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Is there a way to estimate the nutritional calorific value of a chemical
The chemical in question is L Carvone. It is an ingredient of some flavors. Tastes and smells like spearmint.
What's a good way to estimate its calorific value? If any. I cannot see it as a carb / protein / fat etc.
Does it have a calorific value at all?

Calorimetry . Or thermochemistry calculations based on tabulated data. As for fat sugars fuels etc. However from a nutritional view point one should know how the metabolism of the specific molecule work. At extreme, a compound can enter the body and either be excreted intact or accumulated intact.

@curious_cat I agree with Alchimista. It probably isn't too hard to obtain an estimate of the energy of a compound (Nando shows some nice ways below), but that's quite different from the amount of energy you would get from eating it, which would likely require a more detailed understanding of human metabolism.

If carvone ($\ce{C10H14O1}$) is digested by the body using oxygen to generate carbon dioxide and water, you can estimate the heat of this chemical reaction:
$$\ce{C10H14O1 + aO2 -> xCO2 + yH2O}$$
After balancing the equation:
$$\ce{C10H14O1 + 13O2 -> 10CO2 + 7H2O}$$
The required enthalpy can be know from the formation heats of each molecule multiplied by the found factors.

A quick estimation
I just did this exercise using computational chemistry (B3LYP/6-31G(d) in Gaussian09) and found the following energies (in kcal/mol):

carvone: -291456.0217
oxygen: -94322.73333
carbon dioxide: -118326.709
water: -47947.30569

The change in energy of the reaction is then:
$$\begin{multline}10\times (\pu{-118327kcal/mol})+7\times (\pu{-47947kcal/mol})\\-1\times (\pu{-291456kcal/mol})-13\times (\pu{-94323kcal/mol}) = \pu{-1137 kcal/mol}\end{multline}$$
this is $\pu{-1137 kcal/mol}$ (exothermic, as expected).
Since molecular weight of carvone is $\pu{150.2 g/mol}$, we can convert to $\mathrm{kcal/g}$, the negative value indicates it releases heat:
$$\frac{\pu{-1137 kcal/mol}}{\pu{150.2 g/mol}} = \pu{-7.6 kcal/g}$$

Oh god … I would have assumed that someone who has reached the level of calculating with Gaussian had learnt how to use units correctly … To allow me to upvote in peace with myself I have edited your post to do so.

Also re units: SI units would of course be preferred ;p

Sorry, this is just my second answer. I haven't learned much about your strict rules yet. Also, I thought calories would be better since its the common units for food energy. Thanks for edits!

Well yes, the calorie just won’t die any time soon from nutrition tables; that is unfortunately true. No worries, the answer itself was good but you should really make sure each equation including units adds up … Also, you can ping me by adding @Jan if I commented on your question. While SE does implement some auto-pinging mechanism, they may fail. You as the post owner are always pinged.

@Jan just saw your edit and found out how complete and powerful the notation is. BTW, let me ask you here if is it OK to share a picture from a book? I guess it is not due to copyright issues.

Thanks! So, here's a follow up question: The nutritional estimates I see on the packaging of foods, how are they derived? Are they these theoretical numbers? Or experimental ones from a calorimeter? Or something else?

@Nando you can include an image from a book so long as you cite where it came from.

@curious_cat it's usually an estimate based on protein, carb, and fat content. https://www.scientificamerican.com/article/how-do-food-manufacturers/


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Is cis dipole moment is always greater than trans dipole moment?
In trans compound the dipole moment always cancels each other!  Then how can it's dipole moment be greater than cis compound's dipole moment?

What cis and trans compounds are ou talking about?

I took the case of 2,3-dichloro 2-fluoro Ethene ---in this compound trans compound's dipole moment is higher than cis compound's dipole moment.  But why?

In this compound there are no cis or trans. There are Z and E instead.

How is there 3 is an ethene?

No.
The dipole moment of trans-cyclooctene ($\pu{0.82 D}$) is greater than cis-cyclooctene ($\pu{0.43 D}$).
See On the Molecular Geometry of trans-Cycloöctene, J. Am. Chem. Soc. 1958, 80 (8), 1953–1955.

Blomquist, Burge and Sucsy2 have called attention to the fact that while with the cyclodecenes, the cis isomer has the higher density and index of
  refraction as is typical in acyclic systems, with the cyclooctenes, as first noted by Ziegler and Wilms,3 the reverse is true. From these refractive indices, by application of van Arkel’s dipole rule,4 the prediction could be made that the trans-cyclooctene would have a higher dipole moment than the cis.
[...]
The dipole moments of the cis and trans isomers of cyclodecene were previously determined,4 and were found to be normal. The conclusion, drawn from the dipole rule, that the trans-cyclooctene would have a higher dipole moment than the cis represents a situation without analogy.
[...]
The cis isomer was found to have $\mu_{cis} = \pu{0.43 \pm 0.06 D}$ which may be compared with the value found for cis-cyclodecene of $\pu{0.44 D}$.4 The trans isomer was found to have a dipole moment of $\pu{0.82 \pm  0.03 D}$, compared with the value zero found for trans-cyclodecene.4

(Van Arkel's rule is: The isomer with the higher dipole moment possesses the higher boiling point, refractive index, and density.)

This is a very good example that challenges assumptions nicely.

"On the Molecular Geometry of trans-Cycloöctene" Nice o-umlaut you have there ;-) Also, have an upvote!

@paracetamol I was surprised to see the umlaut.  They use it throughout the article.

For general interest, the umlaut here essentially serves as a pronunciation guide---it's a somewhat antiquated way of distinguishing the diphthong 'oo' from two (separate) vowels next to each other 'o' + 'o', and you also see it in words like preëxistence, etc.

And in this context it shouldn’t be called umlaut but trema

@Jan not "trema", but "diaeresis" https://www.theatlantic.com/entertainment/archive/2012/04/we-resist-further-cooperation-cooperation/328832/ and also https://english.stackexchange.com/questions/66126/what-is-the-standard-rule-for-using-or-not-using-hyphen-and-diaeresis-on-the-wor

@DavePhD trema and diaeresis are both correct, the former is more descriptive of the shape the latter of the purpose in this case. Calling the German umlaut diaeresis is about as wrong as using umlaut here but both can be called trema.

@Jan fair enough

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How does adding sodium chloride to a solution of copper sulfate change the equilibrium?
In school we had a lesson about chemical equilibrium. We performed an experiment: in two tubes we poured $\pu{5 mL}$ of $\ce{H2O}$ and placed $\ce{CuSO4}$, the concentration of the $\ce{CuSO4}$ should have been $\pu{0.5 mol/L}$. We homogenize the two solutions in the two tubes and both solutions I think became blue (might have been green, not sure). Then we were asked to put a gram of $\ce{NaCl}$ in one of the tubes and homogenize. The color of the solution changed (also we might have added another $\pu{5mL}$ to the first solution, the one without $\ce{NaCl}$). Can someone tell me what do they think was the purpose of this lesson and how does it correlate to chemical equilibrium?

When you mix $\ce{CuSO4}$ and $\ce{NaCl}$ solutions, the following equilibrium forms:
\begin{align}
  \ce{[Cu(H2O)6]^2+ + 4Cl^- &<=> [CuCl4]^2- + 6H2O},&
  \Delta Η &= \text{positive}\\
\end{align}
Upon heating the mixture, the equilibrium shifts forward (following Le Chateliers principle) to produce more of the yellowish coloured $\ce{[CuCl4]^2-}$ ions, also known as tetrachlorocuprate (II) ion. As the amount of yellow colour in the mixture increases the solution becomes greener due to subtractive mixing of colours, i.e. blue and yellow solutions mix to give a green coloured solution.
The experiment can be found on YouTube, for example: Chemical Equilibrium ($\ce{CuSO4 + NaCl}$)

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What is the exact relationship between the Pi conjugated bonds in organic molecules and fluoroscence of those molecules?
Some compounds are more fluorescent than others, and some of those are Rhodamine, Phenolphtalin, naphthalene, anthracene etc.
I want to know why these exact molecules undergo such phenomena, what is the effect of conjugation in excitation and emission ?
(please be specific about this question)

I do assume that the basic principles and rules of fluorescence are known to the reader. These are provided at a basic but sufficient level by, e.g., the enter "Fluorescence" at Wikipedia.
Being clear that fluorescence efficiency is the output of the fluorescence radiative and several non-radiative de-excitation channels , the question can be put as Can we quantify the weight of the latter?  To my knowledge the answer is not. Except calculations are made for each specific molecule in the specific environment, there are not real easy rules at hand. Designing new fluorescent molecules substantially requires experience and feeling, not to say try and "errors". For instance,  If a fragment is known to be luminescent , we can expect some luminescence from a molecule containing it. But it is not given, at least not in the sense that luminescence efficiency can be predicted within a few percents.
The same is true for several characteristics in material synthesis . There are designing rules but nothing similar to an analytical function. The same is true if you limit the question on how effectively extended the conjugated system you are going to attain is.
I keep this answer qualitative exactly because I am not aware of something not empirical . A physics robust answer will "reduce" to the listing of the various process involved or must be a short treatise on molecular physics and the fate of excited states. Or case by case computations.
A few working rules can be nevertheless given. For example, more rigid is the molecule higher the luminescence yeld , or lower the gap lower is the yeld,  etc. This is particularly true if the system is supposed isolated. It is clear that often a substituent can have no effect in the isolated molecule scenario but even invert the situation in , for example , solution (via interaction with the solvent molecules).

Absorption and emission in the UV/vis range is usually the result of electrons being excited to and deexcited from molecular orbitals that are unoccupied in the ground state. The energy involved in these processes is largely dependent on the energy gaps of the MOs involved (especially the HOMO-LUMO gap).
Many of the molecular orbitals in conjugated molecules form large $\pi$-systems (that's basically the definition of conjugated molecules). Due to the smaller overlap, the energy splitting between $\pi$ and $\pi^*$ orbitals is smaller than in $\sigma$, $\sigma^*$ pairs. This effect is more pronounced in larger $\pi$-systems. One often finds that the frontier orbitals (i.e. HOMO, LUMO and often HOMO-1, LUMO+1 etc.) are dominated by $\pi$-systems, if they are present in the molecule. Due to the relatively smaller energy gap in conjugated molecules, one finds absorption in lower energy ranges and often with higher probabilities.


