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Determine reaction order and reaction rate

In a solution of $\ce{Fe^3+}$ the concentration was measured to be:
$$\begin{array}{c|c}
t/\pu{min} & \ce{[Fe^3+]}/\pu{\mu M} \\ \hline
10\ & 238\\
20 & 227\\
40\ & 206\\
60\ & 187 \\
80\ & 169\\
100\ & 154 \\
\hline
\end{array}$$
The decrease in $\ce{Fe^3+}$ concentration is due to the hydrolysis of the ion and precipitation of hydroxide. Calculate the reaction order and reaction rate.

I am having trouble understanding how to tackle this problem. I began by find the reaction formula:
$$ \ce{Fe^{3+} + H_2O <=> Fe(OH)^{2+} + H^+} $$
So this is a second order reaction. I then did:
$$\begin{array}{c|c|c}
\text{time} & \ce{[Fe^3+]} & \text{H2O} & \ce{[Fe(OH)^2+]} & \ce{[H+]}\\ \hline
t=0 & a\ & a & - & -\\
t=t & a-x & b-x & x & x\\
t = \infty & - & b-a & a & a \\
\hline
\end{array}$$
But I am unsure if I should include water and the hydrogen ion, and also how I should continue if this is correct. Can someone please push me in the right direction here. Thank you!
EDIT
First order equation:
$$ \ln \frac{a}{a-x} = kt$$
Is a then equal to $238$ and $a-x$ is the initial value ($238$) minus the next value? e.g., $238-227 = 11$.
If I plot:
$$\begin{array}{c|c|c}
t & \ln \left (\frac{a}{a-x}\right) \\ \hline
10 & 0\\
20 & 3.07\\
40 & 2.43\\
60 & 2.53\\
80 & 2.58 \\
100 & 2.76\\
\hline
\end{array}$$
I get that the slope is $0.0168$ and that would be the rate of the reaction?

Don't include tuhe water in your calculations of concentrations. The water is $\pu{ 55.555 mol/L}$ at the beginning. It should not be really different at the end of the reaction. And even it is is $\pu{55.554 mol/L}$  at the end, nobody will be able to check that this a different amount from the initial values.

Drawing a table like you have done is no help for answering the question. You better report the logarithm of the concentration $[\ce{Fe^{3+}}]$ vs. time. If the points are alined, the reaction is first order. And the slope of the line is the rate constant of the reaction. Go !

If you have access to a calculator or a graphing application it will be an easier task.
I don't have the ability to calculate and write a long answer at this point in time, but whenever there is a hydrolysis reaction, it is a pseudo-first order reaction. In other words, as @Maurice pointed out, don't include water concentration in the rate order and equation.

I have tried solving it as a first order reaction now, but I am still unsure if the way I did it is correct. I would truly appreciate if someone could just check if my method is correct, thank you! @C_Lycoris

@Maurice I have tried solving it as a first order reaction now, but I am still unsure if the way I did it is correct. I would truly appreciate if someone could just check if my method is correct, thank you!

@confused I typed in an answer; I hope that's better now.

@confused. Calculating log [$\pu{a/(a-x)}$] is a non sense. No. Try first to calculate the natural log of the concentration, on the $Oy$ axis. The first point is a $\pu{t = 10 s}$, and $\pu{y = ln 238 = 5.47}$ . The second point is at $\pu{t = 20 s}$ and $\pu{y = ln 227 = 5.42}$. Do the same for all points. You will see that the points are well alined, except the value for $80$ s. which is a bit too low. But with all other points you will see that the alinement is good. The slope can be calculated with the first and the last point. It is $ 4.89 · 10^{-3}$ par second.

The rate equation is $$\ln[\ce{Fe^{3+}}] = \ln[\ce{Fe^{3+}}]_0 − kt$$ So, $a-x=[A]$. Don't subtract the  next value. Put it as it is. That's just for your understanding, however. Don't try to use the integrated rate form to plot the graph. $[\ce{Fe^{3+}}]=238 \text{ where } t=10 \text{ and } [\ce{Fe^{3+}}]=227 \text{ where } t=20$. Using the above mentioned equation: $$\ln[A] = \ln[A]_0 − kt$$ Calculate the value of $\mathrm{\ln[\ce{Fe^{3+}}]_0}$ by eliminating k. Thereafter, you can graph the above mentioned equation $\ln[\ce{Fe^{3+}}] \text{ vs time t}$ with $\ln[\ce{Fe^{3+}}]_0$ as the y-intercept. The slope will yield the value of the rate constant.

[...]that would be the rate of the reaction?

For some reason unbeknownst to me, a lot of people tend to confuse between rate and rate constant. The slope will yield the rate constant. Next,
$$\textrm{rate}=k[\textrm A]^{a} \quad \text{}$$ where $a$ is the order of the reaction with respect to reactant $\ce{A}$.
As a side-note, since this is a pseudo-first order reaction, the rate law equation, initially this: $$\text{Rate} = k^{'}[\ce{Fe^{3+}}][\ce{H2O}]  $$
now appears as this:$$\text{Rate}=k[\ce{Fe^{3+}}]$$
the variable $'k'$ in the second equation is the slope that you obtained from the graph. Here $k=k^{'}[\ce{H2O}]_o$
For more about pseudo-first order reactions you can check one of my previous answers.


text:
High kinetic basicity of alkoxides compared to alkyls
I'm reading something at the moment where they state:

This reagent has enhanced basicity as it benefits from the high kinetic basicity of alkoxides compared to alkyls.

This is in the context of organometallics so we're talking about a negatively charged alkoxide ligand versus a negatively charged alkyl ligand.
So why do alkoxides have a higher kinetic basicity than alkyls? I thought this might be because the negative charge on $\ce{-O}$ is more accessible for doing the deprotonation than the neagtive charge on $\ce{-CH2}$ due to the bonded hydrogens. Is this correct or are there other factors at play?

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Liquifying a gas above critical temperature
Is there any theoretical/hypothetical way through which a gas can be liquified above critical temperature? For example applying inifinjte amount of pressure to the system?

Already well answered here: https://chemistry.stackexchange.com/a/37179/79678. Also, applying infinite pressure would force the poor gas to become a black hole! ;-)

At the critical point, both liquid and gas have the same density, which is $\pu{0.34 g/mL}$ for water. So at this point the limit between liquid and gas simply disappears. At higher pressures and higher temperatures, your stuff can be considered as made of a liquid or of a gas.

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Why does zinc sulfide have to be doped with silver to phosphoresce?
$\ce{ZnS}$ is a boring white powder and it doesnt glow in the dark, but if its doped with silver it starts to show some phosphorescent properties.How does silver force zinc sulfide to do that?

I remember this question was asked by Birks about $30$ years ago at a symposium about the use of $\ce{ZnS(Ag)}$ as a detector of alpha particules in scintillation counting. I also remember he did not get any answer from the audience to his question.  Birks was specialized in using this substance for counting the number of alpha particules emitted by a radioactive source. He had published plenty of publications about this domain.

Copper and manganese also can serve as phosphorescence activators in ZnS. It is not the ZnS that luminesces: it is the excited (by light) activator species. A similar case, among many, is ruby fluorescence. It is due to $Cr^{3+}$ ions in the sapphire (crystalline alumina) matrix.

Why doesnt a $\ce{Cr^3+}$ solution have these properties then? What is special about the lattice it is doped in?

My understanding is it is remotely similar to semiconductors, where there is specific semiconductor matrix with specific doping atoms. But in the scintillator case, it is about stable excitation, instead of conductivity.

Word "stable excitation" satisfies me to some degree.

It all comes down to the specifics of the photophysics: a crystalline host matrix (e.g., YAG) differs from a glass matrix which, in turn, differs from a liquid matrix. For example, there are $Nd^{3+} YAG$ lasers and $Nd^{3+} glass$ lasers, but I have not heard of any liquid based variants of those. Phosphorescence at room temperature in liquids is rare because it gets quenched, i.e., the excited species is forbidden to emit light, so it has a long lifetime before it does it anyway, but that means it most likely gets “mugged” of its energy (via collisions). In solids, less mugging.

All you say is perfectly correct. Nothing to object. Nevertheless nobody knows why only ZnS has this property to such a high degree of phosphorescence. And nobody knows why silver, and no other atom from the periodic table can make it phosphoresce. This is the same for $\ce{NaI(Tl)}$. This doped substance gives a strong fluorescence when irradiated by gamma rays. Why NaI and not KI ? Why use  Thallium as doping agent ? Nobody knows !

@Maurice     According to the wikipedia article, copper results in the green phosphorescence of ZnS and manganese gives red. I will take that with a grain of salt! But your point is taken: some of these systems still harbor unknowns or are rather poorly understood.


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Among Benzyl carbocation (Ph-CH2+) and Methyl methoxy carbocation (CH3-O-CH2+), why is the latter more stable?
Benzyl Carbocation could form multiple conjugation, then why isn't that stabler? In Methyl methoxy carbocation, Oxygen donates the lone pairs but so does the double bond in the benzyl Carbocation.

https://chemistry.stackexchange.com/questions/37427/confused-about-carbocation-stability

let's see the resoance forms of benzyl carbocation and methoxy carbocation:
-Benzyl carbocation:

You are right, benzyl carbocation have multiple ressonance forms which means that It must be very stable. In fact, benzyl carbocation is one of the most stable carbocations.
-Methoxy carbocation:

In this case, we only observe 2 resonance forms.
so the question is, why methoxy carbocation is more stable despite of having less resonance forms??
Well, the key is the stability of this resoance forms...
In benzyl carbocation, none of its resoance forms have a complet octet, wich mean that the resonance forms are not very stable and therefore the positive charge is not very delocalized.
However, in the case of the methoxy carbocation, one of the resonance form have a complet octet, which means that the resoant form are pretty stable and therefore the positive charge is quite delocalized between the O and de C
In conclusion, number of resonance forms are important, but their stability is also a really important factor.


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How do bond electrons covers both atoms in covalent bond?
I want to know how bonding electrons behave when they have to cover for both atoms. Electrons can spin around only one atom right? If so, then what happens when a covalent bond forms?
Let's say we have two covalent bonds between two atoms (for example: $\ce{C=O}$), so they have two pairs of shared electrons.
Since the double bond clearly exists, how do the electrons form the clouds that cover both atoms?

"They can spin around only one atom right?" No.

"Spin around" is a wrong picture.

If you like images, you may imagine that the electronic cloud looks like an ellipse, with the two nucleus inside. .

I upvoted the comment of @Maurice as for it might serve to OP. However, what is linking the atoms together (bond) is in between them rather than a cloud enveloping both. It is just the spinning around figure that (again, certain useful in several occasions) is wrong. So my comment to OP is "imagine them between the atoms, so that each is under the influence of both nuclei".


