text:
Two compounds (not stereoisomers) have same SMILES code
Are there compounds that have the same SMILES code but are not stereoisomers? I am trying to find inconsistencies with SMILES code and currently I have that many compounds have multiples options for SMILES code but I want to find if there is a SMILES code that can represent multiple compounds.

Did you check the SMILES to be of same type and origin? The three likely most frequently requested SMILES openbabel assigns are canonical, universal, and InChIfied SMILES (documentation); it only is a question of structure complexity of the molecule structures submitted if they differ (within one program), or in comparison to other programs (GUI like ChemDraw/ChemDoodle/Marvin, CLI like RDKit). Both obabel and RDKit allow SMILES to SMILES conversion for a consistent starting point for such a test, too.

Once on a solid ground, comparison with similar benchmarks/checks might be of interest for you, too.  There are (at least) the ones by Noel O'Boyle, John Mayfield, Roger Sale presented to 256th ACS National Meeting Boston Aug 2018 (slides), and the blog post Running a SMILES Validation Benchmark by 2020 by Richard L. Apodaca.

Many SMILES for single structures
There are many, many valid SMILES for most organic structures.  For example, CCO, OCC, and C(O)C are all valid SMILES for ethanol.
Many structures for single SMILES?
There are probably cases where the same SMILES can represent multiple compounds, but most of these probably involve transition metal complexes, which SMILES is (I think) widely regarded as not handling very well, or else cases where what is meant by "compound" is also ambiguous.  E.g. are "ethenol" and acetaldehyde are arguably both represented by C=CO, but in order to see that you need to know about keto-enol tautomerism.

The correct (re)assignment of double bonds in organic compounds indeed can be problematic. We need to talk about Kekulization, Aromaticity and SMILES by Noel O'Boyle, John Mayfield presented to the 254th ACS National Meeting in Washington, DC Aug 2017 provides an insight. For an identification of a molecule in a database, given the computational power available today, use of multiple descriptors (SMILES complemented by a second «key») can be helpful (a bit like two-factor authorization on email).

The SMILES corresponds to the underlying molecular graph (with bonds being edges and atoms vertices). This means the question you are asking is "are there any molecules that are different, not stereoisomers, and have the same underlying molecular graph".
The answer to this is yes. One typical example is atropoisomers (due to conformational barriers), though this can also be considered stereoisomery depending on how strict you are. This is actually relevant, for example in the kinase field some biaryls will show atropoisomerism. Another example is topological isomery. Imagine a trefoil knot C100 cycloalkane vs the ordinary C100 cycloalkane. Same graph, so same SMILES. However, in most typical uses of SMILES, these type of issues can be considered purely curiosities.


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Sodium Citrate / Sodium Phosphate Buffer Calculations
I'm trying to work through the calculations of a specific paper (linked here / below) but, my numbers are way off and I can't get them to match for the life of me. I need to make the same buffer but at a lower concentration. I'm not understanding how the paper was able to get the ratios (found in supplementary information) they used to make the buffer.
For example:
0.9M phosphate/citrate buffer, pH 3
Ratios used:  disodium hydrogen phosphate (0.87g/100mL); trisodium citrate dihydrate buffer (3.51g/100mL)
How were they able to calculate those ratios? I've tried using Henderson-Hasselback equation but maybe I'm utilizing it incorrectly?

Those components are not able to reach pH 3. pH would be > 7. // BTW, questions should be self-contained, answerable without following any link, that should serve just as a reference to content origin, or for further reading for better context.

Those components are not able to reach $\text{pH 3}$ without addition of an acid, as both salts gives solutions with alkalic pH values.

At such high concentrations near $\pu{1 M}$, pH calculations are off, unless going a long, long way of experimental determining activity coefficients (as these cannot be reliably predicted for such concentrations either).
pH of such buffers cannot be calculated from basic principles like via the Henderson-Hasselbalch equation. At the best it can be obtained from published empirical functions or tables of mixing ratios for desired pH.
If no such empirical help is available, the $\pu{pH}$ of such buffers must be adjusted experimentally, using pH meter calibrated on standard pH buffers.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


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Two questions regarding the fourth step of glycolysis
I am studying the 10 steps of glycolysis, and is currently on step 4. In this step, a molecule of fructose-1,6-biphosphate is cleaved into dihydroacetone phosphate (DHAP) and glyceraldehyde 3-phosphate (GAPD). I have two questions relating to the chemistry of this step:

Is this step considered a reverse aldol reaction? Since the products are a ketone (DHAP) and an aldehyde (GAPD), and the enzyme is called 'aldolase', I sense a strong hint that this is related to the aldol reaction.
If this is a reverse aldol reaction, why is the structure of fructose-1,6-biphosphate not similar to that of a typical aldol addition (or condensation) product?

I have tried to answer the these questions by attempting to work out the mechanism of the reverse direction of this step via aldol addition reaction, but was not able to come up with how a ring is formed after the addition product is made. Below is my picture:

I assume the conclusions to my questions could not be made because my mechanism is flawed. If so, which step had I gotten wrong? Would the role of the enzyme aldolase play a crucial role into answering my questions?
Thank you.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Could a PH3 or anything abstract a proton from a positive PH3?
I had this question for an SN2 reaction including 1-bromopropane treated with PH3, I answered that the product would be a propane alkane with a PH2 group. I was wondering whether this would happen.

There is not very much literature about this - not surprising considering how dangerous phosphine is to handle. It appears that RPH2 can be generated from phosphine + alkyl halide under basic conditions but R2PH and R3P are also formed dependent on the conditions paper here. Without the base you will get the phosphonium halide.  This article has more details JACS review


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Why is there a point of orbits along with quantization in case of energy gaps?
As per my textbook and online sources. The definition of Energy bands I have found is this:
A) Energy levels:
( In a simplified view of an electronic structure of a single atom or isolated molecule.
Electrons can exist in quantised energy states each with a discrete energy
value.)
Energy levels refers
to specific or discrete energy values in a quantised energy state that an
electron within a material can occupy.
B) Energy bands:
I)  In case of an isolated atom , electrons move in specific orbits around the nucleus and each orbit correspond to a discrete & definite energy level.In a solid material , atoms are closely packed together. Due to this , electron clouds of neighbouring atoms start interacting with each other.
ii)  This interaction causes the energy levels of the electrons in the same orbit to change slightly from their isolated atom values & electrons in the same orbit of a solid material to not possess a single, definite energy level anymore.
iii) Instead , The energy levels of these electrons spread out and form a continuous range of energies resembling a band rather than a single discrete energy level.
Therefore , Energy bands refers to the continuous range of
energies possessed by electrons in the same orbit of a solid material.
Q1: What the correct definition would be then ?
Q2: Why is there mention of  energy bands in relation to the orbits around the nucleus and also mentioning electron clouds ( a topic of quantum mechanics ) , quantisation. Isn't that wrong ? Because quantum mechanics rejects the theory of electronic moving around the nucleus.

Quantum mechanics applies as well to electron energy levels in solids (particularly crystals). And you may well encounter some solids in chemistry and other fields.

All statements are partially correct if you get rid of the classical term "orbit". Even if you use the term "orbitals", you will have some difficulty understanding solids, especially metals. The obvious transition from molecular orbitals to bands can be described by gradually increasing the conjugation of conjugated molecules. An atom is approximately described by the electrons in a spherical box (without any potential inside), if the box expands and the radius is large enough, the level will be closer and will form a band. An "orbit" is related to a spherical symmetry, while a solid tends to promote a translational symmetry. If you have no notion of QM, try to solve the case of  a particle in a box, you will get bands by increasing the size of the box.


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Why Entropy is not related to work but heat?
We know that entropy of a system only depends on heat energy (supplied  to or extracted from system) not on any other work (expansion  work + non expansion  work) by mathmatical expression ∆S = dQrev / T.
Now suppose we a closed container (whose walls are adiabatically insulated), with a movable piston, is filled with an ideal gas. According to defination to entropy we can't increase or decrease entropy of this system as it's  walls are thermally insulated so we can't supply heat. But if we compress piston , molecules  gets closer and when we pull piston , there is more space available for molecules. Do these changes not affect the disorder or randomness of molecules ?
1.If yes then why entropy doesn't change ?
2.If no then why randomness of molecules is independent of work ?
And also why entropy depend on temperature, means if I supply same amount of heat to system at two different temperatures, why entropy change is different in both cases ?

Note that S=f(T,p) // See also how-does-entropy-change-with-pressure // You should include in questions a short summary of what you already know or found (with eventual quotes/links), so others will not write what you are already aware of, or what you should know if you searched about it. If nothing was found, write at least what you have searched for.

Note that more proper is $$\text{d}S_\mathrm{sys} = \text{d}S_
\mathrm{rev} + \text{d}S_
\mathrm{irrrev} = \frac{\delta Q_\mathrm{rev}}{T} + \text{d}S_
\mathrm{irrev}$$

Do you really think that the only way for the entropy of a system to change is by exchanging heat?

@ChetMiller no I also think we can change entropy of a reactant by converting into product.

What about changing entropy by doing irreversible work?

@ChetMiller could you explain properly what are you saying ?

Are you familiar with the concept of entropy generation, say as a consequence of irreversible transport processes such as viscous dissipation, heat conduction, and molecular diffusion?

No , I just passed high school and the terms you have used I didn't even listen them in my course of thermodynamics in high school.

See also Hyperphysics - tree of Entropy cheat sheets

Try not to think of entropy as 'randomness' but as the number of ways  that energy levels in a molecule can be occupied or of the number of positions a molecule can have in space. (The 'spread' if you wish).
If the ideal gas expands adiabatically and reversibly, (infinitesimally slowly so equilibrium is always maintained) as you move the piston out then the entropy increases due to increasing the volume, i.e. more space available so more possible positions. However, the temperature falls because no heat can enter and work is done so entropy falls. These two effects exactly cancel so the entropy change is zero.
If the change is adiabatic and irreversible, i.e sudden change of piston, this balance no longer applies, the gas does less work and loses less internal energy and so has a higher final temperature than the reversible case, but the volume increase is the same. The entropy change is thus greater than zero.

Could you mathmatically prove that in adiabatic irreversible expansion , net entropy change  is positive ?

We know that it is always the case that the heat $q_{rev} > q_{irrev}$ and because the change is adiabatic then $q_{irrev}=0$ thus $q_{rev} > 0$ and as we use $\Delta S =q_{rev}/T$ to obtain the entropy change, it follows that $\Delta S >0$.

but in adiabatic reversible process also , q will be equal to zero.

Yes quite so, but it is a different experiment. In any adiabatic process $q=0$, suppose when reversible we go from $V_1,T_1 \to V_2,T_2$ and the entropy increase due to expansion and decrease due to gas cooling exactly balance and $\Delta S=0$. In the irreversible case we have $V_1,T_1\to V_2,T_3$ where $T_3>T_2$ because less work is done and the cooling is less and the balance is lost and so $\Delta S>0$ as explained.

