text:
pH calculation of HCOOH +plus HCl solution

Question -  What volume of $0.1 \ce{M}$ $\ce{HCl}$ solution should be
added to a $\ce{500 mL}$ of $\ce{0.5 M }$ $\ce{HCOOH  }$ solution in
order to prepare an acid solution of $\text{pH}$ =$1.5.$ $K_a$ of
formic acid is $2×10^{-4}$ at $\ce{298 K?}$

I am not sure how to proceed with this question. Iam also confused how equilibrium reaction will
I guessed it would be $$\ce{HCOOH}+\ce{HCl \rightarrow salt}$$
Let $V$ be volume of HCl. At $t=\text{equilibrium,}$ $$[\ce{salt}] = \dfrac{0.1V}{0.25-0.1V}$$
Now,
$$\ce{pH}= \ce{pK_a} + \log\dfrac{0.1V}{(0.25-0.1V)}\Rightarrow \boxed{V=2.5L}$$
But answer is $V=0.208 L$
Am I completely wrong somewhere? Can someone give tell me how to proceed?
$\textbf{NOTE:I am not asking full solution.}$
$\textbf{Just telling how to proceed is fine.}$

In a mixture of two acids, nothing happens : no reaction between the acids. No salt !
At $\ce{pH = 1.5}$, the finall concentration of $\ce{H+}$ is $\ce{10^{-1.5} = 0.0316}$ M. This concentration is the sum of the $\ce{H+}$ ions produced by $\ce{HCl}$ and those from $\ce{HCOOH}$. The amount of $\ce{HCOOH}$ is $0.5$ mol/L·$0.5$ L$ = 0.25$ mol. If the concentration of formate ion is $x$, $x$ is also the concentration of dissociated $\ce{HCOOH}$ molecules.  Now the dissociation constant of formic acid is $\ce{K_a = \frac{0.0316·x}{0.25 - x} = 2·10^{-4}}$, From there, $x$ = $1.6·10^{-3}$.
The concentration of $\ce{H+}$ ions coming form $\ce{HCl}$ is $0.0316 - 0.0016 = 0.0300$ M. The total volume of the final solution is : $0.5 + \ce{V}$, it V is the volume of $\ce{HCl}$ to add. The amount in moles of $\ce{HCl}$ to be added is  $0.0300·(\ce{0.5 + V)}$. The volume V where to find this amount of $\ce{HCl}$ is : $\ce{V = \frac{n}{c}} = \frac{0.03·(0.5 + V)}{0.1}$ . From there, one gets : $\ce{V = \frac{0.15}{0.7} = 0.214 L}$
It is "nearly" your expected value !! If you remake the calculations, with more decimals, you should obtain $0.208$ L.


text:
Degradation of 4-MeO-Tetrahydrocarbazole
I recently synthesized 4-Methoxy-Tetrahydroxarbazole from 4-MeO-Phenylhydrazine and Cyclohexanone. The product had a off-white color directly after synthesis but was brown after standing over night at room temperature. I could not find any informations on the sensitivity of the compound toward oxidation or on its stability in general.
Also, other people in our lab reported that their product degraded during column chromatography.
Does anybody have experience with this compound and has also made these observations?

Have you named this material correctly? 4-methoxy-phenylhydrazine + cyclohexanone gives 6-methoxy-tetrahydrocarbazole as reported here  https://www.researchgate.net/publication/358021573_A_review_on_synthesis_methods_of_tricyclic_1234-tetrahydrocarbazoles  The references documented in this review may prove useful.

particularly this reference  Synthetic Communications®. 16 Dec 2008; 39(1): 158-65.

You should first correct your question according to comments of Waylander (above). For your info, these chemicals are susceptible to react woth atmospheric oxygen (as most amines do). For example, see an oxydation process reported in here.

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text:
How can the Kw of water be constant?
I know in neutral solution, $\ce{[H+]}= \ce{[OH-]}= \pu{1.0 x 10^-7}$.
However, let’s say I add an acid like HCl. This will increase $\ce{[H+]}$, therefore it’s no longer $\pu{1.0 x 10^-7}$, but is greater.
So, $\ce{[OH-]}$ would need to decrease to keep $K_w$ at a constant $10^{-14}$, reestablishing equilibrium.
However, how does this $\ce{[OH-]}$ decrease? What does it react with to decrease its concentration? At first I thought it could react with $\ce{[H+]}$ to make water, but that’d be decreasing $\ce{[H+]}$, which doesn’t make any sense since the solution should be acidic if HCl was added.

Note you've forgotten to include the molar units for the concentrations.

Dissociation equilibrium in water is attained when the rates of the following forward and reverse reactions are equal (all species aqueous of course):
$$\ce{H2O ->[k_f] H+ + OH-} $$
$$\ce{H+ + OH- ->[k_r] H2O}$$
The water molecules dissociate with a regular frequency (you can think of each water molecule having a lifetime before it falls apart), and the collision rate between oppositely charged ions depends to a first degree on their concentrations. If you have more $\ce{H+}$ then you need less $\ce{OH-}$ for a "productive" collision resulting in formation of a water molecule, and vice-versa.
The forward an reverse rates become equal when
$$k_f [\ce{H2O}] = k_r[\ce{H+}][\ce{OH-}] \tag{1}$$
In practice activities are used instead of concentrations but for a solution dilute in the ions using concentrations expressions for the ions is ok, as the encounter rate between ions can be assumed to vary linearly with concentration for dilute solutions.
This leads to an expression for the dissociation constant of water:
$$K = \frac{k_f}{k_r} = \frac{[\ce{H+}][\ce{OH-}]}{[\ce{H2O}]}$$
Strictly speaking activities are used and $K_w$ is defined as:
$$K_w = \frac{a_\ce{H+} a_\ce{OH-}}{a_\ce{H2O}} $$
At low concentrations (approx neat water) $a_\ce{H2O}=1$ and the activities can be replaced by concentrations such that
$$K_w \approx [\ce{H+}][\ce{OH-}] \tag{2}$$

I browsed Ref. 1 to double-check my answer. I realized my exclusion of hydronium was asking for trouble. Strict definitions state the reaction as $\ce{2H2O <=>[K_w] H3O+ + OH-}$. But I prefer to leave the answer as is, as I did write aqueous, and  because the definition of pH is closely related to this concept. But the definition of $K_w$ as constant is still imperfect. It is not truly a constant when you start adding other stuff to water. $K_w$ defines the autoionization constant of pure water for a given p and T. But if dilute in other added ions, then the prediction of simple kinetic theory, Eq. 1, leads to the conclusion that Eq. 2 with $K_w$ assumed constant can be used.

Note also: In practice  when you measure pH you are strictly-speaking not measuring the concentration of $\ce{H+}$ but rather its activity.

References
Andrei V. Bandura; Serguei N. Lvov. The Ionization Constant of Water over Wide Ranges of Temperature and Density
In Special Collection: International Water Property Standards. J. Phys. Chem. Ref. Data 35, 15–30 (2006) https://doi.org/10.1063/1.1928231

This explanation is absolutely correct. But it does not give an answer to the question posed by Maria.

@Maurice Well, if you read to the end: "When using activities K becomes a concentration independent constant (concentration dependence of the rate constants being incorporated into the activities)". In addition the beginning of the answer attempts to explain how the concentration of different species is affected by collisions and water dissociation, which is very pertinent to clarifying the problem.

@Maurice BTW your answer addresses how one computes the pH (or how pH depends on concentration of added acid) but not why Kw is constant.

You are right. But there is a strange discrepancy between the title (Is Kw constant ?) and the text itself (How will [$\ce{OH-}$] decrease ?). I have only answered the question in the text.

@Maurice Granted. I guess the answers are complementary in a way.

And then there is still the temperature dependency... so constant-ish is more appropriate.

If you add $\ce{HCl}$ in water, the amount of $\ce{H+}$ increases of course. And part of this supplementary $\ce{H+}$ ions are used to destroy $\ce{OH-}$ ions from pure water. This amount of $\ce{H+}$ required to do this job must be smaller than or equal to $10^{-7}$ M. It is rather small. It is so small that it can be neglected with respect to the amount brought by the supplementary $\ce{HCl}$, which can give a concentration having any numerical value bigger than (let's say)... $10^{-3}$ M or $10^{-2}$ M.

The hydroxide ions will react back indeed, albeit with hydronium ions (not $\ce{H^+}$), but because there are only $10^{-7}$ mol/l hydroxide ions in solution, it can also react with a maximum amount of $10^{-7}$ hydronium ions $\ce{H_3O^+}$, therefore basically not changing the HCl‘s much,much higher own concentration.

Kw is not constant. Kw = Keq [H2O]. As the activity of water changes Kw changes. The activity of water is reasonably constant in dilute aqueous solutions. Elsewhere things get complicated, changes in water activity have effects. This is noticeable in cooking and the behavior of alcohol-water mixes and most likely in cellular chemistry.

This answer is true,but addresses the effects few levels below the trivial basics the OP asks about.

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text:
Why does water present in aqueous solution of CuSO4 ionize into H+ and OH- during its electrolysis?
It’s written in books that during electrolysis of aqueous solutions of $\ce{CuSO4}$ or $\ce{NiSO4}$, the water particles ionize into $\ce{H^+}$ and $\ce{OH^-}$ ions along with dissociation of the ionic salt. Now my question is , why does water ionize along with the dissociation of salt?
I do know that water helps in dissociating the salt by reducing the electrostatic force of attraction between the ions. But why does water break into ions? Is it due to the electricity or the ions of the ionic salt or partial charges on oxygen and hydrogen of a water molecule?

In a solution of $\ce{CuSO4}$ or $\ce{NiSO4}$, water does not dissociate more into $\ce{H+}$ and $\ce{OH-}$ than it does usually. But at the anode, the ion $\ce{SO4^{2-}}$ is not discharged as it may have done when an anion is discharged. Water takes its role and produce electrons instead. The anodic equation is : $$\ce{2H2O -> 4 H+ + O2 + 4 e-}$$ So the cation $\ce{Cu^{2+}}$ or $\ce{Ni^{2+}}$ is discharged at the cathode, and the solution becomes acidic at the anode because of the creation of ions $\ce{H+}$ in solution. The main result is that the ions $\ce{Cu^{2+}}$ or $\ce{Ni^{2+}}$ are progressively replaced by ions $\ce{H+}$ in solution.
It may look as if water has been more "dissociated" into $\ce{H+}$ and $\ce{OH-}$ producing more $\ce{H+}$ and more $\ce{OH-}$, if it is still believed that $\ce{OH-}$ had been discharged at the anode (instead of water) according to $$\ce{? ... 2 OH- -> O2 + H2O + 2 e- ... ?}$$


text:
Why ordinary oil was not recommended in Millikan oil drop experiment?
Why was it recommended not to use ordinary oil in Millikan oil drop experiment? What were the properties of oil used by Millikan?