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Prove that internal energy is a state function
Why is it that heat and work are path functions but their sum i.e. internal energy is a state function? How is it even possible?
$ΔU=q+W$

Either it is a state function, either perpetuum mobile is possible.

See also: https://chemistry.stackexchange.com/questions/22171/why-is-du-an-exact-differential-and-dq-an-inexact-differential?noredirect=1&lq=1

It could go as far as to the Noether theorem about relation of space and time symmetries and fundamental conservation laws.

Just like pressure & volume are state functions but work as you said is a path function. $W = ∫ -P dV$

Your title and your question are asking two different things.  Your title is asking us to prove that energy is a state function.  Your question is asking something different:  How is it possible that the sum of two path functions can equal a state function?  Which is the question to which you want an answer?

I just want a proof as to why internal energy is a state function mathematically.

Check out https://chemistry.stackexchange.com/q/146030/72973

The three laws of thermodynamics (or four, depending on the source) are postulates that scientists have assembled in an attempt to describe the way the universe works. The First Law is a reformulation of the principle of conservation of energy. It says very simply that while energy is indestructible it can be exchanged in various ways. It also implies that a system can exist in different states, each with a set of clearly identifiable properties including a fixed amount of energy. Declaring the state a system is in also stipulates the energy of the system. We can write this mathematically as $U=f(\mathrm{state})$.
One also presumes that if the energy of a system changes, then energy has been transmitted in some measurable form by an identifiable mechanism. In other words, for some process you can always write
$$\Delta U = \sum_x \mathrm{energy~in~transit~via~mechanism~}x$$
where the sum goes over all the ways we imagine energy might be transferred.
Classical thermodynamics suggests partitioning the energy in transit into two basic categories, work and heat. Heat is transferred when objects are at different temperatures (this forms the basis for the Zeroth Law of thermodynamics). If you insulate a system, heat is no longer a viable way to transfer energy. That leaves all other possible ways of transferring energy, which we call work. Alternately, if all identifiable means to do work are inaccessible, then the system can exchange energy only as heat.
The point is that if the initial and final states are known then $\Delta U$ is fixed, but there might be different ways (paths) that change the energy of the system by that amount. For instance, work can be used to change the temperature of a substance just as putting the substance in contact with an external heat bath can.


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Enthalpy vs. delta enthalpy
I haven’t really found a clear answer.
Enthalpy is measured as the sum of internal energy (kinetic and potential energy) and pressure times volume. However, I don’t really see the value of measuring enthalpy, so why is it even mentioned? Aren’t we more concerned with changes in enthalpy, like enthalpy of vaporization, enthalpy of fusion?
In other words, what are the differences of enthalpy and delta enthalpy? Is the former just a formality, and we are really concerned about measuring the energy of a thermodynamic system undergoing change (i.e. chemical reactions)?

If you did not know how is defined particular function, how would you know, how to calculate it's differences, differentials or derivatives ?

Enthalpy is defined, not measured, as H = U + p.V. It is not measurable quantity. Measured are quantities and their changes that affect the enthalpy change.

Do a few actual problems, rather than spending you valuable time speculating.  You will bet the idea of how it plays out.

@ChetMiller I have done a few problems, which is why I came here. I regularly see enthalpy problems being on changes in enthalpy, rather than enthalpy itself. It seems that enthalpy, like Poutnik said, cannot be measured directly; only changes in enthalpy can. So I’m wondering what exactly its value is - for measuring enthalpy in reactions, formations, etc.; or is it more? Because if it’s more, then I wanted to clarify.

If a quantity is not defined, it's changes are not defined either, even if only these changes really matter a/o can be measured, directly or indirectly.

Enthalpy vs ∆Enthalpy
We know that enthalpy represents the energy of a system. So, now think about it? Measuring the energy of a system? Consider all the changes and influences a system is under. This makes it really difficult (we could say impossible) to calculate the enthalpy of a system.
But we do know one thing, that the enthalpy depends on two factors.
A system will experience change in enthalpy if it's internal energy is changed and/or if work is done on/by it.
So what we do is that we assume an initial state where we start our observation and a final state where our observation concludes. And the changes in state cause particular changes in enthalpy and this change in enthalpy of the system is what matters to us anyway.
Why did we define it then if we don't use it?
We just know that a particular quantity depends on two particular factors. It's just that we understand the fact that it is meaningless to calculate the enthalpy of a system - because we do not know where to start from? Where will you start from? You need a point where the enthalpy was zero initially and then gained enthalpy right? But imagine - doesn't that mean there won't be any atomic activity at all if we're starting from zero enthalpy? And if there wasn't atomic activity then that chemical specie might not even be able to exist? It's complicated.
But then we realise: all that matters to us is just change in enthalpy during a thermodynamic process. And that is what thermodynamics is essentially all about - energy in motion. Thermodynamics is but a study of states (this was mentioned in NCERT Chemistry). So we study the change in Enthalpy from one state to another.
We define the enthalpy and what it depends on so that we can observe the factors that it depends on.

Very excellent response! That makes a lot of sense - that because systems are constantly undergoing infinitesimal change, we cannot possibly overconcern ourselves with the enthalpy but rather the change in enthalpy. One remaining question: When a system has undergone a change in enthalpy, we say that it required or exerted energy (endothermic vs. exothermic). That energy becomes part of the system’s new enthalpy. So then, when we talk about enthalpy, why isn’t it made clear from the start that we are concerned with changes in enthalpy and not with the definition itself?

With what I understand from your counter-question, I believe it is just a matter of exposure to resources. Sometimes we read and interpret things in different ways or sometimes overthink about it when it's basically just simple fundamental things. This especially happens in chemistry and we see learners asking about why a particular thing in chemistry is the way it is although it is a good trait to do so. My first teacher of thermodynamics did emphasise that ∆H is what we focus on and not enthalpy and maybe that's the reason I was able to address your confusion.

Enthalpy is a fundamental physical property of the material(s) comprising a system.  It can often be presented for a given material in the literature in tables (e.g., the steam tables for water), relative to some specific datum state (at which it is taken to be zero).  This is analogous to potential energy which is expressed relative to some specified elevation.
In the case of enthalpy, we can't only tabulate changes between states since this would require an infinite number of tables.  So, in order to make and apply enthalpy tables, we need to reference everything to a specified datum state (for that material).  The values in the table can be regarded as "absolute enthalpy."  We can then determine changes between states by subtracting values for state 1 from state 2.


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Why is addition to acid chloride favoured over SN2 at the carbon even though it has a Beta-carbonyl?

I feel that the conditions highly favour $S_\mathrm{N}2$ reaction at the aliphatic center and so the aliphatic chlorine should be replaced before the carbonyl chlorine. But the solution suggests that the nucleophilic substitution of carbonyl chlorine takes place before the aliphatic chlorine.

I would say this is down to basic electron density considerations. Look at which atoms are connected directly to each electrophilic carbon. The $\ce{COCl}$ carbon is connected to $\ce{C}$, $\ce{O}$ (double bond) and $\ce{Cl}$. The $\ce{CH2Cl}$ carbon is connected to $\ce{C}$ and $\ce{Cl}$. More generally, it is known that acid chlorides are more reactive than alkyl chlorides. The additional activation of this specific alkyl chloride by the proximity of the carbonyl does not change that. I am sure a physical chemist can answer more authoritatively, but I believe I am not far off the mark.

I had the same feeling too thanks for confirming it.


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Decoxide vs decaoxide?
Now, I am doing a science worksheet and I came across an opportunity to write either "decoxide" or "decaoxide". Since the latter sounds wrong out loud, I decided to google it before I wrote it out, just to be sure. What I got instead, once I had filtered out all of the references to phosphorus or tetraphosphorus, was a whole lot of mentions of both. Both my phone's autocorrect and my worksheet itself use "decoxide".
So, I did the obvious next thing and went to r/askscience. However, after I'd gotten no responses on there, I realized my mistake and came here.
Update: How do you write "three iodine atoms"? Is it "triiodide" (which my autocorrect says is correct, and which Wikipedia apparently uses) or is it "triodide" (which my autocorrect approves of)? Or even "tri-iodide"?
Thanks for any help you guys can provide!

We generally omit the first vowel.

Concerning your update, you also need to distinguish between "triiodide" and "tris(iodide)".

@NilayGhosh Am I to take that whole thread to mean that if the prefix is monosyllabic, then an elision is used, but if the prefix is bisyllabic, then an elision is not used? If that is the case, then that would seem reasonable, and I would then be inclined to spread the word. Thank you.

Yes, that is right.


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How does evaporation even possible?
Suppose the pressure is 1 atm and we put water (at 25 °C) in an open container. After some time the water will disappear as the water evaporates. We know that water at 25 °C and 1 atm has a vapour pressure. And this vapour pressure is obviously less than the atmospheric pressure or external pressure as no boiling is occurring at this pressure, temperature. If they were equal, boiling would occur which is clearly not happening.
So, if the vapour pressure is less than the atmospheric pressure, then how are are liquid molecules converting into gaseous molecules? I mean the atmospheric pressure above the liquid is exerting a greater force on the liquid and it's vapour pressure is less than the atmospheric pressure.
Then how are the liquid molecules overcoming the atmospheric pressure and changing into gaseous molecules if their vapour pressure is less than the atmospheric pressure? They can only change into gaseous molecules when they can overcome the atmospheric pressure which can occur when their vapour pressure equals atmospheric pressure.

See also: https://chemistry.stackexchange.com/questions/38653/vapor-pressure-and-boiling-point-for-solutions-in-open-containers?rq=1
https://chemistry.stackexchange.com/questions/23566/will-the-water-at-the-bottom-of-the-tube-boil-at-the-same-temperature-as-the-wat
https://chemistry.stackexchange.com/questions/41652/what-is-the-connection-between-vapor-pressure-and-intermolecular-force-explanati?rq=1

Your container is open.

Water molecules at the liquid phase surface know nothing about they have to overcome any pressure, which is macroscopic quantity. So they leave liquid any time if their kinetic energy is greater than energy of bonds holding them.

I remember my days when I started confusing myself in the exact same way. The issue I believe is that you haven't been explained the details of how all this happens.
Evaporation:
All molecules in a liquid have various amount of energy. But the ones at the top, the ones that form the surface - they have the maximum energy in them because they are present there by overcoming any forces of attraction. (The other way to imagine this would be, the layer of molecules at the surface will have mostly downward forces of attraction on them compared to the distributed force of attraction from all directions on a bulk molecule.) Now these surface molecules have the tendency to escape. These molecules will individually leave the liquid by getting rid of the forces of attraction because it is easier for them compared to bulk molecules. And when individual molecules leave the liquid, it is called evaporation. Although this is a form of vapourisation, it doesn't need a constant temperature where the change of state occurs. They are just individual molecules that leave the liquid and escape and cause a vapour over the liquid.
Boiling:
But in boiling, a lot of constant heat supply, constant temperature, constant pressure. All this is required. And when this is obtained, the liquid changes its state and becomes a gas. This gas forms bubbles and escapes from the bulk of the liquid as well and not only from the surface as in case of evaporation.

