text:
Why do alkenes (olefins) burn hotter than alkanes (paraffins), even though alkanes are denser and more hydrogen-rich?
Burning hydrogen releases more total energy than combusting carbon, and all alkanes are denser as well, so why are flame temperatures for alkene molecules higher than their equivalent alkanes?
P.S.: Also, is there a place on the web listing specific energies and energy densities for alkenes (olefins)?

The more water vapor in flame the lower temp.

Actually this is not entirely correct or a complete picture. Flames have been studied for a long time, even intentional injection of steam can increase flame temperature. For example this study concluded that "The results indicate that water vapor does not act as an inert diluent but instead inhibits carbon formation and gives rise to a greater heat release than in its absence." Combustion and Flame, 27 (1976), pg. 205-215

@M.Farooq nice comment. I would have answered like Mithoron, without reading this.

Flame temperatures are not primarily determined by the thermodynamics of the reactions in the flame
Intuitively it might seem that the temperature of a flame is dependent on the thermodynamics of the reactions occurring in the flame (eg total energy released, as the question assumes).
But flames are far more complex than a simple point of view based on equilibrium thermodynamic properties might suggest. Apart from anything else, flames are dynamic. And they depend strongly on the detailed conditions in the burner where the flame occurs. This should be fairly obvious from the known behaviour of flames in a bunsen burner. A well mixed flame where the fuel and air have the right ratio is very hot but the same fuel will produce a much cooler flame when insufficient air is mixed into it. The detailed conditions creating the flame matter.
But, for any given fuel, even well-mixed gases give different temperatures. But even this is not primarily determined by the thermodynamics of the reaction. Kinetics matters. The speed with which the oxygen can react with the fuel depends a lot on the nature of the fuel, so we probably should not expect the temperature to be strongly dependent on the simple thermodynamics of the reaction. If one reaction happens much faster than another, the flame could easily be much hotter even if the other reaction ultimately releases more energy (flames are dynamic and the temperature depends both on the amount of energy released and how fast it can be released).
And observations of how hot flames can be certainly suggest that kinetics might well dominate. Alkenes are more reactive than alkanes (and the flames are hotter) and alkynes more reactive still. So the hottest widely used flame in welding in the oxy-acetylene flame. Many casual explanations online seem to get the nature of this effect wrong, claiming that the acetylene triple bond "has a lot of energy to release" which is just wrong. But it is far more reactive than an alkane so that energy release probably happens much faster leading to a much hotter flame.
The main point is that you have to think about kinetics, flame conditions as well as thermodynamics to understand flame temperatures. And this means the results are more complex than a simple perspective based on just one factor suggests.

Double-bonds are less stable, i.e., their formation is less exothermic, so it takes less energy to break one double bond than two singles.
Or one triple vs. three single bonds. Consider triple-bonded acetylene -- there are not many hotter flames than from oxy-acetylene combustion... well, there is dicyanoacetylene, $\ce{N≡C−C≡C−C≡N}$, with three triple bonds in one compact molecule. Not quite up to plasma torch level, though.
See Derek Lowe's Things I Won't Work With for another stressed-out substance.


text:
Of 2-bromobutan-2-ol and trans-1,2-dimethylcyclobutane which is chiral as well as dissymmetric?
In an examination I was asked to determine the molecules which are chiral as well as dissymmetric. There were four options, and among them, two were achiral, as they had an improper axis of symmetry. The other two options are 2-bromobutan-2-ol and trans-1,2-dimethylcyclobutane:

These two compounds are chiral, as they do not have an improper axis of symmetry. I couldn't find any axis of symmetry as well. The question has only one correct answer, but I don't know which one is correct. Please help me with the answer.

Chiral molecules are always dissymmetric

(c) has an axis of symmetry.

To be honest, at another time of day, this probably would not have been closed. And I don't think it should, while it could still have used some improvements. Anyway, here is something related: https://chemistry.stackexchange.com/q/118985/4945

I assume you already studied molecular symmetry and point groups.
So, first of all, among the existing point groups, you can have chiral point groups (point groups C1, Cn, Dn) and achiral point groups (all the others). As a matter of fact, chirality is unchanged when it comes to operations such as E or Cn (that equal an even number of reflections). On the contrary, this is not the case when it comes to operations that equal an odd number of reflections (σ, Sn, inversion), as they can convert a right-handed object into a left-handed object. So an improper rotation axis (Sn) implies that the molecule cannot be chiral.
Both molecules that you show here are chiral, as you correctly pointed out.
2-bromobutan-2-ol belongs to point group C1 as it has no symmetry other than the identity (which is the C1 axis that every molecule has and that corresponds to a rotation by 360° that of course brings the molecule back to the same initial position). Being the C1 axis the only element, C1 molecules are defined asymmetric.
The other two chiral point groups (Cn and Dn) are dissymmetric.
trans-1,2-dimethylcyclobutane has a C2 axis that allows you to rotate the molecule by 180° (which is 360°/2, that's why it corresponds to a 2-fold rotation axis) obtaining a situation that is equivalent to the initial one:

Let's see this with colors, so it's easier:

So, in conclusion, since (1R,2R)-1,2-dimethylcyclobutane belongs to the C2 point group, it is "chiral as well as dissymmetric".
Note: The above is only true in a first order approximation. A more detailed look will reveal that the cyclobutane ring is puckered and the C2 axis no longer exists.

In first approximation yes, but the cyclobutane ring is puckered, so: no.

@Martin-マーチン Sure, but I think the examination under consideration here assumes the ring to be planar for simplicity.

Yes, I merely stated that it isn't that simple. If some more advanced people come here, then this fact should be included. As a side note: examiners shouldn't pick examples that are only correct at a glance; I don't think they are doing a good job then. I find it very important to not think in schematic drawings, and questions like these trends to reinforce the view that these drawings are somewhat accurate.

I think that this should be re-opened because this answer togheter with the comments can be useful to "teach" a new student or whenever a false simple exercise of this kind is administered. The answer will benefit if the point raised by the comment is integrated.

I took the liberty to do some reformatting and I've included my initial comment as a note. The other one is too personal to be included into a technical answer in my opinion. I'll also reopen the question. cc @Alchimista

source: Stack Exchange

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

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text:
Density and partial pressure
A sample of $\ce{NOBr}$ gas was placed in a rigid flask where it decomposes at $\pu{25 ^\circ C}$ according to the following reaction:
$$\ce{2NOBr(g) <=> 2NO(g) + Br2(g)}$$
At equilibrium, the total pressure and density of the gas mixture are $\pu{0.0515 atm}$ and $\pu{0.1861 g/l}$, respectively. I'm trying to find the partial fractions of each component using $pv=nRT$, but I'm a little confused. I cannot find an expression that allows me to relate the density that was given.

You can get the average molar mass of ideal gas from the pressure and density. From that, you can get partial fractions.

Do you mean partial pressure ??

yes, I wrote it wrong. Can you help me?

@WagnerCoelho You can answer this question did you saw the crucial point the question says its a  rigid flask meaning the volume of the system dose not change also the mass of the total system stays constant in chemical reactions thus the density of the system is constant (initial and final stats) use that hint and try to solve this problem.

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


text:
Conversion of allyl alcohol to 2-chloro propanediol with HOCl
going through the preparation of tri hydric alcohol(glycerol), i came across this reaction i.e. preparation of glycerol from propene. In the first step of this reaction, we did allyl substitution with chlorine followed by nucleophilic substitution with aq. KOH.
In the next step, we used HOCl to form 2-chloro propanediol. Why isn't it forming 3-chloro propanediol?
in other HOCl reactions on alkene, we can simply add HO(negative) and Cl(positive) on alkene in accordance with saytzeff.


text:
Why is it that when dG°<0, dH°<0?
Let me preface this question by saying that I probably made a mistake somewhere. However, I can't figure out what it is.
I was messing around with the Gibbs free energy equation to determine how the value of $K$ (equilibrium constant) varies with temperature.
Given
\begin{align}
K &= \exp\left({-\Delta G^° \over RT} \right). \label{5-7}\tag 1\\
\end{align}
From this it can be observed that when $\mathrm{d}G^\circ < 0$, $K > 1$. Conversely if $\mathrm{d}G^\circ > 0$, $K < 1$. This is also logical from a chemistry perspective.
Now if I differentiate $K$ over $T$,
\begin{align}
  \frac{\mathrm{d}K}{\mathrm{d}T} &=
  \frac{\Delta G^\circ\mathrm{e}^{-\frac{\Delta G^\circ}{RT}}}{RT^2}
  \label{two}\tag 2\\
\end{align}
It can then be observed that the sign of the derivative is dependent only on the sign of $\Delta G^\circ$. That is, when $\Delta G^\circ < 0$, $\frac{\mathrm{d}K}{\mathrm{d}T} < 0$. This also makes sense from a chemistry perspective, because that means that as $T$ increases, $K$ decreases and therefore $[\text{reactants}]$ also decreases.
Now that last part reminded me of Le Chatelier's principle, so now by considering
$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$
and substituting this into equation \eqref{5-7},
\begin{align}
  K &= \exp\left(\frac{\Delta S^\circ}{R}
    -\frac{\Delta H^\circ}{RT}\right)
    \tag 3\\
\end{align}
and once again differentiating with respect to $T$,
\begin{align}
  \frac{\mathrm{d}K}{\mathrm{d}T} &=
    \frac{\Delta H^\circ\exp\left(
      \frac{\Delta S^°}{R}-\frac{\Delta H^°}{RT}
    \right)}{RT^2}
  \label{four}\tag 4\\
\end{align}
And now it seems that the derivative is only dependent on the sign of $\Delta H^\circ$. Therefore, when $\Delta H^\circ < 0$, $\frac{\mathrm{d}K}{\mathrm{d}T} < 0$.
Here's the problem: By considering all the equations, when $\frac{\mathrm{d}K}{\mathrm{d}T} < 0$, both $\Delta G^\circ < 0$ and $\Delta H^\circ < 0$. From a chemistry perspective, this means that if equilibrium $[\mathrm{reactants}]$ increases when temperature increases, I can conclude that $\Delta G^\circ < 0$ and also go further to say $\Delta H^\circ < 0$.
The last sentence above doesn't sit right with me for several reasons.

Given that $\Delta G^° = \Delta H^\circ - T\Delta S^\circ$, simply having a negative $\Delta G^\circ$ value should not necessitate a negative $\Delta H^\circ$ value, because a positive $\Delta G^\circ$ can be offset by a large increase in entropy.

