text:
Isolating calcium acetate from water with electrolysis
That's a long page, so you can optionally jump to the "Thunderstorm it" part, but a bit of context is always neat :)
Context
Say I'm cooking an egg in vinegar. The egg shell is made of calcium carbonate:
$$\ce{CaCO3 + CH3COOH -> H2O + CO2 + Ca(CH3COO)2}$$
After eating the egg, is left in the cookware: a bit of calcium acetate, lots of water, acetic acid and other unidentified stuff.
It'd be neat to isolate that calcium acetate, as it could be of use sometime. I've seen that we can grow cool looking crystals out of it, maybe I could try that! Anyway chemistry is cool, so let's do it :)
But there will also be loads of water (egg cooking). I could store it all, but that wouldn't be very practical...
Getting rid of the water
I see three ways to get rid of that water.
Boil until water evaporates
Energy-hungry process. I'm a kind of environmental-crazy dude, so that's not an option.
Filtering
While boring, that could be an efficient option, if at all possible. It seems calcium acetate is soluble in water. Is there a way to filter it out without expensive equipment, or even at all?
Electrolysis
Energy-hungry as well, but I have a spare solar panel that produces 6 or 12V and anything between 0 and 4W, so that energy comes for free (albeit quite slowly in Belgium).
Thunderstorm it!
No matter how feasible are the other options, I'm here to experiment, so I'd love to do some electrolysis :) I got time anyway!
Constraints
I have very little knowledge of chemistry (but eager to learn!), and no equipment at all, except for household items. Also, this experiment would take place in living space (probably in my room), so it should be safe to do (e.g. do not produce chlorine gas or other nasty products). And if it's not smelly, it's even better :)
How I would do it
I'll need:

container: a drinking glass should suffice
anode
cathode
some electrolyte in the water
electricity: the solar panel will provide what's needed
time: we all have a pool of that in our life

I still need the anode, cathode and electrolyte. And also, I'd love to understand the process that takes place.
In my younger age, I used copper wire as both anode and cathode, table salt as electrolyte, and it would produce tiny bubbles on one end of the wire, and go green on the other end (do I remember properly?).
In this experiment though, the context is a little different:

I need to "dissolve" a lot of water, so just a copper wire won't be enough
It seems that using NaCl as electrolyte produces chlorine gas instead of oxygen, that's pretty bad

Choosing the electrolyte
Since I cooked the egg in vinegar, I'll probably have some acetic acid left in the water. Would that be a good enough electrolyte? Alternatively, I have baking soda readily available, but that would produce other salts when reacting with the remaining acetic acid...
Choosing the anode and cathode
I'm lost here. First and foremost, I remember my copper wire going green when doing electrolysis. Does that mean I'll "consume" my anode/cathode while doing the electrolysis? That sounds pretty expensive. Also, I'll have to separate calcium acetate from copper oxyde afterwards...
Questions

is the above reflection correct at all?
what's the best choice for electrolyte and anode/cathode materials
what's the exact reaction taking place?
any other way to isolate that calcium acetate?

Buy some small reverse osmosis unit and do the thickening with it.

Interesting! I'll have a look at that :)

How to get rid of water to collect calcium acetate?
Evaporate it.
How to do it in an environmentally responsible way?

Electricity from the outlet cost you money and probably used some
combination of coal, natural, gas, and or nuclear to produce it.
Bunsen burner, but it likely uses a fuel that was taken from the ground, made from a crop that that competes against food production for land.
Solar cells to produce electricity. It uses renewable energy but
is relatively expensive and electricity is a high value energy
compared to solar thermal. Using commercial grade solar cells only
converts about 17% of the energy in sunlight into electricity that is then converted into heat to evaporate the water.
Solar thermal is cheap and easy to make with household items like an aluminum or soup can and black spray paint. Further more, it has the potential to be more than twice as efficient as solar cells at providing energy you need to evaporate water.

Electrolysis:
If you insist on electrolysis for experimental reasons, calcium acetate is a salt of acetic acid and calcium from the calcium carbonate in the egg shells. As a salt, like all salts, calcium acetate will dissolve in water forming ions that will conduct electricity. So, there is no need for any electrolyte other than water and calcium acetate.
Furthermore, your end products will be something other than calcium acetate and would depend in part on your anode and cathode, temperature, and applied voltage.

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Wow, I didn't even think of just exposing it to sun. Points for the easiest solution ;) thanks also for the info on the electrolysis, and especially about the "won't be the same part". I'll investigate :)

I'm not completely sure, but I think that the products of electrolysis will be something like: $\ce{H2}$ + $\ce{C2H6}$ + $\ce{CO2}$.