A similar question it was omited the fact that the temperature increases in the compression phase because the gas molecules change direction more frequently after an elastic collisions $T~\propto \langle v^2\rangle$, the motion is disordered $\langle v^2\rangle$ and trivially the entropy increases. It's the reason why the entropy depends on the temperature in this case.
The amount of heat before and after compression has changed the pressure has increased. At the expansion phase this pressure relaxes the piston. The irreversible comes from the fact that after an inelastic collision between the particles and the walls, a portion of the heat is absorbed $Q_{irr}<Q_{rev}$ by the walls, creating new velocity states and an irreversible entropy is created, the work for the expansion is less efficient.
Note than an ideal gas is at high temperture $T\gg$ otherwise even the collision between the particles will be inelastic and an irreversible entropy will be created.


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Galvanic corrosion between copper, silver, nickel, and chrome plated brass
My PC uses a custom liquid cooling loop, in the loop currently I have copper (purity not specified by the manufacturer), silver (Nickel/.925 silver) and nickel plated brass fittings (no specifics from the manufacturer here either). The liquid I use is a liquid cooling "optimized" coolant (safety data sheet). 1 year and so far no issues.
I'd like to add a quick connect fitting to the system but it is chrome plated brass (the word chrome is used by the manufacturer support I contacted). With all these somewhat vague information available is there a way to tell if adding chrome to the system could cause galvanic corrosion? I contacted the manufacturer and this is what they told me:

... is chrome-plated and can not corrode.  You do not need to be worry as long as you do not use aluminium in the system

Good work on getting the coolant SDS! The liquid contains a corrosion inhibitor benzotriazole, and ethylene glycol antifreeze. That decreases risk of corrosion.
Is the new fitting to be soldered into the cooler, to be screwed directly into threaded metal, or to be connected via non-conductive plastic or rubber tubing?

If there is direct metal-to-metal, there is a quite small risk of galvanic corrosion at the fitting, which you might see as a greenish stain on the fitting, after some years.
If the fitting is to be connected via insulating tubing, then there is virtually no risk of galvanic corrosion, if the chrome plating is only on the outside of the part.

Note that there are different alloys of brass and bronze, and that brass, by itself, contains zinc that may slowly react with water. The life of the valve depends on its composition.
BTW, make sure all coolant lines and the radiator are mounted so that, in the event of a leak for any reason (e.g., mouse nibbling), the coolant does not pour onto the working electronic stuff. zap.

Thank you for answering, good news then. Only the screw in fittings will have direct contacts with different materials on the screw threads (nickel to nickel plating, nickel to brass/copper and nickel to chrome plating in my case) but those are unlikely to get into the fluid and cause residue somewhere else, otherwise only the coolant will be the "transmitting" medium with PVC soft tubing.


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Why is pyramidal structure observed in trisilylphosphine?
Why does trisilylphosphine $\ce{P(SiH3)3}$ (ICSD Entry: 72676, [1]) have a pyramidal structure while trisilylamine $\ce{N(SiH3)3}$ (ICSD Entry: 201428, [2]) is planar?

I think since silicon has a vacant d-orbital, it would form a backbond, wherein phosphorus shares its lone pair, as is the case with trisilylamine. This would result in $\ce{sp^2}$ hybridization, i.e. a trigonal planar structure.
However, the crystal structure confirms the pyramidal $\ce{P}$ atom, which is also preserved in the gas phase [1]:

Although IR and Raman spectroscopic results (Davidson, Ebsworth,
Sheldrick & Woodward, 1966) were consistent with trisilylphosphine being similar to trisilylamine in that the central atom was surrounded by a trigonal planar arrangement of $\ce{Si}$ atoms, electron diffraction studies (Beagley, Robiette & Sheldrick, 1968) established pyramidal geometry in the gas phase.
[…]
The crystal structure of trisilylphosphine may be
contrasted with those of trisilylamine and trimethylamine.
As $\ce{N(SiH3)3}$ has a planar heavy-atom skeleton and concomitantly lacks a stereochemically active lone pair (Barrow & Ebsworth, 1984), the absence of discrete intermolecular $\ce{Si\bond{...}N}$ contacts — seen in solid dimethylsilylamine (Blake, Ebsworth & Welch, 1984a) and chlorosilyl-N,N-dimethylamine (Anderson, Blake, Cradock, Ebsworth, Rankin &
Welch, 1986) — is unsurprising.
Like $\ce{P(SiH3)3},$ trimethylamine is clearly pyramidal $[\angle \ce{CNC} = 110.40(7)^\circ]$ but the absence of vacant d orbitals on the $\ce{C}$ atoms removes the possibility of these acting as acceptors
and the crystal comprises isolated molecules (Blake, Ebsworth & Welch, 1984b).

J.D. Lee Concise Inorganic Chemistry [3, p. 77] also specifically mentions that $\ce{P(SiH3)3}$ has a trigonal pyramidal structure:

In case of $\ce{P(SiH3)3},$ the internuclear distance is large and $\ce{P}$ atom has its own vacant d orbital. Hence the tendency to donate lone pair is very less and it adopts pyramidal structure and $\ce{P}$ atom is $\ce{sp^3}$ hybridized.

References

Blake, A. J.; Ebsworth, E. A. V.; Henderson, S. G. D. Structure of Trisilylphosphine, $\ce{P(SiH3)3},$ at $\pu{100 K}.$ Acta Crystallogr C Cryst Struct Commun 1991, 47 (3), 486–489. DOI: 10.1107/S0108270190009593.
Barrow, M. J.; Ebsworth, E. A. V. Crystal and Molecular Structure of Trisilylamine at $\pu{115 K}.$ J. Chem. Soc., Dalton Trans. 1984, No. 4, 563. DOI: 10.1039/dt9840000563.
Sudarsan Guha. J.D. Lee Concise Inorganic Chemistry, for JEE (Main & Advanced), 4th ed.; Wiley India Pvt. Ltd.: India, 2017 (reprint 2020). ISBN 978-81-265-9114-5.

also wikipedia explains the reasons for the geometry in P(CH3)3...https://en.wikipedia.org/wiki/Trimethylphosphine. Could possibly extrapolate to the trisilyl

@lazarus98 but since Carbon doesn't have any vacant d orbitals being a second period element and nor any lone pairs while forming trimethylphosphine I am unable to explain the reason why backbonding doesn't take place in trisilyl since that does have a vacant d orbital  and phosphorus has a lone pair that it can use up

@Substitute_Y The lone pair in phosphorus has mainly 3s character, therefore overlap with the 3d is likely poor due to differences in orbital energies. Therefore, back bonding is not significant for phosphorus.

Just read https://chemistry.stackexchange.com/questions/69568/hybridization-of-nitrogen-in-trisilylamine-sih33n instead of spawning some  "vacant d orbital" nonsense.

https://chemistry.stackexchange.com/questions/152243/when-to-consider-back-bonding-and-when-to-not-when-will-it-change-hybridisation

@Mithoron Nonsense?

@Lazarus98 Just read the linked answers. I dunno why anyone would even know some  trisilylphosphine exist at all, and still buy into stuff deprecated at least 30 years ago.

@Substitute_Y My previous explanationis is wrong. For the case of N(SiH3)3, although there is some amount of back bonding, it is not the dominate "driving force". Rather, polarization of the N-Si bond leads to a stronger electrostatic repulsion between -silyl groups that forces the geometry to be planar. Its not that the silicon d-orbital isn't involved in delocalization of the lone pair, its just that this is not what leads to the planar geometry. Nor is it the dominate mode of delocalization for the N lone pair.

Donating of the Nitrogen lone pair happens more through hyperconjugation to the silicon sigma * orbital.

Let us analyze the three cases:

$\ce{B(SiH3)3}$
$\ce{N(SiH3)3}$
$\ce{P(SiH3)3}$

We are going to look at the back-bonding aspect. It is expected that the structures will be more planer as the extent of back-bonding increases. In this case the $\ce{N(SiH3)3}$, $\ce{Si}$-$\ce{N}$-$\ce{Si}$ bond angle is 119.8 degrees. which is just 120 degrees meaning that the structure is, more or less, triangular planer.
First Observations
$\ce{B(SiH3)3}$
Boron has no lone-pairs to back-donate, or according to VSEPR theory, we expect it to be triangular planer, as predicted by ab-initio calculations.
$\ce{N(SiH3)3}$
Nitrogen has a lone pair of electrons that can back-bond with silicon. In this case the $\ce{Si}$-$\ce{N}$-$\ce{Si}$ bond angle is 119.8 degrees.
$\ce{P(SiH3)3}$
Phosphorous also has a lone pair of electrons that can back-bond with silicon. In this case the $\ce{Si}$-$\ce{P}$-$\ce{Si}$ bond angle is 101.8 degrees.
At RHF level of theory, not only is there more back-bonding in trisilylamine, it is close to triangular planer trisilylboron, or that the electorn pair is completely shared between $\ce{N}$ and silyl groups.
Back-bonding or absence of isolated p-orbitals on the central atom can lead to planarity.


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what does the oxidation/reduction potential actually mean?
The standard potential of a species is actually its tendency to gain/lose electrons and get reduced/oxidized. But what I want to understand : 1- Is the standard potential an absolute quantity, or a relative quantity? If relative, relative to what? , 2- what does the value of the standard potential actually mean. For example, the Standard Reduction Potential for lithium at 25 °C is -3.04 volts, does that mean if a slap of lithium lost a 1 Coulomb of electrons, this quantity of electrons would have, according to the definition of electrical potential, an amount of energy equal to 3.04 Joules?
Note: I'm a physics major trying to understand the redox reactions in a battery, and where did the Electrical potential difference caused by the battery came from.

Describe in the question the explicit summary result of your looking for info or answers before asking. You may then realize during the process you need not to ask any more. It prevents redundant questions/answers and you will not be offered info you (should) already know or do not want.  Well elaborated questions have higher probability to  receive useful answers.  See the guide How do I ask a good question?.

See also chem.libretexts.org - Electrochemistry_and_Thermodynamics and Gibbs_Energy_and_Redox_Reactions

And yes, your intuition is right. Note the charge versus molar amount proportionality constant F = e . Na as the charge of 1 mol of protons.

If you're a physics major you should have better understanding of current...

