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Is there a widespread error/contradiction on whether to use internal or external pressure in computing work done by an expanding gas?
I've been looking at some sample problems involving work done by gas in an irreversible expansion in a piston.  They all say $W = P\Delta V$, but some of them say to use internal pressure for $P$, some of them say to use external pressure, and I'm wondering if a large portion of the sources are simply wrong.  My argument is that you should use (changing) internal pressure to compute work done by the gas, and (usually constant) external pressure to compute work done by the gas plus the piston.  But many sources say to use external pressure to compute work done by the gas, and this sounds incorrect.
Suppose you have gas compressed in a piston at 4 atm, currently volume 1 liter, and the piston is frictionless but held in place by a catch (with normal 1 atm pressure outside).  (Edited to add: let's assume the piston is lying on its side so the mass of the piston does not contribute to the internal pressure.)  You release the catch and the piston expands to twice its volume (and pressure decreases to 2 atm) where it's caught by another catch.  What is the work done by the gas?
Here is my reasoning:

If you consider "the system" to be the gas, then the work done by the gas as it expands is the integral of the internal pressure over the change in volume, because the first thing I learned in physics is that work is the integral of force over distance moved.
If you consider "the system" to be the piston as a whole, then the work done by the piston is the external pressure times the change in volume.  This is the work required to make room in the surroundings for the expanded system.
The difference between #1 and #2 is because when the gas expands, it imparts kinetic energy to the piston, and when the piston hits the second latch and locks into place, the kinetic energy gets diffused as heat energy.  But that should still count as work that the gas did to get the piston moving, even if the piston lost the kinetic energy when it hit the latch.

(This is similar to the question posted at:
https://physics.stackexchange.com/questions/135175/work-done-against-gravity/135194
where someone asked: if I'm lifting a 10 N weight 1 meter by using a 50 N force, why is the work done only 10 Nm and not 50 Nm?  Because if you lift it using a 50 N force, the rest of that work goes into the kinetic energy of the object. Presumably if the object hit something and got locked into place then the kinetic energy would be diffused as heat energy.)
Is my reasoning right so far?
Because then, here are several supposedly reliable sources which all say that the work done by the gas when it expands, is the external pressure times the volume change.  My position is that unless it's specified that it's reversible (i.e. external pressure = internal pressure), this is (subtly) incorrect, and you would use external pressure to compute work done by the gas plus the piston, but not the work done by the gas by itself.  I found these just by googling "pressure volume work" "work done by the gas", and about half of the results say to use external pressure, and that's what I'm arguing is incorrect.  Here are the sources saying to use external pressure; for each document, you can search for the phrase "work done by the gas":
http://www.columbia.edu/itc/chemistry/f1403/lectures_ppt/Wk_10_Mon.ppt
https://facweb.northseattle.edu/hprice/CHEM161/Silberberg%20PPT%20to%20link/ch06_lecture_7e.ppt
https://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_(Atkins_et_al.)/Chapter_2%3A_The_First_Law/Topic_2A%3A_Internal_Energy
why do we use the external pressure to calculate the work done by gas
(the "accepted" answer says to compute work done by the gas, you use the external pressure; the other answer says that you use the internal pressure)
http://www.quantumstudy.com/chemistry/thermodynamics-4/
https://quizlet.com/77523508/chemistry-133-chapter-6-thermochemistry-flash-cards/
http://docfish.com/wp-content/uploads/2014/03/330_06_03a_Energy_First_Law_of_Thermodynamics_HW_KEY.pdf
and many more which come up just googling "pressure volume work" "work done by the gas".
As long as work is the integral of force over distance moved, I'm saying that to compute work done by an expanding gas, you use internal pressure (if it's different from external), and these sources are incorrect.  Am I missing something?
(To avoid confusion, please specify: (1) do you agree that to compute work done by an irreversibly expanding gas, you use internal pressure, or am I wrong; and (2) if I am right about #1, isn't it also the case that the sources above which say otherwise, are incorrect? If not, why not?)

Here is the short answer to the best of my knowledge, after figuring out what was being said in the subsequent comments and in threads on other sites:

In the system under the exact conditions that I specified, where the piston is stopped by a latch, the work gone by the gas will indeed be greater than pressure times change in volume.
In any gas expansion problem where the gas pressure is greater than the outside environmental pressure, generally the work done by the gas, from the moment the gas starts expanding to the moment the gas reaches peak volume, will be greater than pressure times change in volume.
However, what happens next (and the final answer to the question of "how much work is done by the gas") depends on how the expansion is stopped, and it's possible that the work done by the gas is negative, bringing total work back down to pressure times change in volume.  If the piston is stopped by a latch as in my original question, then the gas does no more work and no more work is done on the gas, and the total work done remains greater than pressure times change in volume.  But, if the expansion is stopped by other means (i.e., the frictionless piston moves back and forth until the dampening effect from the gas slows it down and stops it), then work is done on the gas (i.e., work done by the gas during that time period is negative), and the total amount of work done by the gas, after the system has stopped moving, is equal to pressure times change in volume.

By Newton's third law, the force exerted by the gas on the piston is equal and opposite to the force that the piston exerts on the gas.  So to get the work, either force will do.  However, the ideal gas equation of state only applies to equilibrium situations, and, for non-equlibrium situations (such as irreversible expansion or compression), viscous stresses also contribute to the forces and pressures within a gas.  So in irreversible processes, the ideal gas law will not give the correct result for the work done by the gas on the piston.  Therefore, the only alternative is to use the external pressure exerted on the gas to calculate the work.  This is great if you are told what that external pressure is.  Otherwise, you're out of luck.
In the case of a reversible expansion or compression, $P_{ext}=P$, and the system is only slightly removed from thermodynamic equilibrium, so you can use PdV to get the work, where P is the equilibrium pressure of the gas.  However, in all cases, irrespective of whether the process is reversible or irreversible, you can always use $P_{ext}dV$.
ADDENDUM:
A very crude approximation to the compressive stress $P_{int}$ of the gas in the cylinder for both reversible and irreversible processes is given by:  $$P_{int}=\frac{nRT}{V}-\eta\frac{1}{V}\frac{dV}{dt}=P_{ext}$$ where $\eta$ is proportional to the gas viscosity, and dV/dt is the rate of change of gas volume with respect to time.  So you can see that, for an irreversible process, where the gas volume is changing rapidly, the compressive stress in the gas $P_{int}=P_{ext}$ depends not only on the volume but also on the rate of change of the volume.  But, as the rate of change of volume (the second term in the equation) becomes small, the gas behavior approaches that described by the ideal gas law.  The viscous term in the equation is a damping factor, and causes any oscillations of the piston in irreversible expansions to die out (in addition to causing deviations from the ideal gas law).

But even if P(int) is unknown, surely P(int) is what gives the correct answer to the question: "How much work was done by exerting this force to move an object this distance?"  If work done equals force exerted times distance moved.  Again, it's like the question of taking an object that weighs 10N, and applying a 50N upward force to lift it 1 meter, before it locks into place.  The work done by the 50N force is still 50 Nm, isn't it?  And some of that is imparted to the object as kinetic energy, and when it locks into place that gets dissipated as heat energy.

Yes.  Even if it is unknown, it nevertheless gives the correct answer to the question.  And yes, all the rest of what you said is correct too.  Even if it oscillates (rather than locking in place), after the oscillation damps out, the kinetic energy is dissipated, and the work done is still 50N.m.

Um, OK, but if P(int) != P(ext), if P(int) gives the correct answer, then P(ext) would give the incorrect answer, so what did you mean that "you can always use P(ext)dV"?  You can always use it, even though it will sometimes give the incorrect answer? :)

Using $P_{ext}$ never gives the incorrect answer.  For an irreversible process, you can't determine $P_{int}$ using the equation of state, so how can you determine it? What we are saying here is that, at the inner piston face where the work is done,  $P_{int}=P_{ext}$, but we can determine $P_{int}$ only for a reversible process.  I might also mention that, for an irreversible process, the pressure (as well as the compressive stress, which includes viscous stresses) is not even uniform spatially.  So the average value of the internal compressive stress does not equal that at the piston.

That seems wildly counterintuitive that P(int) would be the same as P(ext) as soon as the piston is un-latched, if P(int) was 4 atm while the piston was still latched.  If P(int)=P(ext) then it would seem like the piston wouldn't move.

Also, there are many sources such as https://www.grc.nasa.gov/WWW/k-12/airplane/work2.html which say that to compute the work done by the expanding gas, you integrate P(int) over the volume change, where P(int) is clearly changing and is clearly not equal to P(ext).  This is the approach that I was arguing for from the beginning; however, since it's using P(int) which is different from P(ext), are you saying that page is incorrect?

It depends on what you are calling $P_{ext}$.  If $P_{ext}$ is the force per unit area on the inner face of the piston, then $P_{int}$ (the gas force per unit area) is equal to $P_{ext}$ (the piston force per unit area).  If, on the other hand, you are calling $P_{ext}$ the force per unit area on the outside piston face, then $P_{int}=P_{ext}+ma$ (assuming a horizontal piston), and $W(t)=\int{P_{int}}dV=\int{P_{ext}dV}+KE$, where W(t) is the work done on the piston up to time t and KE is the kinetic energy of the piston.  Finally, the kinetic energy is damped out by viscous forces.

Of course, if the piston is massless, then it doesn't matter which face you assign $P_{ext}$ to.  And, in this case, even the slightest difference between the pressures on the two faces of the piston will cause it to move.

I have edited the question to assume the piston is lying on its side, so there is no ambiguity about "external pressure".  So now, we start from a gas expanding from pressure 4 atm to pressure 2 atm, while the external pressure is 1 atm at all times.  But with or without that clarification, it still doesn't seem right to say that P(int)=P(ext).  Once the piston is unlatched, the piston is moving, so it's not at equilibrium and I don't know why I would assume P(int)=P(ext).

Let us continue this discussion in chat.

Well since you wrote $W(t)=\int{P_{int}}dV=\int{P_{ext}dV}+KE$ , this is what I've been saying all along -- the pressure times the change in volume is not the entire work done by the gas, you need that extra term KE.  When you wrote "You can always use PdV", I assumed you meant, obviously, that PdV was the correct answer, not that you "can use PdV" as part of a larger answer with other terms added to it.

For people coming across this discussion in the future, we continued it at https://www.physicsforums.com/threads/use-internal-pressure-to-compute-work-done-by-an-expanding-gas.951088/  .  My posted and accepted answer was the short version of the information I was able to piece together from all of the subsequent comments.


