text:
Why does HNO2 not have resonance?
A question on the 1996 AP Chemistry Free Response asks:

The $\ce{N-O}$ bonds in the $\ce{NO2-}$ ion are equal in length, whereas they are unequal in $\ce{HNO2}$. Explain.

Since the nitrogen dioxide ion has resonance, the $\ce{N-O}$ bonds are equal as resonance is in reality a hybrid of all of the possible structures for a certain molecule. However, the answer then assumes that $\ce{HNO2}$ has no resonance but rather one single and one double bond ($\ce{N-O}$ and $\ce{N=O}$), and therefore not all of the $\ce{N-O}$ bonds are equal.
How is this possible? Why can $\ce{HNO2}$ not have 2 possible structures where the single and double bonds switch between the two oxygens?

Resonance is a remnant of valence bond theory, which is necessary because it is impossible to describe delocalised bonding within a localised bonding scheme. Resonance structures are not real. One Lewis structure describes one electronic configuration with fully localised electrons. For many molecules it is completely sufficient to describe the bonding situation with one such structure in a first order approximation. For many other molecules this is completely insufficient. For $\ce{NO2-}$ this is the case, hence you need a superposition of more than one structure to get closer to a qualitative description. This caveat is not necessary in molecular orbital theory, since this is designed to treat delocalisation.
Nitrite anion
One question therefore arises from the above: Are the $\ce{N-O}$ bond lengths equal because there is resonance, or is there resonance because the $\ce{N-O}$ bond lengths are equal?
This is obviously more of a philosophical question, but it shows the dilemma of the whole concept. The bonding of any given molecule is a result of many different factors; to name but a few: repulsion of the nuclei, repulsion between the electrons, attraction between electrons and nuclei.
For $\ce{NO2-}$ there is one key question, that gives us a very important clue as to why the bond lengths are equal: Why should one oxygen atom be different from the other? In fact the molecule is symmetric (C2v), therefore the bond lengths must be equal.
If you try to write a Lewis structure that respects this symmetry, you will end up with an electron sextet on nitrogen (and a formal charge of +2), which certainly can only be an insufficient approximation. If you write a Lewis structure with fewer formal charges, keeping symmetry, you will notice that nitrogen will have 10 electrons. That would mean that it would have to use d-orbitals, which simply are not available for nitrogen (more on that another time). There is another way to write a Lewis structure, however, this one has to break the symmetry, and you'll notice that there is a second one, which is equal. In order to keep the symmetry you introduce resonance, to state, that the "true" electronic structure is a superposition of (at least) those two configurations.
TL;DR: The $\ce{N-O}$ bonds in the nitrite anion are equal, because the oxygens are equal.
Nitrous acid
It is false to say that in $\ce{HONO}$ is no resonance. And it is also wrong to say that a lack of resonance makes the bonds unequal. If the solution manual claims this, it is wrong.

Why can $\ce{HNO2}$ not have 2 possible structures where the single and double bonds switch between the two oxygens?

These are valid resonance structures of nitrous acid. You have to keep in mind though, that resonance does not change the geometry of a molecule (or molecular structure). The contribution of the Lewis structure with fewer formal charges to the total wave function will be much higher than the contribution of the other structure in this case. This is not a general statement, there are molecules where the charge separated Lewis structure has a higher contribution to the wave function.
By analogy to the anion case, there is one key question, that gives us a clue about the molecular structure: If you bind a proton to one oxygen, are both oxygen really equal anymore? In this case the symmetry of the molecule is reduced to Cs. Therefore the bonds, cannot be equal.
For this molecule you will have no problem writing a single Lewis structure that satisfies every imposed criteria.
To answer the initial question: The bonds in $\ce{HONO}$ are unequal, because the proton is bound to one of the oxygens. It is a very polar, but covalent bond. The electron density of the $\ce{N-O}$ bond is therefore shifted towards the $\ce{H-O}$ bond, and simultaneously reduced for the $\ce{N-O}$ bond when compared to the anion. Therefore the $\ce{HO-N}$ bond is longer than the $\ce{N=O}$ bond (I implied the most contributing Lewis structure here).
TL;DR: The $\ce{N-O}$ bonds in the nitrous acid molecule are unequal, because the oxygens are unequal.

There frequently is the notion of a most stable resonance structure. While this is very popular, it is completely wrong. I again refer to the question What is resonance, and are resonance structures real? Resonance is used to describe the electronic structure of a molecule, there is no change in the molecular structure allowed. It is necessary to understand, that only the description in terms of all possible resonance structures is in itself correct, i.e. the theoretical limit of valence bond theory, everything else is an approximation. However, not all resonance structures will have the same contribution to the wave function. Whenever someone speaks of the most stable resonance structure, it is likely that they mean the resonance structure with the highest contribution to the wave function. But even if you know what is meant, it is still wrong to speak of a most stable resonance structure, it must be corrected.
(I currently don't have the time to provide a full fledged answer, so instead I was trying to give some pointers in the right direction. I might come back to amend it whenever I have more time.)

If the the electron density of the N−O bond is  shifted towards the O-H bond, shouldn't that decrease the bond length?

@Rick No, usually when you remove electron density, the bond get weaker and longer. The attraction between electrons and nuclei pulls the latter together. If you remove electron density, the relative strength of the nuclei-nuclei repulsion gets stronger.

In $\ce{NO2-}$ ion the resonance structures are equivalent and are stable enough (since there is negative charge on more electronegative element).

In the contrary, canonical structure (II) of $\ce{HNO2}$ is very instable as the +ve charge is present on more electronegative element O (which is more electronegative than N). Also, the positive and negative charges are fairly apart and hence the stability is greatly reduced.

Thus, it can be concluded that due to more stable state of the structure (I), the $\ce{N-O}$ bond length are unequal in $\ce{HNO2}$.

This is almost equally as wrong as the answer previously given. There is no such thing as the most stable resonance structure! Read my comment to the other answer for more information. Also $\ce{NO2-}$ is bent, not linear!

@Martin-マーチン I have answered this question according to my analysis and it might be wrong due to lack of any other considerations. But, what makes you say that there is no such thing as most stable resonance structure??? In many of my books I have found the concept of stability of canonical structures. Why, just google it and you can find ample instance.

@Martin-マーチン By the way, you can answer the question to make us think a little differently.

I know that this is a concept taught in many schools, in many textbooks, it is very predominant on the internet. Unfortunately sheer quantity does not make it correct. In fact it has been taught wrong and it is one of the most common misconceptions in chemistry. It needs to be eradicated.
This question explains very well, why the very notion of a most stable resonance structure is wrong.

@Martin-マーチン Plese say a little bit more details and please help us solve it.

@Martin-マーチン (Note: Please reply to this comment if you see it). What you said in your  answer in the last part is what we mean by stable resonance structure. Actually, in the books it is clearly mentioned that the resonance structures which have more contribution towards the hybrid are considered more stable. By the way, your answer has clearly depicted that. I have found you commenting in many questions in stackexchange stating the conception of more stable resonance structure is wrong. But, please do note that we do indirectly imply what you said, "it is likely that they mean the resonance

It is wrong, and only because I know this misconception I am able to translate this into something that is not wrong.

@Martin-マーチン But, at elementary level, is my answer correct (I'm in Class 11, India)???

No. Wrong is wrong. Just because your answer is accepted in your level, it does not make it correct. Even worse, it does make it sad. If your course is not able to get such an elementary theory right, who knows what other misconceptions are taught as well. It is a tough fight to get these false theories out of the heads of the students, and it is truly awful to know that the next generations of students in India will learn complete and utter garbage. It is even worse that some of them might go on to university and have to learn everything anew. It makes me very, very sad.

@Martin-マーチン No no. My answer might be wrong because we may need to draw other considerations. But, we do have questions asking to compare stability of Resonance Structures.

You cannot compare the stability of resonance structures,resonance structures simply do not exist on their own, why can't you understand that. It simply is wrong. If you learn it that way, you are learning something wrong, complete and utter garbage. Throw that book away. (Anyway, comments are not for extended discussion, so it is unlikely that I will reply any further.)

@Martin-マーチン I agree with you in that. And I do also know that the resonance structure do not exist in that way and only the hybrid is the true structure. In the books also, it is mentioned. But, such questions are asked actually considering their contribution towards the hybrid.

@Martin-マーチン Indian student here. We do have these kinds of questions here. So what if you change the language of the question from "Which of these canonical structures is most stable?" to "Which of these contributes most towards the hybrid?" Will that be correct?

@ballaneypranav yes, that would be better. It needs to be taught still, that this is an approximation and that if taught without the other contributors, it will still lead to wrong conclusions. There really is little point to teaching resonance and then using only one contributor - it kind of defeats the whole purpose. But at least starting it that way, would not be wrong.

The idea is that in a resonant structure, all the structures are equally stable -- for instance, in CO3:
, you can see that the formal charges on the Os initially are -1 on 2 of them and 0 on the last; these charges are consistent in each structure. However, in your example, the structures are:
The formal charges on the Os in this case are 0 and 0 on the left, as compared to 1 and -1 on the right. These charges aren't the same, so this is a case of more stable structure rather than resonant structure.
Image sources: http://www.chem.ucla.edu/~harding/tutorials/resonance/draw_res_str.html
https://s3mn.mnimgs.com/img/shared/content_ck_images/images/HNO2.png

No nitpicking, simply pointing out how wrong the answer is. The idea is that in a resonant structure, all the structures are equally stable [...] There is no most stable resonance structure, resonance structures in itself cannot be stable as they do not exist on their own (compare here). Even if that were not an issue with the statement, the very notion that they have equal contribution is wrong, too. You only display 3 resonance structures, there are many more with minuscule contributions. [...]

[...]  Using equilibrium arrows for resonance is dead wrong. The same applies to $\ce{HNO2}$, formal charges are simply that: formal. A resonance hybrid is a superposition of all possible resonance structures, and it is only necessary in the valence bond framework. Basically everything in this answer is wrong, and it does not explain why the $\ce{N-O}$ bonds are different in length. To be clear: I down-voted this.


text:
Difference between batch and column mode?
I have heard Ion-exchange chromatography can be used in batch or column mode, have you got any idea how are those modes?
I have looked descriptions all over books and web (Skoog for example, both versions of analytical chemistry) but found nothing clear.
Any help?

Probably batch by batch or in continuous mode.

@Alchimista but what is an ion exchange by batch?

First the sample is made to bind than eluted. https://www.slideshare.net/mobile/ParimalaGattupalli/ion-exchange-chromatography-1-4

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text:
Where do the lone pairs go in the "true" resonance structure?
Consider the "true" resonance structure of $\ce{CO3^2-}$:

The bottom diagram leaves out the 2 lone pairs, so 4 electrons disappeared in the bottom image as compared to the top one. Are these disappeared lone pairs partial electrons, or shared? To make the bonds in $\ce{CO3^2-}$ correct, doesn't that mean each dotted line is worth double, so each bond is $1.33$ instead of $1.66$? Why don't we draw a $1.33$ electron lone pair if that's the case?
Image source: Resonance, by  Sharon Wei (UCD) and Liza Chu (UCD), Chemistry LibreTexts.