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Does the slippery feeling of bases always appear at certain pH?
My guess is that the saponification reaction is only dependent on the concentration of $OH^{-1}$ ions and of course the coposition and concentration of fats. I also guess that the fat part of the situation is rather constant. My guess means that the cation would have no or negligible impact.
This would suggest that the degree of slipperyness would be only pH dependant (and not dependent on the cation). Does this mean that there is a certain pH level where the solution starts to feel slippery? If that's the case about which pH level is it?

Why won't you go and check it yourself? Chemistry is an experimental science, after all. I'd say it starts around pH=9. Besides, the feelings are subjective, so we tend to rely on instrument readings whenever possible. Haven't heard of a slipperymeter, though.

@IvanNeretin Yes, but perhaps my question was not that clear about that the question was a bit about if it's only pH that plays a role. That is that if I test with sodium hydroxide for example would that mean that the findings are valid for other bases as well? Also a part of the question is whether saponification occurs only above some certain pH or that it's just that it's below the sensory threshold (I guess that your comment implies the later)

Well, with NaOH, KOH, etc., it is just pH that plays a role. Other than that, chemistry is rich and diverse; one can imagine certain compounds which are not alkaline, yet still slippery.

There probably is no one pH level where things get slippery. Capacity to react as a base has something to do with it.
You can make ammonia and sodium hydroxide solutions both with pH 11, but the sodium hydroxide solution is only 0.001 molar and thus cannot react with much oil on the skin.  Ammonia solution has a lot more base available to react with those fatty acids.  The weak base "hides" most of its reactivity in water alone but reveals when there are fatty acids to neutralize, so it's slip sliding away.
Probably anything that we consider basic and is sufficiently concentrated will slip in the fingers.


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Why can't the electrophile attack the chlorine atom in a halobenzene?
Why can't the electrophile attack the chlorine atom in a halobenzene?
I mean, chlorine is a deactivating group, primarily due to its inductive effect.
If we can say that its ortho and para influence on rise of electron density is why electrophiles occupy those positions, why not they bond with chlorine who does take up those electron density at the same time?
Examples of electrophilic substitution could be chlorination of chlorobenzene, nitration of chlorobenzene, sulphonation of chlorobenzene ...

Could you add an image of what you mean? I'm having trouble with the description.

Apparently, they do have a reaction although they are supposed to be reacting slowly as compared to benzene, which does not have its electron density drawn away from the ring.

Do you mean to ask why do the pi electrons attack the electrophile and not the lone pairs of Chlorine?

No . But why not the electrophile attacks the lone pair on chlorine. That is why not the nucleophile form a bond with chlorine instead.

Given the charge distribution and the shape of the HOMO, I would be very surprised if an electrophile does not attack the chlorine. This is from a calculation on the DF-BP86/def2SVP level of theory.
If you account for the hydrogens effectively shielding the negative charges of the carbons, the only place left to go is the chlorine. (Natural Population Analysis, NBO 6.0)

The largest coefficient of the HOMO is at the chlorine (C1 ipso, C4 para):
Alpha occ 29 OE=-0.224 is Cl12-p=0.29 C4-p=0.24 C1-p=0.19 C2-p=0.09 C6-p=0.09

Generally I would expect it to be rather unreactive towards electrophiles.

In fact, you are right that it is difficult to have a reaction with chlorobenzene; why not is it possible for the electrophile to bond with the chlorine? Is it anything to do with a + charge on chlorine in one of its canonical structures during resonance?

I did not have had the time to go further into detail, but Queiroz et al. J. Org. Chem. 2006, 71, 6192-6203, DOI: 10.1021/jo0609475 (mirror) make quite an effort explaining the mechanism(s).

@V.Karthik Please cite your sources. This statement makes little to no sense to me. The highest electron density is at the chlorine atom, not in the aromatic ring.

I am following this from my high school book (CBSE-Indian curriculum). Grade 12

They have explained it by saying that chlorine would form a double bond with the carbon in the ring and forms + charge in the process. Now, a - charge is developed in the ring that would occur at the ortho- and para- positions.

But due to chlorine's high electronegativity, the not all the electron density is residing over the ring. In fact, chlorine strongly deactivates the ring, but not as much in the ortho- and para- positions due to the - charge originating over there.

@V.Karthik There are many, many factors influencing a reaction, how molecules bind, and what products are formed. At a high school level they cannot sufficiently be discussed. The paper above is very detailed and very specific for only a hand full of molecules and even they come to the conclusion that there is no universal way. However, the claim that elecrophiles do not bind to the chlorine is certainly not entirely correct.

@V.Karthik Without checking out the details, I am pretty sure that while electrophiles do interact with chlorine there is no productive path that leads away from there while there is a productive path if electrophiles attack the pi system on a carbon atom.

Right, but would not it be better if you could elaborate? It will be easier to point out the reasons why such paths are chosen.


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Cheap antifreeze
We bought a house which has an outside boiler that is inoperable due to the tank freezing.  I can repair but filling the boiler with 100 gallons of antifreeze is too expensive.
Sodium chloride would be cheap, but is too corrosive.
Calcium Magnesium Acetate is also cheap (sidewalk deicer), non corrosive, but cannot find much about solubility (partially soluble).
Potassium acetate is soluble, non corrosive, with good freezing point depression; however uncommon.  Can it be made cheaply enough for 100 gallons not to freeze at -20F?  Vinegar + potassium carbonate.
https://en.m.wikipedia.org/wiki/Potassium_acetate

Did you consider that corrosiveness of the solutions you suggested may be larger at higher temperatures (as use in a boiler would imply)? Getting your solution to be pH-neutral or at least tolerable for the boiler may prove difficult given that cheap ingredients may spell uncertainty about purity and content.

What's your budget? That will impact the suggestions.

What type of steel is your boiler made of?

I know that in Switzerland, sugar is used instead of salt to clear snow from roads as being efficient, more ecological and less corrosive than salt.
In Montreal, beet juice is used for the same reasons.
Sugar however costs more than salt.
I let you check if sugar can suit your needs.
See these two articles in French:
http://www.caradisiac.com/En-Suisse-on-ne-sale-plus-les-routes-on-les-sucre-62195.htm
and
http://www.lenouvelliste.ch/articles/lifestyle/techno-et-sciences/quebec-le-jus-de-betterave-revolutionne-le-deneigement-des-routes-485118


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failed simple experiment with surfactant
I wanted to show my son the effect of a surfactant.  So I filled two small glasses with tap-water, added a squirt of RinseAid to one and stirred.  Then I sprinkled powdered flour on both.  My hypothesis was that the powdered flour would sink much more quickly in the water-RinseAid mixture compared to the water because the RinseAid would lower the surface tension.  In fact, the experiment was a flop-- I could not see any difference in how fast the flour particles sank in the two glasses.  What went wrong?

How fast the particles would sink depends on viscosity. Surface tension would keep the particles floating. Try floating a sewing needle using a fork to float the needle. Then touch a finger to some dishwashing liquid and touch your finger to the surface of the water.


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Difference between Phase, heterogenous and state of matter and solution
My question has already  been answered in parts at Difference between state  of matter and phase
But for deep and real understanding I have to ask for the difference between phase and a heterogenous mixture :
Say we have a bowl made of glass. The bowl is filled with a few different solid rocks on the ground, water in the middle and some oil on top.
I try to analyse this situation for understanding:

There is a phase of oil on the top
There is a phase of water in the middle
There is a phase of rocks on the ground
Since the rocks on the ground are not equally the same they are a bunch of single phases that do not mix up to each other
The rocks on the ground are also a heterogenous mixture
The air above the water is a phase of gas
The glass is a solid phase of silicon

Is that correct?

A deep and real understanding?!?
I don't think you'll find any definitions that you can't pick at.
So if in a large bottle you mix some water and benzene that don't fill the bottle then you'll end up with three phases. A gas phase that has air/water/benzene, a liquid phase that is mostly benzene with a tiny amount of water, and another liquid phase that is mostly water with a tiny amount of benzene.
So if I just take some random rocks (different kinds of rocks) and dump then into a bowl of water, then yes there is a solid phase. But as a chemist I wouldn't really think of the random rocks as a phase because there really isn't much of an equilibrium. Picking at this I'd have to admit that a minuscule amount of rock would dissolve in the water. Hence there would be an equilibrium of sorts between the various rocks too.
What we'd normally consider a glass is a mixture of silica and other minerals. However amorphous solids are also referred to as glasses.

"A deep and real understanding?!?" Well yes, I am learning the basics of chemistry right now and I have read some definitions. But coming up with an own example which is a little abstract and checking if this is correct is a good way to improve.


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Should I include water of crystallisation in the product of a precipitation reaction with silver nitrate?
I've deduced that a certain compound has an ionization isomerism. I just want to know which of the following reactions is correct.
I think this one is the more likely:
$$\ce{[Cr(H2O)5Cl]Cl2 \cdot H2O + 2AgNO3 -> [Cr(H2O)5Cl](NO3)2\cdot H2O + 2AgCl}$$
But I would just like to make sure that it's not
$$\ce{[Cr(H2O)5Cl]Cl2 \cdot H2O + 2AgNO3 -> [Cr(H2O)5Cl](NO3)2 + 2AgCl + H2O}$$
Basically the question asked me to find an empirical formula from elemental analysis. Then it said 2 compounds have the same empirical formula but why are reaction I and II different (not the above reactions). I know its cause ionization isomerisation. So I just need clarification that my first equation is correct.
The only thing I know about the reaction is that one mole of $\ce{[Cr(H2O)5Cl]Cl2\cdot H2O}$ reacts with $\ce{AgNO3}$ to form $\ce{2AgCl}$.

So basically, you're asking "Will $\ce{Cr(H2O)5Cl2}$ have a water of crystallization or not?"

To make a long story short yes :)

After the reaction - I know it has one before.

This is not a well defined question because you didn't provide more details on the reaction. If you carry out the reaction with heat, you might drive off the water of hydration. You're implying that the chromium complex and the nitrate are bound but you're concerned about the water of hydration. But this is all in solution anyway...

So it doesn't matter? I can pick either?

Duplicate of https://chemistry.stackexchange.com/questions/84738/empirical-formula-of-2-compounds

That's my question...do you ever answer questions or just look for duplicates.