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Does phosphoric acid remove stainless steel electropolished layer?
Electropolished stainless steel has a high corrosion resistance due to the oxide film that is created during electropolishing.  The film mainly consist of chromium(III) oxide.
Phosphoric acid is used as a rust removing agent:
$\ce{2 H3PO4 + Fe2O3 -> 2 FePO4 + 3 H2O}$
My question is: will 5% $\ce{H3PO4}$ solution remove this protective layer as well? It should, but how fast will be the protective layer removed? Does anyone have personal experience?

The phosphoric acid will remove the coating if it's concentrated enough but the time it takes to do this depends on how concentrated the acid is and how much of the acid it is exposed to.

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deltaS of a decomposition reaction
Suppose we had the following reaction:
$$\ce{Polymer ($n$~units)_{(aq)} -> Polymer($a$~units)_{(aq)} + Polymer($n-a$~ units)_{(aq)}}$$
Without knowing the specific monomer unit of the polymer, and assuming the polymer is straight chain, is there anyway to determine the ΔS of the reaction? I mean, I know we can determine the sign (which will almost certainly be positive), but is there a way to make even a rough approximation of magnitude? We can assume the conformational entropy is negligible, i.e. the polymers are fairly rigid.

I am not a polymer theorist, but I think it would be very hard to estimate this. Consider that you don't know how the units were linked, e.g., what type of bond, how much it can rotate, etc.

Assume it's a polymer small enough that it can be modeled as a rigid Rod.

The largest contribution to the entropy change will be the translational component, which is easy to evaluate using the Sackur-Tetrode equation
$$S^\circ_{\rm{trans}}=R\ln(bM^{3/2}T^{3/2})$$
where $b$ is a collection of constants with a value of 3.75 (mol/gK)$^{3/2}$ $M$ is the mass in g/mol, $T$ is temperature in Kelvin, and the standard state is 1 M.
Due to the slowly varying logarithmic dependence on $M$ and $T$ this contribution is usually on the order of 100's of J/molK.
While the translational contribution is the largest, rotational and vibrational contributions can be sizable so this will give you an order of magnitude estimation at best.

Derivation starting from the wikipedia equation
$$S^\circ=kN \left( \ln \left( \frac{V^\circ}{N} \left( \frac{4\pi mU}{3h^2N}\right)^{3/2} \right)+\frac{5}{2}\right) $$
$V^\circ$ = 1 L = 0.001 $m^3$, $N=N_A$, $U=\tfrac{3}{2}N_AkT$, $M=mN_A$
$$S^\circ=R  \ln \left( \left[e^{5/2}\frac{V^\circ}{N_A} \left( \frac{4\pi \tfrac{3}{2}k}{3h^2N_A}\right)^{3/2}\right]M^{3/2}T^{3/2} \right) $$
The term in square brackets corresponds to $b$ and using SI units $b=1.20\times10^5$ mol(kg/K)$^{3/2}$ which corresponds to 3.75 mol(g/K)$^{3/2}$

Isn't the Sackur-Tetrode Equation derived from an ideal gas? Is it reasonable to make that approximation for a macromolecule in solution?

Since the thermodynamic quantities are state functions you can rewrite the aqueuos phase process as a thermodynamic cycle: 1. De-solvation of the reactants, 2. Reaction in the gas phase. 3. Solvation of the products.  The largest contribution to $\Delta S^\circ$ is likely the translation contribution to the entropy change in step 2.  Also, note that the standard state in aq soln is 1 M ideal solution

Oooooh.
Different question: the equation you have, while simple, does not immediately seem identical to the equation you linked. Is there anyway you can show the derivation from the more complicated one to the simpler one?

I'll try to find time to write it down.  In the mean time some hints if you want to have a go: $U=\tfrac{3}{2}kT$, $\ln(x)+\tfrac{5}{2}=\ln(xe^{5/2})$, $N=N_A$, and $V=1$ L.  The rest is just units.

I've taken a crack at it and got as far as $Rln(e^{5/2}(\frac{2 \pi m}{h^2N_A}k_bT)^{3/2})$

I have now added a sketch of the derivation.  In my previous comment I made a mistake wrt $U$, which is $\tfrac{3}{2}N_A kT$

facepalms I should have caught that. Thank you.

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Why is sodium methanethiolate still a better nucleophile than sodium methoxide when the solvent is ethanol
I'm looking at a problem where sodium methanethiolate and sodium methoxide, both in equal amounts, are being reacted with $\ce{CH3I}$ and the solvent is ethanol.
I understand that sodium methanethiolate is a better nucleophile than sodium methoxide when the two are reacted in equal amounts with something like methyl iodide in a polar aprotic solvent. But why is sodium methanethiolate still a better nucleophile than sodium methoxide when the solvent is polar protic, like ethanol?
I thought the ethanol would reduce the nucleophilicity of both, but would do so at by different amounts, given the different electronegativity of sulfur and oxygen. If ethanol reduces the nucleophilicity of both at the same rate, why even use ethanol as a solvent in this case?