Ordinary oil may contain double bonds that may chemically react in the experiment.

What is "ordinary oil"? There is a huge variety of chemicals called "oils" with a very wide range of properties.

Converting my comment into answer:
Quoting from here:

Ordinary oil wasn’t used for the experiment as it would evaporate by the heat of the light and so could cause an error in the Millikens Oil Drop Experiment. So, the oil that is generally used in a vacuum apparatus which is of low vapour pressure was used.

What type of oil is used then? Please check: Which oil was used by Robert A. Millikan in his oil drop experiment?


text:
Na+ as an electrophile
In inorganic I learnt that s block cations are lewis acids but while learning attacking reagents in organic chemistry I came across the fact that Alkali metals are not electrophiles, why so?

In the case of lithium (and its diagonal partner magnesium from the alkaline earth metals), the cation is very much an electrophile. When either an organolithium compound or a Grignard reagent reacts with a substrate having a polar functional group such as a carbonyl group, not only does the carbanion-like hydrocarbon portion attach to the electrophilic end of the substrate. The lithium or magnesium ion equally acts as an electrophile, attaching to the negative end of the dipole (like the oxygen in the carbonyl group). This electrophilic-attack component actually promotes the nucleophilic attack at the other end of the dipole where the carbon-carbon bond that makes the headlines is formed.

is the reason why li+ can still pose as an electrophile and na+ can't (both being positive and having a stable configuration) because of the charge density/size?

The question was about alkali mrtals, so. LI was fair game. Yes charge density is a major factor. Lithium ion is also known to act electrpohilically in other settings such as displacing carbon dioxide and water when we attempt to precipitate the bicarbonate (so we ger lithium carbonate instead).


text:
How to produce Fe(OH)2?
I'm interested in producing Ferrum hydroxide (II), I did some research on the internet and made a plan for myself:

Conduct electrolysis:
Conduct electrolysis producing $\ce{NaOH}$ and chlorine solution.
Produce $\ce{HCl}$:
Produce $\ce{HCl}$ from chlorine solution by exposing it to UV light.
Make iron and hydrochloric acid react:
Adding iron in hydrochloric acid makes this reaction: $\ce{2HCl + Fe -> FeCl2 + H2}$ producing a solution with $\ce{HCl}$ and $\ce{FeCl2}$, then evaporate it and be left with solid $\ce{FeCl2}$ crystals.
Produce $\ce{Fe(OH)2}$:
As we have $\ce{FeCl2}$ and $\ce{NaOH}$, we can make this reaction: $\ce{2NaOH + FeCl2 -> Fe(OH)2 + 2NaCl}$ producing a participate ($\ce{Fe(OH)2}$).

The information above was taken from internet, and I am not sure that it will work, I asked a question about electrolysis before and I got an answer that when conducting electrolysis of saltwater it generates $\ce{OH-}$ ions that create different chlorine-based substances, maybe they are good for making $\ce{HCl}$ acid as shown in step 2 but I'm not sure.
If this won't work, maybe there is another way of producing Ferrum hydroxide (II), if yes, then please write it in an answer. I also think that I could use baking soda as a electrolyte, but doesn't that mean that it will be same as with salt? Could there be electrolytes that don't react with chlorine? Or is there any way of producing "pure" $\ce{NaOH}$ and $\ce{FeCl2}$?

How about using vinegar and NaOH that can be bought in a shop, which could, well, maybe even work?

This might sound like a frivolous  question, but why do you want to make Fe-(II) hydroxide? ;)

The main problem in the synthesis of $\ce{Fe(OH)2}$ is not the procedure to use. The main problem is how to prevent $\ce{Fe(OH)2}$ from air oxidation. Whatever the technique used, the synthesis of $\ce{Fe(OH)2}$ will be done in contact with atmospheric air. And this contact oxidizes quickly $\ce{Fe(OH)2}$ which is green, into a ferric oxide, which is brown. So it is difficult to synthesise pure green $\ce{Fe(OH)2}$, whatever the reaction to be used.
In my opinion, the best way is to dissolve maybe $1$ mole commercial $\ce{FeSO4·7H2O}$ in $1-2$ liter water (depending on the temperature), and add $2$ moles of $\ce{NaOH}$ dissolved in the minimum volume of water, filtrate the green precipitate, wash and dry it as quickly as possible, if possible in a nitrogen atmosphere. Traces of air oxidation are easy to discover by observing brown spots on the green powder.


text:
How to calculate the resolution of a GC-FID peak using only peak area and height
I think the title pretty much says it all. I'm trying to calculate the resolution of my GC-FID peaks, but the software doesn't report the width of the peaks. Do I need to manually do this by approximating the peak width or should I treat each peak like a triangle and back-calculate from the height and area? I realize the second option is most likely not correct though.
Thanks

"A pure Gaussian peak with a peak height h and full-width at half-maximum w has a total area of 1.064467hw." Source

There is no commercial GC-FID software which will not give peak widths. What is the instrument and what software are you using?


text:
Ring formations catalyzed by piperidinium acetate: Robinson annulation?

Hey everyone, I have a question about the mechanism below. Do you guys think the Robinson annulation is a good starting point for the 6-C ring on the right? If not, what would you suggest me for this mechanism? Thank you

Do you have any thoughts beyond the Robinson annulation?

Welcome to ChemSE! Your question will likely be considered as homework and be voted down. You need to make some effort to answer the question yourself. Take a look here about homework policy. PS: You need to do more than toss in a name reaction. You can resubmit your more detailed analysis. Try tracing the 16-membered ring of the reactant in the product. I have an answer waiting for you.

I started numbering Cs from the one coming after N counter-clockwise. After that I tautomerized b keto ester attached to C12 to obtain an enol, but I do not think OAc is strong enough to pull H from there. So I might need another perspective here. After that, I attacked C7 (double-bond) with the enol pushing double-bond between C5-C6 an also O- at C5. I then tautomerized it again to obtain a ketone which I attacked with amine group in ring. I deprotonated and protonated N and O at C5, respectively and removed water from C5. Later, I tautomerized the ketone at C11 and attacked C5 from C10.

It looks like you have good idea of what is going on! It is also possible that the imine formation is precedes the addition of $\ce{C12 to C7}$. I'll post my thoughts in the future; the hour is late.

Identifying the acidity of the $\beta$-keto ester is correct. That forms a nucleophilic species, so you should look for electrophilic positions. There are enough substituents that it will be really helpful if you can map the carbons in both of the backbones. In general, one does that transpose carbons. Only go that route if you've exhausted simple possibilities. And if it's not obvious, there are multiple steps involved here...

Esra: As I said in my Comment, you have the right idea about what bonds to make to form the final product 2. The catalyst, piperidinium acetate, has both acidic and basic properties that facilitate enolization and dehydration. There are two principle reactions to be considered: a Mannich and a vinylogous Mannich reaction. The possibility of an initial Michael addition may be an alternative but the arguments for that route are the same whether or not the iminium salt 3 is formed first or not. I'm surprised that the macrocycle 1 exists as the free amine because there are too many reactions it could undergo. One of which is the carbinol amine formed by the addition of the amino group to the $\ce{C5}$ carbonyl. Under the influence of the catalyst, the vinylogous iminium salt 3 is formed with the loss of water. This species, which is a better electrophile than an α,β-unsaturated ketone, allows for the enol of the β-ketoester moiety to form the $\ce{C7-C12}$ bond. Given the stereochemistry at $\ce{C8 and C9}$, there are four possible stereoisomers that could conceivably form: two trans- and two cis-isomers. The trans-$\ce{C7-C12}$ isomers will not lead to 2 because the final ring closure is prohibited by the remoteness of $\ce{C5 and C10}$ from one another. Of the two cis-isomers, only 4 leads to product 2. The fate of the other cis-isomer will be addressed later.

The E-double bond in 1 or 3 may also be converted to a Z-isomer by 1,4-addition/elimination of acetate or piperidine. Structures 5a and 5b, ignoring the $\ce{C8-C9}$ substituents for the moment, are enantiomeric. When the substituents are considered, 5a and 5b are diastereomeric conformations. The sequence 5a $\rightarrow$ 6 $\rightarrow$ 2 leads to the the observed product via formation of the "red bonds." However, the pathway 5b $\rightarrow$ 7 $\rightarrow$ 8 has problems. In the transition state for the formation of the $\ce{C7-C12}$ bond, there is a developing steric interaction between the -CO2Et and the -OBn groups. Structure 7 is the other cis-stereoisomer of 4 alluded to earlier. Even if the formation of 7 were successful, the last ring closure, $\ce{C5-C10}$, is fraught with a severe steric interaction between the $\ce{C4}$-methylene and the $\ce{C9}$-methyl group. These interactions are best viewed in the diastereomer 8 that is not formed.


text:
ppm of CaCo3 of an unknown sample
A 1g impure solid sample of CaCO3 was dissolved in 100ml of water. After titration the hardness of the solution was determined to be 900ppm - now, how do I calculate the amount of CaCO3 in the impure solid sample in ppm?

I'm unclear whether this is a homework problem or not. But the site rules say homework problems are off-topic unless you include your attempts to solve the problem and reference the source of the problem. You have not done these, so I'll presume this is not a homework problem.
Calcium carbonate is very sparingly soluble in water. A 900 ppm concentration (which means 9% of your sample dissolved) is implausibly high. So I think your sample is contaminated with either a more soluble calcium salt (if your test measures calcium), a more soluble carbonate salt (if your test measures carbonate), or possibly some other substance that interferes with the test.
If you want to determine the calcium carbonate proportion of your sample, this method won't work very well. You'll detect the soluble salts, and much of the actual calcium carbonate will remain undissolved and undetected. I think you need a different approach. I'd recommend first washing the sample to get rid of the soluble stuff (unless you want to measure soluble calcium and/or carbonate). Then digest it in hydrochloric acid and measure $\ce{CO2}$ evolved. This will get you the carbonate content of the insoluble portion. Finally, you can test the resulting solution for calcium content.

I'm wondering about «A 900 ppm concentration (which means 9% of your sample dissolved)» because $\pu{900 ppm} = \pu{900} / \pu{1E6} = \pu{0.9E-3}$ on one hand, and 9% (or $\pu{90E-3}$) on the other.