Surely having mostly downward forces of attraction is a good reason for them not to escape?

Imagine you are trying to climb up a tree in order to escape zombies. Consider two situations: 1) held from all directions by zombies 2) held only from below by zombies. In which case are you more likely to escape and climb up the tree? Also, the surface area of a molecule exposed to attractive forces is much lower at the surface of the liquid relative to bulk. Just like one zombie pulling you from below versus 5 holding on from all directions. You are just overthinking. This happens alot even to me as well. As Poutnik has already mentioned, once the energy is enough to escape, it escapes.

I'm more likely to move upwards if there are zombies pulling me upwards, than if there aren't.

Yeah ofcourse but there are like five pulling you in all directions why are you biased?

if there are five pulling me downwards, I'm pulled downwards on net, and if there are five pulling me downwards and five pulling me upwards, I'm not.

Yes. So imagine you have an urge to move up with a magnitude of 5 units. But have zombies pulling you from all directions with 1 unit. In the other case, you have just one zombie pulling you from beneath with 1 unit because your area exposed to attractive forces is less at the surface compared to the bulk so just one zombie can get a grip of you, others cant put their hands on you there's no surface area.

so in the first case I move up 1 unit more. Are you aware of vector addition?

Let us continue this discussion in chat.

please explain what causes people to move discussions to chat and then ignore the chat.


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Why does ester hydrolysis occur in basic medium
Ester hydrolysis occurs in basic medium and this is the medium which is usually preferred for the same. But I am facing a very fundamental doubt and not able to understand the reason.
Ester Hydrolysis goes like this:

$\ce{RCOOR' + OH- <=> RCOOH + OR'-}$

Now according to me $\ce{OR'-}$ should be a better base because of +I inductive effect of alkyl group. If it is a better base then why does it not attack the acid back again !? In short, why does the equilibrium not reverse ?

$\ce{OR'-}$ will take a proton from a water molecule as it is a better base and will form an alcohol, in turn creating more $\ce{OH-}$ which will further increase the hydroxide concentration and thus decreasing the probability of $\ce{OR'-}$ to attack the species.

The acid with be present as the carboxylate salt thus not an electrophile

The presence of $\ce{OH-}$ ions destroys $\ce{RCOOH}$ which prevent the reverse reaction, which would produce the ester again.

In basic medium, there is a huge amount of $\ce{OH-}$ and they will deprotonate the carboxylic acid as soon as it forms.
$$\ce{RCOOR' + OH- <=> RCOOH + R'O-}$$
$$\ce{R'O- + H2O <=> R'OH + OH-}$$
$$\ce{RCOOH + OH- -> RCOO- + H2O}$$
The last deprotonation step is almost irreversible and consumes the base. This is why, an excess of base is usually added for ester hydrolysis.
The $\ce{RCOO-}$ is a negative ion, so it repels any nucleophile that comes close to it, so nucleophilic attack cannot happen, and the reverse reaction is prevented (as Maurice has mentioned in the comments). In essence it is this last step that drives the reaction to the product side.


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More ions form pairs when concentration of acid increases, does that decrease the speed of reaction?
Electrical conductivity decreases as an acid becomes more concentrated because more ions form pairs. However, does that affect the speed of reaction of the acid as there is a smaller concentration of Hydrogen ions?

Considering H3O+ activity coefficient steeply raises for high acid concentrations, it is the opposite.

In general, the conductivity of any liquid increases with the concentrations of its ions, and decreases with their dimensions. In pure sulfuric acid, the only ions existing are $\ce{H3SO4+}$ and $\ce{HSO4-}$. They are big and their concentration is low. So the conductivity of pure sulfuric acid is low. But when you add some water, the $\ce{H2O}$ molecules are totally transformed into $\ce{H3O+}$ and $\ce{HSO4-}$ ions which are numerous and sometimes small (for $\ce{H3O+}$). The conductivity must increase.

Does this affect the number/concentration of H3O+ ions available for acid reactions?


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Is there a relation between polarity of molecules and their electric susceptibility?
So it is written in our text that electric susceptibility ($χe$) (the response of say water molecules in the presence of an external electric field) is related to the molecular behaviour of various substances, but it doesnt explore it, and says its out of scope for me. When I tried to find the related equations, I could not find a clear answer.
To be more clear,
We know that the capacitance of a parallel plate capacitor increases when  water(k, dielectric constant = 80) is inserted as a dielectric as opposed to oil (k = 2.2), or any other non-polar substances (with small values of k). Now my question is, whether the dielectric constant (related to electric susceptibility as $k=χe + 1$) depends on the polarisation of the molecule.
Also, given $$P = χe. E$$ where P is the dipole moment per unit volume, called polarisation of the linear isotropic dielectric.
Can I tell someone that the reason Dielectric constant of water (80) is so much higher than oil (2.2) is because water is polar and oil is non polar. If so, what equation should I quote as proof to this statement?
To me it seems empirically true, but I would like to see if the theory supports the fact that polar molecules have a higher electric susceptibility as compared to non polar ones.

I might be totally out of track, but susceptibility is more on the "flexibility" of electrons arrangement,  independently of permanent electric dipoles. Actually, at least at qualitative level, I would perhaps look at polar molecules as not inherently polarisable (though very polar molecules can even undergo internal charge transfer under various stimuli).

Electric susceptibility is equal to relative permittivity (or the dielectric constant) minus 1: $\chi_e=\epsilon_r-1$

@ShoubhikRMaiti I disagree. That's the material polarisability. OP intend molecular susceptibility. If not, or s/he had not a distinction in mind, the question must be improved in order to known what we shall speak about.

@Alchimista What is molecular susceptibility? I have only come across the term susceptibility in the context of magnetism. I got that definition from the wikipedia page for electric susceptibility (https://en.wikipedia.org/wiki/Electric_susceptibility). But yes, I agree the question needs to be clarified.

See also https://en.wikipedia.org/wiki/Electric_susceptibility

@ShoubhikRMaiti yes correct. But scroll down. The fact that the molecule is immerse in its own field is what makes the treatment different. Uno thing is the arrangement of water molecules in response to an external field. Another thing is the arrangement of the electrons within one water molecule in response to an external field. Another difference is that in the first case one can use the vacuum permittivity, while the same isn't true at molecular level.

Am sorry for any confusion caused, but I meant electric susceptibility as pointed out by @ShoubhikRMaiti. I was not aware that any other type of susceptiblitiy existed. I definitely mean the arrangement of polar water molecules in response to an external field.

@KrishnarajPT they are the same quantity. Just if you want to link it to molecular electrical dipole moment than the susceptibility is that of the molecule, not that of the block of matter that those molecules made up. Is is like volume of a sphere and volume of the things that make up the sphere. As usual the microscopic case is more complicated. Here is because you can't make use of the fact that a piece of matter can be treated to be immersed in vacuum, so the use of vacuum permittivity. The same notions is difficult to be applied inside the molecule.


text:
How is this Cope Reaction forming Ethene as major product?
I have been asked to find the major product of the reaction:

As this a case of pyrolysis of $3^{\circ}$ Amine Oxide (or, in other words, Cope Reaction), I figured there are two possible products hexene and hex-2-ene. Therefore, hex-2-ene should be the major product as it is the more substituted alkene.
But the solution to the question states that the major product is Ethene ($\ce{H_2C=CH_2}$). I have no idea how that is even possible.
Here is how I came to the conclusion that hex-2-ene will be the major product (which is basically the standard Cope reaction mechanism):

Please provide clarification on this.

This looks like an error. You can get propene (probably the major product) and the two hexenes you mentioned, but I can see no way of getting ethene.

@Waylander This must be an error then. Errors are not rare in organic chemistry question banks, or at least, that is the case with my book.

Hex-1-ene would be the major product. There are more no. of hydrogens on the 1° carbon. Therefore, at a particular instant, hydrogens of this carbon would be antiperiplanar to O- in more number of molecules of the sample. Hence the major product


text:
Does the rate reaction of clock reaction increase over time?
I'm studing the following version of iodine clock reaction, in which, $[\ce{H_2O_2}]_0$ > [ascorbic acid]$_0$
$$\ce{I^- + H_2O_2 -> I_2}$$
$$\ce{I2 + ascorbic acid -> I- + side products}$$
When the concentration of ascorbic acid is zero, also known as induction period,
$$\ce{I2 + I- -> I3-}$$
$$\ce{I3- + starch -> blue complex}$$
During the first moment I thought that due to the abrupt occurrence of the product, the rate reaction must increase after the induction period. But after doing some research I realized that the occurrence of the product is actually due to the stoichiometric constraint of ascorbic acid.
So, now I think that the rate reaction of a clock reaction decreases over time and that the concentration of products only increases due to the stoichiometric constraint. Is this correct?

Hi, welcome to Chemistry.SE! Could you please edit your question and format the chemical formulae and equations with MathJax, so that they are more readable? A handy guide for MathJax on this site is here: https://chemistry.meta.stackexchange.com/questions/86/faq-how-can-i-format-math-chemistry-expressions-on-chemistry-stack-exchange

Done it, thank you for the advise @ShoubhikRMaiti ^^

You can write a whole formula inside one $\ce{}$ to get this  $\ce{I2 + I- -> I3-}$ is displayed as $\ce{I2 + I- -> I3-}$ $$   $$ gives you the center aligned text, $$\ce{I2}$$

What happens in detail of course depends on the rate constants and concentrations but with similar rate constants for reaction of $I^-$ and $I_2$ the iodine ($I_2$) concentration rapidly rises and then falls more slowly to virtually zero, the $I^-_3$ gradually increases until it reaches a steady value. The reaction stops when all the iodine is consumed by forming the product.


text:
Calculating absolute internal energy of an ideal gas system
In the following derivation, I just calculated the absolute internal energy of an ideal gas as a function of its temperature. But my teacher said that we can never calculate the absolute internal energy in thermodynamics. Then how is the below proof correct?
Proof to calculate absolute internal energy:

We know that from the law of equipartition of energy, average energy ($E_\mathrm{avg}$) can be written as,
\begin{align}
E_\mathrm{avg} &= \frac{f}{2}k_\mathrm bT \\
E_\mathrm{avg}(n\, \mathrm{mol}) &= nN_A\frac{f}{2}k_\mathrm bRT \\
&=\frac{f}{2}nRT \tag{1}
\end{align}
Therefore, average energy for n moles of gas molecules is $\frac{f}{2}nRT$.
In case of an ideal gas,
$$E_\mathrm{total} = K.E$$
Since there are no interactive forces, $P.E = 0$ and other energies are negligible.
Therefore internal energy is equal to the kinetic energy.
From (1), we get:
$$U_\mathrm{ideal} = K.E_\mathrm{ideal} = \frac{f}{2}nRT$$

"other energies are negligible. Therefore internal energy is equal to the kinetic energy." - that's not true, by any means - it's the kinetic energy atom that is negligible in comparison with energies inherent to atom itself.