Thiolates are generally more nucleophilic than alcoholates. So whenever the choice happens to be between a nucleophilic sulphur and a nucleophilic oxygen, the sulphur will win.
Both the methanolate and the methanethiolate will show reduced nucleophilicity due to partial protonation, though. And of course you have introduced ethanolate as an additional nucleophile, too (which should be the least reactive, though).

Nucleophiles can be hard or soft (as can electrophiles); a hard nucleophile tends to react preferentially with a hard electrophile, while a soft nucleophile tends to react preferentially with a soft electrophile.
In your system the methanethiolate anion, $\ce{CH3S-}$, is a soft nucleophile while methoxide, $\ce{CH3O-}$, is a relatively hard nucleophile.  You will also have ethoxide ($\ce{CH3CH2O-}$) in your system.  As to the electrophile, methyl iodide, $\ce{CH3I}$: it is soft and hence it preferentially reacts with the softer nucleophile in the system, methanethiolate.
A more sophisticated explanation of the hard/soft idea can be found in frontier orbital analysis.

Hard and soft nucleophiles and electrophiles very badly describe this case and shouldn’t be pulled into this argument. Remember that the HSAB concept originally intended to explain the formation of stable adducts of Lewis acids and bases.

@Jan thanks for the response but I disagree.  I will try and find some literature to support my argument and amend my answer.


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How do I setup the electrodes for cyclic voltammetry on a capacitor made on dry copper sheets?
I'm trying to measure capacitance of a cell I'm creating. It's composed of two copper-sheets with nano-material coated on one side, sandwiched with a membrane and wetted with $\ce{KOH}$. All of the three electrode setups I've found on the internet are for solution based voltamettry, but this cell is not in a solution. The setup that my labmates had yesterday had the reference electrode, $\ce{Ag|AgCl}$ setting in a solution, and not connected to the cell, which doesn't make any sense to me.

What reference electrode should be used to do this kind of measurement, and what should the schematic look like for our setup?

The reason reference electrodes are needed in 3 electrode solution cells is that you're trying to control the interfacial potential of the working electrode (basically from the surface to a point very near the electrode in solution). The wiki's kinda lousy, but the diagram explains essentially what the potentiostat is doing: It's controlling the counter electrode (the op-amp output), to set the potential from the working electrode (ground) to a point just into the solution and the reference electrode is used to read the potential and provide feedback to the amplifier to correct the output. This way, any extra solution resistance between the counter and reference electrodes doesn't reduce the interfacial potential (known as iR drop in electrochemical circles) because you're measuring the potential near the surface. In other words, we're interested in applying a potential across $Z_2$ precisely, but we don't care about what's happening with $Z_1$.

Now, since you have a capacitor, what you're trying to do is control the potential of one plate versus the other plate, so you want the point where you're getting feedback to be one plate, and the other plate is ground—one plate is the "reference electrode" and the other is the "working electrode". Your capacitor is $Z2$ and there's no solution, so there's no $Z1$. (In some cases, it still makes sense to use a separate cable for the reference input, but it's probably fine to just short them together at the inputs) Many potentiostats will work with the counter electrode and reference electrode inputs shorted together (which is how one typically connects a 2 electrode cell), but a resistor between them can help prevent it from oscillating by increasing the time constant of the system.
You're not really using a reference electrode in this case because you don't care about measuring the applied potential versus some known potential, just in controlling the potential across the capacitor. In a cell in which we're interested in studying redox reactions and such, it's important to know how the potentials relate to one another and to compensate for the various potentials that arise when electrodes are dipped in electrolyte solutions.
If you're really interested in the details of potentiostats, Bank has a good primer with lots of details.

Our setup is basically two copper plates with a membrane between the two. Would we attach the RE and the CE to one plate and the WE to another? I get that you said a resister would be necessary, but where would the resister be attached?  Can you describe what the oscillating would look like?

Yes, the RE and CE go together and WE input to the other. If it needs a resistor, then that can go between the RE and CE, with the RE only attached to the plate. (it probably doesn't need to be that big, depending on the capacitance) Most modern potentiostats are built to tolerate a lot of capacitance, so it may not be needed. There is an example of a marginally potentiostat on page 10 of the link above. If it's completely unstable, you basically just see a sine wave (though if you're not sampling fast enough, you may not see it; if you get weird behaviour, try adding the resistor).

Well, I'm not an electrical engineer by any means, but I do have some experience with this. Basically, the potentiostat is an operational amplifier. The wikipedia page for potentiostat shows the schematic of a test cell as a series of impedances.