9% of a 1g sample is 90 mg. 900 ppm of a 100 mL solution is also 90 mg.


text:
Factor affecting acidity:Hybridization
I have problem here
So I had learn that the percent of s-character of hybrid orbital will affect the bond strength. The higher percentage of s-character,the stronger the bond.For example the strength C-H in ethane(sp3,25%) is the weakest followed by in ethene(sp2,33%) and strongest in acetylene(sp,50%)
The percentage of s-character of the hybrid orbital also affect the acidity.The higher the percentage of s-character,the stronger the acidity. The reason here is that the conjugated base formed is more stable as the electron is in lower energy and pulled nearer to the nucleus of the atom.For example, acidity hydrogen ion in C-H for compound ethane is weakest followed by ethene and acetylene being the most acidic.
But isn't the C-H bond strength in the acetylene is  the strongest.So it should be hard for acetylene to loss it hydrogen. So why isn't the acetylene being the least acidic? Can someone explain to me?

The carbon in ethyne is very much electrophilic compared to ethane/ethene.

Ya, I know about this,but as I stated in the last paragraph, the strength C-H in ethyne is the strongest among the three. So, isn't the H+ is harder to release from C-H in ethyne. Or this effect is not strong and can be ignored?

Take the hydrogen halide as example.The bond length of H-A where A is halogen(Cl,Br and I) increase down the group,hence the H-A bond strength decrease down the group and so the acidity of hydrogen hallide increase down the group.

Yes, it's stronger, but it's means that energy of homolytic cleavage is higher, not heterolytic - that's what happens in acid base reactions.

Ok after seeing the other post ans, I get it now. Tqvm


text:
How to obtain the expression for residual Gibbs energy?
The property involved is Gibbs free energy, the equations that I wrote are presented in my text book Introduction to chemical engineering thermodynamics by Smith and Van Ness, Chapter 6.
The thing is I don't get the meaning of the derivations made to arrive to the expression of residual Gibbs energy. My text book starts with the next identity
$$ d(\frac {G}{RT}) \equiv \frac{1}{RT}dG-\frac{G}{RT^2}dT$$
Where does this expression comes from? Why does it exists in the first place? What does it mean?
Is it related to Ideal gases? By this I mean, is it somehow derived from the ideal gases framework?
Then my book makes use of this expression
$$ d(\frac {G^R}{RT})_{P=0} \equiv J$$
What is J? Is it just a mathematical tool or does it have a physical meaning? My book doesn't mention any name for such constant.
I'm trying to understand these details to get a sense of how to obtain this final expression
$$\frac {G^R}{RT}= \int_{0}^{P} (Z-1)\frac{dP}{P}$$
Every step looks senseless to me, in a physical sense. I just see manipulations and expressions that work but not why they are chosen, they just pop into existence with no justification to be used out of all thermodynamic relations available. And also, what exactly does this integral mean? Why is $P=0$ chosen for ideal gases and no other value such as $0.1$ or $0.0001$ or even $1$?

Everyt capital letter here is written in a per mole basis, so I stick with your nomenclature. E.g., $G$ has units of $\pu{J mol ^{-1}}$, etc.

Where does this expression comes from? Why does it exists in the first place? What does it mean?

Lets put the function $G/RT$ as an exact differential. I see two variables, $G$ and $T$, so we have to differentiate with respect to $G$ and $T$
$$ \mathrm{d}\left(\frac{G}{RT}\right) =
   \frac{1}{RT} \mathrm{d}G -\frac{G}{RT^2}\mathrm{d}T \tag{1} $$
In the first term $T$ was constant when we differentiated with respect to $G$, and in the second term $G$ was constant when we differentiated with respect to $T$. That's all.
Not every expression is thermodynamics is useful. However, this expression exists because we can eliminate the entropy in favor of enthalpy, and that's the reason. Remember that in its fundamental expression $\mathrm{d}G = V\mathrm{d}p - S\mathrm{d}T$, we have $S$ there.
We do this. We combine Eqn. (1) with $G = H-TS$ and also the fundamental equation $\mathrm{d}G = V\mathrm{d}p - S\mathrm{d}T$
\begin{align}
  \require{cancel}
  \mathrm{d}\left(\frac{G}{RT}\right) &=
  \frac{1}{RT} (V\mathrm{d}p - S\mathrm{d}T) -\frac{H - TS}{RT^2}\mathrm{d}T \\
  \mathrm{d}\left(\frac{G}{RT}\right) &=
  \frac{V}{RT}\mathrm{d}p - \cancel{\frac{S}{RT}\mathrm{d}T} -
  \frac{H}{RT^2}\mathrm{d}T + \cancel{\frac{S}{RT}\mathrm{d}T} \\
  \mathrm{d}\left(\frac{G}{RT}\right) &=
  \frac{V}{RT}\mathrm{d}p - \frac{H}{RT^2}\mathrm{d}T \tag{2}
\end{align}
and there you go. Eqn. (2) now is a function of enthalpy rather than entropy.

Is it related to Ideal gases? By this I mean, is it somehow derived from the ideal gases framework?

Eqn. (1) doesn't depend on the model, it is very general. It is valid as long as you are talking about a fluid of only one component.

Then my book makes use of this expression
$$ d(\frac {G^R}{RT})_{P=0} \equiv J$$

I will skip this question and come back after we derived the next one.

I'm trying to understand these details to get a sense of how to obtain this final expression
$$\frac {G^R}{RT}= \int_{0}^{P} (Z-1)\frac{dP}{P}$$

We are going to do this: we restrict Eqn. (2) for a constant temperature process, so the guys with $\mathrm{d}T$ go away. Then, we write what is left for an ideal gas and also for a real gas
\begin{align}
  \mathrm{d}\left(\frac{G^\mathrm{ig}}{RT}\right) &=
  \frac{V^\mathrm{ig}}{RT}\mathrm{d}p \tag{3} \\
  \mathrm{d}\left(\frac{G}{RT}\right) &=
  \frac{V}{RT}\mathrm{d}p \tag{4}
\end{align}
and subtract Eqn. (4) from (3), where we will define the residual property for a general property $X$ as $X^\mathrm{R} = X - X^\mathrm{ig}$
\begin{align}
  \mathrm{d}\left(\frac{G}{RT}\right) -
  \mathrm{d}\left(\frac{G^\mathrm{ig}}{RT}\right) &=
  \frac{V}{RT}\mathrm{d}p - \frac{V^\mathrm{ig}}{RT}\mathrm{d}p \\
  \mathrm{d}\left(\frac{G - G^\mathrm{ig}}{RT}\right) &=
  \frac{V - V^\mathrm{ig}}{RT}\mathrm{d}p \\
  \mathrm{d}\left(\frac{G^\mathrm{R}}{RT}\right) &=
  \frac{V^\mathrm{R}}{RT}\mathrm{d}p \tag{5}
\end{align}
Now we would like the compressibility factor to pop up. For an ideal gas the molar volume is $V = RT/p$, and for a real gas $V=ZRT/p$, so that
$$ V^\mathrm{R} = V - V^\mathrm{ig} = \frac{ZRT}{p} - \frac{RT}{p} =
   (Z-1)\frac{RT}{p} \tag{6} $$
Combining Eqns. (5) with (6) and integrating
\begin{align}
  \require{cancel}
  \mathrm{d}\left(\frac{G^\mathrm{R}}{RT}\right) &=
  \dfrac{(Z-1)\dfrac{\cancel{RT}}{p}}{\cancel{RT}}\mathrm{d}p \\
  \mathrm{d}\left(\frac{G^\mathrm{R}}{RT}\right) &=
  \dfrac{Z-1}{p}\mathrm{d}p \rightarrow \boxed{
  \frac{G^\mathrm{R}}{RT}(p) = \frac{G^\mathrm{R}}{RT}(p \to 0) +
  \int_0^p \dfrac{Z-1}{p}\mathrm{d}p} \tag{7}
\end{align}
There we have derived the expression. I will make the following points to your other questions:

In chemical engineering we sometimes face real gases, so the approach to obtain more accurate values is the followinge. First we calculate all thermodynamic properties as gases behaved ideally, and then we correct them. This makes some derivations easier, hence why residual properties exist. For example, you know that for a constant temperature process, the change in Gibbs free energy when pressure changes is
\begin{equation}
  \Delta G^\mathrm{ig} = RT\ln\frac{p_2}{p_1}
\end{equation}
How can we correct this value for real gases? We write Eqn. (7) two times, one for $p_1$ and the other for $p_2$, and subtract them
\begin{align}
  \frac{G^\mathrm{R}}{RT}(p_1) = \frac{G^\mathrm{R}}{RT}(p \to 0) &+
  \int_0^{p_1} \dfrac{Z-1}{p}\mathrm{d}p \qquad
  \frac{G^\mathrm{R}}{RT}(p_2) = \frac{G^\mathrm{R}}{RT}(p \to 0) +
  \int_0^{p_2} \dfrac{Z-1}{p}\mathrm{d}p \\
  \frac{G^\mathrm{R}}{RT}(p_2) - \frac{G^\mathrm{R}}{RT}(p_1) &=
  \int_0^{p_2} \dfrac{Z-1}{p}\mathrm{d}p - \int_0^{p_1} \dfrac{Z-1}{p}\mathrm{d}p \\
  \frac{\Delta G^{R}}{RT} &=
  \int_{p_1}^{p_2} \dfrac{Z-1}{p}\mathrm{d}p \tag{9}
\end{align}
That $J$ that appears in book has to do with what happens with the quantity $G^\mathrm{R}/RT$ as the pressure goes to zero. Sometimes, we can state how a thermodynamic property will behave in some extreme limits. In this case, we don't know what happens to this quantity in general. It is said to be indertermined However, don't go so deep, because you will always be interested in changes of $G$. As you see in Eqn. (9), the constant is blown off, so we are happy. We could have chosen other lower limit of integration and things would have worked fine.
If you want to have higher precision in your calculations, you need to integrate Eqn. (9). You can do this with measurements of $pVT$ data in the laboratory, or use an equation of state where $Z$ may be function of two variables.

Until this moment I didn't noticed entropy just "disappeared" when applying the identity with $G/RT$, your answer cleared a lot of my doubts, thank you!

