text:
Can acetone be used as a GC–MS solvent for a sample containing ammonia?
Сan acetone be used as a GC–MS solvent/diluent for a sample containing ammonia? Will there be a reaction between these two species?

It could form the corresponding imine, specially at high temperature in GC/MS.


text:
Carbon 11 radiolabeled inhaled Beta-2 agonist or corticosteroids, is it possible?
Background
Positron emission tomography (PET) is a highly sensitive imaging modality that can provide in vivo quantitative information of biological processes at a biochemical level. PET relies upon the administration of a chemical probe, often called a radiopharmaceutical, that is labeled with a short-lived positron-emitting radionuclide, examples of which are $\ce{^{11}C} \ (t_{1/2} = \pu{20.4 min})$ and $\ce{^{18}F} \ (t_{1/2} = \pu{109.7 min})$. Carbon-11 is one of the most useful radionuclides for PET chemistry, since its introduction into a biologically active molecule has minimal effects on the (bio)chemical properties of the compound (Ref.: Dahl, et al., Clin. Transl. Imaging 2017, 5, 275–289)
Biologicals of interest
Beta-2-Agonists are effective bronchodilators due primarily to their ability to relax airway smooth muscle (ASM). They exert their effects via their binding to the active site of b2-ARs, which are densely located on ASM. The presumed cellular mechanism of action involves the canonical signaling pathway via activation of Adenylyl cyclase (AC) and generation of intracellular cAMP,
which in turn can activate the effector molecules cAMP-dependent protein kinase A (PKA) and Epac, a Rap1 guanine nucleotide exchange factor. However, there is still little known regarding these pathways in airway cells (Ref.: Am J Respir. Crit. Care Med. 2013, 187(7), 690–696).
Question
Currently radiolabeled lung studies use labeled drugs that are produced for inhalation are "association” products, that is, the label is associated with the drug but not firmly bound to the drug. Without a firm bond between the active pharmaceutical ingredient (API) and the radiotracer, images must be acquired immediately post-inhalation before the dissociation of the label from the drug leads to a circulating level of absorbed radioactivity that could introduce an unacceptable error in the measurement of radioactivity in the lung.
Carbon-11 radiolabeled ICS or Beta-2 agonists would enhance our understanding of the cellular mechanism of action. Anyone aware of work in this area?

References:

Kenneth Dahl, Christer Halldin, Magnus Schou, "New methodologies for the preparation of carbon-11 labeled radiopharmaceuticals," Clinical and Translational Imaging 2017, 5, 275–289 (DOI: https://doi.org/10.1007/s40336-017-0223-1).
Mario Cazzola, Clive P. Page, Paola Rogliani, M. Gabriella Matera, "$\beta_2$-Agonist Therapy in Lung Disease," Am. J. Respir. Crit. Care Med. 2013, 187(7), 690–696 (DOI: https://doi.org/10.1164/rccm.201209-1739PP).

What does ICS stand for?

@orthocresol: ICS stands for " inhaled corticosteroid"! (from Ref.2 sighted)

It might be possible. But very hard as it would require a very rapid process (ie one taking significantly less time than the half life)  to synthesise and isolate a complex molecule. That's not easy, even less so for steroids than beta-agonists.

Positron emission tomography (PET) has been taken using $\ce{^{11}C}$- and $\ce{^{18}F}$-labelled reagents, although they are not $\beta_2$-agonists (Ref.1). The abstract states that:

Positron emission tomography (PET) provides three-dimensional images of the distributions of radionuclides that have been inhaled or injected into the lungs. By using radionuclides with short half-lives, the radiation exposure of the subject can be kept small. By following the evolution of the distributions of radionuclides in gases or compounds that participate in lung function, information about such diverse lung functions as regional ventilation, perfusion, shunt, gas fraction, capillary permeability, inflammation, and gene expression can be inferred. Thus PET has the potential to provide information about the links between cellular function and whole lung function in vivo. In this paper, recent advancements in PET methodology and techniques and information about lung function that have been obtained with these techniques are reviewed.

It is clear that unto 2007, there are no evidence for the use of radiolabeled $\beta_2$-agonist or corticosteroids for PET in literature. However, there are other radiolabeled reagents have been use for the imaging. For instance, the mixed-venous content is obtained by imaging the right heart during the infusion, and the alveolar concentration is obtained by imaging the lung, after the blood volume is subtracted from the image using a separate $\ce{^{11}C}$-labeled carbon monoxide ($\ce{^{11}CO}$) inhalation.
References:

R. Scott Harris, Daniel P. Schuster, "Visualizing lung function with positron emission tomography," J. Appl. Physiol. 2007, 102(1), 448-458 (DOI: https://doi.org/10.1152/japplphysiol.00763.2006)(PDF).
C. G. Rhodes, S. O. Valind, L. H. Brudin, P. E. Wollmer, T. Jones, P. D. Buckingham, J. M. Hughes, "Quantification of regional V/Q ratios in humans by use of PET. II. Procedure and normal values," J. Appl. Physiol. 1989, 66(4), 1905–1913 (DOI: https://doi.org/10.1152/jappl.1989.66.4.1905).

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Why benzene is more easily reduced by Birch reduction than by catalytic hydrogenation?
As far as I know, H2,Pd is a better reducing agent than Na in liquid ammonia.
However, Na in liquid ammonia can reduce benzene at low temperature conditon of about -30°C.
Moreover, it is hard for H2,Pd to reduce benzene and requires very high temperature and pressure.

The very short version is that these two reactions have substantially different mechanisms with different rate-determining steps and activation energies.

But how can having an anion radical mechanism reduce the activation energy to such an extent?

Because the mechanisms are different, the activation energies are not easily comparable. The Birch reduction starts with a single electron transfer to one of the LUMOs of benzene to form a radical anion that is still partially aromatic. Catalytic hydrogenation starts with a concerted syn addition of the $\ce{H2}$ molecule to form a neutral non-aromatic intermediate.

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How can you qualitatively tell when the hydrolysis of aspirin is over?
I'm doing an experiment on the effect of temperature on the time taken to produce acetic acid and salicylic acid. Are there any qualitative observations I can make about when the reaction has finished?
I am using a 0.01 M aspirin in pure water.

You might want to check this answer and the spectrum therein: https://chemistry.stackexchange.com/a/150095/79678. You might have to run a series of experiments to get a feel for about how long it takes for the hydrolysis to effectively reach completion.

First I was thinking that maybe a chemical indicator could be used, however the pKa values of all three components are relatively close to each other (especially aspyrin and salicylic acid). It might still be a bit of a help to stick a pH electrode in the system and observe the change - however, this is not exactly something you can see easily.
Another thing that came to mind is that salicylic acid (as all phenols) will form a colorful complex with Fe(III). However, the appearance of the red-ish color will only tell you when the reaction has started, since it's borderline impossbile to observe that the solution is not getting any darker anymore. Again, if you have some machinery, you could stick your solution into a cuvette and put it in a photometer.
Apart from the two ideas above, I don't really see an easy way of observing anything.

Why be qualitative when with just a little more effort you can be quantitative?
Acetylsalicylic acid has a pK$_a$ of 2.97, i.e., it is fairly acidic (Ref 1). Salicylic acid, one of the hydrolysis products, is comparably acidic (pK$_a$ = 2.79 Ref 2) and its sodium salt is fairly neutral (pH = 6.0 - 8.5 for a commercial material Ref 3; calculations suggest ~7.4 at 0.01 M).
Acetic acid, the other hydrolysis product, is not so acidic: pK$_a$ = 4.75, and its sodium salt, in solution, has a pH of 8.375 (Ref 4). This is how the calculation plays out:
pH = 7 + 0.5(pK$_a$ + logconcentration)
= 7 + 0.5(4.75 - 2) = 8.375
Approximately. The other ions in solution will probably have some effect.
One more fact: phenolphthalein is colorless below pH = 8.5 and pink/red above pH = 9 (Ref 5).
So, your 0.01 M acetylsalicylic acid will dissolve (soluble in water 3g/L = 0.0166 M). As it hydrolyzes, it will produce salicylic acid with essentially the same strength (pK$_a$ ~same) and concentration (1:1). But at the same time, it will generate acetic acid, somewhat weaker, but still titratable.
If you take 10.0 mL of your experimental solution, add a few drops of phenolphthalein indicator solution and then titrate with 0.01 N NaOH (or KOH), at time = 0, it will require 10.01 mL to turn the phenolphthalein red. As the hydrolysis proceeds, the acetic acid produced requires additional alkali to get the pH high enough to get a red color. In fact, at the end of the hydrolysis, it will require 20.01 mL of 0.01 NaOH to turn red. (The decimal places are for instructive purposes, not to indicate accuracy - you know what I mean.)
In fact, you don't need to use exactly 0.01 M alkali - just realize that whatever you need to turn red at time zero will double when the hydrolysis is complete. And you can weigh the amount of alkali added rather than use a burette. And you can do a reaction rate study by checking every 15 minutes or so.
Ref 1. https://www.chem.purdue.edu/courses/chm333/Spring%202013/Lectures/Spring%202013%20Lecture%202%20-%204.pdf
Ref 2. https://foodb.ca/compounds/FDB000882
Ref 3.  https://labchem-wako.fujifilm.com/us/product/detail/W01W0119-0314.html
Ref 4.  https://tardigrade.in/question/pk-a-of-ch-3cooh-is-4-74-the-ph-of-0-01-m-ch-3coona-is-bus7pexw
Ref 5. https://www.britannica.com/science/phenolphthalein


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Azeotropic mixtures and Raoult's Law
So I find this everywhere:
"Mixtures showing large deviations from Rault's law form azetropic mixtures at specific compositions"
My questions are:
(a)Why is it so?Why deviation from Raoult's law is needed at all.
Wikipedia simply skips over the answer stating its a mathematical consequence.
(Quoting wikipedia:
When the deviation is great enough to cause a maximum or minimum in the vapor pressure versus composition function, it is a mathematical consequence that at that point, the vapor will have the same composition as the liquid, resulting in an azeotrope.)
(b) why can't ideal mixtures form azeotropes?
P.S: (a)A more intuitive answer would be appreciated although maths may be included if necessary.
(b)I did my research but nowhere the reason was to be found. It's presented as a very obvious consequence everywhere.

Can someone pls tell why was it downvoted the moment i posted it...?

About the downvote, it can be said only by the user who did so, as it is fully anonymous. The reason can be you have not paid much effort to think about the topic. Generally, if there is a sloped line, it has to significantly deviate from a line to have maximum or minimum.

Also you can use Raoult Law as a sort of definition for ideal mixtures. And there is no space for azeotropes in it.

@Justcuriois Possible Duplicate Here.

@Everyone... pls see the follow up question i put under #Dario Miric's reply. Thanks(Also i was wondering if i am asking such follow ups how will others be notified...?)

