text:
Multi-solvent recrystallization to extract solute from glycerol?
I have compound A contaminated with impurity B. Recrystallization in water would not yield pure compound A since impurity B is less soluble in it. The only solvent in which substances A and B have opposite solubility is glycerol. Compound A is soluble in it while impurity B is insoluble.
I would propose my first step to be heating the glycerol to minimize the amount of solvent needed to fully dilute compound A. I would then filter the solution to have my purified compound A in glycerol solution (impurity B would remain in filter).
As far as I have heard though, glycerol is not as easy to boil as water, being similar to cooking oil. Unfortunately, I don't have a vacuum distillation apparatus to extract the glycerol without caramelizing it.
I was thinking of slowly add anhydrous ethanol to (purified compound A + glycerol), since compound A is insoluble in it. I would expect multi-solvent recrystallization to occur as I add ethanol and lower the mixture temperature.
Would it work and does it seem to be the easiest method of purification at home?
For the curious: $A = \ce{KNO3}, B = \ce{K2SO4}$.
No, I don't want the $\ce{KNO3}$ to make a bomb, but to make a sugar rocket for model rocketry.

That's a lot of thinking for making sugar rocket...

@Mithoron I have heard it might work well even when contaminated with 7% K2SO4, what I have now. But  I would like to test with pure KNO3 to see if I can extract some extra juice from it.

Well actually you can make a bomb with only sugar... so don't worry about your question.

As an inverse to recrystallisation; you could add varying quantities of water to the same mass of impure solid and experiment with varying temperatures. The goal here is not to dissolve everything, but to dissolve up more A and leave B behind. i. e. A warm reslurry. When you get your best result, filter B off warm.

@Beerhunter Best result would be at 100C, and it has a mass ratio of 91% between KNO3 and K2SO4. My KNO3 is already at 93% in the mixture. I would actually lose purity.
Unless you are supposing the K2SO4 solubility would be reduced by the KNO3 presence, and not the opposite. Could you precise further?

"Recrystallization in water would not yield pure compound A since impurity B is less soluble in it". So maybe try a reslurry where A is preferentially dissolved over B.

@Beerhunter That is what I am trying to do with the glycerol. The problem is that after filtering insoluble B from the glycerol, how do I remove the glycerol to get pure A? I don't have a vacuum distillation apparatus to prevent the caramelization of glycerol to boil it.
I could technically cool it down, but its melting point is higher than 15C. I don't think the recrystalization will yield more than a few grams of KNO3.

I understand what you're trying to do, so try your proposal out on a test sample. Glycerol seems better than water for the separation, but you have the downstream processing issue of obtaining your product. If you used water and sacrificed some yield, you might have a better chance of precipitating and drying (if required) your product. I'm not saying it'd be better, just an alternative. A reslurry is easier to perform than embarking on a recrystallisation.


text:
Determination of magnesium sulfate hydrate formula

A hydrate of magnesium sulfate has a mass of $\pu{12.50 g}$.This sample is heated until no water remains. The anhydrous $\ce{MgSO4}$ has a mass of $\pu{5.30 g}$. Find the formula of the hydrate.

What I tried to do was I divided 12.50 by 5.30 and got 2.3ish. Then rounded it down to 2. After that I times the $\ce{MgSO4}$ by two to get $\ce{Mg2S2O8}$.
I know this isn't right but this is my initial thought process. Can anyone tell me the steps to I need to solve this? This is on a review guide so I just need to know the process.

Do not repost questions, you should have edited original.

You must've confused a hydrate with something else. Just like many other hydrates, magnesium sulfate hydrate dehydrates as follows:
$$\ce{MgSO4 * xH2O ->[\Delta] MgSO4 + x H2O}$$
To determine the formula, you need to find $x$, which can be done using molecular weight of the hydrate:
$$x = \frac{M(\ce{MgSO4 * x H2O}) - M(\ce{MgSO4})}{M(\ce{H2O})} \tag{1} \label{eq:1}$$
Molecular weight of the unknown hydrate, on the other hand, can be found from the amount of anhydrous salt as those two are in $1 : 1$ ratio (see the reaction).

 $$M(\ce{MgSO4 * x H2O}) = \frac{m(\ce{MgSO4 * x H2O})}{n(\ce{MgSO4 * x H2O})} = \frac{m(\ce{MgSO4 * x H2O})}{n(\ce{MgSO4})} = \frac{m(\ce{MgSO4 * x H2O}) \cdot M(\ce{MgSO4})}{m(\ce{MgSO4})} = \frac{\pu{12.50 g} \cdot \pu{120.37 g mol-1}}{\pu{5.30 g}} = \pu{283.89 g mol-1} \tag{2}$$

Plugging the found molecular weight to \eqref{eq:1} will allow you to determine the formula.

$$x = \frac{\pu{283.89 g mol-1} - \pu{120.37 g mol-1}}{\pu{18.02 g mol-1}} = 9.07 \approx 9 \tag{3}$$
This is magnesium sulfate nonahydrate $\ce{MgSO4 * 9 H2O}$.

I think it would be better to just describe how OP should do it, not provide full answer.

@Mithoron I edited the question to address your concern and put the exact answer as well as the key calculations under the spoilers.

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Why is nitrogen dioxide brown?
Nitrogen dioxide (NO2) is a reddish brown molecule, while its dimer nitrogen tetroxide (N2O4) is colorless. Can you give a reasonably detailed explanation for this from electronic structure theory?
Previous queries were not answered because the question posed was deemed to be too broad.

I'm afraid that like the previous question, this will also be deemed as too broad very soon. EST is a very broad and complicated theory and may be hard to explain that in a single answer.

No, I think what made the other question too broad is wanting to include the other coloured gases, which would require an explanation for every single gas.

You are asking why NO2 has low lying energy levels compared to its dimer. (It just happens that these low levels lie in the visible part of the spectrum). To put the question the other way round, why would you expect the energy levels to be similar when the molecules are different?
Perhaps more interesting is to ask why N2O4 is planar yet has a very long 0.175 nm N-N bond whereas the NO bond lengths and ONO angles are similar to NO2.

The color of NO2 is due to presence of single unpaired electeon, while in its dimer N2O4 that lone electron got paired which cause loss of its color.

Unanswered duplicate: 1

To understand this, we must take a look at these molecules one by one.

Nitrogen Dioxide[NO2]:

This here is its resonance structure/canonical structure.

NO2 is a molecule in which a Nitrogen is connected with two atoms of Oxygen and its shape is Bent. Total number of valence electrons is 17 with each Oxygen atoms contributing 6 and 5 electrons are from Nitrogen.
According to MOT (Molecular Orbital Theory), NO2 has an unpaired electron in π* (Antibonding) orbital due to which, orbitals of Nitrogen and Oxygen undergo 2p-2p overlap. Meanwhile this unpaired electrons makes our molecule a paramagnetic one.
This unpaired electron in π* (Antibonding) orbital can react with any external electromagnetic radiations and guess what, visible light is also an electromagnetic radiation. The reaction between this unpaired electron and visible light causes electron transition.

When electrons goes to higher energy orbital from
lower energy orbitals some form of energy i.e LIGHT is absorbed, if the electron goes vice-versa then energy (LIGHT) is released.
The presence of unpaired electrons in paramagnetic specie creates molecular orbitals that are partially filled or has unpaired spins.these electrons occupy higher energy orbitals which are more prone to electronic transitions. So, paramagnetic specie are more viable to absorb light at lower frequency as they have longer wavelength.
Hence, NO2 absorbs lower frequency light in the visible region of the spectrum, specifically in the blue-green range.
As a result, NO2 appears reddish-brown in colour.
2.Dinitrogen Tetroxide(N2O4):

 This here is its resonance structure/canonical structure.

THIS is a dimer of NO2 consisting of two nitrogens and four oxygens. In this molecule due to dimerisation,  the unpaired electrons in the π*(antibonding) orbitals of each NO2 molecule pair up together to form a stable, diamagnetic molecule.
Since it has no unpaired electrons, No interaction with external electromagnetic waves and so, no absorption of light in the visible region. Hence, it appears colourless.

Close to a nice answer,but I think this needs a bit more - why does a molecule being paramagnetic make it able to absorb light at comparatively low frequency? If paramagnetism is the cause of colour why is NO colourless?

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The pyridine/benzene stability 'paradox'?
So, apparently pyridine is more stable than benzene. My lecture notes are saying that pyridine will undergo electrophilic aromatic substitution much more slowly than benzene will, which got me thinking - slower means the compound is more stable (higher activation energy). But isn't benzene more aromatic than pyridine?
I checked the heats of hydrogenation and it checks out - pyridine is indeed more stable than benzene. I'm a little bit lost though, because I don't get where this unexpected stability is coming from. Surely benzene is an ultra-stable compound because of the superb resonance it has (higher resonance energy than pyridine), electron density is (AFAIK) equally distributed around the ring, whereas this isn't the case in pyridine.
Also, the reason I call this a 'paradox' (in my head, at least) is because I've been learning about how benzene is a really stubborn compound when it comes to getting it to react - it doesn't undergo normal alkene addition reactions and all that - yet pyridine is quite a bit more stable than benzene and it has a million different weird reactions going on (acid/base, nucleophilic, EAS, N-oxide formation, metal complexes, etc. etc.)
Apparently pyridine is more stable, yet more reactive due to the lone pair electrons on the nitrogen (which allow lots of crazy stuff). In general, is it true that more stable = less reactive? That's the basic assumption I've been using until now - based on thermodynamics (?) - but I've seen a couple of things that make me think it's not as simple as that. Sorry for the rambling, just blurting my questions out!

Stable does mean less reactive, but it's a relative term. A compound can be stable to one set of reactions and highly reactive towards another set.

I think the ‘paradox’ boils down to you wanting a simple black/white description for a much more complex observation. Note that benzene is not ‘ultra-stable’ or it could not get hydrogenated to cyclohexane.

To build on Tyberius' point: reactivity (reaction rate, determined by activation energy) is determined not only by the energy of the starting material, but also the energy of the transition state (or intermediate, by the Hammond postulate). Obviously, this depends on the exact reaction pathway under consideration. There are some cases where benzene is more reactive than pyridine and some where benzene is less reactive.

I don't think that this is primarily opinion-based; if anything it should be unclear, but what would be most helpful would be to have an answer explaining what the difference is.

I fully agree with @ortho that this is not opinion-based I personally have a problem understanding the question at all. However, it seems some have found it out. I urge these to edit the question and — more importantly — write a corresponding answer. Thus, I am voting to reopen.

