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Why can't an electrochemical cell be a single cell?
If I put two electrodes of two different metals in 1 solution containing ions for both of the metals and connect these electrodes with a wire, do I still have an electrochemical cell ? Or does it have to be made of two half cells in 2 different solutions joined by a salt bridge? If so,why ?

If you mix everything together, the redox reaction happens right then and there. The whole point of the cell is to force the reaction to become circuitous such that the transferred electrons have to go the long away around in a circuit, doing useful work.

I'm sorry if this is a stupid question, I'm just starting out with electrochemical cells, but what do you mean the redox reaction would happen right then and there ? Suppose I'm working with Zn and Cu electrodes, Can the electrons that are a result of the oxidation of zinc atoms move in the solution and reach the Cu ions ?

Welcome to Chemistry.SE!  If you haven't already, please take a minute to look over the help center and tour page to better understand our guidelines and question policies.

If I set up a functional cell, take it apart, and mix it all together, there would have to be a reaction. It's the reaction that creates electricity, but it's only useful to force the process of reacting to happen via moving electrons through a wire.

https://chemistry.stackexchange.com/questions/66123/why-cant-a-galvanic-cell-be-a-single-cell

Side note: this whole industry is pretty much based on single-cell electrochemistry too...

Yes, if you dip two different metals in an aqueous solution containing ions, you will have an electrochemical cell. This is how all lemon batteries work, which consist of zinc or copper inserted into lemons (approximately an aqueous citric acid solution). You can see a potential difference between zinc and copper. The key point is that one cannot determine Nernstian potential in such cases.
Half cell is just a mental construct as it cannot exist alone. If there is a reduction, something must be oxidized. Electrochemists usually separate the "half-cells" in order to conveniently calculate Nernst potential.
Imagine a hypothetical cell, which is partitioned into half cells with a plantinum sheet. Each compartment contains an electrode Cu and Zn, dipped in their respective salt solutions. What will happen in that case?

RE: "Electrochemical usually separate "half-cells" in order to conveniently calculate Nernst potential." -- This just isn't true. Separate "half-cells" are needed to prevent the redox reaction taking place on the electrode itself. If Cu and Zn electrodes were added to a $\ce{Cu^{+2}/Zn^{+2}}$ solution then the $\ce{Zn^{+2}}$ would react at the Cu electrode itself coating it with Zn and there would be very very little to no current flow between the electrodes.

Thanks for adding this point, however, my point was that one cannot calculate or even define Nernst potential for Zn ions when Cu and Zn ions are also present in the solution. It is important to separate half-cells for that reason as well. By the way, one does not draw any current during potential measurements. This is one of the basic requirements of potentiometry.

In order to measure the potential of a half cell you must draw a very tiny amount of current. The idea is that you draw so little that there are no double barriers or diffusion implications.

You miss the point. So long as the reaction continues you can calculate the Nernst potential for Zn ions when Cu and Zn ions are also present in the solution. What you can't do is to reliably measure the potential between the electrodes which requires current to flow between the electrodes.

I am not sure how would one calculate the half-cell potential of Zn when both Zn and Cu are present in the solution. One would see some value on a potentiometer, but it will have nothing to do with equilibrium potential. Do you remember old-school null point potentiometry? Under those conditions, the current flow was indeed zero (or for semantic reasons extremely small).

With the Cu and Zn ions both present in the solution there is still a redox reaction happening. So the half cell potential would be between the Zn cations in solution and the copper electrode itself, not between the Zn and Cu electrodes.

"So the half cell potential would be between the Zn cations in solution and the copper electrode itself, not between the Zn and Cu electrodes."  Not sure what is meant by that. Half cell potential is an imaginary (thermodynamic) construct, and of course, it involves a single electrode in equilibrium with its own ions. There are some papers in Science which do single electrode reduction via electrostatic chemistry by Bard. Anyway, the original poster seems to have lost interest in following up. Comments are getting long, you may wish to post a separate answer addressing the original question.

@MaxW, Why would \ce{Zn^2+} be reduced to Zn at the Cu electrode? I would say that this reaction is not thermodynamically favoured.

@user1420303 perhaps MaxW was referring to an electrolytic cell instead?


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Why zinc sulphate solubility is higher in ethylene glycol than water?
I'm trying to find saturation point for dissolved zinc sulphate in water and ethylene glycol.
Based on my knowledge, water is more polar than ethylene glycol as reflected in its higher dielectric constant value. However, max solubity of this salt in water is roughly limited to 3M and for ethylene glycol is much higher (experiment in progress to determine the right number).
What causes this opposite behaviour ?

It has been a long time since I took a chemistry class, but I'm pretty sure that there is no general theory (over all salts and over all solvents) which allows for a good numeric solution for the question.

If the observation is true, then I would assume that chelation effects are the root cause, but solubility is such a gross and complicated topic that I wouldn't bet a lot on that.

I dont know why either, but it seems to happen, everything when it hits glycol just suddenly liquidates, you dont want to put it near a chalk wall or itll sludge to a sludge pile in no time. But one thing to say, put epsom salt in it, and for some reason it doesnt solute at all, thats funny, I didnt expect it. If something doesnt go into glycol, you can react it in by adding an acid, but I think the acid also affects the glycol not just the salt your adding, then it goes to total liquid for sure, I bet it even works for something really insoluable like ferric carbonate.

Zinc is an amphoteric metal which means it can dissolve both in acidic and basic conditions. However, neutral pH is the transition point. Please see this article Recovery of Rare Earths from Waste Cathode Ray Tube (CRT) Phosphor Powder by Selective Sulfation Roasting and Water Leaching on its behavior in water and eg. I believe the reason is that in eg there is no mechanism for zinc to form insoluble hydroxide complexes which will decrease the solubility of zinc. But in ethylene glycol you simply cannot form hydroxides..

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Determine surface area of heat exchange tube

A heat exchanger is used to cool down a stream of $\pu{20 kg s-1}$ of water from $\pu{80 °C}$ to $\pu{60 °C}$. The cooling water enters at a rate of $\pu{12 kg s-1}$ and $\ce{20 °C}$. The overall heat transfer coefficient is $\pu{2 kW m-2 K-1}$ and the heat capacity of water is $\pu{4.183 kJ kg-1 K-1}$

Calculate the surface area.

I am stuck on what the last step of the calculations should be.
I began with calculating the outlet temperature for the cooling stream using that $q$ is constant, I set IN = OUT:
$$m_hC_pΔT_\mathrm{in} = m_cC_pΔT_\mathrm{out}\label{eq:1}\tag{1}$$
From $\eqref{eq:1}$ I got that $T_\mathrm{2out} = \pu{53 °C}$
Then I calculated the logarithmic temperature difference, which I got to be
$ΔT_\mathrm{ln} = \pu{-33.075 °C}$.
After this I am confused as to how I can get the surface area of the tube.
Edit
I did a heat balance, IN - OUT = 0:
$$(m_hC_pΔT_\mathrm{in} + m_cC_pΔT_\mathrm{in}) - (m_hC_pΔT_\mathrm{out} + m_cC_pΔT_\mathrm{out}) = 0$$
simplified into:
$$m_hC_pΔT_\mathrm{in} - m_hC_pΔT_\mathrm{out} - UAΔT_\mathrm{ln} = 0$$
However when I try to solve for A I get a negative value, which isn't correct.

There is no way to solve the problem with two variables. So the overall heat difference must be directly calculable and the only variable is the surface area of the heat exchanger. So "Hot in" is 80 C and "Hot out" is 60 C. "Coolant in" is 20 C and "Coolant out" is 80 C. (Edited - I had mistakenly stated "Coolant out" was 60 C.)

The outlet temperature  should be 53.3.  The log-mean temperaturr difference should be positive.  Calculate the heat load and divide by the log-mean temperature difference and by the heat transfer coefficient to get the area.
ADDENDUM
This addendum is for the benefit of @MaxW (the downvoter).

The heat load $\dot{Q}$ of the heat exchanger is given by:  $$\dot{Q}=\dot{m}_HC(T_{H,in}-T_{H,out})=20(4.183)(80-60)=1673\ kW$$This heat load represents the rate of heat transfer from the hot fluid to the cold fluid.
The exit temperature of the cold stream is determined from $$\dot{m}_CC(T_{C,out}-T_{C,in})=12(4.183)(T_{C,out}-20)=\dot{Q}=1673$$

Solving for the exit temperature of the cold stream gives  $$T_{C,out}=53.33\ C$$
3.  The log-mean temperature difference is given by:
$$(\Delta T)_{LM}=\frac{(T_{H,in}-T_{C,out})-(T_{H,out}-T_{C,in})}{\ln{[(T_{H,in}-T_{C,out})/(T_{H,out}-T_{C,in})]}}=\frac{26.67-40}{\ln{[26.67/40]}}=32.88\ C$$

The heat transfer area A is determined from the equation:  $$U_oA(\Delta T)_{LM}=2.0A(32.88)=\dot{Q}=1673$$Solving this equation for the heat transfer area A gives:
$$A=25.4\ m^2$$

Questions?

@user5713492 I thought that I had to take the $\ce{T_{c,out} - T_{h,in}}$ but it was the other way around. I now get that the log-mean temperature is $\ce{32.9^oC}$. I hope this is correct.

This is the correct value for the LMTD.

I just saw that @ChesterMiller added the calculations which confirms that I now have the right answer. Thank you both for your help!

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Is there a difference between a chelate ligand and a polydentate ligand?
I'm confused because my lecture notes state that chelate ligands are a subset of polydentate ligands.

That's true. A chelate ligand is necessarily polydentate, but not vice versa.

A chelating ligand should bind more than once to the same metal atom. Therefore, it must be polydentate. However, a regular polydentate ligand does not have to ligate to the same metal, and therefore, it is not necessarily chelating.


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How do I get a 20% ratio of sodium bisulfate with 1 gallon water?
How many lbs do I add to 1 gallon of water to get a 20% sodium bisulfate ratio? I am in the middle of anodizing some aluminum.... I just threw the whole 3 lbs in... hopefully it works... I ended up doing a 2 gallon to 3 lb ratio... is this good? It is very cloudy looking
Picture for proof this is not a homework question like the person in the comments below wrongly assumed vs just helping with the question:

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

actually no its not... i am in the middle of anodizing some aluminum and want to know now... I just ended up throwing the whole 3 pounds in... hopefully it works

What percentage you need to make? Is it $w/w$ or $w/v$?

I do not know... they say just a 20% solution of sodium bi-sulfate to anodize (this method is suppose to be safer than using acid) http://www.observationsblog.com/sciencetechnologyexperiments/anodizing-and-dying-aluminum-without-battery-acid I suck at this math type stuff... I am literally sitting over here with a bucket full of this stuff waiting to make sure I got it right... bout to say F' it and try it anyway and see what happens.