To understand why non - ideal mixtures form azeotropes, we need to understand what is ideal mixture. If you have a liquid mixture of two components A and B, ideal mixture is a mixture in which heterogeneous interactions (A - B)  are the same as the average homogeneous (A - A and B - B). This means when you mix components A and B, they don't "feel" like their enviroment has changed at all and thus no component has higher nor lower tendency to escape the liquid phase and go into vapor phase. When interactions are largely different, in most cases heterogeneous interactions are less favourable than homogeneous (molecules of A in most cases like to be around molecules of A rather than B) than component will have higher tendency to escape from liquid phase to vapor since in vapor interactions are much less pronounced due to much bigger average distance of the molecules. If this scenario happens you got yourself a lower boiling azeotrope since due to higher tendency of molecules to escape liquid phase, vapor pressure is higher than in ideal case (Raoult's Law). In some cases heterogeneous interactions can be very favourable and this is usually due to some specific interactions which mostly originate from hydrogen bonding, if so molecules will tend to stay in liquid and you got yourself higher boiling azeotrope. Example of such azeotrope is HNO3 and water. If you have any questions, fire.

Very well explained. But i have a question, azotropes have constant boiling temperature (so components can't be separated by fractional distillation).//(a)Why is it not possible for an ideal solution that it has constant boiling temperature..? //(b)Say the two components of ideal  binary solution are equally volatile (i think they can be equally volatile...correct me if i am wrong) and so can't be separated by fractional distillation..i.e this ideal solution has constant boiling temperature...will it not qualify such an ideal solution as an azotrope...?

[This was the picture i had in mind while asking why ideal solitions can't be azotropes....] //(c)Finally...all the examples of binary ideal solutions i have seen have taken one of the component(say A) to be more volatile than the other (B)....is it necessary..?

Reason is if mixture is ideal than vapor pressure follows Raoult's Law which means that partial pressure of each component follows a STRAIGHT LINE since Raoult's Law posits linear relation between partial pressure and composition. If we regard vapor as ideal (which you can at low vapor pressure) than vapor pressure is a sum of partial pressure of each component which means that vapor pressure follows a LINE in dependence on composition. Line by definiton doesn't have local maximum or minimum. Take a look on Wikipedia article section Principles https://en.m.wikipedia.org/wiki/Raoult%27s_law

You need to have deviation from ideal behavior to generate local maximum or minimum in phase diagrams (p-x-y or T-x-y). If vapor pressure is non - ideal than temperature will follow it accordingly. Where vapor pressure has maximum, boiling temperature will have minimum. Sources of non - ideality (why vapor pressure deviates from straight line) can be traced in general vapor - liquid equilibrium equations from which you can derive Raoult's law if everything is ideal. You can find these in chemical thermodynamics textbooks and online.

@Putnik and #Dario miric That's fine (i saw wikipedia article)....but can both components of ideal solution have same volatility (pls read parts (b) and (c) in my last comment...)

Yes, they can. In such case vapor pressures of pure components are the same (or approximately the same). This situation happens when you have isomers of some component which have very similiar boiling point and are needed to be separated. If pure components have exactly the same vapor pressure than mixture would boil at the same temperature at any compositon since vapor pressure would always be equal to vapor pressure of pure component. But, such mixtures aren't azeotropes since they would be ideal and because azeotropes don't boil at the same temperature regardless of composition.

Azeotropes have a composition at which they boil at minimum or maximum temperature at which volatility of each component is exactly the same and you can't separate components by simple distillation. This is why they are said to be constant boiling since you can't change composition of neither liquid nor vapor when you reach azeotropic point, but this doesn't mean their boiling point is independent on composition like it is in your example with two components with the same vapor pressure which acts ideally.

#Dario Mirić +1 ....(your emphasis on composition made the difference clear)Thanks :)  // "This situation happens when you have isomers of some component which have very similiar boiling point and are needed to be separated."do you happen to remeber any practical examples of such ideal solutions.

I don't know any component with exactly the same boiling point. Example of isomers can be orto, meta and para xylene separation from reformulated or pyrlosyis gasoline in petrochemical industry.

In his answer Dario Miric has explained the basis of azeotropes clearly, this is a small addition to his answer. Raoult determined experimentally that the partial vapour pressure over some liquid mixtures is directly proportional to the mole fraction $x$ of each component, and thus the total pressure is
$$P=x_1p_1^\text{o}+x_2p_2^\text{o}=x_1p_1^\text{o}+(1-x_1)p_2^\text{o}$$
where $p^\text{o}$ is the vapour pressure of the pure liquid. An ideal solution can be defined as one that obeys Raoult's law at all temperatures. Solutions obeying Raoult's law are, for example, dichloroethane - benzene or ethylene bromide - propylene bromide. As the pressure is the sum of the partial pressures the two species cannot interact to any great extent that would cause their vapour pressures to differ from that expected from the pure solution. Of course this is only true in a few cases.
If the two types of molecule interact with one another then the potential energy of the mixture has an additional term describing this interaction. The result of this interaction is that the total pressure becomes
$$P=x_1p_1^\text{o}e^{\alpha}+(1-x_1)p_2^\text{o}e^{\beta}$$
where $\alpha =(1-x_1)^2)\Delta U/RT$ and $\beta =x_1^2\Delta U/RT$ and $\Delta U$ is a measure of the average difference in interaction energy. When $\Delta U$ is small vs $RT$ only a slight deviation from Raoult's law (pressure $p_{1,2}/p_{1,2}^\text{o}$,vs. $x$) is evident and instead of being a straight line is slightly bent, convex of concave, for example methanol in water.   If the plot of total pressure vs mole fraction passes through a maximum or minimum then this is the Azeotrope and the vapour has the same composition as the mixture. Additionally as $dP/dx=0$ at the azeotrope and as the mole fraction is known an estimate of $\Delta U$ can be obtained. Example of azeotropes are dichloromethane - acetone, carbon disulphide - dimethoxymethane and CS$_2$-CHCl$_3$


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Rate of aromatic nucleophilic substitution reaction on ortho and para fluoronitrobenzene
Why rate of aromatic nucleophilic substitution of Ortho-fluoronitrobenzene is greater than para-fluoronitrobenzene?
Kindly note question is about why rate of ortho is greater than para position. Not meta position.
This is previous year KVPY question but they had not asked reason and just asked comparison of rate.

I think it's probably due to the inductive effect of the nitro group. The effect is strongest at the ortho position

Please do not use formatting on the title.

Reaction mechanism of aromatic nucleophilic substitution is as given -

Source : https://www.masterorganicchemistry.com/2018/08/20/nucleophilic-aromatic-substitution-nas/
The rate determining step is the formation of the C-Nu bond, which causes the development of a negative charge on the carbon atom at ortho position to leaving group (fluorine). Presence of electron-withdrawing groups helps to stabilize the negative charge, and hence increase the rate of reaction.
The -NO2 substituent exerts not only a -M effect (resonance) but also a strong -I effect (inductive effect, which pulls the negative charge away from the ring). Inductive effect decreases rapidly as the distance increases, so its effect is much more pronounced for ortho than for para (Note that the initial negative charge is developed at the ortho position). As a result orthofluoronitrobenzene has a higher rate of reaction than parafluoronitrobenzene.


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Can we compare the vapour pressure of solutions whose concentrations and temperature is known?

If we have some solutions, in which we know the temperature and concentration of the solute particles, can we compare their vapour pressures?

Original question:

Which solution has the highest vapour pressure among the following?
a) $0.02 ~\pu M ~\ce{NaCl}$ at $50^{\circ}$C b) $0.03 ~\pu{M}$ sucrose at $15^{\circ}$C  c) $0.005 ~\pu{M} ~\ce{CaCl_2}$ at $50^{\circ}$C d)$0.005~ \pu{M} ~ \ce{CaCl_2}$ at $25^{\circ}$C

I answered this question by noting that at higher temperature and lower concentration, vapour pressure of solution should be the highest because more temperature implies more kinetic energy which means more volatility and low concentration means that the solute is less solvated so it forms less bonds with the solvent and therefore has more vapour pressure, so answer is c). Is this reasoning correct?
Now I think that these options were set so that it becomes easy to compare.
But what if I had options like $0.02 ~\pu M ~\ce{NaCl}$ at $50^{\circ}$C ,$0.03 ~\pu M~ \ce{NaCl}$ at $60^{\circ}$C,  $0.005 ~\pu M
~\ce{CaCl_2}$ at $20^{\circ}$ C, how shall we compare their vapour pressures without doing experiments?
Now, if there existed a relation between vapour pressure and concentratin or temperature, it would become easier.
For example, relation between vapour pressure and temperature can be given by Antoine equation, but I want something that also incorporates the concentration in the equation also, so that by plugging in the values, it is easy to compare.
I am also not sure, whether van't hoff factor will have a role in it or not.

The change of vapour pressure with temperature is much greater than with salt concentration in the considered scale.

@Poutnik Then should we only see the temperature for comparing? What about their van't hoff factors, If two substances have same concentration and temperature, but different van't hoff factors, would the one with less $i$ have more vapour pressure?

Sure, the van't Hoff's factor does come to play, additionally with hydration of ions. But still, the considered concentration range causes much smaller effect then considered temperature differences. E.g. 20->30 deg C causes about twice vapour pressure. Saturated NaCl ( > 5 M ) cases about 25% vapour pressure drop. And the vapoure drop would be like progressive,as activity of water would go down.

The point of the question is to test your understanding
i. of Raoult's law (that increasing the solute concentration decreases the vapor pressure and that the identity of the solute is not important in an ideal solution; the mole fraction of solute determines $\Delta p$: $\Delta p = p-p^\circ = -\chi_\textrm{solute}p^\circ$
ii. that vapor pressure increases with T
Other things you might want to remember are that molality and molarity are linearly proportional for dilute solutions (ie then $m \propto c_M$), and that ideality is approached for dilute solutions of small solutes (typically at low $\pu{mM}$ concentrations such as in this problem).
You can answer the question based entirely  on consideration of the van't Hoff factor to compute total concentrations, and by assuming that the vapor pressure increases with T.
First compute effective total solute concentrations $im$ (equal to either the solute concentration for a non-dissociating solute such as sucrose, or the total ion concentration for the electrolyte solutions):
$$\begin{array}{|c|c|c|} \hline \text{species} &T (\pu{^\circ C})& c_M (\pu{M})& i & ic_M(\pu{M}) \\\hline  1.~\ce{NaCl} & 50 & 0.02 & 2 & 0.04\\
2.~\text{sucrose} & 15 & 0.02 & 1 & 0.02\\
3.~\ce{CaCl2 } & 15 & 0.005 & 3 & 0.015\\
4.~\ce{CaCl2 } & 50 & 0.005 & 3 & 0.015\\\hline
\end{array}$$
Inspection should quickly eliminate all options but row 4 since they have a higher effective concentration at the same T, or since they are at a lower T at the same effective concentration.
Actual vapor pressures are not stated in the problem. In the absence of that information there is no point in attempting to compare things in more detail, and that's not the point of the problem. However, provided Raoult's law holds it can be used to determine the change in vapor pressure regardless of T. Raoult's law assumes the identity of the solute does not matter, that the effect of added solute on the activity of solvent is entirely entropic (accounted for by ideal mixing entropy) and that the gas is ideal. In general, to predict vapor pressure at any T and solute concentration you need a model of the activity of the solvent as a function of the solute concentration, and of the activity in the gas phase. You can determine the vapor pressure by finding the conditions that render the activities identical.