In order to test your capacitor, simply place your capacitor in for $Z_2$, and a resistor in for $Z_1$. Unfortunately, I can't be much help in selecting a resistance value, but it will depend on your potentiostat and the capacitance of your device.

In my tests, a resistor wasn't necessary at all, so first try connecting the counter and reference to one plate of the capacitor and the working electrode to the other plate.
If it turns out the resistive impedance is necessary, ask an electrical engineer for help, and use "op amp" in your explanations, because they won't know what a potentiostat is (most likely).

All of the electrodes I've looked at look like they need to be in-solution. Is there a reference electrode, that I could use for solid state?

You don't need an actual reference electrode, just clip the lead for the reference to the capacitor.

authors: แฟนทหารพราน Chanel


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Resolution of enantiomers? How to determine chiral compounds apart from each other?
How exactly is the interaction between two chiral compounds so they can tell the difference between S or R?

Are you asking how to separate enantiomers? Are you asking how to figure out the absolute stereochemistry (e.g. R vs S) of a compound? @Ringo read this as asking how to assign R vs S stereochemistry, which I hadn't thought of. Please provide some additional details.

If the compounds have never been made before, then the enantiomers have to be separated prior to analysis. Then analytical and spectroscopic techniques must be used. The R and S configurations cannot even be assigned until the relative stereochemistry had been elucidated. This is usually done by specific NMR experiments. Chiral HPLC can be used to give enantiomeric purity.

The method used to determine whether a specific chiral molecule is the $S$ or $R$ enantiomer is determined according to the Cahn–Ingold–Prelog priority rules. The steps to this are:

Look at the four atoms directly attached to the chirality center, and rank them according to atomic number (1-4, 1 being highest ranking, 4 being lowest ranking). Isotopes of higher mass rank higher than lower mass isotopes.
If a decision can't be reached by ranking the first atoms in the substituent, look at the second, third, or fourth atoms away from the chirality center until the first difference is found.
Multiple-bonded atoms are equivalent to the same number of single-bonded atoms.
Having ranked the substituent groups, draw (or imagine) the lowest ranking substituent group away from yourself. The other 3 substituent groups will then appear to radiate out towards you. If a curved arrow drawn from the highest to second-highest to third-highest ranked substituent ($\ce{1->2->3}$) is clockwise, the chirality center is said to have the $R$ (for rectus, meaning right) configuration. Oppositely, if the curved arrow is counterclockwise, the chirality center is said to have the $S$ (for sinister, meaning left) configuration.


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How to determine the most reactive hydroxyl group of α-ᴅ-glucopyranose?

In the following reaction, only one of the hydroxyl groups in D reacts under the given conditions:

I was wondering which OH group reacts and why? Also I have been trying to determine the structure of E based on which hydroxyl group reacts.
My thoughts are that OH in the side group can react via SN2 with the reactants, but the OH right adjacent to the ether O can react leaving a carbocation that can be stabilised by lone pair from O. This would mean SN1 and a completely different product.
Can someone please explain the correct mechanism?

There is one hydroxyl group that is significantly different from all others, and another one, that is mildly different from the remaining. Does that hint help you? By significantly I mean rea~lly significantly

@Jan Do you mean the hydroxyl on the side group that would undergo SN2? It is a Primary Alcohol..the others are secondary...

That's the mildly different hydroxyl group. Now look for the significantly different one.

You're on the right track with the hydroxyl on the same carbon as the ether. You've described the mechanism (at least the beginning) in words already, with the correct rationale for why that hydroxyl reacts. Write it up and you could answer your own question.

Just so we're clear. The $S_{\mathrm{N}}2$ reaction with an alcohol is crap. Your yields will not be good. When you see these kinds of conditions, you should rule out $S_{\mathrm{N}}2$.

$\ce{EtOH}$ is a rather bad nucleophile and we have the assistance of a lone pair of the neighboring oxygen atom (the positive charge can hence be delocalized over the carbon and oxygen) which greatly speeds up $\ce{S_{N}1}$ in comparison to $\ce{S_{N}2}$. So what will probably happen is protonation by $\ce{HCl}$ of the alcohol next to the ether oxygen followed by nucleophilic ($\ce{S_{N}1}$) attack of $\ce{EtOH}$ on the formed (delocalized) secondary carbocation. In comparison to the possible reactions at the other secondary centers (having the formulation of the question in mind) this primary reaction will be fast enough to justify the ignoring of them.

Might be worth mentioning that the big difference of the $\alpha$ carbon is its oxidation state: Formally it's an aldehyde, not an alcohol. And in the OP's figure, one would call it hemi-acetal.