It is true using classical thermodynamics that $\Delta U = q+w$  so changes only. However, thermodynamics predates our understanding of what molecules are and using Statistical Mechanics we can find the internal energy as you have done without using classical thermodynamics. See the link above for some more info in my comment there.

So, is it actually possible to calculate absolute internal energy or did I go wrong somewhere? Please answer @porphyrin. Yes or No?

I am talking about an ideal gas.

So therefore there is only kinetic energy to consider and for an ideal gas the internal energy is proportional to the temperature as you have found (from kinetic theory of gasses) so 'yes', but not from classical thermodynamics so 'no' from this view point. So your teacher is correct.


text:
Is there a mathematical and rigorous way to prove whether a given function is a state function or a path function?
My textbooks have explained why Enthalpy, Entropy, Internal Energy, Gibbs Free Energy are state functions or path functions only intuitively and with examples. But that never fully satisfied me. Is there any mathematical way to prove whether a given function is a state function or path function?

Yes, perform a path integral over a closed path. The resulting integral should be zero for a state function, independently of path. Not so for the path function. See https://chemistry.stackexchange.com/questions/22171/why-is-du-an-exact-differential-and-dq-an-inexact-differential/22172#22172


text:
Why is degree of dissociation used to estimate strength of an acid?
Problem

Which of the following will give strength of acid?
A) pH
B) Normality
C) Degree of dissociation
D) Dissociation constant

Answer

 C, D

Question
If pH is not considered as a metric to assess the strength of an acid because it depends on dilution (even strong acids can have high pH), then why is degree of dissociation a reliable metric for the same? Doesn't degree to dissociation depend on dilution as well?

It does. Also, some tests look like they were designed by morons. So it goes.

$\ce{HCl 0.001 M}$ and  $\ce{CH3COOH 0.055 M}$ have both a pH value equal $ 3$.

The acid strength could be considered in the absolute way, via dissociation constant, dilution independent, or by the relative way, via dissociation degree, dilution dependent. If at some dilution is for some weak acid dissociation degree approx. 100%, it can be considered at these conditions approx. as strong as strong acids.

I would answer only the constant of dissociation $K_a$, because it is a thermodynamical constant and you can calculate it from potentiometric measurements without dealing with activities and so on. If you have the $\Delta G$ of reaction, which is related with the difference of electric potential of the cell, you can calculate the constant as $\Delta G = -RT \ln K$. The pH is not an indicator of the acid strenght, but only its consequence. If you have a very diluite solution of sulfuric acid, its pH will be near to 7, but it does not mean that sulfuric acid is weak. Also the degree of dissociation can give you incorrect information if you are dealing with very diluite or very concentrate solutions.

Ivan may be right that the question was designed by morons, but an argument can be made that the strength of an acid governs the pH in a solution with a given concentration, rather than the pH governing the strength.  The question apparently asked what properties of the acid can be used to assess its strength and thus the predict the pH you get at a given concentration.

Then again, pH (option A) is a fairly good predictor of itself.

Ka can be calculated from deg. of diss. and conc.


text:
Total concentration conversion
when dealing with waste water and the total values of P, N, C are given in ppm units, what molecular weight is used for converting from ppm to molar?

for Example- what will be 12 ppm of P(T) in Molar unit?

What  (T) in P(T) means ?

T stands for Total

Then it is easy - use the phosphorus molar mass, about 31 g/mol, i.e. about 31 ppm = 1 mol P/L. But consider some P is not in form.of simple phosphate. I remember mineralisation of waste water by peroxodisulphate for further analysis.


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Zinc screw turning black
I tried one experiment. I started with ordinal yellow zinc screws.
The screws were sumberged in bath of two parts of $4.7\%$ solution of $\ce{NaClO}$ and one part $\ce{NaOH}$. The screws turned silverish in color. Then, I submerged them into vinegar. Vinegar turned them black. I am now wondering what actually happened. The bath of $\ce{NaClO}$ and $\ce{NaOH}$ is used to blacken the steel, but it doesn't work with zinc. The vinegar can be used to remove zinc.
But why it was black after that bath?
I was expecting that it would need another bath in $\ce{NaClO}$ and $\ce{NaOH}$ to become black.
Another control screw which wasn't in first bath was just stripped of zinc layer and got a steel color.

Yellow? Zinc is anything but yellow.

E.g. https://www.reliablefasteners.com/documents/docsGr/101/631/7/1016317/1407515_700.jpg

I still stand by my statement. Certain allows of zinc can look like that; zinc alone can't.

Your zinc is probably an alloy called brass, which is yellow, And brass is made of half zinc, half copper. When dipped into a  $\ce{NaClO}$ solution, the copper atoms are oxidized, and transformed into copper oxide $\ce{CuO}$ which is a silverish then black powder

The screws you show are brass. There are many brass alloy compositions, not just half copper and half zinc. You might want to look up brass in wikipedia.

By the way, what is the point of this? Are you testing some hypothesis? If not, there are an enormous number of things you could put the screws to, but to what end?

As has been said, they may be brass or silicon bronze, but similar hardware is often steel with a thin zinc plating. It's hard to give you an answer when the description of the screw is poor.

The screws, with Robertson heads, are brass, not anodized zinc. Galvanized screws are not in need of anodization: they get screwed into wood and are not decorative items. Not worth the cost. Take a file and deeply scratch through the surface to see if the inside looks like the outside. Or use a hacksaw to cut one of the screws in half.

One last thing: are the screws attracted to a magnet? If so, they are steel with coating(s). Otherwise, not.

These screws are steel with thin cover of zinc with colored oxide layer, probably done by anodizing, I think. At first you dissolved zinc oxide with NaOH. This removed the colored layer and exposed zinc, making it look metallic. Then you reacted zinc with vinegar. Zinc itself is more reactive with acids. The zinc acetate that could be forming is transparent, so that doesnt explain the black color. If black color is present everywhere in solution, it could be that zinc was not pure, and as zinc was being chipped away by vinegar impurities colored the solution black. Or, if black was present only on the screw itself it could be that steel formed iron oxide on its surface. In both cases black is likely non zinc metal oxide. Most likely copper or iron.


text:
To calculate total vapour pressure of a solution

Question:
1 mole each of A and B were mixed to obtain an ideal solution of which one mole went to the vapour phase at equilibrium.
If $P°_\mathrm{A} = \pu{100 torr}$ and $P°_\mathrm{B} = \pu{900 torr}$, then $P_\mathrm{S}$ would be?

For the above question, I first calculated vapour pressure of A and B which came out to be $\pu{50 torr}$ and $\pu{450 torr}$ respectively (using $P°_\mathrm{A} × X_\mathrm{A}$ where $X_\mathrm{A}$ is mole fraction of A) and then I calculated vapour pressure for A and B in vapour phase which came out to be $0.1$ and $0.9$ respectively, so the amount remaining in the liquid phase will be $\pu{0.9 mole}$ for A and $\pu{0.1 mole}$ for B, according to which the $P_\mathrm{S}$ should be $\pu{180 torr}$, but the answer provided is $\pu{300 torr}$, am I missing something? (also see the image below for better understanding of my approach).
Any help is appreciated! Thanks in advance!

Could you please type out what you have written instead of posting an image? This is because images are not searchable, and harder to read.

I'm really sorry for that, actually i am pretty bad with message formatting, and i was running out of time so i decided to add an image instead, i'll redo that with a pen and replace the image as i get time. Thanks for pointing out.

Images are discouraged. Please use mathjax/mhchem, see Notation basics and How can I format math/chemistry expressions on Chemistry SE. See also Math SE MathJax tutorial.

The starting point to solve the problem is Raoult's law. For each component we can write that the vapour pressure (related by Dalton's law to the total pressure) is equal to the product of the mole fraction in the solution and the vapour pressure of the pure component:
$$y_iP_S = \chi _i P^{\circ} _i$$
You can exploit the fact that the molar amounts in the problem are particularly tidy:
\begin{align}
  n_{1l} + n_{1g} &=1 \\
  n_{2l} + n_{2g} &=1 \\
  n_{1g} + n_{2g} &=1
\end{align}
So that for the first component:
\begin{align}
  y_1 &= n_{1g} \\
  \chi _1 &= n_{1l}  \\
  \rightarrow n_{1g}P_S &= n_{1l}P^{\circ}_1 \\
  \rightarrow n_{1g}P_S &= (1-n_{1g})P^{\circ}_1  \\
  \rightarrow P_S &= \frac{(1-n_{1g})P^{\circ}_1}{n_{1g}}
    = \frac{n_{2g}P^{\circ}_1}{(1-n_{2g})}
    \tag{1}\label{eq:firstcomponent}
\end{align}
For the second component we can similarly write
\begin{align}
  y_2 &= n_{2g} \\
  \chi_2 &= n_{2l}  \\
  \rightarrow n_{2g}P_S &= n_{2l}P^{\circ}_2 \\
  \rightarrow n_{2g}P_S &=(1-n_{2g})P^{\circ}_2  \\
  \rightarrow P_S &=\frac{(1-n_{2g})P^{\circ}_2}{n_{2g}}
    \tag{2}\label{eq:secondcomponent}
\end{align}
Equating the expressions for $P_S$ in \eqref{eq:firstcomponent} and \eqref{eq:secondcomponent} and rearranging terms we can write a quadratic equation in $n_{2g}$ whose solution can be inserted back into the expression for $P_S$.


text:
Why is sea salt coarse?
What makes sea salt coarser than other salt?
I have tried checking online but no one seems to discuss this. It seems that the grain of sea salt is larger, but I don't understand since the salt is dissolved in the sea water so how does it become coarse.

Hard to see what comparison you make. Is other salts a short for other table salts? By the way this would be an entire treatise of precipitation / crystallisation and they role in table salt production. Generally slower the formation bigger are the solid pieces one get.

Different production methods of salt will give different crystal types and sizes. Sea salt is usually produced by slow evaporation of salt ponds in the open air; some table salts will come from a much more controlled industrial precipitation process probably producing more regular, more compact crystals.