(+1) , Thank you!. Did you want to write row 4 instead of row 3? Also, "However, provided Raoult's law holds it can be used to determine the change in vapor pressure regardless of T" , can you explain how can we determine change in vapour pressure when we don't know the mole fraction and the pressure of pure component ($p^{\circ}$)?

Yes, I meant row 4 has higher $p$.

To use Raoult's law you need some information about either the change in vapor pressure and the vapor pressure of the pure solvent at the T of interest, from which you can determine concentration of solute, or information about concentration of solute (and vapor pressure of pure solvent) from which the change in pressure.


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Lindemann-HInshelwood mechanism difference between high and low concentration
I have a gas reaction which follows the Lindemann-Hinshelwood mechanism and want to calculate the rate constant at low and at high concentrations of the gas but I find myself stuck at what my next step should be and hope someone here could push me in the right direction.
I have that the effective rate constants are $1.01×10^{−5}$ s$^{−1}$  at a concentration of $4.6×10^{−6}$ M and $2.21×10^{−5}$ s$^{−1}$ at a concentration of $8.71×10^{−5}$ M (which are relative low concentrations). Hence, I get that the effective rate constant can be written as:
$$k_\mathrm{eff} = \frac{k_\mathrm{a}k_\mathrm{b}[\ce{A}]}{k_\mathrm{b} + k^{'}_\mathrm{a}[\ce{A}]}\tag{1}$$
and since $k_b$ >> $k^{'}_a [A]$, which then gives:
$$k_\mathrm{eff} = \frac{k_\mathrm{a}k_\mathrm{b}[\ce{A}]}{k_\mathrm{b}} ={k_\mathrm{a}[\ce{A}]}\tag{2}$$
So
$$k_a = \frac{k_{eff}}{[A]}=\frac{k_{eff,2}-k_{eff,1}}{[A]_2 - [A]_1} = \frac{2.21×10^{−5}-1.01×10^{−5}}{8.71×10^{−5} - 4.6×10^{−6}} = 0.145 M^{-1} s^{-1}$$
HOWEVER from what I have seen online it should be that
$$\frac{1}{k_{eff}} =\frac{1}{k_a [A]}$$
So
$$k_a =\frac{1/k_{eff,2} - 1/k_{eff,1}}{1/[A]_2 - 1/[A]_1} = 3.83 M^{-1}s^{-1}$$
Which I don't think this is correct, but that is the only way that I have seen it been calculated. But why does it go from being $k_{eff}$ to being $\frac{1}{k_{eff}}$? Or is my way of doing it correct?
Further, if I then want to calculate the rate constant at high concentration of the gas, $k_a [A]>>k_b$, then  eq. (1) can be rewritten as:
$$k_\mathrm{eff} = \frac{k_\mathrm{a}k_\mathrm{b}}{k^{'}_\mathrm{a}}\tag{3}$$
But since I don't have $k_b$ or $k_{a}^{'}$ how can I calculate $k_a$ at $298$ K?
I truly hope someone can help me with this, I have tried to find similar problems online but I can only find one and I don't understand why the expression for $k_{eff}$ is as it is. I also can't find anything regarding calculation for high concentrations without being given the other rate constants. Thank you!

Without reading everything, aren't your equation (2) and the equation which comes after "however" exactly equivalent? (Provided everything is non-zero, of course).

@LinearChristmas I thought so too but when calculating $k_a$ with eq. (2) and the equation after "however" I get two different answers. From eq. (2) I get that $k_a = 0.145$ and from the other eq. I get $k_a = 3.83$.

If agree that the equations are equivalent, then any forms for $k_a$ derived from them must also be equivalent. Why do you then use a different equation for $k_a$ after "So" than you did before?

Your eqn 2 cannot apply as $k_a$ has only one value and the ratio of $k_{eff}/[A]$ is different for your data. Try taking the reciprocal of eqn 1 and plot $1/k_{eff}$ vs $1/[A]$ to get the gradient $k_a$ (or do so algebraically) and also the intercept. This gives the answer $k_a=3.83$ so this seems to be what you have done here.


text:
Is “double dot” (“··”) notation used for a double radical?
I wondered if this was a typo in the notes given to me. I haven’t come across notation like this before:

Branching — steps in which number of radicals increases (can lead
to explosion)
$$\ce{H^. + O2 -> OH^. + O^{..}}$$

Alone, the oxygen atom has always two separated uncoupled electrons. It never lasts a long time like this, but It has two doublets and two independent electrons in the outer shell. This makes a total of $6$ electrons. On the other hand, the molecule $\ce{O2}$ is a double radical, with one free electron on either Oxygen atoms..

Ahh, that makes sense. Thank you so much for this

It's call diradical or biradical, just saying ;)

TL;DR: The double dot notation is reserved for lone pairs.
Diradicals are depicted either with the appropriately placed superscripted dots within a structural formula (e.g. $\ce{^.O-O^.}),$ or superscripted multiplier–dot pair (e.g. $\ce{O2^{2.}}).$
The recommended notation for monooxygen 1,1-diradical is $\ce{O^{2.}}.$

First, let's sort out terminology.
The recommended name for “double radical” is diradical [1, p. 1333]:

diradicals:
Molecular species having two unpaired electrons, in which at least two different electronic states with different multiplicities [electron-paired (singlet state) or electron-unpaired (triplet state)] can be identified.
E.g. $\ce{H2C^.-CH2C^.H2}$ propane-1,3-diyl (trimethylene). See biradicals.

The term biradical, although being synonymous, is less preferable [1, p. 1322]:

biradicals:
Although this term has been recommended in the past for diradicals, specialists working in the field prefer the latter term.

So, we are dealing with a diradical.
You are probably familiar with a low-energy oxygen 1,2-diradical commonly depicted as $\ce{^.O-O^.}.$
What if we are dealing with a 1,1-diradical?
Note that two dots next to each other signify a lone pair (remember water $(\ce{H-\overset{\Large .\!\!.}{\underset{\Large .\!\!.}{O}}-H})$ or ammonia $(\ce{:\!\!NH3})?)$ and when being used for the depiction of diradicals will inevitably cause confusion.
This seeming contradiction is resolved if you follow guidelines provided in section IR-4.6.2 Formulae of radical from Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) (emphasis mine) [2, p. 66]:

A radical is an atom or molecule with one or more unpaired electrons.
It may have positive, negative or zero charge.
An unpaired electron may be indicated in a formula by a superscript dot.
The dot is placed as a right upper index to the chemical symbol, so as not to interfere with indications of mass number, atomic number or composition.
In the case of diradicals, etc., the superscript dot is preceded by the appropriate superscript multiplier.
The radical dot with its multiplier, if any, precedes any charge.
To avoid confusion, the multiplier and the radical dot can be placed within parentheses.
[…]
Examples:
[…]

$\ce{O2^{2.}}$

[…]

$\ce{NO^{(2.)-}}$
$\ce{N2^{(2.)2+}}$

Here you can also find nomenclature for this species [2, p. 319]:

$\ce{O^{2.}},$ oxidanylidene, monooxygen(2•)

Following these recommendations, I would suggest to rewrite your scheme as follows:
$$\ce{H^. + O2 -> HO^. + O^{2.}}$$
References

Moss, G. P.; Smith, P. A. S.; Tavernier, D. Glossary of Class Names of Organic Compounds and Reactivity Intermediates Based on Structure (IUPAC Recommendations 1995). Pure and Applied Chemistry 1995, 67 (8–9), 1307–1375. DOI: 10.1351/pac199567081307.
IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005. ISBN 978-0-85404-438-2.


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Is there a practical re-use value gained by grinding up activated carbon that is spent/saturated?
I want to buy bulk activated carbon for use in a DIY air-filtration set-up, like this, to improve air quality in my bedroom.
From what I understand, the relatively larger sizes of carbon media on the market are optimal for air filtration to maximize the amount of time the air is in contact with the carbon.
My question is, once my activated large-granule carbon is used up, say after a few months in my air filter, am I throwing away any untapped potential? Specifically, is there value in further crushing/grinding the carbon and then putting it into another application optimized for the new, more fine-grained size-- perhaps a water filter, or an air filter where the carbon is in a long tube, and air is moving slowly through it.
The same FAQ suggests it is impractical and messy to grind up the AC, but it seems to me that mess can be contained with a bit of planning. I'm also happy to be pointed to general resources that are helpful, especially if I'm barking up the wrong intuition trees, here. Thanks!

No. Grinding activated carbon will not change anything relevant.
The reason is because surface area does not have an effective influence on the usefulness of activated carbon. Activated carbon is called "activated" when its grains contain a large amount of large cavities. These cavities are big enough to hold large molecules of undesired substances (oils, smokes) which are present in the air. Only small molecules like oxygen and nitrogen from the air can go through without being retained. So the activated carbon removes pollution from the air. After a long time, there are more and more cavities occupied by these unwanted substances. As a consequence, the effectiveness of the activated carbon slowly decreases. It has to be replaced.
If one wants to reuse this carbon again, one should heat it up to such a high temperature that the substances fixed in the cavities are vaporized. Unfortunately some substances like soot will never get vaporized. So even then, the effectiveness of the activated carbon will never be entirely restored.

Thanks! Just to clarify, all potential cavities/sites to hold large molecules are already accessible, even at a coarse grain? No new, perhaps inner cavities are exposed by breaking down the carbon some? If so, is it because the process of activation only engages the initial surface area?

@MarkK Sorry I do not fully understand what you say. when writing "the process of activation only engages the initial surface area".

Sorry, yes. I'm trying to understand why grinding carbon into smaller pieces is ineffective. to me, after the activated carbon has all it's cavities filled, grinding is a way to expose new cavities of the activated carbon, that are now new binding sites. If grinding is ineffective, then am I to understand that: all initially available cavities on the activated carbon, in the size of granule it is, are the only cavities available, even if new surfaces are exposed by breaking pieces into smaller ones? It makes sense to me if the activation process works only on initially exposed surface of C.

Your answer implies what I'm looking for, can I ask for clarification? The cavities available for capturing unwanted particles, are they only accessible on the exterior surface, or is the carbon structure open / exposed enough that cavities throughout it can also capture particles? If cutting does nothing, then I assume carbon is like a sponge with water analogous to the unwanted particles. In my question I am wondering if the particles behave like a viscous grease that can only be captured by the surface of the sponge. <- happy to update question with this if it is helpful.

The cavities were initially cells. Wood is a living system, made of cells containing a nucleus, cytoplasm, etc. surrounded by a membrane. When overheating wood, water will evaporate, and the membranes are destroyed. Later on, all organic stuffs aretransformed into smoke, tar and other volatil substances. The membrane itself is transformed into charcoal or impure carbon. So the obtained "activated carbon" is made of broken spheres, linked up with tiny channels. The adsorption of pollution is not an effect of surface. It occurs in the bulk of the activated carbon .