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Is it correct to say H2SO4 behaves like an acid in this reaction?
I came across this equation :
$$\ce{ZnCO3 + H2SO4 → ZnSO4 + CO2 + H2O}$$
I wanted to know if $\ce{H2SO4}$ behaves like an acid in this reaction?
How do we generally determine if a compound acts as an acid?

In general, this depends on the definition of acid that you are using.
For this specific case, the Broensted definition would suffice.
A Broensted acid is a reagent that donates a proton ($\ce{H^+}$) to its reaction partner.
The protons of the sulfuric acid are transfered (donated) to the carbonate anion, which gives carbonic acid. Carbonic acid decomposes into water and carbon dioxide, as you will experience when drinking sparkling water.
The key thing now is the affinity to donate said proton, which leads to the concept of acid strength and can be expressed in terms of $K_\mathrm{a}$ values.
If the anions were reversed and $\ce{ZnSO_4}$ would react with $\ce{H_2CO_3}$, one would find no reaction occuring.
This is a very easy explanation of the things going on, and I'd suggest you read a bit into

Acid definitions: Broensted and Lewis acids
Acid strength: Equilibrium of dissociation reactions
Conjugated Acid / Base pairs

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Checking if a reaction is first-order
I saw this question in my book. Some data about a reaction is given and basically I have to demonstrate that the reaction is indeed first order, in order to "justify"the given data.

In the solution given, the integrated rate law for a first order reaction has been used to find the rate constant for different time intervals and it is stated that since the value comes out to be largely constant, the reaction is first order.
Is this enough proof? And am I correct in assuming that this approach was followed because had the values been placed in the integrated rate law for any other order, the values of the rate constant would not have the constancy as shown?
Or is there a better, more concrete approach towards questions of this type?
This is the solution given:

Chemistry isn't like math where a "formal proof" is possible. You can only observe the behavior. So within "experimental error" the reaction is first order.

I understand. So this method too is legitimate.

The integral method is a very good to determine rate constants given that the order of the reaction is known.
For a reaction of unknown order, the differential method is better suited (in my opinion) to determine the order - and then the rate constant can be determined via the integral method (as this gives more reliable results for $k$ than the diff. method).
The generic expression for a n-th order reaction is
$$r = \frac{\mathrm{d}c}{\mathrm{d}t} = k\ce{[A]}^n$$
where n is the order of reaction. Now bringing the equation into a linear form using logarithms:
$$\ln(r) = \ln(k) + n  \ln([A])$$
Now approximating  $\frac{\mathrm{d}c}{\mathrm{d}t}$ as $\frac{\Delta c}{\Delta t}$ and $\ce{[A]}$ as the average between the measurements, one can tabulate the data in the form: $\ln\left(\frac{\Delta c}{\Delta t}\right)$ vs. $\ln ([\ce{A}_{\text{avg}}])$
Plotting this data will give you a linear curve, whose slope will correspond to the overall order of the reaction.
It is important to keep in mind that the approximation $\frac{\mathrm{d}c}{\mathrm{d}t}$ as $\frac{\Delta c}{\Delta t}$ is only valid for a small time interval $\Delta t$ and when the reaction is far from completion.
If you end up with a fractional order of reaction, look at the magnitude of the change of concentration and perform the linear regression only on the data pairs from the beginning of the reaction.

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The effect of the voltage at the Downs cell
I referred the Wikipedia and Advance Chemistry by Philip Matthews, for having a much comprehension on the down cell, but there was a opposition while reading the both references.
Wikipedia states that,

Although theory predicts that a potential of a little over 4 volts
  should be sufficient to cause the reaction to go forward, in practice
  potentials of up to 8 volts are used. This is done in order to achieve
  useful current densities in the electrolyte despite its inherent
  electrical resistance. The overvoltage and consequent resistive
  heating contributes to the heat required to keep the electrolyte in a
  liquid state.

But the Advance Chemistry book by Philip Matthews (page 521) states that,

A large current is passed through the cell, but at a low voltage.
  This has the effect of both discharging the sodium effectively and
  heating the mixture so that it does not crystallize

The both references have a opposition at the fact of amount of voltage to heat generating.
Which one is correct? Also I hope an answer with the reasons of how does the voltage affect on downs cell, with the heat generating.  Thanks

Matthews assertion of "low voltage" is non-specific. Some over voltage will be necessary. The exact amount of over voltage depends on a number of cell design features.


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Is it true that an evaporating molecule has the same kinetic energy as a molecule in a pot of boiling water?
A molecule on the surface of room-temperature water shoots off the surface of said water, or in other words, it "evaporates." It does so because it gained kinetic energy ${x}$, and ${x}$ was great enough to counteract the pressure of the air surrounding it.
Is it true that this ${x}$ would be similar (the same or greater) as the average kinetic energy level of a molecule in a pot of boiling water?
Reasoning for thinking this: boiling is when the vapor pressure = the pressure of the atmosphere.

Are you asking if the minimum kinetic energy to escape the surface is constant, i.e. not temperature dependent?

@MaxW I believe that is one way to word my question. Is it constant given an atmospheric pressure.

Is it true that an evaporating molecule has the same kinetic energy as a molecule in a pot of boiling water?

Well for one the kinetic energy for a given temperature is an average kinetic energy of a distribution of energis. So for the question:

"Is it true that an evaporating molecule has the same kinetic energy as a molecule in a pot of boiling water?"

The answer is maybe or there probably are molecules that have the same energy in each phase, but I think the question you want to ask is:

Is it true that all molecules in steam have the same average kinetic energy as all the molecules in a pot of boiling water?

From a thermodynamic perspective, at the boiling temperature the liquid and vapor phases are in equilibrium and have the same temperature. Since temperature is the measure of the average kinetic energy, if temperature is equal average kinetic energy is equal.
So simplistically speaking yes they are the same. You can stop with reading here or veer with me into super technical semantics and minutia—your choice.

Reasoning for thinking this: boiling is when the vapor pressure = the pressure of the atmosphere.

Now it is commonly taught as you state: "boiling is when the vapor pressure = the pressure of the atmosphere". The thing is though that this is a restatement of "at the boiling temperature the liquid and vapor phases are in equilibrium and have the same temperature" which is similar but not equivalent. You said:

It does so because it gained kinetic energy ${x}$, and ${x}$ was great enough to counteract the pressure of the air surrounding it.

Boiling occurs at the bottom of the pot which has a higher hydrostatic pressure due to the height of the water $(P = \rho gh)$. So the molecule must not just over come the pressure of the atmosphere but also the pressure of the water height. This increased pressure mean boiling occurs at a higher temperature than water at the top of the pot (see: water saturation pressure). As the steam bubble rises, the pressure decreases ($h$ decreases) and the bubble expands the steam will cool due to the work done by expansion (see: internal energy) lowering the equilibrium vapor pressure and condense at the interface of the bubble. The thermal transport is not instantaneous though so the interface inside the bubble will absorb heat from the latent heat of condensation, and in turn be hotter than the surrounding water thus the bubble will be hotter in the center than at the interface with water. Since the vapor is (very) slightly hotter, the average kinetic energy will be (very) slightly hotter, thus (very) technically it would not be correct that all the molecules in steam have the same average kinetic energy as all the molecules in a pot of boiling water

Thank you, this has resolved a year old debate with my cousin :-P

@IsaiahTaylor No problem, but I will note  I didn't even mention super heating.

Seems true just by the definition of boiling point. At boiling point your liquid is in equilibrium between its liquid and gas phases, hence the "average" molecule will possess that minimum amount of kinetic energy to escape into the gas phase (and vica versa the maximum for it to condense back into the liquid phase).
In your earlier example, at "room temperature" (and 1 atm) only a small percent of water molecules will have that much (or more) kinetic energy, and hence the vapor pressure will be less than 1 atm. And that will definitely not be the "average" kinetic energy (as indicated by that less-than-boiling "room temperature") across all of them!

I do not think so since (a) your answer does not take into account the intermolecular interaction energies that are present in the liquid phase water and are lost when reaching the gas phase and (b) if the molecules considered are polyatomic, the energy to be considered also contains vibrationnal and rotational energies.

@PLD care to write an answer to it? I'm curious about these considerations

[OP] It does so because it gained kinetic energy x, and x was great enough to counteract the pressure of the air surrounding it.

A gas particle does not "feel" the pressure of the gas unless it collides with another gas molecule or collides with molecules making up the container. The reason the molecule has to have high energy is that it needs to break the intermolecular interactions with other molecules in the liquid.
On the other hand, a gas bubble does feel the pressure of the liquid in the sense that the size of the gas bubble depends on the pressure (and the number of particles in the bubble, and the temperature of the gas). However, even at the surface of the bubble, individual molecules have to have high energy to break intermolecular interactions. If the molecules thus released into the gas phase (the bulk of the bubble), their number and the temperature of the gas in conjunction with the pressure of the water at that level will determine the volume of the bubble. This in term changes the number of collisions of the gas molecules with the surface, which in turn affects the rate of condensation.
If you look at a pot of water being heated up (which eventually, against the expectations of the "a watched pot" saying, will boil), you can see various stages of bubbles forming and disappearing again, bubbles rising to the surface as they are shrinking, and bubbles rising to the surface as they are expanding. You will also hear the bubbles collapsing, making interesting noises.

The paradox of vapour molecules seemingly needing to have a high kinetic energy to escape from the liquid but not having a high energy in the vapour is resolved if you think about the situation applying to many molecules in both phases.
Yes, to get out of the liquid, individual molecules have to pass a threshold of kinetic energy. So the vapour molecules appear to be selected to be those with high energy.
But this is a very transient process and only in something close to a vacuum would that kinetic energy stay constant over an extended period of time. Molecules in the gas phase interact with other molecules extremely quickly and these interactions redistribute the energy across all the other molecules in the gas phase. The distribution of the individual energies of the vapour molecules will, therefore, equilibrate extremely rapidly with the other molecules making up the gas above the liquid. So the apparent transient selection for molecules with a high average energy (because they need to have enough energy to escape the liquid) will disappear extremely quickly. So the vapour molecules will have–for all practical purposes that are measurable–the same kinetic energy as their fellow gas molecules in the atmosphere above the liquid.
So the apparent selection effect is dissipated so fast that it doesn't carry over to the gas phase.


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Does neopentyl halide undergo SN1?