@Jan Sure it's a hemi-acetal, thanks for pointing that out. I figured that OP's knowledge of chemistry is not very developed yet so I circumvented using the term. Again, thanks for your input.

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Calculate the valence of Fe2O3
$$\ce{FeO}$$
Is called "Iron(II) oxide ".
$$\ce{Fe_2O_3}$$
Is called "Iron(III) oxide ".
The number in the parenthesis is "Valence" of the substance. Frankly, I don't know how to calculate such number. Can you explain to me, with this example, how come $\ce{FeO}$ has valence $2$ and the other one has valence $3$?
The only thing I know how to do is calculate the electron configurations of each element.
For Iron, it is $1s^22s^22p^63s^23p^64s^23d^6$.
For Oxygen, it is $1s^22s^22p^4$.
I am told that based on the electron configuration you can get the valence of each element. And then, you can calculate the valence of $\ce{FeO}$ and $\ce{Fe_2O_3}$. But how?

I made a question before Calculating valence of oxides, but it was wrongly formulated (they corrected me on the definition of Valence, but turns out what I actually need was how to calculate it), so I'm flagging it to be closed.

Edit The question How to get the valencies of elements? doesn't explain how to calculate the valence of a combination of two elements.

@Jan hm, but it doesn't explain how to calculate the combination of two elements. I can tell the valence of $Fe$ and $O$, but what of $Fe_2O_3$?

It's not the combination that has a valency, it is each element in the compound. $\ce{O}$ can always be considered to be divalent. $\ce{Fe}$ can assume multiple valencies, which is what the (II) and the (III) signify.

somewhat related http://chemistry.stackexchange.com/questions/8336/why-can-some-ions-have-different-valence

http://chemistry.stackexchange.com/questions/539/what-does-this-notation-mean

And if you'd want to predict http://chemistry.stackexchange.com/questions/19900/find-the-valences-of-z-26

@Mithoron neither link will explain how to calculate valency when two elements are involved (like $\ce{FeO}$ or $\ce{Fe_2O_3}$).

I'll try to write an answer but for example you know what bonds are?

But well - this question should be closed as you're asking about the same thing, and you even got quite nice answer

@Voldemort you learn the valencies rather than calculate them.  Valency is a fact of experiment whilst electron configuration gives an underlying explanation for the observations from experiment.
Iron has two different valencies (and oxidation states) which are a consequence of the stability of the electron configuration that remains when electrons are lost upon reaction.  So, iron can loose 2 electrons to leave a stable $\ce{Fe++}$ species; in your reaction those 2 electrons are gained by oxygen to give an anion that has 8 electrons in its outer shell which is a stable configuration.
Iron can also loose 3 electrons to leave a stable system and again, in your system, the electrons are picked up by the appropriate number of oxygen atoms to give oxygen anions that have 8 electrons in their outer shell.
If you look at the either side of the periodic table, say sodium reacting with a halogen, say chlorine (to give sodium chloride); you can see the 8 electron rule working.  But you're looking at a system where the iron is found in the middle of the periodic table and is a lot more difficult to predict when compared to the 8 electron rule.

So basically valencies can't be calculated and therefore the only way to know that $\ce{Fe_2O_3}$ is "Iron (III) Oxide" is with memory?

@Voldemort yes.  But they can be explained and they can be guessed at with some certainty.  The alkali metals should have a valency of 1 since that outer electron is going to be easy to loose and the remaining cation has a stable 8 electron shell.  The halogens when reacting with alkali metals should also have a valency of 1 but this isn't always the case.  As for the d-block elements, these can be a bit more difficult.

@Voldemort I'm afraid your question is so terribly basic that user1945827 is overestimating it - do you want to get valence when you already know chemical formula? He's talking about predicting from configuration but if you have the formula you don't have to predict anything

@Mithoron by chemical formula you mean $\ce{FeO}$ and $\ce{Fe_2O_3}$? Then yes, I only want to know how to calculate their valence. So there is indeed a way to calculate it?

Useful links https://en.wikipedia.org/wiki/Lewis_structure, https://en.wikipedia.org/wiki/Valence_%28chemistry%29

@Voldemort Yes that's what I mean. I'll search if there wasn't such question earlier

From \ce{FeO}, I'd guess iron has a 2+ charge, so neutral iron must have lost 2 electrons. While valence electrons are easiest to lose, I'd deduce that the valency of \ce{FeO} is 2.


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Mechanism for anhydride formation
What is the mechanism by which phosphorus pentoxide ($\ce{P4O10}$) and a carboxylic acid form an anhydride? I can't find it anywhere online. Does $\ce{COOH}$ attack twice (I'm assuming it's the nucleophile) because there ends up being two carbonyls in the product? What leaves?