Suppose a platinum plate covered by platinum moss is dipped in a beaker containing a $1$ M $\ce{HCl}$ solution. Suppose now that some bubbles of hydrogen gas $\ce{H2}$ are crossing the acidic solution so that some touch the platinum surface. The whole is an electrode called hydrogen electrode.
Suppose now a metallic plate (like zinc $\ce{Zn}$, copper $\ce{Cu}$, etc.) is dipped into another beaker containing the solution of a salt of the corresponding metal (like $\ce{ZnSO4}$, or $\ce{CuSO4}$). This defines a second electrode, called zinc (or copper) electrode.
If now the two ionic solutions of the two beakers are joined by a bridge (for example blotting paper filled with a neutral solution like $\ce{KNO3}$), a voltage, or an electric potential $E$ can be measured with an electronic voltmeter between the platinum electrode of the first cell and the metal plate of the $2$nd electrode. This measured value is called the reduction potential of the ($2$nd) metallic electrode. For example, if the $2$nd electrode is zinc $\ce{Zn}$, the measured potential is $E_{Zn} = -0.76$ V (if the $\ce{Zn^{2+}}$ concentration is $1$ M). This value is tabulated and is a reference. If the $2$nd electrode is made of copper $\ce{Cu}$ dipping in a $1$ M $\ce{Cu^{2+}}$ solution, the measured reaction potential is $E_{Cu} = +0.34$ V. These values ($-0.76$ V, $+0.34$ V) are considered as constant, as soon as they are measured in a cell containing a hydrogen electrode as first electrode.
But take care. These reduction potentials are only valid when both ionic concentrations are equal to $1$ M. If the ionic concentrations $\ce{[Zn]}$, $\ce{[H^+]}$ and $\ce{[Cu^{2+}]}$ are different from $1$ M, the reduction potentials of $\ce{Zn}$ and $\ce{Cu}$ take the following values :
$\ce{E_{Zn} = -0.76 V + \frac{RT}{2F} ln\frac{[Zn^{2+}]}{[H^+]^2}}$
$\ce{E_{Cu} = +0.34 V + \frac{RT}{2F} ln\frac{[Cu^{2+}]}{[H^+]^2}}$
At $25$°C, the factor $\ce{\frac{RT}{2F}}$ is numerically equal to $0.0295$ V.
Now let's answer your question. The reduction (or standard) potential can be considered as an absolute value, providing it is coupled to a hydrogen electrode, and that the two solutions are $1$ molar.

Thank you so much, this is really helpful!


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Effective pH range in water. What is the definition
https://openchemistryhelp.blogspot.com/2013/02/solvent-levelling.html
I learned the concept of solvent theory and solvent leveling of acid and base,but couldn't understand what the graph below says

I thought since the pKa of ammonium ion(NH4+) in water is 9.24 and pKa of ammonia in water is 38, the line shows the interval between these pKa.
But as I searched for more information I found out that pKa of sulfuric acid in weter is -2.8, which doesn't correspond the graph.
Could someone explain what this graph means?

This graph is hard to understand, apart from water and ammonia. For example, the pKa of hydrofluoric acid is known to be $3.17$. It is not a strong acid. In the graph, the blue domain of HF goes from $-21$ to $-4$, which is typical of very strong acids. These limits have nothing to do with $3.17$. What does it mean? Same thing for acetic acid, whose pKa is $4.75$. In the graph, the blue domain of acetic acid (ethanoic acid) goes from $-2$ to $+12$. What is the meaning of this blue domain ?

@Maurice HF is rather exception. Very concentrated HF solution converges to becoming a superacid and liquid HF definitely is, due 3 HF <-> H2F+ + HF2^-.  H2F+ is acting acidic  agent of fluoroantimonic acid ( HF + SbF5)

@Poutnik. The abscissa does not speak of pH in concentrated solutions. It is "Effective pH in water".

@Maurice It was comment to your comment, not to the question. By other words, pKa of diluted HF in water is very different than pKa of liquid HF, because there is a side fluorine chemistry behind the simple HA <=> H+ + A-.

About the H2SO4, it is not about the acidity of H2SO4 in water, but about acidity of H2SO4 and H3SO4+ in pure sulphuric acid. Hammett function value for sulfuric acid -12 quite matches the range on the chart.

In my understanding "the effective pH in water" is nothing else but the Hammett acidity function, which is extension of the classical $\text{pH}$ quantity to non-aqueous solutions. The blue ranges are then available Hammett function values for the given protic solvent, similarly as there is range of available pH values for water solutions.
The respective dissolved conjugate acid/base pairs can have in other then water protic solvents such activity ratios that would hypothetically belong to the "effective pH in water", if such ratio should be achieved in water solution.
E.g. $\ce{HNO3}$ has reportedly $\mathrm{p}K_\mathrm{a} = -1.5$. If it was dissolved in glacial acetic acid(preferred IUPAC name over the IUPAC systematic name ethanoic acid) and the solution was adjusted in a way $\ce{HNO3}$ : $\ce{NO3-}$ ratio was 1:1, then the solution in acetic acid would have the "effective pH in water" equal to -1.5.
IF the conjugate acid/pair activity ratio belonged to the effective value out of the blue range, it would be the acid is stronger than protonated solvent and it would protonate the solvent. Or, the base is stronger base than the deprotonated solvent and would deprotonate it:
\begin{align}
\ce{HA(solv) + Solv(l) &<=> A-(solv) + HSolv+(solv)}\\
\ce{B(solv) + HSolv(l) &<=> BH+(solv) + Solv-(solv)}
\end{align}
The left edge of blue ranges maps itself to the case $\ce{HA(solv)}$ and $\ce{HSolv+(solv)}$ are equally strong acids.
The right edge of blue ranges maps itself to the case $\ce{B(solv)}$ and $\ce{Solv-(solv)}$ are equally strong bases.
The edge values can be considered as $\mathrm{p}K_\mathrm{a}$ of protonated(left) and not protonated(right) solvent.

@MaxW Thanks you for the edit. I am not sure if glacial acetic acid is the IUPAC preferred name for the pure acetic acid. I have meant that acetic acid as the compound name is preferred by IUPAC over ethanoic acid as the established non-systematic name.

Poutnik - I realize I went overboard on glacial acetic acid but I was trying to make it absolutely clear that you weren't referring to an aqueous solution. When I read your original post I absolutely understood you, but I didn't think the OP would. - Nice answer by the way...

@Poutnik Thank you for you thourough explanation about the graph. I didn't know about hammett acidity function. About the last sentence of your answer, is it correct to consider that the edge values are  pKa of protonated(left) and not protonated(right) solvent "in water"?

Well, yes and no. Yes in sense it matches the idea of pKa in water. No in sense the molecular entity in the solvent has often different pKa than the same in water. Typical cases are those with very different solvent interactions, like for HF. It is also related to the solvent autoionization constant HS + HS <=> S- + H2S+, defining acidity ratio of H2S+ versus HS.


text:
Help explaining drug packaging, "Bromocriptine Mesylate equivalent to Bromocriptine 2.5 mg"
API for Bromocriptine Mesylate is currently quoted by manufacturers as 2x the price per gram that we can get it already in pills and packaged in the retail pharmacy. This doesn't make sense to me... as mesylated salt form it has to be heavier than plain Bromocriptine thus would take more than 2.5mg of Mesylate to equate 2.5mg of Bromocriptine.
Am I missing something? Is Mesylate far more potent so a smaller amount is needed thus the quoted price is accurate? Over 100 quotations so far. My thinking is that the Mesylate salt form is much more highly absorbable so much less is needed. I have no idea how to do the math to determine how much of the Mesylate version to use if I want a 2.5mg equivalent to Bromocriptine itself.
Thanks

Mass ratio of equivalents of different drug forms and their price ratio are highly independent parameters, with the latter being merely business policy.

The one thing is a salt the other is not. They probably have a completely different solubility. So it might just be that. When you eat stuff, quite a bit won't go there where it should. I think from a theoretical and especially financial point of view, this cannot be solved. First you have to look at the active ingredient and the mechanism to get there. Then you look at manufacturing processes. And at the end you'll find that prices are marketing.

Bromocriptine is a natural hormone, produced by the brain. Its formula is $\ce{C32H40BrN5O5}$, with a molecular weight of $612$.
Mesitylate is another way of naming methanesulfonate $\ce{CH3SO3-}$. It is the anion of mesitylic acid $\ce{CH3SO3H}$, whose molecular weight is $96$.
Bromocriptine mesitylate is a compound made from $1$ molecule bromocriptine plus $1$ molecule mesitylic acid, and its molecular weight is $612 + 96 = 708$. The mass of bromocryptine mesitylate obtained from pure bromocriptine is equal to the mass of bromocriptine multiplied by the ratio  $708/612 = 1.16$.
Bromocriptine must be rather expensive, and mesitylic acid should be cheap in comparison.
So bromocryptine mesitylate is expensive because of its content in bromocriptine. The price of the mesitylate ion is small with respect to the price of bromocriptine. So the theoretical price of $1.16$ g bromocriptine mesitylate must be the price of $1.00$ g bromocriptine, plus a few percent due to the price of mesitylic acid and the yield of the synthesis. Roughly speaking,  $1.16$ g bromocriptine mesitylate must cost about $20$% more than $1.00$ g bromocriptine.
Summary : Within less than $10$%, $1.00$ g bromocriptine mesitylate must cost about the same price as $1.00$g bromocriptine

It could be the price of the pharmaceutical active ingredient that bromocriptine mesitylate is sold instead of bromocriptine itself.  The mesylate however may modify the ADMET properties, stabilize the wanted polymorph/crystal phase in the tablet better, fulfill an other requirement of formulation chemistry better, than the neutral compound, a hydrochloride, etc.


text:
Physical meaning of term T∆S
Plot
"I assume the role of an attentive observer, carefully watching a chemical reaction unfold within a sealed container. This enclosed environment, completely closed ( no matter but energy
can exchange) from external influences contains reactants  A and B in gaseous State { which reacts as A(g) + B(g) -> C(g) , and one mole of each reactant is present there}. To ensure a stable environment, a piston is affixed to the container, regulating the pressure, while thermally conducting walls maintain a consistent temperature. The system is connected to a machine which is required 50 kilo joule energy.
Goal
As I am outside of the box , I really don't know what is going inside. So I will try to understand what's happening there based on what I have learnt so far.
Assumptions
• Reaction is carried on constant pressure 1 atm and at 300 k.
• There is no ionization energy , electron affinity , and any other kind of energies involved  in reaction except Bond energies.
• Bond energy of molecule A = 100 kj / mol
• Bond energy of molecule B = 50 kj / mol
• Bond energy of molecule C = 200 kj / mol
• Entropy of formation of molecule A = Sa = 15 joule /mol
• Entropy of formation of molecule B = Sb =  5  joule /mol
• Entropy of formation of molecule C = Sc = 10 joule /mol
Story
Mathmatical Calculation
Okay first I calculate the some mathematical terms to predict that energy released by system is enough or not to start my machine. Let's try to find out enthalpy  of reaction ∆H.
∆H = -(energy released in formation  of C - energy absorbed to break the bonds of A and B)
∆H = -[200 - ( 100 + 50 )] = -50 kilo joule
Feeling relieved, as the machine will work with this energy, but upon measurement ( by some measuring devices ), I discover that the machine only received 47 kilojoules, 3 kilojoules less than what the system released. There seems to be something we missed. I will use my brain to investigate the issue.
Observation
After careful thought, I realize that the amount of energy lost is equal in magnitude to the temperature multiplied by the change in entropy, T × ΔS = 300K × 10 joules/K = 3 kilojoules. Thus, TΔS amount of energy is lost
An Appeal to the Audience
Can someone help me figure out why 3 kilojoules of energy is missing, which is necessary to restart my machine?
Question
"I would like to inquire why we cannot utilize the total enthalpy of reaction (∆H) to perform non-mechanical work (∆G). Specifically, I'm curious about why and where some of its energy (T∆S) is lost. Is it necessary to experience this energy loss (T∆S), or is there a way to prevent it?"
Why so much drama
The reason I asked this simple question in more complected way is that  many questions are unanswered , some  questions are closed and  some previous accounts are also deleted , don't know what is reason behind it so I tried a new way to ask a question , hope I will not recieve downvotes in this question.