Just how would you draw 1.33 dots?

Not sure, maybe just write a dot then put 1/3 or something; it's better than putting nothing there right?

Not necessarily. After all, dot structures are but an approximation of a certain level. They are not supposed to include everything we know, much like Peppa Pig characters are not supposed to be anatomically correct.

I suspect it due to electronic display . At my time everybody simply wrote 2/3 - at each oxygen and it was clear enough.

Resonance is a concept resulting from valence bond (VB) theory. Within this theory it is impossible to describe delocalisation. However, every configuration within this framework can be expressed as a Lewis structure. The congruent molecular orbital (MO) theory describes bonding in a delocalised way, therefore it is impossible to express it as a single Lewis structure.
Often we can find Lewis-like structures representing a resonance hybrid. For example from my own questions:

While it is simply impossible to write a valid Lewis-structure (localised electrons) and imply delocalisation, it is often used to denote how a molecule is "special" in a way. We make many assumptions here, and imply that everybody knows that all the electrons are delocalised. Therefore I omitted all electrons, implying the reader knows where to put them.
What you describe as the "true" resonance structure, is an attempt in the same manner to visualise a resonance hybrid (superposition of multiple valid Lewis structures) in a single Lewis-like fashion. The resonance hybrid, like it is used in LibreText, is an insufficient and inaccurate depiction. It is a futile attempt to jumble together VB and MO theory. Even though they are, at their respective theoretical limit equivalent, i.e. congruent theories, at that very limit they become so complex, that a visual depiction is no longer possible.
In this spirit, these depictions of resonance hybrids are overused in the same manner that resonance is often taught wrong and is an overused concept. (Unfortunately the linked LibreTexts article is point in case.) It is important to understand that these simple concepts have their limitations, and to always question the validity.
As you notice yourself in your question, the depiction of the resonance hybrid omits electrons, you have to be cautious to use these simplifications.

On a related, but slightly different note, I especially dislike the article you linked to on LibreTexts. Just to name a few errors:

it omits formal charges in the ozone example;
the delocalisation and resonance structure rules are complete made up, incomplete (1) or even wrong (4, 5);
in the concept of resonance it is inherently wrong to speak of the "correct"Lewis structure;
rules for estimating stability of resonance structures is complete garbage, there is no such thing as a most stable resonance structure;
Lewis dot structure for sulfate uses $\ce{S-O}$ double bonds, which is completely wrong;
it is not generally true, that the structure with fewer formal charges has a higher contribution to the wave function that one that has more charge separation;
etc..

I suspect it due to electronic display . At my time everybody simply wrote 2/3 - at each oxygen and it was clear enough.

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Electroplating copper yields rough surface?
I am attempting to electro-plate vias in a home brew printed circuit board. The medium I am attempting to electroplate is fiberglass bound together with epoxy. This is an intentional insulator.
I drilled a few holes in this medium using a regular electric drill and covered the holes with a graphite based conductive paint by covering the board in the paint and drawing it through the holes using a vacuum. I repeated this in both directions in an attempt to cover the holes evenly. I then cured the paint at about 100C for 20 minutes. In my experiments, this was far sufficient to dry the paint.
I then created a solution of 1000ml distilled water, 250ml of 38% sulfuric acid (battery acid to improve conductivity) 160g of copper sulphate pentahydrate, and 2 ml of polysorbate 20.
I prepared this solution by adding the sulfuric acid to the water slowly (to avoid rapid exothermic reaction between acid and water) then adding the polysorbate and the  the copper sulfate pentahydrate. I then mechanically agitated the copper sulfate pentahydrate crystals until the solution was homogenous.
I ran .8A of constant current through the solution with the board as the cathode and an anode consisting of two electrically common (14ish guage) copper wire on either side of the board. After about 90 minutes this was my result...

I'm wondering why the copper clad had "beads" on it...the entire surface was sanded with 1500 grit sandpaper directly preceding the electroplating.
I'm also wondering why the holes are so "rough". In my testing, every via has come up as 0.0Oohms...or electrically common to the clad. Aesthetically though, this is not pleasing and makes me wonder if my method is truly reliable.

Electroplating requires either a very uniform current distribution or very uniform cation distribution (which requires very high current density), otherwise irregularities are to be expected. The process is fragile and should be avoided if possible.

There are few things that you can improve.
First, beads are forming because your copper anodes are not pure copper. Beads are forming around impurities. You can try to remove (filter) impurities by firmly wrapping anodes in paper tissue or better with water filtration polypropylene felt filter so only Cu ions can pass into solution. You can after that collect impurities already present in solution by electroplating few hours on some scrap piece of wire.
Second, holes should be drilled cleanly. You should invest into carbide drill bits and good drill stand.
Third, you should also think about adding Cl ions to solution, NaCl or HCl in small quantities. Plated surface will be shinier...
Best regards,
Pavle

I have some muriatic acid laying around for etching. I'll add some of that and see what changes. I ordered a drill press and I'll look into finding more pure copper. Thanks.

Be careful not to add too much muriatic acid.
Please check on: https://www.thinktink.com/stack/volumes/voliii/consumbl/cplatmix.htm
Only 13.3 ml of 35% HCl for 100l of of total solution!

@Pavle I wanted to reiterate your point that’s 13.3ml for 100 LITERS! Also, if OP is using tap water and not deionized water they likely already have too many Cl ions in solution which will lead to passivation of the anode and cause more problems.


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Why does the CO2 gas quantity in Sodastream cartridges differ in different nations?
Just was looking at the $\ce{CO2}$ cartridge for the Sodastream soda / seltzer maker I was planning on buying and this bit of info caught my eye:
Any idea why they may want to pack $\pu{25 g}$ less $\ce{CO2}$ for South Africa? Is it some regulatory issue? Marketing gimmick?
Or more interestingly, some technical factor?

Without doing much more research, I would assume that temperature plays a big role here. The maximum temperatures in South Africa tend to be a bit hotter than in the habitable places in Australia.

Hardly a chemistry question

@Martin-マーチン On the other hand, I’m sure the habitable places in South Africa are also colder than the hottest temperatures in Australia.


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Is significant figure important for atomic radius calculations?
Is significant figure important for atomic radius calculations?

Calculate the radius of a vanadium atom, given that V has BCC crystal structure, a density of $\pu{5.96 g/cm^3}$, and an atomic weight of $\pu{50.9g/mol}$.

What if we solve this question with $3$ significant figures and $5$ significant figures? What will be different?
Regards

The use of significant figures is a poor man's way of doing error prorogation. With a modern calculator I'd use two extra figures in intermediate calculations (so 5 significant figures since the data has 3), then round final answer to 3 significant figures. The problem is rounding errors that creep into the intermediate calculations. Such rounding errors are the bane of calculations. It is particularly nasty if you're calculating small differences between two big numbers (e.g. 100.45 - 100.23).

When I first started chemistry I used a slide rule, so three significant figures were all that was possible. The alternative was a single page of logarithms which was good to three significant figures too. Could use a multi-page table of logarithms which was good to four significant figures.  Interpolation with log tables would let you do a bit better.

Different would be the implied error. You have values with 3 significant figures given, if you report 5 significant figures, then you are reporting over-confidently. (Given that you have to use the hard sphere model in your question, the implied deviation that would be caused by only using 2 significant figures would still be minuscule.)

So what would you recommend me to do? I'm going to give the density, atomic weight and Avogadro's constant. Should I specify the significant figure as well?

How should we know? Perhaps for your use the error estimate is not relevant. I'll doubt it, however.


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Why is triplet carbene sp2-hybridised?
Triplet carbene has a carbon with 6 electrons, of which 2 are unpaired. I would expect the carbon to be $\ce{sp}$ or $\ce{sp^3}$ hybridised and the two singly occupied orbitals to be degenerate. However, Clayden et al. (p. 1011) say that, since the triplet carbenes are almost always bent, they are best considered to be $\ce{sp^2}$ hybridised, with the unpaired electrons occupying a $\ce{p}$ and an $\ce{sp^2}$ hybrids .
The very basic explanation I see why the two SOMOs are not degenerate is simply that $5E_{\ce{sp^2}} + E_{\ce{p}} + 2E_{\ce{sp^2},\ \text{pairing}}$ is less than $4E_{\ce{sp}} + 2E_{\ce{p}} + 2E_{\ce{sp},\ \text{pairing}}$ or $6E_{\ce{sp^3}} + 2E_{\ce{sp^3},\ \text{pairing}}$. However, I wonder if there's a better explanation.

It’s a little hard to tell $\mathrm{sp^2}$ and $\mathrm{sp^3}$ hybrid orbitals apart without exact angle measurements …

The key basically lies in symmetry. Also it is wrong to assume that changing the hybridisation changes the geometry and/or energy. The molecule is bent, because that is the most stable arrangement, everything else follows. Also, atoms cannot be hybridised, only orbitals can.

@Martin, do you mean that you start with the bent geometry and then use symmetry to deduce hybridisation? But what I'm looking for is to rationalise the geometry. Surely in this simple case there should be a way to rationalise the preferred geometry, without resorting to intensive computations?

https://chem.libretexts.org/Core/Organic_Chemistry/Fundamentals/Reactive_Intermediates/Carbenes#Singlet_and_triplet_carbenes

That's actually a good question. I was lucky enough that Dietrich Gudat, who wrote a long chapter on this type of chemistry in the Riedel - Moderne Anorganische Chemie was my professor in inorganic molecular chemistry and we discussed these things in detail back then but it's been a while.
I think it's due to how you want your final structure to look like or VSEPR. You have bascially three ligands, two rests and the lone-pair on the carbon. And by having it sp² you get quite some distance with the 120° angles. This splits your possible energy states in one which is called a1, that is basically the sp² and an b1 (the p-orbital). As the a1 has more s-character it is lower in energy and thus you can, depending on how big the gap is, either get a triplett or a singulett carbene.
This will change however if you move to the isolobal and isoelectronic nitrenes. Here we have no sp² but an sp which is automatically filled with two electrons. The other two electrons occupy the degenerate sets of px and py and thus you always end up with a triplett.

You can put the electrons in whatever orbitals you would like to. The only difference is that a specific electronic structure produce least electronic energy. You can in a relatively simple manner picture this.
Let us consider the sp(linear) and sp2(bent) structures of carbene(Methylene).
Imagine the carbene's triplet state. In the early days it was actually believed that the structure is actually linear (Methylene story).
In such a case the two orbitals will be p and degenerate. Upon bending, the in-plane p orbital starts mixing with the higher lying 2s orbital and gets stabilized (look at the walsh diagram D∞h to C2v). This stabilization reduces the energy of the system stabilizing the carbene. On the other hand, bending increases the electron electron repulsion. A competition between these two factors decides the faith of the carbene's angle.
For example in the singlet state, which has both electrons in the same orbital and electron-electron repulsion between the two orbitals doesn't exist, has a more bent structure.

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text:
What Makes Oils Feel Dry or Oily?
Simple question with a complex answer, certainly:
Why do some oils feel "dry," while others feel "oily"? Is the feel directly related to C-chain length, or hydrogenation, active groups, or something else entirely?
Thanks!

Welcome to chemistry.SE! If you have any questions about the policies of our community, please visit the help center. Best of luck!