It will be very unlikely that you are performing the reaction in solid state. Solid state reactions tend to react only on the surface area and will take very long to complete. (On the upside, they have no entropic terms so the yield is generally $\pu{100\%}$ and thermodynamics dictate the direction.) So likely you will be performing this in aquaeous solution.
If you are, you have previously dissolved the reactant and the product will not precipitate. Thus, you should more accurately write the reaction as:
$$\begin{multline}\ce{[Cr(H2O)5Cl]^2+ (aq) + 2 Cl- (aq) + 2 Ag+ (aq) + 2 NO3- (aq)} \\\ce{-> [Cr(H2O)5Cl]^2+ (aq) + 2 NO3- (aq) + 2 AgCl v (s)}\end{multline}\tag{1}$$
Note that I have neither included the water of crystallisation of the starting complex nor any other water molecules except for the (aq) descriptors showing aquaeous solution. You are unable to distinguish which of the water molecules in solution derives from the salt and which were added as a liquid.
And finally, adding a water of crystallisation to a product in solution is incorrect because there is no crystallisation water associated with it. If you wanted to determine a hydrate content, you would first need to crystallise the product salt and then determine its water content subseqently. There is no way to predict a priori whether it contains one, two, three or more water molecules per crystallisation unit.


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Why doesn't velocity enter into Heisenberg's Uncertainty Principle?
So I came across Heisenberg's uncertainty principle in my 11th grade book, and I'm not convinced. It says $\Delta x \Delta p \gt \frac{h}{4\pi}$. Okay, but how does velocity even matter to determine the path?
Now suppose we have, for the case of my argument, 100 exactly identical atoms of hydrogen. Now we find the positions of the lone electron in 100 different cases, and we can do it with 100% accuracy. Now, if we join the dots, won't we get the path of the electron? Why not? Please explain in simple words if you can.

You're trying to apply your macroscopic intuition to a microscopic event. In your proposal, why do you think you know the path between two measurements? In fact, you have no idea what the momentum at each of those points is at all. And more importantly, you cannot determine position with 100% accuracy. That's the point.

You don't need to be convinced. Nature works this way regardless of whether you're convinced or not.

Colleagues, please refrain from close-voting and downvoting. Neither of us was born with the full knowledge of the subject. The question is completely legitimate and shows some genuine (if mistaken) work of the mind; if this is not welcome here, then I don't know what is.

I was just responding to your phrasing. It seems very arrogant to say that you're "not convinced" of 100 years of physics built by arguably some of the smartest humans who have ever walked the earth and survived every single attempt to prove it wrong.

My bad, I wasn't convinced as I thought from experience that probably 2 years from now in college they'll tell me that it was something dumbed down so that we could understand, like dual nature for example.

No worries. Hopefully, some of the comments and answers have helped you accept more readily that this is a fundamental property of matter.

Possible duplicate of https://chemistry.stackexchange.com/questions/14848/heisenbergs-uncertainty-principle

You will be closer to being convinced after you learn some Fourier analysis.

Some problems with your method: you mention taking 100 copies of the same atom of hydrogen and finding the positions at 100 different times. (i) You actually can't make a copy of a quantum system without knowing its quantum state (the no-cloning theorem), (ii) your measurements of position will disturb the "trajectory" of the electron, so your 100 cases are distinct and cannot represent the path of a single electron, and this is especially true the more accurate you make your measurements.
You should realize, however, that the bound placed by the Heisenberg uncertainty principle is really quite small, and on a macroscopic scale we can essentially determine the trajectory of an electron. A bubble chamber or cloud chamber is essentially a major revision of your proposed method: we fill a vessel with supersaturated vapor, wait for charged ions to pass through the vapor and cause it to condense, and observe the resulting tracks. This describes the ion's motion with a resolution on the order of microns.

No-cloning theorem is an overkill. Let's put it another way: just how are you going to make sure that these 100 atoms are 100% identical?

Okay, so let's say they aren't 100%identical. Now by using our electron microscope we know where the electron "has been" but we can't predict its future because of the energy of the photon. Now even if the hydrogen atoms aren't 100% identical wouldn't my method directly(if the 100pictures are overlapped) give us a picture that looks exactly like the probability density graph for 1s?

@TheLostGuardian0, do note that you'll need to tag Ivan (as I've tagged you here) for him to get a notification about your comment. With regards to your question, you wouldn't get exact agreement, but it will be a better and better approximation as you increase the number of measurements.

@a-cyclohexane-molecule I appreciate all your help, but I'm still not clear as to why we need momentum? I mean isn't position enough if we have large samples? Momentum is only required if we have 1 sample right?

@TheLostGuardian0, I'm not sure what you mean by needing momentum. You don't need to measure (or care about) momentum at all if you're only interested in position; the Heisenberg principle just says that you can't measure both simultaneously accurately.

@a-cyclohexane-molecule my bad I should have worded my question better. If we didn't need momentum to find the trajectory, then why go for quantum mechanical model instead of sticking to Bohr-Sommerfield model? (We are able to determine the path of the electrons by using a microscope so this is similar to Bohr's model)

There is no microscope that would determine the path of the electrons. Also, Bohr-Sommerfield model fails when it comes to 2 or more electrons. So it goes.

Here's a good question: if you have something at point A and something at point B at a later time, what's the path between these two points? Turns out the more certainty you have about those two positions, the less you can tell about which path was taken.

@Zhe But what if we have millions of such points? Wouldn't that solve the problem, as the path would be revealed by the various possible places it was in?

@TheLostGuardian0, how will you obtain these points? Each measurement will affect the future path of the electron, so your points will not be an accurate reflection of the electron's original trajectory.

It doesn't matter how close these points are. You are trying to use your macroscopic intuition of interpolation. That doesn't work at the quantum level. You cannot tell what path was taken between these two points.


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Is a combustion necessarily a redox-reaction or can there be exceptions?
I have a feeling that I have heard of an exception - either based on the material or the oxidizer.

I wouldn't think of combustion as necessarily a redox reaction. For instance burning methane doesn't seem like a redox reaction to me.

@MaxW It kind of is though. You're oxidizing a methane in a relatively uncontrolled setting. You might also consider a putative methane fuel cell which certainly suggests a redox. Still, I agree that it's usually best not to think of combustion in redox terms. There's not much to be gained even if it is a redox reaction.

@Zhe When I learnt redox reactions at school, we took the detour via combustion reactions (of magnesium, mainly) to reach the concept.

@Enthuziast That's not the way things work. Feelings are not studied in chemistry. Bring us an example, then we'll see. Short of that, it's nonsense.

@IvanNeretin Thanks, but the intention of the question was exactly that; are there any examples of exceptions?

No.$\mathstrut$

Yes, every combustion reaction is a redox reaction. Combustion is an exothermic reaction of a fuel and an oxidant (that's a big clue). Most commonly, the oxidant is oxygen, which is reduced to water in the process. The other products will depend on the fuel. Organic compounds will have carbon atoms of various oxidation states that will be oxidized to the +4 state as carbon dioxide product.

And there are no exceptions, for example a reaction with Sulfur as fuel (same group as Oxygen) and Oxygen as oxidant? Any idea why I have the belief that I learnt that there are exceptions?

@Enthuziast burning sulfur gives sulfur oxides, where sulfur has been oxidized and oxygen reduced.

Combustion is also called oxidation of a hydrocarbon, however, this means that a hydrocarbon become oxidized while oxygen atom will be reduced. Almost every reaction involves the transfer and gaining of electrons. So since redox is reduction and oxidation then combustion is a redox reaction.


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Is brass combustible in high pressure oxygen environment?
Some metals are particularly ignitable in high pressure oxygen environments, like stainless steel and aluminum. Copper and nickel, on the other hand will self-extinguish when ignited. Does anyone know about brass?

Don't know what you consider high pressure, but oxygen regulators for 2000 psi tanks are made of brass.

And the tanks themselves are carbon steel.

Metals do not ignite in ambient temperature oxygen , with the possible except exception of alkali metals like lithium. If the question refers to high temperatures, what high temperatures ?


text:
Which diagram shows the effect of catalysis on chemical equilibrium?
My course book has the diagram in figure (1) to represent the effect of a catalyst on chemical equilibrium. I drew the second diagram in figure (2).

When reviewing the diagram in figure (1), I observe the following mistakes:

The rate of catalyzed and uncatalyzed forward reaction  at equilibrium will be the same, which seems to be wrong because the value of rate constant of catalyzed forward reaction ($k_f$ catalyzed) more than the value of rate constant of uncatalyzed forward reaction ($k_f$ uncatalyzed), so the rate of catalyzed forward reaction at equilibrium = ($k_f$ catalyzed) $[A][B] > $ ($k_f$ uncatalyzed) $[A][B]$, so  it isn't the same as the diagram.
The rate of catalyzed forward reaction before equilibrium less than the rate of uncatalyzed forward reaction before equilibrium, which seems to be wrong too.

I need feed back about the diagram or both are wrong and about my reasoning.

OK, I'm bailing on the full answer. I'm 99% sure there is a mistake in the figure you were given and that the y axis should really be concentration, not rate (it was actually Mithoron who figured this out when I brought it up in chat after I found myself stumped). If this is the case, then figure 1 makes sense. To answer your questions in the comment above:  the rate of catalyzed forward rxn is always greater than the rate of the uncatalyzed forward rxn. Of course the same goes for the rates of the reverse rxns (catalyzed is faster). Sorry I can't answer your full Q more directly.

@airhuff thank you ,
1-  I  gave figure1   from course book and from internet, which i consider it wrong because it has the equilibrium at the same rate for the 4 reactions.

@airhuff 2- what about the answer of the second Q in the comment . Does the rate of catalyzed forward reaction before equilibrium will be less or more than the rate of uncatalyzed forward ?

The catalyzed forward reaction rate before equilibrium will be greater than that of the uncatalyzed forward reaction. This is how the catalyzed reaction reaches equilibrium faster than the uncatalyzed reaction.

Can you please include the source of the original figure?

@Martin-マーチンhttps://www.slideshare.net/Kumar4556/chemical-kinetics-55894275     slide Nu. 49 , refer  to  physical chemistry  book  by Atkins

@Martin-マーチンI think slide number 49 not completely right scientifically, because the diagram shows that rates of catalyzed and uncatalyzed froward reaction at equilibrium are equal which is untrue

If (as suggested in the comments) the y-axis is supposed to represent concentration, such that the upper curve is "concentration of $\mathrm{A+B}$" and the lower curve is "concentration of $\mathrm{C+D}$", then Figure 1 is correct, albeit horribly mislabeled.
Your Figure 2 is correct if we want the y-axis to represent a rate-of-conversion instead, i.e., if the upper curve is "number of parcels of $\mathrm{A+B}$ being converted to $\mathrm{C+D}$ per second", and the lower curve is vice-versa. As you point out, a catalyst increases the conversion rate, and does so in both directions. So, exactly as you have drawn in Figure 2, those catalyzed rate curves should both be higher than their uncatalyzed siblings.
("Rate of reaction" is an awfully ambiguous term here – since at equilibrium, the rate of reaction must be zero!)

