }\end{array}
$$
Considering that you work with K orbitals you will have
$$\begin{array}{c}
\sum\limits_{\nu  = 1}^K {{c_{\nu 1}}} \widehat {\mathop{\rm F}\nolimits} (1){\phi _\nu } = {\varepsilon _1}\sum\limits_{\nu  = 1}^K {{c_{\nu 1}}} {\kern 1pt} {\kern 1pt} {\phi _\nu }\\
\sum\limits_{\nu  = 1}^K {{c_{\nu 2}}} \widehat {\mathop{\rm F}\nolimits} (1){\phi _\nu } = {\varepsilon _2}\sum\limits_{\nu  = 1}^K {{c_{\nu 2}}} {\kern 1pt} {\kern 1pt} {\phi _\nu }\\
\vdots \\
\sum\limits_{\nu  = 1}^K {{c_{\nu K}}} \widehat {\mathop{\rm F}\nolimits} (1){\phi _\nu } = {\varepsilon _K}\sum\limits_{\nu  = 1}^K {{c_{\nu K}}} {\kern 1pt} {\kern 1pt} {\phi _\nu }
\end{array}
$$
where $K \ge N$, K is the number of orbitals in the base and N is the total number of electrons.
Multiplying both members on the left by $\phi_1^*$ and integrating over the whole space,
$$
\begin{array}{c}
\sum\limits_{\nu  = 1}^K {{c_{\nu 1}}} {F_{1\nu }} = {\varepsilon _1}\sum\limits_{\nu  = 1}^K {{c_{\nu 1}}} {\kern 1pt} {\kern 1pt} {S_{1\nu }}\\
\sum\limits_{\nu  = 1}^K {{c_{\nu 2}}} {F_{1\nu }} = {\varepsilon _2}\sum\limits_{\nu  = 1}^K {{c_{\nu 2}}} {\kern 1pt} {\kern 1pt} {\kern 1pt} {S_{1\nu }}\\
\vdots \\
\sum\limits_{\nu  = 1}^K {{c_{\nu K}}} {F_{1\nu }} = {\varepsilon _K}\sum\limits_{\nu  = 1}^K {{c_{\nu K}}} {\kern 1pt} {S_{1\nu }}
\end{array}
$$
Repiting this operation for $\phi_2^*$ , $\ldots$, $\phi_K^*$
$$
\begin{array}{c}
\sum\limits_{\nu  = 1}^K {{c_{\nu 1}}} {F_{2\nu }} = {\varepsilon _1}\sum\limits_{\nu  = 1}^K {{c_{\nu 1}}} {\kern 1pt} {\kern 1pt} {S_{2\nu }}\\
\sum\limits_{\nu  = 1}^K {{c_{\nu 2}}} {F_{2\nu }} = {\varepsilon _2}\sum\limits_{\nu  = 1}^K {{c_{\nu 2}}} {\kern 1pt} {\kern 1pt} {\kern 1pt} {S_{2\nu }}\\
\vdots \\
\sum\limits_{\nu  = 1}^K {{c_{\nu K}}} {F_{2\nu }} = {\varepsilon _K}\sum\limits_{\nu  = 1}^K {{c_{\nu K}}} {\kern 1pt} {S_{2\nu }}\end{array}
$$
$$ \ldots $$
$$
\begin{array}{c}
\sum\limits_{\nu  = 1}^K {{c_{\nu 1}}} {F_{K\nu }} = {\varepsilon _1}\sum\limits_{\nu  = 1}^K {{c_{\nu 1}}} {\kern 1pt} {\kern 1pt} {S_{K\nu }}\\
\sum\limits_{\nu  = 1}^K {{c_{\nu 2}}} {F_{2\nu }} = {\varepsilon _2}\sum\limits_{\nu  = 1}^K {{c_{\nu 2}}} {\kern 1pt} {\kern 1pt} {\kern 1pt} {S_{K\nu }}\\
 \vdots \\
 \sum\limits_{\nu  = 1}^K {{c_{\nu K}}} {F_{K\nu }} = {\varepsilon _K}\sum\limits_{\nu  = 1}^K {{c_{\nu K}}} {\kern 1pt} {S_{K\nu }}
\end{array}
$$
where ${F_{\mu \nu }} = \left\langle {{\phi _\mu }(1)\left| {\widehat F(1)} \right|} \right.\left. {{\phi _\nu }(1)} \right\rangle$ and ${S_{\mu \nu }} = \left\langle {{\phi _\mu }(1)} \right.\left| {{\phi _\nu }(1)} \right\rangle$
This operation yields a set of K x K equations known as the Roothaan-Hall equations.
You can write this equation in  compact form
$$
    \sum\limits_{\nu  = 1}^K {{c_{\nu i}}} {F_{\mu \nu }} = {\varepsilon _i}\sum\limits_{\nu  = 1}^K {{c_{\nu i}}} {\kern 1pt} {S_{\mu \nu }}\quad \mu  = 1,2,3, \ldots ,K