Carbocation formed from neopentyl halide is primary in nature. So it is not stable. Although it undergoes rearrangement reaction to form tertiary one, now it can undergo SN1 easily. So if we say it undergoes directly in SN1 will it be okay?

Neopentyl halides are very poor electrophiles

@Waylander - if you can provide examples and references, your comment could be turned into a good answer.

Doesn't the neopentyl carbocation get formed in an E1 reaction?

Neopentyl halides are not good electrophiles in the way that other primary alkyl halides are, primarily due to steric effects but also because of rearrangents and eliminations.
This paper 1 notes in its abstract that "As a rule, the SN2 reactions of neopentyl bromide and tosylate with sodium alkoxides are extremely retarded by steric hindrance, and the yields of alkyl neopentyl ethers are poor because of side reactions such as neopentyl rearrangement, elimination, and decomposition." This is supportive of the view that an SN1/E1 process may be occurring.
This paper 2 confirms rearrangements occur.
I don't think this leads to us being able to say neopentyl bromide undergoes SN1 reactions directly or easily. It is unreactive to nucleophiles; when it does react it does so through mulitple pathways leading to low yields.

Hello, I had some confusion regarding your last paragraph in the answer. First you state that the view of SN1 is supportive due to rearrangements and then you state that it's not a sole argument to decide whether SN1 proceeds . So according to you, we should always doubt that 3° carbocations can proceed with SN1?

The main point is that neopentyl halide , even after re-arrangements or whatever be the experimental conditions maybe, always proceeds with a very very slow rate ; about which the answers don't talk about. Atleast explain that what are the multiple pathways are you talking about and how do they affect.

@Shiva  The multiple pathways are mentioned as " neopentyl rearrangement, elimination, and decomposition". The rearrangements can occur by methyl shift as described in the second reference, other pathways might include dimerisation and cyclopropane formation.

So why don't the rearrangement makes the rate faster?

They do not affect the rate - only the outcome. Everything happens slowly with the neopentyl series. Formation of a primary cation is slow because it is unfavourable. You get mostly unchanged starting material out of these reactions.

Thanks very much, now I understand. What I conclude finally is that formation of the very first carbocation is a very slow process due to it's unstability. Finally due to rearrangements and other processes you mentioned ( of which I don't know much as I am not a chemist) , this time is extended further and though the product will be formed , but very slowly.


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Why is 2-(3-aminomethyl-2-hydroxypropyl) cyclohexan-1-one named so and not 2-(2-hydroxy-4-aminobutyl) cyclohexan-1-one?
What is wrong with naming it as 2-(2-hydroxy-4-aminobutyl) cyclohexan-1-one and taking the side chain to be having 4 carbon atoms with NH as substituent on the last? Would someone please explain this? This is the exact picture given in my book labelled 2-(3-aminomethyl-2-hydroxypropyl) cyclohexan-1-one.

Because -NHCH3 means a different thing from -CH2NH2 which is what you seem to be thinking of.

As drawn the compound does not contain a butyl side chain with four carbon atoms; the side chain is only a propyl chain with three carbon atoms.
The propyl side chain has two substituents: a hydroxyl ($\ce{-OH}$) group and an methylamino ($\ce{-NH-CH3}$) group. Since these groups are not the principal characteristic group in this case, they have to be expressed as prefixes. Note that the correct prefix for the secondary amine is “methylamino” and not “aminomethyl” as proposed in the question.
Therefore, the correct name is 2-[2-hydroxy-3-(methylamino)propyl]cyclohexan-1-one.

I have a doubt, why isn't the -OH group a principal characteristic group of the propyl side chain?

@Hema The entire compound can only have one principal characteristic group that is expressed as a suffix, which in this case is the ketone.


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What are those... crystals?
I have been trying to make muratic acid from salt and phosphoric acid (didn't know that it's made at 150 °C). After waiting a few days, a third of the time it was sealed by placing a container over it, I wanted to check if some HCl was created, by adding peroxide and etching a small PCB, after 2 days I noticed small white crystals and the solution turned green.
What are those crystals and why is the fluid green?

Likely the green solution contains copper(II) phosphate, $\ce{Cu3(PO4)2}$. "It is commonly encountered as the hydrated species Cu2(PO4)OH, which is green."
Try crystallizing it!

Libethenite, $\ce{Cu3(PO4)2}$, crystals.
You have $\ce{H+}$, $\ce{Na+}$, $\ce{Cu+}$ and/or $\ce{Cu++}$, $\ce{OH-}$, $\ce{Cl-}$ and $\ce{PO4{^-}{^3}}$ in solution. You could identify the white crystals by color (white, so probably no Cu in them) shape, birefringence, and solubility; they might be $\ce{NaCl}$ or $\ce{Na3PO4}$. The less soluble should crystallize first, of course.

You almost certainly mean Cu++ (not Cu+++) in the last paragraph. Tried editing it myself but the change is too few characters ...

@IanBush, thanks for catching that typo! Cu(III) is unlikely in that mixture, though it does exist in some rare compounds.


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Which element has the highest binding energy per nucleon
The graph of binding energy per nucleon (Hyperphysics) seems to peak at Iron-56. However, Wikipedia says that Nickel-62 has the highest binding energy per nucleon of any known nuclide (8.7945 MeV). Isn't this a direct contradiction? Why does the binding energy per nucleon peak at Iron-56, but then Wikipedia says Nickel-62 has the highest binding energy per nucleon?

Is this the graph from Hyperphysics? If so it explicitly says on the graph that Ni-62 is the most bound.  If it is not can you quote and give a link to what you are referring to? The most relevant link I can on hyperphysics is find is http://hyperphysics.phy-astr.gsu.edu/hbase/NucEne/nucbin2.html#c1

Ni-62 is not the most common nickel isotope so they have likely plotted the most common isotopes. Of those, Fe-56 is the most stable while Ni-62 is the most stable overall. I suspect there's some very interesting nuclear physics associated with how additional neutrons can stabilize a system and yet make it more unlikely to be created.

https://chemistry.stackexchange.com/a/32123/9961 covers this

@IanBush, yes this was from hyperphysics but doesn’t the statement “nickel 62 has the most tightly bound nucleus” directly contradict the graph, which peaks at iron?

Can't be expertly comment on the plot without knowing exactly how the plot was constructed. But I can make the following observations...

User Mithoron makes a good point that it depends on how the Y-axis of the plot is taken. You can plot average mass per nucleon in which case iron-56 is the heaviest atom per nucleon. However if you plot binding energy per nucleon, then nickel-62 has the highest binding energy per nucleon.
For elements with stable isotopes I'd guess only the stable ones are shown. For elements with unstable isotopes than some selection process is necessary so as not to clutter up the plot too much.
I'll point out that the "peak" is marked as "Iron Group." It isn't entirely clear what isotopes that means. Wikipedia notes that in astrophysics The iron group in astrophysics is the group of elements from chromium to nickel which are substantially more abundant in the universe than those that come after them – or immediately before them – in order of atomic number.
In terms of stellar production much more iron than nickel is produced. The iron  produced in stars is about about 92 % iron-56. Nickel is only about 3.6% Nickel-62.
I'd guess that the plot is a "best fit line." Here is a different plot showing more detail. You can see that not all isotopes fall on "the line."

Here is another plot at link pointed out by user IanBush in a comment. The "best fit" curve doesn't pass through all the data points.

Of some relevance is the course of the silicon-burning process, which represents the final stage of nuclear fusion in massive stars. Basically, the helium-4 nuclei formed by the primary hydrogen-fusion process "polymerize" to give a transient  Be-8 then stable C-12, O-16, and so on. In the silicon burning phase the relatively stable Si-28 can react with helium nuclei photolytically generated from other nuclei to continue this chain, but products survive only as long as the additional fusion is sufficiently exothermic (stabilizing) for the product to hold up in the thermal and photonic environment of the stellar core.
That endpoint happens to be Ni-56, so helium/alpha-particle fueled stellar fusion processes can get only as high as that mass number -- and thus, after positron beta-particle decays, to iron-56. Hence the common occurrence of iron-56 (and where there is intelligent life, steel made therefrom) even though the more difficult to reach Ni-62 would be more stable in terms of binding energy.


text:
If liquid hydrogen fluoride autoionises, how can it be an insulator?
According to Wikipedia, one of the last major hurdles in the isolation of elemental fluorine was the need to mix hydrogen fluoride with another substance in order for its electrolysis to be possible (Henri Moissan eventually hit upon potassium bifluoride, which is still used for this purpose today).
From "Fluorine Isolation":

Edmond Frémy postulated that electrolysis of pure hydrogen fluoride to generate fluorine was feasible and devised a method to produce anhydrous samples from acidified potassium bifluoride; instead, he discovered that the resulting (dry) hydrogen fluoride did not conduct electricity.

And from "Henri Moissan Preparation of elemental fluorine":

Moissan eventually succeeded in preparing fluorine in 1886 by the electrolysis of a solution of potassium hydrogen difluoride ($\ce{KHF2}$) in liquid hydrogen fluoride ($\ce{HF}$). The mixture was needed because hydrogen fluoride is a nonconductor.

However, hydrogen fluoride is a protic solvent, and undergoes autoionisation via the following reaction:
$$\ce{3 HF <=> H2F^+ + HF2^-}$$
Pure liquid $\ce{HF}$ should, therefore, conduct electricity at least weakly, like liquid water does.
What gives?

PURE Water like HF is an excellent insulator - For a simplified version https://www.scienceabc.com/pure-sciences/do-you-think-that-water-conducts-electricity-if-you-do-then-youre-wrong.html . Basically the concentration of the ions in the pure substance is too low to allow a reasonable amount of conduction

Unlike metals which can conduct with free electrons, conduction in non-metallic liquids (ionic and covalent) is dependent on the ion concentration to carry charges. Pure water has a autoionization constant of $\rm K_w = 1\!\times\!10^{-14}$, which corresponds to a total ion concentration of $\rm 2\!\times\!10^{-7} \pu M$. At this concentration the resistivity of water is $\pu{18.2 M\Omega~ cm}$ which corresponds to a conductivity of $\pu{5.5 \mu S/cm}$ and is considered to be a poor conductor. Sea water on the other hand has about $\pu{3.4\%}$ sodium chloride and a conductivity of about $\pu{55,000 \mu S /cm}$ which is a million times improvement in conductivity which is considered conductive.
Hydrogen fluoride has an autoionization constant of $\rm 10^{-12.5} ~~(3.162\!\times\!10^-13)$ which corresponds to a total ion concentration of $\rm 1.125\!\times\!10^{-6}$ which is about five-times that of water. Five times the ions gives us five times the conductivity of $\pu{275\mu S/cm}$ or a resitivity of $\pu{2.28M\Omega ~cm}$ which is still a quite poor conductor. Thus it is necessary to add an ionic substance to increase conductivity to have appreciable conduction of the hydrogen fluoride similar to sea water.