@Poutnik  you are saying right  there are so many gaps not covered by me yet because I am a online student so I have not any option to ask doubts and clarify concepts via connecting to teacher offline and also in online I have not purchased any app I just study from YouTube so I can't ask doubts in online also.  I have already watched so many videos of high school on YouTube but couldn't find anything related to my doubts , I know there are some videos in which answer to my questions are explained but I can't understand anything after watching complete video because they use high mathematics.

@Poutnik sorry I have written isolated system instead of close let me correct it.

"just part of released enthalpy cannot be converted to work but remains as thermal energy" I am just asking about reason that why some part of enthalpy cannot be converted to heat

Hello Bharat Prajapat! I am glad you are an enthusiast. As a suggestion, get some money and buy a good book in thermodynamics. They are infinitely better than any video in YT, designed for entertainment and your attention. I will justify it. I can make a YT account and say nonsense about thermodynamics, and I am allowed. However, publishers that write about science, and write nonsense, they will have serious consequences. There is a reason those books are studied across the globe. I'll see if I can answer this later..

@MetalStorm "I will acquire sufficient funds to purchase a good book , that is not a problem. My main concern lies in selecting the appropriate book for studying thermodynamics at the high school level or a slightly more advanced level. Many recommended books tend to cater to university-level content, making it challenging for me to grasp due to their heavy reliance on complex mathematics. Nevertheless, I am eager to explore books beyond my syllabus and learn new concepts, as long as they provide a clear understanding without overwhelming mathematical complexities." Could you suggest anything?

@BharatPrajapat For high school level... I have no idea, sorry. However, if I were you, I would study real calculus in one variable and linear algebra, and do a lot of exercises. I personally think there is no short-cut, you need to know math.

@MetalStorm okay please suggest some good books of thermodynamics wheather they are beyond the syllabus of high school, I will try to read them.

For this system, we have from the 1st law of thermodynamics, $$\Delta U=Q-W$$From the 2nd law of thermodynamics, we have $$Q=T\Delta S-T\sigma$$where $\sigma$ is the amount of entropy generated by irreversibility in the process.  Combining these two equations, we obtain:  $$\Delta U=T\Delta S-T\sigma-W$$or$$\Delta G=-(W-P\Delta V)-T\sigma$$or$$W=P\Delta V+(-\Delta G)-T\sigma$$So the maximum amount of work that this system can do over and above the PV work is $(\Delta G)=53\ kJ  $.  So, the maximum total amount of work that can be done with the system going between the initial and final state is $$W=P\Delta V-53\ kJ$$If the amount of work is less than this, it is due to process irreversibility.  Part of the process irreversibility is due to the irreversibility of the heat transfer occurring with the average system temperature dropping lower than boundary temperature of 300 C (so there is a finite temperature driving force for heat transfer during the process).  Additional irreversibility is due to the reaction proceeding at an uncontrolled (finite) rate.  And, if the gas expansion and other work occurs very rapidly (due to the finite rate of the reaction), there will be additional irreversibility due to viscous dissipation.

Thank you so much for answer but I have some doubts 1. How did you calculated Gibbs free energy change ∆G = -53 kj.  2. "heat transfer occurring with the average system temperature dropping lower" why temperature is dropping lower ? And note that the process is a ideal and  reversible so viscous and other additional irreversibility makes no sense in this case so try to explain assuming it a ideal  and reversible process.

please answer what I have asked above

$\Delta G=\Delta H-T\Delta S=50-(-10)(300)=53\ kJ$.  I will correct the sign error in my answer.  Temperature in the bulk of the reaction mixture during the change drops lower than the boundary temperature because the change is occurring at finite rate and, in addition, $\Delta H$ is positive (meaning that heat has to be added irreversibly to the system during the process to try to bring its temperature back uo to the boundary temperature).  You can't just make the process reversible by saying that it is.  What are you doing to guarantee that the reaction proceeds at infinitesimal rate?

Hm, did not you get the Delta H reversed? Bond energies 100 + 50 -> 200 would mean reaction is exothermic. Delta_r H = -50 kJ/mol  ( or rather Delta U, otherwise we have to involve delta (pV), being it a reaction in gaseous phase)).

@Poutnik No  I got the same answer as the OP:  $\Delta H=200-(100+50)=50$, unless the bond energies are the opposite sign of the heats of formation?

It is (-200) - (-100 + -50) = -50, as  bond energies, expressed as enthalpies of bond formation, are negative.  By other words, breaking 50 kJ/mol bonds and 100 kJ/mol bonds, forming 200 kJ/mol bonds, releases 50 kJ/mol of energy. Unles I have missed something.

@ChetMiller I think you have misunderstood my question. Bond energy of C is greater than A + B so Amount of energy released in formation of molecule C (200kj) will be higher than it required to break the bonds of A (100) and B (50) and  system will release energy in this case.

So your calculation of $\Delta H$ has the wrong sign? (For an exothermic reaction, $\Delta H<0$).

@Chemistry yes sign of ∆H would be negative, but we are here not discussing about signs , so I wrote it  just 50 kj without adding + or - sign. But I think I should change sign , so that others can also understand it .

Note that at constant $T,p$,  the known formula:
$$\Delta G_\text{sys} = \Delta H_\text{sys} - T \Delta S_\text{sys}$$
divided by $-T$ gives:
$$- \frac{\Delta G_\text{sys}}{T} = - \frac{\Delta H_\text{sys}}{T} + \Delta S_\text{sys} = \Delta S_\text{surr} + \Delta S_\text{sys}=\Delta S_\text{universe}$$
Therefore, $\Delta G_\text{sys}$ related to the maximal amount of possibly done work without violating the 2nd law of thermodynamics which says in one of its formulations that the total entropy cannot decrease.

Response to feedback:

"just part of released enthalpy cannot be converted to work but remains as thermal energy" I am just asking about reason that why some part of enthalpy cannot be converted to heat.

It IS converted to heat. At least that rest of enthalpy that could be converted to work due the 2nd law of thermodynamics.

Further studying:

Libretexts: Entropy and The Second Law of Thermodynamics
Libretexts: Helmholtz and Gibbs Energies
Libretexts - organized source for studying, written in educative form
CH SE recommended resources for learning chemistry

thank you for answer. But you really didn't not understand what I was asking. I was just asking why there is heat transfer in this process in terms of T∆S , what is need of heat generation here ? Like expansion work  is done by or to the system to maintain pressure similarly heat exchange to maintain temperature. But in this case temperature is constant so no need to transfer heat or if temperature is changing in between the reaction then what is the reason behind of this change in temperature?

This is a rather unusual question: if the term 'entropy' had not been coined by scientists, implying a lack of understanding about the concept, how would we then calculate the heat transfer in this reaction?" I want to no the reason for heat transfer without using entropy, is it possible ?

at p,T const, H change means energy released and exchanged as heat  unless part of energy is used for nonvolume work. Theoretially, all that energy can be used for work but the T delta S part. That does not mean it is lost, just it remains as thermal energy and cannot be used.

" Temperature is constant....."  Since the process is reversible let's consider an infinitesimal amount dA and dB of reactant A and B use for production of  product dC. In this process temperature increased by dT and to maintain this temperature constant dQ amount of heat flows out of system.  Until the process is done this dQ heat will continuously flow and at the end net heat flown will be equal to ΣdQ = T∆S. Is it what you want to say ? But why there is a change in temperature dT in infinitesimal change (dA + dB -> dC) ? Could you explain it without using the concept of entropy  ?

Sum of dq = Delta H. You need not the concept of entropy for that.

The symbol dq I have used  here doesn't represent dH ( which represent net amount of energy released or absorbed by the system i.e. thermal energy [ heat dq ] +  chemical energy [ Gibbs free energy]) , which is sum of dq + dg . dq represent here only thermal energy used to maintain temperature , it doesn't  account for change in chemical potential u ( = dg / dξ ). So this dq (= t∆S )is only related to entropy.

G is not generally a chemical energy. It's synonym is free enthalpy. For isobaric process in closed system, enthalpy change is equal to sum of accepted heat and nonvolume work.

Let us continue this discussion in chat.


text:
If I burn a compound, it gives a specific color. So can I find out the elements of the compound using that color it burns with?
For example NaCl burns with yellow-orange colour. So can I find out the elements of the compound using that color(Assuming that I don't know the compound)?

Yes and no. If you put NaCl into a flame, like I did here, you will get the familiar yellow sodium D lines emission. More generally, though, you need to do spectroscopy, e.g., atomic emission spectroscopy or atomic absorption spectroscopy. Flame tests are too crude and not quantitative. As well, the chlorine from the NaCl will not significantly emit visible light (from excited chlorine atoms de-exciting). And note that sodium is an easy element compared to, e.g., uranium: https://chemistry.stackexchange.com/a/175028/79678.

@EdV I was going to post an answer, but I think you should do it instead.

There are a number of related relevant questions here, and answers too, e.g., https://chemistry.stackexchange.com/q/168454/79678. Search at this stack exchange for “flame test” and related terms. Several people here, e.g., @AChem, have posted good answers to relevant queries.

Some elements give very distinct flame colours. But many do not and distinguishing mixtures by colour is not remotely easy without proper instruments that look at the full spectrum of emission lines in the flame.

I wonder what colour osmium + O2 would produce


text:
Simplest existing tricarboxylic acid
I am wondering what is the simplest tricarboxylic acid that can be isolated in a pure form. Would it be methanetricarboxylic acid or ethane-1,1,2-tricarboxylic acid or maybe even longer chain is required to make the compound stable?

Ethane-1,1,2-tricarboxylic acid, $\ce{C5H6O6}$, is sold by Sigma-Aldrich and has experimentally determined properties including a boiling point of 278.5°C.
Honorable mention goes to aconitic acid (propene-1,2,3-tricarboxylic acid), $\ce{C6H6O6}$, which appears as a conjugate base in the biologically important citric acid cycle. Citric acid, $\ce{C6H8O7}$ (actually its conjugate base and conjugate bases of the acids to follow), is dehydrated to give aconitic acid and then rehydrated to give isocitric acid ($\ce{C6H8O7}$, but with the alcoholic hydroxyl group on a different carbon atom), which is oxidized and decarboxylated to continue the cycle.

I have heard esters and salts of methanetricarboxylic acid but I don't think the compound itself actually exists [citation needed?].

Looking up "methane tricarboxylate" on Google -> the triethyl ester turns up. Maybe placing theee carboxyl groups around one carbon atom is possible only if none is specifically COOH, which would decarboxylate.

This report mentions they detected methanetricarboxylic acid and that it's not available commercially. I guess it indeed decarboxylates if you'd try to make it pure, but is quite possible. Even methanetetracarboxylic salts exist, but I guess the thing breaks down very fast in low pH.