Consulting a table of standard thermodynamic values (https://owl.oit.umass.edu/departments/Chemistry/appendix/thermodynamic.html) shows that this isn't true. For example, the process
$$\ce{Cl2 (g) <=> Cl2 (aq)}$$
Has the calculated values $\Delta H^\circ = \pu{-23.0 kJ//mol}$ and $\Delta S^\circ = \pu{-102.1 J//K*mol}$. The given $\Delta G^\circ$ value is $\Delta G^\circ = \pu{+7.0 kJ//mol}$. (Calculating it using the equation $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ gives $\Delta G^\circ = \pu{+7.44 kJ/mol}$, which is similar to the value in the table.) From this example one can observe a process where $\Delta G^\circ > 0$, $\Delta H^\circ < 0$. From equation \eqref{two}, this means that as temperature $T$ increases, $K$ increases and $\ce{[Cl2(aq)]}$ increases.
However, this would contradict equation \eqref{four}. Not only that, it seems to violate Le Chatelier's principle, because given that the forward reaction is exothermic, an increase in temperature should favor the backwards reaction and equilibrium $\ce{[Cl2 (g)]}$ should increase instead.

Overall, I am rather confused. I have a feeling there could be a simple arithmetic error somewhere, which would make this long question rather embarrassing. Either that or there are some implicit thermodynamic conditions that are not being considered. In any case I would appreciate any mathematical corrections but I would also like the answer/comments to use chemistry principles to address any chemistry-based misconceptions.

In (2) you assume that ΔG is constant. In (3-4) you assume otherwise.

In determining (2), you are treating the Gibbs energy as temperature-independent (i.e. a constant), which it is not. If you write $\Delta G(t)$ instead and use the rules of differentiation, it should resolve the issue.

Thanks for the clarifications guys. However @IvanNeretin I still have a point of confusion: Given that $\Delta G^°$ is defined to be the $\Delta G$ at 298K, according to my understanding, it is invariant with $T$, making equation (2) the correct one here. This means that my treatment of the process $\ce{Cl2 (g) <=> Cl2 (aq)}$, where an increase in $T$ leads to an increase in $K$ and $\ce{[Cl2(aq)]}$, is correct. Would this, then, be a violation of Le Chatelier's principle? Because it seems that an increase in T leads to a higher $[reactant]$ even though it is an exothermic process.

"Given that ΔG° is defined to be the ΔG at 298K" No, that isn't right. ΔG° is defined at every temperature, and it has a different value at every temperature.

You might also consider reading FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange?. Seems like you know some of it already, but note that units like kJ/mol should be typed upright, not in italics. The section on "quantities with units" has more info.

Thanks for the suggestions @orthocresol. The edits have been made. Regarding your first comment: I had thought that the standard Gibbs free energy is strictly defined at approximately 100 kPa, 298K, which apparently isn't the case

"Standard" in the context of Gibbs free energy refers to a hypothetical state where activities of all substances involved are equal to unity. That's not the same thing as "standard temperature and pressure" - although it is quite a common & easy mistake to mix the two up. You could see e.g. https://chemistry.stackexchange.com/a/37272/16683 (note the other answer is incorrect...) or if you feel up to it, https://chemistry.stackexchange.com/q/41862/16683.

It was pointed out before, but I went ahead and fixed the MathJax. Here is a boiler plate comment with links: On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax).  If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

Hopefully your problem was clarified with the comments you received. If you still feel like there is an outstanding question which you'd like addressed please edit your post (it might be reopened), or post a new question.


text:
Name and Mechanism of this Reaction
Problem

Find the product X in this reaction -

Answer

Question
After seeing the solution, I came up with the following mechanism for the reaction -

I am not sure if my mechanism is correct, nor do I understand why such a rearrangement takes place. There are no reagents used, the compound is only heated, and I couldn't understand why the old bonds had to break to form new ones. Please clarify -
If my proposed mechanism is correct or not,
If it is, why the reaction occurs this way, and
If the above reaction has a name or not.

What is the source of this question?

This looks like a sigmatropic rearrangement though ( I think)

It is from a FIITJEE AITS paper for JEE Advanced (an engineering entrance exam in India).

https://en.wikipedia.org/wiki/Claisen_rearrangement

@orthocresol Thank you! I was aware of Claisen rearrangement, but I'd seen it only on phenyl allyl ethers till now.


text:
Lead acid battery reduction and oxidation
During charging or discharging a lead acid battery both the positive and negative electrodes will undergo reduction and oxidation the same time.
For instance during discharging process, the cathode will react with the sulfuric acid and will give the electrolyte electrons i.e. oxidation. And simultaneously the cathode will gain electrons from the circuit i.e. reduction.
So, the cathode and anode undergo both process reduction and oxidation during charging and discharging.
Am I right?

The lead acid battery has two electrodes, one made of metallic lead, and the other made of lead dioxide $\ce{PbO2}$. Remember that, whatever the operation (charge or discharge), the anode is always the electrode where oxidation occurs.
Let's consider first the discharge process.
In the discharge operation, the lead electrode is the negative electrode as it produces electrons. And it is working as an anode, as $\ce{Pb}$ is oxidized from Zero to $+2$ according to $$\ce{Pb + SO4^{2-} -> PbSO4 + 2 e-\tag{1}} $$The $\ce{PbO2}$ electrode is the positive electrode; it is working as a cathode, and the $\ce{Pb}$ atom is reduced from $+4$ to $+2$ according to : $$\ce{PbO2 + 4 H+ + SO4^{2-} + 2 e- -> PbSO4 + 2 H2O \tag{2}}$$ At the end of the discharge, both electrodes are similarly made of $\ce{PbSO4}$.
It is time to consider the charging process. In the charging process, the cell is working as in electrolysis and the preceding equations are reversed.
In the original lead electrode, equation ($1$) is reversed, and $\ce{PbSO4}$ is reduced to $\ce{Pb}$. The electrode is not an anode any more. It is a cathode. But the sign of the electrode is maintained negative, as the lead ion $\ce{Pb^{2+}}$ needs electrons to be regenerated into $\ce{Pb}$. So the minus sign (-) can be printed or engraved on the electrode. It stays the same in the charge and in the discharge processes.
In the second electrode, initially coming from $\ce{PbO2}$, equation (2) is reversed. $\ce{PbSO4}$ is oxidized to $\ce{PbO2}$ again. So this electrode, being oxidized, becomes an anode. But it remains the positive electrode. So the plus sign (+) may be printed on the $\ce{Pb/PbO2}$ plate. This (+) sign stays the same in the charge and in the discharge processes.

Remember that, whatever the operation (charge or discharge), the anode is always the electrode where oxidation occurs. This is GOOD

@OMAR While you are generally right, dedicated rechargable cells use convention of fixed anode/cathode labeling according to the galvanic cell  mode. Therefore, if we talk about Li-Ion cell anode, it is clear we talk about graphite-Li intercalate electrode and not about mixed oxide electrode.


text:
What does “Average Width” mean in crystallography?
Mathematician G. Polya noted (the solution for Problem 8, Axes of a Cube) in his book How to Solve it:

... there are three kinds of axes:
(1) 4 axes, each through two opposite vertices ...
(2) 6 axes, each through the mid-points of two opposite edges ...
(3) 3 axes, each through the center of two opposite faces ...
... The desired average is
$$\frac{4\sqrt{3}+6\sqrt{2}+3}{13}=1.416$$
(This problem may be useful in preparing the reader for the study of crystallography. For the reader sufficiently advanced in the integral calculus it may be observed that the average computed is a fairly good approximation to the “average width” of the cube, which is, in fact, 3/2=1.5)

However, I calculated the "average width" of 1.22 to 1.24 in math meaning, which is far different from 1.5. See my post https://math.stackexchange.com/questions/4207524
Is there really such a concept in crystallography?


text:
Can Amphotericin B cross a lipid bilayer membrane in its zwitterionic form when it is loaded in the aqueous core of a liposome?
Can Amphotericin B cross a lipid bilayer membrane in its zwitterionic form when it is loaded in the aqueous core of a liposome? Or must it be uncharged in order to cross the lipid bilyer membrane of liposome?
It has a $\textrm{p}K_\textrm{a}$ of 5.7 for the carboxylic acid group and  $\textrm{p}K_\textrm{a}$ of 10  for the $\ce{NH2}$ end and so its pI (pH at the isoelectric point, when it is present as the zwitterionic form) is 7.85.

Have you tried searching for what makes something able to cross a lipid layer on the internet?

@TRC yes. it must be uncharged but I don't have an idea that neutral charge zwitterioninc can act like an uncharged drug or not for crossing a lipid bilayer membrane.


text:
Why do many p-block dioxide gases dissolve in water to form acids?
I find that many gases of the form $\ce{MO_2}$ like $\ce{SO2}$ , $\ce{CO2}$ and $\ce{NO2}$ dissolve in water to form acids. This observation has many different practical implications, for example, it is due to sulphur dioxide and nitrogen dioxide mixing with water vapour in air that acid rain is formed.
My question is why does mixing of gases into water make them into aqueous acids?

May be due to high electronegative ness of $\ce{S,C,N}$ (compared to metal oxides they make bases) those atoms are combined to $\ce{O}$ even more electronegative atom, Having a $\ce{H}$ to $\ce{O}$ will reduce this effect even better acid give away the proton and makes the acid anion more stable $\ce{O}$ having a minus charge.

That is a good way to think about @Avon97

It isn't true in general. Some, well-known, gases do form acids but many others do not. So the question really should read "why do some gases dissolve in water to form acids".

Usually non-metal oxides are reacting with water to produce acids. Some do not react at all, but none produce an hydroxide or a base, in contact with water. To my knowledge, this can be considered as a definition of non-metals.

As noted by matt_black and Maurice, one example I could think of is chlorine dioxide($\ce{ClO2}$). It doesn't hydrolyze in water. Hence, it is handle as a dissolved gas solution in water.


text:
What is the Gibbs free energy change of physisorption?
In most sites I find online, I see that the Gibbs free energy of adsorption is negative. But, it is also said that physisorption is a reversible adsorption, suggesting that $\delta G=0$ which contradicts the original premise.
Hence, what exactly is the correct free energy change associated with physisorption?
From comments I learn that the reversibility is not used in the conventional thermodynamics sense but rather in the chemical reaction sense (refer). But, I don't see how it applies here since physisorption doesn't involve any bond formation.

NCERT chapter claiming the $\delta G = 0$ and reversibility condition

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text:
Why do mineral particles become wet by oils while the gangue particles by water?
In my textbook, NCERT Chemistry for class 12, it is mentioned that one of the principle of froth floatation method is the following:

The mineral particles become wet by oils
while the gangue particles by water. Source

Definition of wetting:
Wetting is the ability of a liquid to maintain contact with a solid surface, resulting from intermolecular interactions when the two are brought together. Source

My question is, why is it that minerals are wet exclusively by oils and gangue particle by water only?
An idea that I came up with is that gangue particles may be generally hydrophobic and hence repelled into the layer with collector, but is there any reason to why the particles must be hydrophobic? In another way, given an ore, it consists of gaunge and the actual mineral of interest, now why must it be the gaunge part is always hydrophobic?

For further reading, Froth flotation ( flotation process)

Because of the additives put into the solutions.

I understand that part: The additives are the collectors and froth stabiliser according to my book, now I understand that collector is some oily substance and for the water part, that is the original solution.