Sorry for the confusion. I'll rephrase my question properly when I get a chance - I think I get the idea though. I was perplexed that pyridine (more energetically stable) could be more reactive than benzene (less energetically stable). I think what I was misunderstanding was that just because pyridine is more prone to certain reactions does not really change the fact that it's energetically more stable than benzene.

When you think about stability, think of it as a relative value, meaning in comparison to atoms of other elements. Therefore, pyridine is more stable than the benzene but less stable than other elements, in general.
You are correct when you say stability is inversely proportional to reactivity. But because stability is relative, it is very difficult to comment on the relative reactivity.
EDIT
Heat of hydrogenation of an alkene in which the double bonds are conjugated is not additive because such an alkene is resonance stabilized, and, therefore, its Lewis structure, which the calculation would be based on, is a poor representation of the molecule.
I have found 2 aspects :

Pyridine is less reactive, than benzene toward electrophilic aromatic
substitution, because nitrogen is more electronegative, than carbon and acts like an
electron withdrawing group. As a result, the meta hydrogen is substituted.
Pyridine is more reactive than benzene because the presence of nitrogen
enables pyridine to react with nucleophiles. Nitrogen is an electron withdrawing
substituents enables the aromatic ring to participate in such reactions and the
substitution happens in the ortho and meta positions.

Surely it's reactivity that's relative rather than stability? There's an objective measurement for stability (correct me if I'm wrong) - heats of hydrogenation, right? Reactivity is relative seeing as one molecule may be more prone to react than another due to factors like lone pairs, etc.

@Malnurtured stability is also relative. You may call an aluminum less reactive than chorine ,but not compared to oxygen

Fair, that's true in most cases maybe, but for C=C bonds you can measure the energy stored within them using heats of hydrogenation. I'm really just focused on organic compounds (and in this case two resonant, aromatic compounds), and for them I thought stability could be objectively ranked based on the energy content of the bonds?

Please see edit

@Malnurtured like always, happy to help :)

Six membered heterocycles
with an electronegative
heteroatom are generally
electron deficient compared
to benzene. Such compounds
are classified as p-deficient.
Electron-withdrawing heteroatoms decrease
the p-electron density at the carbon atoms
and are thus p-deficient relative to benzene


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Entropy and enthalpy at the melting point of benzene
For the melting of Benzene, since it is melting, it feels intuitive to think that the change in entropy will be positive (more disorder) and enthalpy will be positive (exothermic). However, when I look in my textbook, it says that the sign for delta S will be negative. Why is this? Wouldn't any process of melting from solid to liquid have a positive change in entropy? Here is the original question:
"The normal melting point of benzene, C6 H6 , is 5.5°C. For the process of melting, what is the sign of each of the following? (a) ∆H°, (b) ∆S°"

It would seem that either your textbook is mistaken or that you are interpreting your textbook incorrectly, for benzene has a positive entropy of fusion according to the NIST Chemistry WebBook.
Most substances will indeed have a positive entropy of fusion, as is suggested by your reasoning above. One notable exception, however, is helium at near-zero Kelvin temperatures, where it has a negative entropy of fusion. Presumably this is a quantum-mechanical effect, though I have no further explanation as to why this would be the case.

I'm going to assume that I'm not smarter than my textbook..lol. Here is the original question. "The normal melting point of benzene, C6 H6 , is 5.5°C. For the process of melting, what is the sign of each of the following? (a) ∆H°, (b) ∆S°" The answer says a) positive b)negative. Am I reading something wrong here?

@Joe, it seems to me that your textbook is incorrect, though I could be missing something important---for example, they ask for ∆S° where I would have expected ∆S. I recommend that you update your original post with the question statement and see if someone else responds.


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Calculating Gas Pressure with Water Vapour

$\pu{4.0 L}$ of oxygen was collected by the displacement of water at $\pu{105.0 kPa}$ pressure and $\pu{30 ^\circ C}$. If the vapour pressure of water at $\pu{30 ^\circ C}$ is $\pu{4.2 kPa}$, calculate the pressure due to oxygen alone.

In answering this question, is it correct to just take $\pu{4.2 kPa}$ from $\pu{105.0 kPa}$? I feel as though I must be missing something, considering this was given to us under the title "Challenging questions on gases", but I can't see what other calculations would need to be done.

Your analysis is correct. // Except do watch the significant figures. It is 105.0 kPa, not 105 kPa.


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Ester as a side product in the synthesis of bromoalkane from alcohol. Is it correct?
About 5 months ago, I answered a very interesting question albeit a very simple one.

Why do we need sulfuric acid in creating alkyl bromide and not in creating alkyl iodide from alcohol and hydrogen halide?

The question was why sulfuric acid is used to convert any alcohol to bromoalkane but not iodoalkane. Jan's answer quite pointed out the explanation:

hydrogen iodide is sufficiently acidic to protonate the corresponding    alcohol and drive the reaction. More acids are not
  needed.
hydrogen bromide is not acidic enough to drive the reaction. Adding    sulphuric acid protonates the alcohol partially and allows the
  reaction to proceed.
in the case of weakly soluble alcohols and hydrogen iodide, phosphoric acid may be used instead since it does not oxidize iodide (while sulfuric acid does).

Though I am satisfied with Jan's excellent answer, I further pointed out that sulfuric acid may react with hydrobromic acid to form bromine(aq.) and sulfur dioxide. This bromine may oxidize alcohol to form aldehyde and carboxylic acid and they two will form ester. This was my proposed explanation but I couldn't find any experimental result to back my explanation and heavily relied on theory.
Now recently Nurdrage, a renowned chemist famous for his synthesis of pyrimethamine has uploaded a video on Youtube about the synthesis of bromoethane and bromohexane from ethanol and hexanol using sulfuric acid. He also explained the concern of forming bromine as side product from the reaction of hydrobromic acid and sulfuric acid. He showed a purification technique by adding sodium bicarbonate to react with bromine to form sodium hypobromide.
Now my question is:
Is my explanation of forming ester instead of bromoalkane correct enough? Is there any possibility of forming minute amounts of ester? The reaction may be slow but it can happen before any purification is done (reacting bromine with sodium bicarbonate). If so, what can be done to remove any ester or prevent the reaction of ester formation?
Also related: Why is alcohol boiled with HBr for preparing alkyl bromide?

Technically it could form, but it would be hydrolyzed. If it did form, it would be in negligible quantities. Lots of other stuff could form, and they do, but in miniscule amounts ... bromine oxidized the alcohol to the aldehyde. The aldehyde undergoes aldol condensation. Another alcohol undergoes elimination. The compounds  react in a hetero diels-alder. So dihydropyrans can also form ... Yes, but not in quantities that are observable.

Fair enough but could you back your answer with a reference?

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why lattice energy have to be greater than IE and EA?
Why is lowering effect of the lattice energy not able to be greater than the combined net energy raising effects of the IE &EA?


text:
Can the electrons be above the Fermi level(in a band gap)?
I've been searching this online for some time. It seems to be ignored most of the time but I found this on one website and I'm just trying to confirm it's true.

Note that although the Fermi function has a finite value in the gap, there is no electron population at those energies (that's what you mean by a gap).

From the little I understood the Fermi function gives the probability to find an electron at a specific energy level. At temperature above 0 Kelvin f(E) is as shown above, meaning that there is a chance of finding an electron at an energy level in the gap. But then the website states that electrons can't occupy any state in the gap. If that is true the probability would be zero..

"Occupy any state in the gap" is an oxymoron. There are no states in the gap.

Then why isn’t the probability zero ?

Probability does not care about states. It says something to the effect of "if there were states with such-and-such energy, then their occupation would be..."

Then what is the purpose of the Fermi function? Can we trust it for states in the conduction band, the valence band but not for the gap?

We can trust it for any states. It is just that there are no states in the gap.

Let's say that summing up the probabilities for energy states in the gap gives 1% probability. Then the probability of an electron being in the valence band would be 99%  instead of 100%(the gap 'steals' from it). Now,when trying to find the probability for a specific energy in the valence band I know that it doesn't reflect the real probability since that 1% steals from it.   Do we assume that this 1% is on the top of the valence band?

Summing up the probabilities for energy states in the gap gives precisely 0%, because the number of states in the gap is 0.

That should be the case but the function gives a number, that's what's troubling me. It is also stated in the quote I included " although the fermi function has a finite value in the gap"

So what if it gives a number? Don't use the said number, that's all.

If I make a dice with 5 sides instead of 6 each probability would be 1/5 not 1/6. Fermi is like using 1/6 for 5 sided dice. The results given for the gap are not just wrong, they also decrease the probability for the energies in the valence band, giving me wrong values as well ( 1/6 instead of the real 1/5).

My point is , if the numbers for the gap are not real then we can't trust the numbers for the valence gap either. The gap probabilities 'eat' from the others.

Then normalize the probabilities. You'd have to do it anyway, gap or no gap.

Actual occupancy at an energy is a convolution of the Fermi function with the density of states. No states, no occupancy. This is fundamental solid state physics.


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Diamonds into protective barrier
Theoretically, is it possible to make a large, transparent piece of armor or a shield from a diamond?
Say, a piece of the said armor with a size  $7^\prime \times 7^\prime \times 6^{\prime\prime}$?
What kind of forces could this put up with?

Diamonds on a macroscale are probably one of the worst possible choices for a material that is supposed to be flexible, blend in with the environment and be shock-resistant and inexpensive. Diamonds are hard, but very fragile. Besides, the energy of the incoming projectile needs to be dissipated over a vast surface area quickly, but homogeneous crystalline materials don't provide that – they just crack locally and don't absorb much of an impact after that.

"Diamonds are hard, but very fragile" Herr @andselisk, if you turned that comment into an answer, I'd upvote it for that line alone ;-) Even if you just copy-paste your comment into an answer, it'd make for a beautifully succinct one IMHO. O:)

Though forming films of diamond like carbon onto surfaces such as knives, tools etc is an old idea which is still investigated. One could made nice patents out of that.