Let's start with the first identity (line $5$). Maybe you are not familiar with differentials. Maybe you prefer reasoning with derivates. So transform your first differential expression, namely $d(\frac{G}{RT})= ... $  into a derivate with respect to any other variable, for example the time $t$. So $G$ and $T$ become $G(t)$ and $T(t)$ and you can calculate the derivate of $\frac{G(t)}{RT(t)}$ with respect to $t$. A simple application of calculus gives : $$\frac{d}{dt}(\frac{G(t)}{RT(t)}) =  \frac{1}{RT(t)}\frac{dG(t)}{dt} - \frac{G(t)}{RT(t)^2} \frac{dT(t)}{dt}$$ Hopefully you understand this quite general formula, which has nothing to do with physical chemistry. If not, you should open your book about linear algebra.
Now all you have to do is to multiply this expression by $dt$, suppress the parenthesis $(t)$ and the identity on line $5$ is obtained.
The second expression has nothing special. It is just a definition of the parameter $J$.
The third expression has no meaning for me, as I don't know what is $Z$, and I don't know why the pression $P$ is introduced here.

Z is the compressibility factor.

$\frac {G^R}{RT}$ is related to the fugacity coefficient


text:
Behavior of the Vosco-Wilk-Nusair correlation functional
Assumptions:

interacting system

uniform gas system

In the Kohn-Sham model, the exchange correlation functional is introduced to account for correlation energy $\epsilon_{xc} = \epsilon_{x} + \epsilon_{c}$ is the exchange correlation energy per particle for a uniform electron gas where $\epsilon_{x}$ is the exchange energy per particle for a uniform electron gas and $\epsilon_{c}$ is the correlation energy per particle for a uniform electron gas.
The correlation energy of an interacting electron gas system has analytic solutions only in the limiting cases:
$\lim_{\rho \rightarrow 0}: \epsilon_{c} (\rho) = 0.311ln(r_{s}) - 0.048 + r_{s}(A^{0}ln(r_{s})++C^{0}) \rightarrow r_{s} >> 1$
$\lim_{\rho \rightarrow \infty}: \epsilon_{c}(\rho)=0.5[\frac{g_{0}}{r_{s}} + \frac{g_{1}}{r_{s}^{3/2}} + ...]$
However, for regions within the bounds of the limiting case, the Vosco-Wilk-Nusair correlation functional is invoked.
$\epsilon_{c}(r_{s})=\frac{A}{2}[ln(\frac{x}{X(x)} + \frac{2b}{Q}tan^{-1}(\frac{Q}{2x+b})-\frac{bx_{0}}{X(x_{0})}[ln(\frac{(x-x_{0})^{2}}{X(x)})+\frac{2(b+2x_{0})}{Q}tan^{-1}(\frac{Q}{2x+b})]]$
What does $A, r_{s}, x, X(x), Q$ refers to?

A very quick google shows up https://math.nist.gov/DFTdata/atomdata/node5.html#SECTION00021200000000000000 amongst other hits - note the definition is slightly different from yours and I think you want $x^2$ in the first term above, see also https://www.molpro.net/manual/doku.php?id=density_functional_descriptions. Quoting the first, with images where appropriate as it will be non-trivial to reformat:

The correlation term is related to the function,
where we have $x={r_s}^{1 \over 2}$, $X( x )=x^2+bx+c$, $Q=(4c-b^2)^{1 \over 2}$. The parameters $x_0$, b and c, given in the table below,
are used to create three instances of F, using the table below.

$r_s$ is described on https://math.nist.gov/DFTdata/atomdata/node4.html#SECTION00021100000000000000. Again quoting with images hwere appropriate

The electron gas parameter $r_s$ ... are defined as

... where n is the electron number density (implicitly a function of
the spatial coordinates)


text:
Coarse-grained algorithms
I am fairly new to the field of oarse-grained molecular systems. I am aware that there are a lot of different already developed CG algorithms. Since there are so many, I struggle to find what I'm looking for.
I wonder if there is a CG algorithms that simplifies a molecule drastically, for example a method that describes a mAb with around 20k atoms with only 12 or 20 beads while not assuming the number of beads beforehand, but rather choosing a partitioning that is most favorable according to a certain metric. Is any such method developed and also open-source?

You are probably better off asking this on the Matter modelling forum

BTW I would spell out CG as "Coarse Grained" - I automatically read it as "Conjugate Gradient" and there may be other confusions.

"partitioning that is most favorable": It depends what property of the antibody you want to model. You might be interested in antigene binding, aggregation, conformational change, transport properties, - you name it. Without knowing what you want the model to do for you, it is impossible to evaluate the model or come up with a metric.


text:
Gabriel Phthalamide Synthesis of 1-Phenylethylamine
I had learnt that due to the nature of the Gabriel synthesis reaction being $S_{N}2$, the alkyl halide used must have a primary alkyl group. However, in an exam (JEE Main 2023), it was mentioned that the following compound may be prepared by Gabriel synthesis.

My question is: is it true that 1-Phenylethylamine may be prepared using Gabriel Phthalamide synthesis? If so, how is it possible? I mean, won't the nucleophile in the reaction preferentially act as a base and cause an E-2 reaction instead since the alkyl group is secondary in nature?

This works because of the reactivity of benzyl halides due to the stability of the intermediate benzylic cation. Phthalimide anion is not a good base.

@Waylander But in Sn2 no intermediate cation is formed.

You sure?  The reaction is a mixture of SN2 and SN1 because of the stability of the benzylic cation

Well that's what I'd learnt at least... I may be wrong but the sources I've read from say it's Sn2... however, if you have any resources that suggest otherwise, I'd love to know

@Waylander that's the reason as far as I know, generally Gabriel synthesis only works for primary halides.

I question whether or not the problem is made up or from the chemical literature.

@SrishDutta  I would not use Gabriel for this transformation, Sodium Azide followed by reduction is superior by a long way. However having 40+ years experience as a lab-based synthetic chemist I am confident in saying that the stability of benzylic cations is a real effect.

@Waylander I agree with the fact that Gabriel synthesis isn't the most useful here, however, the problem solution states it is which I don't understand how... as you mentioned stability of benzylic cations can contribute but if so can you explain how that aids the Sn2 process exactly

@SrishDutta  The reaction is a mixture of SN and SN2. The stability of the benzylic cation aids the SN1 reaction. Gabriel is SN2 WITH PRIMARY HALIDES, here it is SN1/2 mix.

But then why does the reaction fail in case of tertiary halides and other secondary halides

In the case of secondary halides the cation formed by E1 is less stable than benzylic cations and eliminates to alkenes faster than capture by Phthalimide anion. In the case of tertiary halides, steric hindrance prevents it. Phthalimide anion is too bulky a nucleophile to efficiently attack the cation.

Oh alright, that's good to know... @Waylander I have a few doubts regarding the mechanism, would it be okay to move it to a chat room?

Let us continue this discussion in chat.


text:
Can heat cause the excitation of argon
Can argon be excited by heating? I work with a process that involves heating titanium to 500c-1000c in an argon environment. On our camera, the material looks purple when heated, the same shade as excited argon. It doesn't show with the naked eye but potentially the camera is seeing something that we cannot. Odds are this is just a camera malfunction but it's interesting to think that there is some process at work here.

How is the titanium being heated exactly? If there is an electrical discharge involved, then excited argon atoms are possible.

I think it's much too low temperature.

Is the camera calibrated? A digital cameras job is to give an image not necessarily accurate.

@EdV the heat source is from friction

Interesting! Can you put a cheap plastic transmission diffraction grating in front of the camera, to look for possible argon atomic emission lines? Or just use the grating and view by eye? I happen to be working with the argon emission spectrum (in the 400 to 700 nm range), so I will test what I just suggested using several argon lamps/discharge tubes.

Stir welding/fusion? At the point of contact, 1. there can be local hot-spots, perhaps enough to ionize Ar, and 2. If there were an oxide coating, there could be triboelectric effect (https://en.wikipedia.org/wiki/Triboelectric_effect), though I doubt there would be in the bulk metal. BTW, triboelectricity also explains earthquake-related lightning: https://www.aps.org/publications/apsnews/201406/earthquake.cfm\


text:
Why is the work done on the system not equal in frame of gas and surroundings?
Consider an isothermal irreversible process for an ideal gas in a cylindrical container closed with a piston (system). Initial and final states are $(P_1,V_1,T)$ and $(P_2,V_2,T)$ respectively. Initially the pressure of surroundings was $P_1$ and which is increased to $P_2$,
Work done by surroundings on the system is $W_1=P_2(V_1-V_2)$ which is in the frame of surroundings. If we see in the frame of gas, work done on system is $W_2=\int -P_{\text{gas}} dv$
Since, process is isothermal and $P_{\text{gas}}V=nRT$ $$W_2=nRT\ln{\frac{V_1}{V_2}}$$
I am unable to figure it out that why is $W_1≠W_2$ because according to the first law it should be same (change in internal energy and heat exchange is same in both the frames) . Can please someone help me out?

You used the ideal gas law to get Pgas for the irreversible non-equilibrium process?  Do you think that is allowed?

@ChetMiller why not?

Are you familiar with the dissipative property of fluids (liquids and gases) known as viscosity?  Have you ever heard the term "viscous Newtonian fluid?"

@ChetMiller Yes.

If you are familiar with both of these, then are you familiar with the 3D mathematical form of the equations for the Cartesian components of stress tensor in terms of the thermodynamic pressure, the fluid viscosity, and the velocity gradients in the fluid?

@ChetMiller No (I am a high school student), but If you give me hint about what is going in the background of the process inside the gas, that would be helpful too.

@ChetMiller, Yes I am familiar with the stress tensor but I am unable to guess the reason using this. And I am desperate to know. Can you please write answers?

It might be helpful to think of a gas in a cylinder, not just as a compressible spring (in terms of pressure-volume behavior), but as a combination of a spring in parallel with a viscous damper (dashpot).  The spring would account for the equilibrium behavior at low rates of volume change, but the damper would come into a play at significant rates of volume change.  So the behavior of the gas would not just depend on volume,  but also on rate of change of volume.

@ChetMiller Do you mean the gas will not be ideal anymore? But how is it possible initially and finally the gas is ideal but not during the process? And are we able to deduce pressure inside to container as a function of volume?

The gas will still satisfy the ideal gas law under equilibrium thermodynamic conditions.  But, just because we call it an ideal gas at equilibrium does not mean that it will satisfy the ideal gas law under non-equilibrium conditions.  A gas or liquid behaves as a Newtonian fluid under all conditions (both equilibrium and non-equilibrium), but  a Newtonian fluid reduces to the ideal gas law (or other real gas equation of state) at thermodynamic equilibrium

@ChetMiller where could I get text regarding this in much more detail? Does Atkins have it? My teacher never taught that ideal gas equation is valid only under equilibrium conditions.

Maybe your teacher isn't aware of this.  See Transport Phenomena by Bird, Stewart, and Lightfoot, particularly chapter 1.

@ChetMiller I think the book you told. I think it isn’t for high school student? Is it? I read the book from library I wasn’t even able to understand the first page. Can you suggest something introductory?