$$
where %$\psi_i (i = 1, 2, \ldots , K)$, so
$${F_{11}}{\kern 1pt} {c_{11}} + {F_{12}}{c_{21}} +  \cdots  + {F_{1K}}{c_{K1}} = {\varepsilon _1}\left( {{S_{11}}{\kern 1pt} {c_{11}} + {S_{12}}{\kern 1pt} {c_{21}} +  \cdots {S_{1K}}{\kern 1pt} {c_{K1}}} \right) $$
$$ \ldots $$
$${F_{K1}}{\kern 1pt} {c_{1K}} + {F_{K2}}{c_{2K}} +  \cdots  + {F_{KK}}{c_{KK}} = {\varepsilon _K}\left( {{S_{K1}}{\kern 1pt} {c_{1K}} + {S_{K2}}{\kern 1pt} {c_{2K}} +  \cdots {S_{KK}}{\kern 1pt} {c_{KK}}} \right) $$
In matrix form you will get
$${\mathbf{FC}} = {\mathbf{SCE}}$$
Addendum: The Fock operator
The Fock operator $\widehat {\rm{F}}$ that appears in the Hartree-Fock equations
$$\widehat {\rm{F}}{\kern 1pt} {\psi _i} = {\varepsilon _{\rm{i}}}{\kern 1pt} {\psi _i}$$
is a one-electron operator.
This operator can be written as
$$\begin{array}{c}\widehat F(1) =  - \frac{1}{2}\nabla _1^2 - \sum\limits_\mu  {\frac{{{Z_\mu }}}{{{r_{\mu 1}}}}}  + \sum\limits_{j = 1}^N {(2{{\widehat {\mathop{\rm J}\nolimits} }_j} - {{\widehat K}_j})} \\ = \widehat h(1) + \sum\limits_{j = 1}^N {(2{{\widehat J}_j} - {{\widehat K}_j})} \end{array}$$
where "(1)" just reminds as that the operator is a single electron operator.
The operator $\widehat {\rm{h}}(1)$
$$\widehat {\rm{h}}(1) = - \frac{1}{2}\nabla _1^2 - \sum\limits_\mu  {\frac{{{Z_\mu }}}{{{r_{\mu 1}}}}} $$
is also called the core-hamitonian since it represents the energy of an electron that occupies orbital $\phi _i$ under the attraction of a nuclear core (i.e. ignoring the interaction with other electrons).
${\widehat {\mathop{\rm J}\nolimits} _j}(1)$ corresponds to the Coulomb operator that accounts for the classical Coulomb electron-electron respulsion
$${\widehat {\mathop{\rm J}\nolimits} _j}(1){\kern 1pt} \,{\phi_i}(1) = {\phi_i}(1){\kern 1pt} \int {{{\left| {{\phi _j}(2)} \right|}^2}\frac{1}{{{r_{12}}}}d{\tau _2}}$$
whose matrix element $J_{ij}$ is the Coulomb integral
$${J_{ij}} = \left\langle {{\phi_i}(1)\left| {{{\widehat J}_j}(1)} \right|} \right.\left. {{\phi_i}(1)} \right\rangle$$
$${J_{ij}} = \int {\phi_i^*(1)\,\phi_j^*(2)\frac{1}{{{r_{12}}}}{\phi _i}(1)\,{\phi _j}(2)\,d{\tau _1}d{\tau _2}} $$
Finally, the exchange operator ${\widehat {\mathop{\rm K}\nolimits}_j}(1)$ is given by
$${\widehat K_j}\left( 1 \right){\phi_i}(1) = {\kern 1pt} \,{\phi  _j}(1)\int {\phi  _j^*(2)\frac{1}{{{r_{12}}}}{\phi_i}(2){\kern 1pt} d{\tau _2}} $$
and its matrix element $K_{ij}$ is the exchange integral (which has no classical counterpart and it is a result that appears when you use a Slater determinant to write the electronic wavefunction)
$${K_{ij}} = \left\langle {{\phi_i}(1)} \right.\left| {{{\widehat K}_j}(1)} \right|\left. {{\phi_i}(1)} \right\rangle $$
$${K_{ij}} = \int {\phi_i^*(1)\,\phi _j^*(2)\frac{1}{{{r_{12}}}}{\phi _j}(1)\,{\phi _i}(2)\,d{\tau _1}d{\tau _2}}$$