I didn't know they did. Can you give examples?

What do you mean by oil? It is a quite broad definition and in a perhaps slang  includes everything insoluble in water for which a solid is not (yet?) attained after synthesis.


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Pople Basis Set Abnormality: 3rd row 6-311G
I noticed when looking through the Basis set exchange website that the 6-311 Pople basis sets don't at all match their formulation once you go past the 2nd row of the periodic table. For example, oxygen's basis set looks like this (in Gaussian input format):
O     0
S   6   1.00
8588.5000000              0.00189515
1297.2300000              0.0143859
299.2960000              0.0707320
 87.3771000              0.2400010
 25.6789000              0.5947970
  3.7400400              0.2808020
SP   3   1.00
 42.1175000              0.1138890              0.0365114
  9.6283700              0.9208110              0.2371530
  2.8533200             -0.00327447             0.8197020
SP   1   1.00
  0.9056610              1.0000000              1.0000000
SP   1   1.00
  0.2556110              1.0000000              1.0000000

Where the core 's' orbital is formed from a contraction of 6 primitive gaussians and the valence 's' and 'p' orbitals are formed from a contracted set of 3 primitives and then two other sets of 1 primitive. Hence, 6-311.
Compare this to Sulfur:
S     0
S   6   1.00
93413.4000000              0.0007430
13961.7000000              0.0057930
3169.9100000              0.0299540
902.4560000              0.1190280
297.1580000              0.3684320
108.7020000              0.5772990
S   3   1.00
108.7020000              0.1431860
 43.1553000              0.6244650
 18.1079000              0.2833660
S   1   1.00
  5.5600900              1.0000000
S   1   1.00
  2.1318300              1.0000000
S   1   1.00
  0.4204030              1.0000000
S   1   1.00
  0.1360450              1.0000000
P   4   1.00
495.0400000              0.0083090
117.2210000              0.0640240
 37.7749000              0.2776140
 14.0584000              0.7450760
P   2   1.00
  5.5657400              0.6137120
  2.2629700              0.4438180
P   1   1.00
  0.8079940              1.0000000
P   1   1.00
  0.2774600              1.0000000
P   1   1.00
  0.0771410              1.0000000

Why would this set of orbitals still get lumped together with the Pople basis sets when it doesn't resemble their formulation at all? And what is the meaning of this particular set of functions?

From the Gaussian[1] webpage:

6-311G: Specifies the 6-311G basis for first-row atoms and the McLean-Chandler (12s,9p) → (621111,52111) basis sets for second-row atoms [McLean80, Raghavachari80b] (note that the basis sets for P, S, and Cl are those called negative ion basis sets by McLean and Chandler; these were deemed to give better results for neutral molecules as well).

I have no idea exactly why they still call it 6-311G: I suspect it is because this configuration provides a more accurate guess for a triple-zeta Pople basis set, and to create an entirely new basis set would be impractical.
What I do know is that when you use the '6-311G' basis set, you're not necessarily getting this configuration. What you're getting is a basis set which is roughly of this quality (computational effort, accuracy).
Whilst this explains the abnormality, it still doesn't explain the exact configuration: the original paper[2] suggests a (621111,52111) set whilst the set used is (621111,42111). I hope somebody can contribute to this and complete my answer as, whilst I recognise the abnormality, further investigation has led me to this confusion.
[1] http://gaussian.com,
[2] J. Chem. Phys. 72, 5639 (1980).

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text:
Do Sulfuric Acid and Calcium Hydroxide Completely Neutralize One Another?
I just completed a unit test at school in which there was a simple fill in the blanks question which asked:

Would 1 mol of sulfuric acid and 1 mol calcium hydroxide in water form an acidic, basic, or neutral solution?

The answer I wrote was neutral.
My reasoning was that sulfuric acid ($\ce{H2SO4}$) has two protons it can donate, and calcium hydroxide ($\ce{Ca(OH)2}$) has two protons it can accept. However, after the test I realized that H2SO4 dissociates into HSO4- which is a weak acid and won't dissociate completely.
After some research I have consistently found that the neutralization equation is written as: $\ce{Ca(OH)2 + H2SO4 -> CaSO4 + 2H2O}$
Do sulfuric acid and calcium hydroxide completely neutralize  one another? Or was I wrong to say that the resulting solution is neutral.
Could someone please explain why I am right or wrong?

Additionally Ca cation in aqueous solution is central atom of aquocomplex which is weakly acidic. Nothing is fully neutral, but if you use such generalisations as acidic/basic/neutral salt, CaSO4 falls under "neutral"

@Mithoron thank you for your response, what I still do not understand is why the second hydrogen in sulfuric acid is able to to neutralize the second OH- ion in calcium hydroxide when HSO4- will not dissociate completely.

HSO4 anion can dissociate as fully as you wish depending on concentration.

@Mithoron, does this still hold true when it is made clear in the question that both calcium hydroxide and sulfuric acid have the same concentration?

Check out https://chemistry.stackexchange.com/questions/36904/does-dissolving-salt-in-water-change-the-ph

also https://chemistry.stackexchange.com/questions/40569/predicting-ph-of-an-acidic-salt-of-an-weak-acid-and-weak-base

The first proton transfer is simple enough and you have no doubts.
$$\ce{H2SO4 + Ca^2+ + 2 OH- -> HSO4- + Ca^2+ + OH- + H2O}\tag{1}$$
Now, you wonder whether hydrogensulphate is a strong enough acid to combine with the remaining hydroxide. Well, hydrogensulphate may be a weak acid but hydroxide is a strong base. So even the second step will go to completion:
$$\ce{HSO4- + Ca^2+ + OH- + H2O -> SO4^2- + Ca^2+ + 2 H2O}\tag{2}$$
Frankly, it does not end there. Calcium and sulphate form a salt with a very low solubility so this salt will precipitate out of solution.
$$\ce{SO4^2- + Ca^2+ + 2 H2O -> CaSO4 v + 2 H2O}\tag{3}$$
So the only thing you are left with in reasonable amounts is water and a precipitate. This, of course, is a neutral solution.
However, you don’t really always have to go that far. Check if there are strong acids and bases (don’t forget hydroxides), cancel them away and then check what remains. If you had done that, you would have noticed that there are as many hydroxide ions as acidic protons, overall giving an approximately neutral solution.


text:
Why isn't the scandium(2+) cation isoelectronic with potassium?
I saw that I got this question wrong on a test, and was very confused.  Why aren't they isoelectronic?  $\ce{Sc^2+}$ with atomic number 21 should have 19 electrons since 21 - 2 = 19, which is the same as the number of electrons as Potassium with 19 protons and electrons.  Does it have something to do with being in different sublevels, one being s and the other d and all?

The neutral $\ce{Sc}$ atom has the electron configuration of $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^1 4s^2}$, and the $\ce{Sc^2+}$ ion has electron configuration $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^1}$.
The neutral potassium atom has the electron configuration of $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 4s^1}$. While both $\ce{Sc^2+}$ and $\ce{K}$ have the same number of electrons, their electron configurations are different and therefore they are not isoelectronic to one another. (Isoelectronic implies same number of electrons as well as the same structure)

thanks.  i remember my teacher teaching us this a while ago now

Could you add a reason as to why they differ in their electronic configuration?

As has already been stated, the two are not isoelectronic because their electronic structures are different: potassium’s single electron being in an s orbital while scandium(II)’s is in a d orbital.
The reason for this difference is the only other thing that is different: the difference in nuclear charge by 2. A higher nuclear charge causes the orbitals to contract and reduces their energy. However, not all orbitals are equally affected by a change in nuclear charge: the lower an orbital’s principal quantum number is, the more it as affected as a general rule. So the 3d orbitals, belonging to the third shell, feel the charge difference more strongly and adapt more than the 4s orbital which already belongs to the fourth shell.
While for neutral potassium and calcium(I) it is very clear that the 4s orbital has a lower energy than the 3d orbitals, the picture reverses itself once you reach scandium — the first d block element. Because the 3d orbitals’ energies have been reduced more than the 4s energy, it suddenly becomes more stable and the single valence electron of scandium(II) will reside in 3d rather than 4s.
In scandium(I), we also need to take electron-electron interaction into account to fully explain the concept. Basically, each electron you remove from an atom will stabilise all orbitals and vice-versa. Again, this effect affects different orbitals differently, mainly depending on the quantum numbers $l$ and $m_l$. So reducing scandium(II) to scandium(I) will shift the orbital energies around to the very interesting result that one electron will occupy a 3d orbital while one will occupy the 4s orbital. (I have glossed over it, but of course spin and spin pairing also play a role.)
In my humble opinion, chemguide.co.uk has done an excellent job at explaining this problem.

This is a very interesting question, I have also experienced this mistake, so should be careful while answering in such questions. You have accept the @ Allison W's answer he has explained it well and I try to give you a technique to answer such question very carefully without mistaking in the future.
1.First you should write the electron configuration of the element which gives you.**
Sc - 1s2 2s2 2p6 3s2 3p6 3d1 4s2
2.Then you should understand how many electrons are left for the given ion. According to this question it should be 2. Then you should leave them from the outer energy level. That means they are in 4s.
Don't misunderstand this with the way of electrons are filling. When the atom is building according to the aufbau principle the first filling electrons are the 4s electrons and then the 3d. So the last electron which is filling is the 3d electron. But you should understand it is not the electron which is most outer of the atom. When the ions are making the leaving electrons which have least attraction of the nucleus. That means the most outer electrons.
3.Then you should see the stability of the made configuration.
This is not affect for this question but you can face such questions. For example,
Sc+-1s2 2s2 2p6 3s2 3p6 3d1 4s1
But this not happening while there is a more stability configuration of below.
Sc+-1s2 2s2 2p6 3s2 3p6 3d2 4s0
So, you should give your attention in such situation to change such like that. (In primary chemistry if you not changed it is not a case, but in advance you should be careful either in this point.)
4.After writing the new electron configuration of the ion now compare it  with the given element's configuration.
K- 1s2 2s2 2p6 3s2 3p6 4s1
Sc2+  1s2 2s2 2p6 3s2 3p6 3d1
Generalize this steps and you will not mistake such question again.


text:
Why dont we use potassium hypochlorite for bleach
I have been reading wikipedia and according to the article on potassium hypochlorite it can be used as a disinfectant. But it is not listed anywhere in the bleach article. I know we typically use sodium or calcium hypochlorite and some others but is there a specific reason we dont use potassium?

Welcome to chemistry.SE! If you have any questions about the policies of our community, please visit the help center.

Because it's more expensive and isn't better.

To expand on Mithoron's comment, because the Chlor-alkali industry is based around sodium chloride,  not potassium chloride.

Well it can be produced without mercury polution and seeing as mercury is toxic and hard (at least for me) to get ahold of i would think it is easier to make.. I accidentally made it trying to convert KCl solution to potassium chlorate.. Only i did it outside in negative temperatures so no chlorates formed.