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Hair styling gels and 'liquid gels' - effect of formulation / composition on viscosity and performance
[Disclaimer: I am aware that this is a very 'first-world problem', 'light-hearted' kind of question, given the health crisis we are still going through. Just to avoid offending anyone, here's an easy solution: no need to continue to read if you disapprove :) ]
I have 3 brands of 'liquid gel', each with rather different characteristics.
Here are their compositions:
Brand 'A': Alcohol Denat., Aqua, Octylacrylamide/Acrylates/Butylaminoethyl Methacrylate Copolymer, Panthenol, Parfum, PEG-12 Dimethicone, Isoamyl p-Methoxycinnamate, Benzoic Acid, Linalool, Aminomethyl Propanol, Citric Acid.
Brand 'B': Alcohol Denat., Aqua/Water, Acrylates/t-Butylacrylamide Copolymer, Niacinamide, Hydroxyisohexyl 3-Cyclohexene Carboxyaldehyde, PEG/PPG-4/12 Dimethicone, Aminomethyl Propanol, Limonene, Panthenol, Benzyl Salicylate, Butylphenyl Methylpropional, Hexyl Cinnamal, Parfum, Fragrance.
Brand 'C': Aqua, Alcohol Denat., VP/VA Copolymer, Polysorbate 20, PEG-12 Dimethicone, Panthenol, Parfum, Linalool, Geraniol, Hexyl Cinnamal, Citronellol, Alpha-Isomethyl Ionone, Hydrogen Peroxide, Phenoxyethanol, Ethylhexylglycerin, Piroctone Olamine, Citric Acid.
My favourite brand is 'C'.
The shops where I used to buy it are no longer stocking it, they replaced it with the same stuff (same commercial name and 'strength'), but in 'gel' form, which I really don't like.
The composition of the gel is similar to the one of the liquid gel, up to a point:
Brand 'C-gel': Aqua, Alcohol Denat., VP/VA Copolymer, PVP, Propylene Glycol, Carbomer, Polysorbate 20, (PEG-12 Dimethicone - removed), Panthenol, Parfum, Linalool, Geraniol, Hexyl Cinnamal, Citronellol, Alpha-Isomethyl Ionone, Hydrogen Peroxide, Phenoxyethanol, Ethylhexylglycerin, Piroctone Olamine, Tetrasodium Iminodisuccinate, Sodium Hydroxide, Citric Acid.
Question 1: does anybody know what specific components, if any, are responsible for 'C' being liquid and 'C-gel' being indeed a rather thick gel ?
Question 2: could one turn the gel into a liquid that can be sprayed, and if so, how ?
This is no academic question: I would actually like to do that. I tried diluting the gel with water, but those formulation chemists did a very good job: it stayed quite gel-like and viscous, until I really diluted it a lot, like 1:3 or more, and even then it was still a viscous liquid that the spray bottle could not spray well at all.
Could a change of pH perhaps loosen it ? By which I mean a small change; not planning to turn this into battery acid or lye :)
Question 3
When using the liquid gels (applied on wet hair and dried either naturally or with a hairdryer), I noticed very different performances.
'A' results in a very unpleasant, 'doll-like' feel and look, and a lot of static electricity, when you comb your hair. And it doesn't smell nice.
'B' is marginally better, it sort of works in terms of disentangling your hair, but it tends to get a bit sticky, and they put something in it that smells absolutely repulsive (at least to me). I hope it's not the 'fragrance'.
'C', as I mentioned, is my favourite (despite being the cheapest): it smells good, it makes your hair very smooth and easy to comb, it does not stick, and even when dry your hair stays soft and tidy (unlike with B and especially A).
So: can you identify any components, or any formulation features, in A, B and C, that cause the performance I am describing ?
E.g. what might be responsible for 'A' turning your hair into individual doll-like plastic-y, electrostatic strands?
Or, do you know if any of the components of 'B' has a particularly offensive smell ?
Or, is there any specific component in 'C' that 'lubricates' hair and prevents it from sticking?
Thanks!

Question 1: Carbomer
Question 2: Not sure about the question, but anyway. Sure you can create a spray (theoretically), but the process is really hard (please, do not make it at your house) because you need a propellant. By the way, such a high concentration of Parfum.
Question 3:
Part1:
According to your observations, in my own opinion, the concentration of these components maybe are too high: Octylacrylamide/Acrylates/Butylaminoethyl Methacrylate Copolymer, Niacinamide, Hydroxyisohexyl 3-Cyclohexene Carboxaldehyde.
Part2:
Maybe because of the oxidation of hexyl cinnamaldehyde to Cinnamic acid you can feel a very unpleasant smell. But it is hard to tell because everything you describe is not in a detail (for example, the scent, what reminds and etc..)
Part3:
VP/VA Copolymer, PVP makes your hair shiny and not sticky.
Panthenol makes your hair not dry.
JUST A REMAINDER:
It does not mean, that those components which I have mentioned are very good or very toxic. It all depends on concentration and experimental data. We use poisons' as drugs, but lethal and effective concentrations differ.
I answered what could possibly be in my own subjective, but a trained opinion based on your poor objective/experimental data. Try to say more about what you smell, what you feel, and there many databases of components like (found some reliable databases for you):
Official Cosmetics ingredients review database: https://www.cir-safety.org/ingredients
Cosmetics and their function (easy to use database): https://www.cir-safety.org/ingredients https://www.ewg.org/skindeep/
Chemical compounds database: https://pubchem.ncbi.nlm.nih.gov/

Thanks Linas! I really appreciate that you took the time to answer, despite the usual sanctimonious, cowardly, backstabbing anonymous PC brigade immediately pouncing on my post to have it closed. For Q2, by 'spray' I mean one of those bottles with a pump, not the classic spray with a propellant. So I would only need the material to be liquid and non-viscous enough to form a mist when sprayed by one such bottle, which is what the liquid gel of brand C was like. Assuming viscosity is the property that determines whether a mist of fine droplets can or cannot be formed by a spray bottle.

I cannot edit my post because of course it is already answered, so if I removed some of the sub-questions in it, parts of the answer would no longer make sense. And apparently I cannot delete it either 'because others have invested time in answering it'. Yes, and that's exactly what proves that the question made sense and could be answered! I said it before and I am still totally convinced of it: giving people the power to shut down other people's posts without any valid justification is absurd, obtuse, anti-scientific, counterproductive and will ultimately turn users away from SE for good.

I have recently registered on this website, unfortunately, my expectations were too high. I could do something more meaningful. So, sorry if I will not reply to this question anymore. For future reference, I suggest the Quora website or LinkedIn.  I would be honest with you, create cosmetic products is a huge try and error work. I could theoretically say, yes you can make your pH a little bit lower. However, the whole formulation could be ruined and the whole product could become a mess. I am not even talking about shorter shelf life.

Nevertheless, you can try to create your own product with the same formulation, but with different rheological properties. Many of my dear colleagues have written a final thesis on a similar topic.

I totally understand you, Linas. Thank you again for your valuable, competent and thorough answer. The only reason I am still using SE from time to time is indeed the fact that there are many highly skilled, open-minded, intelligent people who genuinely care about helping others and sharing their knowledge. All of which is largely ruined by this autocratic, obtuse, blinkered and overly rigid structure, which is in fact just an excuse for very mediocre, frustrated people to anonymously exert power on others. No wonder that many potentially very good new contributors immediately lost interest.


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Why there is a color fading ( to colorless) when Cerium Ammonium Nitrate (CAN) is mixed with ethanol or butanol?
I have mixed CAN in butanol and also ethanol (separately), the solution became dark red immediately but after a couple of minutes the color totally faded and turned into a transparent solution, I am curious to understand the reason behind it.

https://chemistry.stackexchange.com/questions/140471/oxidation-of-alcohols-by-ceric-ammonium-nitrate

And I wonder what are your ideas about it.

Your primary alcohols may be oxidized. You can't use a alcohol as a solvent for CAN.

Recall that Ce(IV) is a pretty strong oxidizing agent. So oxidation of alcohols and fading of the color is expected. The mystery lies in the immediate intense red color. It has been relatively recently studied in great detail and well known long time ago.
See these references:

V. Briois, D. Lützenkirchen-Hecht, F. Villain, E. Fonda, S. Belin, B. Griesebock, and R. Frahm, "Time-Resolved Study of the Oxidation of Ethanol by Cerium(IV) Using Combined Quick-XANES, UV−Vis, and Raman Spectroscopies," J. Phys. Chem. A 2005, 109(2), 320–329 (DOI: https://doi.org/10.1021/jp046691t).

Lewis Brewster Young and Walter S. Trahanovsky, "Oxidation of organic compounds with cerium(IV). VII. Formation constants for 1:1 cerium(IV)-alcohol complexes," J. Am. Chem. Soc. 1969, 91(18), 5060–5068 (DOI: https://doi.org/10.1021/ja01046a021).

Well, you can see Ce(IV), alcohols form transient 1:1 complexes with alcohols, but such complexes don't last long because of oxidation. In ref. 1, one can deduce that alcohol replaces one of monodentate ligand (nitrate), and thus absorption spectrum of this complex changes for a few seconds.


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Why a Li-ion battery, when in storage, is better kept around 85% instead of 100%?
I discovered that in my laptop settings, I can select an option that is best chosen when the laptop is mainly connected to AC power, and apparently keeps the battery at a lower SOC.
I found this feature also in a Sony smartphone (slightly different operation, it would keep the battery at 90% max during the night, and would fully charge it at dawn) and I found out that's actually good for battery health.
Why is that? What happens inside a battery when it is kept at 100%?

This topic is discussed multiple times on internet and I guess this site too. In short, at full charge, there are accelerated side reactions decreasing the cell capacity and increasing the cell internal resistance. As the result, the static cell aging is faster.

Thanks for the prompt answer. Could you be more specific?

It was just a comment, not answer. I was not going to go into details. You can search them out. Related: https://chemistry.stackexchange.com/questions/124781/keeping-li-ion-battery-on-charger-vs-plug-the-charger-in-and-out-regularly/124797#124797 and https://chemistry.stackexchange.com/questions/142688/what-is-the-optimal-voltage-for-storing-a-3-7-v-lithium-polymer-battery/142690#142690

@Poutnik Let me opine that an advantage of SE sites is peer review and clear explanations, something which is frequently lacking from websites and the scientific literature, respectively. So for my money, answers here have a lot more validity than those on arbitrary websites.

@KennSebesta This approach looks for what is convenient for you, on expense of time and effort of others, adding to it cumulation of redundant info on internet or even the same site. SE sites frown upon questions without signs of prior search effort for exactly these reasons. Asking should be the last, not the first option. Preventing redundant info is part of what makes SE sites great.

By other words, question policy is part of what helps to separate "seed and weed".

@Poutnik No disagreement that people should ask last. If this question wasn't valid, though, either because it needs improvement or it's redundant, shouldn't it have been flagged and closed? If I get to this page because I have the same question, it's a bit of a dead-end unless I read the comments. (Great answers in those links, btw!)