The gangue is mainly made of clay, and clay is a mixture of aluminosilicates which are hydrophilic, because their structure is surrounded by $\ce{-O-H}$
bonds. $\ce{O-H}$ bonds can make hydrogen bond with water, so they are easily wet by waters. If minerals are crystallized and not surrounded by these same bonds, they are hydrophobic. It is the case for pyrite $\ce{FeS2}$ or blende $\ce{ZnS}$, etc. They are more easily wet by oils than by water..


text:
Why is the base peak of my MS spectra of 2,3,5-tribromothiophene at m/z=382?
I recently requested a ESI-MS analysis of 2,3,5-tribromothiophene and got this spectrum:

These are the analysis conditions:

Detector: ESI(+)
Source Tem: 150
Capillary voltage: 1.5 kV
Cone Voltage: 20 V
Mass range: 200 - 1000
Flow: 0.2 mL/min
mobile phase: 0.1% formic acid:acetonitrile (50:50)

The sample was dissolved in methanol. Below is the spectrum of methanol:

The molecular weight of 2,3,5-tribromothiophene is 320.83 g/mol. But why does the base peak occur at m/z=382 and what are higher m/z peaks like m/z=423 and 740? I can't interpret the result at all.

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text:
What can happen to overheated aluminum cookware?
I accidentally overheated an aluminum espresso maker, to around 250-300° C in my estimation, but it might be even higher - I didn't measure, though. After research, I found that the material used is an aluminum alloy named Aluminum 1110. Now I wonder if, in general, overheated aluminum cookware is still safe to use. To find out, I would first need to know:
What can happen to an aluminum alloy when overheating? Will an aluminum alloy release (significantly) more aluminum (or other elements/molecules for that matter) after overheating? And out of sheer curiosity, why would that happen?
From the help page, I couldn't quite tell if this is the right place to ask, so excuse me if it isn't.

It reads like you put the Bialetti cooker on the stove and forgot to refill the water. By personal experience (student dormitory, and hot aluminum on the stove just looks like cold one) my concern would be more about their little replaceable safety valve in their foot; of course a question of overall age/wear of the valve, temperature and time of being hot and empty on the stove.  Suggest: https://cooking.stackexchange.com/.

The question is in a grey zone. Sure, there might be some chemistry going on, but this might also be a matter of engineering and food safety. I don't think you can expect an exhaustive answer here.

If it did not melt or distort, then not much happens. 1100 series is more or less pure aluminum, different designations will permit different levels of tramp elements like silicon, iron, copper, etc. It will anneal / soften, if any cold work / strain hardening was present. Without looking it up, very high temperatures like 800 °F (427 °C) could possible cause grain boundary oxidation which could seriously weaken it. The only health risk would be be touching it while hot.


text:
Is there a type of fabric that absorbs smoke and odors, but also washes out?
I don't know where else to post this, and I also saw a question asking why fabric absorb odors so well so I'm hoping it fits here. Any explanation as to why one fabric over another would absorb odor and smoke more would also be appreciated. This question seems to be the opposite of a common one I see, but: which fabric is best for absorbing odor and smoke, but also allows for said particulates to wash out easily?

Each odour is associated with a different molecules. Some fabrics will absorb some molecules well, not others and vice-versa

A comparison of the smoke odor retention and release chartericstics of polyester, cotton (both regular weave and terry cloth*), and wool can be found at: Pozuelos, G.L.; Jacob, P., III; Schick, S.F.; Omaiye, E.E.; Talbot, P. Adhesion and Removal of Thirdhand Smoke from Indoor Fabrics: A Method for Rapid Assessment and Identification of Chemical Repositories. Int. J. Environ. Res. Public Health 2021, 18, 3592.
*The authors give the four fabrics as polyester, cotton, terry cloth, and wool, but don't say what the terry cloth is made from.  Since it's typically cotton, I've made that assumption here.
According to the article, of the three materials, cotton (both regular weave and terry cloth) had the highest uptake of the odor-causing compounds, but also released them most readily (with an aqueous rinse).
The difference in releaseability could be explained as follows:  cotton is polar while polyester is non-polar (that's why cotton absorbs water and polyester doesn't). The authors say odor-causing compounds in smoke are principally non-polar; this would cause them to bind much more tightly to polyester, which would make them harder to release.
As to the difference in uptake, cotton fibers have a rough surface, while polyester fibers are much smoother.  In addition, cotton fibers may be physically porous (relative to the size of the odor molecules, i.e., be capable of physically admitting the odor molecules into the fiber), while polyester fibers may not be.  Thus, for the same mass of material, cotton may offer a much larger surface area for absorption (even though the strength of absorption is weaker).
Perhaps a member who is a textile expert can comment on this latter issue more authoritatively.
Note, however, that this doesn't mean cotton would be a good filter material.  Activated carbon would probably be much better.


text:
Is there a easy way to estimate how quickly something will dissolve depending on its surface area?
How exactly does the surface area of a solid for example a cube of sugar, affect the speed of its solvation. I guess everyone has observed that a batch of fine grounded sugar or salt dissolves much faster than a large chunk.
Is there a simple way to estimate this effect ? Are there kinematic models that include surface area as parameter ?

This provides a nice summary of the historical development of models for the rate of dissolution: https://www.nature.com/articles/s41598-018-25821-x. Since the dissolution of a solid into a liquid happens at the interface between the 2 phases, the surface area of solid in contact w/ the liquid has been considered a key component of dissolution models for over a century: "[In the year 1900] Brunner and Tolloczko determined that the rate of mass removal is proportional to the surface area of the dissolving particle, adding an important new parameter to this diffusion-based model of dissolution".

@theorist Thank you, that article is quite informative. So the proportionality of the dissolution rate (mass removal rate) is linear in regard to surface area, according to the Nernst-Brunner equation.


text:
What is the difference of spectra of EI-MS and ESI-MS/MS?
I want to ask not difference about the principle or applicability to GC and LC, but the peak behavior of spectra. EI is hard ionization method, so there occurs many fragments, sometimes losing molecular ion. In contrast, ESI is soft ionization method, so there are a few fragments. But this can be disadvantage because of lack of molecular information. To complement this, ESI-MS/MS proceeds more fragmentation. Then what is the difference of EI-MS and ESI-MS/MS? Both methods have fragmentation. If my sample is available in both GC and LC, then can I get similar peaks and information in GC-EI-MS and LC-ESI-MS/MS?

Assuming the present question relates to the characterization of a commercial sample 2,3,5-tribromothiophene here, with results used to monitor the progress of a chemical synthesis of simple small molecules (no forensics, no proteins, no polymers, etc.), then EI-MS might suffice (especially if combined with IR, 1H and 13C-NMR spectroscopy) and ESI-MS/MS be an overkill.

GC-MS is a hyphenation.  This means, at first you separate compounds by gas chromatography.  Then these fractions are ionized, shatter into fragments which eventually are separated along the ratio of mass/charge by mass spectroscopy.

In LC-MS (equally a hyphenation), liquid chromatography performs the separation of the compounds.  Because the mechanism of LC differs from the one of GC, you are not able to equate retention times by GC with those of the same compounds of LC.  Because of the large excess of solvent molecules in LC in comparison of your analyte molecules, the chemistry of ionization in the mass spectrometer equally differs from the one by GC.

Contrasting to spectroscopic methods like NMR, the shape of the spectra obtained by mass spectrometry, the relative intensity of the individual peaks experimentally recorded may differ quite a lot from one experiment to an other.  Among the influential parameters are the method of ionization like CI (chemical ionization) with a gas, or EI electron ionization where electrons hit your molecules.  The parameters of the ionization interfere, e.g., an EI with an accelerator potential of $\pu{20 eV}$ is softer (with less fragmentation) than the standard $\pu{70 eV}$; however with the possibility that some molecules are less well ionized at all and subsequently less well seen because of lesser fragmentation.
The conditions of fragmentation in a GC-MS equally differ from those for a LC-MS.  In the case of the later, ionization and fragmentations of your analyte molecules interfere much more with the ionization and fragmentations of the carrier molecules (the excess of solvent compared to your analyte running the LC).

As nicely described @M.Farooq's answer, the combination of MS-MS primarily targets larger molecules where by sheer structural complexity of the molecule's structure (e.g., proteins, polymers) one identifies fragments of first generation by characterization of their subsequently formed fragments.  This is an example of tandem mass spectroscopy.
For the synthesis of small organic molecules, the detailed elucidation of these fragmentation pathways in the mass spectrometer however often are not relevant.  Your time and energy typically are better invested in the characterization of your compounds along the compound characterization checklist by your group or / and the targeted journal (example J. Org. Chem., yet you may use it anyway as a guide).  In many cases, this includes recording
a) a low resolution MS with characteristic peaks assigned ($\ce{M+}$, adducts, important isotopic peaks (e.g., halogen pattern); a few important assigned fragmentations (e.g., $\ce{- H2O}$, decarboxylation, tropylium cation) or pattern (e.g., McLafferty)
and b) one high-resolution HRMS for the newly prepared compound
altogether with a self-consistent set of other (spectroscopic) data (UV, IR, NMR, melting point, combustion analysis, etc.).

Actually my desired product has molecular weight about 500 Da and supposed to be nonvolatile, so I excluded option using GC-MS. When I searched for analytical MS instruments available in my university, there were GC-MS (EI type), MALDI-TOF, and LC-MS/MS (ESI/APCI multimode). In the description of MALDI-TOF, it was written that it is used for samples with relatively high molecular mass. So I finally chose LC-MS/MS and also requested for starting materials to compare with product sample data. I wanted to see data with single MS, but there wasn't except MALDI-TOF.

@KrangLee, Earlier you said it was not an LC MS experiment but direct infusion. Get more information from the analyzer about what he did exactly. You have half information and you are writing even less info in your query. Earlier you stated your M must be around 320 and now you state it around 500. People cannot predict beforehand.

@M.Farooq In summary, 2,3,5-tribromothiophene (previous question, MW=320) is one of the starting materials and purchased, and requested MS as standard. And my product sample is supposed to have molecular weight around 500 (didn't mention in the previous question). Sorry for confusion. I'll be back with more information tomorrow.

There is no comparison of EI-MS as ESI-MS or ESI-MS/MS. They work on different principles, serve different classes of molecules, but the end goal is to identify a given molecule. Just like an airplane vs. helicopter, both serve to carry a person from one place to another, but their lifting capacities are different. EI is used for small molecules (< 400 amu) which can be conveniently vaporized into gas phase and as you noticed it is a hard ionization technique (causes a lot of fragmentation).
ESI was not meant for heavy fragmentation, it was meant to make elephants fly, as the inventor said. You do not want a lot of fragments of a big molecule. Imagine the messy spectrum with hundreds of peaks. ESI was designed for large and huge molecules, where GC fails badly. Nowadays electrospray is routinely used for small molecules too. Since the purpose of both is to identify molecules, ESI-MS does cause fragmentation despite the name soft ionization, but some time the spectra is still too complex or inconclusive. Then you try the tandem approach. You take a certain ion(s) from the first MS, take it to another one, cause fragmentation by various means, and use that mass spectrum for final confirmation of the unknown.
Connecting with your previous query, the mysterious m/z 382 could be chosen and made to break into pieces in tandem. This would help to identify the adduct or the impurity in your analyte. BTW, as your sample contains bromine, you must see bromine isotopic peak near m/z 382 in the expected ratio.