@Jeff the idea of a homework question on the site is essentially a question where you state a problem without context and/or the efforts you have made to solve the problem.

This isn't something that requires 0.1% precision.
The link says "40 grams of sodium bisulfate in 160 ml of tap water."
160 ml of tap water is very close to 160 grams.
A gallon of water is reasonable close to 4 liters or 4000 grams.
$\dfrac{4000}{160} = 25$
$25 \times 40 \text{ grams}= 1000 \text{ grams} \approx 2.2 \text{ lbs/(gallon of water)}$


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Monitoring amino acid esterification
I am trying to monitor the esterification of a non UV active amino acid.I used thionyl chloride and Methanol as reagents.
As amino group is unprotected, Ninhydrin should give positive result for both reactant and products. However, I believe there should be Rf difference between reactant and product.When I performed TLC and stained with ninhydrin, the reactant was strong brown and one spot at very high Rf value showed pale brown coloration.Now can I think that pale brown spot as my target ester?
I am doubtful because reactant and product both supposed to show strong coloration as amino group is not affected.On the other hand,  will just methylating the acid result in high Rf change? Because I feel amino group is also polar, it would not allow spot to raise till high Rf end.

On what medium did you perform TLC? Silica?

Yes Silica.CHCl3/MeOH

You could use a bromcresol green stain which is indicative of carboxylic acids, details here https://www.silicycle.com/faq/thin-layer-chromatography/stains-for-tlc-plates  Do you have access to alumina plates?

Unfortunately I don't have access to Alumina plate. To add some more information the top spot is tailing. Is it due to free amine?

I doubt it is free amine, you normally need to add NH3 to your eluent to get free amine off the baseline. How much MeOH are you using in the eluent?

I add approx. 5:2 (Chloroform:MeOH).Actually I would like to know what usually causes tailing?Also, what is the role of ammonia?I wonder do esters tail?

That's a lot of MeOH! Maybe it is your ester.Try a KMNO4 visualisation.  Adding NH3 neutralises the acidic sites on the silica surface so that the basic amino group binds less and will elute up the plate.

@Waylander So do you think free amine in ester is the cause for tailing? On the other hand,if I use Ammonia during column, is there a risk of Amide formation? In other words,how can I avoid tailing of ester in silica column without allowing hydrolysis/side reactions?

Without NH3 or some other base in the eluent I think the tailing is due to free amine. I have routinely used Ammonia in Methanol/DCM as an eluent for basic species and never seen ester degradation. If it is a concern you could use triethylamine in place of ammonia.

@Waylander Thank you very much for your detailed explanation!


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Work done in an irreversible process
I was given a problem to identify whether the given statement is true or false.
And the statement was:
"Consider an ideal gas which is expanded from $(P_{1},V_{1},T_{1})$ to $(P_{2},V_{2},T_{2}).$ Then, the work done on the gas is maximum when it is compressed back irreversibly from $ (P_{2},V_{2}) $ to $ (P_{1},V_{1})  $ against a constant pressure $P_1$".
I concluded that the statement is false, as:

It is nowhere mentioned what is greater $P_1$ or $P_2$ as there is no information on how the expansion process was done.
Even if I assume that $ P_1 > P_2$, still during irreversible compression I can vary the pressure such that the pressure goes beyond $P_1$ in the middle of the process but finally returns to the required final state $ (P_{1},V_{1})  $ , which would mean that I am doing more work than what I would be doing if I keep the pressure constant.

However the answer says that the statement is true.
Where am I wrong?

In my judgment, this problem is woefully underspecified.

It is tricky wording. Usually when you compress something you want to do so as efficiently as possible. The minimum amount of work would be required if you compressed this reversibly.
If for some reason, as in the question, you want to waste your energy and overkill the work required (i.e., do things the hard way), then compress the gas IRREVERSIBLY. that will make it take the maximum amount of energy to get the desired compression.
The engineer in my cringes at this :)

Is the statement about constant pressure correct?

An irreversible process can be done even if the pressure is kept varying.

pressures dictate whether you are expanding or compressing. Since it says it is compressing, the outside pressure must be larger than the inside pressure. Irreversibilites, for instance friction, are present regardless of the direction or magnitude of the pressure

The statement about constant pressure only means that the external pressure is constant, just as the atmosphere is always 1 bar.

In your problem the external pressure is greater than the pressure inside the volume, hence, the volume is compressed. If it is compressed reversibly the minimum amount of work will be required because there are absolutely no losses i.e. no work is wasted. Conversely, the most amount of work is done if the process is irreversible since there will be losses and you have to both provide energy for compression and also energy for the losses.


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C4v point group complexes and the group orbitals (MO Theory)
Out of curiosity, what would the group orbitals be like in a C4v complex (square planar complexes or octahedral complexes with 5 identical ligands and 1 other) look like? I get a irreducible representation of 2A1 + B1 + E, which raises the question of how the 2 different A1 group orbitals look like.
Can someone explain this to me?

I don't have time for a full answer or derivation right now, but here is a rough explanation. We can just consider s-type orbitals on each of the five ligands.
If the group orbital is to transform under the TSIR, then each of the four "equatorial" orbitals must have the same phase as each other. This is because these orbitals can be interconverted by symmetry operations, and if the character under every operation is to be $1$, then every orbital has to have exactly the same contribution to the group orbital. However, the equatorial ones don't need to have the same phase/magnitude as the "axial" orbital, because the equatorial/axial positions are not symmetry-related.
Therefore, you end up with two group orbitals that look somewhat like this:

Obviously, the exact relative contribution will depend on the system, but this should be the correct form.
For nearly any question of this nature, an excellent reference is Albright, Burdett, and Whangbo's Orbital Interactions in Chemistry, 2nd ed. Unfortunately, I can't find a diagram for C4v MOs, but in general most things can be found there.

There is one for $\ce{C5H5+}$ on p. 263.

I presume projection operator method breaks down here, since by projection operator all the orbitals won’t have the axial ligand contribution. Could I inquire how we derive this then?

Conceptually, you can just start from the Oh ligand groups and remove one ligand. That eliminates one of your t1 orbitals, so the two left are now e.  And the loss of symmetry means that your e orbitals from Oh become a1 and b1. The a1 from Oh remains a1.

@ANZGCFlyingFalcon, I am quite rusty with the projection operator, but my gut feeling is if you start with the axial orbital then the projection operator will simply return the axial orbital itself as the group orbital. That is quite valid, too; because in a real system, you would still be able to construct the MOs from linear combinations of those.

I just ran through the projection method and it worked fairly well. If we number the axial ligand as 1 and the others as 2,3,4 and 5 going around the square, I got 1 alone or 2+3+4+5 for the A1 projections. As orthocresol noted, these can be combined into the two a1 orbitals drawn above. For b1, I get something of the form 5+3-4-2 (ie opposite corners are same phase) and for e I get 2-4 or 3-5 (only two ligands- opposite corners out of phase). The e orbitals match up to the metal px and py, the b1 to d(x2-y2) and the a1's to s and d(z2) or to s and pz

Thanks for all the explanations, everyone. But based on Orthocresol’s diagrams above, one would see that the one with on the left most likely interacts with 4s while the other most likely interacts with 3d(z^2) purely by looking at the shape of the group orbitals. Does this mean that the 4pz orbital is non-bonding?

@ANZGCFlyingFalcon $p_z$ transforms as the TSIR in C4v. It's obvious enough that $p_z$ is capable of interacting with the axial orbital, so I won't labour that point. The four equatorial orbitals are more subtle: if they were exactly in the equatorial plane, then by symmetry they would have exactly zero overlap with the $p_z$ orbital. However, in practice the eq-M-ax bond angle is not exactly 90°, so there is some overlap.

To clarify, does that mean that we have a (mostly) nonbonding pz orbital, and 4 molecular orbitals (from the A1 set) for this case? Thanks a lot for everyone’s answers!


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What is the balanced equation of BH3/THF with an amide to form an amine?
I am doing a green chemistry project as an undergraduate and need the balanced equation to complete the calculations. I cannot find the right information online so any help would be greatly appreciated.

primary, secondary or tertiary amide? It makes a difference

Secondary amide attached to a propyl group.  Ar-C(Me)HCONHPr

Please could you tell the products produced after the reaction other than the target molecule. What does the borane complex become after the reduction? Something like HOBH2/THF or OBH/THF?

The reduction of secondary amides with BH3 requires 3 equivalents, mechanism shown below image sourced from here. The borane becomes HB=O which may transform further depending on work up.

I didn't understand how did the N-H become N-BH2 in the first step?


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Questions on Neutral oxides
Neutral oxides, are non metal oxides which are neither acidic nor basic.
I have noticed a pattern for neutral oxides which I am unsure actually holds true.
$\ce{CO}$, $\ce{H2O}$, $\ce{NO}$ and $\ce{N2O}$ are examples of neutral oxides. It seems that neutral oxides will always have 1 oxygen atom. While non metal oxides with more than 1 oxygen atom     seems to always be acidic oxide. ($\ce{CO2}$, $\ce{SO3}$, $\ce{SO2}$ and $\ce{P4O10}$).
Is this a coincidence? Are there any neutral oxides with more than one oxygen atom?

This is just a coincidence. Besides, H2O is not neutral; it is both acidic and basic, which might seem to be the same, but in fact is quite different.

And, $\ce{NO}$ is not just $\ce{NO}$.  It will contain a little dimer, $\ce{N2O2}$, which is still neutral.

Hmm, Since N2O2 is a neutral oxide, then it disproves the "all neutral oxides have 1 oxygen atom". However, the empirical formula of N2O2 is NO.  Is C2O2 and N4O2 neutral also? If they are neutral also, maybe the empirical formula of a neutral oxide will always have oxygen with a ratio of 1 to the other components.

Do not get hung up on this.  $\ce{Cl2O}$ is acidic.

Cl2O has one oxygen atom and is an acidic oxide. Thus, not all non metal oxides with one oxygen atom are neutral oxide. Also, not all neutral oxides have only one oxygen atom. N2O2, a neutral oxide but has more than one oxygen atom.

As a general rule, where nonmetallic oxides have a relatively low oxidation state of the nonmetallic you are more likely to get a neutral or only weakly acidic oxide. Usually this means only one oxygen atom in the molecule. Adding more oxygen atoms would make the nonmetallic atom centers more electrophilic towards water,which then shows up as greater acidity. Thereby, for instance, $\ce{CO2}$ would be expected to show greater acidity than $\ce{CO}$.
Carbon suboxide. $\ce{C3O2}$, shows am0hoteric character, reacting as an acid with water and ammonia but also deprotonating hydrogen chloride; in these reactions malonic acid or its derivatives are produced[1].
Reference

James E. House (2020). Chapter 13, Chemistry of nonmetallic elements I. Hydrogen, boron, oxygen, and carbon (Third Edition), Academic Press,
2020,
Pages 465-516,
ISBN 9780128143698,
https://doi.org/10.1016/B978-0-12-814369-8.00013-3.