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Why is Coupled Cluster not variational?
It has been noted in several sources (e.g. J. Romero et al. Strategies for quantum computing molecular energies using the unitary coupled cluster ansatz. arXiv:1701.02691 [quant-ph]) that one of the disadvantages of the Coupled Cluster method for the description of electron correlation is that, since $e^T$ in
$$E_\mathrm{CC} = \langle0|e^{-T}He^T|0\rangle$$ isn't unitary, the method isn't variational (the energy obtained by solving CC equations isn't a rigorous upper bound to the exact energy).
Could someone please explain to me the connection between non-unitarity of $e^T$ and the method not being variational.

Related: Is coupled cluster variational for two electrons?

The most popular versions of CC are not variational because they have $\Psi_1 \ne \Psi_2$ in this type of estimate for the energy:
$$ \frac{\langle\Psi_1| \hat{H} |\Psi_2\rangle}{\langle\Psi_1| \Psi_2\rangle}.$$
This is known as a "projective" method instead of a variational method. However there do exist couple cluster methods that are completely variational:
Variational coupled cluster (vCC)
First, you can make a variational CC with a non-unitary T operator in the following way.
Recall the variational principle (taken from Wikipedia):

$$\varepsilon[\Psi] = \frac{\langle\Psi| \hat{H} |\Psi\rangle}{\langle\Psi|\Psi\rangle}$$
The variational principle states that

$\varepsilon \geq E_{0}$, where $E_{0}$ is the lowest energy eigenstate (ground state) of the hamiltonian
$\varepsilon = E_{0}$ if and only if $\Psi$ is exactly equal to the wave function of the ground state of the studied system.

So all you have to do to make a variational CC is set $|\Psi\rangle = e^{\hat{T}}|0\rangle$ in the above expression. This is called "variational coupled cluster" or "vCC", because it literally uses the "variational principle" shown above.
Unitary coupled cluster (uCC)
Unitary transformations preserve eigenvalues, so if $\hat{U}$ is a unitary transformation (unlike $e^{\hat{T}}$), then $UHU^\dagger$ has the same eigenvalues as $H$. Since the energies are eigenvalues of $H$, unitary coupled cluster can recover the correct energies with the unitary operator $U = e^{\hat{T} - \hat{T}^{\dagger}}$. For a long time people thought vCC and uCC were just two ways of writing the same thing, but this was shown to be wrong very recently (G. Harsha et al. On the difference between variational and unitary coupled cluster theories. arXiv:1711.00579 [cond-mat.str-el]; J. Chem. Phys. 2018, 148 (4), 044107.).
Something that a lot of people don't know
Regular CC is actually variational in the full correlation limit. This means that for $\ce{Li2}$ for example, CCSDTQPH is variational. This is because $\ce{Li2}$ has $6$ electrons, so CCSDTQPH is FCI, and FCI is variational. CCSDTQP and CCSDTQ and even CCSDT(Q) are not likely to overshoot the variational limit either since they are very good approximations of CCSDTQPH for $\ce{Li2}$.
Since I was asked by a moderator in the comments to post a Q/A style explanation of how "standard" coupled cluster can be considered variational in the full correlation limit, I have created this: Knowing that CI is variational for any basis set, why is coupled-cluster variational in the full correlation limit?

Thanks! Despite not being variational, does regular CC have any advantages over other types of CC?

In regular CC the wavefunction is not normalised, how can it obey the variational principle if you don't renormalise? The last paragraph also neglects that this can only be true at the complete basis set limit.

That is why I suggested a Q&A style. Just ask a question like "Why are CC(N) and CI(N) variational in any basis set?" and post the corresponding answer there. Then post the link as a comment to this answer; and you'll earn quite a bit of reputation along the way. And if anything gets moved to chat, then the discussion would or could continue there. But for the benefit of many more people, I would prefer the Q&A style.

notes which show that CI is calculated using the variational energy formula (Eq. 3.10). I have also provided now a publication which shows that CC = CI in the FCI limit: https://aip.scitation.org/doi/abs/10.1063/1.1383290, but please can you edit my answer to give the Table number where they show the FCI and CC giving the exact same result in the all-electron limit? This is because I don't have access to journals, but I do remember that they provided the table that showed CC = CI in that specific paper.

@Martin-マーチン But then what's the point of regular CC if neither the wavefunction (not normalised) nor the energy (because the wavefunction isn't normalised) are correct?

Do we have to do the Q/A style again where I ask "What is the point of regular CC?" The point is that CC(2) = CCSD contains a bit of the higher-order excitations (T,Q,P, etc.) thanks to the $e^T$ operator. Therefore CC(N) converges MUCH faster with respect to N than CI(N). This is given in the table comparing CC(N) to CI(N) in that paper I gave you. Can you see it? I do not have journal access but I remember the table. That paper is one of the most important papers in CC, which is why it has more than 500 citations.


text:
Vibrational Self-Consistent Field
There are a few things about VSCF that I don't understand:

Which coordinates do you work in? If you work in normal mode coordinates, how do you evaluate the effect of the average of one normal mode on another one, with a different normal mode coordinate? Why can't you just work in Cartesian coordinates?
Unlike the electronic SCF, the total number of vibrational quanta is not fixed for a particular molecule. How do you select which modes to include in your 'mean field'? Do you assume that all vibrational modes have only one quanta of energy in them?
At the end of VSCF, how easily could you go from the resulting set of vibrational modes described in terms of atomic movement to a vibrational wavefunction? E.g. a one that one could you to directly evaluate a Franck-Condon integral $\langle\psi^{\text{vib}}_0|\psi^{\text{vib}}_1\rangle$?

I feel like I have a lot of misunderstanding here, so I'd be grateful if you could bear with me and give as much detail as possible.


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What is the true depiction of the periodic table?
Normally with the periodic table the lanthanide series is separated out because it's long and would make the table wide. I looked for an expanded version and found this:

I found it kind of strange that scandium and yttrium are above the lanthanide and actinide series on the far left.
Does it make sense for those two elements to be there? I always assumed each column had certain characteristics like same number of valence electrons, that there were certain characteristics that held true for the whole table, but in this format I don't know how to understand the patterns or why the table ends up taking this shape.

https://chemistry.stackexchange.com/questions/7406/why-is-lanthanum-a-lanthanide-and-actinium-an-actinide https://chemistry.stackexchange.com/questions/33614/are-the-elements-lanthanum-and-actinium-considered-to-be-in-the-d-block-or-the-f

https://chemistry.stackexchange.com/questions/8357/why-does-the-3rd-electron-shell-start-filling-up-with-scandium

La and Ac have $d^1$ electrons in their valence shells, rather than $f^1$ electrons.
The long table you found looks like that for a several reasons.
The trends going down Sc-Y-La are like those seen in groups 1 and 2. The trend going down Sc-Y-Lu is like that of groups 4 to about 10. Since lanthanide chemistry is basically that of trivalent alkali or alkaline earth metals, this tips the balance in favour of group 3 as Sc-Y-La.
Also helping is that group 4 is the first in which the really characteristic properties of transition metals (variable oxidation states; colour; paramagnetism) are seen. Ditto, same thing happens with group 12, which is why they're shown as post-transition metals.
And there's the lanthanide contraction, which starts at Ce and finishes at Lu. If you show Lu as being a transition metal it mushes the end of the Ln contraction into the d-block.
And there's the periodic law, which says that the chemical elements, if arranged according to their atomic numbers, show an approximate repetition of properties after certain regular but varying intervals. Here, La is the first element after Y that shows the approximate repetition in properties.
This research supports the periodic law outcome:

Glawe H, Sanna A, Gross EKU & Marques MAL 2016, “The optimal one dimensional periodic table: a modified Pettifor chemical scale from data mining”, New Journal of Physics, vol. 18, 093011, https://iopscience.iop.org/article/10.1088/1367-2630/18/9/093011/pdf
Restrepo G 2017, "Building classes of similar chemical elements from binary compounds and their stoichiometries", in MA Benvenuto (ed.), Elements old and new: Discoveries, developments, challenges, and environmental implications, American Chemical Society, Washington DC, pp. 95-110

Both sources point to La being more distinct from the Ln than is the case for Lu.
Finally, if you count the differentiating electron discrepancies in each block of the periodic table, noting the above table has a split-d block, you'll find that an Sc-Y-La table has 12 such discrepancies whereas an Sc-Y-Lu table has 13. The current agreed-within-IUPAC table has 14 differentiating electron discrepancies!
As many people have a hard time accepting the idea of a split-d block, I personally feel that a better solution would be a Sc-Y-Lu table with group 3 being shown as Sc-Y-Lu-Lr and La-Ac. That would have the extra benefit of resolving long-standing discussions about the composition of group 3, which have been going on since about 1982, when Jensen published his paper in JChemEd, arguing for Sc-Y-Lu:

http://www.che.uc.edu/jensen/w.%20b.%20jensen/reprints/018.%20la%20vs%20lu.pdf

Nothing much has to change. Just add a 3 above La-Ac in the f-block, in addition to the one over Sc-Y-Lu-Lr in the d-block.
A chemistry book chapter on this group 3 would make fascinating reading (a good thing given, to date, that group 3 is supposed to be the least studied group).
Hope that helps.

I'm sure this question will stir up a lot of debate but I agree with your analysis of yttrium and scandium.  Both elements have a $d^1$ electron in their valence shell just like lutetium and lawrencium and belong in the d-block.  Whereas lanthanum and actinium have $f^1$ electrons in the valence shell. Chemically they all act very similarly but electrically I would group scandium and yttrium on the right.
Of note is the representation of zinc, cadmium and mercury as post transition metals. this is another overlooked but debated categorization.

Whereas lanthanum and actinium have f1 electrons in the valence shell. Is that really for certain? The RSC lists them as d1, http://www.rsc.org/periodic-table. #// On the 'debate' about Zn, Cd, Hg: I think there is much agreement that they behave a lot more like 'main group' elements, than transition metals. There are more proposals, which include group 11, but the term itself is not recognised (afaik) by IUPAC, probably because it cannot be rigorously defined.