This is dry, but in the nice way. I think I've learned it differently, but I also think I understood this one. Bravo!

You have explained a lot, but why is it always $F(1)$? Where is $F(2)...F(K)$? Is $F$ a $(K \times K \times K)$ tensor? I will award bounty since it is about to expire! but I really want to know the answer! Either way thank you I think I am closer to understanding!

Maybe my fundamental understanding of what we are doing was wrong. Is the reason there is no $F(2)$ is because we are only creating a function to approximate the probability electron 1 is at $r_1$? And if we want electron 2, then we have to resolve with $F(2)$?

@Frank  The Fock operator is a single electron operator, and we use $\widehat {\rm{F})}(1)$ to mark the fact. It adds no meaning to use $\widehat {\rm{F})}(2)$, etc. It will  make the notation far more confusing. I will edit my answer and include more details of the Fock operator expression.

What are the dimensions of the matrices if $K > N$? I believe matrices would be $N \times K$ but that's not square so maybe you can't solve in this case?

The matrices has the dimensions KxK. You will have N occupied orbitals (actually, spin-orbitals) and N-K unoccupied orbitals (virtual orbitals). The idea is that if you make a linear combination of K atomic orbitals, you will obtain K molecular orbitals,

I think that there is a mistake in the  the Roothan-Hall equations.
The matrix equation should read
$$ \mathbf{FC = SC \epsilon}$$
where
$$
\mathbf{F} = \begin{bmatrix}
    F_{11} & F_{12} & \dots \\
    \vdots & \ddots & \\
    F_{n1} &        & F_{nn}
    \end{bmatrix}
\quad
\mathbf{S} = \begin{bmatrix}
    S_{11} & \dots  & S_{1n}\\
    \vdots & \ddots & \vdots\\
    S_{n1} & \dots  & S_{nn}
    \end{bmatrix}
$$
$$
\mathbf{C} = \begin{bmatrix}
    c_{11} & \dots  & c_{1n}\\
    \vdots & \ddots & \vdots\\
    c_{n1} & \dots  & c_{nn}
    \end{bmatrix}
\quad
\mathbf{\epsilon} = \begin{bmatrix}
    \epsilon_{11} &             0 &  \dots  & 0\\
       0          & \epsilon_{22} &  \dots  & 0\\
    \vdots        & \vdots        &\ddots   & \vdots\\
       0          & 0             &  \dots  & \epsilon_{nn}\\
    \end{bmatrix}
$$
Probably you can find a detailed outline of the solution of the Roothan-Hall equations in many Quantum Chemistry/Computational Chemistry books. You could try section 5.2.3. The Hartree-Fock Equations of  E. G. Lewars, Computational Chemistry 2nd ed., Springer (2011).

I know the correct form I just don't get why it is the correct form. Can you derive $F_{i,j}$? I also edited my post above.

I tried to answer your request in a second answer. It is a bit dry but I hope it helps.

Why two separate answers by the same person?

I think it is a case of a bit of clumsiness of my part. I did not know how to express that my second answer was a response to one of the comments.