When analyzing reaction mixture using LC-MS/MS, if mobile phase condition doesn't work well, there can be bad separation so that there are two or more compounds in one peak in chromatogram. Then in tendam MS, do we pick the most desired product-looking or informative peak in the first MS and continue fragmentation of this peak in the second MS? So in the final data are we looking the MS data for only one ion?

Ah, you answered certain ion'(s)'. Then Do I have to ask what or how many peaks did the analyzer select, to interpret each final peaks?

Your first comment is a separate question. You have to get the exact information from the analyzer. Before that it is useless to think about it.

Ask your analyzer what experiment did he do? First was it LC-MS/MS? Then what ions were chosen in the second MS?


text:
Would there be 6p½ -> π* backbonding in oganesson tetracyanide?
It is known that oganesson, although a group 18 element, is more likely to behave like a group (1)4 element; i.e. with 4 valence electrons outside a flerovium core. This contributes to the fact that the 7s and 6p½ orbitals are likely "inert pairs" and should not affect the geometry of oganesson tetrafluoride, which is thus predicted to be tetrahedral. Logically, the same would hold for oganesson tetracyanide.
However, it is well-known that "inert pairs" can sometimes engage in (generalised) backbonding; the 3 iron lone pairs in the ferrocyanide molecules are "inert pairs" (in the sense that they do not affect the complex's geometry) but nevertheless still are formally valence electrons and are shared, in a Lewis base-acid sense, to the cyanide π* orbitals.
Even though oganesson is not a transition metal, the situation in oganesson tetracyanide is similar to that of the ferrocyanide anion in the sense that,

the 2 oganesson lone pairs, namely the 7s and 6p½ electrons, are "inert pairs" in the sense that they do not affect the geometry of oganesson tetracyanide and

they are formally "valence electrons".

It is known that the 7s orbital is likely highly non-accessible for bonding from moscovium onwards; that would leave only the 6p½ orbital in oganesson potentially available for generalised backbonding(the 7s orbital has the wrong symmetry anyway), via donation into the cyanide π* orbitals, in oganesson tetracyanide.
My question now follows- do theoretical calculations on oganesson, and/or its tetrafluoride, support what I have written here?

Oganesson only known isotope has a half-life of $0.89$ millisecond. It will never find the time to be included in a compound or an ion. So, up to now, its chemistry is non-existant. As a consequence, does it make sense to discuss about the theoretical properties of an unknown cyanide, which would last less than a millisecond, if it could ever be synthesized ?

That's why I wrote every sentence in my post with a doubt and put the "theoretical" part in my final question.

"It is known that oganesson, although a group 18 element, is more likely to behave like a group (1)4 element". Wow, we've learned a lot!

Discussing about the chemical properties of Oganession is like talking about the angels sex, or about the inhabitants of extragalactic planets. Why ?

Then why would there exist legitimate pieces of research on oganesson in the literature?


text:
What is the reason for the boiling point of benzene-1,4-diol being higher than that of benzene-1,3-diol?
What is the reason for the boiling point of benzene-1,4-diol being higher than that of benzene-1,3-diol?

Both have hydrogen bonds. I think it is due to quinol's symmetry but can't point out exactly why this is the case.

Can you take a guess? Have you not been provided with other examples? https://chemistry.stackexchange.com/help/how-to-ask

see https://chemistry.stackexchange.com/questions/112078/what-effect-on-boiling-and-melting-points-does-intramolecular-hydrogen-bonding-h

@Lllt: Not related question. None of these two can make intramolecular hydrogen bonding.

@MathewMahindaratne I think benzene 1,3 diol can do so

symmetry makes it have a lot less dipole moment than resorcinol. I think this has to do with larger surface area of quinol leading to higher Vander Waal forces.

@napstablook By symmetry, I meant its packing and all in liquid state

@RobinSingh packing in liquid state?

@napstablook I'm sorry I don't have a way with words. I mean the ease with which they can form h bonds

Benzene-1,4-diol  has a boiling point of 287°C and Benzene-1,3-diol has a boiling point of 277°C and Benzene-1,2-diol has a boiling point of 245.5°C. This could be attributed to the ease of formation of 2 intermolecular hydrogen bonds with the 2 hydroxy groups of these molecules increasing when the hydroxy groups are placed apart, to minimise steric hindrance and repulsion. Benzene-1,4-diol could form intermolecular hydrogen bonds with more stability than Benzene-1,3-diol or Benzene-1,2-diol as these two will get destabilised by steric repulsion when large groups have to reach nearer to form hydrogen bonds.
Thus, the extent of intermolecular hydrogen bonding in :
Benzene-1,4-diol > Benzene-1,3-diol > Benzene-1,2-diol
To overcome this attraction, more energy is needed. Hence the same is the order of their Boiling Points.

Is that good enough of a reason for a 10°C rise?


text:
How do the qualitative driving forces of a reaction make the reaction proceed?
In high school chemistry, I learned that there are 4 “qualitative driving forces” that cause aqueous reactions to proceed:

Forming a solid (precipitate)
Forming a liquid
Forming a gas
Transferring electrons (redox)

I also learned about the 2 “quantitative driving forces” for general reactions to proceed:

Entropy
Enthalpy

While the two quantitative driving forces can be used to calculate whether a reaction is thermodynamically favorable through free energy calculations, I don’t see why the 4 qualitative driving forces are taught as reasons for predicting whether a reaction will occur.
In sum, my questions are: how are these four forces able to drive a reaction and are they related to entropy/enthalpy?

It's because of the entropic gain during the formation of a product which forms its own seperate phase from the reaction mixture, the first three are good signals. The fourth is a good signal because of its enthalpic implications.

I personally have problems with this idea of driving force. They seems rather descriptors of what is happening / happened rather then driving forces. Take 1. On its bases dissolving stuff would be impossible. This stuff doesn't make sense to me, even when the so called quantitative driving forces are put in the discussion as per Nisarg Bhavsar. I even think this might be your interpretation and not a teached concept. Moreover plenty of reactions as conducted don't see any phase change because most of them happen in solution. Curios about other users commenti...

@Alchimista The four "driving forces" are surely not fundamental but just are rules of thumb to blindly guess if a reaction has any potential to take place. You surely can't rely on them for either confirming or a denying a reaction, you can use them only to justify said reactions

@NisargBhavsar yes they are merely facts. They can happen or not. They are driving "forces" only when they are. Again I won't teach thermodynamics this way, not even at elementary level. I can perhaps use that slang with a senior colleague but not with a young student. So basically OP is right and s/he should proceed on.


text:
What is the meaning of thermodynamic tendency?
While giving a test, I encountered a question which was as follows

Lead reduces $\ce{NO_3^-}$ into $\ce{NO}$ and $\ce{NO_2}$ depending on the concentration of $\ce{HNO_3}$ in solution. Assuming that $\ce{[Pb^{2+}]} = \pu{0.1 M}$ and $p_\ce{NO} = p_\ce{NO2} = \pu{0.001
 bar}$ and use the following data:
\begin{align}
  E^\circ_\ce{Pb^2+/Pb} &= \pu{-0.13 V}\\
  E^\circ_\ce{NO3^-/NO} &= \pu{0.96 V}\\
  E^\circ_\ce{NO3^-/NO2} &= \pu{0.79 V}\\
\end{align}
and at $\pu{298K}$ $\frac{(2.303)RT}{F} = \pu{0.06 V}$, find the concentration of $\ce{HNO3}$ at which thermodynamic tendency for reduction of $\ce{NO3^-}$ into $\ce{NO}$ and $\ce{NO2}$ by lead is the same.

In this question what is the meaning of thermodynamic tendency? Is it represented by $E_\mathrm{cell}$ or $\Delta G$? Quoting a reliable source would help.
The answer is $\pu{10^{0.625} M}$ which comes by equating the $E_\mathrm{cell}$ of the reactions.
My intuition told me to equate $\Delta G$ but it seems I am wrong.

How is equating Ecell different from equating ΔG?

The number of electron that are involved in the reaction are different hence equating $\ce{E_{cell}}$ and equating $\ce{\Delta{G}}$ are two different things.

Guys have tendency to be attracted to ( in their eyes ) the most beautiful girl regardless of the success. Similarly, TD systems have tendency to reach the state with the minimal Gibbs energy, regardless of if they reach it ( e.g. due kinetic reasons ). Diamonds have TD tendency to become graphite, staying being diamonds for billions years.

I am unable to understand that what are you wanting to imply?

The Hess law is applicable on $\Delta G_\mathrm{halfr}$ ( implicitly referring to SHE as the 2nd half reaction), or on its equivalent $-nF.E_\mathrm{redox}$.   // As $\Delta U = q \cdot \Delta E$ (energy, charge, potential difference )

For solving this problem, the $\Delta G$ values cannot be compared, or equalized, because they do not correspond to the same equation. Both reactions start from the same products, $\ce{Pb + NO3^-}$. But the first reaction leads to $\ce{NO}$ and the other one to $\ce{NO2}$, and these two nitrogen oxides do not have the same $G^\circ_\mathrm{form}$ values. So the Gibbs energy values are no use for solving this problem. It is much better to compare potentials through the Nernst equation.

I am not able to understand that if the $\Delta{G^o_form}$ values are different then why can't we compare the $\Delta{G}$ of both the reactions? Also then, under what circumstances the $\Delta{G}$ values of two reactions can be compared?

Are Free Energy changes not directly related to potentials?


text:
Priority among Halogens in Organic Nomenclature
I have been trying to learn IUPAC nomenclature during the past few days.
I have been using the MarvinSketch software to play around with random structures and its generated names and comparing them to my own results.
I stumbled across the following compound:

The nomenclature for the compound is shown as (1Z)‐1‐bromo‐2‐iodoethene
Is there any reason why (1Z)‐1‐iodo‐2‐bromoethene might be wrong? Is there a priority series among the halogens?

@orthocresol (1Z)‐1‐bromo‐2‐iodoethene is different from (1Z)‐1‐iodo‐2‐bromoethene and not just a rearrangement of prefixes since 1-bromo prefix is used in the first name and 2-bromo prefix is used in the second one.