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Which solution has higher boiling point?
Between a 2% (w/v) aqueous solutions of $\ce{NaCl}$ and $\ce{RbCl}$, which will have a higher boiling point.
Here there are two competing factors. First there's the fact that $\ce{NaCl}$ has a higher concentration. So it's solution should have a higher boiling point. Secondly according to Fajan's rule, $\ce{NaCl}$ is more polarised. So it should have a lower boiling point.
So which factor is dominant?

Fajans is irrelevant. Both compounds are going to be fully dissociated anyway.

Which solution has the higher molar concentration?

Molar mass of $\ce{RbCl}$ is $\pu{120.92 g/mol}$, Molar mass of $\ce{NaCl}$ is $\pu{58.44 g/mol}$. Both compounds are highly ionic and dissociate in water solutions. Thus, 2% solution of $\ce{NaCl}$ has approximately double amount of ions than that in 2% solution of $\ce{RbCl}$ (2% is fairly dilute as well, considering their solubility in water). Thus, theoretically, the $\ce{NaCl}$ solution should have higher boiling point (higher solute particles).

Boiling points depend on the strength of the intermolecular forces of the substance.So NaCl has a higher boiling point than RbCl.


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Real gas equation at high molar volume
How would the van der Waals equation of state look like when molar volume of gas is high? Does high molar volume correspond to low pressure?
My textbook says that

A real gas behaves similar to an ideal gas in the limit of large molar volumes.

What is the reason behind that?

What does the van Der Waals equation look like in the limit of  $P<<\frac{a}{V^2}$ and $V>>b$?

Sorry.  I meant $P>>\frac{a}{V^2}$.

The ideal gas equation of state assumes all of the particles are not interacting. This is a bad assumption when particles are close together, such as in a liquid. However as particles get further apart this becomes a more realistic model since particles collide less and less often.
In general low pressure and high temperature leads to conditions where the ideal gas equation will work. We see this in combustion reactions actually. You can model a gasifier fairly well with the ideal gas equation of state because of how high the temperatures are.
At high temperatures, molecules move with higher energy, so they bounce of walls harder. Thus to attain a given pressure, you need fewer molecules if they are hitting the walls harder. Thus, at a higher temperature, there are fewer molecules in your vessel, making them further apart, which makes them more ideal.
In terms of molar volumes (I prefer pressure but won't begrudge your textbook), yes as the molar volume gets bigger that means the same number of particles take up more space, meaning they are farther apart from each other. In the extreme limit they will become infinitely far apart and be exactly like an ideal gas. In practice the molar volume does not have to be extreme for the ideal gas to start modeling the system well.


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Is there a way to search for adducts by exact mass?
We found an unknown of exact mass 217.082. Is there a tool that one can search by an exact mass get a list of possible compounds. We have considered desmethylnaproxen as a possibility, but found its mass as 217.086 for the proton adduct.

What's the error bar on either measurement?

I guess that would be what you are looking for:
http://www.chemcalc.org/mf_finder/mfFinder_em_new
This one gives you theoretical combinations of atoms, that fit the mass.
You could also check the sdbs database for organic compounds.
https://sdbs.db.aist.go.jp
This one gives you existing compounds but it is rounding to 1 decimal so it gives you everything with 217.1 which is not perfect.
I just used the frist calc and then checking all close results (green) at SigmaAldrich to see, which formula matches a compound they sell. There are just two from the List:
$\ce{C10H16ClNO2}$ and $\ce{C11H11N3O2}$
So I would focus on these.

Thanks. I suppose though that without the proton adduct that 216.074 is the actual weight I'm looking for. I'll try that on the sites you provided. Our precision on a Thermo Q-Exactive is at least three decimal places.

@Michael Muehlbauer: Please see the Late addition in my answer.

By your question, I assume you may be taking Spectroscopic Analysis of Organic Compounds course. I'm not sure which book you are using but I recommend Silverstein's book (Ref.1), Spectrometric Identification of Organic Compounds, which is better (I have only 7th edition, but current edition is available as well). That's said, this book lists formula masses of possible compounds using Formula masses of different combinations of $\ce{C, H, N}$, and $\ce{O}$ to four decimal points (You have given exact mass of your compound with only three decimals). These formula masses in the book are based on the exact mass of the most abundant isotope of each element. Well, following are the closest given in Appendix A of Ref.1:
Formula mass: 217.0852 ($\ce{C11H11N3O2}$); Formula mass: 217.0865 ($\ce{C13H13O3}$); and Formula mass: 217.0892 ($\ce{C16H11N}$).
Late Addition:
Formula mass: 216.0774 ($\ce{C11H10N3O2}$) (Ref.1, pp. 62).
Reference:

R. M. Silverstein, F. X. Webster, D. J. Kiemble, Spectrometric Identification of Organic Compounds, 7th Edition; John Wiley & Sons, Inc.: Hoboken, NY, 2005.


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Radial pair distribution function (VMD). How to define it for a water NaCl system with multiple Na and Cl?
I'm struggling to understand a paper in which someone has presented radial distribution functions (RDFs) for a system that contains multiple Na, Cl, and water atoms/molecules.  I don't know exactly how the RDFs have been calculated--there must be some standard method.
I'm a newbie to molecular dynamics and have run my own simulations.  I would like to compare my RDFs to the ones in the paper.  However, I am not sure that I understand RDFs.  One reason I'm worried is that I keep receiving an error when I try to plot my own RDFs using the software, VMD.  The error is something like "cannot measure RDF."  It is either a bug in VMD or a conceptual misunderstanding on my part.  I don't know exactly how VMD calculates an RDF.
Let's say we have an RDF from Na to Cl in my system that contains multiple Na and Cl. Is it correct that an RDF is based on an average of what ALL Cl atoms are doing in relation to ALL Na atoms?  Or do I have to somehow define pairs of Na and Cl (I don't have these--they're all far apart in my system) and create RDFs for each pair?
I'm not sure whether "Chemistry" is the right StackExchange section for this question.  If it isn't and anyone has any advice on a more appropriate section, I'd be very grateful.  Thank you.

I haven't done it in vmd, but a rdf is really kind of like a histogram. For a particle of interest, you count how many particles are between r and $\Delta r$ from it and record that number in a bin, then move outwards, recording numbers in each shell into a bin for that shell. That tells you the rdf of that particle interacting with everything else. If there are three other types of particles in the simulation, you will get 3 rdfs from that one rdf calculation centered at the particle of interest.Repeat for other particles once you are done so you can get a nice average.
So all you need to do is count particles in $\Delta r$'s (and from this you get the volume too!, that is trivial though. I would do this for all Na's and Cl's and then average each of their bins to get one rdf for Cl-Cl, one for Cl-Na, one for Cl-O, one for Cl-H.
So that is how it should work, but I haven't used vmd to do it myself :S
RDF's don't have to be atom-atom, they could be molecule-molecule, or molecule-atom... anything really.

Thanks very much, Charlie. What you've said is very clear.  I'll have to think hard about how to apply it to my system though, and especially about what VMD is actually providing when it is used to calculate an RDF.

It is possible that it only follows one Na around recording the RDF, but it will calculate it at many many configurations to get a good average


text:
Hydrogen Bonding in Water (MO theory)
Valence Bond Theory tells us that each of the “sp3” (in reality, 44% s character) oxygen line pairs in water can act as electron donors, but observing the MO diagram for water tells us that the 2 non-bonding orbitals are of vastly different energies. (One of the “lone pairs” has bonding character too.) As such, how does water even form 2 electron-donating hydrogen bonds?

It is not like one orbital on the diagram corresponds to one lone pair.

I know that. The MO diagram of water has one “strictly nonbonding” (HOMO; b1 symmetry) and one “mostly nonbonding” (a1 symmetry, formed from 2s of oxygen) that has almost all electron density on the oxygen due to the energy difference. However, hydrogen bonding is often shown as the oxygen bonding to 2 hydrogens (Google it up and you’ll know what I mean), so I was inquiring about that.

Then why would you say that one of the lone pairs has such-and-such character, which implies that the lone pairs are different, which in fact they aren't?

@IvanNeretin The two lone pairs of water are of slightly different energy i.e. those two lone pairs are not exactly similar. Those two lone pairs reside in different (non-degenerate) orbitals having different symmetries. In fact, photo-electron spectrum of water also tells about the difference of the two lone pairs. They are not exactly similar.

Thanks, Somnik Das. I don’t think Ivan is referring to the same thing as us.

Indeed, there is an ambiguity here. The two pairs of electrons sitting on different molecular orbitals are different all right. The two protruding clouds of high electron density are similar. If you were referring to "lone pairs" in the former sense, then you may disregard my previous comments.

This website has a number of references to studies of the electronics and orbital interactions of hydrogen bonding in liquid water:        http://www1.lsbu.ac.uk/water/h2o_orbitals.html       Might be a good place to start.

How exactly hydrogen bonding in water looks like in actual liquid is fully known. AFAIK some percent of H2O molecules does accept second h-bond, but definitely not all of them.

related https://chemistry.stackexchange.com/questions/50906/are-the-lone-pairs-in-water-equivalent https://chemistry.stackexchange.com/questions/69370/how-many-hydrogen-bonds-are-formed-by-water-and-by-hf

The assumption seems to strong . The lone pairs are there and this suffices. What makes me puzzling now is if the various H-binds can be treated as the same or not once formed. But shouldn't be, at least in principle, I think. Is where photo-electron spectroscopy should be valuable.


text:
final concentration
Cheers. I am doing a hands on lab for intro to Chemistry. I basically have to do something seemingly simple. I pour 8.0 grams sugar, weighed, into a volumetric flask that is volume of 25mL. I add 15mL of water in then shake it up. I then pour 10 more mL. They want me to weigh the volumetric flask empty without the stopper. I then add the 8.0 g sugar and then the distilled water. After weighing the full flask, I have to determine the initial and final concentration. I have been looking up this formula but I see three different units are being used. This is not in molars. I assume I did the total volume right, which was 25 mL, I assume I did the concentration right, as I basically divided 8.0g by 25mL, then multiplied by 100%, which gave me 32. I am flabbergasted ont his initial and final concentration though. Any feedback on this? I'd appreciate it.

Your procedure as you state it here is different from what you describe in our comments to the answer by LigninPauling, where you say after mixing sugar with 15 mL of water, you filled up the solution to 25 mL with a transfer pipette. So in the end, you have 8.0 g sugar in 25 mL of solution (not 25 mL of water).

You said you measured the mass of the empty flask and the full flask. What were the values?

empty it weighed 22.5 g. sugar was 8.9 g. full with the sugar and remaining water up to the 25mL, the flask weighed 50.4g.