That can't be right. A catalyst does not change the location of the equilibrium. By definition, the rate of the forward reaction has to equal the rate of the reverse reaction.

Yes, but the catalyst here (in my reading of Figure 2) isn't changing the position of the equilibrium, right? The catalyst increases the rate-of-conversion-of-A-to-C, and also increases the rate-of-conversion-of-C-to-A. Our uncatalyzed rates at equilibrium might be one-parcel-per-second, then with the catalyst they might be two-parcels-per-second.

Yes, you're right. I was mixing rates with concentrations. This is confusing.

I think your second assumption is the one we need.

@Jan you mean Figure 2 is correct

Your last statement is incorrect. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, so there is not change in the amount of reactants or products. The apparent rate is zero, but the reactions still happen.

@LDC3 The IUPAC definition of "rate of reaction" has the form d[A]/dt, which at equilibrium must be zero. "Rate of forward reaction" and "rate of reverse reaction" are properly called "chemical flux", though of course everyone uses and understands the "rate of _____ reaction" notation.

To know the exact diagram represent the effect of catalysis on chemical equilibrium, I searched on experimental data but I can't find suitable data.
 so I consider a hypothetical reversible first -order reaction in each direction where
$$\ce{ A <=> B}$$
I choose those two equations for reversible first- order reaction to calculate the concentrations of product and reactant at a different time:
$$\ce{[A]}=\frac{k_b +K_fe^{-(k_f+k_b)t} }{k_f+k_b}{[A]_0}$$
$$\ce{[B]}=\frac{k_f -K_fe^{-(k_f+k_b)t} }{k_f+k_b}{[A]_0}$$
Starting with A at a concentration$[A]_0={0.6}$ and the values of the rate constants of uncatalayzed reaction are $k_f=3s^{-1}$ and $k_b=1s^{-1}$.
and the values of the rate constants of catalayzed reaction are $k_f=6s^{-1}$ and $k_b=2s^{-1}$
I did the calculation using excel sheet and draw the graph of concentration Vs time :

Depending on the rate  law expression$(r_f=k_f [A])\ and\ (\ r_b =k_b [B])$ I did the calculation using excel sheet and draw the graph of rate Vs time

Figure 1 shows the rate as $k_f [A][B]$ whereas Figure 2 shows only $k_f$. This is why Figure 1 has the equilibrium at the same rate for the 4 reactions.

Figure 1 has the equilibrium at the same rate for the 4 reactions,   and Figure 2 show the rate at equilibrium more when catalyzed.so :Which Figure shows the effect of catalysis on chemical equilibrium?

Can you please explain how" Figure 2 shows only $k_f$"

@AdnanAL-Amleh $k_f$ is how fast the reaction occurs. Therefore, the rate of reaction with a catalyst will always be faster than without a catalyst (curves 4 and 5 must be greater than curves 1 and 2 at all times). When you multiply $k_f$ with $[A][B]$ and call that the rate, then the figure looks like Figure 1.

As you said curve 4  must be greater than curve 1  at all times,can I conclude Fig1 is not correct , because it shows curve 4 less than curve 1.

@AdnanAL-Amleh Figure 1 is correct if you take the product $k_f [A][B]$, not just $k_f$ alone.

I take into my consideration the $r_f$ as $k_f  [A][B]$ in figure1 , which means the 2  forward  reactions(catalayzed and uncatalayzed) haven't   the same rate at equilibrium  ,   but figure1 show  the same rate for  all the 4 reactions at equilibrium , so can I conclude figure1 isn't  correct?

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text:
What is IUPAC name of this compound?

Should the benzene group be taken as a substituent, so as to select the longest C atom chain?
In what cases is it taken as as a substituent?

Answer B seems ambiguous to me since the chlorine could be on the phenyl ring and not the methane.

Concerning the name of the $\ce{-CH2-Cl}$ group, see also How to name the following alcohol? or Nomenclature of Halogen substituted Alcohol and longest chain.

The correct answer is

(d) none of these

In the usual substitutive IUPAC nomenclature, simple halogen compounds are always expressed by prefixes (‘bromo’, ‘chloro’, etc.). Hence, the senior parent structure of such compounds is not determined by a principal characteristic group (a characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name).
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the ring or the chain can be the senior parent structure; however, for the preferred IUPAC name, the ring is always selected as the senior parent structure.

P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)
Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).
(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.
(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.
(…) For selection of a preferred IUPAC name, see P-52.2.8.
P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Therefore, the benzene ring is selected as the senior parent structure for the preferred IUPAC name of the compound given in the question.
According to P-29.1.2, The chloromethyl ($\ce{-CH2Cl}$) substituent is a compound substituent group, i.e. a substituent group consisting of two parts, a simple substituent group (the parent substituent group, here: methyl) to which is attached one or more simple substituent groups (here: chloro).
According to P-16.5.1.1, parentheses are used around such compound prefixes. Therefore, the preferred IUPAC name for the compound that is given in the question is (chloromethyl)benzene.

What if it were cyclohexane other than benzene?

However, answer (a) does come close.


text:
Given advancements of computational power an machine learning, how is it still not possible to predict products from chemical reactions?
Given that it seems every answer to this question is that it is "impossible to predict the outcome of a chemical reaction." Is chemistry just trial and error? Given how fundamental and revolutionary a simulator capable of predicting possible outcomes of a reaction would be, why has it not been done yet? Do you think it will ever happen?

A useful computational method like CCSD(T) scales with $N^7$. Even if the Computers get more and more power, those methods will aways be extremely expensive

Possible duplicate of Software for predicting chemical reactions

This isn't necessarily a duplicate if one focuses on the machine learning aspect.

@penta Then that would make the question off-topic for other reasons O:)

@paracetamol I disagree. Machine learning now plays a large role in chemistry, and can predict reaction outcomes. See https://dx.doi.org/10.1021/acscentsci.7b00064 for just one of many examples.

@pentavalentcarbon, Great article! Thanks for sharing.

Machine Learning does not circumvent the computational complexity of the problem. In other words, ML doesn't transform the traveling salesman problem from NP-complete to P, unless P=NP. A product prediction routine has to explore all possible reaction pathways between all possible stable compounds, which grows exponentially with the number of atoms. Or at least find all convex stability regions and evaluate the thermodynamics correctly, which still explodes combinatorially.

It's not an all-or-nothing situation, where either we can reliably prediction reaction products, or we have not tried to.
For instance, drug companies, for many years, have been making extensive use of increasingly powerful, sophisticated, and refined in silico predictive models to choose likely candidates for in vitro testing.  And I would be surprised if they've not also begun to develop, evaluate, and incorporate ML.
At the same time, these models are not perfect.  So the answer is that it is being done, it's been done for a while, we're getting better and better at it, but we still have a long way to go.
It will be interesting to see what happens with quantum computing, which is currently in its infancy.


text:
Conservation of angular momentum in rotational spectroscopy
In spectroscopy, transitions between rotational energy levels obey the selection rule $\Delta J = \pm1$. The rule follows from the fact that the photon absorbed has an angular momentum of $\pm \hbar$, so the molecule must change its angular momentum for the total angular momentum to be conserved.
However, the angular momentum of a molecule is given by $\hbar \sqrt{J(J+1)}$, so the change in molecular ang. mom. is $\hbar\sqrt{(J+1)(J+2)} - \hbar\sqrt{J(J+1)}$. As $\hbar\sqrt{(J+1)(J+2)} - \hbar\sqrt{J(J+1)} \neq \hbar$, it seems that the total angular momentum is not actually conserved.
Questions:

Does the conservation of angular momentum apply to quantum systems?
If yes, what is the source of the contradiction above?

To the first question: Yes

@user1420303 thanks! What about the second one!

Angular momentum is a vector and the addition of the angular momentum of the photon to that of the molecule has to follow the rules for vector addition. That means that you first have to add these two vectors together before you can take the operator of the total angular momentum.

@Paul, I see why $\hbar\sqrt{(J+1)(J+2)} – \hbar\sqrt{J(J+1)}$ can be smaller than $\hbar$ because of vector addition, but it's in fact greater than that, so I don't see how this helps -- could you please clarify? Also, what do you mean by "add vectors together before your can take the operator of the total angular momentum" -- do you mean add vectors, then get the magnitude of the sum? If so, this still doesn't explain why the increase is greater than what one would expect even if vectors were aligned. And thanks for your help!

Before the absorption takes place the total angular momentum in the system is given by $\vec{J}=\vec{j}+\vec{j}\text{ph}$ where $\vec{j}$ and $\vec{j}\text{ph}$ are the angular momenta of the molecule and photon respectively. The total angular momentum operator is thus $\hat{J}^2=(\hat{j}+\hat{j}\text{ph})^2=\hat{j}^2+\hat{j}\text{ph}^2+2\hat{j}\hat{j}_\text{ph}$. In your treatment you have ignored the last term (you only add the angular momentum of the individual molecule and photon and ignore their interaction). A proper evaluation involves quite some angular momentum algebra.

@Paul, ok, thanks! Could you please recommend where I can find a rigorous treatment of this?