Related: Order of citation of alkyl and halide prefixes in IUPAC names

You have figured out the Z part of the nomenclature correctly. Now as to the question of why 1-bromo-2-iodo... and not 1-iodo-2-bromo..., we look at section P-14.4 (g) of the Nomenclature of Organic Chemistry. IUPAC Recommendations
and Preferred Names 2013, which states the following:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the
following decreasing order of seniority:
[...]
(g) locants for the substituent cited first as a prefix in the name;

According to this rule, the priority in numbering depends upon where each prefix is present within the name. You cannot have 2-bromo-1-iodo... or 2-iodo-1-bromo... as a possible name when the substituents are simple substituents.
Now, how do we decide which comes first? Is it the bromine or iodine group? For this we look into section P-14.5 which deals with alphanumeric order and states:

P-14.5 ALPHANUMERIC ORDER
[...]
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically;
multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Therefore, since bromo- comes first alphabetically, we place it first before iodine and by P-14.4 (g), the bromine gets priority.
Therefore, the PIN of the given compound becomes (Z)-1-bromo-2-iodoethene

For a PIN you need 1Z not just Z

Is it necessary. since the there can only be one location for the double bond to exist?

@SafdarFaisal Check P-14.3.3 of IUPAC Blue Book 2013, for a name to be a PIN, dropping of a locant like '1' is generally permissible if without ambiguity, it is not allowed in PIN.


text:
What molecule is this. Pretty sure its not ethanol?
This is the molecule in question
This in advance.

Humulene is a compound found in aromatic plants, which has insect-repellent properties in such plants. Currently it is also being researched as an anti-inflammatory agent.

The compound of the structure on the cup is humulene, which is naturally occuring sesquiterpene (three isoprene units) isolated from the cones of Humulus lupulus (hops), thus its name:

Although Oscar Lanzi gave the brief description good enough to identify the compound, my intention to give extended answer is to clarify the OP's comment of "pretty sure its not ethanol." I believe OP made the comment because the structure is on the cup from a brewing company. The reason the structure to be on the cup is humulene and its reaction products in the brewing process of beer are known to  give many beers their “hoppy” aroma (Wikipedia and The essemtial oil of Hops). The company described it as "[...] juicy hop forward beers brewed with a New England yeast strain with lots of flaked oats, barley, and wheat, and generous late and dry hopping"! The company's symbol says its all: It is a cone of the Hop plant.

I'm no organic chemist but why is the green circle an isoprene, the others have two carbons at one end and one at the other?

@ porphyrin: Regardless of where is the double bond, isoprene unit is 2-methylbutene or 2-methylbutane. The place of double bond(s) depends on the biochemical process during the biosynthesis. For example, see how many monoterpenes $(\ce{C_{10}})$ in literature.

@ porphyrin: Two units in blue and red are traditional head-to-tail attachment but one in green is not, again, due to relavent biosysnthesis.


text:
Do positive results of a lead paint testing kit imply that a dipped door is now toxic?
We have had some 1930's doors dipped in a caustic solution to remove the paint.
Before dipping, we found that a layer of yellow paint contained lead.  It tested positive with: https://www.amazon.com/Testing-Results-Seconds-Suitable-Surfaces/dp/B07NBH7KJJ/
(The testing kit requires the swabs to be dipped in vinegar, where they change color from white to yellow.  Rubbing for 30 seconds against safe paint results in no change from the yellow color, while lead paint causes the swab to turn red.)
What appears to be the bare wood of the dipped doors now tests positive when we use the paint-testing kit.  (We were testing a small patch of unremoved paint, and initially tested the bare wood by accident.)
Not only does the bare wood test positive, but it turns a much brighter purple than the red color-change produced by the original lead paint.
Is the door now impregnated with lead as a result of dipping?
Or is this just a reaction between the caustic substance in which the door was dipped and the testing chemical?

You're using a sodium rhodizonate kit. The deep violet color likely comes from lead that was absorbed by the wood. There is at least one alternate source of the color you're seeing, which would be a reaction at high pH with the indicator, which would in turn form croconic acid which can form colored complexes. I'd let some more knowledgeable contributors chime in, as I'm not the best source for organic/inorganic chemistry. Also, you might find a definitive answer in the DIY StackExchange site.

@ToddMinehardt Thanks, I've cross-posted to https://diy.stackexchange.com/questions/231577/do-positive-results-of-a-lead-paint-testing-kit-imply-that-a-dipped-door-is-now

@fadedbee have a read: https://chemistry.meta.stackexchange.com/questions/594/if-from-comments-or-answers-it-appears-that-the-question-is-more-fitted-to-anoth


text:
Finding half life of a given decay path given percentage decayed in each path

$^{64}_{29}\ce{Cu}$(" half life" =$12.8$ "hours" ) decay by $\beta^{-}$- emission (38%), $\beta^+$- emission(19%), and electron capture (43%). Write the decay products and calculate partial half lives for each of the decay processes. Source: JEE Advanced
Soltn:
Half life of beta-= $\frac{}{}=\frac{12.8}{.38}$, beta+ =$ \frac{12.8}{.19}$ and electron capture$ \frac{12.8}{.43}$

From googling I find, that the term partial half life is popularly known as branching fraction:

the branching fraction (or branching ratio) for a decay is the fraction of particles which decay by an individual decay mode with respect to the total number of particles which decay
Source

Question: I don't get why the half life of each individual path is the total half life divided by the fraction of nuclei decayed that way. I foudnd this related post but there it involves a the actual reaction being given.
I also found this wiki article on half life, but I am not sure how to use it to derive this.

Deriving the relation of the half life, partial half lifes and probability percentage of the particular decay path is not complicated algebraic exercise.

What equations should one begin with? @Poutnik I'll try for a self answer then

Equations for the reaction kinetic of the first order - all radioactive decays follow it.

The solution given does seem strange but maybe something common in study of radioactive decay. The Cu decays with a single half life and each product appears with the same half life, 12.8 hours even though the rate constants for each product can be different.  If the rate constants are (in order as above)  $k_1, k_2, k_3$ then $\beta^-$ appears with fraction (branching ratio)  $k_1/(k_1+k_2+k_3)=0.38$ and so on and the sum is the rate constant for decay of the Cu which is easily related to the half life. I suspect  partial half lives means get k's for each product and convert to half lives.

Note that "partial half lifes" is not very useful concept, it is better to stay with the overall half life, decay constants and decay mode probability.

How is the percent of product equal to fraction of rate constant over total ?? @poryphrin Secondly if the equation is true, how would I find rate constants anyways

It is easier to use rate constants than half lives. You have three equations and three unknowns, $k_1,k_2,k_3$ you know the sum of these given by $\ln(2)/t_{1/2}$ and each ratio, 0.38 etc.


text:
Is $ Q $ dependent on the ϵ which is chosen in the expression for Gibbs free energy of a chemical reaction?
In this lecture by MIT open courseware (*), Moungi Bawendi states that we can have some change in our chemical system containing reactants and products scaled by some $ \epsilon$ amount. Using this set up and the equation which gives chemical potential as a function of pressure (**) he derives the $\Delta G$ as a function of $ \epsilon$ as shown below:
$$ \mu_{i} (T , P_{tot} )  = u_{i}^{o} (T) + RT \ln \frac{P}{P_o}$$
and, with the expression for $ \Delta G $ in terms of chemical potentials:
$$ \Delta G  = \sum_{i=prods} \mu_i \nu_i - \sum_{i= reacts} \mu_i \nu_i$$
He gets,
$$ \Delta G(\epsilon) = \epsilon [ \Delta G^{o} + RT \ln \frac{ (P_d)^{v_d} (P_c)^{v_c} }{ (P_b)^{v_b} (P_a)^{v_a} }]$$
Now, this expression I have a few questions about:

Is the pressure ratio term on the right hand side dependent on  $\epsilon$? Like I think that as the reaction proceeds, would each pressure term fluctuate?

How does one correctly interpret the final equation ? Is it related to calculus of variations ? I ask this because it looks very similar ot the derivation of fundamental theorem of calculus of variations (found here)

Usually we assign a Gibbs free energy to a whole reaction, but here there is an aspect of this $\epsilon$ paremeter, so how does this idea of parameterizing chemical reactions relate to the regular values of Gibbs free energy we see in textbooks? The professor equates bracketed term to $\Delta G$ but the connection is not a 100% clear to me. Perhaps an explicit explanation would help.

*: Reference video
**: page-2 of these notes from mit ocw
Note: I had posted this question more than three months back on physics stack exchange without finding any explanatory answer. Hopefully the question finds more home here.

I'm not able to see the notes, and the video doesn't have anything involving $\epsilon$ as far as I can tell.  Can you please provide a precise definition of what $\epsilon$ represents or a better reference?

$\epsilon$ is an infinitesimally small change in amounts defined to see whether the Gibbs energy change is positive or negative if starting with a certain set of partial pressures. This means the partial pressures do not change significantly..

Pls check the edited link @ChetMiller

Check out this link. Page 623 on Chemical Potential and Thermodynamics
of Spontaneous Chemical Change, has a pretty clear derivation and explanation. http://www.pearsoncanada.ca/media/highered-showcase/multi-product-showcase/petrucci_ch13.pdf

The other two answers interpret $\epsilon$ as the extent of reaction $\xi$ and make factually accurate statements under this assumption. I disagree that $\epsilon$ should be interpreted as $\xi$; instead, $\epsilon$ seems to be used as a (small) perturbative parameter that describes how the system will change under a small perturbation from its current state. (In general, this does not seem to be a useful way of analyzing chemical reactions, since parametrizing the extent of reaction by $\xi$ is more versatile and provides more information besides.)

Is the pressure ratio term on the right hand side dependent on ϵ? Like I think that as the reaction proceeds, would each pressure term fluctuate?

It will be approximately independent of $\epsilon$ because $\epsilon$ is taken to be small. However, it would not be independent of $\xi$ for the reasons you describe in your second question.

How does one correctly interpret the final equation ? Is it related to calculus of variations ? I ask this because it looks very similar ot the derivation of fundamental theorem of calculus of variations (found here)

The change in the Gibbs free energy due to this small perturbation is given, to first order, by $\Delta G(\epsilon)$. It is related to the calculus of variations in the sense that they both deal with infinitesimal analysis, but lots of things deal with infinitesimal analysis and there is no specific connection to be made, so this analogy is neither very precise nor very useful.

Usually we assign a Gibbs free energy to a whole reaction, but here there is an aspect of this ϵ paremeter, so how does this idea of parameterizing chemical reactions relate to the regular values of Gibbs free energy we see in textbooks? The professor equates bracketed term to ΔG but the connection is not a 100% clear to me. Perhaps an explicit explanation would help.

This question is really asking about $\xi$ rather than $\epsilon$. In the notes you link, the equation $\Delta G = \Delta G^\circ + RT\ln Q$ is derived by setting $\epsilon = 1$, which does not seem appropriate to me. Certainly the final result is correct, but there are clearer and more mathematically responsible derivations that you should be able to find in any standard physical chemistry textbook.

I wrote one of the other answers, and failed to clarify this. Good answer and nice catch of the misconception the OP is struggling with.

Could you suggest a source which gives a better derivation?

@Buraian, maybe Atkins' or McQuarrie's Physical Chemistry?

I had tried Atkins but in the edition I had, I was more confused because they used the other epsilon like looking symbol in it and I thought it was the same. I'll check it out again when I get time

Is the pressure ratio term on the right hand side dependent on  $\epsilon$? Like I think that as the reaction proceeds, would each pressure term fluctuate?