Water is about 1.0 g/ml. 50.4 - 22.5 - 8.9 = 19.0 Thus you could only added about 19.0 ml of water into the 25 ml volumetric flask, not 25.0 ml.

@Karsten Theis: According to OP's answer to your comments, OP has 8.9 g sugar in 25 mL of solution or 35.6% ($w/v$) concentration (not 32% as OP calculated).

Let's try and break this down.
We will assume that the mass of sugar was exactly 8.0 grams and that the volume of water was exactly 25.0 mL. Weighing the flask before and after the addition of the water would infact give you the volume of water added. 1 mL of water has a mass of 1 g and so the mass of the flask should have increased by 25 g.
For the concentration, it is unclear from your question what units were requested. If it is percent by mass then 32% sugar looks reasonable.
If it is requested in mass units per volume then perhaps g/L will suffice. You can calculate this for yourself quite easily.
Lastly, if molarity is required for your answer, this is slightly more involved. 'Common table sugar' (presumably what was used in this practical) is sucrose and a quick search will reveal its molar mass to be 342.3 g/mol. Therefore 25 g of sucrose equates to 0.0730 mol. Using this, the molar concentration may be worked out, taking care to note the units of molarity.

Yes my bad. Basically, it was just a 25mL volumetric flask, I shook the distilled water and sugar til it fully mixed and then after entering the volume before and after adding the solution mix(8.0 g sugar and 25mL water) the final question is to give the initial concentration and the final concentration. The initial has a % sign next to it g/mL. The final has a percent sign next to it as well. I'm assuming later I will learn why they ask for an initial and final concentration but for now I guess idk why they would want the same thing written twice.

In your question you mentioned adding 15 mL first and then afterwards adding another 10 mL. Was this included in the lab instructions? If so then perhaps the first could be the initial concentration they are asking for?

They said 15mL because it would be easier to stir the solution before adding the other 10mL. I also thought that same thing but, the result was filling the 25mL flask with the 8.0g of sugar and then using the pipet to fill up to the 25mL mark after mixing the solution better.

Since it is a lab for intro to Chemistry, this may directing to find basic knowledge of density and/or concentration, % concentration, etc. However, what puzzled me was the asking of intial concentration and final concentration. After reading all of comments and OP's answers, I thought this (see below) may be the question:
Weigh empty 25-mL volumetric flask first without the stopper. Then weigh $\pu{8.0 g}$ sugar and pour into the volumetric flask. Then, add $\pu{15 mL}$ of water to it and shake it up. Then, pour more water to the $\pu{25.0 mL}$ mark. Weigh the full flask again without the stopper. Determine the initial (after addition of $\pu{15 mL}$ of water) and final concentration (after addition of water to the $\pu{15 mL}$ mark).

that is exactly right. Also, after I weigh everything, I pour the solution 0 into a 100mL empty beaker, then take my pippete serological and I take 2.5mL and put it into the emptied and cleaned volumetric flask, and then add 15 more mL of distilled water and re-weigh everything and then repeat the steps again with different denominations of distilled water. Then after, I make a graph and show the x and y axis for the mass and volume.


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How to measure UV index?
I am a high school student so please apologise if I say something very dumb or irrelevant. I am conducting an experiment in which I want to find UV index at different points. Such as under shade, inside house, near beach etc. Anything will be greatly appreciated, also even if there is a device thats very expensive please tell me about that as well.

Finding 'UV index' at a point might be difficult. UV index is an index created by I think was WHO, and is calculated using
 - Thickness of ozone above
 - Cloud density of the area
 - Altitude of location
 - Sun's height
more about it here - https://www.who.int/uv/publications/en/UVIGuide.pdf
So I think you are better off finding the UV light intensity instead, using some measurement tools. There's few different options like photo reactive dyes, and digital scales, but probably the most cheap and quantitative data can be obtained using arduino or other similar development board and creating a small module yourself. Good Luck!

Hey first of all thanks for your help. Is there a way that I can convert the UV index into UV light intensity, vice-versa. I need to compare my results with the UV index in my city. Thanks

Sorry I am not aware of conversion of UV index to any physically measurable units. It's more of a guideline of how harmful is the UV at that location at that time, and is multi-component. While light intensity is a physical measurement (But there's multiple ways of measuring, each with different meaning). Moreover, comparing your measurement directly with the city's UV index makes no physical sense, it's like comparing apples and oranges. On the other hand you could decide on how to measure UV and make your own index.

Thanks again!!! I did some research and I found this could you please tell me if this device is what I think it is. It converts the UV intensity into UV index measurement. - https://www.arpansa.gov.au/user-guide-uv-index-meter

I guess it works. It could provide you with similar index value to what city announces. That devices seems to use SiliconCarbide photodiode, so do a little research about what’s being measured and include that info in your project


text:
Oxalate Anion MO Diagram
I figured I’d study $\ce{C2O4^2-}$ as an organometallic ligand, and found that it is weak field (pi donor). I hence decided to construct a MO diagram.
The oxalate anion is of D2h symmetry, so one can immediately use that, form irreducible representations and find the C-C orbitals:

Following which, I tried to use projection operator to look at all the group orbitals.

Unfortunately, this means that there are a lot of orbitals with Ag and B2u symmetry, and a couple with B1g and B3u symmetry. How would these orbitals interact?

You've pretty much done the hard part already. I would focus on the sigma framework first - exclude the O px orbitals as nonbonding and all of the pz orbitals as pi and then match up what's left. For each pair that you match, make a bonding and an antibonding version (or bonding/nonbonding/antibonding if there are three going in). Bear in mind that the orbitals of same symmetry can mix, but for the first pass ignore that and just find symmetry matches.

Thanks! Are the oxygen 2s orbitals mostly nonbonding too?

No, the s orbitals mix with the py orbitals, so both contribute to bonding and nonbonding. If you want to simplify things, you could represent the Os as having an sp-like orbital (oriented with large lobe toward C) that participates in sigma bonding and an sp-like orbital (oriented with large lobe away from C) that is nonbonding.

Also, for the pi framework, you can simplify by treating each CO2- separately, since nonplanar conformations of oxalate are common.

If it’s correct, is there any way for me to identify the ordering of the MOs?

The ordering is quite difficult to get exactly right without quantitative approaches, but you can get pretty close by assuming that sigma interactions are more stabilizing than pi and that generally more nodes means higher energy.  So group your orbitals as sigma bonding, pi bonding or nonbonding and then try to differentiate energy within each group. I might have time later to write up a longer answer with some of that info, but I'm not sure.

@Andrew could you take a look at the 2 imgur links I put above? I’m not sure if they are correct, so I’d like you to take a look.

Update: I re-did my drawing, and got this. How accurate is it? https://i.stack.imgur.com/ClWK7.jpg

Here's how I would approach it. I would start by making some (over)simplifying assumptions:
1) O $p_x$ orbitals are mostly nonbonding, so I'm going to ignore interactions of those orbitals with carbon orbitals
2) Due to $sp$ mixing, the carbon $p_y$ and $s$ orbitals can be merged into $sp$ orbitals aligned on the $y$ axis.
3) Same for the O $p_y$ and $s$ orbitals. In this case, the mixed $sp$ orbital with its large lobe oriented away from C is mostly nonbonding, so I'll ignore interactions of it with carbon orbitals
With those assumptions in place, I focus on building and ordering the occupied molecular orbitals. The highest energy occupied orbitals will be some of those oxygen-based nonbonding orbitals, so let's start by putting all eight of them at the top of the diagram. The relative energy levels among them will be affected by the number of positive and negative interactions, so the highest energy one will likely be the $b_{1g}$ where all of the $p_x$ orbitals have anti-bonding interactions with their neighbors. I mention this orbital in particular because it is the HOMO, and I think that the $\sigma$ interaction between oxalate and a metal center would come from the interaction of this orbital with the metal $d_{x^2-y^2}$ or $d_{xy}$, for example.
Mixed in with the $p_x$ and $sp_y$ nonbonding orbitals will be the two $\pi_{nb}$ orbitals, also centered on the oxygens. These are the $b_{2g}$ and $a_u$ orbitals in your set above. Since there isn't a carbon $\pi$ orbital with the right symmetry to interact with either of these, the resulting MO's have a nodal plane aligned with the C-C bond. These are the orbitals that will have $\pi$ interactions with a metal center if the oxalate is acting as bidentate ligand.
Below the nonbonding $\pi$ orbitals, we have the two $\pi$ bonding orbitals formed from the $b_{1u}$ and $b_{3g}$ central and oxygen $p_z$ orbital combinations. The $b_{3g}$ is higher energy of the two because of the C=C $\pi^*$ interaction.
At the bottom of the occupied orbitals, we have the five primarily $\sigma$ orbitals. I think the highest energy of the group is the $\sigma_{C-C}$ formed from the mixed $sp_y$ carbon orbitals oriented towards each other, ie a mix of the two $a_g$ central orbitals. Since the density is primarily between the carbons, I'm going to ignore the interactions with the O orbitals.
The four $\sigma_{C-O}$ orbitals come from the pairings of the central $b_{2u}$, $b_{1g}$, $b_{3u}$ and $a_g$ orbitals with their ligand orbital counterparts. In energy, from low to high I would guess $a_g$ ($\sigma_{C-O}$ and $\sigma_{C-C}$), $b_{3u}$ ($\sigma_{C-O}$ and $\pi_{C-C}$), $b_{1g}$ ($\sigma_{C-O}$ and $\pi^*_{C-C}$), then $b_{2u}$ ($\sigma_{C-O}$ and $\sigma^*_{C-C}$).

Thanks so much, Andrew! That definitely helped :D


text:
Quasi-static implies that T aprox equals Tsurr?
A few days ago I re-visited my undergraduate thermodynamics notes and (as you may have seen from the amount of questions I have been posting) I have encountered many difficulties that I overlooked at that time.
And I think most of my problems come from how unclear is the concept of quasi-static process in books.
Many textbooks define a quasi-static process as:

A process that has its macroscopic variables well defined and in the system it is at each instant of time in a state infinitesimally close to the state of equilibrium.

The first part of that definition is logical. But the second part, related to thermodynamical equilibrium confuses me.
What is clear is that if it is stated that a certain process is quasistatic, then the pressure of the system is infinitesimally similar to the external pressure, which allows us to easily calculate the volumetric work.
My concise doubt comes from the relationship of the external temperature with the internal temperature. We know that for a system to be in true equilibrium with the outside, both pressure and temperature must be the same (without considering material equilibrium).
If this is the case, T system should be infinitesimally equal to T of the environment which would completely eliminate the generation of entropy by transfer between finite temperatures.
Now, when one studies theoretical cycles, it is generally stated that they are internally reversible ( quasi-static internally and without internal friction) but irreversible at last because the unstopabble heat transfer between finite temperatures in many steps.
I see a contradiction here:
As I understand it, many define a quasistatic process as:

"An extremely slow process where internally the P and the T are equal, homogeneous". (Completely agree)
"The system  at each instant of time in a state infinitesimally close to the state of equilibrium". Dont' get it since many problems considers Pexternal=Psystem but T System different to T surroundigs.