Sorry for the late answer, but I could not find the time to write something earlier. A concept that you encounter often in quantum mechanics is the so-called coupled and uncoupled basis. Your question is a variation on that theme: before the absorption the uncoupled representation is the more natural where the angular momentum of the photon and the angular momentum of the molecule (and their projections) are good quantum numbers, whereas after the absorption the coupled representation is more suitable and the total angular momentum of the system (and it's projection) is a good quantum number. In Dirac's braket notation we have
$$
|jm_j\rangle |j_\text{ph}m_{j_{\text{ph}}}\rangle
$$
for the uncoupled representation and
$$
|JM_J\rangle
$$
for the coupled representation.
Your question concerns the conservation of the total angular momentum of the system $\hat{J}^2$ and in the coupled representation the expectation values are easily obtained because $J$ is a good quantum number
$$
\hat{J}^2|JM_J\rangle=\hbar^2 J(J+1)|JM_J\rangle
$$
In the uncoupled representation the total angular momentum is given by $\hat{J}^2=\left (  \hat{j}+\hat{j}_\text{ph} \right)^2=\hat{j}^2+\hat{j}_\text{ph}^2+2\hat{j}\cdot \hat{j}_\text{ph}$
The first two terms of this expression are again easy, but the thrid term is a bit tricky. Using the ladder operators you can write it as
$$
2\hat{j}\cdot \hat{j}_\text{ph}=2(\hat{j}_x\hat{j}_{\text{ph},x}+\hat{j}_y\hat{j}_{\text{ph},y}+\hat{j}_z\hat{j}_{\text{ph},z})=\hat{j}_+\hat{j}_{\text{ph},-}+\hat{j}_-\hat{j}_{\text{ph},+}+2\hat{j}_z\hat{j}_{\text{ph},z}
$$
where
$$
\hat{j}_\pm|jm_j\rangle=\hbar\sqrt{j(j+1)-m(m\pm1)}|jm_j\pm1\rangle
$$
and it thus couples states of different $m_j$. Let us, for example, couple $|jm_j\rangle=|1m_j\rangle$ and $|j_\text{ph}m_{j_\text{ph}}\rangle=|1m_{j_\text{ph}}\rangle$ so that we have $(2j+1)(2j_\text{ph}+1)=9$ basis states:
$$
|1,-1\rangle|1,-1\rangle,\\|1,-1\rangle|1,0\rangle,\\|1,-1\rangle|1,1\rangle,\\
|1,0\rangle|1,-1\rangle,\\|1,0\rangle|1,0\rangle,\\|1,0\rangle|1,1\rangle,\\
|1,1\rangle|1,-1\rangle,\\|1,1\rangle|1,0\rangle,\\|1,1\rangle|1,1\rangle
$$
We determine all expectation values of the total angular momentum operator using the 9 basis states which gives us a total of 81 values that we can order in a 9x9 matrix (in units of $\hbar^2$)
6   0   0   0   0   0   0   0   0
0   4   0   2   0   0   0   0   0
0   0   2   0   2   0   0   0   0
0   2   0   4   0   0   0   0   0
0   0   2   0   4   0   2   0   0
0   0   0   0   0   4   0   2   0
0   0   0   0   2   0   2   0   0
0   0   0   0   0   2   0   4   0
0   0   0   0   0   0   0   0   6

We can diagonalize this matrix and find the eigenvalues 0, 2, 2, 2, 6, 6, 6, 6, 6. These are the same eigenvalues (including the $2J+1$ degeneracy) of $\hat{J}^2$ in the coupled representation!
So the expectation value of the total angular momentum is the same in both cases, the difference is that in the uncoupled case the matrix has off-diagonal elements and it is a bit more work to find its eigenvalues. In the coupled case, the matrix for $\hat{J}^2$ is diagonal and you get the expectation values directly. You might ask if there is an easier way to convert from the coupled to the uncoupled case and back. As you might have guessed there is such a way and it involves the so-called Clebsch-Gordan coefficients.


text:
How do I arrive at the α-trimethylsilyloxy ketone from an unmodified ketone using TMSCl, triethylamine and mCPBA?
I'm confused about what is going on here. I gather that step 1 is typically supposed to protect an OH group, but there isn't any in the starting material.

http://www.organic-chemistry.org/namedreactions/rubottom-oxidation.shtm

What does MCPBA usually do with an electron rich double bond?

Since step 2 occurs after step 1, epoxidation only occurs after the protection of the OH group. By implication, enolization must have occured before step 2 ( I suspect enolization catalyzed by NEt3). By logical extension the carbonyl oxygen in molecule A is also the oxygen within the -OTMS group. But since the carbonyl oxygen in meta to the methyl group in molecule A, and para to the methyl group in molecule B, I am unsure of what happened.

If this was something like TBDMS, I'd expect it to resist hydrolysis, but OTMS normally hydrolyses under acidic conditions as shown. Unless there's stabilisation from the carbonyl?

@Beerhunter- the TMS would go. Possibly a typo in the question, or possibly just left there as a hint as to what's occured

Yeah, TMS is more of a disappearance than a protecting group ;)

In connection with a reasonable base (triethylamine works fine for this), ketones are known to be enolised by silyl chlorides and trapped as silyl enol ethers. This is not only a key process in Mukaiyama aldol reactions but also in your structure.
In general, the combination of $\ce{NEt3}$ and $\ce{TMSCl}$ has three possible ways to enolise your compound:

γ-deprotonation to the methyl group, creating an exocyclic double bond
γ-deprotonation of the ring proton to give a 1-silyloxycyclohexa-1,3-diene system
α-deprotonation to the other side, giving a 2-silyloxycyclohexa-1,3-diene system

By the general rules one is taught to follow, the second option should be identified as the thermodynamically most favoured. However, the third option is typically the kinetically favoured one, so given the correct conditions (low temperatures), we typically expect enolisation by deprotonation the α proton. It helps here that $\ce{TMSCl}$ is a rather small and reactive silyl chloride (Lewis acid), meaning that once it has attached to the keto oxygen deprotonation will be a rather rapid process.
If you left the solution to equilibrise with $\ce{NEt3}$ before quenching you would probably get much more of the γ-deprotonation products.
The next two steps, a Rubottom oxidation, are basically the generation of an epoxide by $\ce{$m$CPBA}$ (as an isolated reaction also known as the Prilezhaev reaction) followed by acid-catalysed epoxide opening which reforms the $\ce{C=O}$ double bond and initiates a Brook migration of the silyl group. The overall mechanism is given in the scheme below.

Scheme 1: mechanism of the Rubottom oxidation.[1]
Reference:
[1]: L. Kürti, B. Csakó, Strategic Applications of Named Reactions in Organic Synthesis, Elsevier Academic Press, 2005, pp. 388–9.

Brilliant! Though I think there should be an arrow from the O-TMS bond to the O so as to restore the neutral charge, in the final step.


text:
Ylides nucleophilicity comparison
Why are phosphonate-stabilised ylides used in HWE reaction more nucleophilic than phosphonium ylids used in Wittig reaction? Explanations without rationalisations using d orbitals would be welcome.

Which ylides are you comparing? Can you give us generic structures? ‘unstabilised ylides’ could be a lot of things …

related https://chemistry.stackexchange.com/questions/65233/which-is-the-currently-accepted-mechanism-of-a-wittig-reaction https://chemistry.stackexchange.com/questions/5313/why-does-the-unstabilised-wittig-reaction-selectively-form-cis-alkenes https://chemistry.stackexchange.com/questions/48239/does-the-synthesis-of-beta-keto-phosphonates-from-esters-have-a-name

As per @Jan's comment, it might be helpful if you expanded your question to back up the statement with structures or data.  As it stands, I don't think it really holds true.

I will try to find in what textbook I've seen the statement. Anyway, the same is stated on wikkipedia https://en.wikipedia.org/wiki/Horner%E2%80%93Wadsworth%E2%80%93Emmons_reaction

I have found in Clayden et al. that the phosphonates being anionic are more nucleophilic than the neutral phosphonium ylids. On the other hand, they are less basic than the phosphonium ylids

It might not be the only factor, but I think that the steric effect of the huge phenyl groups in Wittig ylides might contribute in making them less nucleophile.
If you give a fast look at a "space filling" model of the two molecules, you see that the negatively charged carbon in a Wittig Ylide is much more masked by the phenyl groups
Note that in order for a nucleophile-electrophile interaction to occur, an electrophile should approach to the carbanion's HOMO orbital (approximatively orthogonal to the screen in both the pictures).


text:
ΔH of reversible expansion of ideal gas
Since its isothermal, $\delta U$ is zero. But enthalpy change $\delta H$ is $\delta U + \delta n(\text{gaseous})RT$ or $\delta H = \delta U + \delta (pV)$.
For reversible isothermal expansion of ideal gas, what is the $\delta H$, is it positive, negative or zero.
Ref: I've checked my study material and it says positive, which makes sense but my exam worksheets state it to be zero.

Recall that, for an ideal monatomic gas, $$U = \frac{3}{2}nRT.$$ (Diatomic and polyatomic gases have a larger coefficient.) For an isothermal expansion, reversible or otherwise, we therefore have $$\Delta H = \Delta(U+PV) = \Delta(U+nRT) = \frac{5}{2}nR\Delta T = 0,$$ since $\Delta T = 0$.

So for an ideal gas its zero. Is it a any different for non-ideal gas?

@Cinnamaldehyde, yes, certainly. Exactly how it changes will depend on your equation of state; I don't recall any further detail off the top of my head.

For a non-ideal gas, enthalpy is a function of pressure (in addition to temperature).


text:
Why can't I dry/concentrate ethanol using calcium chloride ... or can I?
My (very unreliable) school textbook tells me:

Lower alcohols form a solid derivative with certain metal salts.

and,

It is for this reason that ethanol cannot be dried/concentrated using anhydrous calcium chloride.

Nowhere else in the book, is there any reference to this supposed reaction between a lower alcohol (ethanol) and a metal salt (calcium chloride). So I couldn't read more about this from there...
Googling wasn't much help; though I did find this paper that seems to concur with my textbook (albeit, with more detail) ... the paper is dated all the way back to 1923, so I suspect the modern Chemist's view of the matter would be different (the paper makes use of the term "alcoholate", which no longer carries the same meaning).

Question:
Does ethanol really react with calcium chloride to form some "solid derivative"?
Additionally,
Do lower alcohols (methanol and ethanol to name a few) really react with "metal salts" (textbook didn't specify, so i would use the following examples: copper sulfate and magnesium chloride)?

You understand what happens with water when CaCl2 absorbs it? Why wouldn't ethanol do the same?

http://www.sciencedirect.com/science/article/pii/0040603181801475

@Curt That confirms the ethanol- $CaCl_2$ "solid derivative". Know anything about ethanol/methanol reacting with other metal salts? Thanks  again for the link! O:)

Strongly related: https://chemistry.stackexchange.com/questions/54878/methanol-adducts-with-cacl2

@Nilay Yep, very strongly related :D

I am guessing that the correct answer to this question would be "CaCl2 is rather soluble in dry ethanol". Like so many other inorganic salts that are used to easily dry unpolar solvents.

https://youtu.be/RGG2GNvpwA4?t=438 also claims that calcium chloride forms a complex with ethanol.

The apparent answer is yes, to quote from a source (composed from extracts of old chemistry journal):

Calcium chloride forms addition compounds with the alcohols. On evaporating a solution in ethyl alcohol at a low temperature rectangular plates of $\ce{2CaCl2.7C2H5OH}$ are deposited. The compounds $\ce{CaCl2.3C2H5OH}$ and $\ce{CaCl2.CH3OH}$ have also been separated, as well as a mono- and a di-acetone compound, and compounds with isobutyl and amyl alcohols.