What would the carboxylic acid attack twice? What bonds do you need to make and break to convert two acids to an anhydride? What is the (theoretically at least) byproduct of anhydride formation?

Would it attack P? Wiki says "when combined with a carboxylic acid,  the result is the corresponding anhydride:  P4O10 + RCO2H --> P4O9(OH)2 + [RC(O)]2O. I cannot find the mechanism anywhere for this equation.

I highly suspect that this reaction does not have a known mechanism, and $\ce{P4O10 + 2\ RCO2H -> P4O9(OH)2 + [RC(O)]2O}$ is just a formal reaction meant to balance atoms.

David Klein (in O.C. As A Second Language) says that the mechanism is rarely shown in texts so he invites the reader to ponder it (but does not show it.) I cannot figure it out. And can't find it anywhere!  Should I email David Klein? Lol


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Why doesn't the SeO2 oxidation of ketones stop at the hydroxyketone stage?
When $\ce{SeO2}$ is used to oxidise aldehydes and ketones, $\ce{SeO2}$ becomes hydrated in water, and the active reagent is selenous acid $\ce{H2SeO3}$ which oxidizes aldehydes and ketones to furnish 1,2-dicarbonyl compounds such as 9 in the scheme below.1 The α-hydroxyketone 10 is not observed. Why is that so?

1 Sharpless, K. B.; Gordon, K. M. Selenium dioxide oxidation of ketones and aldehydes. Evidence for the intermediacy of .beta.-ketoseleninic acids. J. Am. Chem. Soc. 1976, 98 (1), 300–301. DOI:10.1021/ja00417a083

As noted in the cited paper by Sharpless and Gordon$^{[1]}$, the problem with the Corey$-$Schaefer mechanism is the existence of the selenous ester (7) which, due to its instability, would undergo hydrolysis prior to any elimination reaction.
A study by Qingjiang Li and Tochtrop$^{[2]}$ supports Sharpless and Gordon's counterargument. The former pair give a good overview of the hypotheses put forward up to that point (i.e., 2014).
$\hskip 1 in$

$\beta$-hydrogen transfer (Speyer mechanism)

Based on the logic of the proposed mechanism, we reasoned that replacement of $\beta$-hydrogens of the substrate with deuterium would result in a ﬁrst-order
isotope eﬀect in the reaction rate.

Consequently, a $\beta$-deuterated fatty acid was prepared ... and converted to triﬂuoromethyl ketone to react with selenium dioxide. The reaction rate was monitored by NMR and compared with that of the analogous fatty acid without deuterium.$^{[2]}$

Based on this experiment, Li and Tochtrop conclude$^{[2]}$,

No signiﬁcant diﬀerence in reaction rates was observed, which suggested that the removal of $\beta$-hydrogen to form a carbon−carbon double bond is not likely to be the rate-limiting step, and we therefore excluded the $\beta$-hydrogen transfer mechanism.

Verdict: not the best explanation.

Sharpless$-$Gordon

While solving the selenous ester controversy, Li and Tochtrop found other problems introduced by Sharpless and Gordon.

[A]dding acid should accelerate the Sharpless mechanism by accelerating the enol−ketone tautomerization, as the rate limiting step is electrophilic attack of selenium dioxide to the enol. Since no signiﬁcant increase of the reaction rate was observed when adding acetic acid to the reaction mixture, the Corey−Schaefer mechanism seemed more reasonable.$^{[2]}$

To additionally verify this result,

$1.2$ equivalents of $\ce{Ti(O^iPr)4}$, a strong Lewis acid to coordinate with the carbonyl group, was added to the triﬂuoromethyl ketone for $1\ \mathrm{h}$ prior to selenium dioxide addition. This experiment resulted in no reaction, which subsequently served as further evidence that the reaction initiates with the ketone instead of the enol.$^{[2]}$

The Sharpless$-$Gordon mechanism also fails to explain some products detected in Li and Tochtrop's work with carboxylic acids.$^{[2]}$
Verdict: not the best explanation.

Corey$-$Schaefer

Verdict: not the best explanation due to the selenous ester intermediate.$^{[1]\ [2]}$
New mechanism proposed by Li and Tochtrop
Li and Tochtrop came up with a fourth mechanism to account for experimental evidence. Here, the reaction initiates via the ketone. The further pathway is dictated by whether $1$,$4$-elimination is restricted.$^{[2]}$
Li and Tochtrop examplify their mechanism with a trifluoromethylketone that has varying radicals attached to the non-fluorinated $\alpha$-carbon.$^{[2]}$
$\hskip 1 in$

As shown in Scheme $5$, we propose that high valence selenium reacts with
triﬂuoromethyl ketone to form the enol ester $(\bf{17})$. The enol ester undergoes $1$,$4$-elimination directly instead of nucleophilic attack of the carbon−carbon double bond to form selenium ester.$^{[2]}$

[---] When $1$,$4$-elimination is restricted, nucleophilic attack at the $\alpha$-position occurs and the subsequent hydrolysis of selenium($\mathrm{II}$) ester $(\bf{18})$ aﬀords alcohol as the product instead of enone.$^{[2]}$

In the context of the Li$-$Tochtrop mechanism, we can thus answer OP's original question.