La and Ac do indeed have d1 electrons.


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Obtaining calcium iodide from calcium and iodine
According to:
https://www.webelements.com/calcium/chemistry.html
Calcium reacts with iodine upon heating as follows:
$\text{Ca}_{\,(s)} + \text{I}_{2\,(g)} \to \text{CaI}_{2\,(s)}$
I am effectively attempting this by dropping elemental calcium and iodine into a flask and putting it under a Bunsen burner. I have a tube leading from the flask to a beaker of water to safely condense escaping iodine gas.
However, this reaction is not occurring at any appreciable rate. I am wondering if I should be using powdered calcium instead of these larger pebble-sized chunks or if a Bunsen burner is insufficient to breach the energy barrier. Are these measures or others sufficient to increase the rate of reaction?

The trouble with having a big lump of calcium is that the reaction will only occur on the surface. And, if you heat the mixture, the iodine will sublime out of the mixture faster than the reaction occurs leaning you with little product.

Your reactor should be sealed tight and probably fluidised bed would give fastest reaction. So flask with tube doesn't do it.

Looks like the Bunsen burner is not providing enough energy for the reaction to occur. AFAIK, you need around 200-400°C of heat for the reaction to occur. Yes, you can always powder the calcium metal to increase surface area and see if the reaction is occurring or adopt simple(chemical) methods. You can react any calcium salt with hydroiodic acid to form calium iodide. I am using calcium carbonate as example[1]:
$$\ce{CaCO3 + 2 HI → CaI2 + H2O + CO2}$$
Or alternatively make a iodide of iron and make it react with a calcium salt[2]. There are many more ways to make calcium iodide in [2]. You can look at them but these two methods is by far the most simple methods.
References:

https://en.wikipedia.org/wiki/Calcium_iodide
Caty J. Braford;  H. A. Langenhan; A preliminary report on the composition of syrup of calcium iodide N. F. V. DOI:https://onlinelibrary.wiley.com/doi/pdf/10.1002/jps.3080180218
http://albumen.conservation-us.org/library/monographs/sunbeam/chap10.html

Hey Nilay! Thanks so much for your help. Your second link seems to not work. Also, I know I am getting WAAAY above  200-400C with my burner.

@DavidReed I fixed the link. Hope it will work. The temperature seems to be fine in your case. Looks like the calcium needs to be powdered. You are asking to obtain HI from calcium iodide + any acid so you can't work with it. No problem. Make a iodide of iron and react with any calcium salt. Look in the 2nd link. They gave many ways of obtaining calcium iodide without using HI. If you say, I will add that part in my answer.

@DavidReed good enough

--self answer--
Using the (quite ancient) links provided above by Nilay I was able to successfully isolate calcium iodide via an alternative method. I have posted it below for those who are curious.
1.) Mix excess iron filings with elemental iodine in water. Stir and heat. At first the solution will redden, indicating the formation of triiodide, and hence the oxidation/reduction of iron/iodine and the formation of iron (ii) iodide. Right around boiling, the solution will change color from red to light brown indicating completion.
2.) Slowly add calcium hydroxide to the solution, taking care to stay below saturation. Iron hydroxide will precipitate.
3.) Filter the precipitate, which should be bluish green. (This is consistent with the color of air-oxidized iron hydroxide).
4.) Recrystallize the resulting solution via evaporation.
Why I believe this worked---
1.) The resulting white compound was soluble, but a precipitate formed when mixed with sulfuric acid (calcium sulfate).
2.) The solution turned purple when electrolyzed, consistent with the presence of iodide.


text:
Atomic weight = expected weight?
The atomic weight of an element, is it accurate to say that another way to think of it is the expected value of that element's weight if we were to sample one at random from the environment?
Are man-made versions counted as part of the weight? I'm new to chemistry but for example let's say we had a typical carbon-12 and we just started removing/adding neutrons at will, resulting in a whole bunch of various isotopes of carbon. I don't know if it's even possible to do that but say we had carbon-6, carbon-7, carbon-8, ... etc etc etc and so on until whatever isotope of carbon can't hold any more neutrons. Would those be counted?

You can't just "add" neutrons at will. This is hard to do and most of the products for any given element will be unstable and decay to something else. What gets counted in normal atomic weights is the natural abundance of the compound.

See also Atomic weights of the elements 2013 (IUPAC Technical Report) and Interpreting and propagating the uncertainty of the standard atomic weights (IUPAC Technical Report)

Atomic weights in any periodic table are based on the natural abundance of different isotopes of the element. But sometimes the source matters and the results will be different.
The most important thing to note here is that creating new isotopes is hard. You can't just "add" neutrons to an atomic nucleus and often when you do it decays to a different element.
So most periodic tables quote atomic masses using the natural abundance of the elements as they are found in nature.
When there are exceptions, different atomic masses might be quoted for samples know to deviate from the natural abundance (or, more commonly, the specific ratios of the isotopes will be mentioned). Most elements abundance will be determined by the natural processes in stars (or the Big Bang) that created them in the first place. Occasionally natural nuclear processes will alter that ratio. For example, the radioactive isotope of carbon, 14C has a half-life of 5-6 thousand years but is produced naturally in the upper atmosphere. It is well distributed in the air and incorporated into plants. But if those plants die, the amount of the isotope will decay over time lowering the ratio of that isotope in the plant. This is both the basis of carbon dating and the reason why it is possible to tell whether alcohol is produced from fossil fuel sources (old enough that all the 14C has decayed) or from fermentation of sugars (produced from living things so incorporating the atmospheric level of the isotope).
It is even possible for nature to contrive conditions that mimic a nuclear reactor causing havoc with the natural abundance of heavy isotopes of uranium. In the uranium deposits in Oklo (in Gabon) the ratio of 235 was found to be far lower than the normal levels elsewhere because of a natural nuclear reactor that ran 2bn years ago.
But we don't change the numbers in the periodic table because of this. When we talk about specific samples of elements where the ratio is not normal, we specifically mention what the ratio is compared to the natural level.

The atomic weight of an element should reflect the natural isotopic abundance. However there are statistically significantly different sampling variations. If you look at the nuclides the individual isotopes have masses known with much greater precision that the overall atomic weight for the particular element. For example Wikipedia lists atomic weight for lead as 207.2(1). However the individual isotopes of lead have masses known with much greater precision.
$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\hline
\text{Isotope} & \text{% Abundance}&\text{Atomic Weight} \\
\hline
\ce{^{204}Pb}   &  1.4  & 203.9730436(13) \\
\ce{^{206}Pb}   & 24.1  & 205.9744653(13) \\
\ce{^{207}Pb}   &  22.1 & 206.9758969(13) \\
\ce{^{208}Pb}   &  52.4 & 207.9766521(13) \\
\text{Avg Atomic Wt.} & & 207.2(1) \\
\hline
\end{array}
$$


text:
Anyone know if this kind of Polyethylene is safe to use?
I'm planning to make a weighted blanket and I've looked around the internet and most people use poly pellets or glass beads for the weight.
Glass beads is kinda hard to find in where I live, so I went with poly pellets. I realize that there are many type of poly such as PE, PP, HDPE, etc.
I ended up buying 4 pounds of these, I asked the seller and he said it's PE so I'm wondering if these are safe to use? Because the smell is quiet strong (smells like oil to me). They have a white transparent version but it costs more, this only costs 1/3 compared to the white transparent version.
Any help will be appreciated.

We really don't have enough information to provide useful answers here. Many types of polyethylene are used safely for consumer products, including food storage containers, but if nothing else, the strong smell will make whatever you've got undesirable. It may be possible to wash the beads with some solvent, but without knowing the grade of plastic or what they may be contaminated with, it will be hard to give any answer about how suitable they are for your use.

I wouldn't recommend a petroleum-based product anywhere near an area you are going to sleep in - full stop as it were. The main use of many of these products seems to be to help get a barbecue going. Try a craft site or
veterinary supplier for bulk supplies of plastic pellets? And good luck with the blanket.


text:
Mismatch in the trends of atomic size and atomic radius
In the periodic table, the atomic radius decreases as we move from left to right in a period. However, the atomic volume should also decrease as a result of the decrease in atomic radius, as volume is given by $\frac43\pi r^3$. Instead, the atomic volume initially decreases upto group 14 and then increases, as is shown by the Lothar Meyer curve:

I am aware that the atomic volume is given by gram atomic mass divided by density. However, these form two contradictory theories - according to the Lothar Meyer curve, should atomic radius not follow the same behaviour? Or conversely, according to the measurement of atomic radius, should the Lothar Meyer curve and atomic volume not decrease across a period?

You have to take all such iron clad trends with a grain of salt. For instance there are a number of ways to measure atomic radius and here is a data page.


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Total SCF Energy with Pseudopotentials
Is there a physical meaning to the SCF energy of a molecule obtained using pseudopotentials to represent the core electrons of heavy atoms (for example gold)?
I've been running calculations in Gaussian on a system with a number of gold atoms and I became curious whether my use of pseudopotentials for the gold core electrons was skewing the results I obtained. So, I attempted to run a smaller test comparing two identical systems where one used pseudopotentials (specifically the LANL2DZ basis set) for the gold atoms and the other was all-electron with the DZP basis set. When I performed thess calculations, I got very similar orbital energies for the frontier orbitals, but the total energies were completely different, with the total energy of the all-electron calculation approximately 50 times more stable than the pseudopotential calculation (e.g. -100000 vs -2000 au).
Perhaps I've missed some obvious shortfall of pseudopotentials, but I just don't see how the energies obtained could be so different. I would expect an error in the calculation if the orbitals hadn't made sense or there was any unusual messages in the log file, but I didn't notice any such problem. The geometries and route sections used in each case are the same (besides the one having pseudo=read) and the only difference was the basis sets, where LANL2DZ is built in and I found DZP through the basis set exchange.

How many electrons are in the all-electron basis set? How many electrons when using pseudopotentials? Knowing this, how might the Hamiltonian between the two calculations differ, and how might this change the total energy? (Remember, for a given method/basis the total energies are seldom meaningful, but differences between energies are.)

@jjgoings There are 992 with all electrons and 746 with pseudopotentials. For the Hamiltonian, there should be less electron-electron repulsion terms and less electron kinetic energy in the pseudopotential case, but I assumed the purpose of pseudopotentials was to give a rough estimate of these for the core electrons which are relatively indifferent to the surrounding molecular environment and so should give roughly the same answer. Two different functionals give differences of about .1 au. These are different by about 100000 au.