I'm not a gun owner or military expert, but from what I understand the requirements for the armor material are more or less uniform; it should:

Be able to withstand an impact and quickly dissipate the kinetic energy from a projectile. Diamonds are are hard, but very fragile. They cannot redistribute incoming impact from a projectile in the opposite or perpendicular direction. Once they crack, there is no more protection, and they cannot self-heal.
Have low weight. The density of diamond is $\pu{3.51 g cm-3}$, which is nearly 2.5 times heavier than kevlar (ca. $\pu{1.44 g cm-3}$).
Be easily molded in desired form, and/or be flexible (relevant for body armor). Process of crystal growing is rarely precisely controlled and often requires mechanical post-processing, which is ineffective due to hardness of the diamond and can potentially weaken the structure.
Have adequate thermal interface. Due to exceptionally high phonon thermal conductivity, the person wearing this armor will suffer from cold in winter and from heat in summer time, not to mention what damage an opponent with a flamethrower can cause.
Be modular and easily repairable. Diamond plates can only be replaced, so one needs to carry a stockpile of the shards.
Be suitable for camouflage (absorb light and other EM radiation, optionally). Diamond surface is hard to modify, it has low adhesion, so there is no easy way to tarnish the object if the diamond layer is the outer one.
Be optically transparent and don't cast any distortions (relevant for the bullet-proof windows). Due to high refraction index it's a questionable choice for a transparent shield.
Be inexpensive in terms of production and maintenance. No comments here.

To wrap it all up, for now classical macro-scale diamonds seem to be a very expensive and ineffective material for use as an armor material due to their physical properties. However, at a nanoscale the situation can be different, as nanodiamonds show somewhat different properties, yet not nearly fully investigated for military applications.
Also, you might want to check this topic on Worldbuilding.SE: Is diamond armor better than traditional armor?

Your original comment alone was excellent, really... this answer  is brilliant!! \o/ Kudos for the nanodiamond and worldbuilding links too O:)


text:
Do the amino groups in EDTA deprotonate first, in contrast to amino acids?
We know that in amino acids, carboxylic acid groups deprotonate first than the amine groups. For example, in alanine, $\mathrm{p}K_\mathrm{a,\ce{COOH}} = 2.34$, $\mathrm{p}K_\mathrm{a,\ce{NH3+}} = 9.69$ according to this page.
But according to the structures of EDTA as searched from Google, and from this extract from p.325 of Analytical Chemistry, 8e by Christian, et al.:

At sufficiently low pH, the nitrogens can also be protonated and this diprotonated EDTA can be considered a hexaprotic acid. However, this occurs at a very low pH and EDTA is almost never used under such conditions.

the hydrogens of these amino groups seem to be deprotonated first before those of the four carboxylic acid groups. Why is this so?

Before you ask "why", ask "if".

How many covalent bonds can N form with other elements?

There used to be a place on the internet where you could enter compounds and things such as acidity constants were calculated, including which proton deprotonated in which order. Alas, I can’t find the site any more.

What do you mean @IvanNeretin? It doesn't make sense to me since carboxylic acid groups are usually more acidic than amino groups in the examples I encounter, which is due to the electronegativity of the protic oxygen, resonance and inductive effects felt by the same oxygen atom. Then I see this hydrogen held by the amino groups getting deprotonated first before the other hydrogens held by the carboxylate groups, so I asked why.

I mean you should not blindly trust some random images brought to you by Google image search. Trust your logic; it is sound.

@IvanNeretin, okay, I added another reference that supports this ionization behavior of EDTA.

Dissolved in water, EDTA can have net charges varying from -4 to +2, from fully deprotonated to all four carboxylic acid groups and two amine groups protonated.
It is the fully deprotonated form that binds to divalent cations (see e.g. https://chemistry.stackexchange.com/a/81988/72973). If necessary, deprotonation occurs upon binding (see e.g. https://doi.org/10.1016/S0301-4622(99)00047-2).
The intermediate protonation states occur as zwitterions both in aqueous solution and in some of the crystal forms. Disodium EDTA is an example:

Source: https://pubchem.ncbi.nlm.nih.gov/compound/57339238#section=2D-Structure

[OP] the hydrogens of these amino groups seem to be deprotonated first before those of the four carboxylic acid groups. Why is this so?

In aqueous solution, this carboxylic acid deprotonates before the ammonium. If you see other statements, they ignored to mention the prevalent tautomeric state. The situation in organic solvents might be different. The neutral species might have the carboxylic acid groups protonated and the amines deprotonated when EDTA is dissolved in a no-ionic organic solvent. The cationic or anionic species would probably not be soluble, so for those species, the question is moot.

The carboxylates deprotonate first followed by the amines.  The pKa values for the carboxylates ar 0.0, 1.5, 2.0, and 2.66.  The amines have pKa values of 6.16 and 10.24.  Ref. Harris.  Quantitative Chemical Analysis.

Yeah, that was wrong, I edited it now, but my question still stands.


text:
Designing a device to separate water from water-ethanol mixture using dry ice baths
I've been trying to design a device to increase the percentage of ethanol and water, or to at least concentrate the water-alcohol mixture without using distillation.
The way I'm trying to do this is by using a dry ice bath (dry ice+propanol). I then insert a stainless steel vessel in the bath with the said water-alcohol mixture (between 30-60% ethanol to begin with). Going by the ethanol-water phase diagram and the temperature of the ice bath (-78C), I should, in theory, be able to concentrate the mixture up to ~80% ethanol.
Where this goes wrong is that instead of cleanly separating into water ice and concentrated water-ethanol mixture, what I get is a kind of ice-alcohol slushy.
Any suggestions on how to get a clean separation and get ice that I can scoop out?
Thanks!

If I remember correctly,  3Å molecular sieves haves been used for the separation of ethanol–water mixture for decades. I suspect you might want to work at the ambient temperatures to utilize the full capacity.

You’re probably not going to get anywhere near a clean separation and definitely not going to get scoopable ice. I guess there will be people who can point out why. I personally don’t like ‘any suggestions’ questions, they do seem overly broad in my point of view.

One suggestion: try cooling the alcohol-water mixture more slowly. Slower crystal growth may produce larger ice crystals that occlude less of the ethanol and are easier to separate by filtration. How to do it? Place the steel beaker in the outer bath before adding dry ice to the outer bath, then add the dry ice incrementally to gradually bring the temperature down to where you first observe ice crystals, then keep the temperature constant for a while as crystals form. Then lower the temperature a degree or two and repeat the cycle. Probably better not to stir the inner mixture. It will probably help to use a thermometer to carefully monitor the temperature of the alcohol mixture.


text:
How to find all published Pharmacophore Structure of a Drug target?
Other than literature review how do you find the structures of Pharmacophore of a drug target? I am searching Pyruvate Dehydrogenase Kinase (PDK) inhibitors and I have reviewed lots of literature but I've only found few structures. I have tried searching google by "PDK inhibitors patent reviews" "PDK inhibitors general review" but I am not finding any concrete literature. There are 4 sub classes: PDK1, PDK2, PDK3 and PDK4.
I am looking for drug candidates which are potential Pharmacophore for this drug class.


text:
Uv vis and fluorescence spectroscopy: sensitivity
Is there any reason why fluorescence spectroscopy is said to be more sensitive than Uv-Vis?

Fluorescence is a 'zero background' or absolute type of measurement meaning that single photons can be measured against a 'dark' background so the sensitivity is huge, and limited by the fraction of light caught and efficiency of the detector.
Absorption is a relative measurement as it is the difference in two large numbers, one the light intensity with no sample $I_0$ and the other when the sample is present I. In terms of photons these intensities are huge, say $I_0=10^6$ and a 1% change would reduce the number of photons to $10^6-10^4$; still a large number and so the absorption measurement is looking for the small difference in two large numbers. This measurement is limited by noise in detectors and other electronics.   (Its a bit like trying to weigh the captain of a ship by weighing the ship with him standing on the deck and then when he is not there.)

I'll just add that the correct mathematical abstraction for sensitivity is the signal to noise ratio.

yes I should have added that point.

In fluorescence, what is detected is really photons emitted.  The measure is done in absolute term (versus 0 , the dark) and the response can be easily amplified.
Conversely, in absorption is a difference that must be measured. And this can of course be arbitrarily tiny.
It is like massing a sample. Is easy to weight a mg than detecting a mg difference among two " about 10 kg " samples.
This general consideration is exacerbated by other practical details, as listed in the  answer by Dan.
Also your consideration about geometry applies, which again is dictated by the above.
Edit: Porphyrin answers appeared after I submitted this. Else I wouldn't have answered as this is basically a duplicate.

Absorbance may occur from non-analytes, while fluorescence is often quite specific to an analyte or a fluorophore-tagged analyte. Also, instrumental setup could play an effect. UV-Vis specs are often horizontal in terms of incident beams, sample, and the detector. Fluorescence specs are often at 90 degrees to "extract" the fluorescence from the sample to the detector. If you are only "extracting" what you are interested in, I'd imagine the S/N would be better, yielding higher sensitivity.


text:
How to find out the state symbol of a compound by looking at its chemical formula?
Like, in the equation between rubidium and water:
2Rb(s) + 2H2O(l) → 2RbOH(?) + H2(?)
How do we know what state symbol to use in 2RbOH and H2? I'd like to do this just by looking at the chemical formula, but it's OK if I have to memorize a list of stuff to be able to do this.

If you don't get that compounds can be in different states depending on temperature...

You can't do this by:

just by looking at the chemical formula

But instead look up the properties of the compounds in question (Wikipedia is a good source of data).
$\text{RbOH}$ is highly soluble in water, so assuming there's an excess of water present, it will end up being dissolved in the water. So its state symbol will be $\text{(aq)}$.
Hydrogen on the other hand is a gas (at room temperature and above) and highly insoluble in water: it will leave the reaction as gas and its state symbol will be $\text{(g)}$.
So without knowing the physical/chemical properties of the reagents/reaction products, the state symbols cannot be determined.
Note that the state symbols are often idealisations: solid or liquid compounds are rarely completely soluble or completely insoluble. The same is true for gases. Even water is a case in point: liquid, $\text{(l)}$, at room temperature but with significant vapour pressure.

In many cases, it helps to know either something about the type of reaction occurring, or the temperature at which the reaction is taking place.  For example: water is generally a product of combustion reactions, but since these take place at high temperature (we are burning something after all) the water will be present as a vapor (g).


text:
Why butane exist as a liquid in a lighter ?
This is my go at explaining after researching ...
Butane is light and normally would be a gas at standard atmospheric temperature and pressure due to Low boiling point (-1degrees)
The butane gas is in liquid state in the lighter . This is because , when it is placed in the body of the lighter , it is placed there under very high pressure as compared to the open air in standard atmospheric temperature and pressure .
Increasing the pressure of the butane gas forces the molecules closer together . This is because increase in pressure will also lead to increase in average kinetic energy meaning molecules need to 'hold' onto each other more to prevent themselves from breaking their bond. (LDF)
With that theory, that means increase in pressure leads to increase in boiling point of the molecules because more heat is needed to break the bonds and move apart .
Hence , When the butane gas are compressed at high pressures , (at a certain pressure for butane) , the alkane will undergo a phase change from gas to liquid.
Now , am I right to say that the vapour pressure is equal to the pressure of the butane gases in the small container , That's why they undergo a phase change immediately ?