To me, it should be called partially quasistatic process if Tsys does not equals to Tsurroundings.
Regards.

I will try to word a full answer later, but quasi-static does not mean Text=Tinternal. It means if you make a change, the change happens so slowly as to be continuous. This is convenient if you want to calculate the change by integration. For instance the work done is $\int PdV$ and in a quasi static process P is continuous and thus, a differentiable and integrateable function. Assuming ideal gas you can replace P with $\frac{nRT}{V}$ and integrate wrt V.
Quasi-static does not mean that the temperature inside the system is equal to the temperature of the surroundings during a change.

As they explained to me (which does not mean that it is correct) the volumetric work always, no matter the nature of the process is equal to Pext.dV. And the pressure of the system is considered equal to the external pressure in a quasi-static process, "since it is so close to equilibrium".
Now, * casually * temperature issue is always omitted at this point.

pressure is not equal inside and outside. If that was the case, nothing would move.

quasi-static is not a measure of the difference between inside and outside for anything. It is a statement about how changes occur. Changes only occur when there is a driving force. There is no driving force when things are equal.

Ok, no doubt about it. Let's say that quasi-static only refers to the evolution is "slow" and "continuous", easily integrable.
Now, I understand that the first law of thermo is related to energy transferred to the system  by external agent, so Volume Work =Pext.dV . Is not obvious the relation of Pext to Psys. And in many many places, with no further explanation that is considered quasiestatic they do Psys=Pext. :(

For example, on "Work on a quasistatic process" by Sevilla University: When we have a system, typically a gas, on which we perform a quasi-static process, based on performing differential work, let it return to equilibrium, perform another differential work, let it return to equilibrium, etc. we have that the system is always approximately in a state of equilibrium with the outside. This means that the system pressure is defined and will be (approximately) equal to the external pressure applied

Yes, in this case the pressure is constant, and the change occurs by applying work to the system, it does not occur because of a pressure gradient, although you could argue that after the work has been done, a pressure gradient exists... however it is infintisimally small. The work is $P_{ext} dV$ and is the sum of all little changes, which were all done at the same P. If you did the change in one big step, there would be a pressure gradient created, which would create irreversibilites and pressure would not be a constant. It would take more work to compress in one big step.

I think the same would apply to temperature and entropy, they go hand in hand, like pressure and volume do.

But when exactly we can calculate work as Pext.dV=P.dV? In many examples is used when the process is stated only as quasi-stati, so we should suposse pExt=constant?

In many cases, reversible and quasistatic are taken to be synonymous adjectives, even if quasistatic more accurately takes on the meaning that you've described.

@a-cyclohexane-molecule that's the problem in many textobooks, and then suddenly, in cycles the throw at you some so-called quasistatic process but irreversible, internally reversible processes... so confusing when you see it at pespective.

See the following thread from Physics Stack Exchange:  https://physics.stackexchange.com/questions/443527/entropy-generation-in-an-endoreversible-engine

In my judgment, when they are talking about irreversible cycles, they are referring to idealized cases in which (a) the process is internally reversible with respect to the working fluid but (b) irreversible with regard to the region outside the working fluid where there is a finite thermal resistance (e.g., a cylinder wall with finite thermal conductivity) between an (external) ideal constant-temperature reservoir and the working fluid.  So the rate of heat transfer to the working fluid is very low (quasi static), and all the entropy generation takes place within the (higher) thermal resistance wall.  At least, this is how the irreversibility is often modeled.
To illustrate all this, see the following thread from Physics Stack Exchange:  https://physics.stackexchange.com/questions/443527/entropy-generation-in-an-endoreversible-engine


text:
Conversion from a PPB value to µg/m3 of Isobutylene
I have a quite simple problem, found a lot of information about it, but I am not sure anymore if I do my calculations right.
I own a sensor, which reports measurements in isobutylene units as PPB. So if I understand this right, the measurement would be the number of isobutylene molecules per one billion (x/1'000'000'000).
Therefore to convert this value into the more common µg/m3 unit, I would just use this formula:
$$ \text{Concentration}\ \frac{µg}{m3} = \frac{\text{Concentration}\ \text{PPB}\times\text{Molecular Mass}\ \frac{g}{mol}}{\text{MolarVolume}\ l} $$
Therefore to convert 400 PPM of isobutylene, with a molecular mass of 56.106 g/mol or 0.0005879 g/l, in the molar volume of a gas at STP with 22.4 l, the calculation would be:
$$ \frac{400\times56.106}{22.4} = 1001.9 \frac{µg}{m^3} $$
This seemed very high to me, and the Range of the sensor is 0-1056 PPM.
Is this the correct formula for the conversion?

This is much simpler than you think, even though the value you've obtained looks reasonable to me.
Part per billion ($\pu{1 ppb} = \pu{1e-9}$) on its own is meaningless.
Based on a context, it can refer to anything: mass fraction, mole fraction, particles etc.
The thing is, for gaseous mixtures $\pu{ppb}$ always refers to the volume fraction $\phi_i$:
$$\varphi_i = \frac{V_i}{V}$$
where $V_i$ is the volume of the $i$th gas (here, isobutylene) and $V$ is the total volume.
In order to convert volume fraction expressed in $\pu{ppb}$ to desired mass concentration $\rho_i$ expressed as
$$ρ_i = \frac{m_i}{V}$$
where $m_i$ is the mass of $i$th component in a mixture, you have to use density $d_i$:
$$ρ_i = \frac{m_i}{V} = \frac{m_i}{V_i}\varphi_i = d_i\varphi_i$$
To conform with your units of choice ($\pu{μg m-3}$) we need to juggle the units a bit as the density is usually given in $\pu{kg m-3}$, e.g. for gaseous isobutylene at NTP $d(\ce{C4H8}) = \pu{2.3959 kg m-3}$ (Source):
$$
\begin{align}
\pu{1 g} &= \pu{1e6 μg} \\
d(\ce{C4H8}) &= \pu{2.3959e6 μg m-3}
\end{align}
$$
Now, using this density value you can multiply your volume fraction in $\pu{ppb}$ or $\pu{ppm}$ and get an answer, e.g. for $\pu{400 ppm}$:
$$ρ_i = d(\ce{C4H8})\varphi (\ce{C4H8}) = \pu{2.3959e6 μg m-3}\cdot\pu{400e-6} = \pu{958 μg m-3}$$
Note that all notations involving "part per something" are deprecated and should be avoided.
It also wouldn't hurt if you ask manufacturer whether $\pu{1 ppb}$ the sensor reports is indeed $\pu{1e-9}$ and not $\pu{1e-12}$ (just in case) if you think the reported values are too high.

@downvoter I'd be grateful if you leave a comment about what's wrong with the answer and how it can be improved.

400ppm means if you have million molecules, 400 of them would be isobutylene in this case. To convert ppm into µg/m3:

Find the number of gas molecule in $1m^3$ (using PV=nRT and Avogardo's const.)
Find the number of molecule of Isobutylene in $1m^3$ (400ppm)
Convert the number of Isobutylene molecules into moles to find the mass (Using molar mass)
Change the mass unit from g to µg

By the way, your formula is directly multiplying grams/liter with ratio of number of molecules, which gives non-physical number.

According to Environmental Science and Technology Briefs for Citizens (USA), concentrations in soil can be reported as parts per million (ppm) or parts per billion (ppb) where $\pu{1 ppm} = \pu{1000 ppb}$. For concentrations in soil, $\pu{1 ppm} = \pu{1 mg/kg}$ of contaminant in soil, and $\pu{1 ppb} = \pu{1 \mu g/kg}$.
For concentrations in water, $\pu{1 ppm} \approx \pu{1 mg/L}$ (also written as $\pu{mg/L}$) of contaminant in water, and $\pu{1 ppb} \approx \pu{1 \mu g/L}$ (also written as $\pu{\mu g/L}$).
Since, $\pu{1 L} = \pu{1 dm^3}$ and $\pu{1 m^3} = \pu{1 dm^3} \times \frac{\pu{1 m^3}}{\pu{10^3 dm^3}}$, your conversion for $\pu{400 ppb}$ should have been:
$$\pu{400 ppb} = \pu{400 \mu g/L} = 400 \times \frac{\pu{1 \mu g}}{\pu{1 dm^3}} \times \frac{\pu{10^3 dm^3}}{\pu {1 m^3}} = \pu{400 \times 10^3 \frac{\mu g}{m^3}}= \pu{400 \frac{mg}{m^3}}$$


text:
Cross Cannizzarro reaction for ketones
I want to ask that cross cannizzarro could be possible between any aldehyde + ketone mixture(without alpha H) and we get products as aldehyde get oxidised and ketone get reduced(along with self oxidations and reduction products of aldehyde)? Is it correct?
I think it is possible after going through the mechanism but searched in different books and doesn't get anything about it...

Possible duplicate: https://chemistry.stackexchange.com/questions/86573/how-can-this-ketone-show-cannizzaro-reaction/86592#86592

@user55119 I think it isn't a duplicate of that question. In that question it is asked for a special type of ketone and not following in same mechanism as of Cannizzarro. But I am asking in general they give reaction or not (any ketone)


text:
Will the motion of particles always increase with the rise of temperature?
I came across this following question while poring over my textbook.

Predict the change in the motion of particles you would expect in the following if the temperature was increased:
a) Solid
b) Liquid

I think if the temperature increases, solid or even liquid particles will receive more energy and therefore their motion will speed up. I'm not sure why there are two parts for this question. Aren't they supposed to be similar?
Thank you so much for helping! Since this is my first post, please pardon me if I accidentally violate any rules of the forum.

related: https://physics.stackexchange.com/q/113643

@A.K. Thanks for the link, Physics.SE delivers big time :) I don't think there is a straightforward answer to this one (well, at least not if treating this as a HW question where a student is supposed to give a simple answer), and it would be interesting to receive some answers first, and not close the question as an off-topic.

I am surprised the no respondent mentioned sulfur. A little above its melting point it's a mobile liquid but as you heat it up it undergoes a chemical reaction that results in an allotrope with long chains that is more like thick tar in consistency. Thus although the individual sulfur atoms may be oscillating more strongly about their average positions as they heat up, their capacity to move freely through the bulk material becomes sharply restricted.

@andselisk I was not implying that it was a duplicate, I only wanted to link a question that was related to the current one.

@A.K. Neither was I, I haven't voted to close the question, I suggested not to close the question. Sorry if it wasn't clear.