@Mithoron yes, I see this. As you mentioned, whether the acid can be purified (as the C5H6O6 compound is in the product quoted by Sigma-Aldruch) remains a question.


text:
The boiling point of carbon monoxide (CO) at 1 Torr
The boiling point of carbon monoxide (CO) at one atmosphere (760 Torr) is -191.5 degrees Centigrade or 81.65 degrees Kelvin. What is its boiling point at 1 Torr ? Or perhaps there is no liquid phase at 1 Torr ??

Try to find a detailed phase diagram for CO or N2, their properties are similar. The triple points are about 65K ; the vapor pressure of the solid falls drastically below the triple point.

Jim, That's a good Idea. There is probably no liquid phase at one Torr anyway. Stan

Google vapor pressure of carbon monoxide.


text:
Why does carbanion rearrangement not take place in aldol condensations?
I tried to find Aldol products for two molecules of propanal.
While doing the mechanism, I observed the formation of a $2$° carbanion, so I did a $1,2$-hydride shift to convert that into a 1° carbanion (as $1$° carbanions are more stable than $2$° carbanions)
but, while continuing with the $1$° carbanion, I get the wrong result for the product (hex-3-enal), and the $2$° carbanion forms the actual product (2-methylpent-2-enal). Why is this so?
Below attached is the image of my mechanism for reference showing both the 1° and 2° carbanion reactions:

Your equilibrium to a primary carbanion does not take place, only the secondary carbanion is formed. The carbanion formed is better drawn as the enolate anion.

It is an alpha carbanion not a secondary. The left column is correct. Since the carbanion reacts with the alpha carbon the enol structure might not be correct either. The structure is a resonance hybrid with dispersed negative charge. Aldol reactions are reversible equilibria and can make a mess if not run carefully.

You are comparing an isolated carbanion to a carbanion next to the carbonyl. The latter is much more stable and easily formed because of the partial +charge on the carbonyl carbon and resonance with the double bond. The alpha hydrogens are much more acidic. This is the property of carbonyl compounds such as aldehydes, ketones, esters, anhydrides, even nitriles, nitro compounds and the 'onium salts that enables aldol type reactions. These reactions take several chapters in most Organic Chem texts. [Get one, I recommend Morrison and Boyd]
Stability of carbanions: Your very first equations get you in trouble! There is no equilibrium between the primary and alpha carbanions; the energy difference is too large. The alpha carbanion has partial negative charge on the carbon and oxygen by resonance. [There is chemical equilibrium between the keto and enol forms of the molecule]. The primary carbanion is difficult to form but is stable and does not rearrange because the sp3 hybridization exposes the electron pair and it is strongly bonded to its counter ion. There must be no nearby Lewis acids for it to react with. If there were the carbanion would not have formed [your case above]. This lack of rearrangement extends to other electron rich transition states such as E2 and SN2. However, carbanions that are resonance stabilized lack the strong ionic or ion-paired bonding and can be multidentate.
Carbocations are the opposite. They are generally formed in a manner that solvates the anion leaving an intense +charge on the carbon. The nearest Lewis base is on an adjacent carbon atom. The electron pair will migrate in accordance with its basicity and the stability of the new carbocation with some geometric considerations.
Conclusion: Isolated Carbanions are difficult to form but their ionic bonds are stable if no Lewis acids are present; they do not rearrange. Conjugated carbanions as in carbonyl compounds are much more readily formed and can be polydentate. Carbocations are formed poorly solvated and subject to attack by adjacent electron pairs causing rearrangement to a more stable carbocation.
Finally these are my ideas on this. Study some texts for some specific examples and further discussion. look up the difference in acidity between alkanes and various carbonyl compounds.

Thanks for the answer. This is the book you're referring to, right? I think I've seen it in my school's library, I'll check it out.

The older editions might be just as good in describing mechanisms.

My library has the 7th edition, it's good. Thank you so much.


text:
Solubility of SrF2 in aqueous solution of NaF
This might be really simple question but I have no idea how to proceed to solve such kind of question.

The solubility product of $\ce{SrF2}$ in water is $\pu{8E-10}$. Calculate its solubility in 0.1M $\ce{NaF}$ aqueous solution.

I do have solution to this question, but I don't understand it.

$$\ce{NaF -> Na+ + F-}$$
Since $\ce{NaF}$ is strong electrolyte, the concentration of $\ce{Na+}$and $\ce{F-}$ is 0.1M.
$$\ce{SrF2 -> Sr^{2+} + 2 F-}$$
Let the concentration of $\ce{SrF2}$ be S. Then, the concentration of  of $\ce{Sr^{++}and F^-}$ is S and 2S respectively.
Now, $$K_\mathrm{sp} = \ce{[Sr^{2+}][F-]^2} \\ \pu{8E-10} = S (2S+0.1)^2 \\ \pu{8E-10} = S \times (0.1)^2 \\ S = \pu{8E-8 mole/L}$$

I don't understand why 0.1 is added to 2S and squared and later 2S in removed and how doing this exactly gives solubility of $\ce{SrF2}$ in $\ce{NaF}$. Can anyone help me in making me understand what is being exactly done in the solution and why does it gives the solubility?

OK, then do you understand the line before that? I mean, $K_\mathrm{sp} = \ce{[Sr^{++}][F^-]^2}$?

I read that it is ionic product of $\ce{Sr}$ and $\ce{F}$.

In the expresion $\ce{S(2S + 0.1)^2}$ the first $\ce{S}$ is for the concentration of $\ce{Sr^{2+}}$ which is dissolved. There are two ions of $\ce{F^-}$ for each atom of $\ce{Sr^{2+}}$. So $\ce{2S}$ is the concentration of $\ce{F^-}$ from the dissolved $\ce{SrF_2}$. However there is also 0.1 moles per liter of $\ce{F^-}$ from the $\ce{NaF}$. Thus the total $\ce{F^-}$ is $\ce{(2S + 0.1)}$.
Now you have to consider significant figures. Very little $\ce{SrF_2}$ is going to dissolve, so all the $\ce{F^-}$ essentially comes from the $\ce{NaF}$. Thus the simplification from $\ce{(2S + 0.1)^2}$ to $\ce{(0.1)^2}$.
You have to realize that in chemistry math there are limited significant figures and making reasonable assumptions greatly simplifies the math. Without the assumption that all the $\ce{F^-}$ essentially comes from the $\ce{NaF}$ you'd be trying to solve a messy cubic equation. Not a problem for a computer, but very very messy to do by hand. So in any kind of book problem it would be very very rare to need anything beyond a quadratic equation.

Thank you very much for the clear explanation. I would like to ask one more thing. Is $K_\mathrm{sp}$ for any electrolyte is constant for a given temperature in any solvent or is it different for different solvent? Like in this case, does the $K_\mathrm{sp}$ of $\ce{SrF2}$ changes in $\ce{NaF}$?

In general, the $\ce{K_{sp}}$ will change with temperature.  // The $\ce{K_{sp}}$ for aqueous solutions is only good for aqueous solutions. For example if the solvent was changed to methanol, then there would be an entirely different $\ce{K_{sp}}$. // This is getting more advanced but different fluoride salts would effect the equilibrium. That is because using concentrations of the species is a simplification. The equilibrium should actually be calculated using the chemical activities of the species.

Solubility and Ksp are different in different solvents. Ksp does change with changes in ion content in a solvent [or the activities of the ions change] Most salts become more soluble in higher ionic strength solutions than Ksp predicts.


text:
Liquification of ideal solution of two liquids from vapor phase
A and B form an ideal solution. In a cylinder piston arrangement, $\pu{2.0 mol}$ of vapor of liquid A and $\pu{3 mol}$ of vapor of liquid B are taken at $\pu{300 torr}$ and $T~\pu{K}$. At what pressure $30\%$ of the total amount of substance of vapor will it liquefy? Given $\ce{Pa^{.}}=\pu{300 torr}$,  $\ce{Pb^{.}}=\pu{600 torr}$, ($\sqrt{48.25}=6.95$). Kindly provide a conceptual and brief solution with an explanation.
The answer comes out to be $\pu{445 torr}$. This question is from a preparatory-test for The indian jee Advanced exam, And i think it can be a discussion. As i put my input into it already, i dont think this should be a homework-question, But instead, it raises a thought and creates new perspectives

You first. What have you done so far?

i tried raoult`s law, daltons law. Didnt get answer. I actually dont know what to do with external 300 torr presurre given in question.And Should i consider that after 30% of vapours are liqufied, now equillibrium between external pressure and internal pressure is established??

No.  The 300 torr initial pressure is increased to a higher final pressure.  You need to find out what that final pressure will have to be to end up with 30% liquid at the same temperature T.  I think that the initial pressure of 300 torr means that the initial vapor is not saturated.

got that..How can i proceed with equations now. Because thats where i lack now. My equations earlier had no use of given numerical estimation, which indicates i might be at wrong path. Can u please write them?

Please [edit] in your attempt of solving the exercise. Please also explain what ist actually given.

@om Makadia. 1) What is the meaning of Pa* and Pb* ? 2) What is the meaning of the number 48.25 which appears as a square root without any explanation ?

hey maurice, pa* and pb* are vapour of pure A and pure B. And there is no meaning to any number. These are approximations one has to use while solving, So that u dont have to calculate the sqaure root yourself. Also, u aint getting a calculator in indian exams. So these approximations are given to u.

I have tried for a week everyday to solve this question and then i have came here. Why does it close my question??

And i think it can be a discussion.  // Note that, in contrary to various web forums and other interactive platforms, the format of the StackExchange sites is being question + answers site, they are not primarily intended as discussion platforms. There do exist chat rooms, but they are separated form  the particular Q/A pages. // Comments are not intended for discussion either, they can be deleted or moved to a chat room.

$L$ is the amount of substance of liquid
$V$ is the amount of substance of vapor
$x$ is the mole fraction A in the liquid
$y$ is the mole fraction of A in the vapor

Total amount of substance:
$$L+V=5\tag{1}$$
Amount of substance of A in liquid plus vapor:
$$Lx+Vy=2\tag{2}$$
Raoult's Law:
$$y=\frac{300x}{300x+600(1-x)}=\frac{x}{2-x}\tag{3}$$
Fraction of total amount of substance that is liquid:
$$\frac{L}{L+V}=0.3\tag{4}$$
So, from Eqns. 1 and 4, we have $L = 1.5$ and $V = 3.5$.  Substituting these results and Eqn. 3 into Eqn. 2 gives:
$$1.5x+\frac{3.5x}{2-x}=2,$$
which leads to the quadratic equation $$1.5x^2-8.5x+4=0.$$
Solving this quadratic equation for $x$ using the quadratic formula:
$$x=\frac{8.5-\sqrt{48.25}}{3}=0.517$$
This gives $$P=300x+600(1-x)=300(2-x)=\pu{445 torr}.$$

you Are a GOAT ...... I was stuck for almost a week....

thanks martin too.✌


text:
Does seawater foam more when its warm? And if so, why
I work at an oil rig out in the sea.
We are pumping seawater into the rig, and back out again after it has gone through the system.
Its really warm outside now, and when we are pumping seawater into the ocean it foams much more than usual.
The technical department think something is wrong with either the equipment we use, or an imbalance in the chemicals.
The old veterans onboard says this always happens when its warm and its nothing to worry about.
Can anyone elaborate?