Total enegies in general are not very meaningful. They are very basisset dependent and you cannot compare directly compare calculations using different basis sets, e.g. when calculating binding energies (BSSE). Also ECPs may include scalar relativistic effects.

you have to use something like ECPs, or you should use something like DKH to account for relativistic effects etc. Furthermore it seems that your SCF is not converged.


text:
The difference between CASSCF and MCSCF methods
I'm not sure the exact difference between the MCSCF and CASSCF methods. I know that the CASSCF method proceeds by optimizing both the CSFs and MOs (I mean the coefficients), but I don't know if this is also the case in MCSCF calculations.
My guess is that in MCSCF only the coefficients of CSFs will be optimized, but the HF reference MOs would be the same.

Both methods optimize orbitals and CI coefficients. MCSCF is the more general term, while CASSCF is a special type of MCSCF.
As the name Multi-Configurational Self Consistent Field already implies, it combines the Configuration Interaction method with the SCF method (Hartree-Fock). Since this combination increases complexity of the wave function, one usually applies restrictions to both sets of parameters (CI and orbital coefficients). For example core orbitals may be kept fixed (frozen) or only a small subset of configurations is considered.
Regarding the selection of configurations, there are various possible approaches. For example:

hand selecting configurations using chemical intuition or some other a priori knowledge
selecting a subset of orbitals, which should at least cover the HOMO and LUMO. In this active space one considers all possible configurations, hence Complete Active Space Self Consistent Field (CASSCF)
restricting configurations only to certain excitation levels (e.g. Singles and Doubles) within the active space: Restricted Avtive Space Self Consistent Field (RASSCF)

So CASSCF is a special variant of MCSCF, therefore there is not really a difference between the two. Although not every MCSCF is necessarily  a CASSCF, the later is the most common type.

I still cannot understand the difference. You are saying that MCSCF calculations are different in the orbitals, but in what way? I completely understand that CASSCF is based on orbital selection and you do a full-CI in the active space, but what MCSCF is based on I don't get it. It would be great if you could elaborate more on the MCSCF. thanks

Your question is like asking about the difference between an apple and a fruit. There is no difference, because an apple is always a fruit, but there are also other kinds of fruits. I hope I made this more clear now in my answer.

MCSCF: "Pick some orbitals, and inside that, pick some configurations." It could be all the configurations, or not. CASSCF: "Pick some orbitals, and inside that, do EVERY configuration". It's specifically all of them.

MCSCF is just "pick some configurations", you don't need to explicitly pick orbitals  first (although you do implicitly).

CASSCF is a particular type of MCSCF in which there is a set of orbitals (active space) for which all possible configurations (complete set) are included.
There are types of MCSCF that are not CASSCF:

RASSCF (restricted active space SCF): some of the CAS configurations are restricted in a particular way, such as allowing single and double excitations into and out of a smaller CAS.
FCIQMC-SCF: like CASSCF but rather than doing FCI on the CAS you do FCIQMC, which includes configurations with all possible levels of excitation, but only the most important configurations survive a Monte Carlo sampling process.
DMRG-SCF: like FCIQMC-SCF but with deterministic DMRG used instead of Monte Carlo FCI.
GASSCF (generalized active space SCF).

MC just means multi-configurational, so it is like HF-SCF, but with not just one configuration.


text:
Benzene from cyclopropane-1,2,3-triyltrimethanol in acidic medium
What is the mechanism of the conversion of cyclopropane-1,2,3-triyltrimethanol to benzene in in acidic medium?

After performing E1 thrice I was expecting 1,2,3-trimethylidenecyclopropane to form instead:

I know it's highly strained, but I couldn't adopt any other pathway to reach any other product. Is the triene product not formed at all? If cyclopropane-1,2,3-trione can exist, I think even the triene product can.
This reaction is from my exercise book, couldn't find it on OrgSyn.

This reaction does not show up in Scifinder. The triol is known though: Org Lett, 2007, 9, 2617 DOI:10.1021/ol070707r

No hits on Reaxys either. I strongly doubt it is real.

I have not written the steps for carbocation formation, simple dehydration.

The mechanism is plausible, but I think it is still important to emphasise that this reaction is not known. There are zero reports of anybody getting benzene by treating the starting material with acid. Probably somebody made it up.

@ortho With a little foresight into this whole JEE business, you'll know that it's almost definitely made up ;-)

It looks nice on the paper.

Hey primary cations are too unstable.

@Tech kidda: They may be unstable but that doesn't mean they don't form. I also checked Chem. Absts. NADA!

The diene could very well lose hydrogen and become aromatic; that would be favourable to ring cation I think.

As per my knowledge, in the second step there should be migration of Hydrogen to form a 3° carbocation and then the double bond will form between the two methanol branches.This will entirely change the product but as per rule the 1° cation has to be further stabilized by shifting of hydrogen.


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Why is direct addition not preferred in this reaction?

According to HSAB, direct addition is more preferrable in this reaction with the hard carbonyl carbon electrophile and the hard grignard reagent. But answer given in my book is that with 1,4- addition. I would like to know why 1,4-addition is done here? Thermodynamic control? But I don't think the reaction is reversible.

strong nucleophile gives 1,2 addition only

It seems that the answer in your book may be wrong. Refer here: https://chemistry.stackexchange.com/questions/31618/does-a-grignard-reagent-react-with-enones-to-give-the-1-2-or-1-4-product?rq=1


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Why does 2,4 DNP give orange colour crystals?
Why does 2,4 DNP give orange crystals on reacting with solution of aldehydic or ketonic group? Why does this test give orange crystals? Is this related to the electron emissions?
This is just a question from a curious student after 2 hours of lab work.

2,4-dinitrophenylhydrazine is orange and remains like that after reaction with simple carbonyl compounds

But why the colour change in that .I think red signifies aromatic and a more yellow colour suggests aliphatic.

And there is notable difference in colour and a ppt is clearly obtained.

@starunique2016 https://en.wikipedia.org/wiki/2,4-Dinitrophenylhydrazine#Brady's_test This gives the same information, your observations are correct

Hydrazines react with ketones and aldehydes to make hydrazones. For example, see Chapter 11 of Organic Chemistry by Clayden, Greaves, and Warren.
For the case of the product from 2,4-dinitrophenyl hydrazine, the resulting hydrazones have an extended network of conjugated π bonds. That makes them good chromophores for absorption in the blue to blue-green range of visible light, which is why they appear orange to red-orange. Extending the conjugated π system (e.g. substituted aromatic aldehydes) leads to longer wavelength absorption, making the crystals appear redder.
Nitro aromatics are often crystalline. Hydrazones resulting from 2,4-dinitrophenylhydrazine turn out to be highly crystalline. This is the basis for the classical qualitative analysis method for identifying unknown ketones and aldehydes by the melting point of their 2,4-DNP hydrazone derivatives.

Generally, you will notice that compounds containing chromophore and auxochrome groups have a colour. Chromophore groups are groups that are responsible for colour of the molecule once they are “attached” to it: benzene is colourless, but nitrobenzene is pale yellow. Apart from a chromophore group there are groups called auxochrome groups which alter this colour and so the intensity of absorption ( example: nitrobenzene is pale yellow, but p-nitroaniline is a bright yellow to red, the auxochrome enhanced the colour ). Among the chromophore structures, there are groups such as nitro, azo, nitrile, thioketone, conjugated double bonds ( like in retinal, carotene ). Some auxochrome structures are: the aldehyde group, the methyl mercaptan group, hidroxyl group, amine group. To sum it up, functional groups such as these will account for the colour.
Now, onto analysing DNP. This compound has 2 nitro groups and a hydrazo group ( 2,4-dinitrophenylhydrazine ). From the start, we have a colour: red ( if it is pure ). When it reacts with an aldehyde or a ketone in a condensation reaction, a double bond forms between the carbon on which oxygen was attached and the unsubstituted nitrogen of DNP. The formation of this double bond will alter the colour, shifting it to an orange. It also depends on the carbonylic compound you use. If the carbonylic compound contained conjugated double bonds ( like propenaldehyde ), the condensation product will contain conjugated double bonds which are chromophoric ( so in this case I can only guess what the resulting colour would be  because we have more chromophoric groups ). So, DNP is red in its purest form. Its colour is the result of the existing functional groups. A reaction with an aldehyde or ketones will shift its colour to orange.

Colour in Organic compound arises due to absorption of energy of excitation when visible light falls upon them. The colour which is visible to us is the complementary colour of the colour absorbed by the compound.
Now, due to presence of high conjugation in any molecule, its energy of excitation comes down. As extent of conjugation increases, the difference in Energy between excitation energy levels decreases and molecule thus absorbs light of higher wavelength (lower energy) for getting excited.
 Observe the two compounds, (without the Nitro group ) i.e $2,4$ DNP  and itscondensation product with a carbonyl compound.
Conjugation in the right molecule is more than the left one due the extra conjugation of the lone pair on Benzylic $N$ atom with the double bond. Nitro group does nothing but increases the extent of delocalisation in same amount in both the compounds.If you have observed before doing the experiment, $2,4$ DNP is a yellow coloured compound. So, it must absorb at the violet colour region of the visible light. So, the right one should absorb at a lower wavelength. In fact, that is the case we are also observing. The right one absorbs in the blue region of the visible region, and we are observing its complementary colour i.e. Orange.


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Explain the effect of intramolecular hydrogen bonding on solubilities in cold and hot water?
There is a statement given in my textbook (Cengage, Organic Chemistry (Part 1),  page 4.33) which is as follows: (I edited the statement to make it simpler)

One can account for the solubilities of o-nitrophenol, salicylaldehyde,and salicylic acid in cold water, in which they are insoluble, while in hot water they are soluble, using the concept of intramolecular hydrogen bonding. However, their para- and meta- isomers are soluble even in cold water. This can be explained by intermolecular H-bonding.

I have difficulty in understanding this statement.
How does hydrogen bonding influence the solubility in hot and cold water and moreover can we draw any conclusions about boiling and melting points of the compounds, in which one has intramolecular hydrogen bonding and the other having intermolecular hydrogen bonding?