Potassium hypochlorite can be used as a bleach as an alternative to sodium or calcium hypochlorite. It is not that it is not used but its uses as bleach is not common as compared to sodium or calcium hypochlorite. Enviro Klor™ is a potassium based bleach (12.5% Potassium Hypochlorite Bleach alternative). As written here:

Enviro Klor™ potassium hypochlorite is a very versatile EPA registered
  (#63838-10, #63838-CA-01) biocide with all the power of a hypochlorite
  bleach and is approved for use wherever typical sodium hypochlorite
  bleach is, including:

Recirculating cooling water systems
Sewage and wastewater treatment
Disinfection of drinking water
Public water systems
Pulp and paper mill process water systems
Swimming pool water disinfection
Sanitization of food contact services
Laundry sanitizers
Agricultural applications

ADVANTAGES
Bleaching and Laundry

Allows claims for a more environmentally friendly household bleach
Provides a sodium-free solution to laundry discharges

Industrial & Institutional (I&I)

Chemical Compounding – allows creation of sodium free formulas
Replacement chemical in CIP cleaners
More stable in certain formulations

Agriculture

Useful as a disinfectant for wastewater applied to land for    irrigation, conserving water
Since potassium is one of the three required soil nutrients, there is    little objection to the introduction of potassium ions onto
  land.

Here's a table of comparison:

You can see that available $\ce{Cl2}$ (wt %) is more in case of sodium hypochlorite for which sodium hypochlorite is preferred as bleaching agent. Also, byproduct of potassium hypochlorite is potassium chloride for which potassium hypochlorite is used in agricultural purposes where soil needs potassium. Another thing is that potassium hypochlorite may disproportionates to potassium chlorate which is a strong oxidant and may react vigourously with water and ignite or explode.


text:
Formation of silica from TEOS
Usually tetraethyl orthosilicate/tetraethoxy silane (TEOS) is mixed with both ethanol and water to produce silica in a sol-gel process. I am curious what happens if one adds TEOS to water at a high pH, without any alcohol. Would TEOS simply not mix and not react with water? Or it would hydrolyze too rapidly and not condense at all?

It depends. if your pH is really high, silicate $\ce{SiO3^2-}$ or $\ce{SiO4^4-}$ ions will form and will become a stable dissolved solute in the solution. if it is moderately high the rate of hydrolysis will increase greatly and the silicon will crash out in the form of orthosilicate hydrates [$\ce{Si(OH)4}$], which may agglomerate to form silica particles. These results are a bit of a spectrum and there is no pH limit where the reaction gives a binary change in result, as it is also sensitive to extrinsic properties of the solution such as the amount of base.


text:
Why does alkyl group withdraw electrons in Grignard reagent?

R group has electron donating tendencies. Why is it withdrawing?

Is sodium potentallly electron withdrawing?  Given a suitable complexing agent and solvent one sodium atom can withdraw an electron from another that is complexed (https://en.wikipedia.org/wiki/Alkalide).
Whether something is electron releasing or electron withdrawing depends on its environment.  An alkyl group is slightly electron releasing in some hydrocarbon systems, as when it is attached to a phenyl group.  But that does not carry over to all compounds carrying an alkyl group.


text:
Why does the Second Law of Thermodynamics hold true?

The Second Law of Thermodynamics states that the state of entropy of
  the entire universe, as an isolated system, will always increase over
  time. The second law also states that the changes in the entropy in
  the universe can never be negative.

According to that, $\Delta S$ of the universe should always be positive for a spontaneous reaction to occur. But what is the reason for that?
In Advanced Chemistry, Philip Matthews states:

Pg-287  This result cannot be proved; rather it is a statement of
  experiment

While there is true reasons for every point in chemistry, I feel there should be a real reason for it rather than experimental statement. But I searched it on Google but can't to find an answer. So, can you help me to give the reason?

The author is right. Anything referred to as a law cannot be proven. All the other things we know are derived from what we call laws. The second law is an experimental observation. If you wish, however, you could say that we figured there was some property that increased during any process, and it turns out this is entropy. Similarly, we wanted there to be some property whose value was constant for all processes and this we call energy.

I would say to look into the Fluctuation Theorem.

There is no such a law that the "the entropy of the universe" increases, (does the universe have an entropy?); but there is a law derived from experience that the "entropy of a closed thermodynamic system can only increase". There is no experience that the universe is a closed system in the thermodynamics sense, and in no sense, especially not a mathematical one, that the former can be derived from the latter.

According to second law of Thermodynamics,
During spontaneous process:
$$ ∆S_{total} > 0 \ or ∆S_{total} = 0 $$
So total entropy is positive.
But $ ∆S_{system} \ or \ ∆S_{surroundings} $ may be less than 0.
Which $ \implies $ any one of $ ∆S_{system} \ *OR* \ ∆S_{surr} $ may be less than 0.

Now, according to Big Bang theory, the Universe, soon after existence , is expanding more and more and so - its entropy is increasing.

 Why entropy should be positive to occur a spontaneous reaction?
An example first:
• Heat flow is unidirectional. That is, from higher to lower temperature.
Similar to it, all Physical or Chemical Process will tend to proceed spontaneously in one direction only(unless altered by some external agency).
Hence,
Spontaneity is the potential to proceed without assistance of external agency.
Now, Entropy means randomness or disordered behaviour.
And there is always a tendency for the system's energy to become more disordered. An example would be diffusion.
Where gases diffuse and become more chaotic. And during the expansion of our universe, its entropy increases.
So, we interpret that entropy for a spontaneous process increases, till it reaches maximum. And our universe having a positive $∆U$ and gaining entropy in the direction of spontaneity.
$∆S=0$ at equilibrium.
If $∆S$ is negative that would mean that reaction or process is going opposite to the spontaneous direction, and which is not possible in case of our universe.

From your answer I understood why it is expanding  of the delta S universe more and more. But could you say the reason why do it always be positive to occur a  spontaneous reaction?

@Osal read my further answer.


text:
Material that changes solubility when electric potential is applied
I am looking for a material witch is easy to deposit in thin films and which changes its solubility if a electric potential is applied.
Do you know of any such materials ?
EDIT: This is for what I need it.
I have a Si Substrate on which there is sputtered SiO2.
Then I have the material (to be found) and then a Si on top. Some parts of the Si are connected electronically to the bottom Si and create a electric potential. The orange material underneath the top "electrodes" should be removed. The aim is to remove the "orange" material underneath the top "electrodes" but keep it where the Si is not connected to the bottom electrode.

How could you do this ?
EDIT2: please Igore the spelling error in the bottom electrode. It should be "connected"
EDIT3: based on Alchimista's answer:

That is an incredibly broad question. Have you tried any literature research before? If so, include it, if not, do it, then include it.

This is a very interesting question.

Solubility in what? Can deposition solvent / electrolytic solvent / any third solvent be different?  If you want a fixed insoluble film I know many ( especially if electrochemistry count as deposition techniques)

@Alchimista By applying a electric potential it should be able to remove the material from the substrate. Preferably the electric potential should be small.

In this case I have to think

I would also need more info on the purpose. I mean the material should detach from the electrode but remain intact in  essential?

@Alchimista I have a Si substrate with approx. 50nm SiO2 sputtered onto it. Now I place this material (to be found) on top of the SiO2 and place another electrode also made of Si on top of this material. Now the aim is to get rid of the sandwished material (between the SiO2 and the top electrodes, but leave it where there is no top electrode. So I don't care if the material is not intact afterwards.

So all this should be solid state and dry, no electrolyte?  Can the remaining material be a conductor?  Probably yes as you have SiO2. A conducting polymer could work.  It can be reduced to insulator were the top second  electrode is and stay in conductive form every where else....

@Alchimista Yes, solite state is prefered. Concerning the conductive polymer, I do not understand how applying a electric potential (maybe together with a solvent ) will make it disolve. Can you give an example or illustration ?

Let us continue this discussion in chat.

In principle it can happens but I have no clear example . The reverse is straightforward. What I was thinking is to attain a pattern based on different conductivity.  But in solid state I have no weaponry sorry

I am in app. Just a smartphone here. What I do not understand if why to place a material then remove it. Anyway I was thinking of electrochemistry that won't work in compact dry configuration. No to mention that Si is not very robust as electrode

@Alchimist. The material is there due to the manufacturing process, but it should be discarded at specific places, given by the top electrodes. I was thinking of maybe using an e-beam resist, but it is probably not sensitive to the electric field.

Are you looking for a material or the way to remove a specific material.  These are two different questions.  I think you will have indeed to look for some nanolitography process.

@Alchimista I am looking for a material.

Saw but it is really tricky ....

This is a tentative that perhaps does not fit with restrictions posed by the setup of the question, but does offer an example of a class of materials whose solubility (integrity, to be more precise) is easily altered by an electrochemical process.
A number of conjugated polymers can be deposited atop an electrode by
A: electrochemical polymerisation ( this resulting in a film of the polymer in a more or less conductive state ) or
B: as pristine material if available and soluble in some solvent ( spin coating, dipping, dr blading. ..).
The state of the polymer can be electrochemically controlled by the applied potential. ( however the complete neutralisation to its insulating/semiconducting state is generally difficult ).
Overoxidation of these films results in chemical damage associated with removal of the material, directly or by subsequent washing.
3-methylthiophene and poly(3-hexylthiophene) are well known examples for an electrochemical depositable polymer and for a polymer which can be directly cast from its solutions.
In both cases the poly(3-alkylthiophene) (P3AT) film is insoluble in polar solvents and electrolyte solutions. Applying an electrochemical potentially higher than, say, 2 V, overoxidize it damaging the skeleton to such an extent that the material fall out the electrode and / or can be washed out.
Specifically to the question: only case B is viable.
Looking at the figure in the Q, bottom, the orange part will be consisting of neutral P3AT.
Resolution will be not too precise as overoxidation will take place at the edge of the electrode, too.

Thank you very much!! Is my drawing in the answer correct ? How would you deposit P3AT onto the wafer: Can you do it with spin coating, sputtering PECVD or anything similar ? What electrolyte would you suggest ? Where should the positive and where the neagtive potential be applied ?

I would go by spin coating. Yes your drawing is correct just a reference electrode would be necessary to control the process.  In principle you could let flow current and destroy the material in contact with the electrode.  A controlled potential and moderate current will help you no to dope the adiacent material and clean everything.  you would likely need the help of somebody with a minimal knowledge of electrochemistry as well as a setup.  Plus test of the behaviour on electrodes similar but different from your final stuff. I said poly3hexylthiophene as readily available commercially.

It is not that the material will migrate to the counterelctrode but rather it will be oxidized to an extent disrupting not only the pi system but the sigma skeleton too. Ergo low mechanical properties / enhanced solubility. Just for a more chemically precise picture.

Thank you so much ! This is really helpful. May I ask what electrolyte you would suggest for the process ? One thing that is still not clear to me is what to do after the oxidization has taken place ? Should I dissolve P3AT in a specific solvent ?

(I also changed the last picture)

Check out this question:
Do indigo and leuco-indigo have fully conjugated pi systems?
Leucoindigo is a soluble, colorless compound which can be oxidized to dark blue indigo, which is insoluble.  Such oxidation is accomplished electrolytically through anodic polarization (or reversed through cathodic polarization).  You get a color change too.