The $\alpha$-hydroxyketone (10) is not observed. Why is that so?

The intermediate (4) has a small hydrogen attached to the original $\alpha$-carbon. In other words, $\ce{R2 = H}$. The $1$,$4$-elimination is not restricted. Hence, nucleophilic attack of the $\ce{C=C}$-bond does not take place.
The mechanism by Li and Tochtrop is also useful because it explains, to an extent, the dependence of the end product upon the starting material. Solvent and $\mathrm{pH}$ effects still have to be considered, and the pathway that yields diketones is not covered.

The nature of the starting material seems to have a great influence on the course of the reaction but usually one of the two types of compounds can be specifically obtained by using the right solvent. For example, dioxane favors the formation of the dione whereas t-butanol favors the unsaturated carbonyl compound.$^{[3]}$

The amount of diketone (or even triketone) will further depend on $\mathrm{pH}$ and the quantity of $\ce{SeO2}$.$^{[2]\ [3]}$ Depending on the mixture (e.g., with $\ce{H2O2}$), carboxylic acids and $\alpha$-ketocarboxylic acids are also possible$^{[3]\ [4]}$ with various degrees of saturation$^{[2]}$.

Some information on how to eliminate sidereactions that lead to diketones can be found here:
0 Hans J. Reich, James M. Renga, Ieva L. Reich. 'Organoselenium chemistry. Conversion of ketones to enones by selenoxide syn elimination'. Journal of the American Chemical Society 1975, 97(19), 5434$-$5447. DOI: 10.1021/ja00852a019.

1 K. Barry Sharpless, Kenneth M. Gordon, 'Selenium Dioxide Oxidation of Ketones and Aldehydes. Evidence for the Intermediacy of 8-Ketoseleninic Acids,' Journal of the American Chemical Society 1976, 98(1), 300$-$301. DOI: 10.1021/ja00417a083.
2 Qingjiang Li, Gregory P. Tochtrop, 'New Methodology toward $\alpha,\beta$-Unsaturated Carboxylic Acids from Saturated Acids,' Organic Letters 2014, 16(5), 1382$-$1385. DOI: 10.1021/ol500127u.
3 Alain Krief, Laszlo Hevesi (1988), Organoselenium Chemistry I. Functional Group Transformations, pp 115$-$180. ISBN: 978-3-642-73243-0 (print) 978-3-642-73241-6 (online).
4 Thomas Wirth (2012), Organoselenium Chemistry: Synthesis and Reactions, p 212. DOI: 10.1002/9783527641949.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Finding cubic ice CIF files in crystal database
Could anyone please tell me where I could find the the structure of cubic ice (ice Ic) as a CIF file?
I have been searching the cambridge structural database using "H2O", "water", "dihyrogen oxide" and "ice" as search key words, but cannot find anything!

I don't have access to CCSD from home. Sounds more like a problem in how you're searching. Nevertheless, ICSD has it. If you perform an advanced search "Structure Type" search, it's "H2O(Ic)" as a pre-defined structure. There should be a few hits.

Ok, thanks. I did that, but it doesn't look like any of the files have the hydrogens attached. Do you know where I could find the file with the hydrogens included?  One of the files ( Neutron-diffraction study of ice polymorphs. III. Ice Ic) seems to have three hydrogens attached to each oxygen.

CCSD is probably wrong place for the search. It's organic DB, e.g. the molecule should have at least one C-H bond. Of course there are some exceptions but you will not find a lot of unsubstituted-boranes there either.

One resource you can use is the Crystallography Open Database. Search for structures with two elements O and H, and named “ice”. You will then find several experimental structures for cubic ice:

with downloadable structure files in CIF format: 1, 2. These were done from X-ray diffraction studies and only feature heavy atoms (oxygen) positions.