Pure water can’t foam, but seawater is far from pure and maybe warm seawater has more plankton and other organic debris in it that would facilitate foaming. Just my two cents.

Pumping seawater into the ocean? Then it's not exactly seawater any more, just some waste?

As @EdV states, check the coastal water report for algae blooms. The foam is likely from organic byproducts. If you're near the estuary of a river, it could also be due to organics released from industry, agriculture and other sources after heavy rains, more likely in warmer weather.

How long does the water reside on the rig before it gets pumped out. Is it kept at room temperature or used for cooling purposes? Are you pumping it in from the surface (where the temperature would vary more) or from deeper down (where the temperature would vary less)?


text:
Temperature-dependent pressure equilibrium constant of the heterogeneous water-gas shift reaction
In the course of my research, it became necessary to perform thermodynamic calculations involving chemical reactions. One of the chemical reactions is the heterogeneous water-gas shift reaction
$$\ce{C(s) + H2O(g)<=> CO(g) + H2(g)}$$
and I need the temperature-dependent (pressure) equilibrium constant for my calculation. However, I could not find any information about it on the net or in the literature at hand. If anyone has an experimental or empirical equation, I would appreciate it if you could provide me with the source as well!

Look for standard formation ΔG, ΔH,  and ΔS and calculate the constant from it for given T. For more accurate calculations, you may involve molar heat capacities, as the above partly depend on T too.

Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.

@Poutnik Thanks for the edit and comments! I tried to find each value you provided but could not find the value for each temperature.

The point is, you need not to find it for each temperature.

@Poutnik My understanding is insufficient. How can I get the equilibrium constant?

this might help: https://www.researchgate.net/figure/Equilibrium-constants-for-WGSR-Twigg-1989_tbl5_228506095 It does trigger on the keywords.

Related: https://chemistry.stackexchange.com/q/154276/4945

@Martin-マーチン Thank you for your comment! I looked at the URL you provided, but could not understand the calculation procedure for my problem...

You may [edit] your question (as long as there is no answer) with any and all relevant details. What have you tried what have you read, where did you get stuck. Be precise and specific about your problem and other may help. I'm probably not the one, as I only searched superficially. This comment section is getting a bit chatty for its purpose, so better enhance the questions with the things you've learned.


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Why have my recent boiled eggs contain water within the membrane?
My wife recently boiled eggs from the grocery. Every single one of the eggs has contained water with in the membrane where the hollow portion of the egg is normally. I have only seen this if maybe the egg is cracked and exposing the inside of the egg to water. I did read from this question here that the normal water that is inside of the egg is mixed with the broken down proteins of the egg and made into a gelatinous-type structure that is the boiled egg.
My concern is that there is a chemical composition change that happened that is preventing the egg  from doing this. To what end, I have no idea, but it is concerning enough for me to propose this question.

If it's fresh, then I see no problem. Also Cooking.SE would be more appropriate here.

Only reason I did it here is the linked question is here as well

… normal water that is inside of the egg is mixed with the broken down proteins of the egg and made into a gelatinous-type structure

The raw egg white is, aside of dissolved salts and small organic molecules, a colloidal solution of water soluble protein from the albumin protein group. This protein gets denaturated during egg boiling by regrouping its 3D structure, forming wit water a white gel.
I assume the eggs got boiled properly, so the most of egg while formed the gel as usually, save the cavity usually filled by air. I also assume you have excluded possibility of macroscopic or microscopic ruptures, allowing filling it by external water.
The option of changing chemical structure of the egg white is not probable, unless the egg is really spoiled.
The remaining option is the membrane, separating the egg content and the air cavity is corrupted, weakened or more permeable, especially during boiling and increased internal pressure.  This may lead to a semipermeable membrane transfer of the albumin serum – solution of ions and small molecules, but without big molecules as proteins – that may resemble water.


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How to distinguish silver chloride, bromide and iodide?
When we test for chloride, bromide and iodide ions we use the obvious silver nitrate and dilute nitric acid test. However distinguishing between the precipitates is often quite tricky. How can we tell the difference between a white, cream or yellow precipitate respectively. I understand we could use aqueous ammonia for the white precipitate, and that if it dissolved it would indicate the presence of chloride ions. However, the next two are much more difficult given my circumstances. With my practical test only aqueous ammonia is supplied, so I do not have access to concentrated ammonia. So what should I do now to distinguish between the remaining bromide and iodide ions?
Are there any tips such as holding white paper behind it to help me distinguish the two?

Run the reactions on known samples and carefully observe. Then compare the unknowns. Science is based on careful observation.

See these images: https://upload.wikimedia.org/wikipedia/commons/thumb/5/5f/AgBr_Sample.jpg/330px-AgBr_Sample.jpg and https://upload.wikimedia.org/wikipedia/commons/thumb/c/c0/Chlorid_st%C5%99%C3%ADbrn%C3%BD.PNG/330px-Chlorid_st%C5%99%C3%ADbrn%C3%BD.PNG . Also note solubility info in Wikipedia.

All lab supplied ammonia is aqueous ammonia around 25%, unless you mean gaseous ammonia, that turns to liquid  in deep frost or under pressure.

Aqueous ammonia should do the trick. Silver chloride is soluble in dilute ammonia solution. Silver bromide is sparingly soluble in dilute ammonia solution. Silver iodide is very slightly soluble in concentrated ammonia solution (so insoluble in dilute ammonia solution). Silver bromide/iodide is off-white as compared to silver chloride, but you can easily distinguish them if put side-by-side. Silver iodide has a slightly darker tone than silver bromide.
If you are confused with the silver nitrate-ammonia test, you can opt for sulfuric acid which gives you a more definite result. Chloride will give hydrogen chloride gas (pungent smell), bromide will give bromine gas (reddish-brown vapor), iodide will give iodine gas (purple vapor).

Nilay Ghosh's method is excellent, but it does not work if the original deposit is a mixture of the three precipitates $\ce{AgCl, AgBr, AgI}$. In this case, it is first necessary to get rid of the silver from the mixture of precipitates, and look for $\ce{Cl-, Br-}$ or $\ce{I-}$ in the residual solution.
The best way to do it, is to add $3-5$ mL $1$ M $\ce{H2SO4}$, then about $0.5$g zinc powder. Wait for half an hour, filtrate the mixture $\ce{Zn + Ag}$. Save the obtained solution and test the presence of $\ce{Cl-, Br-, I-}$ ions in solution.
Anion $\ce{I-}$ can be detected by adding $1$ mL hydrocarbon (like cyclohexane) and adding $1-2$ drops $0.1$ M $\ce{NaNO2}$. The hydrocarbon turns violet or purple in the presence of iodide.
Bromide $\ce{Br-}$ can be detected by first destroying iodide ions by first neutralizing the acidic solution by $\ce{Na2CO3}$, then adding enough $\ce{KMnO4}$ solution (which produces $\ce{MnO2 + I2}$ precipitates, but $\ce{KMnO4}$ gives no reaction with bromide), filtrating, and testing the filtrate by adding two drops of $\ce{Cl2}$ in solution (chlorine water). Bromide is transformed into bromine $\ce{Br2}$. Adding now $3$ mL cyclohexane, it becomes brown if bromine is in solution.
Chloride ions are detected in the original message by Gotfried Leibnitz.

I have fond memories of doing a lab like this back in high school. Our teacher said we could bring whatever we wanted from home to help with the lab. My idea was to bring in a 9V battery and two pieces of graphite from a pencil. Stick the graphite leads in a tiny 96 well plate of your test solution and run current through them. If chlorine is present you will observe the solution turning slightly yellow-green due to the presence of chlorine formation. If there is bromine or iodine present you will see a brown color change due to the presence of bromine and tri-iodine ions. It is possible to tell the difference between the electrolyzed bromine and iodine on their own, but if you want to be extremely sure add some starch to complex with the tri-iodine and you'll see a much darker blackish blue color if there is iodine compared to bromine. Hope this helps!

I believe that silver bromide, chloride and iodide have significantly different colours with silver chloride is white crystalline solid , silver bromide is somewhat pale yellow and silver iodide is bright yellow  , therefore physical examination must be enough for the required distinction


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What is the free radical mechanism of Ruff's degradation?
There are very few resources on the mechanism of Ruff's degradation of carbohydrates, which adds $\ce{Br2}$ to oxidize aldehydic group of reducing sugars like glucose to carboxylic acid and then adds ferric ions $\ce{(Fe^3+)}$ with hydrogen peroxide $\ce{(H2O2)}$. Most sites simply show a concerted mechanism involving salt of $\ce{Fe^2+}$ and the sugar, however this isn't the proposed mechanism I'm searching for.
I want to know the free radical mechanism of Ruff's degradation involving OH free radicals. The mechanism should probably be similar to Fenton's reagent which uses ferrous $\ce{(Fe^2+)}$ ions with $\ce{H2O2}$.

If very few resources then you should refer to them for other not to duplicate your prior effort.

Please quote (if allowed) and cite resources for the concerted mechanism. Why do you believe it needs to be a different mechanism?

Oxidation to the aldonic acid with bromine and water is an ionic redox mechanism.
Then due to the basic conditions from the counter anion of the iron3 salt, the carboxylic acid is deprotonated.
The Iron 3 can then oxidise the carboxylate oxygen to a radical, enabling the decarboxylation via free radical mechanism to leave a tertiary carbon radical.
As the peroxide O-O bond is weak it spontaneously undergoes homolytic cleavage and so the OH radical can  hence terminate with the carbon radical to produce the alditol.
Chemoselectively, the hydrogen peroxide will oxidise the C1 alcohol to produce the corresponding aldose.


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How does a silver chloride electrode (Ag/AgCl) convert the potential in aqueous solution to the wire?
I am aware of the redox chemistry that happens at the electrode surface, but I'm not sure how exactly a change in electrical potential in the solution would change the potential in the electrode?

The best explanation I found is the beginning of chapter 5 of the textbook "Medical Instrumentation: Application and Design".

Before raising a question, check if the initial premises are true, otherwise it would be climbing a ladder leaning at a wrong roof.

Note that the potential of any electrode is considered relative to present or assumed SHE, both floating wrt the the potential of the solution.

It all come down to attractions of electrons to various nuclei. In the case of electrodes there must be intimate mixing of all components so that the transfer of electrons is kinetically facile. It tends to be an art. A potential develops between the cathode, reduction, reaction and the anode, oxidation, reaction and if there is electrical connection current will flow allowing the reactions to occur. The single electrode potential between the Ag and Ag+ ions has been difficult to measure. [So far, I think, impossible.]


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Why are HFOs like R-1234yf considered okay (for now) but PFAs are under attack? Don't they both contain strong Carbon-Fluorine bonds?
The popular (and professional) press is awash with alarms about some so-called 'forever' chemicals, the PFAs...
The older, ozone-destroying CFCs and HCFCs, as well as global-warming-inducing HFCs, and now the world-saving HFOs, all contain strong C-F bonds, just like the poly- and per- fluoroalkyls....
So why are only the latter considered problematic for possessing long-lasting CF bonds?
Am I missing something?