Thanks for your answer. I am not so familiar with oxidisation, therefore would you mind to further elaborate your answers. Especially how it should be done for this question. The best would be, if you give me a step by step answer, such that I can clearly follow what you think. Thank you very much!

Also PANI in one of its various forms might help. The problem in answering is that one should fully know what you are aimed to. .....Thus A is correct but it will let you remove the material where the electrode is not connected.

@Alchimist The elcectrode which are not connected should stay.

Yes therefore it is tricky. Look at my answer and try that.  It seems you need help from every one somehow familiar with el chem and some tests. It really depends on your final expectations.


text:
3-index integral schemes for density fitting
Density fitting allows us to approximate the 4-index 2-electron integrals using 3-index integrals:
$$(ij|kl) \approx \sum_{Q}^{N_{aux}} (ij|Q)(Q|kl)$$
My question is, what schemes are used for the evaluation of the 3-index integrals? Do we simply modify existing schemes such as Obara-Saika and McMurchie-Davidson or are completely separate schemes given for this purpose?

What do you mean with schemes? The algorithms used to solve the integrals?

@Martin-マーチン yeah

DF is never done like this...

@Fl.pf. how so? That's not very helpful.

Why don't you take a look at these notes by David Sherrill (main prof. behind Psi4). From what I can gather, there are many possible schemes because one uses the auxiliary basis to minimize some functional which gives the coefficients in an expansion of an electron density. There are multiple possible functionals one might choose. From here, the 4-index integrals are approximately reconstructed from the 3-index integrals which one gets from the fitting process. This is the bird's-eye view. The details are probably... detailed.

https://chemistry.stackexchange.com/questions/87622/3-index-integral-schemes-for-density-fitting

I think in the early days one really used just the normal integral schemes Obara-Saika, McMurchie-Davidson, Rys where for one exponent just a Gaussian with exponent zero was used. Later the people examined the schemes for this special purpose and modified them for calculating 3 index coulomb integrals. A quite recent paper on this topic can be found in Gyula Samu and Mihály Kállay, J. Chem. Phys. 2017, 146, DOI: 10.1063/1.4983393.
One minor thing I want to point out is that you presented RI/DF in a slightly unusual way (notation). Using RI/DF the integrals are evaluated as
$$
(ij|kl) \approx \sum_{PQ}^{N_{aux}} (ij|P)[V^{-1}]_{PQ} (Q|kl)
$$
where
$$
V_{PQ} = \left(P|Q\right) = \int \int {{\phi_{P}({r_1}) \frac{1}{r_{12}}  \phi_{Q}({r_2})}}d{r_1}d{r_2}
$$
Of course you can rewrite it a bit with forming $V_{PQ}^{-1/2}$ to arrive at a similar expression as given by you.

I am giving you its solution from this paper by Gyula Samu and Mihály Kállay:
Abstract-
In this study we pursue the most efficient paths for the evaluation of three-center electron repulsion integrals (ERIs) over solid harmonic Gaussian functions of various angular momenta. First, the adaptation of the well-established techniques developed for four-center ERIs, such as the Obara–Saika, McMurchie–Davidson, Gill–Head-Gordon–Pople, and Rys quadrature schemes, and the combinations thereof for three-center ERIs is discussed. Several algorithmic aspects, such as the order of the various operations and primitive loops as well as prescreening strategies, are analyzed. Second, the number of floating point operations (FLOPs) is estimated for the various algorithms derived, and based on these results the most promising ones are selected. We report the efficient implementation of the latter algorithms invoking automated programming techniques and also evaluate their practical performance. We conclude that the simplified Obara–Saika scheme of Ahlrichs is the most cost-effective one in the majority of cases, but the modified Gill–Head-Gordon–Pople and Rys algorithms proposed herein are preferred for particular shell triplets. Our numerical experiments also show that even though the solid harmonic transformation and the horizontal recurrence require significantly fewer FLOPs if performed at the contracted level, this approach does not improve the efficiency in practical cases. Instead, it is more advantageous to carry out these operations at the primitive level, which allows for more efficient integral prescreening and memory layout.

@obackhouse-The source of this paper is $\color{aqua}{J. Chem. Phys. 2017, 146, DOI: 10.1063/1.4983393.}$,which I have used here only to solve your problem.


text:
How to calculate the spin of atomic nuclei?
I have recently been learning about nuclear magnetic resonance spectroscopy. It is mentioned that the spin of the $^{1}$H nucleus is $\frac{1}{2}\ $and the spin of the $^{2}$H nucleus is $1$. This made me think that this likely has something to do with the number of protons and neutrons and this is indeed the case. This site  reveals that:

If the number of neutrons and the number of protons are both even, then the nucleus has no spin.
If the number of neutrons plus the number of protons is odd, then the nucleus has a half-integer spin (i.e. $\frac{1}{2}\ $, $\frac{3}{2}\ $, $\frac{5}{2}\ $).
If the number of neutrons and the number of protons are both odd, then the nucleus has an integer spin (i.e. $1$, $2$, $3$).

However, I was unable to find a site which stated the exact formula needed to determine the spins of atomic nuclei. Thus, I would like to ask if such a formula exists or is the determination of nuclear spin only done through experiment?

There is no formula.

@IvanNeretin So the nuclear spin of an atomic nucleus can only be known through experimental means?

@Ivan what happens if we try to use nuclear shells?

@OscarLanzi It works poorly. The system is extremely computationally complicated and involves relativistic effects and strong spin-orbit coupling. Add that nuclon-nuclon interactions are a shadow of quark-quark interactions and things become... sketchy.

I'm afraid you could as well ask how to get spin of arbitrary molecule and it's too broad. Also are you interested in spins of ground states? Nuclei can be excited just as molecules can be.

@Mithoron I am mainly interested in the nuclear spins. However, if you would like to include how to obtain spins in general in your answer, I would be more than happy to accept it.

By checking out table list like http://easyspin.org/documentation/isotopetable.html


text:
Naming of halogen group in ethane
Can we write "1-Chloro-ethane" when a single chloro group is present in the ethane? I think only one position is possible for the chloro group in ethane.

I do not remember if is totally wrong but surely locator is redundant here. Saving character / space and aiming to simplicity should be a kind of rule.

In general nomenclature, the practice of omitting locants when there is no ambiguity is widespread. For preferred IUPAC names, however, the current recommendations given in Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) are prescriptive about when locants are omitted. In particular:

P-14.3.4.2 The locant ‘1’ is omitted:
(…)
(b) in monosubstituted homogeneous chains consisting of only two identical atoms;
(…)

Therefore, the preferred IUPAC name is indeed chloroethane and not the (correct, but not preferred) complete systematic name ‘1-chloroethane’.
(Nevertheless, note the correct capitalization and hyphenation of the names.)

The position can be omitted because there indeed is only one possible position for the chloro group to be.
Therefore the name Chloroethane is unambiguous.
It is however not wrong to specify the position because the chloro group is in fact in position 1.
Therefore it's redundant but correct.


text:
Volume of gas in a system
With two closed vessels of equal volume that are connected to each other through a narrow tube, the temperature of one of the containers is increased.
Why does the final volume of gas inside each container remain the same as earlier?
This is a solved example that I was trying to understand where they have taken the final volume of gas in each container to be the same as before. (Refer to equations 2 and 3)
(https://i.stack.imgur.com/Q7UD8.jpg)

The volume stays the same, because the containers stay the same. If the volume of one container would increase, that would mean it would expand. (BTW Note that variables like $T$ are printed in italics and units like K are printed with an upright font.)

@mhchem but isn’t it possible that the volume increases or decreases corresponding to an increase or decrease in pressure ?

No, that's the point. If in a stable container, the gas cannot take more volume. It cannot take less, because a gas always fills the whole container. Temperature can change, pressure can change (for instance by forcing more gas in the container), but volume cannot change (if the container does not change, expand, explode).

The volume stays the same, because the containers stay the same. If the volume of one container would increase, that would mean it would expand.
If in a stable container, the gas cannot take more volume. It cannot take less, because a gas always fills the whole container. Temperature can change, pressure can change (for instance by forcing more gas in the container), but volume cannot change (if the container does not change, expand, shrink, explode).


text:
Energy needed for electrolysis of aqueous NaBr
I'm trying to find the amount of energy it would take to electrolyse sodium bromide to get 1 mole of Br2. I'm not sure how to go about calculating this figure. If anyone could help me out, I'd be very grateful.


text:
NaHCO3 required to raise pH of weak HCl solution from 3.5 to 7.2 (doubting my calculations)
I'm a dentist, working on a medical device that will make shots at the dentist office hurt a lot less. The device mixes a drug (2% Lidocaine) that has been acidified with $\ce{HCl}$ to pH 3.5 for storage, with $\ce{NaHCO3}$ to bring it to physiological pH (about 7.2) just prior to injection.
The current method is by manually mixing an 8.4% $\ce{NaHCO3}$ solution with the drug in a 1:9 ratio V/V. In medicine, an 8.4% solution typically means 84mg of drug / ml of solution. So, the current protocol of mixing 100 ml of bicarb solution with 900 ml of drug solution, means the doctors are delivering 8400mg of $\ce{NaHCO3}$ /L , which is a LOT more than the 27mg that my current calculations (please see below, and please let me know if I'm off on the wrong track) say they need to do the job. You can see why I was (and am) skeptical of my calculations.
Assuming the calculations below are correct, then the next problem is:
If I want to deliver the $\ce{NaHCO3}$ in the form of a fluid (whether a solution, slurry, or any other way to make it flow), what is the best way to do that, and how much $\ce{NaHCO3}$ can I fit into a vessel with volume of 35 cubic millimeters (pretty small, and all the room I have available in my device)? In other words, what is the maximum concentration of $\ce{NaHCO3}$ that will stay in solution at a standard temperature and pressure with no head space in the vessel? Or, can I make a slurry that will hold an even higher concentration that remains fluid, and what would that concentration be? I only need to fit 27mg/L x 2ml = 0.054mg of $\ce{NaHCO3}$ into the space, and it would be a lot easier to manufacture if it were a liquid rather than a powder.
Also, I'm in need of some physical testing of the device to confirm that mixing has properly occurred at the tip of the needle when the prototype is built in a few weeks. There is some thought that we could use excess $\ce{NaHCO3}$ in a chamber, and have bypass mixing occur as the drug passes through the chamber and into the needle, instead of trying to get the whole volume (2ml) mixed prior to moving any drug out of the needle. If there is someone on here that might be interested in performing those validation experiments to help pass FDA gates, then please let me know how to find you. Thanks, Scott
The chemical reaction: $\ce{HCl + NaHCO3 → NaCl (aq) + CO2(g) + H2O(l)}$
The initial hydrogen ion concentration is: (pH = 3.5) [H+]i = 10-3.5 = $\pu{3.16228 x 10^-4 M}$
The required final hydrogen ion concentration is: (pH = 7.2) [H+]f = 10-7.2 = $\pu{6.309573 x 10^-8 M}$
moles $\ce{H+}$ reacted = moles of $\ce{HCl}$ reacted = $(\pu{3.16228 x 10^−4} − \pu{6.309573 x 10^−8}) = \pu{3.1616467 x 10^−4 mol}~\ce{HCl}$
moles of $\ce{NaHCO3}$ required = moles $\ce{HCl}$ reacted = $\pu{3.1616467 x 10^−4 moles}$
mass of $\ce{NaHCO3}$ required = $\pu{3.1616467 x 10^−4 moles}~\ce{NaHCO3} \times \pu{84g}~\ce{NaHCO3} / \pu{1 mole}~\ce{NaHCO3} = \pu{0.027 grams￼}$

I'd also wonder about the solubility of Lidocaine as a function of pH and ionic strength.