You choose to make $\epsilon$ very small so you can neglect changes in partial pressures, concentration, or activity.

How does one correctly interpret the final equation ?

The Gibbs energy is an extensive quantity. The more reacts, the bigger the change in Gibbs energy (e.g. you can do more work with a D battery compare to a AAA battery, even though they have the same voltage).

Usually we assign a Gibbs free energy to a whole reaction, but here there is an aspect of this $\epsilon$ paremeter, so how does this idea of parameterizing chemical reactions relate to the regular values of Gibbs free energy we see in textbooks?

For the Gibbs energy, you can state the change at a given set of concentrations, but you can't just assign a Gibbs energy to the entire reaction (you would have to integrate over the extent of reaction from initial to final state). In contrast for enthalpy, it is trivial to do the analogous calculation because the enthalpy is not concentration-dependent in the first approximation. It is clear that the Gibbs energy changes with concentrations, otherwise reactions would not attain equilibrium from either side of the equilibrium (e.g. starting with all reactants or all products).
The next step in this derivation would be to define the intrinsic quantity $\Delta_r G$, the Gibbs energy of reaction, as
$$ \Delta_r G = \frac{dG}{d\epsilon}$$
In textbooks, it is more common to use $\xi$ (xi) as symbol of the extent of reaction, i.e. how the concentrations changed. $\Delta_r G$ is dependent on concentration, but in the limit of an infinitesimal change in $\xi$, you can neglect the changes in concentration and just use the initial concentrations (or activities or partial pressures or fugacities).

"You choose to make ϵ very small so you can neglect changes in partial pressures, concentration, or activity." I have a hard time believing this, because the professor showed how varying epsilon runs us through the whole gibbs free energy curve for the reaction in the video

You keep epsilon small and use different sets of partial pressures along the reaction path to explore the slope of Gibbs energy change at different stages in the reaction. @Buraian

I can see what the professor is doing here.
First of all, he implicitly asks us to accept the idea that the free energy of an ideal gas mixture that is not at chemical equilibrium can still be calculated for the mixture using the usual expression for a gas mixture at equilibrium:  $$G=n_a\mu_a+n_b\mu_b+n_c\mu_c+n_d\mu_d$$
I have no problem with his doing this if you don't.
He then looks at the effect of a small change in reaction conversion $d\epsilon$, such that $$dn_a=-\nu_ad\epsilon$$$$dn_b=-\nu_bd\epsilon$$$$dn_c=+\nu_cd\epsilon$$$$dn_d=+\nu_dd\epsilon$$This causes a change in G given by
$$dG=-\mu_a\nu_ad\epsilon-\mu_b\nu_bd\epsilon+\mu_c\nu_cd\epsilon+\mu_d\nu_dd\epsilon=d\epsilon[-\mu_a\nu_a-\mu_b\nu_b+\mu_c\nu_c+\mu_d\nu_d]$$
The rest is straightforward algebra.

How would you explain the point about pressure fluctuations? Would  it be the same as @Karsten Theis

Not exactly.  You are aware that, because G is an extensive property, the changes in the partial molar quantities such as chemical potentials must satisfy the Clausius Duhem equation, right?  That would allow us to write the change in G without including terms for the changes in chemical potential (which would ordinarily follow from the product rule for differentiation).

You can work it out algebraically, and you will find that those pressure variation changes (as a result of changes in mole fraction) cancel out.

Clausius Duhem, I checked it up online, seems to be a result relating to Chemical Engineering and the math back ground for understanding the derivation directly seems a bit out of my reach. The algebra is indeed simple as you said, but the problem for me is fully comprehending the premise which the derivation begins with. Oh well, this turned out more complicated than I had initially expected.

By the way quick question, why havne't you considered differential variations in chemical potential as you kick up the moles?

Like I said, those cancel out as a result of the Clausius Duhem equation.  Do the algebra, and you will see this for this particular problem...

I understand, I had missed the part since I got stuck right after the C-D eqtn part. I'll check it up in time.

All you need to do is show that the CD eqtn is satisfied by this mathematical form of the chemical potential.

I believe you're referring to the Gibbs-Duhem equation rather than the Clausius-Duhem inequality, which might explain some of Buraian's confusion.

@a-cyclohexane molecule Yes.  Senior moment.


text:
Can a Lindlar catalyst reduce aldehydes, nitriles, and carbonitriles?
The major product obtained in the following reaction is:

Question source: JEE Main 2019
The molecule shown as product is the official answer given by JEE from the given options.
I have written the possible intermediate molecule formed by reagents. On Google I have obtained information in this PDF document (page 3) that a Lindlar catalyst is poisoned, so that it cannot reduce an aldehyde. Also there is no information of a Lindlar catalyst being used as the reducing agent for nitriles and carbonitriles.
So, is the answer by JEE incorrect or am I missing something?

Note: $\ce{Pd/C}$ is not similar to $\ce{Pd/BaSO4}$ or $\ce{Pd/CaCO3}$. So the reagent in your question isn't Lindlar's catalyst in the first place. It behaves like normal $\ce{Pd/H2}$ as far as I've learnt.

From MasterOrganicChemistry Reagent Guide - Pd/C reduces alkene, alkyne, nitro, cyanide, C=N, azide - and probably most other reducible groups too.

From Organic Chemistry for JEE Advanced by KS Verma, Cengage publications - It reduces all of the above but they haven't mentioned alkyne and C=N. Most likely missed mentioning those two.

Re "carbonitriles": Just "nitriles"?

@PeterMortensen Essentially they are the same, but in some nomenclature rules, carbonitrile is used instead of nitrile: https://www.acdlabs.com/iupac/nomenclature/79/r79_566.htm

There is just a silly mistake. As TRC told, $\ce{Pd/C}$ is not a Lindlar catalyst and it behaves as a normal $\ce{Pd/H2}$. Refer to the link for more information.


text:
Dative Bond in Nitrogen Dioxide
I learnt that a coordinate covalent bond forms only when

Donor atom has a complete octet and also has at lest 1 lone pair.
The receiving atom must have an incomplete octet and possess at least 1 empty orbital.

So how does nitrogen in Nitrogen Dioxide forms a dative bond with one oxygen atom while the octet of nitrogen is incomplete?
Also is this the reason why it dimerizes?
i.e.

I mean by doing so it completes the octet.
And the equilibrium constant for the reaction is about 29 orders of magnitude at 77.15 K.

There are no dative bond in $\ce{NO2}$ and in $\ce{N2O4}$ In both molecules, the nitrogen atom is charged positively and one of the oxygen atom is charged negatively

I think the charge is carried  by the molecule as a whole. What you might be referring to is the formal charges. I tried checking other sources also, but they mentioned presence of dative bond as well

The molecules $\ce{NO2}$ and $\ce{N2O4}$ are not charged as a whole.

Yes I mean the positive and negative cancel. So the molecules are neutral.

@Maurice I think the OP is using the term coordinate bond to refer to the N-O single bond... just like nitric acid can be shown with a coordinate bond too.

@Barry Where did you see those two conditions?

Yes I mean the single bond between Nitrogen and Oxygen.

I was taught about those conditions from my teacher.

@TRC. OK. The concept of a single dative bond between $\ce{N}$ and $\ce{O}$ may be accepted.  It is not the way I teach it, but why not ? In my concept, when an electron pair is given from $\ce{N}$ to $\ce{O}$, it produces a negative charge on the oxygen atom, and a positive charge on the nitrogen atom.

@Maurice True, even with a dative bond (the way I have been taught) the formal charges will be +1 on N and -1 on O, same as your depiction.

@Barry Can you ask your teacher for a reference? Or better still, ask them directly about $\ce{NO2}$ and how it fits with the conditions.

I'm extremely sorry but as the current circumstances stand, I can't contact him.


text:
Why does CO2 have a greater affinity for hemoglobin than oxygen?
I know about CO but I am wondering if CO2 has similar mechanism? do you guys have any scientific reference

These species help form interactions between amino-acid residues at the interfaces of the four subunits in hemoglobin.  These interactions are called "salt bridges," because they are between positively-charged and negatively-charged amino-acid residues on different subunits of the same protein.  When "salt bridges" form, the subunits are held in a position that "tugs on" the histidine that is attached to the heme iron.

This is taken from here
http://www.chemistry.wustl.edu/~edudev/LabTutorials/Hemoglobin/MetalComplexinBlood.html

What is the purpose of this second paragraph of text?

@orthocresol♦ this is all what I found about it but I want to know if they compete on the same binding site? do they have a effect on if CO2 present or not when Oxygen is being transferred to the blood?

You can't just quote a paragraph of text without any context. Where is it taken from? What is it trying to describe? It's not even clear what the first two words, i.e. "These species", refers to.

@orthocresol♦ above a link

Before asking why something happens, it is usually worth asking whether it happens. Though carbon dioxide is bound by haemoglobin, oxygen displaces it (look up the Haldane Effect) so the premise of the question is wrong.

See this paper: https://www.jbc.org/article/S0021-9258(19)45105-3/pdf


text:
Query Regarding Activity of Pure Water
While calculating $K_i$ i.e. ionization constant of water, why do we put $[H_2O]=55.5M$ in the expression below:-
$K_i=   \dfrac{\ce{[H^+][OH^-]}}{\ce{[H_2O]}}$
$$\ce{H_2O <=> H^+ + OH-}$$
although we know that activity of pure liquids is 1.
So going by that should we not put $[H_2O]=1$ in the expression for $K_i$
Thanks

I understand why the concentration of pure water is 55.55 mol/L but my doubt is, in the expression for $K_i$ we use activities and not concentrations of substances and hence shouldn't we be using $[H_2O]=1$ as activity of pure liquids(and pure solids) is taken as 1

[H2O] is concentration, not activity.

As per the law of Mass action (using which we derive the expression for Equilibrium constant), it is more precise to use Activities instead of concentrations.
The same can be read from the link below
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Solutions_and_Mixtures/Nonideal_Solutions/Activities_and_their_Effects_on_Equilibria
So I still have the same doubt.

Then do not use concentration instead of activity. For compound $\ce{A}$, concentration is expressed as $c_\mathrm{A}$ or $\ce{[A]}$. Activity is then  $a_\mathrm{A}$. Remember that activity for water is defined differently than for ions.

Please overlook my incorrect use of notation, what I meant was $a_A$ where A is $H_2O$. Can you please elaborate on how I can calculate activity of $H_2O$ for use in such expressions.

As you have said yourself $a_{\ce{H2O}}=1$ for pure water or diluted enough solution, by definition. As in contrary to concentration, it depends on convention what state is taken to have $a=1$.


text:
Why don't ant bites kill us?
So, methanol is supposed to be toxic even in minute quantities because it is oxidised in our body to methanal and then methanoic acid. This can cause blindness and even death.
Interestingly, ants eject the very same methanoic acid inside us when they bite us, and I haven't heard anyone die of that.
Is it because of the minuteness of the quantity in which it is secrete, or is there some other explanation to this?