Now, am I right to say that the vapour pressure is equal to the
  pressure of the butane gases in the small container, that's why they
  undergo a phase change immediately?

You seem to have answered your own question.
The pressure inside the lighter container is indeed the vapour pressure of the mixture of n-butane and iso-butane that it contains, at ambient pressure.
Here is the pressure-temperature dependence for iso-butane:

(Source)
The high vapour pressure inside means that when you open the valve, a gaseous mixture of both isomers will come streaming out and some of the liquid mixture will spontaneously evaporate to replace it.


text:
Why is the probability for the signal at 160 in the bromine mass spectrum twice as high as for the signals at 158 and 162?
The mass spectrum in Bromine, with the molecules $\ce{^{158}Br2+}$, $\ce{^{160}Br2+}$ and $\ce{^{162}Br2+}$:

As you can see, the $\ce{^{160}Br2+}$ is almost double in intensity compared to the $\ce{^{158}Br2+}$ and the $\ce{^{162}Br2+}$ peak.
The book I am reading simply states that this is because

The probability of two different isotopes occurring in a $\ce{Br2}$ molecule are twice that of the same isotope appearing in a $\ce{Br2}$ molecule.

This is supported by the $\ce{^{160}Br2+}$ peak, formed from the $\ce{^{79}Br}$ and $\ce{^{81}Br}$ isotopes.
Likewise, $\ce{^{158}Br2+}$ peak is formed from two $\ce{^{79}Br}$ isotopes and $\ce{^{162}Br2+}$ is formed from two $\ce{^{81}Br}$ isotopes.
However, I am confused by the explanation given by the book above.
Why is the probability of two different isotopes occurring in a $\ce{Br2}$ molecule twice that of the same isotope appearing in a $\ce{Br2}$ molecule?

All possible arrangements of $\ce{Br2}$ molecule:

$\displaystyle 79 + 79 = 158$
$\displaystyle \color{red}{79 + 81} = 160$
$\displaystyle \color{red}{81 + 79} = 160$
$\displaystyle 81 + 81 = 162$

The amount of $\ce{^{79}Br}$ and $\ce{^{81}Br}$ in nature is roughly the same, thus each permutation is equally probable. There are two arrangements that lead to $160$. While $158$ and $162$ each have only one arrangement. Therefore $160$ is twice as likely to be found compared to other masses.

Because there are only ever two isotopes of Bromine, the outcome of 160 amu is currently illustrated as more likely than 162 or 158.

A way to understand this that may be familiar is that of the Punnett square from biology, since the two isotopes have nearly 50/50 split in nature.
\begin{array}{c|cc}
              &  \ce{^{79}Br}   &  \ce{^{81}Br}   \\\hline
\ce{^{79}Br}  &  \ce{^{158}Br}  &  \ce{^{160}Br}  \\
\ce{^{81}Br}  &  \ce{^{160}Br}  &  \ce{^{162}Br}  \\
\end{array}
When breeding two hybrids (Aa x Aa), it's twice as likely to get a hybrid (Aa) than to get either homozygote.  Similarly, here you have twice the chance to get a 'hybrid' $\ce{^{160}Br}$ than a particular 'homozygote' $\ce{^{158}Br}$ or $\ce{^{162}Br}$ .
However, I would disagree with the wording of the statement:

The probability of two different isotopes occurring in a $\ce{Br2}$ molecule are twice that of the same isotope appearing in a $\ce{Br2}$ molecule.

The probability is actually identical of two different isotopes occurring and any pair of identical isotopes occurring.  This could be worded better:

The probability of two different isotopes occurring in a $\ce{Br2}$ molecule are twice that of a particular same isotope appearing in a $\ce{Br2}$ molecule.

@Mithoron I just thought it might be a good (different) way to visualize it for someone who might be used to seeing this from high school biology, but not used to it in this context.

I see what you did, well OK, I just wanted to say that this question shouldn't be here at all.

Your better wording uses some not great english: "of a particular same isotope" just doesn't read well. Better might be "twice that of a particular isotope appearing twice in a $\ce{Br2}$ molecule".


text:
Exactly what happens when nitrogen is converted into C14
When a cosmic ray ( neutron) strikes a nitrogen atom in the atmosphere, does the cosmic ray knock one of the nitrogen electrons out of its orbit and into one of the protons in the nitrogen nucleus? Does the cosmic ray just continue on its way through the atmosphere? Is the electron that got knocked out of orbit absorbed  into the proton making it become a neutron? And when radioactive decay occurs in the c14, does the neutron that used to be a proton spit out an electron that becomes part of that atom or does it go out into space?

https://en.wikipedia.org/wiki/Carbon-14#Natural_production_in_the_atmosphere . Not really chemistry, IMO.

@Gert - Glenn Seaborg might beg to differ! But, yes, nuclear chemistry has been ceded in entirety to nuclear physics. Plus, I would add that the cross section for the 14N(n,p)14C reaction rises again above 100keV, so it is not purely thermal neutrons that take part.

Please bare with me as I know very little about chemistry and physics. So , when 14C is formed in the upper atmosphere, the random negative neutron that strikes the atom of nitrogen is then absorbed into one of the protons in the nucleus along with one of the atoms electrons, this resulting in an atom that has 6 electrons, 8 neutrons, and 6 protons ( one of the protons having absorbed an electron of the nitrogen atom and a free neutron from space) ?

Also, when the 14C nucleus erodes and turns back into 14N , does the proton that became a neutron emit an electron that then orbits that atom and also emit a neutron that then exits the atom and becomes a free negative neutron?

First, the reaction I wrote in my comment means you start with a 14N, a neutron goes in, a proton comes out, and you are left with 14C. For you most recent comment, try to write out that reaction and see what the result would be (hint - you appear to be trying to create nucleons from nothing).

First of all, thank you very much for taking the time to answer my questions. I am just curious as to what exactly happens to all the particles involved in this event . I have read dozens of articles online but am curious if the negative neutron from space that collides with the nitrogen atom becomes part of one of its protons along with one of the electrons and if both of those particles are then spit back out of the neutron they created when the nucleus erodes and becomes nitrogen again

I guess what I am trying to say is what happens to the electrons of 14N and Where do the electrons come from when 14N is recreated at beta decay. I understand that the proton becomes a neutron when a neutron is added and I understand that a neutron becomes a proton in Beta decay but where does the 7th electron come from in the newly formed atom and where does it go when nitrogen becomes carbon?

It's a nuclear reaction. The normal electrons around the original nucleus do not take part in it, except for very heavy elements where the inner electrons (1s) are very close to the nucleus. Neutrons are not negative, but neutral, and beta decay is when one gives off an electron and thereby turns into a proton.

So does an atom of 14C have 7 electrons or 6?

@Dale An atom is electrically neutral and has as many electrons as its core has protons.

So the newly formed 14C drops one existing electron from the atom’s Shell and does it spit out an up quark from one of its protons or does the impact from the electron from space cause the bosons that already exist in the quark to change the quark to a down quark (is the quark changed in the proton or is it replaced). In beta decay do the bosons just decide they want to change a down quark into an up quark so that the nucleus can be stable and when this is done do these bosons actually create a new quark? I am assuming that they must because an up quark has to come from somewhere .


text:
Why does an OH bond absorb a higher frequency than a CH bond?
I'm wondering why a bond between an O and an H is going to absorb a higher frequency than a bond between a C and an H (any hybridization).
O is heavier than C in terms of atomic masses, and I'm assuming the O-H bond will be weaker because of the electronegativities, so why would it absorb a higher frequency?
Thank you.

The O–H bond is stronger.

@orthocresol: you might want to elaborate a bit on why that results in a higher frequency.

See https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/spectrpy/infrared/irspec1.htm

First, let me just put down the approximate $\ce{X-H}$ stretching frequencies (where, $\ce{X = O, C}$) just as a reference.
$$\begin{array}{c}
 & \ce{X-H}\ \text{stretch}\ \mathrm{(cm^{-1})}  \\ \hline
\ce{X = O} & 3600  \\ \hline
\ce{X = C} & 3000 \\ \hline
\end{array}$$
Now, as a simplification I will first discuss a diatomic molecule, thought of as a two balls of some mass $\mathrm{m}$ linked together with a spring. Solving, Schrodinger's equation for this system (called a harmonic oscillator), yields the following relation:
$\mathrm{\bar{\nu}} = \frac{1}{2\pi c}\sqrt{\frac{k}{m}}$
For, dissimilar masses, (say, $\mathrm{m_1}$ and $\mathrm{m_2}$), we can switch to the centre of mass coordinates, and the $\mathrm{m}$ in the preceding equation is replaced with the reduced mass ($\mu$) defined as,
$\frac{1}{\mu} \equiv \frac{1}{m_1}+\frac{1}{m_2} $
Consequently, $\ce{C-H,O-H}$ bonds in general have much higher stretching frequencies than do corresponding bonds to heavier atoms other than $\ce{H}$.The mass effect becomes evident when deuterated isotopes are examined. The stretching frequency of a free $\ce{O-H}$ bond is ca. $3600\ \mathrm{cm^{-1}}$, but the $\ce{O-D}$ equivalent is lowered to ca. $2600\ \mathrm{cm^{-1}}$. Since deuterium has a mass twice that of hydrogen, the mass term in the equation changes from 1 to 1/2, and the frequency is reduced by the square root of 2.
Naively, one would expect the frequency of the $\ce{C-H}$ bond to be higher, based on the preceding discussion, but there is another factor at play, the force constant $k$. This, is different for $\ce{O-H}$ and $\ce{C-H}$ bonds, and depends on the bond strength (stiffness of the spring)
Bond strengths can be estimated from bond disassociation energies (BDEs), which is defined as the standard enthalpy change when a bond is cleaved by homolysis, with reactants and products of the homolytic cleavage at $\mathrm{0\ K}$. This values are frequently tabulated, and can be easily found online (here for example; caveat the values here are at $\mathrm{298\ K}$)
$$\begin{array}{c}
 & \ce{X-H}\ \text{BDE}\ \mathrm{(kJ\ mol^{-1})}  \\ \hline
\ce{X = O} & 428.0\ (21)  \\ \hline
\ce{X = H} & 337.2\ (8) \\ \hline
\end{array}$$
Anyway, this tells you that the $\ce{O-H}$ bond is stronger; Why? Well, simply look at the approximate atomic radii of Oxygen and Carbon. The Oxygen is about $\mathrm{48\ pm}$ while carbon is about $\mathrm{67\ pm}$. The longer atomic radii implies a longer $\ce{C-H}$ bond length, ergo a weaker bond.
Stronger bond, implies larger $k$, which in turn means greater frequency of vibration.
So, the $\ce{O-H}$ bond is NOT weaker. Remember, when we define bond disassociation energy, we are looking at homolytic cleavage. I really don't understand the point you try to make about electronegativities.