Search about hydrogen bonds influencing solubility on Google! :) It'll give you some great learning resources. It also helps you to learn more about the phenomenon itself, which will catalyse your ability of applying the concept to various questions, rather than just learning the theory out of a textbook

This is a really good related question: https://chemistry.stackexchange.com/questions/5503/what-exactly-is-hydrogen-bonding-and-does-it-really-need-fluorine-oxygen-or-n?rq=1

With intramolecular hydrogen bonding, the substance would generally be less soluble in water and would thus require a higher T for dissolution. This is because the molecules are not as able to interact with the solvent water molecules, via H bonding, as their polar functional groups are involved in an intramolecular interaction. Thus, there would be less mixing, resulting in a much smaller entropy increase upon dissolution. Thus, a higher T is required to increase the value of TdeltaS to make delta G negative.


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Finding empirical formula
When determining the formula of a compound, why do we divide by the smallest number of moles once we already have calculated the number of moles for each element in order to find the empirical formua of a given compound?

Because we don't like to have too many decimal fractions hanging around.

The empirical formula is the simplest positive integer ratio of the atoms that are  present. Thus, by dividing by the smallest number, we obtain the simplest whole number ratio of atoms.


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Why doesn't graphite dissolve in organic solvent?
The graphene layers are held by weak van der Waals' force, this should be comparable to the force holding solvent particles. Why doesn't graphite dissolve in organic solvent and separate into graphene layers?

An individual van der Waal's bond is weak but graphite sheets have zillions of such bonds, hence overall there is a strong attraction between the individual sheets.

That said, you may separate layers all right, but it does not mean that graphite would dissolve.

You don't need to go to graphite. Already tetracene (four C6-rings) is only soluble in boiling benzene, and not very well. pi-stacking makes quite strong bonds. https://en.wikipedia.org/wiki/Stacking_(chemistry)

@Karl According to NIST's compilation, pyrene dissolves at 295.35 K up to 13.47 mol%  in THF (https://doi.org/10.1063/1.4775402 or https://digital.library.unt.edu/ark:/67531/metadc152456/m2/1/high_res_d/Pub-BBC-22.pdf p. 013105-179, left column).  Not terrific, but not none.  Here, https://en.wikipedia.org/wiki/Coronene coronene is claimed to dissolve well in ordinary organic solvents. While I agree with the big picture (the more C, the more difficult, e.g. some of the perylene dyes) I want to point out that occasionally there are some compounds not that dead precipitates (e.g., helicenes).

@Buttonwood From that wp article on Coronene : Centimeter-long crystals can be grown from a supersaturated solution of the molecules in toluene (ca. 2.5 mg/ml). Not what I would call "well soluble". ymmv ;-) 1mmol/L (.3g) in chloroform even much less.

While you cannot directly make a suspension or dispersion of graphene nano-platelets from graphite, it is possible to perform very similar process, called exfoliation, on graphite oxide, as graphite oxide has functional groups decorating it that reduce the functional strength of the van der Waal's forces between layers.
The exfoliation process can create a dispersion of graphene nanoplatelets in an organic environment, however, outside energy is often (if not always) necessary to promote the exfoliation of the individual layers. Most often, this energy is provided by sonicating the solution.
A very quick search of suggests that this method has been shown to work using NMP, DMF, THF, and ethelyne glycol. (see Paredes, Villar-Rodil, Martínez-Alonso, and Tascón, 2008 for more info)

Because lots of weak interactions lead to large attractive forces
The reason why graphite (or graphene) won't dissolve in common solvents isn't to do with the nature of the chemical forces holding the layers together: it has more to do with the sheer number of such weak forces in each small flake of the substance.
Each hexagon has only dispersion forces to bind it into a solid crystal. Compare to benzene which has just a single carbon hexagon. Those forces between hexagons mean that benzene crystallises at around 6 C. Disrupt them even slightly (by adding a methyl group to give toluene) and the solid doesn't form until you have cooled the substance about 100 degrees. The forces are weak, but notable in their effect. Benzene is soluble or miscible with many organic solvents because you only have to disrupt single benzene-benzene interactions to get a solution.
So why are layers of graphite not soluble? The same forces that makes benzene crystallize at such a high temperature are present but, in each tiny flake, there are millions of hexagons in the same plane. When even a microscopically tine such flake interacts with another flake the total force between them will be millions of times stronger than the forces between benzene molecules. That is a lot of attractive force to overcome if you want to get a flake into solution. You can't overcome the hexagon-hexagon interactions one by one as you do for benzene, you have to disrupt the whole sheet all at once and that is monumentally harder than doing one hexagon at a time.


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Why fire spreads out, depending on burning substance, with different speed?
In Action movies, I have seen a couple of times that if flammable gases or liquids like benzene are ignited, the fire spreads out within fraction of a second and sometimes there is also an explosion. However, if a Piece of wood in a grill is lightened, it may take minutes if the fire has the full Magnitude. Why there is a difference in ignition Speeds (what is the physical cause)?
Are there kinetic equations how to determine the Speed how fast a fire would spread out?
I am interested in the viewpoint of thermodynamics how to derive the Speed of ignition.

Frankly, thermodynamics doesn't help understand the kinetics much at all. While a piece of wood gently burning in the fireplace and a propane tank undergoing a BLEVE (boiling liquid expanding vapor explosion) are both doing something thermodynamically preferred, the kinetics matter a lot.


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Orthogonality of wavefunctions for different normal modes

Vibrational wavefunctions corresponding to different vibrational levels of the same normal mode are orthogonal because they are eigenfunctions of the harmonic oscillator Hamiltonian, which is hermitian.
Displacement vectors corresponding to different normal modes are orthogonal because they are eigenvectors of the Hessian matrix, which is symmetric.

QUESTION: given the above, how can one show that vibrational wavefunctions corresponding to different normal modes are orthogonal?

@A.K. but then how can you tell what is chemistry and what is physics? One might say that all of theoretical chemistry is basically physics...

I don't know if agree with this being moved. While it certainly could be asked on physics, I think it also falls well within the normal bounds of computational chemistry.

This question absolutely fits under the category of chemistry.  There is absolutely no reason to redirect computational chemistry questions and would only serve to hurt our community by prohibiting this field of study.

given the above, how can one show that vibrational wavefunctions
  corresponding to different normal modes are orthogonal?

The Hamiltonian for a system with two (or more) normal modes can be written as a tensor product of (Hermitian) Hamiltonians for each normal mode. For example for two modes (stretching and bending):
$$
H_{\rm{total}} = H_{\rm{stretching}}\otimes H_{\rm{bending}}
$$
The tensor product of two Hermitian matrices is also Hermitian.
Therefore the wavefunctions are orthogonal for the same reason that you figured out on your own for the single mode case.
More generally, all Hamiltonians (Hermitian operators) have orthogonal wavefunctions (eigenfunctions) for distinct energy levels (eigenvalues). So wavefunctions corresponding to different energy levels are orthogonal as a consequence of the Schroedinger equation which is based on a Hermitian object called the Hamiltonian, and is the basis for the existence of all distinct energy states (vibrational or not) in quantum chemistry.

Thank you. But then why are the vibrational wavefunction necessarily eigenfunctions of a single total Hamiltonian? Because (1) one almost always assumes that vibrational are independent or, at most, interacting via an averaged self-consistent field, which isn't true in $H_{total}$, (2) anharmonicity effects are commonly neglected.

@munchnone: In your question you say it works because the Hamiltonian is the harmonic oscillator Hamiltonian, but actually it works for anharmonic vibrations such as the Morse/Long-range potential as well. You ask why the wavefunctions are eigenfunctions of a single total Hamiltonian? Actually it is a postulate of quantum mechanics that the universe can be described by $H|\psi_n\rangle = E_n|\psi_n\rangle$ where $H$ is the Hamiltonian of the universe. Then if you want to look at smaller parts of the universe (like a water molecule) you consider a sub-matrix of the full $H$.

Thanks. So is what you mean is to consider the total harmonic Hamiltonian matrix, which is a tensor product of the Hamiltonian matrices for each normal mode because the total harmonic Hamiltonian operator is the sum of Hamiltonian operators for each normal mode? Note that this would not be the total Hamiltonian, because it contains higher order terms that couple all normal modes.

Is there an easier way to think about this? E.g. if we have $f_1(x_1)$ and $f_2(x_2)$, where $x_1$ and $x_2$ are orthogonal directions in space, then  I assume it is always true that $f_1(x_1)$ is orthogonal to $f_2(x_2)$? What is a good way to show it?

@munchnone: No the total Hamiltonian (harmonic or anharmonic) is Hermitian, because all Hamiltonians have to be Hermitian. In your question you say the harmonic Hamiltonian is Hermitian and that's why wavefunctions are orthogonal, but actually all Hamiltonians are Hermitian (harmonic or anharmonic doesn't matter)


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Synthesis of branched alkanes
The question might seem kind of trivial, although it is even harder than I've thought. How can we synthesise branched alkanes from linear one? For example, I want to obtain 3,4,7,8-tetramethyldecane from n-decane. I've already found Stille coupling, but I guess I'm searching for easier method. I'm just a high school student, studying chemistry by myself as well.

You have to functionalise first, introduce a substituent that you can manipulate e.g. halide

Yeah, that's what I know, but what methylating agents should I use?

The easiest way is to make the Grignard from the halide then react with MeI

There is an odd photochemical route to many hard-to-get branched alkanes described by Crabtree. It is unclear how much it is used (https://chemistry.stackexchange.com/q/84777/81).

If this is a homework problem, it is an exercise in futility. No knowledgable chemist would consider n-decane as a starting material to obtain your tetradecane, stereochemistry notwithstanding. A better question is to prepare the tetradecane from commercially available compounds --possibly with some restrictions-- with no regard for stereochemistry.

It's not a homework problem, I am just interesed in method (doesn't have to be easy) that can branch a linear alkane.

Homework or not, @user55119 is right, alkanes are not a sensible starting point to synthesise alkanes.

I did add my educational background @Karl

Hmm, on this level, you may not realise the difference between lab synthesis and industrial process. I think answer for you should be about isomerisation, like one used for better fuel.