The method of production of sea salt usually involves bulk evaporation of seawater. This allows the salt to crystallise and due to it's bigger size it is sold without much processing. On the other hand, the table salt has much smaller size and is processed a lot more than we could assume before it is sold.
This could be one reason why sea salt has coarse texture.
See also: Maxine Builder and Maxine Builder. What's the Difference Between Kosher Salt and Sea Salt? (via the Internet Archive)
I was interested to answer this question because I have always thought that both the salts are just the same. The tiny size of salt grains just seems less coarse but it essentially is having the same texture is what I would believe.
But then I looked at sugar as well. The tiny grains of sugar or a little bigger sugar diamonds seem smooth compared to the rough textured bigger blocks of sugar. And so I was convinced that it was due to it being processed. But then think of powdered sugar? That appears to be smooth.

Thanks, but I have seen that river salt that is also made by bulk evaporation seems to have a different texture.  For instance, river salt from Australia is I quote "Produced using the pristine mineral brine water from the ancient saline aquifers of the Murray Darling Basin in Australia. These are light, flakey pink salt crystals. Not coarse Himalayan pink salt. So there seems to be something that makes the salt coarse.

@Avrohom I would suggest you should place your observations like you mentioned in this comment in a new structured question. Place your observations, possible differences between the two salts and a community member with expertise in this could help? We didn't realise the level of the question. I would believe the difference in texture is due to various minerals that different sources of salt could contain and hence alter the crystal lattice.

Please do not link through google, but instead provide a link to the target page itself. I have replaced the link with a more stable version to the Internet Archive.

Sugar can be obtained in various grain sizes, from 0.02 mm to 0.75 mm. https://www.whatsugar.com/refined-cane-sugar And then, of course, there's rock candy

Because it is not ground as fine , unless a customer wants finely ground sea salt. It is passed over screens to separate it into sizes after grinding, just like table salt.  Fun fact ; table salt that is too fine for traditional marketing ,is sold as "popcorn" salt.

Yes. Short answer to question is 'because nobody ground it up'.

I think it is a little more complex than this. For instance, river salt from Australia is I quote "Produced using the pristine mineral brine water from the ancient saline aquifers of the Murray Darling Basin in Australia. These are light, flakey pink salt crystals. Not coarse Himalayan pink salt. So there seems to be something that makes the salt coarse.

Sea salt is usually sold as coarse crystals because, as others have noted, it is separated by crystal size before packaging.  Sea salt, along with Kosher salt, is desirable by chefs and food enthusiasts because it allows one to see how much salt is actually put a piece of food, let's say a steak, without dissolving as smaller crystals would do.  And it gives a desirable crunchy mouth feel which many people like. With some foods it is desirable to add salinity to saliva as the food is about to be chewed.  This  causes more saliva secretion and enhances the food's flavor as it's broken down by chewing. It's important to know if you're using Kosher or sea salt or regular table salt when measuring salt for a recipe because salt with large crystals will take up the same volume but weigh less than salt with small crystals, such as common table salt. So a teaspoon of one contains less salt than a teaspoon of another.

But what makes the crystals to be larger or smaller


text:
Turning oil into solid for candles
I added stearic acid to hot vegetable oil and saw it turned to a sludgy, slush material in about 1.5 hours. Is there any other natural additive that I can add to make it more solid in the same amount of time?
I'm currently trying soy wax + stearic acid.
Are there any other natural alternatives?

i think hydrogenation is basically what im doing here, but I cant seem to find a way to strengthen the bonds beyond a slush. I dont have access to any heavy equipment

For hydrogenation you need to add hydrogen. Hydrogen gas. And then some.

And you need to describe your experiment better. What temperature, approximately? Still hot after 1.5h? What vegetable oil? Plus: what gave you the idea stearic acid could solidify oil? What are you trying to achieve? As is, I vote to close, an answer requires too much speculation.

What you did and especially what you want to do is totally unclear. Turn something into solid means, in everyday context, a lot of things. In chemistry almost nothing. So hard to say. Unless you want a margarine or a soap.

I dont think hydrogenation is at play in here. More likely you just get chunks of solid in liquid, thus sludge. Otherwise you would get more smooth material.
Similar processes that can make more solid-like material are gellatine and agar-agar, that make gel that holds everything, emulgators like egg that help to hold different materials together by making small droplets.
You could be expecting this process to happen, but stearic acid is already a result of this process, and you dont seem to add NaOH:
https://en.m.wikipedia.org/wiki/Saponification


text:
Why do I see two peaks for methyl salicylate in my GC–MS spectrum?
I ran a neat sample of methyl salicylate (CAS: 119-36-8) on the GC–MS and used $m/z~120$ as the quantifier and $m/z~92$ and $152$ as qualifiers. Below is the SIM spectrum for these ions, with two distinct peaks.

Both peaks are identified by the NIST spectrum database as that of methyl salicylate. Considering this is a neat sample and is not contaminated or diluted, why do I see two peaks in my spectrum?
GC parameters: $\pu{60 °C min^-1}$ ramp to $\pu{230 °C}.$

+1 But I don't understand if both peaks are identified by NIST as methyl salicylate, then what is the problem?

Methyl salicylate has a molar mass $152$. I don't see how to justify the mass $120$. $121$ corresponds to a loss of $\ce{CH_3O}$. But $120$ ?

I offer the following explanation for the fragments of the mass spectrum of methyl salicylate (1, m/z 152). Loss of methoxy radical affords acylium ion 2 (m/z 121). Subsequent loss of a proton from 2 gives rise to the base peak 6-oxomethylidenecyclohexa-2,4-dien-1-one (3). Loss of carbon monoxide from 3 produces carbene 4, which rearranges to cyclopenta-2,4-dien-1-ylidenemethanone (5).

Is there a reference or are you proposing it ?

@M. Farooq: This was my proposal. It is in agreement with this paper. DOI:10.1021/jo5011087

The short answer: Do not rely on the SIM trace alone.  Amplify the representation of the trace about the TIC (total ion current) until you see a reasonable peak  intensity at the same retention time as in your SIM traces (perhaps a time spectrum, but not a mass spectrum).  Then integrate with the peak-shaped cursor (currently shown in the magenta coloured trace of one SIM) in the trace of the TIC and equally consult the then displayed EI scan at this time interval (in part already seen at the lower part of your photo) to assess the mass spectrum.  Compare this to a trustworthy reference spectrum; e.g. an authentic sample; your spectrometer's library; NIST, SDBS, etc.
The longer answer:
Your experiment coupled a separation technique, either GC, with an analytical, MS.  Often, the mass resolution of these simpler, less-boxy bench-top MS hyphenated to the chromatography (i.e., not a tandem MS-MS detector) is low compared to, e.g., a sector field mass spectrometer.  Thus, observing the SIM traces alone is more like asking «tell me if whatever ion passes the detector with a $m/z = 119.2$», or even worse «tell me if any ion passes the detector in the range of $m/z = 119 - 121$.  You will see this approach often used (sometimes misused) in HPLC-MS if one is not sure about the setup's sensitivity for $\ce{M+}$ vs. an adduct like $\ce{[M + H]+}$.  Or because the software allows you to check for an $\ce{NH+_4}$-adduct after the chromatography run is recorded because your eluent included an ammonium buffer.
Only if you couple the information by the SIM with the integration in the TIC to yield a complete mass spectrum you may collect additional evidence that the peak in the SIM may belong to your target compound.  Otherwise, it could be (but need not to be) caused by an ion which accidentally shares the same ratio of $m/z$ within the setup's resolution in time (think elution of chromatography, scan rate of the MS), and mass (the MS) to your fragment in question; but shows up regardless if it belongs to the defragmentation pattern of your target compound, or not.  Without additional inspection of the spectra it equally may be an impurity, e.g. the sample just before yours was too concentrated and some is eluated (and subsequently characterized by the MS) now altogether with your sample.
Without the separate inspection of these integrated spectra around $t_R = \pu{4.1 min}$ and $\pu{5.1 min}$, the present form of your question does not allow to answer if one of these peaks (and if so, which one) safely may be attributed to methyl salicylate.  A mixture of isomers of compounds, e.g. (cis)/(trans) dienes may elute at different times and -- depending of the $m/z$-ratio you set for the SIM -- equally yield two peaks in the SIM but distinctly different mass spectra.

Neat or pure for organic chemists has a different meaning for analytical chemists. What does your manufacturer label say? 100% pure methyl salicylate? That is not possible. It must have closely related impurities. Did you try a thin layer chromatography experiment with this sample?
There is no concept of 100% pure substance in analytical chemistry, especially if you are doing GC.  MS is so sensitive that it can detect parts per billion level of impurities.
I gather you are concerned about the two peaks in the total ion chromatogram (TIC, top), between 4 and 6 min, almost invisible in your picture?
It seems that a very closely related impurity is generating the same fragments and this is the source of two peaks.
Also did you notice what what m/z 152, 120 and 92 correspond to?
Some more things to worry about:
a) Did you wash the injection needle very well?
b) Did you run the blanks, inject air (possibly)?
c) Are the peak positions reproducible in retention time? Sometimes there are carry over problems or peaks appear from the previous run if you did not wait long enough.

While I like this answer for its logic, it contains seven questions for OP that must've been asked in the comments before posting an answer. SE sites are not forums, so please refrain from asking anything that requires an external input in your answers. The answer should be self-contained and never require additional input from OP. Failing to do so yields an avalanche of comments below your answer, which are eventually either deleted or moved to chat, decreasing the value of the answer.


text:
Why is the ether linkage breaking here?
I have been asked to find the major product of the following reaction:

This is my thought process: the lone pairs of the oxygen atom connected to $\ce{-Et}$ are less delocalised than the other oxygen atom. So, that will abstract the $\ce{H^+}$ from $\ce{HI}$ leaving behind $\ce{I^-}$ and producing a positive charge on the oxygen atom. Now the oxygen atom will take up the bond pair electrons and $\ce{I^-}$ will form a bond in $\ce{S_{N}2}$ mechanism, producing $\ce{I-Ph-O-Ph}$ and $\ce{EtOH}$ as products.

But, the answer in the book states that:

The ether linkage will break as it does not have a double bond character due to the absence of resonance.

Why is the ether linkage breaking here? What does it mean by "it does not have a double bond character"?

Having double bond character makes a bond stronger so the O-CH2R bond is weaker than the Ar-O bond. There are also steric factors to consider - Iodide is a bulky nucleophile and the approach to the O-CH2R is less hindered.

$\mathrm{S_N2}$ reaction does not happen on $\mathrm{sp^2}$ centres, that is why your proposed reaction scheme would not work. See the accepted answer for this question: https://chemistry.stackexchange.com/questions/14482/why-do-sn1-and-sn2-reactions-not-occur-at-sp2-centres

I have used the image you used $2nd$, see carefully the markings.
The $C-O$ bond with benzene ring is stronger than the $C-O$ bond of ethyl, as there is resonance with benzene making it partially a double bond. So, that $C-O$ is difficult to break compared to the $C-O$ bond in ethyl branch.