We know that C-H, O-H, and N-H bonds absorb at a higher frequency due to Hydrogen's lighter mass (using Hooke's Law, we know that a lighter mass means it has a higher vibration, and therefore a higher frequency). Now although Oxygen may be heavier than Carbon, remember that Oxygen-Hydrogen bonds are stronger than Carbon-Hydrogen bonds (as there is a higher electronegativity difference between the two, therefore it has a higher ionic character than C-H which instead has a higher covalent character). Again using Hooke's Law, a stronger bond means there is higher vibration and therefore a higher frequency (just as triple bonds have higher frequencies than double bonds). Think of the effect of this stronger bond as 'overpowering' the effect of the heavier Oxygen atom. Don't worry, OChem got me messed up too.


text:
How do I solve oxidation numbers for unknown elements?
I am doing very well with oxidation numbers but I have a few problems. I know all the rules but when I get an element where I can't use any of the constant oxidation numbers what do I do?
For example: $\ce{HCNS,Ag2S,Bi2S3,KSCN}$ etc. ?

related https://chemistry.stackexchange.com/questions/47369/how-to-assign-oxidation-number-of-sulfur-in-thiocyanate-anion?rq=1

You shouldn’t really need to use the shortcut rules (fluorine having $\mathrm{-I}$, hydrogen having $\mathrm{+I}$ and oxygen having $\mathrm{+II}$ ‘except in peroxides’) with the exception of really simple compounds where it all falls into place immediately. Instead:

If the compound is molecular or contains a molecular ion, draw the Lewis structure.
Assign each bonding electron pair to the more electronegative bond partner.
Count the electrons each atom ends up with, substract from the number of electrons it should have according to its position in the periodic table, invert that number’s sign.
If at step one you are only left with ions, just take the ion’s charge as oxidation state.
You did it right if the overall sum of oxidation states matches the molecular ion’s charge or zero for the overall structure.

(Note: I am simplifying here by assuming that compounds are either molecular or ionic. That, of course, is most certainly not the case but the very crude generalisation gets us very far in this case.)

To work through one of your examples:
$\ce{KSCN}$ is an ionic compound consisting of potassium cations and thiocyanate anions. According to step 4, we can assign potassium the oxidation state of $\mathrm{+I}$.

The structure of thiocyanate is $$\ce{S=C=\overset{-}{N} <-> \overset{-}{S}-C#N}$$
In both cases, the terminal atoms have enough lone pairs to complete their octet.
The electronegativites are $\ce{N}>\ce{S}>\ce{C}$. Thus, we assign the bonds so that:

sulphur and nitrogen each have eight valence electrons
carbon has zero

This is independent of the resonance structure we used.
Calculating:

Sulphur:
eight electrons, should have six.
$\displaystyle-(8-6) = \mathrm{-II}$
Carbon:
zero electrons, should have four.
$\displaystyle-(0-4) = \mathrm{+IV}$
Nitrogen:
eight electrons, should have five.
$\displaystyle-(8-5) = \mathrm{-III}$

Has been done before.
To check:
$\displaystyle \mathrm{+I + -II + +IV + -III = \pm0}$ (for $\ce{KSCN}$)
$\displaystyle \mathrm{-II + +IV - III = -I}$ (for $\ce{SCN-}$)


text:
Can citric acid create a polyester?
Citric acid has both alcohol and carboxylic acid groups, so could it "esterify" with the other citric acid molecules in solution?
I notice that it is a tertiary alcohol. I don't suppose that would significantly hinder esterification, would it? Then again, it's not as if citric acid simply polymerizes with other citric acid  molecules in solution anyway.

Tertiary alcohols tend to lose water and produce alkenes, if you press a little too hard. Obtaining the ester you want seems to be possible, but not simple.

The answer is yes, but you can't make it by normal means because citric acid decomposes to easily. All I did was search on Google for "poly citric acid." You'll find stuff like this: http://www.sciencedirect.com/science/article/pii/S0011916416301813

The esterification of carboxylic acids with tertiary alcohols usually occurs by an SN1 mechanism wherein the oxygen of the alcohol is protonated and a tertiary carbocation is formed. It in turn reacts with the carboxylic acid to form the tertiary ester. The hydroxyl group in citric acid, while tertiary, is alpha to a carboxyl group. A cation alpha to a carboxyl group (-CO2H) is a destabilized entity. This mechanism would be precluded. Traditional Fischer esterification is likely sluggish owing to steric hinderance.


text:
Atomic mass number and mass number in atomic symbol
In chemistry classes or even many different branches of chemistry textbooks, I am told that the atomic symbol
$$\Large ^A_Z\ce{X}$$
includes the following notations:

$A$ – "atomic mass" or "mass number";
$Z$ – "atomic number".

But, because I want to make myself to remember easily, I would prefer to call $A$ "Atomic mass number" instead of simply "mass number". I think many say that I am making a big deal and unnecessary things. But as a non-native learner, if I can confirm that using the term "atomic mass number", it would make me easier to remember the term.

The IUPAC preference for A in $\ce{^A_ZX}$ is mass number. The term would also apply to a molecular fragment, for instance in a mass spec. So depending on the context atomic mass number or molecular mass number might be used to be more specific.
The problem with you arbitrarily defining A in $\ce{^A_ZX}$ as atomic mass number is that various references might just use mass number. So depending on context you'll need to be able to decide which possibility is the correct one. Of course in spite of IUPAC definitions there are no doubt multiple variations in the past and current literature.

Maybe I'm missing something, but I don't see how atomic mass number is easier to remember than just atomic mass. If it helps with distinguishing A from Z, Z is also called the proton number.
Not exactly a proof, but wikipedia includes "atomic mass number" as a synonym for mass number, so it is probably valid terminology.

In nuclear physics one would call A the number of nucleons. Of course this ties to the mass pretty directly (ignoring binding energy).


text:
Selecting parent chain considering substituent (prefix) functional groups

I am attempting to derive the name of the above compound. Based on the IUPAC rules, I can tell that the carboxylic acid group should be taken as parent i.e suffixed and numbering should start from there. However, since the lengths of both possible choices are equal, there is some ambiguity in my mind as to the choice of parent chain.
Further, what would happen if the -OH were shifted to the carbon adjacent to its current position? Or the amine?

What do the IUPAC rules say about the priorities of hydroxyl and amino groups?

Yes, I know that if I were selecting parent among -OH and -NH_2, it would be -NH_2. But that is not what we are doing here, hence my confusion.

According to the usual nomenclature rules, two equivalent names are possible for the compound that is given in the question:

6-amino-4-(2-hydroxyethyl)hexanoic acid
4-(2-aminoethyl)-6-hydroxyhexanoic acid

In order to identify the preferred name, we have to dig deeper in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), where we finally find two potentially relevant rules:

P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.

and

P-45.5.1 The preferred IUPAC name is the name that is earlier in alphanumerical order (see P-14.5). Alphabetic letters are considered first in the order that they appear in the name; all Roman letters are considered before any italic letters, unless the latter are used as locants or are a part of a compound or composite locant, for example, ‘N’ and ‘4a’. Then, if still there is a choice, numerical locants are considered in the order of their appearance in the name.

Since P-45.2.3 comes first, the preferred name is 4-(2-aminoethyl)-6-hydroxyhexanoic acid and not 6-amino-4-(2-hydroxyethyl)hexanoic acid because the locant set ‘4,6’ is lower than ‘6,4’.

By way of comparison, a similar example given in the (corrected) Blue Book is 2-(2-bromoethyl)-4-chlorobutan-1-ol and not 4-bromo-2-(2-chloroethyl)butan-1-ol since the locant set ‘2,4’ is lower than ‘4,2’.
Remarkably, also according to the subordinate rule P-45.5.1, the preferred name would be 4-(2-aminoethyl)-6-hydroxyhexanoic acid and not 6-amino-4-(2-hydroxyethyl)hexanoic acid because ‘aminoethyl-hydroxy’ is earlier alphabetically than ‘amino-hydroxyethyl’. However, the preferred name for the example given in the Blue Book would have been 4-bromo-2-(2-chloroethyl)butan-1-ol and not 2-(2-bromoethyl)-4-chlorobutan-1-ol since ‘bromo-chloroethyl’ is earlier alphabetically than ‘bromoethyl-chloro’. This example was actually given in the Blue Book by mistake but it has been corrected as part of the official Corrections to Nomenclature of Organic Chemistry. IUPAC Recommendations and Preferred Names 2013.


text:
How to derive Pauli Exclusion Principle without assuming anti-symmetry?
So, it appears that the statement of the Pauli Exclusion Principle is equivalent to the statement that fermions are anti-symmetric. That is, if you assume that fermions are anti-symmetric, then you can derive the Pauli Exclusion Principle. If you assume the Pauli Exclusion Principle, you can derive anti-symmetry in the wavefunction.
At some point, one of these must have been proved independently of the other. I can't seem to find such a proof anywhere. I think this confusion also comes from the fact that spin is a difficult thing to conceptualize, let alone measure. Does some experimental data to necessarily imply one of the two? I guess, some insight on how spin was discovered would probably help.