I think you should try a divide-and-conquer approach. Lets begin with a top-down view  to avoid getting tangled up in details.
The Roothaan-Hall equations can be stated in the following form
$$
\mathbf{ FC = SC E }
$$
where $\bf F$ is the Fock-matrix which we obtain by expanding the Fock Operator in a discrete basis set of functions. $ \bf C$ is the coefficient matrix that contains the eigenvector coefficients with respect to this basis. $\bf S$ is the overlap matrix of the basis set and $\bf E$ a diagonal matrix with the eigenvalues.
First, lets assume that our basis is orthogonal such that $\bf S$ can be dropped since it becomes a unit matrix, as this is irrelevant to your question of the dimensionality of $\bf C$.
The form  of the equation* becomes then that of a regular eigenvalue problem
$$
\bf FC = CE
$$
and the question how the dimensionality of F and C are connected is the same question as if applied to any other eigenvalue equation.
Eigenvalue problems are often given with a single column vector, like this
$$
M\vec v = \lambda_v \vec v
$$
Now if $M$ is quadratic and diagonalizable, with dimension $N \times N$, then we know that we have $N$ eigenvectors. Lets say $N=2$, then we can write the whole system of equations down in a form like this
$$
M(\vec v, \vec w) = (M\vec v, M\vec w) = (\lambda_v\vec v,\lambda_w\vec w,)=(\vec v,\vec w)\left( \begin{matrix}\lambda_v & 0 \\ 0 & \lambda_w  \end{matrix}\right) =\left( \begin{matrix}v_1 & w_1 \\ v_2 & w_2  \end{matrix}\right)\left( \begin{matrix}\lambda_v & 0 \\ 0 & \lambda_w  \end{matrix}\right)
$$
It is clear that we can denote the whole system of equations using a coefficient matrix, instead of using a label for each individual eigenvector.
$$\begin{align}
MC &= C\Lambda \\
(M \left(\begin{matrix}C_{11} \\ C_{21} \end{matrix}\right), M \left(\begin{matrix}C_{12} \\ C_{22} \end{matrix}\right))&= (\lambda_1 \left(\begin{matrix}C_{11} \\ C_{21} \end{matrix}\right), \ \lambda_2 \left(\begin{matrix}C_{12} \\ C_{22} \end{matrix}\right))\\
&=\left( \begin{matrix}C_{11} & C_{12} \\ C_{21} & C_{22}  \end{matrix}\right)\left( \begin{matrix}\lambda_{1} & 0 \\ 0 & \lambda_{2}  \end{matrix}\right)={C\Lambda}
\end{align}$$
Note that all matrices involved have the dimension of $N\times N$. The overlap matrix would also have the same dimension. You get $N$ eigenvectors and each eigenvector has $N$ coefficients. "Putting" does eigenvectors side-by-side into a matrix yields the coefficient matrix with dimension $N \times N$. The details how the matrix elements are obtained are irrelevant to any of these points.
* It only looks like a linear eigenvalue equation, but the problem is nonlinear since the matrix elements of the Fock operator are functions of the coefficient matrix elements, which is why we have to solve the problem iteratively until self-consistency is reached.


text:
Is there an algebraic form for the textbook reaction coordinate curves?
So here is a curve often used in textbooks to illustrate a reaction coordinate.

It's a nice short-hand device to learn the relationship between kinetic and thermodynamic parameters that control the progression and equilibrium of chemical reactions.
I was thinking whether there is an analytical expression for that curve, one that I could use to draw several curves like that (using, say, Desmos) for distinct values of free energy contents of reactants and products and activation energies.

You know they are just some "pretty lines"? OK, there might be somewhere a case where curvature there is meaningful, but probably not in any textbook.

@Mithoron, yes I know. But I was looking for a way to systematically draw the pretty lines. Perhaps in a way that could be a tad more useful in an educational setting.

As nice as such a reaction coordinate looks like, it only is a projection (hence, a simplification) of a potential surface. Only with two degrees of freedom one can draw the later like a map.

See Wikipedia article on reaction coordinates, "https://en.wikipedia.org/wiki/Reaction_coordinate"

As has already been pointed out, the exact shape of these curves is usually meaningless. In practice you should calculate it (by calculating the energy $E$ as a function of some reaction coordinate $r$, which represents how a combination of bond lengths/angles changes over the course of a reaction), and then you can plot $E(r)$ versus $r$. But, If you just want the shape in this picture, I think you can define the energies $E(r)$ at the three points, and using that $\mathrm{d}E/\mathrm{d}r = 0$ at these points, fit it to a polynomial curve.

If the reaction is a simple one such as F+H2 then a LEPS potential can be used. This gives a 3D plot of the energy surface from reactants to products and the lowest path between them gives the potential curve you ask about. The LEPS potential is based on the Morse potential with values taken from experimental data. See this post for a plot of a potential https://chemistry.stackexchange.com/questions/170556/why-does-a-symmetric-stretch-mode-not-have-an-imaginary-frequency/170578#170578 Naturally there are any number of empirical mathematical functions that have the shape you show.