Given that Lidocaine is protonated at pH 3.5, don't you need more base to neutralize the Lidocaine than to neutralize the HCl? // 2% solution is 20 mg/ml or 20 grams per liter. MW of Lidocaine =234, moles = 20/234 = 0.085 moles/liter. 1/6 of that is 0.0142 moles per liter.

The approach I would suggest is a bit different.
First I would calculate the nominal concentration of HCl (i.e. the number of moles of HCl added, divided by the total volume of the solution after the addition, regardless of what happens to the resulting chloride ions and protons) in your initial solution, using all the necessary equations (with the approximation that activity equals concentation, although the figures are a bit large here).
1) dissociation constant of lidocaine:
$K_{a,lido}=\frac{[L] \cdot [H^+]}{[LH^+]}$
2) the total nominal concentration of lidocaine must be the sum of all its forms:
$C_{lido}= [L]+[LH^+]$
3) charge balance:
$[Cl^-] + [OH^-] = [H^+]+[LH^+]$
4) the total concentration of chloride is the same as the nominal concentration of HCl, as the dissociation is complete and $Cl^-$ does not react with anything else:
$C_{HCl}= [Cl^-]$
5) the ion product of water:
$K_w=[H^+] \cdot [OH^-]$
The system of these 5 equations can be solved for $C_{HCl}$, after eliminating all chemical species (except $[H^+]$, so in total 4 unknowns):
$C_{HCl}= [H^+] - \frac {K_w}{[H^+]} + \frac {C_{lido}}{1+ \frac {K_{lido}}{[H^+]}}$
Plugging in the numbers:
$C_{lido}=\frac {20 g/L}{234.34 g/mol} = 0.085346 M, K_{a,lido}=10^{-7.9} M, [H^+]=10^{-3.5} M, K_w = 10^{-14} M^2$
results in:
$C_{HCl}= 0.08566 M$
i.e. to reach a pH of 3.5 starting from freebase, we used just slighly more than the stoichiometric amount of $HCl$. Here we are assuming as well that the added volume of HCl has not changed the nominal concentration of lidocaine very much. If it has, then you need to account for that.
Now, when you add sodium bicarbonate, all the equations above except 3 are still applicable; you need to add the dissociation constants for carbonic acid (I'll consider it like hydrated $CO_2$, and that none of it is escaping as gas, to keep things simple) and hydrogenocarbonate, and the mass balances for sodium and the various carbonate species.
Here they are:
$K_{a,carbonic acid}=\frac{[HCO_3^-] \cdot [H^+]}{[H_2CO_3]}$
$K_{a,hydrogcarbonate}=\frac{[CO_3^{2-}] \cdot [H^+]}{[HCO_3^-]}$
$C_{bicarb}= [Na^+]$
$C_{bicarb}= [H_2CO_3] + [HCO_3^-] + [CO_3^{2-}]$
The charge balance now includes more ions:
$[Cl^-] + [OH^-] + [HCO_3^-] + 2 \cdot [CO_3^{2-}] = [Na^+]+[H^+]+[LH^+]$
We have now 9 equations (5-1+5). Like before, eliminating all chemical species except $[H^+]$  (i.e. 8 unknowns), gives a single expression. It is a bit complicated, but essentially it contains only $[H^+]$, $C_{bicarb}$ and $C_{HCl}$, and known constants.
In our case, as we know what the nominal concentration of $HCl$ is from the previous calculation, and the final pH we want to reach (7.2), we just need to solve for $C_{bicarb}$ and plug in the numbers:
$C_{lido}=\frac {20 g/L}{234.34 g/mol} = 0.085346 M, K_{a,lido}=10^{-7.9} M, [H^+]=10^{-7.2} M, C_{HCl}=0.08566 M, K_{a,carbonic acid}=4.3 \cdot 10^{-7}, K_{a,hydrogcarbonate}=5.6 \cdot 10^{-11} $
resulting in:
$C_{bicarb}=0.1142 M$
So for each litre of your initial solution you should add 0.1142 mol of $NaHCO_3$, which is 9.59 g.
This is closer to the 'official' quantity you mentioned (even quite a bit more in fact), although I have trouble following all the figures. Maybe they did this experimentally and found that some $CO_2$ escapes, making the pH rise without the need for more base. Just a hypothesis.
In a way, even without all these involved calculations, one could say: if you have a 0.085 M solution of lidocaine hydrochloride and want to react all the $HCl$ with sodium bicarbonate, use $0.085 mol/L \cdot 84 g/mol = 7.14 g/L$ of $NaHCO_3$. The excess used in your official method compared to this figure is probably due to the slight excess of $HCl$ in the initial solution, and the need to take the pH closer to the desired 'physiological' value you mentioned.
EDIT: addendum to answer further questions by the OP
The OP asked what would happen by adding more sodium bicarbonate.
The answer is: as long as there is no solid precipitate of sodium bicarbonate (and always assuming that lidocaine stays in solution), the above equations remain valid; as soon as the nominal concentration of sodium bicarbonate is sufficiently high for it to precipitate, new equations apply, where all the above equations still apply, except for a correction of the mass balances related to $C_{bicarb}$, in particular by adding the contribution of the solid:
$C_{bicarb}= [Na^+] + [NaHCO_3]_s$
$C_{bicarb}= [H_2CO_3] + [HCO_3^-] + [CO_3^{2-}] + [NaHCO_3]_s$
(where $[NaHCO_3]_s$ represents the solid salt expressed in the units of concentration, for homogeneity with the other terms).
And the equation for the solubility product of $NaHCO_3$ must be added:
$K_{s,bicarb}=[Na^+] \cdot [HCO_3^-]$
From the data table in this site it seems that $K_{s,bicarb}=10^{-0.55} M^2 = 0.2818 M^2$ (although from my own calculations based on a solubility of 96 g/L at 20°C it should be $1.277 M^2$, but OK, I will use the 'literature' value).
First, the limiting value of $C_{bicarb}$ after which residual solid bicarbonate is observed at equilibrium must be calculated.
This can be done by solving (numerically) the system of all 10 equations described above for all the chemical species plus $C_{bicarb}$, after setting $[NaHCO_3]_s=0$. The result is:
$[C_{bicarb}=0.5467 M, [Na^+]=0.5467 M, [HCO_3^-]=0.5155 M, [H^+]=2.504 \cdot 10^{-8} M, [H_2CO_3]=0.03002 M, [CO_3^{2-}]=0.00113 M, [Cl^-]=0.08566 M, [OH^-]=3.9936 \cdot 10^{-7} M, [LH^+]=0.0568 M, [L]=0.02855 M]$
meaning that this system is only valid for $C_{bicarb} > 0.5467 M$.
Then the system can be solved using any value of $C_{bicarb}$ greater than the above threshold, and, interestingly, it gives always the same values, except for $[NaHCO_3]_s$, that increases accounting for the undissolved salt.
$C_{bicarb}=0.7 M$
$[[NaHCO_3]_s=0.1533 M, [Na^+]=0.5467 M, [HCO_3^-]=0.5155 M, [H^+]=2.504 \cdot 10^{-8} M, [H_2CO_3]=0.03002 M, [CO_3^{2-}]=0.00113 M, [Cl^-]=0.08566 M, [OH^-]=3.9936 \cdot 10^{-7} M, [LH^+]=0.0568 M, [L]=0.02855 M]$
$C_{bicarb}=7 M$
$[[NaHCO_3]_s=6.453 M, [Na^+]=0.5467 M, [HCO_3^-]=0.5155 M, [H^+]=2.504 \cdot 10^{-8} M, [H_2CO_3]=0.03002 M, [CO_3^{2-}]=0.00113 M, [Cl^-]=0.08566 M, [OH^-]=3.9936 \cdot 10^{-7} M, [LH^+]=0.0568 M, [L]=0.02855 M]]$
$C_{bicarb}=70 M$
$[[NaHCO_3]_s=69.453 M, [Na^+]=0.5467 M, [HCO_3^-]=0.5155 M, [H^+]=2.504 \cdot 10^{-8} M, [H_2CO_3]=0.03002 M, [CO_3^{2-}]=0.00113 M, [Cl^-]=0.08566 M, [OH^-]=3.9936 \cdot 10^{-7} M, [LH^+]=0.0568 M, [L]=0.02855 M]]$
So it appears that, once solid sodium bicarbonate is present, the system is determined in a specific state, where all chemical species have a fixed concentration, regardless of how much solid there is.
In particular, the pH seems to be fixed at $7.60$. This is significantly lower than the pH of a saturated solution of $NaHCO_3$ ($8.31$) in water, probably because of the $H_2CO_3$ liberated in the reaction with $HCl$ (which, we are assuming, is remaining in solution).

If the NaHCO3 were to be delivered in excess to the acidified drug solution, am I correct in saying that the pH would wind up in the 9.6 range?  And I'm guessing that wouldn't feel too good for the patient either?

And would that free up even more unionized Lidocaine for pharmacological action on the nerve membrane?

As for the second question: sure, the higher the pH, the more non-protonated lidocaine there will be in solution (the dissociation constant of lidocaine is always holding true), although after a certain pH the difference becomes practically insignificant, I reckon. As for what will happen in the patient's gums, no idea, it's more a physiology question. I'd have thought that the body is pretty good at buffering whatever you inject in it, but how fast that will happen in different tissues, and the effect of injecting a rather alkaline liquid parenterally are outside my field of expertise.

For the first question, I removed my earlier comment, which was imprecise, and I will edit the original answer instead.

Okay, this next question is more chemical engineering than chemistry, I suppose, but I think (hope, actually) your answer suggests: If we take 2ml of 2% lidocaine HCl solution (contains 40mg of lidocaine and 0.08566M of HCl to reach pH 3.5), and then push it at a rate of about .25ml/second through a "mixing" chamber of volume 0.05 ml  that contains excess NaHCO3 (let's say, 38mg, which is about 2x what's necessary), the output from the mixing chamber should remain < pH 7.69, and the reaction should proceed rapidly enough to get the effluent solution > pH 7.2?

In theory yes (about the pH). But you're right to question the kinetic aspect: all my calculations assume equilibrium, and although acid-base reactions are usually quite fast, 0.25 mL/s is a pretty fast flow, and with $NaHCO_3$ not being very soluble in the first place, I am not sure. Maybe you need a long cylindrical cartridge with a very large excess of solid $NaHCO_3$ (more than 87 mg, say twice that at least), through which  the acidic solution must flow. Frankly, like others already said, this may be best tried out in practice than theorised. I would also watch out for $CO_2$ buildup.