I'm reading through this brief article that explains how the exclusion principle came about: https://www.aps.org/publications/apsnews/200701/history.cfm

The Pauli Exclusion principle is determined from experimental facts alone.
First, its consequences were observed and then it was incorporated in the theory. See Kaplan´s article for a more erudite answer. Also, see Towler´s exposition. Although, he dwells in foundational issues.
Important: they are not equivalent unless you consider only totally symmtric/antisymmetric functions! The antisymmetry is assumed, it cannot be derived rigorously from particle independence nor tested with experiments.
As for getting to the conclusion that the wavefunction of fermions has to be antisymmetric here is a weak argument without requiring the Pauli Exclusion principle.
You could also say that since you cannot follow the trajectory of particles therefore they are indistinguishable -- at least, if they are of the same kind. That is, the permutation of any two of them
$$ \Psi(\pmb x_1, \pmb x_2, ...) \to \Psi(\pmb x_2, \pmb x_1, ...) $$
maintains the same physical description
$$ |\Psi(\pmb x_1, \pmb x_2, ...)|^2 = |\Psi(\pmb x_2, \pmb x_1, ...)|^2.$$
The permutation can be described as the action of an operator that satisfies $P^2 = 1$.
There are only two ways to acomplish this. Either the wavefunction is symmetric (the eigenvalue of $P$ is $-1$) under the exchange or antisymmetric (eigenvalue $-1$).
$$ \Psi(\pmb x_1, \pmb x_2, ...) =  \Psi(\pmb x_2, \pmb x_1, ...) \quad \text{  Symmetric: bosons}$$
$$  \Psi(\pmb x_1, \pmb x_2, ...) = -\Psi(\pmb x_2, \pmb x_1, ...)\quad \text{Antisymmetric: fermions}$$
Therefore, if $\pmb x_1 = \pmb x_2$ then $\Psi(\pmb x_1, \pmb x_1, ...) = -\Psi(\pmb x_1, \pmb x_1, ...) = 0$. That is, Pauli Exclusion principle.
The reason as to why this argument, that is the one that was elaborated to justify Pauli Exclusion principle and is seen nowadays in most textbooks, is weak is neatly explained in the reference mentioned above.


text:
Magnesium combustion heat
When you are going to burn magnesium you need to wait a few seconds before it actually starts to burn. To get to this point it needs a certain amount of energy/heat. How much does it need?

Ummm, it's not a matter of specific amount, just at least a small bit should get very hot and then reaction gets self-sustaining.


text:
Given this liquid-liquid phase diagram, is the green line the nitrobenzene-rich phase?

  The temperature-composition diagram for hexane and nitrobenzene at 1atm. The region below the curve corresponds to the compositions and temperatures at which the liquids form two phases. The upper critical temperature, $T_\mathrm{uc}$,  is the temperature above which the two liquids are miscible in all proportions.

Am I right in assuming that the right side (green line) is the nitrobenzene rich phase? Is that always the case? Whichever one has the side with a mol fraction of 1 that side is more abundant? If you don't understand my question just tell me, and I'll clarify

Here is the same diagram with more annotation from the physical chemistry textbook by Atkins:

It shows that above the temperature $T_{uc}$, there is only one phase (the substances mix in any ratio). Below that temperature, there are two phases, with compositions shown by the solid lines. The horizontal line shows that if you try to make a mixture with a mole fraction of $a$ at the given temperature, you will get two solutions with mole fraction $a'$ and $a''$ instead. You can calculate the approximate ratio of the volumes using this technique.

Am I right in assuming that the right side (green line) is the nitrobenzene rich phase?

Yes, because the x-axis is the mole fraction of nitrobenzene, the phase with higher fraction of nitrobenzene will be that on the right, as you say. It does not necessarily mean that the fraction is higher higher than 50% on the right and lower than 50% on the left, though.

first of all, the picture should have two arrows on top of the bars that are perpendicular (ie parallel to each other), because those represent the two components. Lets just call one A and the other B. If you choose a point on the right hand side of the curve, it means that you have little of component B that is "rich in" component A.


text:
Is there a way to calculate bond energies using computational chemistry?
I'm currently trying to see if a particular set of ligands will react with a metal ion to form a complex. While I'm waiting for the wet lab results to arrive, my adviser suggested we look into computational chemistry to see if we can model the proposed structures theoretically.
Neither of us (or anyone in the department for that matter) have any experience in computational chemistry. I got my hands on a copy of Avogadro and GAMESS-US and have optimized the geometries of the complexes we think might arise. Is there a way to determine if the complexes would form? Thanks.

Is there a way to calculate bond energies using GAMESS?

Yes. You would normally model the bound and unbound states
$$\ce{ML_{n - 1} + L -> ML_{n}}$$
The definition of bond energy is based on the enthalpy of reaction in the gas phase (thanks DSVA!), so you would calculate it as follows:
$$\Delta H^\circ = H_{\ce{ML_{n}}} - (H_{\ce{ML_{n - 1}}} + H_{\ce{L}})\text{.}$$

While not strictly bond energy by definition, a value for the reaction Gibb's free energy might also be useful in some contexts:
$$\Delta G^\circ = G_{\ce{ML_{n}}} - (G_{\ce{ML_{n - 1}}} + G_{\ce{L}})\text{.}$$
This takes entropy into account.
For the calculation of Gibb's free energies, most codes calculate the molecular partition function from translational, rotational and vibrational contributions and get enthalpy and entropy from it.
Due to the vibrational contributions, you'll need to diagonalize the Hessian matrix of the system. This will give you an association (minus dissociation) energy value.
(You might take solvation into account through explicit/implicit solvation, use decent level of theory, validate/benchmark, etc.)
There are a lot of approximations and considerations in the process. More can be found here (some equations) and here (more practical).
Observe that
$$K = e^{- \frac{\Delta G^\circ}{R T}}$$
is the equilibrium constant associated with the process above, which is called the stability constant, formation constant or binding constant.
This is widely used to "measure the strength of the interaction between the reagents that come together to form a complex" (source).
Bear in mind that $\Delta G^\circ$ goes inside an exponential, i.e., small errors in $\Delta G^\circ$ become (exponentially) large errors in $K$, so $\Delta G^\circ$ might be more useful in a computational context than $K$ itself.

Is there a way to determine if the complexes would form?

This is complicated.
Your reaction might be too slow though, i.e., actual formation might depend on transition states. Furthermore, there might be faster processes in competition that you might miss.

Going back to this after a few months, how exactly do you get the Free Energy of the Metal? I understand modelling the Ligands and getting their SPE and Vibration Energies, but I'm not sure where to get the data on the metals.

You would get data from the complex and the complex minus a ligand (I edited the question to highlight it). You normally don't need the metal alone to do it for complexes.

Aren't bond energies usually defined as the difference in enthalpy rather than gibbs free energy? See: https://goldbook.iupac.org/html/B/B00701.html

Thanks @DSVA, you're right! I edited the answer to make this definition clearer.

Thanks @FelipeS.S.Schneider! So you're saying I get the energies of the ligands, multiplied by the coordination number and subtract that from the energy of the bound state and I'll get the enthalpy of formation?
Is there a way, then to measure the entropy of the system so I can find the Free Energy Change? Or would the enthalpy suffice to say if the reaction is feasible? Thanks again for all the help!

This and this might help.


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CH2 group in alkyl glucoside gives two separate HNMR signals

Zoom in:

It seems like the CH2 next to the sugar gives two different NMR signals which is very interesting. Could someone explain me why?

The real question is why all the other $\ce{CH2}$ groups don’t ;)

Without the sugar, those two hydrogen atoms are enantiotopic, since substitution of those protons would lead to different enantiomers. Achiral measurements still will appear identical though.
Once the sugar is attached, the two hydrogens are diastereotopic, and are chemically and magnetically different. Especially so close to the chiral environment of the sugar, these two hydrogen see enough of a difference in their local environments that their shifts are different.

I would like to expand on Zhe's excellent explanation. The concept of why these hydrogens are in different environments is one that often befuddles people who are just learning about NMR spectroscopy. There is an example in the literature that I can no longer find in which the authors were unaware that the methylene hydrogens in diethyl α-methyl malonate were non-equivalent. [One of the co-authors won a Nobel Prize in Chemistry. So don't feel bad!] Imagine that you are chemistry's version of Maxwell's demon. You sit on the red hydrogen of ethyl bromide and look at the bonded carbon. What do you see? A clockwise (CW) environment in the Cahn-Ingold-Prelog (CIP) sense [Br>CH3>H]. When you sit on the blue hydrogen the environment is counterclockwise (CCW). Now you are more capable than an NMR spectrometer. You can distinguish between enantiotopic hydrogens. But the spectrometer can't tell the difference, the same result that was observed by the questioner. In a chiral environment, either by doctoring the sample with a chiral chelating species or by having a chiral environment proximate to the methylene group in question, a distinction can be made. Notice that the methylene group close to the glucose moiety displays diastereotopic hydrogens (circles) while the ones next to the phthalimide group (triangles) are not resolved in spite of the fact that they, and every methylene group in the chain, bear diastereotopic hydrogens!

Now consider (S)-1-bromo-2-methylbutane (CW-S). Sitting on the red hydrogen you see the same order of substituents Br>C>H (CW) as on ethyl bromide but the carbon has an S-configuration. Now jump over to the blue hydrogen of the S-bromide. The local environment is CCW but C2 is still of the S-configuration. That is to say, CW-S and CWW-S have diastereotopic hydrogens! We can use the same arguments with the R-enantiomer where CW-R and CWW-R have diastereotopic hydrogens. The racemate displays the same diastereotopic hydrogens as do the enantiomers. Finally, the 400MHz NMR spectrum of cholesterol [http://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi] displays the two methyl groups at the end of its chain as separate diastereotopic groups separated by a mere 2 Hz even though there are three methylene groups between them and the nearest center of asymmetry. Resolution at low field is not to be expected but yet the hydrogens inherently remain different.

The way people do this is to replace the atom of interest with a special one that has higher priority than the other. For example, you'd replace one of the hydrogens with $\ce{H}*$. Checking both hydrogens in the OP's question gives two different enantiomers without the sugar and two different diastereostereomers with the sugar.

One way to look at it.

The detailed explanation was meant for the novice and not those in the know.


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Kinetics of nitric oxide production
Several articles affirm that NO can enhance the glutamate release from the presynaptic site; moreover, glutamate has been recognized as a major brain neurotransmitter and is considered to mediate excitatory transmission at most synapses of the central nervous system.
First point is the following: is the chemical reaction reported here the equation that describes how NO is produced? If yes, is there a way to model this reaction in such a way it can be simulated by using MATLAB? And finally, do you know what are the kinetic properties of this reaction?

I believe NO is nitrous oxide , not nitric oxide ( NO2)

NO stands for nitric oxide (or equivalently nitrogen monoxide), but nitrous oxide is N2O.

It's enzymatic reaction, so I guess it probably follows Michaelis-Menten kinetics.

@blacksmith37 Ah nope, $N_2O$ is nitrous oxide (nitrogen in the +1 oxidation state) and $NO$ is nitric oxide (nitrogen in the +2 oxidation state). I had trouble wrapping my head around it too, when I was first introduced to those terms ;-)


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Dependence of Van der Waal's constants with distance
Shouldn't the constant change as the intermolecular forces vary at different distances?
Or is it the case that the constants have nothing to do with the forces because we are using the 4 gas laws i.e Charles,Boyles, Avogadros and Gay-lussac laws?