@KennSebesta Question closure is based on a voting system. When I get involved in commenting, I frequently withdraw myself from voting, letting others to decide. In time of search engines, it cannot be a dead end, you could find those links  too, using site:stackexchange.com and few chosen keywords.


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Calculating pH of methanoic/formic acid (HCOOH)
When calculating the pH of methanoic/formic acid (HCOOH) why do we not multiply the given molarity of the acid by 2 as there are 2 moles of H+ ions, as seen in chemical formula HCOOH?
However when it comes to calculating the pH of sulfuric acid (H₂SO₄), we do multiply the given molarity by 2 as there are 2 moles of H+ ions.

The reason for this is that the hydrogen atom attached to the carbon will not be lost. Only the hydrogen attached to the oxygen will be ionized. In fact, in a solution of formic acid, not all molecules will be ionized anyways. Only very strong acids will fully ionize. Even in the case of sulfuric acid, not all molecules will fully ionize and lose the second proton. It is actually a simplification.

Hmm, philosophically speaking, if it's incorrect, is it actually an assumption? Maybe "simplification" is more appropriate?

I say assumption because it was stated that 'there are 2 moles of H+ ions' given off per mole of sulfuric acid which technically isn't fully true because ka2 of sulfuric acid is only 1.2*10^-2. I will change it though.


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Condenser shooting out hexane in a soxhlet extraction

Im running a soxhlet extraction using hexane as the solvent. Once the hexane has filled up the soxhlet, a considerable amount of hexane will come out on top of the condenser. Is this an issue with pressure or defects in the glasses?

That's not supposed to happen. Are you running the bath at the proper temperature and is your water cold enough? Are you using boiling chips? You also might be overloading the filter cup, which would reduce flow back into the reservoir.

no im not using any boiling chips

Heat up slowly.

The glass joint between the top of the Soxhlet and the condenser looks broken. It is probably leaking hexane vapour profusely. You can't get a good seal with broken glass joints. Solution: don't use broken glassware.

Heating without boiling chips or stirring will cause vapor to burst. Your condenser is not the efficient type either. Suggestion: add boiling chips (after it is cooled down); change a more efficient condenser; heat up slowly.

As noted above by matt_black (eagle eye!) the female joint on the top of Soxhlet is broken.  That joint most be replaced ASAP by a qualified glassblower.
Otherwise, you can either injure yourself when moving around the joint, or, due to repeated heating / cooling as well as downward pressure from the weight of the condenser above, the crack will propagate deeper into the Soxhlet itself, potentially resulting in a much bigger accident involving a flammable solvent and a heating element.

Plausible cause: OP built this skyscraper, put on top the condenser behind the non-moveable (typically less frequently cleaned) part of  the sash.  Then the less than lucky clamp (looks like the only one from the stand ...) was not able to sustain the leverage when directing the hose outward (instead of using a clear sink in the rear of the hood) + closing the sash once; the noise of the AC in the hood was louder then the chirping sound of the joint broken, the fracture out of sight if standing in front of the hood. Thankfully it did not topple over (yet) with all the splashes possible.

It is not a defect of pressure (you rightfully leave the system open, because the condenser is open), nor a defect of the glasses used.  With the details provided by you, I would reconsider the setup and the scale of the setup.
Instead of the single large (and very tall* [you can't access the top quickly]) apparatus shown in your question, I recommend to use multiple smaller ones (there are smaller Soxhlets, too) than the one used.  It looks like even if all of the solvent were in the round bottom flask, it never will be more than half filled.  See if you can get two round bottom flasks of $\pu{500 mL}$ (or, two of $\pu{1 L}$) instead which then each would fill up to two thirds at max, topped by their Soxhlet and a good Dimroth condenser.  At this size, it is easier to find suitable magnetic stirring bars for a slow, but continuous agitation of the liquid phase to lower the risk of a sudden «shooting», as you name it, especially if using a heating bath easier to monitor and manage than the heater you use now.
For convenience, these magnets should be the shape of an elongated rugby, because a straight cylindrical bar bears too much danger of hitting (and eventually, breaking) the glass.  Equally, because of their weight, do not drop them into the flask, but slowly glide them into the container; it is possible to slow their move to the bottom if you tilt the empty flask and held a magnet on the outer side of the wall.

(credit)
These are available in different sizes (example entries for these eggs) to match the shape of the round bottom flask's bottom.  Start with a rotation set at low rpm and gently increase the rate.
Like for a distillation, you may wrap the parts supposed to be hot (upper part of the round bottom flask, ascend of the extractor but not the beaker, nor the siphon) with aluminum foil).  This equally helps to run a gentle extraction.
Consider mechanical support with a clamp for the flask / extractor in addition to the one currently only seen for the condenser.  A two / three necked round bottom flask (instead of the single necked one) offers you to refill of hexane (if necessary), to monitor the inner temperature, and to sample the extract for TLC / GC to monitor the advancement of the extraction, too.  (The necks at the side need not be of the same large diameter as the one in the centre.)
*) It looks like you use a (an already bent) lab stand to balance the condenser on top of the mount.  Despite of its metal plate, given the high centre of gravity of the setup, it would be safer to use a grid of rods mounted to the wall of the fume hood instead (see examples here).  Ensure the drain of the hood is clean and working well, and then direct the hose from the condenser to this sink.  This allows you to close the sash without worries that the extraction will topple over plus increased lab safety (you don't topple over the hose, less solvent vapours in the lab, less energy consumption (air condition), less noise, etc.  Similar for hoods where the sash closes automatically.


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Finding the valency of transition metals in coordination complexes
I understand how to find the valency of atoms in small organic compounds. For instance, for CH4, the valency of carbon is 4, while the valency of each hydrogen is 1. But for a coordination complex, for instance:

How do I find the valency of the center metal?
And if the center metal was bound to another metal:

How would the valency calculation change, if at all?
I need the valency of these transition metals because I wish to ultimately find the d^n configuration of the transition metal in the complex via the formula:
n = group number - valency
EDIT: If you have a different approach (apart from using valency) for finding the d^n configuration of the transition metal, feel free to suggest that instead.

This formula will work for most cases of transition metals but to make it universal you should use n= ultimate s electrons + penultimate d electrons - electrovalency

What do you mean by "valency"? The number of bonds? The oxidation state? Or something else?

Umm, well, valency has pretty specific meaning, it's largely forgettable beyond school, though, I guess ;)

According to my lecture slides..."Valence (or valency): The valence of an atom in a covalent molecule is the number of electrons that it has used in bonding."

Metal complexes are not exactly covalent, they are coordinate compounds. Generally valency isn't used for them.

For the first example, you are given that there are $6\, \ce{CN}$ ligands which are -1. That means that there is a total of 6- from the ligands. We also know the overall charge of the complex is -3. That means the charge of the central metal must be +3. In the second example, we can look at the charge again. Because the $\ce{CH3-}$ has a  charge of -1, the molybdenum metals must have a +3 oxidation state each to counteract the -6 total charge from the $\ce{CH3-}$ groups. To do these problems, you should look at the charges of the ligands/metal(s) and the total charge of the complex and know that they have to match. After you figure out the oxidation state, you can find the $\mathrm d^n$ configuration by removing electrons from the $4\mathrm s$ subshell first and then 3$\mathrm d$. For example, For $\ce{Fe^{3+}}$, a neutral $\ce{Fe}$ atom has $6$ $3\mathrm d$ electrons. You first remove 2 electrons from the $4\mathrm s$ subshell and then a final one from the $3\mathrm d$ subshell. Finally, you will find that $\ce{Fe^{3+}}$ is $\mathrm{d}^5$. Again, this is based on the oxidation state of the metal.

It looks like you calculated the oxidation state for each metal. However, I was curious about the valency, not the oxidation state.

Does the primary valance not indicate the oxidation state of the metal? Once you find the oxidation state, you can just use the formula you provided. n = group number - valency. Since the oxidation state of Fe in the first question was 3+, n = 8-3 = 5. We then know that it is d^5. I'm sorry if it is me misunderstanding the topic.

Ah, I am actually not familiar with the term "primary valence." So, does that mean I can use the oxidation state in place of "valency" for the two examples?

Honestly, I am not too sure. I thought you were talking about Werner's theory which I myself am not too familiar with and I literally just learned about while looking at this question. I just saw that you were trying to find d^n configuration with that formula and it works if you just use n = group number - oxidation state.


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Periodic acid oxidation of carbonyls?
The following reaction seem to be valid;
$\ce{CH2OH-CHOH-CH2OH ->[HIO4]CH2O + HCOOH + CH2O + 2H2O}$
Apparently, there are similar reactions by 2,3-dihydroxypropanal, and also by the straight chain form of fructose.
The first two reactions involve doing the Malaprade reaction on molecules with adjacent alcohol and aldehyde groups, and the last one involves doing it on a ketone. Both times, it seems sufficient to just cleave the carbon-carbon bonds and oxidise each resulting molecule one step up the ladder.
As far as I know, the Malaprade reaction is for vicinal diols, and I can't stretch the mechanism to work for carbonyls.
Can anyone help me with how these reactions work?

Hope this help you.

@MathewMahindaratne: "This reaction has been further extended to the cleavage of α‐hydroxy carbonyl compounds, 1,2‐dicarbonyl compounds,..." - so I know it works, great. But I can't find the mechanisms for these variations. Can you provide them?

I have to do some research. It's fast midnight in this neck of the woods. I'll try to give you answer tomorrow morning. No guarantees though! :-)

@Harry Holmes A carbonyl exists in equilibrium with its geminal diol, thus it would undergo oxidative Cleavage in a similar way.

More on Malaprade reaction. https://chemistry.stackexchange.com/questions/29601/periodic-cleavage-of-carbohydrates/102529#102529

In 1928, Malaprade demonstrated that periodic acid reacted with ethylene glycol to produce iodic acid and formaldehyde (Ref.1). Hence, the oxidation of adjacent diols with periodic acid or its salt in aqueous solution is now generally known as the Malaprade reaction, the mechanism of which is depicted below:

The reaction proceeds faster under acidic conditions and can be applied to higher polyhydric alcohols, which behave similar to ethylene glycol. The necessity for the reaction is to have 1,2-dihydroxy functionality $\left(\ce{>C^1(OH)-C^2(OH)<}\right)$, and upon exposer to periodic acid, the middle $\ce{C^1\!-C^2}$ bond would oxidatively cleave to give two compounds with gem-diol functionality $\left(\ce{>C^1(OH)2 \ and \ (OH)2C^2<}\right)$, which are essentially carbonyl groups in aqueous solutions. The reaction has also been further extended to the cleavage of α‐hydroxy carbonyl compounds, 1,2‐dicarbonyl compounds, α‐amino alcohols, α‐amino acids, and polyhydroxy alcohols since then, and successfully applied for structural analysis specifically of sugars:

(source, Ref.2)
As shown in the figure, each $\ce{C-C}$ oxidative cleavage of $\ce{C-C}$ bond gives $\ce{OH}$ group to each carbon. For example, 1,2-bond chevage between $\ce{-CH(OH)-CHO}$ gives aldehyde carbon another $\ce{OH}$ group, making it formic acid as shown in the diagram. Meanwhile, 1,2- and 2,3-bond cleavages gives $\ce{C}$2 carbon two extra $\ce{OH}$ groups so that resultant molecule is formic acid again. Thus, complete oxidation of $\pu{1 mol}$ of D-glucose gives $\pu{5 mol}$ of formic acid and $\pu{1 mol}$ of formaldehyde.
The requirement is to have 1,2-diol functionality. So, how ketone and aldehyde with $\alpha$-hydroxy groups get oxidized? In aqueous solutions, ketones and aldehydes are in equilibrium with their corresponding gem-diols, thus providing 1,2-diol feature.
The 1,2-diol does not have to be cis-orientation. The trans-diol will also get oxidized but with slower rate (see here and Ref.3).
References:

Zerong Wang, "Malaprade Reaction (Malaprade Oxidation)," In Comprehensive Organic Name Reactions and Reagents; John Wiley & Sons, Inc.: New York, NY, 2010 (https://doi.org/10.1002/9780470638859.conrr406). ISBN: 9780471704508.
Fathia Mohammed Ibrahim,  Mubark Elsayed Osman, "Elucidation of Sugars Structure through Periodic Acid Oxidation Cleavage," International Journal of Science and Research (IJSR) 2018, 7(1), 1152-1155 (https://www.ijsr.net/search_index_results_paperid.php?id=6121702)(PDF)
G. J. Buist, C. A. Bunton, J. H. Miles, “149. The mechanism of oxidation of α-glycols by periodic acid. Part V. Cyclohexane-1 : 2-diols,” J. Chem. Soc. 1959, 743-748 (https://doi.org/10.1039/JR9590000743).