Typical supposed lethal dose of methanol for humans is 20-30 mL. Formic acid is not dramatically more toxic than acetic acid. See LD50 here https://en.wikipedia.org/wiki/Formic_acid

Famously, the dose makes the poison.

Because we never met "them".

Of course, being staked out atop an anthill might have adverse health effects...

Fire ants: https://abcnews.go.com/GMA/Health/story?id=2143091&page=1


text:
Nature of π-bonds in permanganate ion
I was doing a question which asked about the nature of $\pi$ bonds between the $\ce{Mn}$ and $\ce{O}$ atoms in $\ce{MnO4-}$. According to the answer all the three $\pi$ bonds were formed by $p\pi-d\pi$ overlap. However, I know that the hybridisation of $\ce{MnO4-}$ is $d^3s$. Which means three of the $\mathrm{3d}$ orbitals are involved in hybridisation and only two are  left for making $\pi$ bonds. Which means that all the three $\pi$ bonds cannot be formed by  $p\pi-d\pi$ overlap. According to me two of the three $\pi$ bonds are formed by $p\pi-d\pi$ overlap and the third is  formed by $p\pi-p\pi$ overlap using the unhybridised $\mathrm{4p}$ orbital (to which the electron from $\mathrm{4s}$ gets excited from the ground state).
Where is the flaw in my reasoning?

From what I have learnt on this site, hybridization is a very inadequate and often incorrect concept for describing such compounds.

Why is it wrong to use the concept of hybridization for transition metal complexes?

The primary flaw is assuming that there are three full pi bonds like you might draw in a Lewis structure. In reality, there are pi interactions, but not three bonds worth. The Mn has positive charge and the O's negative, so there is a strong ionic component as well.

The hybridisation of permanganate is d³s for a duodoctet rule and sp³ for a decaoctet rule. To quote Wikipedia, the academic consensus on this matter is that neither rule is mandatory for  transition metals but the rule has to lie somewhere between, with a seperate rule for each transition metal complex.

For the people who spam comment how "hybridisation is not appropriate for transition metal complexes"- this is true FOR MO THEORY ONLY and not many people, predictably including the OP,  are that obsessed with, or even familar with, MO theory.


text:
Why should hydrogen peroxide be added to sulfuric acid and not vice versa?
When preparing aqueous solutions of sulfuric acid through dilution of the pure acid, it is well known that we should add acid to water and never the reverse.
To prepare a solution of hydrogen peroxide and sulfuric acid it appears that the standard operating procedure is the opposite: the hydrogen peroxide is added to the acid, and never the reverse.
Given the similarity between water and hydrogen peroxide, this is surprising. Like water, hydrogen peroxide is less dense than sulfuric acid, has a lower boiling point, and a higher volumetric heat capacity. Therefore, all the reasons why adding acid to water is supposed to be safer than adding water to acid would seem to indicate adding acid to hydrogen peroxide.

There is no similarity between H20 and H2O2 except that both are made by O and H only. There is nothing surprising here. What you have listed are marginal reasons. The main reason (on which your list works on) is the partial heat of mixing. And there is no reason to expect the same.

@Alchimista As 30% H2O2 still contains 70% of water, there is still similarity. I suppose it must have something to do with the overall H2O2 concentration and perhaps also relative H2O2 stability alone and in H2SO4. Adding the acid to concentrated H2O2 may be more vulnerable to sudden H2O2 decomposition, while adding H2O2 to the acid never leads to high  H2O2 concentration but very locally and temporarily. So adding H2O2 to acid may be the choice of the lesser evil.  Note that I have no personal experience with "Piranha solution".

@Poutnik all right but the premise is false. 30% x in water isn't water. Now for the precise reason I don't know as well. I just wanted to say that in principle the order of mixing 30% H2SO4 and pure water might be reversed as well (though it does not pose much problem, just talking about of heat).

In either case, I'd be extremely cautious, since boiling, effervescing piranha acid, evolving nascent oxygen, might be somewhat hazardous. https://www.ehs.harvard.edu/sites/default/files/lab_safety_guideline_piranha_etch.pdf

One possibility is that with either ordering the solution becomes hot enough to destabilize the hydrogen peroxide, and the resulting risk of violent decomposition apparently outweighs the "acid spattering" risk that you have adding just water to acid. So the decision is made to accept the "add water to acid" spattering in exchange for lowering the concentration of heated hydrogen peroxide, and also doing the addition slowly (recommended anyway for diluting strong acids) to minimize that.


text:
Can resonance structures account for NMR spectroscopy chemical shifts?

I am unable to understand why the alpha carbon to the carbonyl group is of lower chemical shift compared to the beta carbon based on the effects of electronegativity and hybridisation. Since if electronegativity was a major factor, it would not explain why the beta carbon, which is further from the carbonyl group, has a larger chemical shift value than the alpha carbon. It is also not possible for this to be due to hybridisation effects as, both alpha and beta carbons are $\mathrm{sp^2}$ hybridised.
Is it possible that this difference in chemical shifts between the alpha and beta carbons can be explained by drawing the resonance structures?

I first learnt of this possible resonance structure when studying 1,4-Michael addition and thought that it could help explain why the chemical shifts of alpha and beta carbons are in this way. Since the resonance structures show that the electrons are unequally shared, and has resonance contributors where the carbonyl carbon and beta carbon have positive charges. The positive charge thus makes them deshielded since they would have "less electrons on them," hence explaining their higher chemical shifts compared to the alpha carbon.
I would like to know if this is a valid or correct explanation as to the different chemical shifts of carbon. If this is a possible explanation, why then is it not taught along with electronegativity and hybridisation effects?

You should always be wary of trusting simulated NMR spectra, especially those from such simple tools as the ChemDraw one - proper NMR spectrum simulation is very computationally demanding. Whenever possible, look for an experimental spectrum instead, such as this one. That said, in this case the match is good (it's a very simple molecule after all).

Just keep in mind, resonance structure is not a "structure". It is a hypothetical construct. NMR sees only one structure- which is real.

As for the above comment... Your title question can be answered by yes and not. But it depends on the meaning one gives to it! It is not that more res structures - - - > more peaks! Hope this is clear to all future readers (I see that your meaning is the correct one).

You are talking about the product of the aldol condensation of benzaldehyde and acetone, which is (1E,4E)-1,5-diphenylpent-1,4-dien-3-one. Compared to other more complex organic compounds, this is pretty simple one for $\mathrm{NMR}$ analysis (see Nicolau Saker Neto's comment above). As a consequence, ChemDraw has done pretty good job on suggesting the carnon resonances. For instance, below is the actual $\ce{^{13}C}$-$\mathrm{NMR}$ spectrum of the sought compound (Ref.1):

Since you indicated that you have studied 1,4-Michael addition, I'd like to ask that did you ever wonder why is the nucleophile always attacks $\beta$-carbon during Michael addition reactions? That is because it is the most electron lacking carbon on the system due to the conjugation with more elecrton demanding carbonyl group. That means it is more deshielded than neighboring $\alpha$-carbon (meaning it is having less electron density compared to that of the $\alpha$-carbon). Thus, the carbon chemical shift of $\beta$-carbon $(\approx\pu{144 ppm})$ is higher than that of the $\alpha$-carbon $(\approx\pu{126 ppm})$.
The principles of $\mathrm{NMR}$ spectroscopy is not directly dependent on electron density of individual atoms. Although it is a major contributing factor, but there are other factors such as anisotropy and steric factors. For example, in a aromatic system, proton at outside the ring current shows higher chemical shifts (more down field) compared to those in the inside ring current (more down field). You will learn those facts when you are about to learn advanced spectroscopy. I'd say for your level of education, you have made a good suggestion to explain the difference in chemical shifts on $\alpha$- and $\beta$-carbons.

Late edit:
Since  Azamat has pointed out a valid argument, I like to extend my discussion here:

The $\ce{^{13}C}$-$\mathrm{NMR}$ spectra of above image are the stimulated one by ChemDraw (top spectrum) and actual one from reference 2 (bottom one) of the compound, (3E,5E)-hepta-3,5-dien-2-one. The reason to choose this compound is it lacks the phenyl nucleus and also have an extended double bond. You may see that ChemDraw has given an excellent assignment, mainly due to the simplicity of the compound. As shown in the structure, $\beta$- and $\delta$-carbons ($\ce{C}$4 and $\ce{C}$6) are more deshielded compared to that of $\alpha$- and $\gamma$-carbons ($\ce{C}$3 and $\ce{C}$5). That is what exactly would have suggested by the conjugation that would show partial positive charges on $\beta$- and $\delta$-carbons as well as on the carbonyl carbon (less electron densities than that on $\alpha$- and $\gamma$-carbons). Here, it is a methyl group next to $\delta$-carbon compared to phenyl group next to $\beta$-carbon in the original compound. Yet, additional shielding or deshielding effect is minimal. Therefore, it is safe to conclude that the major factor here is high electron demand by the carbonyl carbon compared to phenyl group.
Note also that, in original compound, mesomeric effect would overcome inductive effect due to extended resonance. It is also worth noting that carbonyl carbon is more shielded ($\pu{189 ppm}$ vs $\pu{199 ppm}$) in original compound due to cross resonance by both directions.

References:

Mari Vellakkaran, Murugaiah M. S. Andappan, and Kommu Nagaiah, “Oxygen as Single Oxidant for Two Steps: Base-Free One-Pot Pd(II)-Catalyzed Alcohol Oxidation & Arylation to HalogenIntact β-Aryl α,β-Enones,” RSC Adv. 2014, 4(85), 45490-45494 (DOI: https://doi.org/10.1039/C4RA07478E) (PDF of authors’ version).
Zhenbo Gao, and  Stephen P. Fletcher, “Construction of β to carbonyl stereogenic centres by asymmetric 1,4-addition of alkylzirconocenes to dienones and ynenones,” Chem. Commun. 2018, 54(29), 3601-3604 (DOI: https://doi.org/10.1039/C8CC01201F)

can it also be because of the adjacent benzene ring? It should also withdraw electrons from $\beta$-carbon, isn't it? I wonder if there's a way to know which factor affects the most.

@ Azamat: I have edited my answer so I hope it'd be clear for you. Your question is well directed.


text:
Mission: Impossible? Hydrocarbons in an aqueous liquid (e.g. humulene in beer)
Your mission, should you decide to accept it, is to describe how hydrocarbons can be stably mixed into a water-based liquid such as beer, as in this question. According to the usual textbook wisdom, hydrocarbons which are essentially nonpolar and have little or no hydrogen bonding with water, should be the classic hydrophobic material, surely repelled from water like, well, oil. How is a hydrocarbon such as humulene ($\ce{C_{15}H_{24}}$) stably incorporated into a liquid as watery as beer?

Humulene + detergent + beer. It's pretty much the recipe for Coors Light.