So, if you are working with a cold solution of $\ce{C2H5OH}$ with $\ce{CaCl2}$ exposed to air undergoing evaporation, it is possible to form an addition compound (taking the form of rectangular plates) with the formula, $\ce{2CaCl2.7C2H5OH}$ or $\ce{CaCl2.3C2H5OH}$.

Yes, I think that lower alcohols cannot be dried using $\ce{CaCl2}$ as they form stable complexes with it but higher alcohols can be dried using $\ce{CaCl2}$ as they form unstable complexes while water forms stable complex. On the other hand,
aromatic alcohols like phenols can't be dried using $\ce{CaCl2}$ as they are somewhat acidic in nature.
At the same time, I assume that all salts wouldn't do the same thing with lower alcohols as each salt-alcohol complex has its own degree of stability and the salt-water complex may be more stable than the former. So, I don't recommend $\ce{CuSO4}$ as a drying agent.

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text:
What is the name of the diagram in this question? How do you read it?

My teacher calls it an 'orbital diagram' but every time I search for information something completely different pops up. I assumed it was a fancy Bohr diagram, but the electron placement is always weird. I understand electron configuration , but I cannot understand why each electron goes where on this diagram. This example is really simplistic, but other ones that I've seen make no sense in terms of electron placement.

So, Bohr's diagram is a super-simplified version of this, almost to the point where it is arguably incorrect.
This diagram, which shows the orbitals of an electron, shows where an electron has a >90% chance to be. That means that if an electron is in any of the orbitals in your picture there, and you take many snapshots of the atom, it will show a higher concentration inside the orbital. You will find quite a few outside the orbital, but the biggest concentration will be inside those "bubbles".
Basically it shows where you are most likely to find the electrons around an atom.
If you google "different orbitals" you will find picture similar to that of yours. (I tried just now). Hope that clears most of it up, if not I really reccommend you to watch Khan Academy's videos on the subject. That really helped me.


text:
Dilution solution
Please help me, I am a student.
Here is the question:
b)  You have a stock solution of 2 M HNO3  (nitric acid).   You need to prepare a 0.1M HNO3 solution.
(i)  What dilution do you need to do on the stock solution?
I think the fomulas for this question is:
volume of solution required to be diluted = concentration of solution you want / concentration of original solution  x volume of dilute solution you want.
But the questions did not give the volume of dilute solution wanted, so I could not work out this question.
Please help.
Thank you.

You need to dilute it to 1/20th of its current strength so add 1ml of the stock soln to 19ml of water

The fundamental concept at the core of dilution is conservation of matter. The amount of substance (number of moles) before and after the dilution is constant.
$$n_{1} = n_{2}$$
If you express the amount of substance in terms of solution concentration and volume, you have:
$$M_{1}V_{1} = M_{2}V_{2}$$
In your problem, what you have are values for $M_{1}$ and $M_{2}$. That gives two unknowns, but you can solve for the ratio $\frac{V_{1}}{V_{2}}$. This provides the ratio of between the two volumes which is how much you need to dilute by.
Notice that the ratio between concentrations and ratio between volumes are inversely proportional. Since the concentration is 20 times smaller at the end of the dilution, the volume must be 20 times larger.


text:
(Hydro)Carbon organic compounds with only 5 carbon atoms only and beyond
How many different organic structures (from the pure theoretical viewpoint) are there with carbon and hydrogen atoms and atoms without hydrogen can be made with 5 carbon atoms only? Is there any way to count them all using Polya enumeration theory for "n" (hydro)carbonatoms?
This is a follow up question from this Hydrocarbons with only 4 carbon atoms

ron was probably hooked to this earlier one, but  for 4 C atoms it could be already considered too broad, and for 5 I wonder how it could be not.

I think it is not. Indeed, I was wondering about the mathematics behind the structure using chemical graph theory, and if...A mathematical formulation could be better....I mean, is there any known series for the number of structures with n-carbon atoms and hydrogen?

Oh, you wanna calculate it not list them? There was quite a bit of such questions https://chemistry.stackexchange.com/search?q=is%3Aquestion+number+of+isomers

Isn't this fundamentally a maths question?

This is simpler version of your question, then: https://chemistry.stackexchange.com/questions/16135/how-to-determine-number-of-structural-isomers

https://chemistry.stackexchange.com/questions/33356/is-there-a-formula-to-determine-the-total-number-of-constitutional-isomers

What makes this a bit difficult to calculate is that some theoretically possible compounds may be too unstable due to steric hindrance or bond strain. E.G. cyclopentane (C5H10) and cyclopentene (C5H8) are common industrial chemicals, 1.1.1-propellane (C5H6) exists, but is highly strained, and C5H5 is the cyclopentadienyl radical, not a stable molecule.

BTW, spiropentadiene, or bowtiediene  [no, I did not make up that name], C5H4, was first synthesized in 1991 and is highly unstable.

I know those names, I will answer the question in a few weeks, I did it as exercise this summer but I want to see what manages people here...


text:
Newman projection of 3-methylpentane
How would you draw the Newman projections of 3-methylpentane, sighting along the C3-C4 bond? I am unsure as to how to position the $\ce{-CH3}$ and $\ce{-CH2CH3}$ groups on the C3.

You can put them anywhere you want unless there is a specific conformation you are preferring.

I thought the biggest groups always had to point upwards? For example, I drew it with my front C3 attached to the ethyl (biggest grp) pointing straight up and the H and CH3 on the left and right. That way for my anti configuration, I had the methyl group on the back C4 180• to it. Am I correct in my reasoning?

The Newman projections of 3-methylpentane are:

This is a staggered conformation.
Another one would be

This is also a stggered conformation and is the most stable of conformers.
You can generate other conformers by rotating the C-3 or C-4 atom along the C-3 C-4 bond axis.

You should add that you only considered one enantiomer and that the mirror image of your drawings is equally a solution unless stereochemistry is specified.

@Jan , 3-methylpentane is achiral compound. How can you find enantiomer?

Yes, I later realised that how I wrote it didn’t make much sense but what I meant is: the two $\ce{C-C}$ bonds should be considered enantiotopic; looking down one will give you the opposite image from looking down the other. So in your pictures, you can exchange methyl and ethyl and still get a valid representation. That’s what I meant

@Jan , I kind of get what you mean but I am not sure how I can include them in my answer. As far as I see, exchanging the methyl and ethyl groups in one structure will result in a structure which can be obtained by rotating the original structure. However, feel free to edit my answer, in case I have missed your point.

Oh no, the key about these different versions is that they cannot be obtained by rotation, because switching the positions of only two substituents on one side of the Newman bond is equivalent to mirroring, not rotating. I can’t add any pictures right now (and at least until Monday) because I won’t have access to ChemDraw.

@Jan , consider the eclipsed structure at the bottom, and swap the positions of ethyl and methyl group. So, in the new structure, the overlapping groups are CH3-CH3, C2H5-H, H-H. This is exactly the same structure as the eclipsed structure on the upper right, and it has been generated by rotating the central bond. As you can see, they can be obtained by rotating the bond


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Is sodium acetate strong or a weak electrolyte?
Since it is formed of a strong base and a weak acid it should be a strong electrolyte but a chemistry teacher said that it is a weak electrolyte.....

Sodium acetate is a strong electrolyte in water.  The status of sodium hydroxide as a strong base triumphs; it makes even a weak acid such as acetic acid dissociate essesitally 100% into ions (by stealing the protons).  And when something is essentially 100% dissociated into ions, by definition it's a strong electrolyte.

A misconception, I feel, that maybe some other students have made after tackling equilibria for the first time. Let me explain by first giving the equilibrium expressions regarding acetic acid and its conjugate base:
\begin{align}
  \ce{CH3COOH + H2O &<=> CH3COO^- + H3O+}&
  \mathrm{p}K_\mathrm{a} &= 4.76\\
  \ce{CH3COO^- + H2O &<=> CH3COOH + OH-}&
  \mathrm{p}K_\mathrm{b} &= 9.24
\end{align}
I essentially deduced that since both equilibria constants were quite small that both acetic acid and, hence, sodium acetate would constitute weak electrolytes. I have since found out that though my reasoning was correct for acetic acid it wasn't so for sodium acetate.
The reason why is because, in an acetic acid solution, acetic acid is the sole electrolyte, and, hence, the equilibrium expression dictates its dissociation into ions (since it is low there will be slight dissociation). However, as for the sodium acetate solution, the equilibrium expression considers acetate ions completely dissociative from their cations, i.e. it assumes 100% dissociation of the salt. This is indicative of the sodium acetate constituting a strong electrolyte.

Strong or weak electrolytes will depend on the hydration enthalpy of products in water. That's why any salt like $\ce{NaCl}$ will completely dissolve in the water, though $\ce{NaCl}$ has high bond energy in powder form. In the above case, the hydration of the $\ce{Na+}$ ion will be so dominating that the salt will dissociate (hydrated) completely despite less hydration enthalpy of the anion.
So, as a rule of thumb strong acid/base salts will have more hydration enthalpy but it is just a correlation and will depend on the solvent. The reason for partial or full dissociation is in hydration enthalpy and entropy change.


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How to calculate Debye length of a 2-1 electrolyte
I am given the Debye length to be 10nm for 0.001M KCl at 298K. I want to know its Debye length in 0.01M CaCl2. How can this be calculated?
I know that for 1-1, the Debye length is inveresely proportional to the root of its concentration. But how can I work out for electrolytes other than 1-1?

If we follow the derivation of the Debye length model, we would arrive at the following adapted model (doesn't assume uniform charge distribution):
$$\lambda_{D}^{2}=\frac{\kappa k_{B}T}{4\pi Z_{1}Z_{2}e^{2}c_{0}k_{c}}$$
In our equation:

$\lambda_{D}$ is the Debye length;
$Z_{1}$ represents the charge on the cation;
$Z_{2}$ represents the charge on the anion;
$\kappa$ is the dielectric constant;
$k_{B}$ is the Boltzmann constant;
$T$ is the absolute temperature; and
$c_{0}$ represents the number of ions per a cubic metre of solution.

Note that the general equation assumes that $Z_{1}=Z_{2}$, hence why this is usually written as $Z^{2}$.
Keep in mind this is only applicable for calculating the Debye length for a single dissociated species in solution! Further alterations would need to be made in order to consider a mixture.

what is kappa in the numerator? isnt kappa/lambda the debye length?

Kappa represents the relative dielectric constant, $\lambda_{D}$ is the Debye length.

Another way of writing it might be $\epsilon_{r}$.