Indeed they are similar, but there is a difference. For solids, particles can only vibrate about their own position. Heating results in a stronger vibration. For liquids, particles can wander around the whole volume of the liquid. Heating increases the speed of the particles so that they can move more quickly inside the liquid, hence a higher diffusion constant.


text:
Why is graphite brittle?
Carbon exists in more than one allotropic forms, and the allotropes have different physical properties due to different arrangement of atoms. I am going to compare their hardness for now.
Below is the structure of graphene:

Source: Wikipedia article on Graphene
From what I read, graphene is hard, but graphite is brittle.

If graphite is made up of many layers of graphene, then why is graphite not stronger than graphene?
I suppose when you say graphene (or diamond) is hard, it means that, for example, when you take a hammer to strike the layer of graphene, the layer is not easily broken, that is, the covalent bonds are hard to be broken by physical means.

By this reasoning, the layers in graphite are easily separated because they are held together by weak van der Waals forces, whereas it should be difficult for us to cut across the layers (in direction perpendicular to the plane of layers, cf: image below)
Does this mean that we can only cut the graphite in one dimension? I thought the graphite being brittle already implies that graphite can be cut in more than one dimension?

Original image source: Tirupati Graphite

By the way, you should not have problem crushing a large graphene  sheet. ....

A large graphene sheet, if not held in form by an underlying hard surface or otherwise fixed, crumbles up by itself.

@Karl my point was to warn OP not to mix thinghs at molecular/nanoscale and material properties. Likely a pile of graphene sheets should be graphite. By the way, it would be nice to know the behaviour of graphene under is own weight.  I think in principle it could extend quite far.

@Alchimista Of course, the only thing "hard" about graphene is the E modulus. Btw. graphite has a 3D crystallographic structure. Randomly stacking sheets does not do it. And the first picture above is very misleading. A free graphene sheet is not flat. It forms undulations (a "bumpy" surface), and if not fixed at the edges, it crumbles up.

Hardness and brittleness are not contradictory properties. There are plenty of substances that are both hard and brittle (cast iron is a good example, glass is another).
And "strength" is different as well. It depends on context. Tensile strength is not necessarily related to ability to survive a blow: some compounds with high tensile strength (which basically means the compounds resist being stretched) are incredibly fragile (or brittle if you prefer that term). Some compounds with low tensile strength are very hard to break (nylon rope, for example).
Part of the problem is that many of the properties of materials are based on the structural properties of the bulk material not the idealised properties of the chemical bonds making up the material. Some metals, for example, are tough (not brittle) and are hard to break but this property arises because defects in the crystal structure can move around absorbing external stress. Brittleness is often caused because no such mechanism exists and, once a surface bond breaks, the stress on the next bond gets worse, releasing energy and propagating a crack, tearing the material apart. Glass often has many small surface cracks which, under tension, concentrate stress making them brittle and subject to falling apart. But very pure, carefully grown glass fibres can be nearly crack-free making them much stronger under the same stress. In short to explain brittleness, toughness or strength, you need to know the macroscopic makeup of the bulk material not just the chemical composition.
So it is with graphite. There is no easy mechanism in a single graphite crystal to dissipate stress if a bond breaks, so it will tend to be brittle. But most samples of graphite are not single crystals: in most cases they will consist of a bunch of small units with the layers oriented in different directions. So bulk graphite isn't just brittle but also soft as some orientations will slide easily around. And it is hard to even talk about bulk graphene because by definition it is a single layer or a very thin bundle of layers. When people say graphene is strong they usually mean that it has high tensile strength along the plane of the layers. This doesn't mean it isn't either brittle or hard and even if it were in some orientations, it probably would not be in others.

(That graphene is "hard" is a somewhat dubious (or ambiguous) statement. It's true that it is has a high ($E$) modulus when you pull evenly in the direction of the plane. Under all other attempted deformations, it breaks almost instantly.)
The planes of graphite are easily shifted against each other. That means if you try to bend it, the individual planes cannot support each other and distribute the load. In a solid 3D structure like diamond or wood, the outer layers would just feel a linearly increasing extensional force, and the inner layers an increasing compression.
Not so in graphite, the few innermost layers at the bend get all the load, and snap. And then the next few, and so on. Graphite, despite being rather soft, is brittle.


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Strange "far out" orbitals in HF and DFT calculations. Why do they occur?
I calculated some orbitals and I get quite strange results with orbital geometries I wouldn't expect from MO theory. A more experienced colleague just said: "That's always the case with big basis sets, just ignore those and search for the ones that make sense" which is a very unsatisfying answer and he couldn't explain what's the cause or what they mean. I also noticed that these occur more often/with lower energy when using HF but they also occur using DFT.
Can anyone explain what happens here and why?
Here is one example with the first two looking like I would expec.
The first 5 unoccupied orbitals using HF (6-311+G(d,p)) including energies in eV:

Don't put too much thought into virtual orbitals. They do not contribute to the Fock matrix, hence they don't have any effect on occupied orbitals or the ground state energy.

@Feodoran but they are used in (F)MO theory besides other applications, that's why we look at them after all. And that's why I would like to understand what's going on here and not just ignore it or don't think much about it.

@DSVA That is why I said "don't put too much thought them". They are valid eigenfunctions of the Fock matrix, and can be used for interpretaions, e.g. the LUMO energy in Koopmans' theorem. But they are still virtual, as in not really there.
Can you try what happens when you remove the diffuse basis functions?

Maybe they do exist.  But if they do, we can ask:  (1) how much of the electron density do these far-out lobes carry?  (Probably less than it looks like in the drawing) and (2) do they affect the observedelectronic structure of the molrcule?  (If they require antibonding interactions to form, then no).  We may be observing a tempest in a teapot.

@OscarLanzi Virtual orbitals don't carry any of the electron density in HF and DFT calculations, and AFAIK they have no effect whatsoever on the ground-state electronic structure of the molecule.

In addition to PAEP's answer, I want to shed some light on how much meaning virtual orbitals may have.
First of all, there is the whole discussion on whether orbital in general are meaningful or not. Without going into details here: they are an approximation and as such they may give meaningful interpretations occasionally. If those interpretations do not reflect reality, we need to go to more sophisticated theories.
Why do we even have virtual orbitals?
As already mentioned, they show up because in linear algebra (which we use to numerically solve the Hartree-Fock-Roothan-Hall equation), we get as many eigenfunction (MOs) as basis functions (AOs) are supplied. Since usually more AOs are supplied, than electrons are in the systems, we get unoccupied or  virtual orbitals. And if we want to improve accuracy by adding more basis function (e.g. diffuse/augmented or polarization functions) we do get more virtual orbitals. (At this point one could ask what the additional virtual orbitals even mean. Probably not much, that is why I suggested to compare your result to a calculation without diffuse basis functions.)
Are the virtual optimized?
What happens during the HF algorithm is, that we iteratively optimize the MOs based on the current (approximate) wave function (or electron density, if that helps imagining it). In HF this corresponds to the Slater-determinant, which is constructed from all occupied MOs. So all MOs (occupied and virtual) are optimized, but the optimization is with respect to the occupied MOs only.
Another perspective would be to say that the occupied MOs are coupled together via the electron-electron interaction. Thus they somehow need to "work together" to minimize the ground state energy. The virtual orbitals however, do not contribute to that energy and are not coupled by the electron-electron interaction (because they are empty). Maybe one could say, that the virtual orbitals are not optimized with respect to each other. But maybe this statement does not really make sense, since due to the missing coupling there is nothing they could be optimized for. However, they are still valid eigenfunctions of the Fock operator and are therefore part of the orthonormal basis set of molecular orbitals.
Summary
So there is something well defined about virtual orbitals which may justify interpretations based on them. But this is more vague/not as rigid as for occupied orbitals. Therefore a meaningful interpretation of virtual orbitals may not work as well as for occupied orbitals (and even those are limited).

As already mentioned in the comments you have to be careful when interpreting the meaning of the virtual orbitals in an ab initio or DFT calculation. In the calculation the only orbitals that are optimized are the occupied orbitals while the virtual (unoccupied) orbitals give you a guess of the shape you expect from them if they were occupied. To look at it from a simple perspective, somehow you are using the LCAO method, you obtain as many MO as AO you are using in your basis set, but you only optimize the MO that are occupied.
There is a second point that may be relevant to your question. In many programs the default orbitals you get in your calculation are the canonical orbitals (delocalized orbitals) that even if occupied may not have an evident chemical interpretation on its own. For those purposes it may be better to use localized orbitals (e.g. NBO) that may be easier to interpret.
Two book recommendations:

C. J. Cramer, Essentials of Computational Chemistry. Theories and
Models, Wiley (2004).
Frank Weinhold, Clark R. Landis, Discovering chemistry with natural bond orbitals, Wiley (2012).


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Bond rotation in buta-1,3-diene
Why don't all the atoms in s-cis-1,3-butadiene lie in one plane?
All carbon atoms are $\mathrm{sp^2}$ hybridised and I feel that there would be no $\ce{C-C}$ bond rotation due to the conjugation that leads to a double bond on $\ce{C2-C3}$ making them all lie on a plane.
Is it any different for s-trans-1,3-butadiene?

Consider flagpole repulsions by the hydrogens of the terminal carbons of the molecule,C1 and C4

Oh so due to sterric repulsion, the C-H bond twists and causes the hydrogen atoms to go to a different plane than the molecular plane(like Ortho effect). But in the case of trans 1,3-buta-diene there would be little to no sterric repulsions so all atoms would lie in a single plane. Am I correct?

1,3-Butadiene (Preferred IUPAC name: Buta-1,3-diene) is the organic compound, which is industrially important as a monomer in the production of synthetic rubber. It is also the simplest conjugated diene among organic compounds. Since all carbon atoms are $\mathrm{sp^2}$ hybridized, orbital overlap between double bonds is possible and, theoretically, two different conformer shows maximum overlapping, which are s-trans conformers with $\ce{C2-C3}$ dihedral angle of $180^\mathrm{o}$ and s-cis conformer with $\ce{C2-C3}$ dihedral angle of $0^\mathrm{o}$. However, due to steric hindrance, this s-cis conformer is approximately $\pu{16.5 kJ/mol}$ higher in energy than its s-trans counterpart, making this geometry a local energy maximum. Thus, in contrast to the s-trans geometry, which is global minimum, s-cis geometry is not a conformer (Wikipedia).
The double bonds of the s-cis geometry are twisted to give a dihedral angle of around $38^\mathrm{o}$, is considered to be a second conformer (called gauche-rotamer) that is around $\pu{12.0 kJ/mol}$ higher in energy than the s-trans conformer. The high level ab Initio calculations, utilizing coupled cluster theory with quasi-perturbative triple excitations and augmented quadruple $\zeta$ level basis sets, done by Feller and Craig in 2009 showed that the transition state separating the trans- and gauche-rotamers lies $\pu{26.8 kJ/mol}$ above the trans global minimum (Ref.1). These calculations further showed that the gauche rotamer lies $\pu{12.6 kJ/mol}$ above the trans-rotamer and the s-cis-form is a transition state $\pu{2.0 kJ/mol}$ higher than the gauche-rotamer. Wikipedia state that:

This increased rotational barrier and strong overall preference for a near-planar geometry is evidence for a delocalized π system and a small degree of partial double bond character in the C-C single bond, in accord with resonance theory.