Yes C-O bond has partial double bond character, but I don't believe that is the main reason. Even if you have a leaving group like $\ce{I-}$ you would be hard pressed to achieve the reaction, because $\mathrm{S_N2}$ doesn't work with $\mathrm{sp^2}$ carbon centres. On aromatic rings, only $\mathrm{S_NAr}$ can happen.

Yes, SN2 do not occur at sp2 Carbon because vinylic substitution is hindered due to π electrons but with benzene there is more, which is lone pair donation to benzene ring from Oxygen, that is why i focused on double bond character


text:
Why does malleability and ductility of interstitial carbides decrease?
Why does the malleability and ductility of interstitial carbides decrease? I read the following in my book, but I am not able to find what is the reason for the same. Why does this happen? Is it to do with the change in lattice properties?
I read this in JD Lee (adapted version by Sudarshan Guha), it stated after a section on interstitial carbides

The hardness and density are increased for these (interstitial) carbides while malleability and ductility are decreased.

I read that in JD Lee (adapted version by Sudarshan Guha), it stated after a section on interstitial carbides, "The hardness and density are increased for these (interstitial) carbides while malleability and ductility are decreased"

In order for a metal to be malleable or ductile, metal atoms must move relative to each other. The carbon atoms intercalate into the metal lattice and form more bonds between the metal atoms (via the bridging C). Can you see why that might affect malleability and ductility?

@Andrew Yeah makes sense to me!


text:
What happens inside salt bridges?
I don't understand how a salt bridge work exactly: what happens inside it and how it affect the cell's lifetime.
I have already read:
Salt Bridges/Porous Disks
Salinic bridges in voltaic cells
Difficulties understanding how salt bridge works
What I understand:
The salt bridge's goal is to maintain electro-neutrality in each of the half-cells so that the voltaic cell keeps working and it helps avoiding the mixing of the half-cells' ions in solutions.
When the salt bridge is being used in the cell, its ions (anions and cations which constitute the salt bridge's electrolyte) are moving towards the half-cell that attracts them.
What I don't understand:

Do the salt bridge's ions leave the salt bridge to go all the way within the hall-cells' solutions or do they stay at the border ? Is it the case in all bridges ?
Do the half-cells' ions move inside the salt bridge or do they stay in the salt bridge at the border solution/salt-bridge ? Is it the case in all bridges ? And is it in the same proportions as the salt bridge's ions ?
How do the ions move exactly in the salt-bridge ? It seems like the propagation in, let's say agar-agar, is much slower than the displacement of ions in the liquid solutions in the half-cells, is it not a problem for the whole cell ? Is it like a longitudinal wave, ions pushing others ? If so, are they two speeds of propagation (the wave speed and the ions speed) ?
If ions are moving inside it, we can consider that there is a current in the salt bridge, correct ?
Are they families of salt bridges depending on how they work ?
Finally, can the salt bridge be exhausted before the half-cells (whether the solutions or the electrodes) ? It seems the salt bridge is quite small in basic cells examples compared to the sheer volumes of the solutions and sizes of the electrode, thus containing much fewer ions.

Ions leave the salt bridge and move further, without accumulating at the bridge end. For all salt bridges where there is no other reactions involved, so all simple salts in solution or molten salts.
Ions move all the way through. There is some concentration difference in salt bridge and main half cell volume. It is true for simple salt solutions and molten salts. Same ions move through half cell and salt bridge.
The speed of a wave is on the order of 100 000 km/s. The speed of ions moving is on the order of 0.001 mm/s. You can estimate the ion speed from calculating amount of ions participating in a reaction per second and dimensions of the bridge and ion density in the solution or molten salt.
Yes. But it is important to remember that there are two cancelling each other currents in the salt bridge. Thus magnetic field is close to zero.
Most important is a number of ion types. If there are just two types, such as in solutions and molten salts, then this intuition applies. If more ion types are present, and especially if ions can convert between types, things will get more complex. Another option to consider is a fuel cells where only one type of ions passes through, and electrons use outside route to complete the loop. So, cases with one negative and one positive ion types are considered here. All other collections of ions will behave very differently.
No. Ions are constantly being replenished by the bigger volume of the both half cells.

Electroneutrality strictly speaking is not conserved when the bigger volume than an atom is considered and precise measurments are available. There is a gradient of potential due to different concentration of ions. Concentration of ions is the reason for the effect you are interested in. Important parts to consider are electrodes, where ions react, thus dissapearing from what we calculate, and narrow areas, where ions' speed increase, such as in salt bridge.

Are you sure about #2 and #6 ? Salt bridges are prepared with $\ce{KNO3}$ or $\ce{KCl}$ or some inert salt like that, in a way that both cation and anion has similar mobility through the agar-agar. If the ions from solution could get into the salt bridge, there would be no need for all of that.

@ShoubhikRMaiti I am sure. How else would conductivity work? Gel is beneficial because ions also move a lot of other material around, as they are pulled through the liquid. Without gel ions would grab a lot of neutral solution nearby. Ratio of ion movement and neutral matter movement can be as high as x1000, due to friction.

@SurprisedSeagull, thank you for your excellent answer, it explains almost everything. However I have two small issues before validating the answer: (a) if the ions from one half-cell go all the way in the salt bridge, they may eventually end up in the other half-cell (11,5 days for a one-meter bridge at 1μm/s): is it not what we are trying to avoid with a salt bridge: the half-cells' ions producing unwanted reactions together ? (b) Also, could you edit your answer (I cannot yet) to use the Greek letter μ instead of u for the speed so it does not confuse people reading it later ? Thanks.

Changed to mm, that I hope everyone knows. We would want to avoid side reactions, but we cant. We just accept that some contamination will inevitably occur. If you want to reduce contamination, you use inert electrodes.

Here are the answers :

The ions form the salt bridge must get out of this bridge. But they usually stays at the border, because they push other ions already present in the solution to compensate for the new charge having appeared on the electrode, like a locomotive or an engine pushes its trucks on the other end of the train. It is the same in all bridges.
The same movement appears in the salt bridge. The "train" of similar ions are pushed as a whole through the bridge.
There is a current of ions through the bridge, exactly like the current of electrons in the outer circuit.
The bridge cannot be exhausted, because it gets the same number of ions from one electrode as what it looses on the other electrode.
The bridge must be as small as possible, if the cell has to deliver a large current.

Thank you for your answer. I will however probably validate @Surprised Seagull's answer since it is a more exhaustive.

If we have $\ce{ Zn/Cu}$ cell with $\ce{ ZnSO4/CuSO4}$ electrolyte and $\ce{ KNO3}$ salt bridge and we run the cell . 1- Do $\ce{ K^+}$ travel from the  salt bridge to the solution around the cathode to neutralize $\ce{SO4^{2-}}$  2- Do  $\ce{ Zn^{2+}}$  travel from the  solution around the anode to the solution around the cathode through the salt bridge 3-  Do $\ce{ SO4^{2-}}$ enter the bridge salt? 4-Do $\ce{NO3-}$ migrate from salt bridge to the solution around the anode to neutralize $\ce{Zn^{2+}}$

In the $\ce{Zn/Cu}$ cell (made with sulfates of these metals), the cathode loses positive ions ($\ce{Cu^{2+}}$. As the number of negative ions are not modified, something must be made to compensate this excess of negative ions. First these ions $\ce{SO4^{2-}}$ may pass into the bridge was they are attracted by the excess of positive charges on the other side of the bridge. Second, some positive ions ($\ce{K+}$ from the bridge) may move into the cathode zone. Both phenomena are occurring simultaneously. It is difficult to state which is the most important. it is a question of ionic mobility.

On the other hand, $\ce{Zn^{2+}}$ ions do enter the bridge, pusj^hing in front of them $\ce{K+}$ ions. So it is highly improbable that the $\ce{Zn^{2+}}$ ions can cross the whole bridge, at least in a reasonable amount of time.  Then, as I told you before, $\ce{SO4^{2+}}$ do also enter the bridge. And finally,  nitrate ions form the bridge do get out of the bridge to compensate for the excess amount of $\ce{Zn^{2+}}$ ions appearing in the anode region.

@MauriceThanks for your useful and plausible answer . Do $\ce{SO4^{2-}}$ ions travel from the cathode zone  the anode zone to compensate for the excess amount of  $\ce{Zn^{2+}}$ ions appearing in the anode  through the salt bridge?

Not exactly. The $\ce{SO4^{2-}}$ ions releasing the cathode zone and will enter the bridge. But here, they first push the nitrate ions which are already here. On the other side of the bridge, one or two nitrate ions will enter the anode zone. Later one, another sulfate ion make the same and push the precedent sulfate ion plus the other nitrate ions. After a very long time, all nitrate ions have been pushed into the anode region. And then, but only then, the sulfate ions will be able to enter the anode region.

This is just about question number 6. The answer by Maurice does not address it, and the answer by Surprised Seagull is a bit short.

[OP] 6. Finally, can the salt bridge be exhausted before the half-cells (whether the solutions or the electrodes)?

Most salt bridges start out with a higher concentration of ions than that of the electrolytes in the half cells. This ensures efficient charge transfer, and minimizes electrolyte traveling through the salt bridge. Even if you run the cell for a long time, the concentration of ions will never go to zero; if the concentration of ions in the salt bridge is lower than in the half cells, diffusion will transport ions into the salt bridge. You could even start with a salt bridge that contains no ions, and you would still be able to detect a current.

[OP] It seems the salt bridge is quite small in basic cells examples compared to the sheer volumes of the solutions and sizes of the electrode, thus containing much fewer ions.

I don't think salt bridges are used in applications where you want a lot of charge transport (i.e. such as a battery). They are used to study electrode potentials, with very low currents sufficient for measuring a voltage.
Why is a large concentration of inert ions good to have in a salt bridge?
If the ions in the salt bridge are effective in transporting charge, you won't have a large electric potential at the salt bridge that would speed up the mixing of the two electrolyte solutions. Eventually, they will mix anyway (through diffusion). Once electrolytes start mixing, the redox reaction can happen directly (not sending electrons through the wire), so that will lower the voltage measured between electrodes. Having a high ion concentration in the salt bridge and having a gel-based salt bridge (to minimize convection mixing) slows down the process of redox-active ions making their way through the salt bridge.

"you won't have a large electric potential at the salt bridge that would speed up the mixing of the two electrolyte solutions"
meaning that ;you will have a small  electric potential at the salt bridge that would slow the speed of  the mixing of the two electrolyte solutions"

Explain" convection mixing"

"redox-active ions" meaning that decrease the movement of anions at the cathod toward the anod through the salt bridge

Convection is when an entire part of the solution moves, like it does in a stirred solution. Even without stirring, there might be temperature fluctuations resulting in density fluctuations that cause bulk flow that is faster than diffusion. The agar prevents this from happening.