One thing to consider:  the Lidocaine is itself a base that gets protonated in the acidified solution.  You may need to convert some of the Lidocaine back to its deprotonated, basic form, along with neutralizing the "free" acid that the pH meter sees, when you add the bicarbonate.  According to the comments the pKa of Lidicaine is about 7.9 (comments) or 7.7 (https://www.google.com/url?sa=t&source=web&rct=j&url=http://www.frca.co.uk/documents/anaes.5.5.158.pdf&ved=0ahUKEwiEs6nPvZfYAhUl7IMKHc7FBe0QFggmMAE&usg=AOvVaw0RhVYVPEby1Mgu8nIUy48v), so to get to pH 7.2 you must in fact deprotonate at least one sixth of the Lidocaine.

The literature says the pKa of Lidocaine is 7.9, and as you point out, at pH 3.5 it exists predominately (99.996%) in it's protonated form, which has no ability to cross the lipid membrane of nerve channels.The body can do this over time, but the NaHCO3 makes it almost instantaneous.  This is another good reason to buffer anesthetic, to save time getting numb.

So we deprotonated a limited amount of the Lidocaine, about 1/6 of it.

Yes, that is consistent with what I have read.  Most of it remains pharmacologically unavailable, even after the body fluids bring the ambient pH to about 7.4

Still need to add it into the calculations though.

No argument there, but if I thought I could do the calculations (even the basic, no pun intended, calculation), I wouldn't have posted the question in the first place.  Do they look correct to you?  Or did I mess things up somewhere?  I can't believe that the convention in medicine is so far away from what I calculated, it makes me think I missed something big, on order of 100 times or more.

Your numbers a ree OK you need to add the Lidocaine.  Also check the actual pH of the mixture.  A few tenths is a lot in the buffering range.


text:
Are all highly shock sensitive chemicals highly temperature sensitive too?
I found that some highly shock sensitive explosives like lead azide, mercury fulminate have a relatively high auto-ignition temperature. I want to know if it's a general property or not and why so? Doesn't shock sensitivity means that the molecular bonds can be broken easily by supplying only a small amount of energy?

Yes -- but only to a degree
I assume you are talking primarily about explosives. All explosives are sensitive to both shock and heat to some degree. They are also sensitive to friction and spark. In most jurisdictions, as part of the regulatory approval process for new explosives the initiation sensitivity to each of these stimuli is determined with standardised tests.
In general, the lower the activation energy of the decomposition reaction, the greater the sensitivity to all stimuli. However, there is no simple relationship between sensitivity to different stimuli. Comparing two explosives, one may be more sensitive to impact than the other, but less sensitive to heat; a third may be less sensitive to either heat or impact, yet very sensitive to friction.
These sensitivities are not only a function of the chemical nature of the explosive, but also its physical form, purity, and the other ingredients in a composition. For example – not surprisingly! – addition of oily or greasy ingredients, even in small amounts, tends to reduce friction sensitivity. Most famously, liquid nitroglycerine is extremely shock sensitive, but is greatly desensitised by adsorption into a porous medium.
Again, for some important crystalline primary explosives the heat sensitivity is more or less fixed, but the shock sensitivity depends on the particle size distribution.

Shock and heat sensitivity are only weakly related
The shock sensitivity of an explosive is the degree to which a physical shock can set off an explosion in the compound. The heat sensitivity is the equivalent for heat. Both of these properties depend not just on the chemistry of the explosive but also its physical form and the device that contains it. This is because the physics of an explosion depends on how fast shock waves propagate in the substance. Whether heat alone triggers an explosion depends on whether heat can initiate a shockwave.
There are some explosives that are exquisitely sensitive to shock like nitroglycerine as a liquid. But even liquid nitroglycerine can be burned without triggering an explosion (not, to be sure, an experiment I'd recommend as a class demonstration).
Useful explosives should not be either too heat or shock sensitive or they will be impractical in real world uses. Many usable explosives are not very heat sensitive as a matter of design. For example, TNT can be melted by steam making it easy to fill munitions. C4 plastic explosive can be set on fire without risking an explosion (and was widely used as a source of cooking heat in Vietnam).
And excess shock sensitivity is also undesirable in useable explosives which is why Nobel made so much money from finding a way to tame nitroglycerin by absorbing it in clay. Modern explosives need fairly strong shocks to set them off so the accidental explosions are minimised.
The lack of a simple relationship between thermal and shock sensitivity is an indication that explosive properties are more complex than just having a compound where some bonds are easy to break. The difference between something that burns easily and an explosive involves not just the bonds but whether a physical shock wave can propagate through the material with enough energy to break more bonds. Heat alone will often not initiate such a wave, which is why many explosives can be safely burned.


text:
Thermodynamic Functions

Indicate the correct statements:
a. $\mathrm{d}A = -S\mathrm{d}T$ for an ideal solution undergoing an adiabatic process.
b. $\mathrm{d}G = -\mathrm{d}(TS)$ for an ideal solution undergoing an isobaric process.
c. $\mathrm{d}A = \mathrm{d}U - T\mathrm{d}S$ for an ideal solution undergoing an isothermal process.

The answer is statements (a) and (b).
How, though?
$\mathrm{d}A = \mathrm{d}U - \mathrm{d}(TS)$. Since it is adiabatic, $\mathrm{d}q = 0$, or $T\mathrm{d}S = 0$.
So $\mathrm{d}A = -P\mathrm{d}V - S\mathrm{d}T$. What happened to the $-P\mathrm{d}V$ term?
$\mathrm{d}G = \mathrm{d}H - \mathrm{d}(TS)$
$\mathrm{d}G = V\mathrm{d}P - S\mathrm{d}T$, so $\mathrm{d}G = -S\mathrm{d}T$.
Why did they add the $-T\mathrm{d}S$ term?
$A$ is the Helmholtz free energy and $G$ is the Gibbs free energy.

Are you sure it wasn't indicating the incorrect statements? Because c is the only one at a glance I can can see justification for. Can you put a source for where you found this question?


text:
What is the relative size of the (M+2) peak?
The $(M+1)$ peak is often considered in the high-resolution mass spectra of organic molecules as it reveals the number of carbon atoms in the sample. In general, it is known that the ratio of the size of the $M$ to $(M+1)$ peaks is $98.9 : 1.1 \times n $ since the relative abundance in nature of $^{13}$C is $ 1.1$% for the mass spectrum of an organic molecule containing $n$ carbon atoms and no heteroatoms. It is mentioned that the $(M+2)$ peak is statistically insignificant on this site. However, I believe that only applies for organic molecules with a relatively small number of carbon atoms and this peak would become significant when considering the mass spectra of larger organics. Using simple mathematics, I derived that the ratio of the $M$ to $(M+2)$ peak is $98.9^{2} : 1.1^{2} \times _nC _2 $. I would like to verify if this is correct. If it is not, could someone then correct it by posting an answer?

You are correct on all accounts.
To a very good approximation, molecules can be thought as made of elements (with their respective isotope distributions) combining completely independently. You can think of it like rolling multiple die at once. This means that a simple multinomial distribution will describe this problem mathematically.
Let's start with something easy and consider the hypothetical molecule $\ce{C_5}$. Furthermore, let us consider that the only carbon isotopes with significant natural occurrence are $\ce{^{12}C}$ (98.9%) and $\ce{^{13}C}$ (1.1%). We can find all of the isotopic peaks and their relative abundances by then expanding the binomial $(0.989\times m[^{12}C] + 0.011\times m[^{13}C])^5$, where $m[^{12}C]$ and $m[^{13}C]$ denote the exact masses of the carbon-12 and carbon-13 isotopes, respectively. Expanding the binomial yields:
$\begin{equation}
\begin{aligned}
(0.989\times m[^{12}C] + 0.011\times m[^{13}C])^5 ={} & \ \ \ \ \ \binom {5} {0}(0.989\times m[^{12}C])^5 \\
      & + \binom {5} {1}(0.989\times m[^{12}C])^4 \times (0.011\times m[^{13}C]) \\
      & + \binom {5} {2}(0.989\times m[^{12}C])^3 \times (0.011\times m[^{13}C])^2 \\
 & + \binom {5} {3}(0.989\times m[^{12}C])^2 \times (0.011\times m[^{13}C])^3 \\
 & + \binom {5} {4}(0.989\times m[^{12}C]) \times (0.011\times m[^{13}C])^4 \\
 & + \binom {5} {5}  (0.011\times m[^{13}C])^5 \\
\end{aligned}
\end{equation}$
Calculating the coefficients in each term:
$\begin{equation}
\begin{aligned}
(0.989\times m[^{12}C] + 0.011\times (m[^{13}C])^5 ={} & \ \ \ \ \ 0.946 \times (m[^{12}C])^5 \\
      & + 0.0526\times  (m[^{12}C])^4 \times m[^{13}C] \\
      & + 0.00117 \times (m[^{12}C])^3 \times (m[^{13}C])^2 \\
 & + 0.0000130 \times (m[^{12}C])^2 \times (m[^{13}C])^3 \\
 & + 7.24×10^{-8}\times (m[^{12}C]) \times ( m[^{13}C])^4 \\
 & + 1.61×10^{-10} \times m[^{13}C])^5 \\
\end{aligned}
\end{equation}$
From the expansion, we see that 94.6% of all $\ce{C_5}$ molecules contain only carbon-12 (the lowest possible mass for the molecule), and almost all of the rest (5.3% out of the remaining 5.4%) is accounted for by molecules that contain a single carbon-13 atom. Only about 0.1% of $\ce{C_5}$ molecules contain two or more carbon-13 atoms.

But what happens in very large molecules? Intuitively, if there are many atoms, you would expect a higher chance of there being at least one less common isotope in the mix. Let's see the first few terms for the molecule $\ce{C_100}$:
$\begin{equation}
\begin{aligned}
(0.989\times m[^{12}C] + 0.011\times m[^{13}C])^{100} ={} & \ \ \ \ \ \binom {100} {0}(0.989\times m[^{12}C])^{100} \\
      & + \binom {100} {1}(0.989\times m[^{12}C])^{99} \times (0.011\times m[^{13}C]) \\
      & + \binom {100} {2}(0.989\times m[^{12}C])^{98} \times (0.011\times m[^{13}C])^2 \\
 & + \ ...\\
\end{aligned}
\end{equation}$
Calculating the coefficients:
$\begin{equation}
\begin{aligned}
(0.989\times m[^{12}C] + 0.011\times m[^{13}C])^{100} ={} & \ \ \ \ \ 0.331 \times (m[^{12}C])^{100} \\
      & + 0.368 \times (m[^{12}C])^{99} \times (m[^{13}C]) \\
      & + 0.203 \times (m[^{12}C])^{98} \times (m[^{13}C])^2 \\
      & +\ ...\\
\end{aligned}
\end{equation}$
Well that's interesting. Now only 33.1% of the molecules contain only carbon-12 atoms, and in fact more molecules contain exactly one carbon-13 atom, at 36.8% of the total. Even molecules with two carbon-13 atoms are quite abundant, at 20.3%.
Indeed, peaks containing rarer isotopes eventually dominate. For the huge molecule $\ce{C_10000}$, the strongest mass spectrum signal would come from molecules contaning 110 carbon-13 atoms, corresponding to 3.8% of the total, while a measly $9.2\times 10^{-47}\%$ of molecules contain only carbon-12. This happens because when $n$ is large, the term $\binom {n} {k}$ grows very quickly as $k$ rises from zero, overwhelming the increase in the exponent of the rarer isotope. You can see this behaviour quite nicely in this sequence of mass spectra of molecules with increasing size.