You were looking at page 1 of Chapter 1, right; not Chapter 0.  Well, this is kind of a complicat subject, and is built on gradually.  The math requires calculus and partial derivatives.  I don't think they cover partial derivates yet in HS.  I don't know what else to suggest.  Part of Chapter 1 (section 1.2) gives the physical motivation for the Newtonian fluid derivation.

@ChetMiller okay I will try to. I meant chapter 1 to chapter 0 and thanks for your suggestion.

This is for @Chesx
For non-polymeric fluids (liquids and gases), Newton's law of viscosity very accurately describes the state of stress within a fluid as a function of the local fluid deformation and local rate of deformation.  It also reduces to the ideal gas law (or other thermodynamics equilibrium equation of state) in the limit of low rates of deformation (i.e., low velocity gradients).
In Cartesian coordinates, the components of the compressive stress tensor are related to the thermodynamic pressure p(v,T) and the fluid velocity gradients by:
$$\sigma_{xx}=p(v,T)-\frac{2}{3}\mu\left[2\frac{\partial u_x}{\partial x}-\frac{\partial u_y}{\partial y}-\frac{\partial u_z}{\partial z}\right]$$
$$\sigma_{yy}=p(v,T)-\frac{2}{3}\mu\left[-\frac{\partial u_x}{\partial x}+2\frac{\partial u_y}{\partial y}-\frac{\partial u_z}{\partial z}\right]$$
$$\sigma_{zz}=p(v,T)-\frac{2}{3}\mu\left[-\frac{\partial u_x}{\partial x}-\frac{\partial u_y}{\partial y}+2\frac{\partial u_z}{\partial z}\right]$$
$$\sigma_{xy}=\mu\left[\frac{\partial u_y}{\partial x}+\frac{\partial u_x}{\partial y}\right]$$
$$\sigma_{xz}=\mu\left[\frac{\partial u_z}{\partial x}+\frac{\partial u_x}{\partial z}\right]$$
$$\sigma_{yz}=\mu\left[\frac{\partial u_z}{\partial y}+\frac{\partial u_y}{\partial z}\right]$$where the u's are the components of gas velocity, v is specific volume, and $\mu$ is the gas viscosity.
In the limit of low gas velocities, these equations reduce to the isotropic thermodynamic equilibrium state values of the stress components:$$\sigma_{xx}=\sigma_{yy}=\sigma_{zz}=p(v,T)$$and$$\sigma_{xy}=\sigma_{xz}=\sigma_{yz}=0$$
For a fluid satisfies the ideal gas thermodynamic equation of state $$p=\frac{RT}{v}$$
We also see from all this that the isotropic thermodynamic pressure p does not capture the state of stress in a gas that is deforming at finite rate. In fact, what we call the pressure is not the force per unit area on an internal surface  of arbitrary orientation, but rather it is merely the isotropic part of the overall stress tensor.


text:
Is my CO2 in liquid form in a mixed gas cylinder?
I recently acquired a 5-liter mixed gas cylinder that has 70% nitrogen and 30% carbon dioxide, used for dispensing beer. The cylinder pressure is 200 bar (2900 PSI) and my room temperature is around 20C to 22C (68F to 72F).
I've started to wonder whether the gas ratio can be disturbed by the CO2 changing its thermodynamic state due to room temperature fluctuations. Given the parameters, should I expect the CO2 to be in gaseous or liquid form?

To a first approximation, it seems the carbon dioxide is near its liquid/gas equilibrium curve, though it's hard to be precise because the ideal gas approximation probably isn't very good in these conditions. If any of it is liquid, then yeah the ratio is going to be affected. If you gently swirl/rock the cylinder, can you detect any sloshing?

Without particular gas data it is hard to say the gas condition. For illustration: Nitrous gas N2O has comparable critical temperature as CO2. While it is liquid at room T at 100 bar, the mixture 1:1 (V/V) with oxygen is at 200 bar gaseous.

I can only assume that the store that sold you the tank knows what they are doing and that it will work for the intended purpose. So use it and don't worry.

Calculated CO2-N2 Pressure-temperature phase diagrams, albeit at lower N2 concentrations, can be found in an Energia paper. With the addition of N2, the line that was the gas-liquid boundary (terminating at the critical point) now splits into a vapor saturation line and a liquid saturation line which meet at a critical point. You are concerned that your gas mix may fall into the two-phase (liquid+gas) area between those two curves.
In Figure 4 of the paper (below, annotated by Karsten) they authors show the calculated phase diagram for a 80% CO2 - 20% N2 mixture. For pure CO2 the critical point is at (30.978C, 73.773 bar). With only 20% N2 in the mix the critical point has already moved to (14.55C, 105.47 bar).
With 70% N2 in the mix, you will not be in the two phase region for any conditions you are likely to encounter.

@Karsten - thank you! I put that in. I guess I'm just too used to phase diagrams... That is much clearer.

Thank you a ton, both of you! I also assume that the CO2 does not settle to the bottom of the cylinder even though it's denser than nitrogen? I'm just wondering because the cylinder has a dip tube in it, drawing the gas from the bottom.

There is enough mixing to keep it uniform throughout the cylinder. I used to run a Van de Graaff accelerator with a giant pressure vessel holding an N2-CO2-SF6 mix to hold off voltage. There was no indication that the SF6 separated out, much less CO2 vs N2.

Got it. One concern though. In the diagram, the percentage of gas is reported as mol%. I was wondering if it would make a difference if the gases were measured by volume or mass? I see some information floating around that the CO2 partial pressure would be 60 bar, which would mean that my margins for being either in liquid or gas state are very fine, given my room temperature is 20C.

The margins would be fine if you had 80% CO2 in the mix. This just shows how quickly the critical point (and the 2-phase region) move with adding a bit of N2. At your mix, the phase diagram will look more like N2, so you are far away from any issues.


text:
Is Ksp measured for covalent compounds as well?
A quick search on Google tells me that it is only used for ionic compounds but I think it should be used for some covalent compounds given that some of the covalent compounds also dissociate into ions when they dissolve (like most acids)

You can use it because there is always a mix of characteristics. Only that it will be extremely small.

It can be used with any equilibrium process.

Ksp is an equilibrium constant assuming that the dissolved compound dissociates completely. For compounds that are only partially dissociated (eg weak acids), it is not applicable. In those cases a total observed solubility at a specific pH is necessarily used instead.


text:
Can someone help with GAMESS input files for heavy atom calculations (MoS2, WS2, etc.)?
I'm doing computational research on transition metal dichalcogenides for my chemistry class and I want to calculate a transition state optimization for the reaction $\ce{MoS2 +N^- -> MoSN + S^-}$. However the geometry optimizations refuse to converge after 200 iterations. This is being done with a MCP-TZP Data Set within GAMESS on WebMO. Any help would be appreciated.

You should give some more details on the job you are running so anyone acquiented with GAMESS can give you some advide.

Sibling site matter modelling may be an audience more suitable because your question then may be tagged by gamess for either GAMESS(US) or GAMESS(UK).

Generally I agree that it is more suitable for [mattermodeling.se], but I'm not sure the question is complete enough. Maybe add an input for what you currently doing. Also, it could be a chemical issue and not a program issue. Is this a known state? What is you initial guess for the TS based on?

MоS2 isn't a molecular compound, you should use a solid-state modeling software like quantum espresso instead


text:
In a few species such as dicarbon (shown below) why are the pi bonding orbitals of lower energy than the sigma bonding orbitals?

Normally the pi bonding orbitals are of higher energy because pi orbitals experience less direct overlap than sigma orbitals however in a few species this is not the case and I could not find any websites directly addressing this topic.

The molecule $\ce{C2}$ is extremely instable. To my knowledge, it only exists in the famous blue cone to be seen in the Bunsen flame. Its lifetime should probably be less than one millisecond. I would be pleased to know how and who somebody has decided that the first pi level is lower than the first sigma level. Is it not a mistake ?

I should note that dicarbon is in fact very unstable but this pehnomenon is observed in a few other species such as N2 as well.

How to justify that the first pi level is lower than the sigma bond made with 2p orbitals ?

C2 is a very weird molecule. I don't think all theoreticians agree how does it really bond.

https://chemistry.stackexchange.com/questions/594/bonding-in-diatomic-c2-a-carbon-carbon-quadruple-bond

https://chemistry.stackexchange.com/questions/85223/in-c2-molecule-both-bonds-of-the-double-bond-are-pi-bonds-but-isnt-the-rule-t https://chemistry.stackexchange.com/a/32552/9961


text:
Why doesn’t the reaction between 3-mercaptopropionic acid and methyl iodide yield a methyl ester?
Problem

Answer

Question
I don’t understand why 3‐methylthiopropionic acid (X) is the correct answer. Wouldn’t sulfur and carbon have similar electronegativities?
As the consequence, isn’t it more likely that the second step involves hydroxyl’s oxygen forming a dative covalent bond with the partially positive carbon of methyl iodide?

You don't really have an alcohol. With the carbonyl group also there you have a carboxylic acid instead. The combination of a carbonyl group directly bonded to hydroxyl is in fact considered a separate functional group called "carboxyl".
With this coupling to the electron-withdrawing carbonyl group, the hydroxyl portion of a carboxyl group is not nearly as nucleophilic as it would be in a simple alcohol and also not as nucleophilic as the hydrosulfide function at the other end of the W molecule. The methyl iodide then attacks at the more nucleophilic site to give the X product.


text:
Fractional Distillation Fractions
Currently, I am trying to remember the order of fractions within the fractionating column. However, I have found a contradiction.

Consider the above diagrams, both from reputable sources. The first diagram places lubricating oil above that of fuel oil. However, the bottom diagram places them the other way around. Which one is correct? Looking for a definite answer.
EDIT: I have found this contradiction so many times between various sources.

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

You may be in trouble with getting a definite answer. I don't think that either thing is really a fraction - they are rather products made from various stuff got in distillation, or even otherwise.

https://en.wikipedia.org/wiki/Fuel_oil  https://en.wikipedia.org/wiki/Motor_oil

Mention the sources because the text is not readable.