The clarification should be included in the answer.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. While it may seem strange at first we do not use salutations in our Questions and Answers. Please use the upvote button to say thanks, or leave a comment how the post can be improved. Please also do not include edit statements in your post, the history can be usedto track changes.

how could I use the information given to work this out though? This equation doesnt make use of 10nm for KCl does it?


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Is it easier to extract CO2 from seawater than from air?
It takes some effort to extract $\ce{CO2}$ from air in order to produce synthetic and potentially 'green' fuel. Doesn't seawater provide this environmental $\ce{CO2}$ more easily?

It is not the extraction of CO2 that poses any difficulty (after all, there are many sources of it, such as fossil-fuel fired electric generators and cement production). It is the process of changing CO2 into fuel that requires energy. The question is non sequitur.

@DrMoishe Yeah... except the title says "Is it easier to extract $CO_2$ from seawater than from air?", and the body of the post essentially asks the same thing. The OP does mention the potential use of $CO_2$ as a green fuel, but I only see it as a helpful pointer/example (Before reading the body, I was curious to know what the OP had in mind), not the central idea of the post. So yeah, title and body corroborate (perhaps not obvious at first glance?). ;-)

As for closing the question, I'm unsure as to whether opt for "Primarily opinion-based" or "Too broad"... because the question is somewhere in the middle. A good question, but it might help if the OP restricted answers to specific sets of methods.

How would you propose to concentrate CO2 from sea water in a way that is more efficient than air distillation?

@airhuff I would not. I am asking wether or not such a method exists.

@paracetamol What opinion may it be that makes me ask this?


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Is the crystallization process of aqueous solutions of substances such as lithium chloride endothermic?
$\ce{LiCl}$ has an enthalpy of solution of $\pu{-37.1 kJ/mol}$, so its dissolution process is exothermic. Does this mean that it's crystallization process is exothermic? I'm asking since I've looked up "exothermic crystallization" and "the crystallization of endothermic solutions" and couldn't find anything.
So my question is: Is crystallization always an exothermic process, or is it endothermic for substances such as this whose enthalpy of solutions are negative?

There is also a heat of dilution that you need to account for.

Oh okay. Unfortunately I'm not really familiar with what that is, but what effect does it have on the energy released or absorbed by the crystallization?

You don't just form the crystals from a 1 molar solution, so you'll need to account for the fact that you're presumably doing so from a more concentrated solution. Since enthalpy is a state function, you can just sum the enthalpy changes from the two processes to determine the total enthalpy change.

I see. So can a sum of those two enthalpies ever result in a value that makes crystallization endothermic?

Depends on the substance, but it could be.


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Misleading features of Lewis Dot Representation
Today I had an embarrassing experience when I sat to help my nephew learn Lewis' Dot Representation and realized that I actually don't get it at all. I have a physics background and always thought that the Lewis Structure in chemistry mimics the physical picture, but soon realized that it doesn't.
First of all, the "sockets" where the dots go on Lewis Diagrams do not correspond to actual physical orbitals. They are neither 4-fold symmetric, nor are equivalent (s-orbitals are spherically symmetric, p-orbitals are lobed, etc), so I couldn't give any explanation to my nephew what their physical interpretation is.
Also, I noticed that the rule for filling these "sockets" do not follow the Aufbau rule. For example, in the ground state of carbon, Nature first fills the $\ce{2s}$-orbital and just after that inserts an electron in two of the $\ce{2p}$-orbitals. In contrast, the rules of Lewis' Dot Representation say that C should have a single dot in each of its 4 "sockets", and just after they get exhausted, one starts pairing them. (I wonder how this principle explains compounds like methylene, in which carbon has its $\ce{2s}$-shell pair present... but anyway.)
All of this prompted me to ask, what is the story behind Lewis' Dot Structure and what goals Gilbert Lewis pursued when he introduced it? I guess if I learn this, I'll be able to explain its misleading features and not get embarrassed in front of a 14-year-old.

The Lewis dot structure is just an electron counting device.  It is not intended to indicate anything about the physics of the electrons.

Carbene exists in both singlet and triplet forms and the singlet carbene’s lone pair can be in either in an s-type or in an $\mathrm{sp^2}$-type orbital.

Lewis dot structures predate the advent of modern quantum theory and quantum chemistry. There is no Schroedinger equation associated with them -- in fact placing a dot on an atom (seems to me that it) violates the concept that electrons must be indistinguishable.
There are no slots, there are no "octet rules".  There are only particles, and kinetic and potential energy that depend on their masses and charges. They take a configuration that minimizes the total energy with respect to their location in space.
So my advice is to learn Lewis dot structures as a heuristic device that can be figured out for the purposes of passing General Chemistry and speaking with the vast majority of chemists who have continued to suffer with learning about octet rules, resonance structures, etc. and the "astonishment" when these rules fail. They are hard to learn, because one can not draw on experiences learned about other physical systems that depend on e.g. QM, electrodynamics, etc.
There is a subject known as Valence Bond theory, that uses (spin-adapted) Lewis structures as a basis. But if you knew what went into constructing a proper quantum mechanical valence bond wave function, you might as well say "sure, anything so long as it forms a complete basis."  i.e. we might as well talk about molecules being made up of millions of plane waves.
You did mention orbitals: "Orbitals" only arise due to the fact that they are complete bases of Spherical Harmonics and Laguerre Polynomials that were used by Schroedinger to solve the case of the hydrogen-like atom -- a spherically symmetrical system.  They do not really apply to molecules, and it is common to use other bases to describe molecules such as Gaussian-Type orbitals, plane waves, etc.  Atomic orbitals, molecular orbitals (constructed from linear combinations of AO's) and plane waves have never been observed.  Mathematically, orbitals may be complex, and leaves me wondering why we place so much stock in them except as useful complete bases with convenient boundary conditions, e.g. periodic with a unit cell or decaying in the limit of infinite distance.  (In contrast, the electron density is real and observable.)

I thought I remembered something about having been able to observe the orbitals of a water molecule but I could be badly wrong.

How do we observe molecular orbitals?  It's not like they're tiny glass globules attached to the molecule.

@OscarLanzi In the same way that we have pictural representations of measured atomic orbitals, for example? At least that one is known.

@Jan I don't think they have been observed, even if there is a paper where they say that they observed them. Even one case would imply that it's possible to observe Psi and Psi*, and I don't see how its possible to observe things that are represented as a + b i and a - b i.

Lewis published his theory in "The Atom and the Molecule". April 1916.
J. Am. Chem. Soc., 38, 4
It means that no orbitals and electron density were present. Therefore his arguments were not related to quantum mechanics.
By the way, orbitals can be obtained without any basis set.


text:
What's the best way to interconvert between molarity and percentage?
Let's say that I want to convert concentration of an acid from % to molarity.
One way I found to do this is by using the following formula:
percentage = (molarity * molar mass) / 10.
Another way is to use an equivalency chart like this one (see bottom of page): http://www.ochemonline.com/Acid_and_Base_Concentrations
For example, to convert 38% HCl to molarity I calculated:
38% = (molarity * 36.46) / 10 and molarity = 10.42 = 10 M.
However, the equivalency chart says that 38% is actually 12.4 M.
The same discrepancy happened when I tried to convert 68% nitric acid molarity. Using the equation, I got molarity to equal 11 M, however, according to another source 68% nitric acid is closer to 15 M.
So my questions are:
1) Is there something I'm doing wrong that I get such a huge discrepancy?
2) What's the best way to figure out molarity from % and vice versa?

Yes, you are doing it wrong. The equation you are using is not based on known principles. It may be from an observation for 1 acid (or base). As for the best way, there is only one way to do it correctly. Take a known amount of your solution (like 1 kg), determine the mass of the components, convert the mass to the number of moles, and then determine the concentration. You will need to make an adjustment for the volume using the density.

It looks like your values are off by a factor of the density of the conc acid. See LDC3's comment.

What do you mean by percentage? https://en.wikipedia.org/wiki/Concentration https://en.wikipedia.org/wiki/Mass_concentration_%28chemistry%29#Usage_in_biology

Oh my goodness! Thank you, @LDC3! You've answered my question, which I wish I could accept as an answer. I'll post the answer to my own question but if I can give credit to you instead let me know how!

Thanks to LDC3's comments, I understand that the best way to convert from % to Molarity is as follows:
1) multiply density by volume at hand to get weight
2) solve ratio for x, where x is grams of solute (in volume at hand)
(given% / 100%) = (x / weight from step 1)
3) multiply x by molar mass to get moles (in volume at hand)
4) if needed, multiply resulting moles/volume in hand to get Molarity
Example: Let's say I have 100 mL of 38% HCl.
1) 1.19 g/ml * 100 mL = 119 g
2) (38%/100%) = (x/119) where x = 45.22 g
3) 45.22 g * 1mol/36.46g = 1.240 moles
4) 1.240 mol/100 ml = 12.40 mol / 1000 mL = 12.4 M
Now, the result matches the equivalency chart!

There is no easy way to transform the (mass) percentage of a solution into molarity. This is because percentage relies on the relative mass ratios while molarity is a measure per volume. These two values are connected via the solution’s density; while we can approximate the density of an acidic solution to be approximately that of water for low concentrations, highly concentrated acidic solutions have very different and unpredictable densities.
For example, $\pu{38\%}\ \ce{HCl}$ has a density $\rho=\pu{1.189g/ml}$. This allows us to calculate:

$\pu{1.000l}$ of $\pu{38\%}\ \ce{HCl}$ contains $\pu{1.189kg}$.
$\pu{38\%}$ of this is $\ce{HCl}$, so $m(\ce{HCl}) = \pu{0.45kg}$
Therefore, the amount of $\ce{HCl}$ in $\pu{1.000l}$ is $n(\ce{HCl})=\pu{0.012kmol} = \pu{12mol}$.
Therefore, $\pu{38\%}\ \ce{HCl}$ is $\pu{12M}$.

Your formula does not contain any reference to a density and is thus doomed to fail.


text:
Nucleophilic substitution in aromatic compounds

Predict the major product in following reaction:

In my view this reaction will proceed via benzyne intermediate and $\ce{NH2-}$ will act as a base and substitute one of the chlorine. After benzyne formation as $\ce{Cl}$ is ortho/para directing my view was that $\ce{NH2-}$ will attach to the meta position.
However, the answer key says overwise. Any ideas?

Think about which species can act as electrophiles in EAS, would ammonia work?

@AS_1000 in my view $\ce{NH_2 ^{-}}$ acts as a nucleophile

Look up Meisenheimer intermediates rather than Wheland intermediates.  The former are valid for nucleophilic aromatic substitutions.