Here is a function that works: https://www.desmos.com/calculator/cqwgfj2nzt
The general function would be
$$ a \left( \frac{b}{1 - e^{5-x}} - \frac{c}{1 - e^{10-x}} \right)$$
I'm sure you could rearrange this such that the three parameters directly correspond to initial, activation and final energy, but as this is conceptual only anyway, I did not.
Here is an example graph (smoother than the one posted by the OP):

And here is a Desmos file with sliders.

Well now that's just showing off. I thought a quartic would do the job! :)

The classic would be a Bézier spline.

You can simply use matplotlib to plot a smooth line connecting points
import numpy as np
import matplotlib.pyplot as plt
from scipy.interpolate import interp1d

x = np.array([0.1, 0.3, 0.5, 0.7, 0.9])
y = np.array([0.57, 0.85, 0.66, 0.84, 0.59])

x_new = np.linspace(x.min(), x.max(), 500)

f = interp1d(x, y, kind="quadratic")
y_smooth = f(x_new)

plt.plot(x_new, y_smooth)
plt.scatter(x, y)
plt.show()

No so simple - maxima aren't in the right points.

@Mithoron Well, this is just an example that the OP can adjust later. If one wants the right maxima shown on the line, then just return the coordinate of the maxima from the interpolation and add text manually.

And by this way, we don't need to know any mathematical function.


text:
How does hydrogen and oxygen combine to make water?
When two hydrogen atoms fuse with one oxygen atom to create a molecule of water, each hydrogen atom donates its single electron to the oxygen atom, resulting in 10 electrons for the oxygen, instead of eight. But if the hydrogen atom no longer has its sole electron, is it still a hydrogen atom. Are we saying that there are some hydrogen atoms out there with zero electrons and some others with one electron? Is that allowed?

I took the liberty to edit the title. It seems to me your confusion is largely about a misunderstanding of what defines "hydrogen".

@BuckThorn I think it made it even worse. There was a big mismatch between the title and the rest.

I took a question from the body and turned that into the title. Currently the title is one very unfocused (broad) question and the body contains false premises and multiple additional related but at least more focused questions. I tried to make the post narrower and clearer. I think editing it back to the original title doesn't help at all.

You make some false assumptions, for instance that in water hydrogen atoms give away their electrons and oxygen ends up with 10 electrons. In molecules electrons are usually "shared". They don't "belong" to anyone. It is a meritocracy of sorts. Electrons may display a preference for some parts of a molecule, such as the regions between nuclei, so you could say they "hang out" there more.
If a hydrogen atom loses its electron then it is then usually referred to as a proton, although it might be something else depending on the hydrogen isotope involved.
IUPAC gave some attention to the issue of hydrogen nomenclature and provided guidelines:

Under IUPAC's nomenclature proposal, the prefixes hydro- and hydri- would be used to refer to the mixture of hydrogen isotopes that occurs in nature. Proto- and proti- would be reserved for the $\ce{^1H}$ isotope, either alone or as a group. Thus, the word proton would mean the $\ce{^1H+}$ ion exclusively and not $\ce{H+}$ in natural abundance. The $\ce{H+}$ cation would be called the hydron. Similarly, the $\ce{^2H}$ isotope would always be referred to using the prefixes deuteri- or deutero- (as in, for example, potassium deuteride,  $\ce{K^2H}$). The $\ce{^3H}$ isotope would always have the prefix triti- or tritio-.

The following summarizes the recommended nomenclature, adapted from a webpage on the subject (see also the authorative references [2-3]):

General
$\ce{^1H}$
$\ce{^2H}$
$\ce{^3H}$

Atom (H)
hydrogen
protium
deuterium
tritium

Cation (H+)
hydron
proton
deuteron
triton

Anion (H-)
hydride
protide
deuteride
tritide

Group (-H)
hydro
protio
deuterio
tritio

Transfer of cation to substrate
hydronation
protonation
deuteronation
tritonation

Replacement of hydrogen by a specific isotope

protiation
deuteriation (or deuteration)
tritiation

I remember a lecture from which my then supervisor left enraged because the speaker had repeatedly referred to hydrogen atoms as protons (I believe it was in the context of NMR, although the talk was about formulation). Despite the difference, the terms are used to mean roughly the same thing in some contexts, specifically in NMR, where it is the behavior of the spin associated with the $\ce{^1H}$ nucleus - the proton - that is most commonly interrogated, although deuterium NMR is very much possible.
References
(1) Nomenclature for Hydrogen Proposed. Chem. Eng. News Archive 1986, 64 (29), 27. https://doi.org/10.1021/cen-v064n029.p027.
(2) IUPAC-IUBMB Joint Commission on Biochemical Nomenclature (JCBN) and Nomenclature Committee of IUBMB (NC-IUBMB). Journal of Molecular Biology 1998, 275 (3), 527–537. https://doi.org/10.1006/jmbi.1997.1433.
(3) Bunnett, J. F.; Jones, R. A. Y. Names for Hydrogen Atoms, Ions, and Groups, and for Reactions Involving Them (Recommendations 1988). Pure and Applied Chemistry 1988, 60 (7), 1115–1116. https://doi.org/10.1351/pac198860071115.