To calculate the specific $M/M+2$ ratio for a molecule containing only $n$ carbon atoms, all you need is to get the ratio for first and third terms in the binomial:
$\begin{equation}
\begin{aligned}
(0.989\times m[^{12}C] + 0.011\times m[^{13}C])^n ={} & \ \ \ \ \ \color{#0000ff}{ \binom {n} {0}(0.989\times m[^{12}C])^n} \\
      & + \binom {n} {1}(0.989\times m[^{12}C])^{n-1} \times (0.011\times m[^{13}C]) \\
      & + \color{#0000ff}{\binom {n} {2}(0.989\times m[^{12}C])^{n-2} \times (0.011\times m[^{13}C])^2} \\
      & +\ ...\\
\end{aligned}
\end{equation}$
The ratio is then:
$$\frac{\binom {n} {0}0.989^n}{\binom {n} {2}0.989^{n-2} \times 0.011^2}=\frac{2\times 0.989^2}{n(n-1)\times 0.011^2}$$
Technically this only holds if there are no other elements which contain multiple isotopes, though it will hold approximately if the other elements only have very rare alternate isotopes, such as hydrogen (99.98% hydrogen-1, 0.02% hydrogen-2).

As a last curiosity, all of the above extends to analysing more complicated molecules. For example, glucose ($\ce{C6H12O6}$) will have a mass spectrum exactly described by the expression:
$$(0.989\times m[^{12}C] + 0.011\times m[^{13}C])^6 \times (0.9998\times m[^{1}H] + 0.002\times m[^{2}H])^{12} \times (0.9976\times m[^{16}O] + 0.004\times m[^{17}O] + 0.020\times m[^{18}O])^6$$
Happy expanding!

Let’s assume your compound is $\ce{C_nH_xO_y}$. Thankfully, both hydrogen and oxygen are elements that only have one significant naturally occuring isotope. Therefore, we can treat the entire contribution of $\ce{H_xO_y}$ as a constant $c$. All we need to answer is how large the $M+1$ and $M+2$ peaks are given the number of carbons, $n$. (This treatment is not entirely correct as highlighted by orthocresol, but it is close enough for my purposes.)
Assuming you do not have any isotope enrichments, each carbon will independently have a chance of either being $\ce{^12C}$ or $\ce{^13C}$ (again, we will ignore all other isotopes such as $\ce{^14C}$). The independence is the big key word here. We can use general principles of stochastics to calculate the result.
The probability of all carbon atoms of one molecule of your compound are $\ce{^12C}$ is:
$$P(\ce{^12C_nH_xO_y}) = 0.989^n$$
The $M+1$ peak is represented by one single atom being $\ce{^13C}$. Again, the principles of stochastics apply:
$$P(\ce{^12C_{n-1}^13C1H_xO_y}) = \left ( n\atop 1\right) \times 0.989^{n-1} \times 0.011$$
And finally for the $M+2$ peak:
$$P(\ce{^12C_{n-2}^13C2H_xO_y}) = \left ( n \atop 2\right) \times 0.989^{n-2} \times 0.011^2$$
Using this, we can plot what the probability of each peak is – and use that as their relative heights.

We see that from $n=90$ the $M+1$ peak is actually larger than the $M$ peak. From $n=128$, even the $M+2$ peak will be larger than the $M$ peak. And from $n=181$ the $M+2$ peak becomes the largest of these three.
Of course, further peaks will also start appearing, meaning that the mass spectra of very large molecules will become difficult to analyse.
Now the numbers from which $M$ is no longer te principal peak are quite alrge. Unless you are synthesising maitotoxin, you will probably not need to resort to any analysis that does not only involve $M+1$.

Oxygen-18 has 0.2% abundance. In a molecule with six carbons (for example) the probability of having two 13C is 0.17%. Even if the molecule only has one oxygen, the probability of having one 18O is comparable and not negligible.  (Obviously the significance of 18O will depend on the C/O ratio.) Now, I’m thankful that HRMS exists :)

Thanks for showing the graphical analysis and the trend. They certainly helped me to understand it a lot better, which is why I'm awarding you the bounty. Thanks! However, I accepted the other answer as I preferred something more mathematically rigorous.


text:
What is the correct name of this compound?

What is the correct name of this compound?

1. 2-Chloro-5-isopropoxy hexane;
2. 2-Chloro-5-(1-methyl ethoxy)hexane;
3. 5-Chloro-2-isopropyl hexane;
4. 5-Chloro-2-(methyl ethoxy)hexane.

Isopropyl or methyl ethoxy. Which one is correct?

Both are correct, but only one is systematic.

Please see our guidelines on how to ask nomenclature questions.

Strictly speaking, the answer should be ‘none of these’ since all proposed names include unnecessary spaces. Anyway, this question seems to be based on obsolete IUPAC recommendations from 1993 since ‘isopropyl’ is no longer a preferred prefix in the current recommendations from 2013.

The answer is 5-chloro-2-isopropoxy hexane. When you write an IUPAC name of the compound, you need to write the substituents’ names in alphabetical order I.e., C comes first, hence 5-chloro should be written first and then follows I, hence -2- isopropoxy comes next. This is the IUPAC(systemic) name.
The common name is 5-chloro hexyl isopropyl ether

【EDIT: The group encircled has been wrongly mentioned as isopropoxy group. It's isopropyl group. But the encircled part along with the O forms the isopropoxy group.
[Note: I am not the original answerer of the question.]】

The correct answer is D.
First of all, you have to know that numbering in this compound starts from the carbon, which is nearly connected to ether, so accordingly you find $\ce{Cl}$ is at 5th position.
Now, according to your doubt how you think that it is isopropyl. It must be either isopropoxy or methyl ethoxy. So, in the following option there is only methyl ethoxy is present, holding all the rules of IUPAC nomenclature it is the correct answer.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. Please see our guidelines on how to answer nomenclature questions. Please back up your explanations with corresponding citations of the rules your have used. I believe the name should be 2-chloro-5-isopropoxyhexane, according to alphabetic ordering and lowest set of locants.


text:
Where does Oxygen come from in KO2 reaction with CO2?
$\ce{KO2 + CO2}$ gives $\ce{K2CO3 + O2}$
I think it should be from $\ce{KO2}$ as $\ce{2KO2}$ can split to $\ce{K2O}$ and $\ce{3/2O2}$. This $\ce{K2O}$ being unstable reacts with $\ce{CO2}$ to form $\ce{K2CO3}$.
Only problem is, is the first step possible?
Is $\ce{O2}$ released from $\ce{KO2}$ or $\ce{CO2}$?

Homework is a deprecated tag!

KO2 can decompose, but as Nilay Ghosh said, the product is not K2O; also, even that requires heating, which is simply not there. Also, K2O is perfectly stable by itself (though it would surely react with CO2, given a chance). Other than that, you are right.

Yes, the oxygen indeed comes from the decomposition of potassium superoxide ($\ce{K2O}$) but the other decomposition product is potassium peroxide ($\ce{K2O2}$) and not potassium monoxide($\ce{K2O}$).(Here)
$$\ce{2KO2 ->[290°C, vacuum] K2O2 + O2~~~~~~~~~~(1)}$$
Now, this potassium peroxide reacts with carbon dioxide to form potassium carbonate and oxygen.(Here)
$$\ce{2K2O2 + 2CO2 → 2K2CO3 + O2~~~~~~~(2)}$$
Adding $\ce{(1)\times2}$ and $\ce{(2)}$, we get full reaction. (Here)

$$\ce{4KO2 + 2CO2 → 2K2CO3 + 3O2}$$

If reaction proceeds in aqueous medium, it gives potassium hydrocarbonate and oxygen.(Here)
$$\ce{4KO2 + 4CO2 + 2H2O → 4KHCO3 + 3O2}$$

chemiday.com is by no means a reliable source of information.

"If the reaction proceeds in an aqueous medium"? Doesn't potassium oxide react violently with water?


text:
In 2-bromo cyclohexanone,why bromine prefers to occupy axial rather than equatorial position?
I read that the substituents generally tend to occupy equatorial position rather than axial because diaxial interactions make the molecule unstable. But why is 2-bromocyclohexanone an exceptional case? I also read that it’s because of the dipolar interactions. What actually are these dipolar interactions and how do they stabilise 2-bromocyclohexanone with bromine being on axial position. How do they outweigh the effect of diaxial interactions.

I think there must have been some misunderstanding. This stabilization is working for all 2‑halocyclohexanones (but does not always cause the axial conformer to predominate), where bromo- one is not a special case, but has been used just as an example.
There is a combination of various effects, but the dipolar one is supposed to cause a destabilization by dipolar repulsion in the equatorial-X conformer. See the difference of partial charges distribution:

Measured or calculated molar fractions (%) of the axial conformers in different phases/solvents (taken from Yoshinaga, Fabiana, et al. "Conformational analysis of 2-halocyclohexanones: an NMR, theoretical and solvation study." Journal of the Chemical Society, Perkin Transactions 2 9 (2002): 1494-1498. doi:10.1039/B204635K):
$$
\begin{array}{c*}\hline
& \text{vapor} & \text{liquid} & \ce{CCl4} & \ce{CHCl3} & \ce{MeCN} & \text{DMSO} \\
\hline
\ce{F}  & 64 &  5 & 28 & 13 &  2 &  2 \\
\ce{Cl} & 86 & 26 & 62 & 42 & 15 & 13 \\
\ce{Br} & 92 & 53 & 81 & 66 & 36 & 33 \\
\ce{I}  & 96 & 87 & 92 & 85 & 65 & 63\\
\end{array}
$$
Overall stabilization effects in axial-X conformation of 2‑halocyclohexanones clearly increases in the order of F < Cl < Br < I.
However, the ‘gauche effect’ participates, which is stabilizing the equatorial conformation in largest extent for F, also for Cl, has almost zero effect for Br, and has a negative effect for I.


text:
Colloids and Ions
Suppose I have this reaction:
$$\ce{3NaOH(aq) + FeCl3(aq) → 3NaCl(aq) + Fe(OH)3}$$
My questions:

$\ce{Fe(OH)3}$ doesn't have the (aq) suffix because it isn't separated in water as opposed to $\ce{NaCl}$?
Will the $\ce{3NaCl(aq)}$ be separated in positive and negative ions that will aggregate to the $\ce{Fe(OH)3}$ producing colloids ?

"Separated" is a wrong word; use "dissociated" instead. Fe(OH)3 isn't soluble and therefore precipitates, hence no "(aq)".

Question 1: You are correct. $\ce{Fe(OH)3}$ is not soluble in water meaning it will exist as a separate solid phase and thus it has no (aq) designation.
Question 2: It is correct that $\ce{NaCl}$ is soluble in water and will dissociate into $\ce{Na+}$ and $\ce{Cl-}$ ions in aqueous solution. Thus it has the (aq) designation. Also, aqueous $\ce{NaCl}$ can stabilize colloidal $\ce{Fe(OH)3}$.