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Can the solvent fraction of a sol-gel process containing tetraethyl orthosilicate (TEOS) be analyzed by GCMS?
Can the solvent fraction of a sol-gel process containing tetraethyl orthosilicate (TEOS) be analyzed by GCMS? I am worried that the easy polymerization of TEOS to form silica (SiO2) will ruin the GCMS system.

See the conditions and detector in https://www.sciencedirect.com/science/article/pii/S0021967307015385


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On which side is the intermediate carbocation formed during acid-catalyzed opening of epoxide?
I found an organic chemistry question which asked the major product of the following reaction -

The mechanism for the same in the answer was -

My questions regarding this are as follows -

The molecule has two oxygen atoms, one as an epoxide and the other as ketone. How do we decide which oxygen atom accepts the proton? Seeing the mechanism provided in the answer, I reasoned that the epoxide oxygen is sp3 hybridized, hence less electronegative than the ketone oxygen, which is sp2 hybridized. Hence as the less electronegative atom it donates its lone pair to the proton. Am I correct? If not, what is the real reason?

Acid-catalyzed opening of epoxide follows SN1 mechanism, which implies more stable carbocation must be formed. In the given mechanism, however, the carbocation is formed on the right side, where the carbonyl group exerts -I effect and destabilizes the carbocation. I could not make sense of this.

Are there any specific rules as to which site is attacked by the proton if multiple nucleophilic sites are present in the molecule? Also, is my reasoning in (2) correct?
I am also open to the fact that the given mechanism in the image may be wrong, in which case, I would also like to know the correct mechanism and product.

The second point is incorrect. The cation gians stability due to +R effect of the conjugated carbonyl.

@NisargBhavsar Thank you for your comment. You mean the oxygen atom on the carbonyl group acquires a positive charge by +R effect, while stabilizing the cation with a double bond between the carbonyl carbon and the cation? Positive charge on an oxygen atom seems a bit counterintuitive...

Yes the +ve charge on oxygen is not the best but the delocalisation it provides is definitely better than nothing.

@NisargBhavsar Thank you so much for the clarification. It would be helpful if you could please let me know if my reasoning in (1) is correct or not

Yes your reasoning in (1) is correct.ヾ(＾-＾)ノ

@NisargBhavsar Thanks a lot for your help!

I think for your point 1, you are overcomplicating stuff. If you protonate the ketone, then you can't draw any mechanism from that point, but if you protonate the epoxide, you can open the ring. Minor differences in electronegativities of oxygens is probably a secondary factor.


text:
First step in a reaction between an acyl chloride and sodium acetate in acetic acid

What would the first step be in a reaction between acyl chloride and sodium acetate in the presence of acetic acid?

I'm a bit confused because I'm not quite sure which step would occur first. I was thinking that perhaps the sodium acetate will first attack the carbonyl carbon, but I'm also wondering if it would be possible that sodium acetate would first react with acetic acid, turning the latter into an anion that could react with the acid chloride forming an anhydride.

If the sodium acetate reacts with the acetic acid then the result is sodium acetate and acetic acid. You have only exchanged sodium ion between acetic acid molecules.

@Waylander I guess the OP means by "acid chloride"  an acyl chloride, particularly acetyl chloride and the reaction $$\ce{CH3COCl + CH3COONa ->[CH3COOH] (CH3CO)2O + NaCl}$$.

Hi! Thank you so much for answering
Yes @Poutnik, that's the reaction I'm referring to, thank you for this!

My other question though is, what is the carboxylic acid used for? Will it be fine to just use any other group with an alcohol group like, say, butanol?

Acetic acid serves as a solvent. Butanol would IMHO react $\ce{CH3COCl + BuOH -> CH3COOBu + HCl}$


text:
Lack of polysubstitution in Friedel–Crafts t-butylation of phenol
According to our lab manual:

In general, Friedel–Crafts alkylation is not employed to make monoalkyl benzene derivatives because of the polysubstitution.

But our professor told us that while we are doing this experiment, polysubstitution is quite impossible.

I know the when the electrophile is bigger and amounts are exact, it is harder for polysubstitution to occur but not impossible.
So I wonder if the reason why she said that, is about chemical properties of the reagents (like being activating) or is it something about the method we use in the experiment?

Phenol first forms a complex with AlCl3 so the ortho sites are hindered for approach by a large electrophile

@Waylander That really makes sense thank you!


text:
On the stabler tautomer in amide-imidic acid tautomerism
In the section on acid-catalysed hydrolysis of nitriles, this webpage has the following line of reasoning;

...the more stable amide tautomer predominates the (following) equilibrium.

(credit)
How are the relative stabilities of these tautomers compared? I made up a reason of the resonance structure for the amide one having negative and positive charges on oxygen and nitrogen respectively, and vice versa for the imidic acid one, but I'd like clarification on that.
There's also another hand-wavy thing I thought about of the double bond being stable on the more electronegative atom, oxygen. Is that correct, too?

For future reference: If there is an illustration you refer to, take a moment to include it here on ChemSE altogether with a note of credit.  The reason:  Perhaps the address of the webpage you refer to changes (or is closed for whatever reason).  This would yield your question less easy to access, than necessary.  Regardless if «amdie» looks like a typo, or not.

The relative stability of the two tautomers (two isomers differing how protons are bound) may be assessed from the equilibrium constant you may formulate for the two.  Deriving from above illustration, it would be

to indicate that this is dynamic, with a back and forward reaction at microscopic level.  Once formulated as such, you probably recall $\Delta{}_RG = -RT \ln{}K$ to relate between the sample temperature $T$, the universal gas constant $R$, the free reaction enthalpy $\Delta{}_RG$ and the equilibrium constant $K$.
Depending on the substrate, you may use spectroscopic evidence (e.g., $\ce{^1H}$-NMR spectroscopy) to assess the concentration of the two forms.  Equally, based on prior work by others, there are databases to either list these values, or provide a prediction (i.e., an educated guess) about $K$.  Knowing these equilibria is important to anticipate the likelihood of chemical reactions. It equally offers insight regarding the pharmaceuticals and their uptake, anabolism, interaction with the body (metabolism), and catabolism and eventual elimination, too.
One of the databases available to the public is Tautobase (2020JChemInfModel1085). It is implemented in the freely available DataWarrior program allowing you a search-by-(sub)structure.  With some luck, one of the 1680 pairs matches your substrate closely enough, too.  In the simplest case one may imagine, however, the amide is strongly favoured:

The database backs this particular entry with the publications 1998JAmChemSoc10359 and 2001JAmChemSoc2681.

+1, Thanks for the amazingly well-researched answer. But I was asking about the reasons you make up to explain the observation that we already know, not the values that tell you which the preferred tautomer is themselves. Could you clarify whether my guesses at these reasons are valid?

I understood your question «How are the relative stabilities of these tautomers compared?» as a search on how experimental evidence is collected.

@HarryHolmes Neglecting influence by solvent / reaction conditions, for the isolated molecule the total energy including a carbonyl $\ce{C=O}$ may be a different one compared to $\ce{C=C}$ (as in keto-enol equilibria), or $\ce{C=N}$ (as for the case of imidic acid). It would be nice if there is a second perspective including this aspect (and maybe tabulated data like these), too.

Okay, is there any resource from where the molecule energies are accessible? And is there any weight to the resonance thing I was going on about?

Last picture shows different type or tautomerism.

@Mithoron: yes, but I don't see how that explains it. Moreover, it looks like the product-side of the equlibrium has something different from the imidic acid in the question.

@Buttonwood: the double bond in the original equations was to the nitrogen in the imidic acid, but the last pic has it into the organic fragment.

@HarryHolmes True, the screen photo is about the putative equilibrium about formamide and amino-ethenol.  For the current version of DataWarrior (April 2021) and tautobase (2020), this is the most similar pair with a formide-like pattern. The pair with an imidic acid equally has an amino-ethenol on the other side of the equation.  For the energies, I hoped one of the quantum-chem savy readers either a) could point to a compilation, or b) offer an insight / result of computation.


text:
Finding the mole fraction of nitrogen in a mixture of nitrogen and oxygen
Question

The density of a mixture of Nitrogen and Oxygen is $\pu{4.7 g L-1}$.
Its given that $P = \pu{4 atm}$, and that $T = \pu{298 K}$. Assuming both gasses are ideal gasses, what is the molar fraction of nitrogen in this mixture?

Answer

 Mole fraction of nitrogen is 0.82.

My Attempt
I tried using the $PV=nRT$ equation, having $V = 1$ and $n$ being the number of moles of both gasses together.
Then I used a second equation for the density being $D=\dfrac{16n_1+14n_2}{V}$.
However I got a negative answer that is also larger than $1$.

You can get the effective molar mass from equation $M_\mathrm{ef}=\frac{\rho \cdot R \cdot T}{p}$ and from that the molar fraction.

Oxygen and Nitrogen exist as $\ce{O2}$ and $\ce{N2}$ respectively.