Googling fractional distillation of petroleum and going on images, you find the contradiction many times


text:
Determine potential and electric field distribution between electrodes
I want to analytically determine a simplified potential and electric field distribution around planar electrodes, and an 40nm square electrode in an 50nm insulating well.
I think that the Poisson-Boltzmann equation should be appropriate but I’m unsure about a few things:

how do I account for the presence of the stern layer?

how do I find the electric field distribution when the potential applied is high (e.g. 2V? Where the Debye Huckel approximation cannot be used)

Cheers,
Bryan


text:
Is a realistic chemistry system viable for video game design (restrictions apply)?
I assume this might be an unusual question, and I'd like to first make a note as to why I'm asking here, on Chemistry Stack Exchange and not somewhere more relevant to video games. This is because we'd like to get an answer from people with knowledge in the field of chemistry since we want to approach the issue from the direction of verisimilitude.
We have a simulation/adventure game with the capacity for complex item manipulation. This means that we can already handle things like containers for substances, laboratory equipment, item combinations (including combinations which require multiple inputs and specific tools), etc. These systems were developed for other reasons than specifically a chemistry system, but the idea of building one cropped up.
The issue is that we are not informed enough to properly assess the feasibility of building such a system, especially under the limitations described below. And if such a system was possible, we are unsure where to start looking up information about it.
The restrictions that we're working under are:

We make a game where the player character is non-violent and thus have no interest in including things such as explosives or poisons. Instead, we are interested in puzzle/environmental problem solving and possibly some medicine;
We'd be simplifying things to some degree, partly for quality-of-life reasons and partly so that people can't follow the in-game processes in real life. I will additionally note we work on a horror game for adults;
If it were to exist, it would need to be an actual system rather than including specific recipes for crafting without multiple uses for the components;

We'd appreciate any opinions on the matter, as well as any more specific suggestions for material we should study.

I doubt it.
There is a sophisticated system to learn about genetics through a game: https://concord.org/teaching-genetics/dragons/
There is a game to try to figure out protein folding: https://en.m.wikipedia.org/wiki/Foldit
The chemistry domain that might fit is organic synthesis, with rule-based or AI based prediction engines available: https://www.sciencedirect.com/science/article/pii/S2095809922005665
All of these have development costs that are probably beyond your budget.
There are also pandemic-borne simulations of teaching labs, but those are script based and boring.
So I doubt it’s possible to make something fun within the chemistry domain on a small budget, but it would be great if I’m wrong.
You could also look I to board games for inspiration, e.g. covalence, https://www.teachersource.com/product/covalence-molecule-building-game.

Thank you for your response. In broad terms, I absolutely agree with you.
However, my post was specifically related to everyday chemistry, and as such, what you are referring to has little to do with what we were considering doing. We don't want to gamify anything like organic chemistry or genetic. We're looking to add to already existing systems, in a game that is fundamentally not focused on chemistry.
So for example, environmental problem solving means we're looking for the system to be useful when the character traverses the environment or interacts with objects within it.

You want "multiple uses for the components". You could have different miniaturized analytical instruments (like the Star Trek tricorder) to help solve environmental problems. There could be one for organic volatiles, one for elemental analysis, and one for water-soluble organic substances. And there could bet a set of tools and a set of reagents to sample to environment and get samples into a state ready for analysis. Otherwise, all I can think of for everyday chemistry are different solvents (cleaning, dissolving evidence), and ?pool chemistry? (measure pH, chlorine, and add acid, base, ...)

The "pool chemistry" idea is actually very interesting. Thank you. I'll look into that.


text:
At 80°C a solution having pH = 7 is?
When I came across this question I couldn't understand whether, they are asking the nature of solution at 25°C which has pH = 7 at 80°C  or nature of solution at 80°C which has pH = 7 at 25°C ?
Because in the first case answer would be acidic and in second case answer would be basic

for sure it refers to some exercise about Nerst Equation and dependence of pH on temperature but please give more context and try to reformulate the question. https://chemistry.stackexchange.com/help/how-to-ask

The neutral $\pu{pH}$ is half the $\pu{pK}$ value for water dissociation, and of course the latter is almost exactly $14$ at $25°$C and one bar. That's where we get $\pu{pH}=7$ for a neutral solution — under ambient conditions.
The open source reference [1] gives an expression for $\pu{pK_w}$ as a functoon of temperature and pressure. Because water autoionization is endothermic, $\pu{pK_w}$ will go down on heating and the neutral $\pu{pH}$ will follow suit. For instance, at low pressure and $75°$C $\pu{pK_w}=12.70$, so the corresponding neutral $\pu{pH}$ would be $6.35$. A solution with $\pu{pH}=7$ at this temperature would be basic; if cooled to ambient temperature the solution would autoionize less and we'd see the $\pu{pH}$ rise above $7$.
Reference

Andrei V. Bandura; Serguei N. Lvov (2006). "The Ionization Constant of Water over Wide Ranges of Temperature and Density" . J. Phys. Chem. Ref. Data 35, 15–30.
https://doi.org/10.1063/1.1928231


text:
Do conformations arise from vibrations?
I was wondering how a molecule transforms from one conformation to another. Surely, since the conformations are separated by a potential barrier, some activation energy must be required. The energies related to a molecule are translational, rotational, vibrational, and electronic. Out of these, translational and rotational motion do not change the intrinsic coordinates of a molecule and won't bring about a conformational change. The electronic transitions are too fast on the conformational time scale. This leaves, at least for an independent molecule, vibration as the only candidate for bringing about conformational transformations. Although collisions may also be considered, my focus is on independent molecules. I tried searching for a relation between conformations and vibrations but couldn't find any relevant results.
Further Clarification
Thank you everyone who has answered the question. It is clear that vibrations do result in conformers by changing intrinsic coordinates and many computational models utilize this in conformational analysis. Qualitatively, I am able to grasp the way this occurs. However, none of sources talk explicitly about physical emergence of conformations. If, for example, I could bring about a conformational change through radiation corresponding to a normal mode, then that would answer the question in affirmative. Moreover, knowing about the mechanism by which such a change would occur would be deeply illuminating.

You probably need to define conformation for this kind of question quite rigorous. And then what is a vibration. And then you might want to look at transition states. I guess it is again not that simple a question to answer. Or it's very simple to answer, but then the answer becomes meaningless.

@Martin-マーチン, as I understand, conformations arise from (un)restricted rotation of bonds. (3N-6) vibrations correspond to various symmetries of the molecule not assigned to translation or rotation.

Here is a source: https://www.biorxiv.org/content/10.1101/2022.11.30.518608v1.full.pdf

And another one: https://arxiv.org/pdf/2103.11959.pdf

@Karsten thank you for sharing sources.

I am confused because the update in the question takes the question in one new direction, while the bounty text takes it into two other new directions (directions which I don't quite understand).

@Karsten sorry for the confusion. Let me summarize the question more clearly. (1) Do conformations physically arise from vibrations? If so, what is the physical mechanism by which conformers are generated by vibrations? (2) How to explore the conformation space via connectivity or an initial structure? Question (2) has, more or less, been answered. Question (1) is what is bothering me. Knowing the physical mechanism by which vibrations result in conformers, would be insightful to answering question (2) as well. Hope this helps.

Cross posted on Matter Modeling SE. Its generally encouraged to either not cross post questions or include links to all versions of the question to avoid repeated work

@Tyberius noted.

Cross-posting is highly discouraged up to the point where questions get instantly closed. Then one question can survive on the platform that is best suited for it, the other one is deleted. The bounty makes things more complicated here. Questions can't be closed merged, etc., they are basically locked. There are of course exceptions: But these are more like follow-up questions then duplicates.

On another issue: A question should not morph this much, especially after it has gotten already a decent amount of interaction.

@Martin-マーチン, I don't think the question is morphed much because of the Further Clarification. If you notice, I was asking about the physical nature of the emergence of conformations in the original questions, and this was unanswered because most answers focused on the computational generation of conformations.

Perhaps the question should be How do conformations arise from vibrations? I think also this post could have been closed due to a lack of clarity, although I now see from the comments what the OP meant. The existing answers explain that YES! vibrations do drive formation of new conformations, but the OP wanted to know how this happens. I wonder if there is an easy way to explain this in a concise way that does not employ advanced stat mech?

In conformational analysis, it is customary to describe a molecule by its internal coordinates, namely bond lengths, bond angles and torsion angles (or dihedral angles, with a slightly different definition compared to torsion angles).
Typically, two states of a molecule with different conformations are characterized by a difference in torsion angle. Often, there are multiple energy minima along a torsion angle coordinate. At low temperature, you can imagine the molecule being trapped in one of those (have "that conformation") while at high temperature, it might have access to multiple conformations, spending more time in the low energy ones.
Vibrations are often discussed in terms of normal modes, where all atoms move in some concerted manner. For the water molecule, we know the symmetric and asymmetric stretch (bond lengths change) and the bending mode (angles change). Water does not have a torsion angle, and you might not thing of water as having different conformations.

Source: PChem at Sam
You need at least 4 atoms for a torsion angle. Hydrogen peroxide (above) is a good example. Of the 3N - 6 = 6 normal modes, one is on a path that will change the torsion angle (the rightmost, labeled Au and called dihedral bend). At low temperature, hydrogen peroxide has a gauche conformation, with the torsion angle varying slightly (but staying either in the gauche+ or gauche- conformation). At higher temperatures, it can show rotation around the central bond to switch between gauche- and gauche+ via planar arrangements of the atoms.
In an isolated molecule, the energy to change conformation comes from the other normal modes through coupling (which might or might not be symmetry-permitted), like in a double pendulum.

I was wondering how a molecule transforms from one conformation to another.

In the context of normal modes, the molecule stays in the same conformation when you consider harmonic vibrations (vibrations around a single minimum). To go from one conformation to another (i.e. change at least one torsion angle substantially), you have to have sufficient activation energy to "escape" the local minimum, i.e. have to consider anharmonic vibration models.

Do conformations arise from vibrations?

In the normal mode picture, conformational change is a vibration along those modes that change torsion angles. So you could say that a gauche+ to gauche- conformational change in hydrogen peroxide is linked to the out-of-plane vibration labelled "Au" in the image above.

This leaves, at least for an independent molecule, vibration as the only candidate for bringing about conformational transformations. Although collisions may also be considered, my focus is on independent molecules.

Yes, for an isolated molecule, you would have coupled vibrations. Sometimes, the normal mode associated with a given torsion angle (or set of torsion angles) would have sufficient energy to break out of the local minimum to reach another conformation. If at that time, the energy is transferred back to the other modes, the molecule would remain in the new conformation until their is sufficient energy to break out of the new local minimum again.

Just a note on the example of peroxide. That mode Au has both Hs moving into or out of the plane in concert (I can't tell which way the arrows are supposed to be pointing). Therefore it isn't really analogous to a torsional mode. In a torsional mode the two atoms would move in opposite directions, one into, one out of the plane. I think the motional mode analogous to a torsional mode about the OO bond would be a rotational mode, not vibrational, with arrows in opposite directions.

I disagree. The figure implies that the oxygen atoms stay in place. The torsion angle changes if both hydrogen atoms move up. The molecule rotates around the O-O bond if one hydrogen goes up and the other goes down.