If a copper ion is supposed to be reduced at the copper electrode but travels to the zinc electrode via the salt bridge instead, it could react directly without electrons going through the wire.

THANKS  What about the first comment ? Is it the same meaning ?

@AdnanAL-Amleh Yes, you are trying to minimize the gradient of net charge between the electrodes by making ion transport fast. If you remove the salt bridge alltogether, the redox reactions stops very quickly.

Let us continue this discussion in chat.


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Why does Red Phosphorus react so violently with an oxidizer like Potassium Chlorate?
In Armstrong's Mixture, mixing those two chemicals results in a friction sensitive explosive which produces a loud bang. According to wikipedia, it is made by adding a strong oxidizer to red phosphorus, but what is the explanation as to why it reacts so strongly? Thank you

Because Phosporus is very easily oxidised and the reaction releases a lot of energy. P-P bonds are weak (200 Kj/mol) and easily broken to initiate the reaction

$\ce{P-P}$ bonds are very weak due to high angle strain. This makes phosphorus very reactive. White phosphorus is the most reactive but might not be used because it can release too much energy.


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How would sodium amide react with a non-terminal alkyne?
The reaction of a terminal alkyne with $\ce{NaNH2}$ leads to the loss of hydrogen ion from the terminal carbon.
But what will happen if the alkyne is a non-terminal one? How would $\ce{NH2-}$ attack then? Would any reaction even happen?
As an example, what would happen in case of this following reaction:
$$\ce{H3C-C#C-CH3 + NaNH2 ->[\text{liq.}~NH3] ?}$$

I don't believe if any major reaction would take place. This species isn't acidic enough.

One of the questions I did asked which of the following compounds would give optically inactive compounds with the set of reagents I gave above, and this was one of the options. It had marked this option as one of the correct ones. So essentially there is no reaction taking place and the product will have the same stereochemistry as the original reactant right? This is what I thought so, a reaction didn't seem reasonable

The answer depends upon the conditions of the reaction. In the presence of NH3 as solvent it is unlikely that alkyle-allene-alkyne isomerization will occur. Under the correct conditions the  propargylic hydrogen is acidic enough for NaNH2 to form the anion reversibly and in low concentration and effect isomerization to the allene and ultimately to the anion of but-1-yne. This zipper reaction is a variant of the Varrentrapp reaction. http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/Varrentrapp/varrentrapp.html

@NilayGhosh thanks! Seems like a possible dupe, I'll close this question once I confirm the answer

@user55119 possibly that is so the case. How does $\ce{NH3}$ as a solvent affect it though ?

@Ashish: Probably adversely.


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Why use Km in catalytic efficiency?
The catalytic efficiency of an enzyme is given by $k_{cat}/k_M$ where $k_{cat}$ is the turnover number, or the number of molecules that can be produced per second per active site of an enzyme.
$K_{M}$ is a measure of the affinity of the enzyme with the substrate, or the likelihood of binding.
Why bother dividing the $k_{cat}$ by $K_{M}$? Isn't the affinity of the enzyme already encoded into the quantity of $k_{cat}$? How could you be an enzyme that has low affinity, but still have a huge turnover? To me this doesn't seem possible, and thus it is redundant to divide by $K_{M}$. Likewise, could there be a situation where $k_{cat}$ is low, but $K_{M}$ is high?

The most basic kinetic scheme for enzymes is represented as
$$\ce{E + S <=>[K_m] ES ->[k_{cat}] E + P}$$
As should be clear, the $k_{cat}$ is the rate constant for the reaction that occurs after substrate is bound to the enzyme. The resulting rate (kcat[E]tot) is only achieved when every molecule of enzyme essentially always is in the act of converting substrate to product. That is, every time a product molecule is released, a substrate molecule immediately binds. Even if the substrate molecule dissociates before reacting, another immediately takes its place. We describe this situation as the enzyme being "saturated" with substrate.
The Km is a measure of how tightly the substrate binds to the enzyme, approximately equal to the equilibrium constant for the dissociation of the substrate from the enzyme. If the substrate is at a low concentration relative to this Km value, then many of the enzyme molecules will not have substrate molecules bound to them and will be unproductive as a result. The overall rate will be substantially below the maximum  kcat[E]tot.
All of this is captured in the basic Michaelis-Menten kinetics equation:
$$Rate = \frac{k_{cat}[E]_{tot}[S]}{K_m + [S]}$$
You can see that the impact of $K_m$ on the rate increases substantially as [S] decreases relative to $K_m$.
Since most substrates exist physiologically at concentrations below what is required for maximum rate, it is the combination of $k_{cat}$ and $K_m$ relative to [S] that determines the in vivo rate of reaction (along with the amount of enzyme of course).

how did you get kcat[E]tot? Is that just by assuming [ES]=[E]tot?

But besides that, I think I get it. Km and Kcat are controlling different aspects of the reaction. Km is for the part 1 of the reaction and Kcat is for the second part. Is this a correct understanding?

Yes, when the enzyme is saturated, [ES] is effectively equal to [E]tot.

The turnover number or catalytic constant $ k_{\mathrm{cat}}$ in the Michaelis-Menten model is the rate constant for the productive dissociation of intermediate $\ce{ES}$:
$$\nu = k_{\mathrm{cat}}[\ce{ES}]$$
The constant  $k_{\mathrm{cat}}$ says how much product forms from intermediate but does not say how much intermediate forms in the first place. It is assumed that there is a rapid equilibrium between enzyme, substrate and intermediate:
$$\ce{E + S <=>[fast] ES}$$
that can be described by either an association or a dissociation constant for the equilibrium between intermediate, apo enzyme and substrate:
$$K_\mathrm{a}=K_\mathrm{d}^{-1}=K_\mathrm{M}^{-1}=\frac{[\ce{ES}]}{[\ce{E}][\ce{S}]}$$
At low substrate concentration $[\ce{E}]\approx [\ce{E}]_0$ and
$$[\ce{ES}] \approx \frac{[\ce{E}]_0[\ce{S}]}{K_\mathrm{M}}$$
so that
$$\nu \approx k_{\mathrm{cat}}K_\mathrm{a} [\ce{E}]_0[\ce{S}] = \frac{k_{\mathrm{cat}}}{K_\mathrm{M}} [\ce{E}]_0[\ce{S}]$$
In summary, the reason the ratio might look strange is because $K_\mathrm{M}$ is a dissociation constant, and $k_{\mathrm{cat}}/K_\mathrm{M}$ is the rate constant for low substrate concentration.

I think when you wrote “early in the reaction” you meant “at low substrate concentration”? Otherwise it doesn’t really make sense.

Yes, you are totally right, thank you.

[OP] The catalytic efficiency of an enzyme is given by $k_\mathrm{cat}/K_\mathrm{M}$ where $k_\mathrm{cat}$ is the turnover number, or the number of molecules that can be produced per second per active site of an enzyme.

The last part is not quite accurate. $k_\mathrm{cat}$ is the rate of the reaction under saturating conditions divided by the enzyme concentration. The dimensions are one divided by time (a first-order rate constant).

[OP] $K_\mathrm{M}$ is a measure of the affinity of the enzyme with the substrate, or the likelihood of binding.

The Michaelis-Menten constant $K_\mathrm{M}$ has a rigorous definition based on rate constants, and its dimensions are the same as those of a concentration. If you interpret $K_\mathrm{M}$ as the affinitiy of the enzyme to the substrate, you have to know that higher values of $K_\mathrm{M}$ correspond to lower degree of binding. The likelihood of binding strongly depends on the concentration of substrate. For discussion of the catalytic efficiency, we are interested in substrate concentrations lower than $K_\mathrm{M}$.

Why bother dividing the $k_\mathrm{cat}$ by $K_\mathrm{M}$? Isn't the affinity of the enzyme already encoded into the quantity of $k_\mathrm{cat}$? How could you be an enzyme that has low affinity, but still have a huge turnover? To me this doesn't seem possible, and thus it is redundant to divide by $K_\mathrm{M}$. Likewise, could there be a situation where $k_\mathrm{cat}$ is low, but $K_\mathrm{M}$ is high?

Here are three examples showing rate vs. substrate concentration. Let's say the red curve shows kinetics of a given enzyme. If we compare the red enzyme to one (in green) that has the same $K_\mathrm{M}$ but a $k_\mathrm{cat}$ smaller by a factor of two, the rate is half the "red" rate at all concentrations. On the other hand, the blue enzyme has the same $k_\mathrm{cat}$ as the red enzyme, but twice the $K_\mathrm{M}$ (remember, this means it is hard to get the substrate to bind). At high substrate concentration, red and blue show the same miaximal rate, but at low concentrations, the "red reaction" is twice as fast as the "blue reaction".
In fact, the green and the blue enzyme show identical behavior at low substrate concentrations because $\frac{k_\mathrm{cat}}{K_\mathrm{M}}$ for the two enzymes is the same. That is the idea of catalytic efficiency.

image source: https://www.desmos.com/calculator/ue4eeg4xyh
Try it yourself
Here is a tool where you can change $K_\mathrm{M}$ (abbreviated as "K" in the tool) and $k_\mathrm{cat}$ (via $v_\mathrm{max}$ abbreviated as "v") to explore how the graph of rate (abbreviated as "y") vs. substrate concentration (abbreviated as "x") changes while varying those two parameters: https://www.desmos.com/calculator/egmxfzxct8


text:
HOMO and LUMO Conjugated System

The particle on a ring is a useful model for the motion of electrons around the porphine ring, the conjugated macrocycle that forms the structural basis of the heme group and the chlorophylls. We may treat the group as a circular ring of radius $\pu{440 pm}$, with 20 electrons in the conjugated system moving along the perimeter of the ring. Assume that in the ground state of the molecule quantized each level is occupied by two electrons.
(a) Calculate the energy and angular momentum of an electron in the highest occupied level.
(b) Calculate the frequency of radiation that can induce a transition between the highest occupied and lowest unoccupied levels.

How do you calculate HOMO and LUMO given that 20 $\pi$ electrons exist? Writing the electron configuration of the ground state molecule seems too lengthy.

You don't need to look at the molecule at all, except to count the number of pi electrons, which you've already done. You only need the solutions to the particle on a ring system. It's not asking you to calculate the HOMO and the LUMO of the molecule, it's asking you to calculate the $n$-th and $(n+1)$-th stationary state of the particle on a ring.

if I want to use the following formula, then I need to know the quantum number of rotational motion. It is given that with 22 pi electrons it is supposedly five, but I do not know where this value came from.

Do you know the particle on a ring? And how to solve the Schrodinger equation to get the states and energies for that particular system?


text:
What is the difference between "vapour" and "gas"?
I have read the definition of vapour on Wikipedia, but I don't seem to have a clear idea about the difference between a vapour and a gas. I know what it means when it says, "gas at a temperature lower than its critical temperature".