Note that halides are weak EWG's. So, para product is preferred


text:
Formation of Hydrogen - from a proton and an electron
If I place a proton and an electron close to each other, can they form an atom of hydrogen? Why or why not? Please explain the scenarios where this is not likely/unlikely.
Personally I doubt it, since the electron would have to spin around the proton to call it an atom.
Can it be well-explained by classical mechanics?
How does quantum mechanics explain this sort of a possible bonding? I mean,an electron and a proton can't collide with each other because they would have had infinite attraction between them (that is they would act like magnets), and we know that this does not happen.
Is there something in QM that defines a minimum distance of some sort that is relevant to the proton-electron forming hydrogen?

A proton by itself IS already a hydrogen atom - the electron doesn't matter in that regard at all. The proton is an atom that (also) is a positively charged ion, $\ce{H+}$

@StianYttervik That is quite wrong.

@Mithoron I might have been too generalizing, at the detriment of accuracy, in the second sentence, I'll admit that. BUT - even IUPAC recognizes, and calls a lone proton for "hydron" - cationic hydrogen ion. I won't admit defeat on that one, just yet...

@Mithoron Elements are defined by the number of protons they have.  Unless there is some crazy technicality thing that I'm missing, I am going to agree with Stian.

@LordStryker Long time, no see; I could suspect this would draw you, though. BTW this question shouldn't be here, it's hardly about chemistry and rather broad/unclear.  For a chemist proton may seem to be ionised hydrogen, but this here is rather physics and bare proton is no more hydrogen then single electron.

@LordStryker An element is defined by the protons, an atom is defined as uncharged with the same amount of electrons and protons. So a proton is not an atom, it's an ion. https://goldbook.iupac.org/html/A/A00493.html

@DSVA I'm talking about whether or not a proton is an element (hydrogen; H+).  Is Na+ not sodium?

@LordStryker you were clearly saying that a proton is an atom and nowhere in this comment is the word element: "A proton by itself IS already a hydrogen atom" and "The proton is an atom that (also) is a positively charged ion,  H+".  Na+ is a sodium ion, not a sodium atom but yes, it's sodium.

Sorry for the confusion @DSVA.  My comment states 'element' but I suppose it was in the context of Stian's comment which states "atom".

@LordStryker oh sorry, I somehow confused some of the commenters.

Both in classical and quantum mechanics you have to get rid of a (rather large) amount of energy, i.e. the difference in potential energy from zero (=infinite distance) to the bound state (= e.g. Bohr radius), also called the binding energy.
Nature has no way to do that without a third particle taking the excess energy with it. In quantum mechanics there is the additional problem with the spin, but also classically that doesn't work if you assume elastic interaction of proton and electron.
So the answer is no.
(Inelastic interaction requires a zoo of additional assumptions, which I let fall victim to Occams razor here. Inverse photodissociation is a possibility, if overall energy and momentum are conserved.)

Assuming initially stationary nucleus and electron and full conversion to kinetic energy, my back-of-the-envelope calculation gives 1.6 km/s for $\ce{H}$, which is a lot compared to 0.4 km/s for $\ce{O2}$ at r.t., but not completely outlandish?

It actually doesn't matter how large the energy is, if you can't give it to a third particle, there is no way to reach a bound state.

The answer and question are rather old, but in QM this can happen via a process called radiative recombination in which a photon is formed which takes away the binding energy (this is actually how elements became neutral after the Big Bang). This is basically the reverse process of photoionization.

@Paul You are (obviously, any microscopic process is reversible) right. However the collision must occur in a way so the photon (the third particle) is emitted in exactly the right direction so the overall energy and momentum is conserved. This has a rather low probability.

If you put an electron and a proton together you get a hydrogen atom, assuming there's nothing else around it will stay a hydrogen atom.
The positive proton and the negatively charged electron will attract each other to form a stable atom. However, classical mechanics would predict that the electron would collide with the proton. That is not the case and to explain what is happening we need quantum mechanics. This also teaches us that electrons aren't spinning around the nucleus.

"this also teaches us that electrons aren't spinning around the nucleus". You mean that classical mechanics does not actually prove that an electron actually moves around the nucleus? But Bohr made an assumption that it is revolving?

Yes, Bohr was wrong. There are pretty good explanations all over the internet why the electron does not fall into the nucleus, why Bohr was wrong and why the Bohr radius still works (it's the average distance between the electron and the core rather than an orbit).  See for example here: https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Why_atoms_do_not_Collapse

yes
This is exactly what happens when hydrogen is ionised in the gas phase and then returns to an unionised state.
When it is less likely to happen is when the proton has very high energy (eg. having been emitted by nuclear decay) such high energy protons are  either absorbed by proton capture into another atomic nucleus or the lose their energy by inelastic collisions until they are able to capture an electron.
In the same way, most of the helium on earth was formed by alpha particles from nuclear decay capturing electrons.


text:
What is the correct approach to obtain PES using CASSCF?
I am new to CASSCF. I want to draw a PES for the ground state and the first excited sate of methyl amine for learning purpose. I want to verify if I am correct using the correct approach.

Do a RHF/STO-3G opt pop=full ... orbital calculation.
Use the checkpoint file obtained in step 1 and use CASSCF (provided I know the active space) to optimize the structure again and then use the scan keyword to map the PES by increasing the N-H bond distance in a different Gaussian job.
Take the optimized geometry of ground state obtained using CASCF and then use NROOT=2 and opt to get the the optimized geometry of 1st excited state and then scan to get the PES, similar to what was done in step 3.

Any correction would be greatly appreciated.

Yes in principle that is a correct approach, but you should be aware of a few pitfalls here and there.
The basis set you chose for your optimisation might not be large enough. I assume you are not going to use STO-3G for your CASSCF calculation. It is better if you run your initial optimisation with the same basis set that you are using for the CASSCF calculation. You will most likely be closer to the minimum and will safe a few cycles when optimising the molecular structure on the larger level of theory.
Picking the right active space might not be trivial. You may have to run a couple of single points to find the right balance of what you must include and what you may neglect.
The SCAN produces a rigid scan of the molecule, i.e. the backbone of the structure does not react with the induced change. That might not lead to the correct behaviour. More sophisticated (and more costly) is a relaxed scan via the ModRedundant keyword.
Be aware, that CASSCF calculations are quite resource demanding and you probably won't get results overnight.

Thanks for the reply. I am well aware of SCAN and ModRedundant keywords. I will also use another basis set which will yield values matching with the experimental ones.
However, the problem is choosing the right active space. Can you refer any book or article explaining this in detail and maybe a few examples,if possible? It would be really helpful.

@Chemist figuring that out is by some considered an art form. Unfortunately I don't really know a concise guide for that. Maybe you can find some help from one of the books in our resource thread. I'm sure I have written a few words on this site, too, but I'm on mobile and can't find it right now. You can always ask more questions here, when you have exhausted your other resources. You can also find me and other computational chemists in the periodic table ([chat]), feel free to drop by.

@Chemist Multiconfigurational Quantum Chemistry by Björn Roos et al https://g.co/kgs/hp86Uo ... also, Gaussian is by far not the nicest CAS program around.

To expand on Martin's comment about STO-3G, I know it's usually my first "test" basis as well, but the virtual space may be too poor to ever converge. You can almost certainly afford def2-SV(P) or pc-0. If those don't work, def2-SVP or pc-1. These are inexpensive but much better basis sets, and not so much larger that you can't count MOs by hand.

You also have to be careful if you scan over a large range of geometries.  It is very common for states of different symmetry to cross, so that e.g what was the first excited state when the molecule is planar is not the excited state when it is twisted.


text:
How to determine whether inversion takes place?

Predict in which of the following reaction inversion of configuration takes place:

In my view, C) involves $\mathrm{S_{N} i}$ and therefore doesn't involve inversion of configuration. In D) the syn addition of $\ce{H2}$ across the double bond takes place and doesn't change the configuration. However I am not sure about A) and B).  Any ideas?

A) is $\mathrm{S_N2}$ due to the secondary $\ce{C}$ and the sufficiently good nucleophile $\ce{I-}$. So inversion takes place.
B) is $\mathrm{S_N1}$ due to the tertiary $\ce{C}$. The solvent is also more polar than in A). So the stereochemical information would be lost.


text:
Quickly deduce block (s, p, d ...) from atomic number
A question from previous year papers of IIT JEE. There's not enough time to write the electronic configuration in the exam. Please suggest a quick, objective approach.

Need you to show your workings fast, please.  We have to see where you need help in this problem.

If you don't want to write the electronic configurations, have a reference point (usually noble gases). For example, knowing that atomic number of Xenon is 54, atomic number 55 would have to be from s-block.

Other than memorizing (and writing down for visualization) some atomic numbers, I cannot think of how to do this. I suggest the atomic numbers of the noble gases for this, which are: $\ce{He}$ 2, $\ce{Ne}$ 10, $\ce{Ar}$ 18, $\ce{Kr}$ 36, $\ce{Xe}$ 54, $\ce{Rn}$ 86. (Since noble gases are kool with a k, one should know them anyway. Use a mnemonic, if necessary.)
Then, one finds:

55 is one above $\ce{Xe}$, so s-block.
45 is nine down from 54, that is surely d-block.
81 is five down from 86, that is p-block (barely).
64 ten up from 54, that is f-block.

I have another question.How do you calculate the magnetic moment of Hg2+?I know the formula,but again-you need the unpaired electrons for which EC is necessary.What should i do here?

@GKA Comments are not for unrelated questions.

It wouldn't let me ask another question,anyway this is kind of related to electronic configuration isn't it?


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Oxidation by DMSO

1,2-Butadiene is obtained by oxidation of which compound with DMSO?

I thought that (A) will form a aldehyde and (D) a ketone. (C) will not react will DMSO, and so (B) should be the correct answer, but I couldn't get the mechanism and how 1,2-butadiene is formed from (B). Any ideas?

I am completely at a loss of how this is supposed to happen at all. Which is weird.

1,2-butadiene??

1,2-Butadiene is an allene (CH3CH=C=CH2); or was the name supposed to be 1,3-butadiene? Both are C4 compounds. C and D are eliminated because they are both C5 compounds. Compound A and B are both C4. Compound A, 1-bromobutane, would yield, in the presence of NaHCO3/DMSO at elevated temperatures, n-butyraldehyde (Kornblum Oxidation, 1959). Conceptually, compound B could provide 2-butanone by the same oxidation. While both of the carbonyl compounds are at the same oxidation level as either diene, I have severe reservations that either diene is formed.