The question is how much humulene are you expecting in 5% ethanol-95% H2O? Everything dissolves in everything (again at a low level). The concentration of water in toluene is about 500 ppm. That is a lot and detectable by GC. Benzene dissolves about 1000 ppm water. Pure water dissolves about 0.17 % benzene.

So we have detergent action in one comment and a small quantity of the humulene in another. Interesting to see how these ideas are incorporated into an answer.

A good starting point is the ampersand brew mentioned in the linked question. It's list of ingredients should hopefully contain a little more than 'stabilizing agents'

The more straightforward answer, similar to "is the dose that makes a poison" is in the comment by M. Farroq. It points to the relative amounts and this brings along that beer is a complex solution. Everyone knows the joke "what having sex on a kayak and Xoors have to do each other? “? Ans: fucking closed to water. The latter is OoT :))

Is this a joke question?

@kanghun I am offended.

If all you're looking for is a stable-ish suspension of a hydrophobic oil in water, you can shake (or even better, sonicate) the liquids in the absence of gasses. See this video for reference.

What's the reward for completing the mission @Oscar? ;)

@nilay because of the mixed voting record on this question I have only +44 rep from it, so I have to settle for accepting an answer and thus +15 for the winner. I have an answer but typically wait a few days to "shop around".

The foaminess of a poured beer is a good indication of how much surfactant activity is present. Clearly a very different environment than pure water

There are many industrial problems similar where one would like to incorporate a hydrophobic ingredient into a hydrophilic matrix (or vice versa).
Luckily there are a few main strategies that can be used to incorporate dissimilar chemistries with each other.
The first is co-solvancy where you introduce a secondary solvent to help dissolve your target solute. This is a complex area of chemistry and is generally done empirically or semi-empirically in industry. The basic idea is that your solvent A is a bad solvent for your solute and solvent B is also a bad solvent for your solute. But a mixture of A+B is a good solvent. The chemical concept underpinning this is the Hansen Solubility Parameters.
In the case of beer, you have about 95% water, but you also have about 5% ethanol which acts as a co-solvent and will help dissolve more polar solutes. I can't say this is the dominant effect without doing some experiments, however.
What is much more likely is the second main strategy for mixing dissimilar chemistries: surfactants. Surfactants have a polar and non-polar end can dissolve organics in water (or vice versa). Your primary exposure to surfactants would be in cleaning as detergents are surfactants and they effectively 'dissolve' oil and dirt into water to take it away.
Beer is fermented which means it has all kinds of biomolecules in it that can act as a surfactant and dissolve the humulene. The most likely culprits will be metabolites from yeast or broken-up yeast cells. Yeast cells membranes (and all other cell membranes) are made up of a phospholipids bilayer and individual phospholipids are surfactants. There are likely to be plenty of phospholipids floating around the beer (in addition to proteins and nucliec acids) which will act as a surfactant and dissolve the humulene.


text:
Can chemical composition be determined from elemental composition?
I have the energy dispersive X-ray spectroscopy results from a sample of salt that I produced by collecting and processing ocean water by filtering, boiling, evaporating, and additional steps to try to reduce calcium and magnesium content in the final salt crystals:

Atomic %

25.70
C
carbon

20.68
O
oxygen

20.30
Na
sodium

2.71
Mg
magnesium

14.25
Al
aluminum

1.13
S
sulfur

14.46
Cl
chlorine

0.77
Ca
calcium

I know that I can discount the aluminum content because the mount within the scanning electron microscope is made of aluminum and that's most likely what is being detected, as well as some of the carbon content because the carbon tape holding the salt sample to the mount is contributing to that detected amount.
If there were only one compound present, then I imagine the ratios in the atomic percentages / elemental composition would give us a pretty strong hint at what that compound would be, but evaporating ocean water leaves multiple compounds such as calcium carbonate $\ce{CaCO3}$, calcium sulfate aka gypsum $\ce{CaSO4}$, sodium chloride $\ce{NaCl}$, magnesium chloride $\ce{MgCl2}$, and potassium chloride $\ce{KCl}$ (https://manoa.hawaii.edu/exploringourfluidearth/chemical/chemistry-and-seawater/salty-sea/weird-science-types-salts-seawater).
Can I know with confidence that these same compounds make up my produced salt? Can I know with confidence that the salt doesn't contain other combinations of the ions present in ocean water, such as sodium carbonate $\ce{Na2CO3}$ or calcium chloride $\ce{CaCl2}$ or magnesium hydroxide $\ce{Mg(OH)2}$?

Rather than speculating about the carbon and aluminum, you could do a control run with distilled water and subtract.

@BenCrowell   Distilled water in scanning electron microscopy? This is not possible.

@M.Farooq I suspect Ben means process distilled water in exactly the same way the seawater was processed (i.e. evaporate it all off), and analyse it.  That may be tricky if the processing resulted in a decent amount of mixed salt powder, but starting with a solution of (a known concentration comparable to the seawater) of pure NaCl in distilled water  would also provide a control sample - as you would know, now I've seen your answer

What were the additional steps to try to reduce calcium and magnesium content? How could they affect other ions?

This is OP, but I posted as guest so I don't think I have edit or comment privileges. This answer is merely to provide more detail on the "additional steps to try to reduce calcium and magnesium content" if it will benefit the discussion at all. I do appreciate everyone's responses. The specific steps that I did to produce the salt were: 1. collect water from the ocean 2. pass water through coffee filter to strain out particulates and organisms 3. boil water to concentrate salinity and kill any organisms that made it past the filter 4. solar evaporate the water until about ~10% of original vol

@KevinSu, In this way you did not reduce calcium or magnesium content. You just removed insoluble stuff from sea water.

The short and sad answer is "no"- you cannot determine the compounds and their formulae just from energy dispersive spectrometry. You have identified the problem correctly as you are not dealing with a single compound but plently of mixtures. There are other deeper issues with EDS. First you should remember that no analytical technique is absolute with two or three exceptions, i.e., the analytical signal is useless in the absence of standards. So you do not have standards data, i.e., signals from known materials.
The next problem is that it your choice that you include or remove a certain element. If you remove carbon, all the atomic % will change. If you remove Al, it will cause further changes in numbers. This adds a lot of uncertainity and of course errors. Is the carbon from the tape or from carbonate? See these types of uncertainities exist.
Third problem is that did you gold coat your sample or not? You must have removed gold peaks too. Insulators are not that great in SEM. Most of it is addressed here Is scanning electron microscopy/energy dispersive X-ray spectrometry (SEM/EDS) quantitative?.
Since seawater is a solution, you cannot say what compounds were originally dissolved in it. You can talk about what ions were present and then you must account for charge balance to guess what were the starting compounds. If you were actually given the task to determine what salts and compounds were actually present, you will need to do a lot of wet chemistry and elemental analysis. The good news is that you do not have to re-invent the wheel. A lot of people have reported and studied seawater composition.
Coming to your isolated and purified salt. I feel the charge balance does not make sense in your data. You need to have a similar number of Cl atoms as sodium. If possible, do a pure NaCl EDS and see what you get.
The proper analysis of this salt will require quantitative ion chromatography for anions. Also then you need elemental data from ICP-OES to see which metals are present and in what ratio (Li, Na, K, Ca, Mg etc.), only then you can estimate what compounds you have in your isolated salt.

Why wouldn't the charge balance make sense? There's likely a whole lot of carbonate anions in there (perhaps added in the "additional steps to try to reduce calcium and magnesium content"). If all the oxygen came from carbonates, it would account for over 13% (as monovalent), which would be plenty to balance all the cations. Of course some of that oxygen is water, but there's still a lot of room there.


text:
Does Phenylacetylene undergo side chain oxidation with hot alkaline Potassium permanganate?
This question was asked in my book:

To my knowledge, side chain oxidation with KMnO4 occurs only when a benzylic hydrogen is present. However this doesn't contain a benzylic hydrogen and hence according to me it wouldn't give any reaction.
However the answer in my book states that Phenylacetylene will be oxidized to benzoic acid.
How does this take place then?

Action of hot KMnO4 is akin to that of oxidative ozonolysis on alkynes.

Brief explanation
Hot alkaline solution of potassium permanganate oxidises a terminal alkyne according to the following process:

Triple bond between the first and second carbon atoms (they are marked in the figure above) converts into two carbonyl groups $\ce{C=O}$. The resulting molecule consists of a ketone part (blue) and an aldehyde part (red). The subsequent oxidative cleavage divides the molecule into a carboxylic acid with the $\ce{R1}$ radical and formic acid. But formic acid is oxidised in this conditions to carbon dioxide.

Reaction
Since ethenylbenzene is a terminal alkyne, it can be oxidised with a hot alkaline solution of potassium permanganate to form a carboxylic acid and carbon dioxide.
The oxidation of carbon atoms is described in half-reactions (all the oxidation states are marked in the figure below):
$\ce{C^0->C^{+3}}+3e^-$
$\ce{C^{-1}->C^{+4}}+5e^-$
Here is an equation of oxidation of ethenylbenzene with a hot alkaline solution of $\ce{KMnO4}$:

References

Oxidative cleavage of alkynes:
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(McMurry)/09%3A_Alkynes_-_An_Introduction_to_Organic_Synthesis/9.06%3A_Oxidative_Cleavage_of_Alkynes
Oxidation of alkynes:
https://chem.libretexts.org/Courses/Sacramento_City_College/SCC%3A_Chem_420_-_Organic_Chemistry_I/Text/10%3A_Alkynes/10.07%3A_Oxidation_of_Alkynes

I was expecting $\ce{HCOOH}$ rather than $\ce{CO2}$ based on the manner of oxidative cleavage $\ce{KMnO4}$ shows.. why was this pattern broken for terminal alkynes

I have supplemented my answer with more information about the process (not a detailed reaction mechanism) of oxidation of a terminal alkyne with a hot alkaline solution of potassium permanganate.

According to this source Chemistry Libre Texts

Alkynes, similar to alkenes, can be oxidized gently or strongly depending on the reaction environment. Since alkynes are less stable than alkenes, the reactions conditions can be gentler. For examples, alkynes form vicinal dicarbonyls in neutral permanganate solution. For the alkene reaction to vicinal dialcohols, the permanganate reaction requires a lightly basic environment for the reaction to occur. During strong oxidation with ozone or basic potassium permanganate, the alkyne is cleaved into two products.

oh so while oxidation of alkyl groups attached to benzene to benzoic acid is unique to aromatic systems, oxidation of alkynes with KMnO4 to carboxylic acids takes place whether or not they are attached to benezene, right?

@Mia  that is correct

One possibility: the alkyne gets its $\alpha$ hydrogen through formation of a hydrate. The triple bonded carbon is electrophilic with its relatively high electronegativity, with the attack occurring on the remote carbon to form a stabilized (benzylic) carbanion:
$\ce{C6H5C#CH\overset{H_2O/OH^-}{<=>}C6H5\overset{-}{C}=CHOH<=>C6H5CH=CHOH}$
The form with the $\alpha$ hydrogen (or the tautomerized keto form which has two such hydrogens) then reacts in the usual way with alkaline permanganate.