You should substantiate your first sentence, remove some of the snark, introduce abbreviations and explain what "the latter" actually refers to. And a yes or no question isn't a good fit for our site.

Perfluoroalkoxy alkanes (PFA's) are generally liquid or solid at STP, and stay in ground water where they can be absorbed. Organism concentrate them in tissues (bioaccumulation), and the concentration is further increased up the food chain (bioamplification). The fluorocarbons used as refrigerants (and previously as aerosol propellants) are gaseous, and don't hang around at Earth's surface to be swallowed; they are quickly diluted by air (where they cause other issues in the upper atmosphere).

Since PFA's contain bonded oxygen and/or sulfur, the molecules are more likely to have biological activity than the more inert halocabons. They have known health effects, as evidenced by scientific research. e.g., a study at Harvard, the IARC, and these articles cited by the NIH.
a.  They have been implicated in hypercholesterolemia by acting as peroxisome proliferator-activated receptor (PPAR) agonists.
b. They have been implicates in hypothyroidism by decreasing thyroid peroxidase, resulting in decreased production and activation of thyroid hormones.
c. They have been implicated in  liver adenomas, Leydig cell tumors of the testis, and pancreatic acinar cell tumors, possibly due to increased PPAR alpha activation in the liver which increases hepatic aromatase concentration and subsequent serum estrogen levels.


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Discrepancy between calculated and experimental enthalpy change of formation? Hydrogen Iodide
Afterthought: See edits below!

$$\ce{1/2 H2(g) + 1/2 I2(g) -> HI(g)} \qquad \Delta_\mathrm f H^\circ=26.5\ \mathrm{kJ\ mol^{-1}}$$
source: https://atct.anl.gov/Thermochemical%20Data/version%201.122/species/?species_number=47
However, from bond enthalpies (exact! though I do not know at what temperature)
$E(H-H) = 436 \ \mathrm{kJ\ mol^{-1}}$
$E(I-I) = 151 \ \mathrm{kJ\ mol^{-1}}$
$E(H-I) = 299 \ \mathrm{kJ\ mol^{-1}}$
Therefore, $\Delta \mathrm H = \{436 + 151 - 2(299)\}/2 = -5.5 \mathrm{kJ\ mol^{-1}}$
This page also gives a negative enthalpy of reaction....
https://chem.libretexts.org/Courses/University_of_Kentucky/UK%3A_General_Chemistry/13%3A_Fundamental_Equilibrium_Concepts/13.3%3A_Shifting_Equilibria_-_Le_Chateliers_Principle
Numerical agreement is not expected, however, the difference in signs would lead one to wrongly conclude whether the reaction is endothermic or exothermic, a problem when trying to apply Le Chatelier's principle....
What's wrong?
Is this significant: Does the bond dissociation enthalpy vary with temperature?

EDIT:
Sorry! Wrong equation! Let me clean up my mistake!
$$\ce{1/2 H2(g) + 1/2 I2(s) -> HI(g)} \qquad \Delta_\mathrm f H^\circ=26.5\ \mathrm{kJ\ mol^{-1}}$$
Enthalpy change of formation: iodine starts off as a solid!
$$\ce{I2(s) -> I2(g)} \qquad \Delta H_{sub} = 13056 \ \mathrm{cal\ mol^{-1}} \approx 54.63 \ \mathrm{kJ\ mol^{-1}}$$
Determination of the heat of sublimation of iodine: A physical chemistry laboratory experiment
https://pubs.acs.org/doi/pdf/10.1021/ed064p551
for $\ce{1/2 H2(g) + 1/2 I2(g) -> HI(g)}$, $\Delta H = \Delta H_f - \Delta H_{sub} = 26.5 - 54.63/2 \approx -0.815 \ \mathrm{kJ\ mol^{-1}}$
In fact, the reaction is exothermic!
$$\ce{H_2 (g) + I_2 (g) -> 2HI(g)} \qquad \Delta H_r (0K)  \approx - 9.5 \ \mathrm{kJ\ mol^{-1}}$$
https://atct.anl.gov/Thermochemical%20Data/version%201.122/reaction/?&species_number[]=1&species_number[]=47&species_number[]=60&c1=1&c2=-2&c3=1

By the way, this question was inspired by the following CIE A-levels problem: Chemistry 9701 October/November 2021 Paper 2 Variant 2 Question 2 (9701 Oct Nov P22 Q2)

There are playing roles component heat capacities, and their different quantization due different masses.

By other words, H2, HI and I2 have at the same temperature the same mean translational energy, but due rotation and vibration quantization the different mean rotational and vibrational energy. I do not say it counts for all the difference, but it is a contributing factor.

Based on a 1983 PhD thesis, the first value (positive) seems more likely.

A 1977 paper has similar values for HI in the vapor phase, and also gives the (negative) value for HI in the aqueous phase. The internet has a wide range of values, i.e. probably did not catch up but is copying from older textbooks and problems sets.


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Creating Chloramines
I'm currently conducting some testing related to indoor pool environments.
Here I believe, airborne chloramines are responsible for stress corrosion cracking in austenitic stainless steels.
Unfortunately, I do not know how to generate the said chloramines for testing purposes.My setup so far includes a sealed container in which I can suspend samples over hydrous chlorine (~ pH 7.4) similar to pool areas. This container also has a clear lid with reflective panels aimed at it, to act similar to a box-type solar cooker (adding heat to increase humidity and accelerate corrosion).
From my basic research chloramines are measured as free (elemental) chlorine plus ammonia compounds (urea/carbamide). In my basic (pardon the pun) understanding, I thought the addition of hydrogen peroxide (urea) or Cloudy Ammonia would create chloramine(s). However, I've now found a thread that states hydrogen peroxide REMOVES chloramines from water.
Send help! (I know chloramines in pools are created by bodily fluids, please don't make me sh!t in a bucket at work...)

Sorry I can not help with your queston, but what you mean by cloudy ammonia?

'Cloudy' ammonia is an off-the-shelf product used for cleaning purposes. Honestly have no idea what percentage "ammonia" or how it works as a solution. For the purposes of this test, we trying to make things corrode, so a little rough-and-ready is acceptable, so long as we know we've got some of the right gasses going.

Keep an eye on safety: https://chemistry.stackexchange.com/questions/42121/is-mixing-ammonia-and-bleach-spray-in-the-same-parcel-of-air-considerably-danger ... https://chemistry.stackexchange.com/questions/47121/mixing-bleach-and-ammonia/126824 ... https://chemistry.stackexchange.com/questions/161945/does-hydrogen-peroxide-remove-chloramine-from-water/162006

@PAEP https://www.differencebetween.com/difference-between-ammonia-and-vs-cloudy-ammonia/amp/

I think I've figured it out, whilst not a complete answer it's got all the important bits. This summary of chlorine action in pools has really pulled my admittedly simple research together. Looks like I just need to get some aqueous ammonia and we should be off to a good start.

Seems like I was barking up the wrong tree on the Hydrogen Peroxide front. Or I've just confused myself and they're kind of the same (still unsure how 'urea' fits in TBH)


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If I make CO2 by exposing CaCO3 to HCl, how much of it would be dissolved in the water
For an experiment we're trying to measure the amount of gas produced in this reaction. But we always find less gas then expected, then a colleague told me that it could be because some of the CO2 produced might get dissolved in the water/HCl solution. This reactions takes place at 310K. The reaction takes place under normal pressure in closed containers, when the pressure in the container reaches over 100mbar the container will release ~50mbar of pressure.
We add 0.2 grams of CaCO3 to 100ml of 0.1M HCl. The 0.1M HCl is dissolved in pure water. This should generated 0.002M of CO2, and if I did my calculations right that should be about 51.02ml of gas. But everytime we measure we get around 30% less. Is the difference there because of the CO2 that has dissolved in the water or is there anything wrong with my calculations? Tips and advice are welcome.

You have not shown any calculations yet... It should not be a problem to involve easily found solubility of CO2, which is  1.45 g/L at 25 °C (77 °F), 100 kPa.

I have found that solubility, but I couldn't figure out how to calculate the solubility for 37C and pressure between 50 and 100 mbar. I found something about Henry's Law, but I couldn't get my head around it. I'll try to formulate my problem better next time. Chemistry is not my subject of expertise.

I assume there can be found data of solubility at different temperatures see this. Even if not, you can see significant portion of CO2 would dissolve. // Exact portion cannot be calculated, as due reaction dynamic, solution can be locally saturated, forming bubbles, while rest is yet not saturated. // Gas solubility is proportional to pressure.

Thank you, that is a useful table. I think the best I can do is try to saturate the water before the reaction to make sure it's as accurate as possible in gas production. Thank you for your help.

Certainly $\ce{CO2}$ dissolves in water, and the experiment design allows an over-pressure of 50 mbar, so cannot go to completion, since $\ce{CO2}$ is still present. You've made club soda, or seltzer, if you prefer. Considering the temperature and dissolved solids, this might be a fair model of $\ce{CO2}$ in mammalian body fluids, e.g., ~25 millimoles per liter (mmol/L) in human blood. Ah! Read Michael Crichton's Andromeda Strain.
To find how much is dissolved in the solution, look at tables such as this. Note that the bicarbonate ion is important: "Calcium carbonate reacts with water that is saturated with carbon dioxide to form the soluble calcium bicarbonate.  $\ce{CaCO3 + CO2 + H2O -> Ca(HCO3)2}$.  $\ce{CO2}$ solubility is influenced by the carbonate-bicarbonate buffer system.
However, check molar quantities of reactants and see if one limits the reaction!

There is 10 mmol HCl and 2 mmol CaCO3, 6 mmol HCl, resp 0.06 M HCl would left. Not much of HCO3- there.

Thanks for the sources, I wil look those up. The experiment is designed to have an abudance of HCl so that most/all CaCO3 will react. As I'm not a chemist but a Biotechnician, can I ask if the Calcium Bicarbonate will also eventually react and produce CO2, or are those innert?

Unless one or more of the products are removed, e.g., CO2 going into air, or a solid precipitating out, reactions can proceed in both directions, eventually reaching equilibrium where the amount of reactants going equals the amount of product reverting to the reactants. So, yes, the bicarbonate only exists in solution, in equilibrium with carbonate, bicarbonate and dissolved CO2. See https://en.wikipedia.org/wiki/Calcium_bicarbonate That 50 mbar CO2 pressure prevents it's complete loss. This seems like a good but simple model of body CO2 chemistry!


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Efficiency of Gas Liquid Reactions Using Different Methods
I am looking to react MEA (monoethanolamine) with $\ce{CO_2}$. I was planning on doing this reaction by bubbling atmospheric air through water by using an air pump. I have researched that a smaller bubble size will lead to increased efficiency, but how efficient can this method get (like roughly how much of the $\ce{CO_2}$ will react with the MEA)?
Alternatively, I have also considered misting the MEA, then using a fan to blow air through the mist. I would think this method would be more efficient, because the fan can blow more air than the air can pump, and the particle size of the mist would be low. How efficient will this method be?
I would prefer to use the bubbling air through water method, but it depends on how much efficiency can be gained by using the mist/fan.