@Radi: Critical changes to original post! I will guide you through a reasonable approach. You have to realize that you have not specified the stereochemistry of 11 -- of which there are 16 possible stereoisomers (8 racemates) -- but I will ignore the issue. You will recognize the symmetry in 3,4,7,8-tetramethyldecane (11), so we will look for a coupling reaction of two 7-carbon entities. (E)-2-Buten-1-ol (crotyl alcohol) (4) and triethylorthopropionate 5 are commercially available. The reaction of 4 and 5 is a Johnson-type Claisen rearrangement (Google it!) which produces racemic esters 7 via transition state 6. Reduction of the double bond of unsaturated esters 7 leads to saturated esters 8. Reduction of the ester group to an alcohol followed by conversion to a bromide gives structure 9. The Wurtz reaction (9 --> 11) is rarely used but here is an opportunity.
Alternatively, esters 8 under the same conditions of the Wurtz reaction produces the α-hydroxyketones 10 (acyloin condensation). While the Clemmensen reduction (10 --> 11) of aliphatic ketones does not work as well as with aromatic ketones, under these conditions you would lose both oxygens. If the ketone is produced (structure 10 less the hydroxyl group), a Wolff-Kishner reduction would lead to the tetradecanes (plural) 11.Read about these name reactions which will be of more value to you than your original proposal. Good luck!

Thank you for guiding me through really interesing synthesis, but the question was  if there is a general method (simpler that Stille's reaction) that allows us to obtain branched alkanes from linear. Synthesis of 3,4,7,8-tetramethyldecane from n-decane was just an example since they both have 10-carbon chain.

No, there are no general methods. Waylander suggested a coupling -- like your thought of the Stille -- between an organometallic and halide. The organometallic would be better as the methyl unit. To halogenate (Cl or Br) n-decane at the 3,4,7,8 positions, the yield calculated is 0.02% of all tetrahaloisomers. And this is optimistic amid  a myriad of other polyhalogenated compounds. Absolutely impractical.

This problem has two parts: One is industrial chemistry. You typically halogenate alkanes, which gives you a wild mixture, and then fractionate and purify the products. You want to make a lot of different chemicals, so you can use most of the products, and you try to fine-tune the process so the composition matches your customers' demands, more or less.
Now you sell these n- and sec-haloalkanes (and other relatively simple, functionalised compounds)  to a chemical laboratory. They now make some fancy metal-organic coupling reaction(s) like Stille (after perhaps making alcohols or ketones out of your haloalkanes) with some specific pair of compounds, possibly a following reduction step, and get a defined, branched alkane. A straightforward example is the https://en.wikipedia.org/wiki/McMurry_reaction .
There are a lot more possibilities for both parts, but the guy making the end product will always buy pre-products that each already contain sizeable parts of the structure he aims for. To plan such a sequence of reaction for a certain target is called retrosynthesis.
OR
If you don't want some specific compound, but just more branching,  https://en.wikipedia.org/wiki/Catalytic_reforming would be your horse. You also get some aromats then, or cracking, depending on hydrogen pressure. This is also an industrial process, hard to and sort of pointless to do in the lab unless you are developing catalysts. I've visited a lab doing that, and it looked like a shrunk-down refinery with a roof and three outside walls, not at all like a chemistry lab.


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Why are some strucutres isomers while others are not?
Why are some structures considered isomers but others are not? Take C3H9N for instance, according to https://www.chegg.com/homework-help/questions-and-answers/four-constitutional-isomers-molecular-formula-c3h9n-q17859870   , only the four checked boxes are isomers, but why are the other 4 not? The other four seem to also satisfy all Lewis rules and do have same formula but with different arrangements. Also, how do we determine which structure is the "original" while which are the isomer(s)? Thanks for all help

Count the electrons and that should show the difference. Focus on whether bonds are single or double.

All of the (valid) structures are a group of isomers. Among, other example, pentanes, you could say n-pentane is the most basic compared to the two branched isomers, but that's just an arbitrary definition. The isomers of C3H9N don't even have a common name, except the subgroub of the two propylamines.

The supposed answers are wrong. The Lewis dot structures are based on forming an octet of electrons around each atom using the s and p orbitals, typically hybridized to sp3. The four isomers are:

Ethylmethylamine

Isopropylamine

Propylamine (which isn't even shown on shown key...)

Trimethylamine

Error 1 - This structure isn't valid since the double bond shown in red box gives carbon and nitrogen both 10 electrons instead of 8.

Error 2 - This structure isn't valid since the double bond shown in red box gives both carbon atoms 10 electrons instead of 8.

Error 3 - This structure isn't valid since the double bond shown in red box gives carbon and nitrogen both 10 electrons instead of 8.

Error 4 - This structure isn't valid since the double bond shown in red box gives carbon and nitrogen both 10 electrons instead of 8.


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Why is the standard atomic weight of chlorine not a whole number?
Why is the standard atomic weight of chlorine, 35.5, not a whole number? Like for example, it could be exactly 35 or exactly 36. Please show the solution of formulae on how u reach to 35.5) with some English words so I can write the answer in my homework assignment and please make it small. Thanks

You might have heard of isotopes. There are two kinds of chlorine: one is 35, and another is 37.

yes I've heard of it but other elements also have 2 or more than 2 isotopes.

Well, if one isotope is clearly dominant (like 99.9%), you will have an integer mass within two-digit precision. Not the case with Cl, though.

The ratio for chlorine-35 to chlorine-37 is somewhere around 3:1, due to which this value of 35.5 arises.

related https://chemistry.stackexchange.com/questions/99226/atomic-weight-expected-weight https://chemistry.stackexchange.com/questions/9016/the-formula-for-finding-the-percentage-of-isotopes-in-an-elements-atomic-weight https://chemistry.stackexchange.com/questions/38082/what-is-the-difference-between-molecular-mass-average-atomic-mass-and-mola

@Mithoron I would agree with any of those being a dupe. The previous one I wouldn't because it specifically recognizes in the question that isotopes are factored in, but asks why elements with significant isotopes still aren't integers.

Because it has various forms so-called as isotopes and also it is not required for the weight to be a whole number. Is anyone's weight a whole number?

The process is explained (for silicon) here:

The calculation is exemplified for silicon, whose relative atomic mass is especially important in metrology. Silicon exists in nature as a mixture of three isotopes: $\ce{^28Si}$, $\ce{^29Si}$ and $\ce{^30Si}$. The atomic masses of these nuclides are known to a precision of one part in 14 billion for $\ce{^28Si}$ and about one part in one billion for the others. However the range of natural abundance for the isotopes is such that the standard abundance can only be given to about ±0.001% (see table). The calculation is
$$A_r(\ce{Si}) = (27.97693 \times 0.922297) + (28.97649 \times 0.046832) + (29.97377 \times 0.030872) = 28.0854$$

For chlorine, the idea is the same, but you only have two isotopes, $\ce{^35Cl}$ (76%) and $\ce{^37Cl}$ (24%).


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Rietveld refinement "fronting" of peak signal
I was trying to refine a crystal structure using Rietveld refinement yesterday. The crystal structure is already known for a related compound and the powder pattern seems to show a phase pure compound. An automatic peak indexing and refinement gave a possible cell that matches the literature values. Using the Pseudo-Voigt function however did not lead to any good results as the function did not match the asymetric peak profile.
The profile itself shows something I'd call 'fronting', so there is a slope at the front of the peak but then the measured curve just falls off directly after the peak. I was looking for that in literature but I could not find the term 'fronting' here. There was something similar called 'transparent' or 'transparency' but I wasn't sure whether this is referring to the profile or thr setup.
Does somebody know what this effect is called and what type of refinement is used here instead of Pseudo-Voigt?

Could it be that each of the visible peaks is actually composed of 2+ peaks? What's the radiation source?

Pretty sure it is K-alpha lines of Cu which is typical for x-ray diffraction.

I think the term you are looking for "asymmetric peak." I'm guessing low angles are more asymmetric. One cause is instrument geometry. See second paragraph, second column on page 896 of paper Finger et al  "A Correction for Powder Diffraction Peak Asymmetry due to Axial Divergence " where it discuses zeolite peaks. Constants H and S are discussed on lower right of first page of the paper.

It is a molybdenum source due to the presence of Europium in the sample. That links seems to summarize the problem but is there a different polynomial that solves the issue besides the Pseudo-Voigt one?


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Knowing that CI is variational for any basis set, why is coupled-cluster variational in the full correlation limit?
This is the question asked in Martin's comment to my answer to: Why is Coupled Cluster not variational?
In the comment I was also asked to provide a publication as a reference:

This doesn't seem to be proper form. Questions are supposed to be generally useful. You can say it's a follow-up. but making a picture of comment was total waste of time.

I agree with Mithoron. While I like the question and answer, I think including the comment itself is unnecessary; The link to the previous question is enough for a curious person. The question itself should be edited to be more general.

Maybe I put too much effort making the screenshot and adding it. Martin asked this question in a comment to a different question, and then when I answered it he said to make it a question and answer. I am new to the chemistry stack exchange.

The CI (configuration interaction) energy, given by Eqs. 3.9-3.10 of this, is an upper bound to the exact ground state energy (i.e. CI is variational). Coupled cluster energies are not generally variational, as explained in the above linked question, except in the case of CC($N$) for an $N$-electron system. The reason is because CC($N$) for an $N$-electron system is equivalent to FCI (full configuration interaction) which is an example of CI, and all CI energies are variational.
Since I was asked for a publication as a reference, a good one is this, in which it is shown that CC and CI become the exact same in the full-correlation limit.


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Why don't these Entropy equations contradict each other?
My textbook says that $\delta S_{rev}= \frac{q_{rev}}{T}$. It also says that $\delta S_{surr}= \frac{-q_{rev}}{T}$. Logically, this should mean that the change in entropy in both the surroundings and the system are equal in magnitude but opposite in sign.
This is NOT true, though, as it would go against the Third Law of Thermodynamics. Also, my book proves the Law by comparing the equation for $\delta S_{surr}$ to the Entropy of the system found when subtracting the products' entropies from the reactants'.
Where am I going wrong? Why can't we compare the two $\delta S$ equations and disprove the Third Law?
Note: The first equation is $q_{rev}$ due to something to do with a reversible, isothermal reaction that I don't really understand, which might be the source of my confusion...

In fact it is "$\geqslant$" in both equations, not "$=$".

No real world process is really thermodynamically reversible.

@Mithoron-I know...textbooks build on theoreticals.

Did you really mean the 3rd law of thermodynamics, or did you mean 2nd law of thermodynamics?

What are the exact initial and final states of the system in this development, including temperatures, pressures, and chemical composition.

I can help you with this, but I need a little more information.  Is the irreversible process you are describing one starting with pure reactants at T and P, and going to pure products at T and P, with the system in contact with an ideal reservoir at T, and with surroundings pressure also held constant at P?