I understand that the Os don't move. The Au mode moves the Hs into a staggered conformation (towards being eclipsed). I was thinking that peroxide was being used in the example as a placeholder for a more complex molecule with a soft mode about the bond being rotated. My guess is the barrier for the out of plane mode is much larger in comparison to say a CC rotation in an alkane? I guess I should look that up.

It is reasoned in the source that some vibrations have the same symmetry as rotations. But these are molecular rotations, and not rotation around chemical bonds resulting from change in intrinsic coordinates. The various rotations in character tables refer to rotation of coordinate systems.

Your broad intuition is right, conformational changes–like the flipping of a cyclohexane ring–happen because of combinations of vibrations and rotations in the molecule.
But it helps to put this in context. Even at room temperature there is a lot of thermal energy available to drive vibrations and rotations in many molecules. And that thermal energy is well distributed across many of those modes (and molecular motion of the whole molecule, molecules banging into each other is happening very fast and redistributes vibrational and rotational energy).
Flipping a cyclohexane ring (swapping equatorial for axial hydrogens or other substituents) requires coordinated changes across all the bonds in the molecule. (Remember, If the ring were "frozen" in the chair conformation, there would be two distinct types of hydrogen in a cyclohexane but at room temperature we see only one which implies the chairs flip rapidly). The lowest energy conformation is the "chair". But other conformations (eg the "boat") are also seen but have higher energy. But there is enough thermal energy at room temperature to drive the specific series of vibrations and rotations that convert chair to boat and to the opposite chair conformation.
The activation energy barrier to the flip is ~40kJ/mol which is small enough to overcome from normal thermal energy at room temperature. But it isn't just a single vibration or rotation that achieves this, it is a combination of many. Since the thermal energy is well distributed the sequence of vibrations and rotations is readily explored quickly at room temperature.
It is also worth noting that your claim that "translational and rotational motion do not change the intrinsic coordinates of a molecule and won't bring about a conformational change" is clearly not true for the cyclohexane example. The combination of bond rotations do change the spatial relationships of the atoms in the molecule. The major rotations around the C-C bonds, for example, swap the orientation of the hydrogens attached to the carbons.
Many other molecules also have enough thermal energy to cause vibrations and rotations to change their conformations at room temperature.

Strictly speaking there are no rotations "in" the molecule. There are rotations "of" the molecule, but not "in".

@IanBush To be clearer, the rotations I mention are relative internal rotations of bonds in the molecule (eg in cyclohexane, HC-CH bond angles change).

Yes, I understand that, but given the OP says "Out of these, translational and rotational motion do not change the intrinsic coordinates of a molecule and won't bring about a conformational change" I think in the context of answering this question it is best to be clear quite what is a rotation and quite what is a vibration.

Yes, different conformations do arise from vibrations. And you've almost answered your own question. Conformations are different relative spatial arrangements of the atoms. Pure translations, rotations and electronic transitions by definition do not change the relative spatial arrangement, only vibrations can do that. Hence conformations arise from vibrations.
Now we often talk about rotating about a single bond to generate conformations. But strictly speaking that is a vibration, not a rotation, as it changes how atoms moved by that rotation are placed relative to the rest of the molecule - as chemists we often like to decouple the "interesting part" of a molecule and focus on the properties of that, similar is talking about the C=O stretch in a molecule when we really should be discussing normal modes, but in this case it leads to slightly sloppy language. That rotation is really a vibration. Hence conformations arise from vibrations.

A vibration is of dynamic (time dependent) nature, as such similar to a movie.  On the other hand, a particular conformation is static (frozen), similar to a still frame, or single photo.
In this analogy, a vibration is a sequence of events where a molecule passes multiple conformations.

So, some of those movements might result in a conformational change, in the sense that the new 'static' state is different from the previous one.


text:
How can I read an Anschütz/Bennert manometer and how can I make sense of it?
Recently I have come across a mercury-filled manometer for vacuum distillation that somewhat resembles a U-tube shaped manometer, but with one arm sealed and completely filled with mercury. Searching through the Internet, I have found that such manometer is being called Anschütz/Bennert manometer. Here is a drawing showing what I am talking about:

Also, the page "Vacuum or reduced-pressure distillation" leads to a website showing the whole vacuum distillation set up with such a manometer (Fig. 15.5):
Since one arm of this manometer is sealed and completely filled with mercury, it doesn't seem like the difference in height of mercury in both arms could be related to pressure difference between vacuum in the system that the manometer is connected to and air—as it is being described most commonly for U-tube shaped manometers that have one arm open to air. Thus it makes me a little confused what is the working principle here. I have been searching for an explanation, but I didn't find anything for this particular case.
How does such an Anschütz/Bennert manometer work and how can I read the vacuum pressure from it?

All operations should be gentle:

open the valve to the system evacuated
if the vacuum applied is good enough, the level of mercury in the right hand (open) leg rises, the level in the left hand (closed) leg decreases
after stabilization of the two levels, move the sliding rule behind the U shaped tube to read the total of the difference of the two levels in mm, which by filling with mercury equates to pressure in expressed in mmHg (link to a pressure converter, but no particular endorsement for this one).  Because the left leg is closed, the recording is about absolute pressure.
if no further monitoring is needed, close the valve to the system evacuated.

To return the «ground state»

check the valve, which initially should be closed
detach the manometer from the system to monitor
gently open the valve to return to atmospheric pressure; the level of mercury in the left hand leg should rise so that there is no empty head space left
close the valve again, and stash it in the shelf

If you return too rapidly to atmospheric pressure, the momentum of mercury can be enough to shatter the glass tube.

Depending on local law applicable, finding mercury manometers / thermometers becomes more difficult/impossible, though this short leg manometer as well as the swivel Mc-Leod type (the later more suitable for oil pump vacuum) still have their point beside the electronic ones.  But they can break (then there can be some trouble [OSHA]), are not so handy in an already crammed fume hood, have no data link etc.


text:
Student Teaching of RP and NP HPLC - Labor practise
I am teaching pharmacy students in the instrumental analytics course and my main topic is HPLC. currently we have a (in my opinion) boring students experiment were the student determine impurities in a benzocain sample. for this we use a typic RP-18 HPLC.
We have also access to an SFC System which works very similar to normal phase.
I want to add an experiment where I can show them the differences between NP and RP HPLC and if possible show of the advantages and disadvantages.
Do you know any protocol which we can adopt into our lab for the students course? For you information: the experiment can last up to 4h (measuring time not included).
Maybe you had some good courses in your university and know some good experiments.

The Journal of Chemical Education presents often protocols known to work (tested in classes) and appealing for an introduction into a technique.  By keyword HPLC there are for instance 164 records tagged as lab experiment. Back at school, the demos (on TLC) were extracts of carrots (carotenoids), and dyes (could be really something for RP!).

Hi, I had a look it. but I had the hope, that someone already did a similar experiment and could tell me his/her experience or give some recommendations (analytes, mobile phase etc)

If there are literature databases about (HP)TLC (example CAMAG) then possibly the manufacturers of the NP/RP HPLC columns equally have ones.  Mark Perri's chemcompute.org runs an ajustable simulation of a HPLC of caffeine/acetaminophene/aspirine (link, described in 2020JChemEduc2950) so there should be enough data out there, maybe some did the comparison NP vs RP.


text:
How does adding sulfuric acid to water affect the specific heat capacity of the solution?
How does adding diffrent concentrations of sulfuric acid to water affect the specific heat capacity of the solution?

Review the guide How to ask and Asking FAQs to prevent clarification requests, objections,  down-voting or closure.

A, S. Socolik, Z. Phys. Chem. $158$A, $305 - 312, (1932)$ gives the following heat capacities for the mixtures $\ce{H2SO4}$ + $\ce{H2O}$ at different mass percent at $20$°C:
% $\ce{H2SO4}$ .... $\ce{C_p}$ in cal/g/K
..$0$% -----------  $1.000$
$12$% -----------  $0.900$
$24$% -----------  $0.812$
$36$% ----------- $0.715$
$48$% ----------- $0.611$
$60$% ----------- $0.525$
$72$% ----------- $0.464$
$84$% ----------- $0.441$
$92$% ----------- $0.372$
$100$% ---------  $0.332$


text:
How is the central atom in SNF3, Nitrogen?
Shouldn't it be Sulphur as it is larger and has ability to make more bonds due to its 6 valence electrons?

It is sulfur https://en.wikipedia.org/wiki/Thiazyl_trifluoride

The formula is not written in structural order. It's written in electronegativity order.

Thiazyl trifluoride $(\ce{NSF3})$ is a chemical compound of nitrogen, sulfur, and fluorine, first reported in 1956. Its structure has been suggested to have the framework $\ce{F2S=NF}$ based on hydrolysis experiments. However, it was revised in 1961, based on infrared and fluorine nuclear magnetic resonance data by Richert and Glemser who have correctly interpreted the structure in favor of $\ce{N#SF3}$ (Ref. 1), which was confirmed unambiguously and quantitatively by mass and microwave spectroscopies (Ref. 2). According to Ref.2, the SN-bond length $(\pu{1.416 Å})$ is short, which is in the range expected for a triple bond of $\ce{N#S}$ from theoretical calculations:

The above structure is confirmed later again by crystallographic data (Ref.3). Accordingly, bond distances and angles in $\ce{N#SF3}$ in the solid state (X-ray diffraction, Ref.3) and gas phase (microwave spectroscopy, Ref.2) are well correlated.

References:

Hans Richert and Oskar Glemser, "Schwefel-Stickstoff-Fluorverbindungen. XI. Struktur einiger Schwefel—Stickstoff—Fluorverbindungen," Zeitschrift für anorganische und allgemeine Chemie 1961, 307(5-6), 328-344 (DOI: https://doi.org/10.1002/zaac.19613070512).
William H. Kirchhoff and E. Bright Wilson, Jr., "The Microwave Spectrum and Structure of $\ce{NSF3}$," J. Am. Chem. Soc. 1962, 84(3), 334–336 (DOI: https://doi.org/10.1021/ja00862a003).
Tobias Borrmann, Enno Lork, Rüdiger Mews, Simon Parsons, Jan Petersen, Wolf-Dieter Stohrer, and Paul G. Watson, "The crystal structures of $\ce{NSF3}$ and $\ce{(NSF2N(CH3)CH2–)2}$: How short is the ‘Crystallographic’ $\ce{N#S}$ triple bond?," Inorganica Chimica Acta 2008, 361(2), 479-486 (DOI: https://doi.org/10.1016/j.ica.2007.05.016).


text:
How to properly determine number of vibrational modes of CO in carbonyl complexes?
it's been a few years since I took inorganic chemistry and was starting to refresh some of the concepts. What I didn't really understood when I reread was the following: How to properly determine the number of vibrational modes of CO in carbonyl complexes.
I always understood the vibration of CO as the streching of the C-O-bond as in the diatomic molecule, disregarding the M-C bond; so only the internuclear distance of C and O changes and so does the dipole moment. That is the premise; is that correct?
Also, I recall that one can only see the asymmetric strech of CO when bound in a trans-fashion to the metal center, because the symmetric strech won't change the dipole moment. How come then that this strech on the left is IR-active:

As the C-O bonds both enlargen, the dipole moments should cancel? Or am I missing something?