While CO2 slightly reacts, the process is called absorption of CO2. Using amines is standard, afaik, so you should be able to find plenty of data.

Perhaps use a very fine filter on the outlet for the CO2 gas...


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Rate of SN1 reaction in derivative of furan and thiopene
When I was practising questions on the rate of SN1 mechanisms, I stumbled across this:

I could eliminate options B and D because they can stabilize by only cross conjugation (if I am correct) and options A and C show extended conjugation. However, as $\ce S$ donates electrons more easily than $\ce O$, I picked C but the answer given is A. Upon discussing, my tutor said that electrons are initially delocalized in empty orbitals of $\ce S$ and hence $\ce O$ donates better.
Is there any experimental data to support this?


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Exothermic Rapid Multistage Reactions
I asked a similar question on SpaceExploration, but I think it might be better answered here:
Most chemical reactions in rocket combustion sequences are fairly simple, even with "exotic" fuels like Diborane etc.
For instance: N2H4 + NTO, which forms a relatively small number of light, fast-moving products.
(For background; Rockets usually depend on multiple tanks (2-3), to store propellants, but this adds mass.)
To solve this and still have a fairly high-performance engine, a single storable propellant is required.
Such a propellant would likely be organic, and would need to be able to decompose upon contact with a catalyst into products that then combust hypergolically (upon contact, and relatively quickly.
I can't think of any reaction quite like this, but maybe I'm missing some clear and obvious reason for this being a pointless goal...
Also, I'm not asking for the sake of making such a thing, this is purely theoretical.

H2O2 was used for this, but the performance is below binary fuel.

You're missing that it's hardly a big deal and should have just read up on (mono)propellants. BTW " into products that then combust hypergolically" in there isn't making much sense - what decomposed is already "burnt" - that's essentially the point of explosives and monopropellants.

gun powder, ammonium nitrate, nitroglycerine. The only effective method is antigravity but thinking about it your destination would repel you! Possibly the effective use of dark matter; it seems to be accelerating everything else. It is frustrating trillions of galaxies, all that gravity and we are stuck on a small planet or a solar system.


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Residue left after evaporation of a salt solution
When a salt solution is evaporated (bunsen burner, evaporating dish/beaker), for example CuSO4 aq, there will be residue left over, as with all salt solutions.
However, the common form of copper (II) sulfate is in copper II sulfate pentahydrate (and I'm sure some other ones that are less common which I am not familiar of). It has a bluish tint.
Pure copper II sulfate is white.
When the copper II sulfate solution ( for example 1 L, 0.5M) is evaporated, is there a mathematical way to work out at which points copper II sulfate pentahydrate would be the residue, as well when the pentahydrate would evaporate as well to form pure copper II sulfate.
Furthermore, is there such case with a salt solution where hydrated salt version would not occur as the residue after evaporation, or vice versa with the anhydrous version?

Yes there is a mathematical way, and quite simple at that. When the solubility limit is reached, you will have a solid precipitate. When a certain temperature is reached, your hydrated salt will dehydrate.

Wikipedia references https://en.wikipedia.org/wiki/Copper(II)_sulfate with:

Copper(II) sulfate pentahydrate decomposes before melting. It loses two water molecules upon heating at 63 °C (145 °F), followed by two more at 109 °C (228 °F) and the final water molecule at 200 °C (392 °F).

A quote from the question:

Furthermore, is there such case with a salt solution where hydrated salt version would not occur as the residue after evaporation, or vice versa with the anhydrous version?

I had a related question about Aluminum Chloride https://en.wikipedia.org/wiki/Aluminium_chloride:

Reactions with water

Anhydrous aluminium chloride is hygroscopic, having a very pronounced affinity for water. It fumes in moist air and hisses when mixed with liquid water as the Cl− ligands are displaced with H2O molecules to form the hexahydrate [Al(H2O)6]Cl3. The anhydrous phase cannot be regained on heating the hexahydrate. Instead HCl is lost leaving aluminium hydroxide or alumina (aluminium oxide):

$$\ce{[Al(H2O)6]Cl3 → Al(OH)3 + 3 HCl + 3 H2O}$$
The point being that the Bunsen burner may not be hot enough (or there alternatively may be too much humidity in the room) to completely drive out the water at "200 °C (392 °F)" for Copper(II) sulfate pentahydrate.  After all, the water solvent boils nearly at 100 °C and after that, additional heating results in popping of the salt from the beaker.
Has a different oven than the Bunsen Burner been possible to use to get the Copper(II) sulfate pentahydrate decomposition temperature after full evaporation of the solvent, and slow heating to the required temperature (with some additional margin) to prevent popping of the Copper(II) sulfate pentahydrate at a lower temperature from the Bunsen Burner?

You're asking about a number of points..

"Massive" copper sulfate pentahydrate is blue. However it can have a    white coating if partially dehydrated.

In geology there is a streak test where a mineral is rubbed on a  alumina plate to create microscopic particles. The streak test often    yields colors that are different than that of the "massive" specimen.    In this case copper sulfate pentahydrate will have a white steak because of the small particle size and light coloring.

If you try to dry a solution of copper sulfate over a Bunsen burner it will splatter. Whether it will be the anhydrous form or the pentahydrate depends on how hot you heat the residue as you evaporate the solution.

As you evaporate the solution you'll reach the solubility limit and copper sulfate pentahydrate will start to crystalize. The solubility limit does depend on temperate. Copper sulfate behaves "normally" in that it is more soluble in hot solutions than cold. You can look up the solubility as a function of temperature.

If you just air dry the copper sulfate solution you'll get the pentahydrate unless you're in an unusually dry climate.

I'll add that growing a large copper II sulfate pentahydrate crystal from a supersaturated solution using a seed crystal on a string is a common experiment. I don't know about now but copper sulfate used to be sold in drug stores to treat infected (fungus?) toes.


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How does the heat-activated blowing agent in foaming plastic filament survive the manufacturing process?
Special 3D printer filaments are available with blowing agents mixed in. These blowing agents cause the filament to foam as it is extruded, causing the resulting structure to have a much lower density.
https://hackaday.com/2020/02/17/bubbly-filament-works-better-than-you-think/
How do these blowing agents survive the manufacturing process? Since the filaments are spun out of almost-molten plastic, would that not trigger the agent? Or is the agent's reaction temperature able to be tuned so narrowly that it survives the filament extrusion process but not the printer melting process?

This thing you linked hardly tells any details.

@Mithoron I'm not sure I've ever seen any more details than user reports. https://www.cnckitchen.com/blog/testing-colorfabb-varioshore-tpu-foaming-3d-printing-filament mentions sometimes baking soda is used, and https://onlinelibrary.wiley.com/doi/full/10.1002/advs.202105701 just mentions the general process. So it's possible this is a poorly posed question, if it's the case that there's not a single blowing agent which is typically used.


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Transform raw atomic coordinate data into SDF file
I have raw data of the atom types and their coordinates. For example, water molecule is:
O -0.034 0.977 0.007
H 0.064 0.020 0.001
H 0.871 1.300 0.0006

Here, I would like to transform this atomic coordinate data into SDF file. How can I do it with a software or library such as RDKit?

You can use openbabel for this. https://openbabel.org/wiki/Main_Page

Your work ahead requires two steps

completion of the file to be a valid .xyz file
transformation of the .xyz into a .sdf file, a task for which openbabel already is one of the tools suitable (in my experience, easier to set up than RDKit facing the IT department)

For the first one, the snippet shared looks like the block about atomic coordinates in a .xyz file.  To be syntactically complete, two leading lines are missing: one about the total number of atoms (an integer number), and one which either remains blank empty, or optionally may carry a comment.  In your case, the minimal form of your snippet to pass the next step hence would be a file
3

O -0.034 0.977 0.007
H 0.064 0.020 0.001
H 0.871 1.300 0.0006

You do not share the origin of the snippet, nor the operating system of the computer used.  Possibly, the quantum chemical software used to provide you with said geometry offers the export of the result as a complete .xyz file which already contains the number of atoms.  (Else, because this type of programs often run in Linux/Unix, wc -l my_file.txt will report the number of lines in file my_file so that you don't have to manually count the lines in your original snippet either.)

The second step is the transformation into a format richer in information, .sdf.  It is richer than .xyz because in addition to the atomic coordinates (as provided by the .xyz) there is a chemical table.  This is to report which atoms are connected with each other, the assigned nature of the bond (bond order single, double, etc.), information about stereochemistry (CIP rules about R vs S, or E vs Z configuration), assignment of charges, etc.  And similar as the .xyz, additional information optionally may be provided in comment-like entries.  The English edition of Wikipedia has a dedicated entry, including a permanent link to .pdf here which specifies the .sdf formats in greater detail.
Assuming the above, now already complete .xyz file was saved as file example.xyz, and openbabel successfully installed, the command on the command line would be
obabel example.xyz -O output.sdf

to write a .sdf (by default in version V2000) of name output.sdf in the pattern of
example.xyz
 OpenBabelDDMMYYhhmmXX

  3  2  0  0  0  0  0  0  0  0999 V2000
   -0.0340    0.9770    0.0070 O   0  0  0  0  0  0  0  0  0  0  0  0
    0.0640    0.0200    0.0010 H   0  0  0  0  0  0  0  0  0  0  0  0
    0.8710    1.3000    0.0006 H   0  0  0  0  0  0  0  0  0  0  0  0
  2  1  1  0  0  0  0
  3  1  1  0  0  0  0
M  END

This record states the name of the file initially read, followed by the creator of the new file and a time stamp.  After setting the ground for the information to follow, there are the atoms and their coordinates; followed by said now additional block of chemical information.
Side note:  By default, openbabel provides the conversion into .sdf (V2000).  Depending on your wanted use down the road, you might want to use the more recent dialect of .sdf (V3000).  You can obtain this by a conversion with the additional -x3 flag, i.e.
obabel example.xyz -O output.sdf -x3

to yield a record in pattern of
example.xyz
 OpenBabelDDMMYYhhmmXX

  0  0  0     0  0            999 V3000
M  V30 BEGIN CTAB
M  V30 COUNTS 3 2 0 0 0
M  V30 BEGIN ATOM
M  V30 1 O -0.034 0.977 0.007 0
M  V30 2 H 0.064 0.02 0.001 0
M  V30 3 H 0.871 1.3 0.0006 0
M  V30 END ATOM
M  V30 BEGIN BOND
M  V30 1 1 3 1
M  V30 2 1 2 1
M  V30 END BOND
M  V30 END CTAB
M  END
$$$$

instead (see the program's documentation here).  This format dialect becomes advantageous to you if extended stereochemistry features e.g., to describe racemates and diastereomeric mixtures are needed (instead of either (R)- or (S)-mandelic acid, you describe both in one model) in larger sets of data* and databases, rather than to depict the conformation of a specific molecular structure only.  Contemporary viewers (e.g. avogadro, Jmol) are ready to display either .sdf (V2000), or .sdf (V3000).
* Because a .sdf can serve as a container for multiple, individually separate structure descriptions.