I will elaborate on a previous comment. Once you find the effective, you find that  $28.02∗0.82+32∗0.18=_{}$ as 28.02g/mol is the molar mass of $N_2$ and 32g/mol for $O_2$

OP has given a good effort to solve the problem using correct path. Only loose point was not considering the diatomic nature of the gases as Safdar Faisal pointed out in a comment. Suppose the amount of $\ce{O2}$ in the gas mixture is $n_1 \ \pu{mol}$ and the amount of $\ce{N2}$ in the gas mixture is $n_2 \ \pu{mol}$ in volume of $\pu{1.0 L}$ container. Thus, two parallel equations can be setup as follows:
$$PV = nRT \ \Rightarrow \ n_1 + n_2 = \frac{PV}{RT} = \frac{4 \times 1}{0.082 \times 298} = 0.164 \tag1$$
$$\frac{32n_1 + 28n_2}{1} = 4.7 \ \Rightarrow \ 32n_1 + 28n_2 = 4.7 \tag2$$
To resolve two parallel equations, take $(1) \times 32 - (2)$, which gives you,
$$(32-28)n_2 = 0.164 \times 32 - 4.7 = 0.538$$
$$\therefore \ n_2 = \frac{0.538}{4} = 0.135$$
Thus mole fraction of $\ce{N2}$ is $\chi_\ce{N2}$, then
$$\chi_\ce{N2} = \frac{n_2}{n_1 + n_2} = \frac{0.135}{0.164} = 0.820$$
Therefore, the mole fraction of $\ce{N2}$ in the mixture is $0.82$ (hence, the mole fraction of $\ce{O2}$ in the mixture is $0.18$).


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Orbital Interaction for electrophile and nucleophile

Clearly, the above diagrams depict the interaction between empty orbital of electrophile with filled orbital of nucleophile. We may assume that the HOMO of  nucleophile being more electronegative will have energy less than that of less electronegative electophile as above.
As on the diagram on page 111 on Organic chemistry by Clayden, it is stated that the empty orbitals of Nu are too high in energy to interact with the electrophile. But if the HOMO energy is higher than the vacant orbital of electrophile will the reaction not be feasable?

If the HOMO energy was higher than the vacant orbital of the electrophile, it would transfer those electrons to the electrophile. If electrons get a lower energy level that they can occupy, they will do so.

Why is the electrophile lumo at a higher energy than nucleophile filled orbital?

https://chemistry.stackexchange.com/questions/115150/why-is-the-homo-always-lower-in-energy-than-the-lumo


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Comparing retention time of acetic acid, salicylic acid and acetylsalicylic acid in HPLC
I want to compare retention time of acetic acid(AA), salicylic acid(SA) and acetylsalicylic acid(ASA) in HPLC. What are the factors do I have to consider when comparing? Considering dipole moment only, retention time would be AA(1.74D) < SA(2.65D) < ASA(6.49D). Is this correct? (Addition : Mobile phase would be water)

It very depends on character of the stationary mobile phase, that can point out particular type of interaction ( dipole, e. p. donor or acceptor ) or suppress it. By other words, at the same elution time of chosen analyte, elution times  of other analytes change with changed MF, or they can even switch the order.

HPLC does not do separations on the basis of dipole moments or so-called polarity. In fact, you will not see any retention of these very hydrophilic molecules on a typical C18 column. If you add more details, I can try to address your query.

See related HPLC MP triangle on context of aromatic carboxylic acids on C18 column ( acetic acid would have probably just minimal retention).


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SARS-CoV - relative size of the spike protein
I am taking an online course in image analysis and have been asked to use the attached image to determine the size of the SARS-S protein of the SARS-CoV relative to the total of its proteins. However, I have been given no explanation of the image, and with a physics background, I simply do not understand it. Can someone please explain the image (specifically the SARS-CoV column).

Biology is more suitable for this, I think.

@orthocresol, I tried but they said that they don't help with homework (although I am not asking you to solve the problem - simply to help me understanding the image).

Hmm. This is a tough case. After all, Chem.SE has a similar 'homework" policy; and although I can understand your point, it may or may not be that other users will be sympathetic towards it. I think it could also logically be closed as being off-topic (i.e. not about chemistry). To be entirely honest, the root cause of the problem is that you're given an image without any context whatsoever, and asked to interpret it. That seems like an unrealistic expectation to me, and even though you may happen to find an answer on SE, it seems to me that that'd only be a one-time, superficial fix.

It's been a long time since I did biology, but I think I can at least point you in the direction of the Western blot and gel electrophoresis. Wikipedia might not be the best resource on this. You might be able to find other easier explanations elsewhere on the Internet.

The Western blot looks similar to the image, so I will try and read about that. Thank you for trying to help, @orthocresol.

Measure how far over the background the first band in lane 8 is, and divide it by how far over the background all signal in lane 8 is.

@KarstenTheis, when you say 'the first band' in lane 8, do you mean the black line in the top at the SARS-S row?

Yes, the one at the same height as the main bond in the SARS-S lane.

Although this is a homework question, I also think OP deserves a help as orthocresol suggested. I'm not going to go into details but leave it at orthocresol's suggestion of this being a Western Blot of protein analysis, which is absolutely correct. Hence, OP should start reading of electrophoresis and western blots techniques since, I assume, s/he taking molecular biology class.
In brief, the given image is a X-ray photograph of the nitrocellulose membrane after transferring the proteins from the SDS-PAGE gel (after electrophoresis). I'm suggesting that because the image looks like visualized by radioactivity (I suggest OP to read about these techniques in details to understand). In SDS-PAGE gel electrophoresis, the proteins separated mainly by molar mass of each protein while heaviest protein go first since electric current run from top to bottom as sepicted in following diagram:

As marked in the diagram, SARS-CoV consists of SARS-M ($\pu{20 kD}$, SARS-N ($\pu{54 kD}$, and SARS-S (~$\pu{175 kD}$ proteins. I may give you your answer but I strongly suggest to read about the subject for better understanding. I do not support online courses, but there are no choice you can made during a bad pandemic era.

I am not taking a class in molecular biology; it is supposed to be a class in image analysis with a focus on using the software, not understanding biology (we have also analyzed pictures of faces, traffic, etc.). It is online because the university is shot down.
You say SARS-CoV consists of the three proteins you indicated with a red line, but what are all the small lines in between?

Those are some other proteins, of which authors either didn't understand or have no interest. SARS-ME have identical or similar proteins with different concentrations. If you pay attention, SARS-CoV also have protein similar to Vero protein with $\pu{50 kD}$ (just below SARS-N protein). However, authors didn't mark it meaning they are not interested (I keep telling authors because I assume the image is from a publication).

Okay, so I believe the image analysis part is that I need to identify all the little lines since I need to consider the total of its protein. Thank you very much.

@JKalle it is still surprising that the exercise has been given. How one can analyse an image of that type in relation to protein size (obscure in a way in itself, it is number of amino acids or the physical V) without knowing about proteins and the technique used? Really puzzling. Unless you arbitrary resume the actual quiz text.

@Alchimista, yes and it is frustrating to be given such a task when the image analysis part is not that difficult once I know what he wants me to do. But maybe the teacher thinks that a Western blot is second nature to anyone in science, which is not the case.
The task does say the 'approximate' size of the S protein relative to the total of the proteins, but that would be implicit either way since it is impossible to determine the exact value from that image.


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Regarding stability and reactivity of m-xylene
I came across the following two details about m-xylene:

m-Xylene is thermodynamically most stable compound among o-xylene, m-xylene, and p-xylene.

m-Xylene is most reactive towards nitration and sulphonation among the others.

I did some research and was able to find valid reasons for both the statements, but they still seem to be contradicting each other. How can m-xylene be most stable as well as most reactive at the same time?
Could someone please clarify my doubt and also, add more specific details if required.

UPDATE :
"In m-xylene, the two methyl groups reinforce one another in their activation
of the ring, while in o- and p-xylenes the methyl groups cancel the ring
activation to some extent."

I totally agree with this statement and it clearly explains the reason behind the the reactivity of m-xylene towards nitration and other such reactions.
But I realized that I was considering this statement as the reason behind the stability of m-xylene as well. Reference:
Thermodynamic stability of meta-xylene over ortho- and para-isomers
The reasoning seems to be fitting for both the cases (I do realize now that stability and reactivity are not opposing terms necessarily).
So, is the reason why m-xylene is most thermodynamically stable the same reason why m-xylene is most reactive towards nitration? Could someone please confirm this or add their inputs if I am making some mistake in correlating terms or in my reasoning.

Stability and reactivity have only a superficial relationship. Stability is about energies. Reactivity depends on activation energies which in turn depend not only on the energy of the starting material, but also the energy of the transition state.

@orthocresol Thank you for your explanation, I understand it now

@Pal If you got your answer, feel free to self-answer it. Self-answering is always encouraged and it will also help future readers.


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What toxic gases are emitted during the combustion of potassium ferrocyanide?
I have read that Hydrogen Cyanide is emitted if Potassium Ferricyanide is combusted, but would this be the same case for Potassium Ferrocyanide? I am preparing to perform some case hardening on steel using a compound that contains 30-50% Potassium Ferrocyanide, so I need to be sure what measures are needed for protection from the fume.

I thought "combustion" only applies to something that can burn. Let's say "heating" instead. Also, heating or no heating, K3Fe(CN)6 contains no H, hence it can't possibly give off HCN. (Unless reacted with something else, that is.)

See my answer here: https://chemistry.stackexchange.com/questions/32189/how-to-prepare-potassium-cyanide-from-potassium-ferrocyanide/83702#83702

I contacted the manufacturer of the case hardening compound. Their chemist replied:
Specific hazards:
Contact with acids liberates very toxic gas. Hydrogen cyanide (HCN).
Thermal decomposition or combustion products may include the following substances:
Hydrogen cyanide (HCN).

Potassium ferrocyanide cannot burn, because it is the product of a combustion. When meat or blood is overheated, carbonized and burnt, the remaining ashes contain potassium ferrocyanide. And this product was initially recovered by washing these ashes and evaporating the obtained solution. In German, this compound is called Blutlaugensalz, which means "salt from washing blood" This is even the way this product was originally discovered in the old times of beginning chemistry.
So potassium ferrocyanide resists high temperature, and will not produce any $\ce{HCN}$ by heating or by combustion. The only way of producing HCN out of it, is to heat it with diluted sulfuric acid, according to the equation : $$\ce{K4Fe(CN)6 + 3 H2SO4 -> 2 K2SO4 + FeSO4 + 6 HCN}$$This reaction is difficult to carry out properly, because, if the sulfuric acid is more concentrated, it will destroy $\ce{HCN}$ according to  $$\ce{H2SO4 + 2 HCN + 2 H2O -> (NH4)2SO4 + 2 CO}$$

Potassium ferricyanide comes with water of hydration, $\ce{K4Fe(CN)6}\cdot\ce{3H2O}$ (See here). Decomposition in the presence of this water of hydration could give hydrogen cyanide gas.