In this graph, the blue curve, means the part where gaseous and liquid phase coexist, so did the scientists coined the term Vapour for only this reason so, that we can have orderly fashion of different states of matter ?
My question is Are we naming vapour and gas to the same thing just at different temperature or they are distinct in some other way too?
Are there some other things which justify that vapour and gas are actually different things than just a gas below and above critical temperature ?

To OP, yes vapour refer to a gas in equilibrium with its liquid (Indeed we have vapour pressure...) or -  as you said -  can be brought in that condition by mean of compression only. That is.

Beside the most upvoted answer (gas which is liquifiable by pressure), I heard another usage of vapor: that is colloid liquid in gaseous media. But I am not sure, if it is a correct terminology on English.

@peterh, Colloid consisting of a liquid dispersed in a gas is called aerosol. Vapor is a single phase.

There is even the complication of steam. Sorry to bring it up :(

Seems like that phase-diagram is mislabeled: the entire vapor/gas areas should be "gas", and then the gas-side of the phase-transition lines could also be called "vapor".  I get the feeling that whoever labeled that graph was thinking roughly that in labeling the entire area beneath the phase-transition line "vapor".

tl;dr–  "Gas" and "vapor" aren't mutually exclusive.  Generally:

a gas is any material that'd fill a volume to its boundaries; and

a vapor is a gas-like material that's associated with a condensed-state transition.

It's a bit misleading for a state-diagram to label a region "vapor" in a manner that might imply that a vapor's not a gas.

State-diagrams should refer to "gas", not "vapor".
Definitions:

A gas is material that would fill any volume up to the boundaries, ignoring gravity and the like.

A vapor is a gas or gas-like material that's associated with a condensed-state transition.

These aren't alternatives; something can be a gas or/and a vapor.  Examples of each combination:

Not vapor
Vapor

Not gas:
A rock.
Suspended droplets in fog.

Gas:
Helium.
Steam.

State-diagrams are primarily about states-of-matter, i.e. the gas/liquid/solid thing, so usually a state-diagram should label things as "gas", not "vapor".

We can talk about "vapor" to stress a relationship with a condensed phase.
Often we talk about

vapor pressure,

vapor/liquid equilibrium,

vaporization,

etc., in terms of "vapor", rather than "gas", to stress the relationship with the condensed phase.
To be clear: once something vaporizes, it's a gas.  That said, we might continue to refer to it as a vapor as well, stressing the physical closeness to the condensed state.

Discussion: Where would vapor be on a state-diagram?
Since "vapor" isn't a state-of-matter, it doesn't really have its own region in a state-diagram.  Instead, it'd overlap with the gas-state near boundaries with condensed states (such as liquid and solid states).
A gas would be most vapor-like right at a boundary with a condensed state.  The further a state is from a transition between a condensed-state and gas-state, the less vapor-like it'd be.

Comments are not for extended discussion; this conversation has been moved to chat. Please note that further comment may be deleted without further warnings.

Vapor is a much older word alluding to dampness and it was not coined by scientists. It is in use since the 1300s. The actual meaning of meaning of vapor is "Matter in the form of a steamy or imperceptible exhalation; esp. the form into which liquids are naturally converted by the action of a sufficient degree of heat. This is the original 13th century meaning of vapor. The German equivalent of vapor is Dampf (which is relatively clearer in terms of feeling- imagine entering a basement which is damp from water vapor). Gas is relatively a new word as per the unabridged Oxford English Dictionary, and it entered English in the 1600s.
In fact, the OED mentions the distinction "Gas: A substance in a state in which it expands freely to fill the whole of a container, having no fixed shape (unlike a solid) and no fixed volume (unlike a liquid); spec. (distinguished from a vapour) such a substance above a critical temperature such that it cannot be liquefied by the application of pressure alone"
You can also think of the general meanings of vapor in scientific usage:
A substance which is a liquid at room temperature or under ambient conditions, and if it is heated enough so that the liquid evaporates (same root of word), you can call that as vapor of that substance.
Examples: Mercury is a liquid metal, when it is heated enough in a an electric bulb, it becomes a gas, hence the name mercury vapor lamp. It is not called a mercury gas lamp.
It is not necessary that heat be involved. You will find the usage "cold vapor method for Hg analysis" again for mercury which means that Hg was a liquid in a sample but by blowing a gas over it, it has been evaporated.
The recent meaning of vapor from OED, "Vapor: In modern scientific use, a fluid that fills a space like a gas but, being below its critical temperature, can be liquefied by pressure alone."
This is clear enough. Now apply this distinction to your phase diagram and notice the position of the critical point. The vapor "state" is below the critical point.

But nobody calls CO2 nor N2O vapour.

Well, perhaps because nobody usually works with liquified carbon dioxide and liquid carbon dioxide does not exist under ambient conditions. But imagine if you had a transparent chamber full of liquid CO2, below its critical point, it would be fine to call it CO2 vapor in equilibrium with liquid CO2. Google Scholar shows the usage of carbon dioxide vapor in equilibrium with its liquid.

IMHO, usage of the term  vapour is context dependent and is used for gas in context of liquid evaporation, condensation or equilibrium. So atmospheric CO2 is not considered vapour, even if it is below its critical point. OTOH, atmospheric gaseous water is considerer vapour, as out is related to water evaporation and condensation.

You are right, again, I think the reason is that carbon dioxide does not exist in liquid phase under ambient conditions. The context is always important.

In general "gaseous" state of substances which are liquid/solid at room temperature is called a "vapour" in the Layman's term. Scientific definition might involve critical point but the normal usage doesn't consider it. For example, fluorine gas and chlorine gas vs Bromine Vapours and Iodine vapours.

@M.Farooq I think what you wrote pretty much sums it up:  Vapor is generally used to refer to the gaseous form of a substance that is liquid under ambient conditions.  Correspondingly, it's a substance for which it would be routine to talk about its "vapor pressure."

The normal use distinguishes "vapour" from permanent gas
At normal lab conditions there is a (fairly obvious) distinction between things that could exist as liquids and things where no liquid phase is possible. Oxygen, for example, is a permanent gas, but dichloromethane is not. But the vapour pressure of dichloromethane is pretty high and there will be a fair amount of it above the liquid in a flask of the substance.
In this sense it is useful to distinguish between a vapour and a gas.
But the distinction is  sometimes a little blurry. Butane would normally be called a gas but will liquefy under moderate pressure; Nitrogen is a gas but is commonly used in laboratories as a very cold liquid.
But, under normal lab conditions both butane and nitrogen would be described as "gas" but the "gas" above boiling dichloromethane during a distillation would normally be described as a "vapour".
So the terms are a little useful, but not very precise in any technical sense.

Clarification note
The chart in the question uses a more stringent criterion for the regions called gas and vapour that only labels a region gas at a temperature above the critical point where there cannot be any distinction between the two at any pressure. I think conventional conversational use by many chemists is broader than that and considers normal atmospheric pressure as well as temperature which is more useful and pragmatic at lab conditions.

I would appreciate if you could be more clear as to what you are trying to convince because it did not strike me directly.

@Goarkz I'm arguing that the vapour/gas distinction is not a precise technical term but a convenient way to differentiate between substances that depends on their normal state under lab conditions. Things that are normally gases are "gas'; things that are normally liquid have their gas phase described as 'vapour'.

Alright got it now, But then you see the graph that I have used from Wikipedia, they have a distinct place for vapour, then whats the explanation for it?

@Goarkz As far as I can tell their distinction is less useful but more precise: they only label the region where there is no distinction between gas and liquid at any pressure. My definition is closer to normal usage which considers both pressure and temperature.

Yes. Actually after reading all the answers, now I see why you used the examples which you used.

I'm surprised the OED has such a strict definition for gas. I could not find a strict definition in the IUPAC color books (certainly not in the gold book). Presumably these words are in such common use that their definition is assumed understood or easily found. The analytical compendium (orange book) and physical chemistry book (green book) mention vapour (or vapor) in entries for the meaning of the abbreviation g:

g     gas            or            vapour

The green book (physical chemistry) also has an entry for vapour pressure and for the abbreviation vap:

vaporization, evaporation (liquid→gas)  vap

and definitions for a variety of acronyms that include the word vapor.
There is no suggestion there that gas should be restricted to the description of substances in their supercritical phase.
Curiously the IUPAC books do not adopt a uniform spelling convention. Some choose the spelling vapour, others vapor.
In common usage, a vapor is a gaseous substance emanating from a condensed phase. The scientific meaning is roughly consistent with common use. Vapor implies the existence of a condensed phase that is the source or destination of the gas, or with which the gas may be in equilibrium; while gas does not make such an assumption.
The origin of vapor is presumably Latin with earlier roots in Ancient Greek.
The origin of the word  gas  appears to be the Greek word chaos by way of Dutch:

Coined by chemist Van Helmont. Perhaps inspired by geest (“breath, vapour, spirit”) or by chaos (“chaos”), from Ancient Greek χάος (kháos, “chasm, void”).

"spec. (distinguished from a vapour), ...", offset with a semicolon, is OED's way of saying that this distinction isn't necessarily made, but that if someone is treating "gas" and "vapor" as two different things, then this is the distinction that they probably mean.

In college, I had a thermodynamics teacher who was awesome. He had a way of explaining things that were accurate and easy to understand. He explained this difference to us this way:
A gas will not condense into a liquid with an isothermal compression. (i.e. an Ideal Gas)
A vapor will form liquid when isothermally compressed.
His example:
When you boil water, you get vapor. Hold a spoon over the kettle and it will get water droplets on it.
Steam is heated to a much higher temperature. "If you hold a spoon in from of a steam jet, there will be no condensation. Although, I suggest you not do that because you will be burned."   (He tried to have a sense of humor.)

That's consistent with the definition of permanent gas

Back when I was taking thermo (and when I was a practicing Chemical Engineer), the distinction between gaseous water and water vapor was similar.  Dry steam (https://www.thermal-engineering.org/what-is-dry-steam-definition) is a gas; it's superheated beyond a point of being in equilibrium with liquid water at that pressure.  Wet steam is vapor.

Isothermal compression seems like a reasonable rule-of-thumb, though it'd seem a bit off to me.  For example, we wouldn't tend to say that a nebula is composed of vapor, regardless of its temperature.  Isothermal-compression points in the right direction, but taken literally it would over-extend to near-vacuum conditions.

Except, we aren't talking astrophysics. We're talking about earth at 1atm. In a vacuum, things behave differently. Seems like, if that's what the OP was referring too,, that we be stated in the question.

Vapor (or steam for H2O) can be used for the gaseous form of a particular substance which at normal T and P are solid or liquid.  They are all gases although not always ideal (steam is certainly not ideal below its critical T).