This is illustrated in the following scheme with the most stable conformer of 1,3-butadiene, which has the s-trans conformation (Ref.2):

Quantum chemical calculations on Π-Electron Delocalization in Butadiene has also done and published in 2006. The resultant bond lengths were illustrated in the following diagram and the total abstract of the paper was reproduced here for readers with interest:

Abstract: Equilibrium structures have been determined for s-trans-1,3-butadiene and ethylene after adjusting the rotational constants obtained from rotational spectroscopy by vibration-rotation constants calculated from the results of quantum chemical calculations. For butadiene, the formal $\ce{C=C}$ bond length is $\pu{1.338 Å}$, and the formal $\ce{C−C}$ bond length is $\pu{1.454 Å}$. For ethylene, the $\ce{C=C}$ bond length is $\pu{1.3305 Å}$. These values appear to be good to $\pu{0.001 Å}$. It is shown for the first time that π-electron delocalization has the structural consequences of increasing the length of the formal double bond by $\pu{0.007 Å}$ and decreasing the length of the formal single bond by $\pu{0.016 Å}$. Comparisons are made with structures computed with several quantum chemical models. The MP2/cc-pVTZ results agree best with the new $r_e$ structure.

References:

D. Feller, N. C. Craig, “High Level ab Initio Energies and Structures for the Rotamers of 1,3-Butadiene,” J. Phys. Chem. A 2009, 113(8),  1601–1607 (DOI: 10.1021/jp8095709).
N. C. Craig, P. Groner, D. C. McKean, “Equilibrium Structures for Butadiene and Ethylene:  Compelling Evidence for Π-Electron Delocalization in Butadiene,” J. Phys. Chem. A 2006, 110(23), 7461–7469 (DOI: 10.1021/jp060695b).


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Chemical bonding based on hybridisation model
A snippet from a textbook:

Therefore, the hybridization model predicts that an $\mathrm{sp}$-hybridized carbon atom is more electronegative than an $\mathrm{sp}^3$-hybridized carbon atom. Evidence for this effect is that the positive end of the dipole moment in $\ce{N#C-Cl}$ is on the $\ce{Cl}$ atom. We conclude that the carbon atom in the cyanide group is more electronegative than a chlorine atom.

The paragraph suggests that the $\ce{N#C}$ bond has a different hybrid orbital compared to the $\ce{C-Cl}$ bond, doesn't it? I am a bit confused since there is only $1~\ce{C}$ atom but somehow it has $2$ hybrid orbitals of different state (namely $\mathrm{sp}^3$ and $\mathrm{sp}^2$). Is this due to different ligands with which the central atoms connect?
And point of interest is why the conclusion isn't

the nitrogen atom in the cyanide group is more electronegative than a chlorine atom

instead of carbon? Because I would imagine that the electron density of both $\ce{N#C}$ and $\ce{C-Cl}$ bonds would lie dominantly on $\ce{N}$ and $\ce{Cl}$, respectively. Hence, it is more reasonable to me to compare the electronegativity of $\ce{N}$ and $\ce{Cl}$ instead that of $\ce{C}$ and $\ce{Cl}$.

There are no such things as hybridised atoms, that is very sloppy language. Please add a citation to your quote, otherwise one could consider it plagiarism.

The usual explanation for $sp$ orbitals being more electronegative than $sp^3$ is the fact that more $s$ character means a shorter average distance from the nucleus, thus experiencing a greater effective nuclear charge. This partially explains the decreasing magnitude of the dipole moment of $\ce{CH3-CH2-Cl}$ to $\ce{CH2=CH-Cl}$ to $\ce{HC#C-Cl}$ and the increasing acidity of ethane to ethylene to acetylene.
I agree with you that the example they give doesn't really demonstrate that, since the orientation of the dipole (as you pointed out) relies on the electronegativity of the nitrogen, not the carbon. Specifically, the stability of the left resonance structure below:
$[\ce{^-N=C=Cl+  <-> N#C-Cl}]$
which clearly shows the orientation of the dipole, even if this resonance structure is only a minor contributor.
As a counterexample, chloroacetylene has a dipole with the (-) end on the chlorine, even though the carbon is still $sp$ hybridized. The corresponding resonance structure with a negative charge on C is much higher in energy than in the nitrogen case and so contributes negligibly to the overall structure, so the inductive electron-withdrawing of Cl dominates over the resonance electron-donating.

can you explain why in $- \mathrm { N } = \mathrm { C } = \mathrm { Cl } ^ { + }$  $N$ is partially negative and $Cl$ partially positive?

I added the starting structure to make it more clear. In order for Cl to form the second bond to C, it has to share electrons with C and so it becomes (+) charged. The N takes electrons from the third N-C bond, so it gains electrons and becomes (-) charged.


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Stability of carbocation on doubly bonded carbon

What can be said about the stability of the following carbocation?

I think that it would have something to do with hybridisation as, clearly one carbon is sp hybridised and the other is sp2.
Can anyone confirm this or elaborate on this?
Also what would happen if the Hydrogens are all replaced by a carbon chain?
And what if there's positive charge on both the given carbons?

stability can only be compared not determined

Look here https://en.m.wikipedia.org/wiki/Vinyl_cation. Note: either you present an empirical formula or you place the + on the C without octet.


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Hybridization of carbon atoms present in the smallest ester?
I found this question in the Resonance DLPD Organic Chemistry book on Page 55:

What is the hybridization of carbon atoms present in the smallest ester?

The smallest ester as far as I know is methyl formate ($\ce{HCOOCH3}$):

The first carbon atom (on left) has 3 σ bonds and the one on the right has 4 σ bonds. Hence, the hybridization is $\mathrm{sp}^2$ and $\mathrm{sp}^3$, respectively.
However, according to the answer given in the book, all carbon atoms in the smallest ester are $\mathrm{sp}^2$ hybridized.
What did I do wrong? Is there a smaller ester? Or the hybridization of the one I have above not correctly determined?

Consider the hybrid obtained after resonance: This is only a canonical structure

@YUSUFHASAN sorry I don't know about resonance as I am only a high school student and have only recently started studying organic chemistry.

Ok...See this could be either one of two cases: the book wanted you to consider only this structure,and so there is a misprint, as then the answer would be one sp2 and one sp3 carbon. Or, they wanted you to consider resonance, and understand the fact that the hybridization will be somewhere between sp2 and sp3(atleast I would have done that), but the answer doesn't show that.So,most probably,it is a misprint.Although, resonance should be taken into consideration as the actual existence is of the hybrid Check it out here

You are right. Methyl formate has one $\mathrm{sp}^2$ and one $\mathrm{sp}^3$ hybridized orbital.
Your book probably has a misprint.

@YUSUFHASAN Atoms are not hybridized, orbitals are. Please avoid misleading edits with questionable improvements. Also, orbitals are denoted with upright roman letters.


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How to show a band structure for water?
This question is generated by my attempt to answer this one.  I thought I could demonstrate that electronic conduction in water is possible because of the impact of an electric field on the band structure, but I am being hit with comments questioning the existence of band structure in liquids.  I thoughtot was self-evident, but I was asked for references showing water has a band structure and what I am coming up with does not appear to be working (the comments are getting more upvotes than my answer).  So I need help.  Can anybody come up with more clear-cut references demonstrating that water, or liquids generally, has such a band structure?

do Couto et al have published an article [1] characterizing the density of states (band structure) of water using DFT theory combined with MC sampling. Extrapolating from clusters to bulk water, the HOMO-LUMO gap is 8.55 eV (Table 3).
From the abstract:

By extrapolating the results for larger clusters the threshold energy
  for photoelectron emission is $9.6±0.15 eV$ free clusters and
  $10.58±0.10 eV$ embedded clusters. Our results  for  the  electron
  affinity $V_0=−0.17±0.05 eV$ and  adiabatic  band  gap
  $E_{G,Ad}=6.83±0.05 eV$ of liquid water are in excellent agreement
  with recent information from theoretical and experimental works.

From the introduction:

It is generally accepted that water canbe  described  as  a  very
  large  band-gap  amorphous semiconductor.11,17 However, it is not
  obvious that the band-gap of liquid water can be uniquely associated
  with an optical vertical excitation process, where the gap is
  defined simply as the highest occupied molecular orbital-lowest
  unoccupied molecular orbital HOMO-LUMO energy difference. As recently pointed out
  by Coeet  al.,3 the reactive nature of electronically excited water
  molecules,18 the reorganization of water molecules around charged
  species in liquid phase, and the observed photophysics of anionic
  defects known as the anion problem 4 strongly indicate that an
  adiabatic route can be defined for accessing the conduction-band edge
  in liquid phase. Moreover, given that the time scale ofthe solvent
  relaxation is much larger than the vertical process, the adiabatic
  band gap of liquid water cannot be determined vertically.

Reference
[1] Cabral do Couto, Estácio, and Costa Cabral J. Chem. Phys.123, 0545102005

Great, just what I needed!


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How can an electron shield another electron of the same subshell?
While I was preparing for my upcoming exams, I stumbled upon this sentence which is bothering me quite a bit:

The contraction of the lanthanoids is due to the imperfect shielding of one electron by another in the same subshell.

Though I am well aware of lanthanoids contraction and the poor shielding effect of the f- and d-orbitals, but still according to what I have studied and understood so far is the shielding effect is the shielding of valence electrons by the electrons between the nucleus and the valence electrons from the nuclear charge.
Then how can an electron shield another electron of the same subshell?

@Zhe So does this actually effect the atomic radus of the lanthanides

Normally, as you add electrons, after adding several, you'd need to expand to a higher energy -- and larger -- orbital. But with lanthanides, you're basically sticking all of the electrons in the same energy level where shielding is imperfect. So the shielding does not grow commensurate with the increase in nuclear charge.

Shielding is simply the cancellation of some of the effective charge of the nucleus.
It's a lot easier to envision an electron to shield another electron that, by time average, is likely to be farther from nucleus, e.g., $1s$ shielding $3p$.
But for 2 electrons in the same subshell, on average, you can expect one to be closer to the nucleus than the other roughly 50% of the time. While it's true that a good deal of the time that other electron may be on the other side of the nucleus, for d- and f- orbitals, we're looking at more than 1 electron, so there's some partial shielding. Partial, because the shielding electrons don't actually shield a good portion of the time due to relative distances from the nucleus.