The said laws have nothing to do with forces and constants, because they are strictly valid for ideal gases only. Constants do have something to do with forces, but not in the way you seem to think.

the constants are a value which show the trend of a proportionality. Eg.
P(total) is given by --> P(real) + (an^2)/V^2
here a is a van der waals constant. it actually specifies the trend or the rate of increase of decrease in the expression (an^2/V^2) and not some force. n^2/V^2 provides us the necessary pressure change.
thus the constants 'a' and 'b' remain constant, of course, for a particular gas.
this is what my understanding is. Hope this helps


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Lever rule confusion

At point a3, I know the ratio of the liquid phase to the solid phase, but I don't get the highlighted statement which says b3 gives the composition. I thought a3 already gave the composition? Do I need to apply another lever rule where b3 is the center? If so, where would the endpoints be?

The statement refers to the composition of the liquid phase, which is simply the x-value of $b_3$.

To more fully flesh it out: the overall composition remains at (a2,a3,...). However, the composition of the liquid is read off the liquidus curve, so is b2, while the composition of the solid is read off the solidus curve. For this diagram there is no mutual solid solubility in the solid, so the solid is pure B.


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Is the kinetic energy turning into activation energy when the reaction takes place?
So I want maybe a simple question that I had on a test today. For a reaction to take place it needs a minimum energy called activation energy. At school I heard that this energy is coming from the kinetic energy of the particles that collide. When the reaction takes place, is the kinetic energy small because it turned into activation energy?
If you need any clarification just ask me and sorry for any misspell or grammar error.

You got it right: when the reaction takes place, the kinetic energy of the reacting particles turns smaller because some of it is (sort of) spent on activation energy. It is just that the "when" in question is incredibly short.
Imagine two tennis balls flying at great speed and hitting each other head on. Sure enough, they will bounce off each other and fly in the opposite directions with the same speed. Some time into the collision, there must be an instant when they are not flying anywhere; their speed is 0, and so is kinetic energy. All energy they possessed has converted into the potential energy of elastic deformation, only to return into kinetic energy again really soon.
Same thing with molecules, except they are way smaller and fly way faster. Also, there might or might not be a reaction, and if it happens, then the kinetic energy of the molecules after the collision will not be the same as before. It may be greater, or it may be smaller; that's how exothermic and endothermic reactions happen.
So it goes.

Could not some of the kinetic enery also be transferred into deformations and subsequent vibrations in the molecules, as well as rotational energy?

@Arthur It could all right.

@IvanNeretin Can we say that for exothermic (the reverse for endothermic) reactions because potential energy is lower than kinetic then the kinetic energy is higher (in order to be conserved) so heat is produced?

@ado Yes.$\mathstrut$

@IvanNeretin When I said "potential energy is lower than kinetic" I wanted to say that poetntial reactants is lower than products.


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Relationship between solubility and enthalpy of solution
I'm reading my lab manual and there is a part where it goes through Van't Hoff's isochore. That's fine but then it states

The variation of the solubility of a substance with temperature
  may thus be given by the relation
$$\dfrac{d \ln Sol}{dT}=\dfrac{\Delta H}{RT^2}$$

where do they derive this from. Is it from chemical potentials and mole fractions. I can't really find a good derivation and explanation online.

For equilibria such as $\ce{[C6H5COOH]_{(s)} <=> [C6H5COOH]_{(aq)} }$.  The equilibrium constant $K_{eq}$ can be written in the very simple form:
$$K_{eq}\equiv \ce{[C6H5COOH]_{(aq)}} \equiv \text{solubility}\,  (sol)$$
That's more than likely where the equation comes from. Then you get
$$\dfrac{d \ln Sol}{dT}=\dfrac{\Delta H}{RT^2}$$
when you separate the variables and integrate with respect to both sides you get:
$$\int_{\ln sol_1}^{\ln sol_2}d(\ln sol) = \int_{T_1}^{T_{2}}\dfrac{\Delta H}{RT^2}\, dT$$
$$\ln sol_2 - \ln sol_1 = \dfrac{\Delta H}{R}\Bigg{(}\dfrac{1}{T_1}- \dfrac{1}{T_2}\Bigg{)}$$
Which gives the useful result
$$\ln \dfrac{sol_2}{sol_1}= \dfrac{\Delta H}{R}\Bigg{(}\dfrac{1}{T_1}- \dfrac{1}{T_2}\Bigg{)}$$

Take this as a particular case of the relation between $\Delta G$ and equilibrium constant:
$$K_{eq}=e^{-\frac{\Delta G}{RT}}$$
Really, for a process of dissolution, the solubility basically is the equilibrium constant. (That is, unless the compound dissociates in the solution.) Then $\ln K=-{\Delta G\over RT}$, and $\Delta G=\Delta H - T\Delta S$, and I'm sure you can plug one into another and take the derivative.
As to how the said relation came about, that's another question, and a great deal more difficult. It comes from the statistical mechanics. I'd rather not run ahead of your curriculum, but if the sudden appearance of natural logarithms out of nowhere will bother you to the point of insomnia, go read about the Gibbs paradox.


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What is the methodology to discover thousands of compounds from a complex mixture?
So this is something that has always amazed me,

More than 1,000 chemical compounds in coffee,[1] and their molecular and physiological effects, are areas of active research in food chemistry.

https://en.wikipedia.org/wiki/List_of_coffee_chemicals

How is it possible to identify so many separate compounds from just a single matter? What methods are used?

Sorry to interrupt the row. Just to answer the original question: mostly chromatography. Concerning the (somewhat off-topic, isn't it?) discussion above, yes, a scientist will indeed use intuition from time to time, but the final answer will always come from rational observation and reproducible methods. You may 'think' that coffee does not contain thousands of chemicals, but if you put it through GCMS, LCMS, etc, and find thousands of different peaks, well, unfortunately you can't but accept that you were wrong.

Most compounds can only be analytically quantified individually, so the first step in analyzing a complex mixture is typically to separated the different compounds from each other. For something like coffee, this would typically be done via different types of chromatography (i.e. gas (GC), liquid (HPLC), ion (IC)).
As the separated compounds exit the chromatography column, they can then be analyzed. This is usually done in real time using a number of techniques; the most relevant in this case is mass spectrometry (MS) following GC or HPLC.
In short, the primary techniques used to analyzed the compounds found in an organic sample like coffee would be GC-MS and HPLC-MS.

+1 Add a large database to identify your mass spectra.


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Dalton's law for a gas mixture

If $\pu{200 mL}$ of $\ce{N2}$ at $\ce{25 ^\circ C}$ and a pressure of $\pu{250 torr}$ are mixed with $\pu{350 mL}$ of $\ce{O2}$ at $\pu{25 ^\circ C}$ and a pressure of $\pu{300 torr}$, so that the resulting volume is $\pu{300 mL}$, what would be the final pressure in $\pu{torr}$ of the mixture at $\pu{25 ^\circ C}$?

I understand that to find final pressure, I need to add the partial pressure of $\ce{N2}$ and $\ce{O2}$. I'm not sure how to find the partial pressure using Dalton's law. To find partial pressure of $\ce{N2}$ first:
$$p(\ce{N2}) = X(\ce{N2}) \cdot \pu{0.3289 atm}$$
where $X$ is the mole fraction. How do I find it?

Ideal gas equation PV=nRT can be used to find out "n" of both N2 and O2 and then use X=(n1)/(n1+n2) for both N2 and O2 and you are done.

@SourabhYelluru what does the "resulting volume is 300ml" mean ? Do I need that for any of my calculation for partial fractions and hence final pressure ?

You get total no of moles using ideal gas equation first.You can then use it again to find final pressure after volume is decreased to 300 mL

There is no need to use ideal gas law in $pV = nRT$ form or find molar fraction $X$. The process is isothermal ($T_1 = T_2 = \pu{25 ^\circ C}$, hence $p_1V_1 = p_2V_2$), so it's a matter of finding the sum of partial pressures of each gaseous component $i$ in the system:
$$p = \sum_i{p_{2i}} = \sum_i{\frac{p_{1i}V_{1i}}{V_{2i}}}$$
Also, converting the pressure back and forth between $\pu{torr}$ and $\pu{atm}$ is counterproductive as you are explicitly asked to provide an answer in $\pu{torr}$.

 $$p = p_2(\ce{N2}) + p_2(\ce{O2}) = \frac{\pu{250 torr} \cdot \pu{200 mL}}{\pu{300 mL}} + \frac{\pu{350 torr} \cdot \pu{350 mL}}{\pu{300 mL}} = \pu{575 torr}$$

This method is quicker and more efficient than using ideal gas equation.


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Elements which do not form oxides
I had thought that this question would be easy to answer but my searching has been unsuccessful in finding a clear answer.
For the metals, I expect that they all form oxides.  Some very readily, e.g. sodium and magnesium, and some reluctantly, e.g. gold and platinum, but I cannot find a metal which does not form an oxide.
The metalloids also all appear to form oxides.
The noble gases are obvious candidates for exceptions but xenon has a known oxide.  I think that it is expected for radon but I don't know whether that is confirmed.  A fluoride is known for krypton but no oxide yet.  No oxide is known or expected (?) for any of the lighter noble gases.
This leaves the other non-metal elements which are sufficiently few to check easily.  I find no exception except for the possible pedantic case of fluorine.  It forms a binary compound with oxygen but since it is more electronegative the compound is called oxygen fluoride rather than fluorine oxide.
Some elements are too radioactive to study e.g. francium and astatine but I expect that both would form oxides if we could manage to study them.
Edit: a good point that came up in the comments is oxygen itself.  A pedantic exception together with fluorine.
Is that correct?

@andselisk I may have misunderstood.  Oxygen is a good answer.  A subtle exception alongside fluorine.

https://chemistry.stackexchange.com/a/64699/9961

Eh, this bothers me enough that I have to ask: Should the subject read "Elements that" instead of "Elements which"? (Not a native speaker, it just looks wrong somehow)

@JollyJoker A good point.  It's a question for a rather different stack exchange but this Oxford Dictionaries article seems to be on your side: https://en.oxforddictionaries.com/usage/that-or-which.  It says: "This common British construction is not strictly incorrect in American English, but it is generally avoided, especially in formal writing" which may explain it.  I am British, are you more familiar with US English?

@Mithoron Thank you.  That's interesting though my question was not primarily about the chemistry of noble gases but more whether I had missed anything else.