Isn't that because the left molecule is polar, and right one is not. Because the first molecule is polar, any vibration will be IR active. Just like in water.

https://chemistry.stackexchange.com/questions/177246/how-to-properly-determine-number-of-vibrational-modes-of-$\chem{CO}$-in-carbonyl-complexes

That is the same question..

@Mäßige-Yes that may be right my friend, but here you have to make one thing, there is one part of this particular title which has to be changed into $\ce{CO}$ ASAP

https://chemistry.stackexchange.com/questions/177246/how-to-properly-determine-number-of-vibrational-modes-of-co-in-carbonyl-complexes

@Mäßige-You have to change the link given in the comment which I posted here on Jan 15 at 7:08 to https://chemistry.stackexchange.com/questions/177246/how-to-properly-determine-number-of-vibrational-modes-of-$\ce{CO}$-in-carbonyl-complexes prior to deleting the other comment that was posted on this website in the same day on 12:13

You are completely right in your thinking. Only assymetric stretching can be seen in IR spectroscopy, and this is exactly reflected by the diagram. The left part shows stretching above and below Me center, these cancel out; but on the rear there are 2 CO molecules contributing to the change of dipole moment.


text:
How does a fragment with a m/z ratio of 94 form when benzoic acid breaks down?

Hello, I don't understand how when benzoic acid forms a fragment with a m/z ratio of 94 during breakdown. I understand it is [C6H6O]+ but how does that happen, does the ring breakdown and McLafferty rearrangement occur. I can't even think of how it could rearrange itself for that to occur or if there is a different mechanism.

Ok I got it a phenol forms by Homolytic cleavage, that also explains the low intensity since alcohols are not stable.

The peak, $\ce{[C6H6O]^+ }$ could be an artifact of impurity ($\lt 0.1 %$). It could be $\ce{M+}$ peak of phenol, for example.

Yes that is the first thing I thought so as well at first, but this is the theoretical spectrum and it is the same everywhere online. I'm starting to get it a C=O fragments, it's not McLafferty, but it's done by none of the mechanisms that I know.

Loss of CO is a known way of fragmentation of aromatic carboxylic acids in mass-spectra. I didn't find a good description of the resulting ions, but I will point that mass-spectrometry is a high-energy process, so common chemical sensibilities may not apply.

A subtle hint may lie in the presence of the peak at $28$, which would likely represent carbon monoxide. Carboxylic acids decarboxykate to give $\ce{CO2}$, which can lose an oxygen atom under mass spec conditions to form $\ce{CO}$. The displaced oxygen plus a proton plus a phenyl radical (or oxygen plus a hydrogen atom plus a phenyl cation) could then form the apparent phenol cation.


text:
Can polychloroprene further react
Chloroprene undergoes radical polymerisation to form polychloroprene.
Chloroprene is a special structure since it has two carbon-carbon double bonds and it turns out that polychloroprene also has one carbon-carbon double bond.
In both addition polymerisation and radical polymerisation, the carbon-carbon double bonds in monomers are broken to form the polymers.
I wonder, if the carbon-carbon double bonds in polychloroprene can further undergo:
(i) Addition reactions
(ii) Addition/ radical polymerisation

It certainly can, that's like the main point of using it in the first place.


text:
Calcium oxide in self-heating packs
I've read some articles on the internet about self-heating meals. Heat released from an exothermic reaction is used to heat up the food. In fact, several reactions are recommended for this purpose including the dissolution of calcium oxide $\ce{CaO}$ in water. However, after doing some experiments, I've got into some problems with this method.
$\ce{Ca(OH)2}$ as a product is only slightly soluble in water. Therefore, the reaction stopped quickly after adding water as the remaining reactant is covered by this insoluble compound.
I've tried using acids to dissolve insoluble $\ce{Ca(OH)2}$. This method seems to be effective firsthand. However, choosing which types of acid that best fit this situation turned out to be a dilemma. As the pack is in white powder, solid form, some popular strong acid like $\ce{HCl}$ and $\ce{H2SO4}$ as liquids are not suitable. Other weak organic acids are also not considered as the corresponding calcium salts hardly dissolve into water.
My questions are:

I've heard that metals like Al, Mg and Fe are added also, however what is the function of these in the mixture with CaO? Absolutely Al can react with $\ce{Ca(OH)2}$ but this seems to be slow and therefore not the main reason as it cannot explain in case of Mg and Fe.
Is there any proper methods that is used in the commercialized, $\ce{CaO}$-based flameless heater? I feel like my approach is rather unpractical and somewhat expensive if it is formulated to an industrial scale.

Use fine CaO powder, or add some vinegar to water.

$1$) $\ce{CaO}$ expands when reacting with water. $\ce{Ca(OH)2}$ does not cover $\ce{CaO}$, and  does not prevent new water from getting in touch with "deeper" $\ce{CaO}$. $2$) The best way to dissolve $\ce{Ca(OH)2}$ is a  $\ce{HCl}$ solution. Why don't you want it ?

Not sure if this is used commercially, due to the danger of the reactants. For example, concentrated HCl would be a hazard, and dilute HCl would "draw off" too much heat, as the added water absorbs the energy. Same with acetic acid - vinegar is only5 or 6%, so you'd have a lot of water, and glacial acetic acid is hazardous.

If soluble products are what you want, try one mole of magnesium oxide (which is more stable in typical environments than calcium oxide) to two moles sodium bisulfate. These react to give sodium and magnesium sulfates, which are both soluble, and no gases. Excess sodium bisulfate may improve solubility (by buffering against magnesium hydroxide formation) but may make the solution about as acidic as vinegar.


text:
Does a chemical reaction depends on the amount of reactants?
A chemical reaction depends on many things like the temperature, pressure etc. i wanted to ask if It also depends on the amount of reactants for example;
If A + B gives C and assuming this equation is balanced, all other conditions are same but less amount of B is there so will the same reaction occur where B will be the limiting reagent and less amount of C will be produced or to compensate the lesser availability of B another product will be formed  like it happens in case of incomplete combustion of Carbon?
If yes then doesn't the concept of Stoichiometry kind of fails?
If reading this question makes you feel angry at how stupid i am please forgive me

Do you mean ratio of reactants, or amount of reactants? The amount could be significant if one takes the environment into account. For example, mixing one mg of glycerol and "mixed acids", without cooling, might produce a few mg of nitroglycerin, but mixing in 1,000 kg with mixed acids without cooling would produce various nitrogen oxides, CO2, and a large hole in the ground.

@DrMoishePippik you forgot water vapor maybe?

@OscarLanzi, yes, and some other products. Perhaps I should have picked a simpler example -- making a little NI3, or drying a kg of it.

The concept of stoichiometry says that molecular entities mutually react in ratio of small integer numbers.
That applies on particular reactions.
It does not say there cannot be different preferred reactions for different reagent ratios.
E.g. two elements may form several binary compounds and the product ratio is closely related to the molar ratio of reagents.
But each of such reactions follows the stoichiometric requirements.

The example you use doesn't show that the laws of stoichiometry fail, it merely shows that the equations you write have to consider all the components both before and after the reaction.
In the simplest case of A + B -> C the equation is fine if the amounts of A and B are equal. If side reactions are unimportant but you have a deficit of B, you might think that the equation A + B -> C is no longer true; but, in reality, the correct equation (obeying the laws) would be 2A + B -> C + A (with half the starting amount of A unreacted). The laws hold but only if you count all the components before and after the reaction.
In cases where observing all the products or reactants is hard, chemists usually assume the laws hold (and maybe find better ways to measure them to check as early chemists did to understand what "burning" meant when they burned things and carefully measured the mass of both solid products and the emitted gases).


text:
Why doesn't the reaction between molecule "B" and diethyl oxalate yield a diester?
Problem

Answer

Question
Why isn't this structure the answer for the second reaction?

From my understanding, the resonant structure (B) with the full negative charge on the oxygen is the most stable as the negative charge is on the most electronegative atom.
If the resonant structure (B) with the full negative charge on the oxygen attacked diethyl oxalate I presume this would be the intermediate.

Is my suggested structure for "C" not the answer as the best leaving group would contain the benzene ring due to resonance?

The enolate of p-chloropropiophenone --namely, B-- is being generated from ketone A by a kinetic base such as lithium diisopropylamide (LDA) in a polar, aprotic solvent, typically tetrahydrofuran (THF). You've showed the correct mechanism for obtaining the O-oxalylation product E. If E is formed at all, or if it were exposed to LiOEt independently, it would revert to enolate B and diethyl oxalate (DEO) via microscopic reversibility. The intermediate in this process is the one you have presented. Oxalylation of B can also occur on carbon, which leads to C. This acidic β-diketone is irreversibly deprotonated to form the enolate D and ethanol. Treatment of D with aqueous HCl would yield β-diketone C.

There are other subtleties in this reaction mechanism. If enolate B deprotonates β-diketone C forming ketone A and enolate D, can LiOEt deprotonate ketone A under the reaction conditions? If not, because the reaction requires stoichiometric base, one would require a 2/1 ratio of B/DEO or LDA/A.
But this part is probably TMI. Let's move on to pKa's.

The larger the pKa, the weaker the acid and the stronger the conjugate base. Approximate pKa values (logarithmic scale) of the "acids" in this reaction are listed above in decreasing order of conjugate base strength. This means that LDA deprotonates ketone A; LiOEt deprotonates β-diketone C; and enolate D deprotonates HCl. This process is known as the leveling effect.

A chemist would not do this reaction this way. Too much work! The low temperature preparation of LDA using n-butyllithium with syringes in an inert atmosphere is a bit of a hassle. It is much simpler to use sodium ethoxide in ethanol as the base and solvent. No inert atmosphere required and it is amenable to scale-up. Take a look at this Organic Synthesis procedure (https://DOI:2010.15227/orgsyn.011.0042) for the oxalylation of cyclohexanone. Take note of the stoichiometry of the reaction.