I am guessing you mean a fused salt bath at 1400 to 1600 F to diffuse both carbon and nitrogen into the surface of the steel. The other salts in the bath likely have an affect on fumes. I never worked with molten salts but it is definitely not for amateurs. Find a commercial heat-treat shop that performs this process.


text:
Relation between enthalpy and internal energy for chemical reaction at constant volume
Question

The combustion of benzene(l) gives $\ce{CO2(g)}$ and $\ce{H2O(l)}.$ Given that the heat of combustion of benzene at constant volume is $\pu{-3263.9 kJ mol-1}$ at $\pu{25 ^\circ C}$; heat of combustion (in $\pu{kJ mol-1}$) of benzene at constant pressure will be: $(R = \pu{8.314 J K-1 mol-1})$

Answer

 $\Delta H = \pu{-3267.6 kJ}$

Explanation of Solution

 $\ce{C6H6(l) + 15/2 O2 (g) -> 6CO2(g) + 3H2O(l)}$
$\Delta n_\mathrm g = 6- 7.5 = -1.5$
$\Delta U$ or $\Delta E = \pu{-3263.9 kJ}$
$\Delta H = \Delta U + \Delta n_\mathrm g RT$
 So $\Delta H = -3263.9 + (-1.5) \times 8.314 \times 10^{-3} \times 298 = \pu{-3267.6 kJ}$

My question
When a reaction occurs at constant volume they equate enthalpy with internal energy which I think is wrong because according to the equation:
$$\mathrm dH = \mathrm dU +\mathrm d(PV)$$
At constant volume:
$$\mathrm dH = \mathrm dU + V\mathrm dP$$
In the question, heat of combustion is given as $\pu{-3267.6 kJ},$ while solving for constant volume $\mathrm dU$ is given as $\pu{-3267.6 kJ}.$
Why is $\mathrm dH = \mathrm dE$ at constant volume?

why is dH=dE at constant volume?

It isn't, and nowhere in the problem or answer is this implied.
First of all some definitions.
For a combustion reaction, the enthalpy change can be equated with the heat of combustion at constant pressure, whereas the internal energy is the heat of combustion at constant volume:
$$\Delta U = q_V ~~~~~\text{constant volume} \\
\Delta H = q_p ~~~~~ \text{constant pressure}$$
There is a simple integral form for the relation between internal energy and enthalpy that follows from the definition of the latter:
$$\begin{align} \Delta H &= \Delta (U+pV) \\ &= \Delta U + \Delta (pV) \end{align}$$
This relation always holds (it is a definition).
For this particular problem, the key to the next step is to use the ideal gas law, $pV=n_gRT$. This is allowed only if that equation of state holds for the substances in question (here carbon dioxide and oxygen). Inserting the ideal gas relation gives
$$\begin{align} \Delta H &=  \Delta U + \Delta (n_gRT) \\ &= \Delta U +  RT\Delta n_g\end{align}$$
where T is assumed constant. This is the expression used to find the solution. Clearly it shows that the change in enthalpy and internal energy are different.
You could derive this last expression slightly differently to emphasize the differences in the energy and enthalpy. The enthalpy change includes the heat input required to perform work against a constant pressure atmosphere and isothermally. In the case of an ideal gas, this heat input is
$$q_p-q_V = \Delta q = -w_{pV} = p\Delta V = p \Delta (n_g RT/p) = RT \Delta n_g$$
Note that for an ideal gas the enthalpy is a function of T only, so it's true then that a change in enthalpy is the same under constant V or p, provided you converted equal amounts of substance at same T. The same holds for the energy (since for an ideal gas it is also a function of T only).


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Why are all of the group 16 hydrides/hydrogen chalcogenides except H2O toxic?
I mean ($\ce{H2O}$) is non-toxic and is needed everyday but all of the other group 16 hydrides (hydrogen sulfide ($\ce{H2S}$), hydrogen selenide ($\ce{H2Se}$), hydrogen telluride ($\ce{H2Te}$) and hydrogen polonide or polonium hydride ($\ce{H2Po}$) are toxic and flammable. We observe that water is an outlier. Why is this the case?

Because life was born in an ocean of water.

It is just confusing because you would think since a chemical is related to one of the other you would think that it is similar,but chemistry is obviously more complicated then that

https://chemistry.stackexchange.com/questions/86540/why-h2o-is-a-liquid-water-and-h2s-is-a-gas

Chemistry within a group is not as similar as the Periodic Table might lead one to believe (except, perhaps, in groups 1 and 2). You should already know that first-row elements (B, C, N, O, F) stand out in terms of e.g. hydrogen bonding. I'm not trying to say that toxicity is caused by the lack of hydrogen bonding, I'm just trying to point out that there are substantial differences between O and S (and beyond), and perhaps it's not all that surprising after all.

Evolution means that we are resistant to all of the substances that are abundant on Earth. For example, nitrogen, orygen, water, salt etc. If those compounds were toxic then we wouldn't survive would we?

Tangentially related question: https://chemistry.stackexchange.com/questions/4636/why-do-chalcogens-group-vi-stink-so-badly/ One of the answer says: "The bio-degradation of sulfur compounds mostly leads to some sulfides which very strongly bind transition metals, i.e. they can destroy/block Metalloenzymes". The same might occur with selenide and telluride.

Continuing Mithoron's comment: H2O is liquid others are gas. So, how do they contribute to toxicity? Because of its volatility and smell. Due to them being gas, they have the tendency to have higher exposure to human body than water, so they are more prone to be to smelly and toxic. See: https://chemistry.stackexchange.com/questions/81935/how-does-the-smell-of-a-compound-come-about-and-is-it-possible-to-define-a-smel/81937


text:
Should uranium hexoxide exist?
This compound of uranium called uranium hexoxide or uranium (XII) oxide is a theoretical chemical compound. The uranium should have  an oxidation state of +12 ,the highest of any known chemical. The closest is the $\ce{[IrO4]+}$ cation with iridium having an oxidation state of +9 so this is obviously mind blowing. Does this compound exist or has it been synthesized yet,without it spontaneously decomposing?

See: https://pubs.acs.org/doi/abs/10.1021/jp102107n ... https://www.sciencedirect.com/science/article/abs/pii/S0009261400009581

The oxidation number may not be +$12$, if the oxygen atoms are part of peroxy groups ·O-O·, where their oxidation number is -$1$.

@maurice it's more complicated than that.  The references describe oxygen in multiple ligand states, including not only peroxide but also superoxide and even monovalent radical oxygen ($\ce{-O•}$).


text:
Can the RDS be different in different direction in a multi-step reversible reaction?
The following mechanism is given in Peter Sykes when they are "talking about de-esterification" of an ester where the alcohol part is bulky-

Here the second step is given as RDS. According to me, in the reversed reaction(esterification), this step will remain RDS and so the reaction is bimolecular. Someone argues that RDS will be the formation of carbocation then and so the reaction is unimolecular.

One reaction is esterification and the other is acid catalysed hydrolysis.. does this answer your question?

@SafdarFaisal I'm not sure I get what you mean

But then how can we be sure that formation of carbocation is RDS. Determination of RDS is experimental, right?

Yes, What is the RDS in acid catalysed hydrolysis of esters?

It's as given in the above picture. Second step, right? I want to know what will be RDS in esterification.

Let us continue this discussion in chat.

After discussion with OP in chat, the issue in logic can be rectified as follows. Simply put:

$\ce{A<=>B<=>[slow]C <=> D}$ doesn't imply that $\ce{B <=> C}$ is the RDS in both $\ce{A -> D}$ and $\ce{D -> A}$.

Rather than this, we need to consider $\ce{A->D}$ and $\ce{D-> A}$ as two separate reactions. Doing so for the two reactions given in the question (acid catalysed hydrolysis ($\ce{A ->D}$) and  esterification ($\ce{D -> A}$)), we see the mechanisms are as follows:
Acid catalyzed hydrolysis

Here the RDS is the formation of the carbocation $\ce{^+CMe3}$.
Esterification of an acid and a bulky alcohol

Here the RDS is the formation of carbocation but it happens at a different step and not the same step.
Therefore the two reactions have different RDS steps as they are different reactions

I don't understand the logic of your answer. The forward and reverse reaction must have the same rate determining step, unless there are low energy intermediates in the reaction coordinate (and I think this doesn't have that). Forward and reverse reactions aren't different reactions. https://chemistry.stackexchange.com/questions/49548/is-the-rate-determining-step-the-step-with-the-largest-ea

The product in the RDS remains the same - formation of the trimethyl carbocation. It just happens at a different step in the reaction. the trimethyl carbocation is the intermediate here.

If they are the forward and backward version of the same reaction, it's not possible to get the RDS at a different step of the reaction, due to microscopic reversibilty. If you agree that the forward and reverse reaction has the same reaction coordinate, then they really must have the RDS at the same step.

That's my point. They aren't the same reaction, both require slightly different conditions. tbh, the arrows are misguiding, the slow reaction in the first one becomes a fast reaction in the second. the formation of the carbocation leads to a valley between two peaks in the energetics graph. One step has the same rate forward and backward at equilibrium but in the whole reaction each step is distinct

@ShoubhikRMaiti Sorry, that reasoning is not correct. Forward and reverse definitely have the same reaction coordinate, but that doesn't mean they have the same RDS. Consider A -> TS1 -> B -> TS2 -> C, with relative energies: A = 0, TS1 = 5, B = -20, TS2 = -10, C = -15. This means that the RDS in the forward direction is the one with TS2, but the RDS in the reverse direction is the one with TS1.

@SafdarFaisal Your claim that $\ce{A <=> B <=> C}$ does not imply $\ce{A <=> C}$ is not correct due to microscopic reversibility and the definition of equilibrium.

@Zhe I understand that. What I meant was what I gave as the logic in this answer..

@Zhe I already mentioned that caveat in my first comment, "unless there are low energy intermediates in the reaction coordinate". Your example has B as the low energy intermediate. Does this esterification reaction have a low energy intermediate?

@ShoubhikRMaiti That's fair. I should've read your comment more carefully. I think you can however come up with tricky situations where the intermediate is intermediate in energy between the reactant and product, i.e., set C = -25.

@Zhe Yes of course. If there is any intermediate that's lower than the reactant in energy, then the rds would be different for forward and reverse.

@ShoubhikRMaiti Sorry to drag this out, but that's not necessarily true. It depends on the relative energies of the transition states even in that scenario. I think the point here is that the number of scenarios increases quite rapidly with the number of steps, and it's hard to make generalizations for all classes of reaction coordinate diagrams.

@Zhe Ok, but my point is that for the usual organic reactions, starting from the reactant we have a series of high energy intermediates before the product. In those cases, the highest TS is definitely the rate determining TS, which means the same step has to be the rate determining step in both directions. Which is why I think this answer is probably wrong. Do you agree?

@ShoubhikRMaiti The problem here is that the equilibrium, under practical conditions, lies on one end or the other. Under that treatment, the forwards and reverse reactions (run to generate predominantly either one or the other product) are different and therefore have different rate determining steps, which addresses the OP's actual question. But for the given equilibrium, the RDS ought to involve the same transition state. So, I think people are arguing about different things here without specifying the underlying assumptions.


text:
Significance of p functions in compounds of the lighter alkaline and alkaline earth metals
It is often said that lithium follows the duet rule instead of the octet rule, unlike the more "usual" main-group elements.
My question is the following:
Does this hold for $\ce{Be, Na}$ and $\ce{Mg}$ as well?

Which sort of lithium compounds are you speaking of ? And which calculation ?

That's not true even for lithium - it's usually tetracoordinate.

So-called lithium duet compounds are not so.  An organolithium compound, for instance, is oligomeric.

Btw, what exactly is "p function"?