@JollyJoker "Elements that" would be slightly better, but it's obvious what the question means. ("Metals are elements that form oxides" would be defining metals to be the elements having oxides; "Metals are elements which form oxides" is just giving an incidental property of metals. The question title is using "do not form oxides" as a defining relative clause, so "that" is better. My impression is that American English cares more about the that/which distinction than British English does.)

@DavidRicherby Yes, in this case, Brits seem to be rather more relaxed than  Americans.

The first thing that came to mind is oxygen itself, as the term "oxide" suggests one other element in its chemical formula. Merriam-Webster suggests a more strict version: "a binary compound of oxygen with a more electropositive element or group". So technically pure oxygen is not an oxide.
Putting oxygen aside, helium $\ce{He}$, neon $\ce{Ne}$ and argon $\ce{Ar}$ are probably the only three conventional elements that come to mind are not yet proven to form oxides. I find it rather tricky to back up this statement properly as the lack of knowledge about an object doesn't prove its non-existence; ideally one might want to do a stability simulation at extreme pressures and temperatures to prove the existence of $\ce{(He|Ne|Ar)O_x}$ or lack thereof. There are algorithms like USPEX that are capable of that – by the way, with the help of USPEX a family of krypton oxides has been recently predicted.

Thanks.  I also considered fluorine as an exception since as you quote "a more electropositive element"  whereas fluorine is more electronegative.  The rule I knew was that in binary, inorganic compounds, the more electronegative element is named second.  $\ce{O}\ce{F}_2$ rather than $\ce{F}_2\ce{O}$.

Yes, oxygen can by definition not form oxides. To the best of my knowledge, no compounds of helium or neon are known to date and only $\ce{HArF}$ is known for argon, but that could be dated already.

" lack of knowledge about an object doesn't prove its non-existence" - of course, but  I am happy with answers of the form: "no oxide of X is known".  So, krypton might be in the category "no oxide known but that might change" whereas neon is "no oxide known and none expected either".

@jan FYI disodium helide.

@OscarLanzi Thank you, I love chemistry!

So, combining all of the responses.
All metals and metalloids have known oxides.
Among the noble gases, xenon and radon have oxides, one has been predicted but not observed for krypton, and one is not expected for helium, neon, or argon.
Among the other non-metals, all have oxides except the technical exceptions of oxygen itself and fluorine; as andselisk says, an oxide is "a binary compound of oxygen with a more electropositive element or group".  A binary compound of oxygen and fluorine is known but it is oxygen fluoride rather than fluorine oxide.

Would that include astatine and francium?

A good point.  I expect that if we could move quickly enough, we could create francium oxide and astatine oxide.  The problem is the physics, I doubt that there is any chemical obstacle to these compounds.

The intense radioactivity of all isotopes of astatine and francium would prevent collecting any macroscopic samples of any oxides, for the heat generated by radioactive decay would vaporize such samples.  Wikipedia reports several empirically known compounds with astatine-oxygen bonds in solution.


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Lowest Michaelis constant Km
I want to find a lower limit of the Michaelis constant for some evaluations of Michaelis-Menten enzyme kinetics.
What is the lowest $K_m$ you ever encountered? Is there a theoretical limit?

https://en.wikipedia.org/wiki/Michaelis%E2%80%93Menten_kinetics#Applications

The link does provide some examples, but no lower limit for k_m.

It is generally believed that evolution places a theoretical limit on kcat/Km, but not necessarily on Km. I know of one reported Km in the nanomolar rnage (3.2 nM to be precise): https://aem.asm.org/content/85/13/e00438-19. There are probably others that are lower.

Since the function under consideration is catalytic (accelerating the reaction rate), you might consider the natural rate of the reaction (in absence of enzyme) as setting a limit on kcat. From the diffusion limited on rate (as explained in an answer) and setting the off-rate kr to zero you can  estimate a lower meaningful (theoretical) bound on Km.

The other way to set an upper limit on kcat is by the lifetime of the enzyme. Most cellular proteins are degraded within hours or days. So you could set a conservative lifetime as 100 days and say that at least one turnover must happen (on average) within that time, so minimum kcat is 3 per year ~ $1\times 10^{-8}$ s$^{-1}$. Dividing by kf = 10^9 M-1s-1 gives a limit on Km of $10^{-17}$ M give or take a few orders of magnitude.

Take this generic reaction representing the Michaelis-Menten enzyme kinetic:

You can derive the Michaelis-Manten constant $K_m$:
 (for derivation see https://en.wikipedia.org/wiki/Michaelis%E2%80%93Menten_kinetics)

The quasi-steady state hypothesis was used, therefore the rate of formation and breakdown of $ES$ are the same; this is a good approximation if the enzyme concentration is much less than the substrate concentration $[S]$ or $K_m$ or both. That means that $K_m$ can be extremely small (almost zero) and the Michaelis-Menten equation still be valid proven that the enzyme concentration is much less than  $[S]$.
$K_m$ is very small if:
$k_r$ and $k_{cat}$ are both very small compared to $k_f$, thus if the complex $ES$ is formed very quickly compared to its breakdown.
In this case the reaction proceeds at its maximum velocity $V_{max}=k_{cat}*E_0$, with $E_0=[ES]+[E]$:

About the theoretical limit on how small $K_m$ can be there are two considerations:
 First, $k_f$ cannot be cannot be faster than the diffusion-controlled encounter of an enzyme and its substrate. This means that $k_f$ cannot be higher than $10^9 s^{-1} M^{-1}$  (ref.https://www.ncbi.nlm.nih.gov/books/NBK22430/#:~:text=This%20rate%20cannot%20be%20faster,s%2D1%20M%2D1.)
Second, $k_{cat}$ cannot be too small or the product $P$ formed is produced at a rate not suitable for the cell survival.
Generally, I haven't encountered $K_m$ smaller than $10^{-6}M$ in biological enzymes (e.g Triosephosphate isomerase, TPI) but possibly there are. In TPI, $ k_f$ ∼ $10^9 s^{-1} M^{-1}$, therefore  $k_r$  ∼ $k_{cat}$  ∼ $10^{3}s^{-1}$ that is still quite high. It would be very interesting to know if there are enzymes with an even lower $K_m$. Very intresting question.


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Sense of radii quotient to determine coordination number in crystals?
Every common textbook explains the method of the radii quotient between the cation and anion of a selected salt to determine to coordination number(s) in that salt. For example deduce "Inorganic Chemistry" (Huheey) the coordination number of BeS by following calculation:
Radii quotient $k = \dfrac{r_{\text{cation}}}{r_{\text{anion}}} = \dfrac{59   \text{ pm}}{170 \text{ pm}} \approx 0.35 $
This quotient $k$ indicates a tetrahedral coordination (coordination number for both ions = 6), because $k$ lays between the two limits 0,255 and 0,414 for a energetically favorable tetrahedral coordination (limits are derived through a simple mathematical problem with pythagoras).
Now my actual question or problem with this way to determine the coordination number: The used crystal radii $r_{\text{cation}}$  and $r_{\text{anion}}$ of Be2+ and S2- depend on their coordination; on their coordination number. For example are the crystal radii for Be2+:

Coordination Number = 6: $r_{\text{cation}} = 59 \text{pm}$
Coordination Number = 4: $r_{\text{cation}} = 41 \text{pm}$
Coordination Number = 3: $r_{\text{cation}} = 30 \text{pm}$

The difference between these values are tremendous and change the result of the $k$ calculation above, e.g. if I choose $r_{\text{cation}} = 30 \text{pm}$ then $k=\dfrac{30 \text{ pm}}{170 \text{ pm}} \approx 0.18$ and therefore BeS should be linear coordinated...
In Short: How can we use a method to determine the coordination of a ion in an (simple) ionic structure if the used radii in this method are again depended on the coordination?
Crystal radii data from: http://abulafia.mt.ic.ac.uk/
Edit: I am aware that this method has exceptions and is just simplified (there are many other factors that determine the coordination number)...


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How to avoid oxidation of Si using a sub nanometer layer
Does anyone know how and which material to use to protect a Si layer from Oxidation by a  small layer of thickness less than 1 nm ? (probably using atomic layer deposition)

Hydrogen terminate the surface. Not permanent though...

@JonCuster Thanks for the note. I am looking for a permanent solution.

Doesn't SiO2 layer protects from further oxidation?

See Mithoron comment.  If you need to avoid SiOx or OH groups, I really think that a subnanometer layer won't suffice. Unless there is a special technique/material known to those working with wafers

Fluorine termination might be worth a look, if you have the facility for it. Sunada et al., in 1990 Jpn. J. Appl. Phys. 29 L24082, write in the abstract:

The principal role of silicon-fluorine bonds in the chemical nature of HF-etched Si surfaces has been investigated by angle-resolved X-ray photoelectron spectroscopy. The native oxide growth thickness and the fluorine coverage have been systematically measured as functions of HF concentration, pure water rinse time, air or N2+O2 gas exposure time and gas phase H2O concentration. It is shown that the native oxide growth is strongly suppressed by the existence of Si-F bonds of about 0.12 monolayers on the surface. This is explained by a model in which Si-F bonds chemically stabilize the surface reactive sites such as atomic steps as supported by the result of the layer-by-layer oxidation of Si.

(My bolding).
I haven't read the full text, and I suspect that "strongly suppressed" really means that there was a very small amount of oxidation, rather than none at all. And note that while silicon surfaces may be stabilised by fluorination, scientists tend to get violently destabilised by the same reaction. This is a potentially hazardous process and not to be undertaken lightly.
In any case, "stability" really depends on what the surface is going to be exposed to. F-termination may oxidise only very slowly in air, but if you get water in there, it's a different story. In 1993 Jpn. J. Appl. Phys. 32 3125, Nakamura et al. write that:

The 1.0 monolayer of Si-F bonds was found to be replaced with the 0.7 monolayer of Si-H bonds and the 0.2-0.3 monolayer of Si-OH bonds after dipping into water for 1 min.

So there may not be a "permanent" solution, just one that resists further oxidation for a while. What are you actually trying to achieve?
One other thing you could investigate is growing a self-assembled monolayer (SAM), probably on an H-terminated Si surface. It's been a while, but IIRC some pretty stable coatings can be generated that way. A SAM forest may offer a decent steric barrier to further reaction.

Thank you very much. This is very interesting! The aim is to build a Si reflector, but the reflection is reduced, if there is a oxydation surface at the top because of the change in refraction index. But if the surface is smaller than 1nm in width then it can be neglected.  I was also thinking of using atomic layer deposition and deposit one layer of Ag... but I don't know if it is possible, or even useful ? I just supposed that Ag does not oxydise easily.