Consider that there are two electrons in two different but adjacent circles. Assume that these circles encircle the nucleus.
Now also assume that another circle is encirciling the nucleus and this circle is currently vaccant. Now an electron is passing nearby this whole system  This electron gets attracted by the positevely charged nucleus, and gets accomodated initially on the middle circle. This electron starts experiencing repulsive forces from the nearby electrons and get repelled away to the outermost circle. Even though repulsion is still there, it is reduced to some extent currently. At present the electrons in the inner circles are shielding the electron in the outermost circle or they are preventing the electron in the outermost circle from getting a more effective nuclear attraction.
That is an electron is being shielded by another electron. If we assume that the three circles mentioned above form a single subshell then it becomes clear that an electron in a subshell is being shielded by another electron which is present in the same subshell.


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Formulas for the number of spectral lines
While reading through Narendra Avasthi's Problems in Physical Chemistry, I came across two formulas on p. 64 (Scanned page):

When electrons de-excite from higher energy level ($n_2$) to lower energy level ($n_1$) in atomic sample, then number of spectral line observed in the spectrum is
$$\frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$$
When electron de-excites from higher energy level ($n_2$) to lower energy level ($n_1$) in isolated atom, then number of spectral line observed in the spectrum is
$$n_2-n_1$$

Can anyone explain how the process of emission differs in these two cases?

the formula gives the total number of transitions provided the cascade of all possible transitions are added together, thus Balmer+ Paschen+Brackett+Pfund gives (6-2)(6-2+1)/2 =10 transitions.

To my understanding, it is simply single atom versus many number of atoms. For example, suppose one atom with an electron at energy level 7 ($n_2=7$). That electron can "de-excite" from $n_2=7$ to $n_1=6, 5, 4, 3, 2,$ or $1$. All those transitions give one spectral line for each. Thus, total of $1 \times 6 = n_1(n_2-n_1)$ (foot note 1) spectral lines would be present in the spectrum.
Similarly, when there were more than one atom in the sample, excited electrons ($n_2$) would be in different states $(n_2=2, 3, 4, 5, 6,....,\infty)$. For example, suppose we have aom population having electrons in all levels up to energy level 8 ($n_2=8, 7, 6,...$). Suppose those electrons "de-excite" to energy level 2 ($n_1=2$). Thus, electrons in $n_2=8$ can "de-excite" to energy levels $7, 6, 5, 4, 3,$ and $2$ meaning total of 6 spectral lines $(8-2=n_2-n_1)$. Some atoms with electrons in energy level $n_2-1=7$ can also "de-excite" to energy levels $6, 5, 4, 3,$ and $2$ meaning total of 5 spectral lines $(7-2=n_2-1-n_1)$, etc., etc. Thus, total numbers of spectral lines ($s$) in this case would be:
\begin{align}
s&=6+5+4+3+2+1=21=\frac{42}{2}=\frac{7\times6}{2}\\
&=\frac{(8-2+1)(8-2)}{2}\\
&=\frac{(n_2+1-n_1)(n_2-n_1)}{2}
\end{align}
Foot note 1: Total number of spectral lines for single atom where $n_2=7$ should be: $1 \times 6 = (n_2-n_1)$ in the spectrum, not $n_1(n_2-n_1)$ as I originally suggested (Thanks @porphyrin for careful reading).

How does your $1 \times 6 =n_1(n_2-n_1)$ work with $n_2=7; n_1 =4$?

@porphyrin: It was just mental mistake, which I have corrected. There was no $n_1$ in the expression in the question, anyway. Now, to answer your question, total number of spectral lines for $n_2=7$ to $n_1=4$ transition in single atom would be $1 \times 3=7-4=(n_2-n_1)$, representing transitions of $7\rightarrow 6$, $7\rightarrow 5$, and $7\rightarrow 4$. Thank you for your careful reading.

Mathew Mahindaratne: I am unable to understand in your answer how did you write s = 21 = 42/2. What was the basis for that?

@Somdeb: I have it as an example. Read the question and my answer completely to understand my concept.


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What neutralizes the active ingredients (napthalene and/or paradichlorobenzene) in moth balls?
Question: What is effective in removing residual moth ball smell from inside a car?
Issue: We used moth balls in our car to rid it of mice. (Never do this; it does not work.) We removed the moth balls after about a month. The car, free of moth balls for three months, still reeks of the moth ball smell. After driving in the car for 10-15 minutes, our clothes also smell like moth balls and require washing to rid them of the smell. Would airing the car out be effective or do I need to be more drastic?

Let air circulate for a long time (days) if outside temperature allows it. Naphtalene sublimes (goes from solid state to gas) and the bad odor should vanish. Otherwise is internal washing possible?

In summer, expose your car to the hot sun until it's heated up. Then open the car doors (don't let the mice in again) to exchange the air. Repeat the procedure until the smell is gone.


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In search of a safe and inexpensive solvent that can dissolve depolymerized polypropylene waste plastic
I work at a place where we depolymerize post consumer waste polypropylene waste plastic through a thermal catalytic conversion process. During process shutdown we have to clear up the vessels which becomes a really difficult job as the depolyermized material which we call wax hardens without heat and have to hammered and scraped out which takes a lot of time. Therefore to make this cleaning process easy I am looking for a chemical clean up technique. Is there a reasonable and economical solvent that can dissolve/soften this wax?

Economically and ecologically would be difficult as then you'll have to dispose out of a liquid waste.

@Alchimista Na, this "waste" can directly be fed into a refinery after you have filtered out the solids (which you can feed into the OPs reactor a second time, where they are hopefully broken up a bit more).

@Karl. I am speaking of a DIY scenario as for the question drives in that direction. Where is the refinery, for instance. This is unlikely a chemical plant diagram scenario. I see that your answer basically says the same. If there would be a scenario as you are thinking of, probably the Q wouldn't be asked.

@Alchimista Well, yea. I don't exactly get that either. You can't sensibly do this process outside of a rather large-scale chemical plant, which is ideally located directly next to a refinery. I'm guessing OP is working there, but his employer doesn't tell his workforce why things are run the way they are?

@Alchimista you've guessed that correctly. The plant isn't a large scale chemical process and we don't have any provision to utilize this 'waste'. If we are able to find a solvent that can dissolve this, we'll have to find a safe way to dispose that off.

@Sasha What size are we talking about here, in terms of reactor size and tonnes per day?

@Karl Reactor size is 1200 litres, and we are able to produce approximately 5 tonnes per day from it

Short answer: Not really, unless you can do the cleaning still at high temperature. Solid hydrocarbons tend to only become soluble rather close to their melting point.
The only reasonable solvent is petrol ether, i.e. a mixture of alkanes. Polar solvents obviously don't work, chlorinated solvents are ecologically problematic and only have the advantage of an increased boiling point. Toluene or xylene might be alternatives.
So basically what you should do is to switch your feedstock from polyolefines to liquid alkane (petrol, basically) before shutdown. I would however be amazed if your process engineers don't do exactly that anyway. Maybe they could do that for a little longer, or maybe your process should be run (at least from time to time) at a slightly higher temperature so solidified (i.e. crystallised) residues in the reactor can melt and be washed out.
And whatever remains after shutdown and cooling, you will have to scrape off by hand, sorry.
(A sloppily designed or maintained reactor with bad flow profile, insufficient/broken insulation/heating devices of course might or might not be fixable.)

Thanks for your recommendation. I'll give that a try by using petrol ether or toluene/xylene on a small sample of that waste wax to see how it goes. So do you suggest to raise the temperature of the hardened wax and then try these solvents. However raising the temperature and exposing it to oxygen is not safe, as this material (wax) catches fire at its autoignition temperature of 240 deg C.
I am not sure if we can change the feedstock to liquid alkane prior to shutdown as it might affect our catalysts (not sure), but that's an interesting proposition. Will ask my process engineer. Thanks


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Shea Stearin Extraction by Dry Fractionation
I have been able fractionate palm oil and extract palm stearin using the following method:

Melt the palm oil (over 80 °C)
Keep the melted oil at 33.5 °C for 24 hours (using an immersion circulated water bath)

At this point I was able observe crystals being formed. I assume they were palm stearin crystals.

Centrifuge to separate the crystals to obtain palm stearin.

I was able fractionate almost 30 % crystals by weight.

However the same process did not really hold true while processing the shea butter to fractionate shea stearin. After Step 2, it was still in liquid form and there was no visible crystal formation. I'm further considering seeding (with crystals) or playing with the time and temperature variations.
How can I modify my approach to promote stearin crystal formation in shea butter?

As I’m planning to use food grade shea butter, I’m looking for ways to dry fractionate it.

Shea butter is a complex combination of esters:

(Wikipedia). The overall ratio of oleic to stearic acids is about 2/1, but statistically, there will a little of the high melting stearic, more of the 2 stearic to 1 oleic; much more of the 2 oleic to 1 stearic, and a lot of all oleic ester. The melting point of shea butter is low (body temperature) and so if it does not precipitate out the tristearin even at room temperature, you will have to do something different.
A fractional crystallization involving higher temperatures would use a poor solvent for the tristearin (mp ~ 72C), like ethanol. The oleic esters will be more soluble. Make a 20-30% solution in hot ethanol (be careful!!) and let it cool. Experimenting with similar alcohols (propyl, isopropyl, and perhaps with a bit of water added (~5-10%) could give better results.
Oiling out is a common problem in trying to crystallize organic compounds.

Considering ethanol boils at 78C and the tristerain melting at 72C. I would guess the “hot ethanol” temperature would be between 72C and 78C? Or should I get a temperature lower than 72C?

Tristearin is dissolved in (or mixed with) the other components of shea butter at room temp. If you add 20-30% shea butter to ethanol at RT, it will probably not all go into solution, the tristearin least of all. Warm the mixture until it is all dissolved (~clear), then let it cool, or cool it in a refrigerator or freezer. If it oils out, more ethanol, or a different alcohol will be needed to keep the oleic esters in solution so the solid tristearin can crystallize out. To get solid tristearin, the final cool temp will have to be below 72C. Way below!

I’ve tried 30% shea butter and it oiled out. I’ll be trying 15%. Btw the ethanol I’m using is 96.2% pure. Hope that’s good.

Sounds like you are a good experimentalist. Keep trying.

The industrial processes uses acetone for shea fractionation. The crude oil is degummed first by a water degumming process and then fractionation is carried out using acetone as a solvent for separating oleic fractions.
Palm oil melts at approx. 36 °C. And you provided the temperature for crystallization as 33.5 °C.
Decrease the temperature. Shea butter melts between 32 °C and 37 °C. I would recommend decreasing temperature and slow agitation at a temperature below 20 °C if possible. If you cannot agitate, try it at 30 °C.
If it solidifies at 30 °C, try at 32 °C.