text:
Is there any chemical that can react with two different chemicals to generate heat or cold?
For example. There are three chemicals. Chemical A,B,C

Chemical B react with Chemical A = heat
Chemical B react with Chemical C = cold

Is there any chemical available like this? And is it safe to use? Easy to create OR available at home/market?

Is this suddenly a lifehacks.se? https://en.wikipedia.org/wiki/Heating_pad https://en.wikipedia.org/wiki/Ice_pack#Instant_ice_packs

Yes there is, chemical B is water.
If we take ammonium nitrate as chemical A and then react it with chemical B by dissolving it then the temperature of the mixture will go down. The chemical reaction which transforms either crystalline ammonium nitrate or choline chloride into an aqueous solution of the ions is an endothermic reaction.
One of my academic interests revolves around strong aqueous solutions of sodium chloride and also strong solutions of choline chloride. I have found when I have been making solutions such as 6 M choline chloride in water that when I add the water the mixture cools far below room temperature.
What is happening is that the reaction is being driven forwards by the fact that the randomness (entropy, $S$) of the system greatly increases. This overcomes the energy (enthalpy, $H$) demand of the reaction.
$$\Delta G=\Delta H-T\,\Delta S$$
When $\Delta G$ is the change in Gibbs free energy for the reaction, when $\Delta G$ is negative then the reaction is thermodynamically able to go ahead without anything outside the system driving it forwards. The $\Delta H$ and $\Delta S$ terms are the changes in the enthalpy (energy) and entropy (randomness) of the system.
For an exothermic reaction between water and something else we have quite a lot of choices, both sodium hydroxide and concentrated sulfuric acid when mixed with water undergo exothermic reactions which release lots of heat.
So I would say that C should be concentrated sulfuric acid.
When I was young my grandfather told me a horror story about a man who attempted to dilute concentrated sulfuric acid in the hand basin in a toilet. The exothermic reaction caused the basin to crack and fall off the wall onto the floor. I would chalk this one up as something not to try either at home or in the "Gents". I also consider it something which female chemists should not attempt in the "Ladies".

I appreciate your response and in-depth knowledge in Chemistry. Thanks for the answer. I just want you to know that I want to create a pouch (those heat/cold shake pouches) but you can choose that you want heat OR cold pouch inside single pouch. No need two different pouches for different temperature. Well, I understand the risk behind dangerous chemicals and specially H2SO4 (experience lol) so I will check and research more about this. Thanks again


text:
On Le Chatelier's principle
According to my textbook,

Increasing the concentration of the reactants increases the rate of forward reaction only to reestablish equilibrium (Le Chatelier's principle)

But doesn't that also increase the rate of the backwards reaction as products are produced?

It does, to an extent. As both rates become equal again, the new equilibrium is established.

Yes, and both product and reactant concentrations are higher at the new equilibrium concentrations.  That means that both the forward and reverse reaction rates are higher at the new equilibrium position but are still equal.

Thanks I got it now

Please always include a proper citation when you quote a source.

Yes, you are correct. The equilibrium constant is just that i.e. constant (at a given temperature), thus if $A=B$ then $K=[A]/[B]$ so if A increases so must B.

@Martin-マーチン sorry I am new and still learning

@porphyrin I was thinking of how to proof that equilibrium constant doesn't get affected by increasing the concentration not the other way around

Ahmed, the equilibrium constant $K_e$ is the ratio of rate constants and these are determined by the properties of the molecules involved. $K_e= k_1/k_{-1}=[A_e]/[B_e]$ where $k_1$ is for  A forming B and $k_{-1}$ for the back reaction B forming A. The subscript 'e' indicates equilibrium concentrations. Have a look for 'kinetic approach to equilibrium' in a textbook.

What textbook did you quote from?

@BuckThorn the textbook of the Egyptian curriculum

@porphyrin thanks