Although I could not find in any open database a CIF file for cubic ice with hydrogen positions, it is easy to make one. From the paper “High-resolution neutron powder diffraction study of ice Ic”, Le Journal de Physique Colloques, 48.C1 (1987), C1-631:

you can create the CIF file yourself:
data_Ice_Ic
_audit_creation_method         'generated by CrystalMaker 9.2.7'
_publ_section_comment
;
Chemical formula (sum): H2 O

Systematic name: H2 O

Crystal structure data from: W.F. Kuhs, D.V. Bliss, J.L. Finney,  High-resolution neutron powder diffraction study of ice Ic, Journal de Physique Colloques, 1987, 48, C1-631

;
_cell_length_a                   6.358(3)
_cell_length_b                   6.358(3)
_cell_length_c                   6.358(3)
_cell_angle_alpha               90.000(0)
_cell_angle_beta                90.000(0)
_cell_angle_gamma               90.000(0)

_symmetry_space_group_name_H-M     'F 41/d -3 2/m'
_symmetry_Int_Tables_number         227
_symmetry_cell_setting             cubic
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loop_
_atom_site_label
_atom_site_type_symbol
_atom_site_occupancy
_atom_site_fract_x
_atom_site_fract_y
_atom_site_fract_z
    H1    H  0.5000      0.4670(19) 0.4670(19)  0.4670(19)
    O1    O  1.0000      0.3750     0.3750      0.3750

If you see the subsequent comment, this actually didn't answer the OP question - the files in the COD don't place the hydrogens, only the oxygens. To my knowledge, there are no open CIF files for cubic ice with hydrogens present.

@GeoffHutchison but protons are disordered in cubic ice… so if you had hydrogens present they wouldn't be in crystallographic sites with full occupancy anyway

Indeed. I remember some sort of chat with the OP, since CCSD does have a neutron-diffraction structure with disordered protons (and IIRC the occupancies?). That was sufficient once the occupancies were clarified, but the COD structure was problematic, since you have to guess O-H bond lengths and occupancies.

@GeoffHutchison I've made a CIF file from a 1987 neutron paper, and I've added it to the answer, in case it can help someone in the future


text:
Distillation Heating Mantle Setup
In my lab courses, we were taught to use textbooks off the shelf as elevating props to adjust the heights of heating mantles to the right height during distillations, but I was wondering if there was a safer and better way to do so? I recently used a different type of heating mantle that was like a nest and charred the cover of a textbook as well as a heat glove that I put on the second time to prevent further damage to the book.

The use of any flammable material like that is a hazard. You are querying the practice because you've made a correct assessment of the situation. Why haven't those in charge done the same? I hate heating mantles anyway as they tend not to have as appropriate safety cut-offs as stirrer-hotplates.

You need to use something called a lab jack ... just search google to see what they look like.  They are quite cheap too.
You should tell whoever told you to use books about lab jacks, too.

Wow, didn't even know these things existed. I'll notify our chemistry department materials services person.

One supplier is www.fisher.co.uk but there are many others.


text:
What is the amount of nitrate ions in 20 g of Fe(NO3)3?

What is the amount of nitrate ions in $20.0\:\mathrm{g}$ of $\ce{Fe(NO3)3}$?

The chemical formula for the nitrate ion is $\ce{NO3}$, I think.
The molar mass of $\ce{Fe(NO3)3}$ is $242~\mathrm{g~mol^{-1}}$ of which $186\ \mathrm{g\ mol^{-1}}$ belong to $\ce{NO3}$. That means that around $76.85\:\%$ of the substance is nitrate ion.
$20.0\ \mathrm g$ of the substance are equivalent to $20.0\ \mathrm g/242\ \mathrm{g\ mol^{-1}} = 0.0826~\mathrm{mol}$ of $\ce{Fe(NO3)3}$.
Since roughly $76.85\:\%$ are nitrate ions, there are about $0.06~\mathrm{mol}$ of nitrate ions in those $20\ \mathrm g$ of substance.
One mole is $6.02\times 10^{23}$, so if I multiply $(0.06\ \mathrm{mol})(6.02\times 10^{23}\ \mathrm{mol^{-1}}) = 3.612\times10^{22}$.
My Answer: There are $3.612\times10^{22}$ nitrate ions in $20.0\:\mathrm{g}$ of $\ce{Fe(NO3)3}$.
However, that is wrong. The options in the website are:

$1.49\times 10^{23}$
$4.98\times 10^{21}$
$60.0$
$8.25\times10^{21}$

And my answer is not even close to any of them.
What did I do wrong?

The molar mass of iron (III) nitrate is $m(\ce{Fe(NO3)3}\approx 242~\mathrm{g\, mol^{-1}}$. Pay close attention to the unit.
You correctly calculated the amount of substance of iron (III) nitrate to be $n(\ce{Fe(NO3)3}= 0.0826~\mathrm{mol}$
Now you should ask yourself the question: How many nitrate ions are in one formula unit of iron (III) nitrate?

 There are three $\ce{NO3^-}$ per every $\ce{Fe(NO3)3}$.

What does that mean for the number of moles of nitrate ions?

 It means that $n(\ce{Fe(NO3)3} = \frac13\cdot n(\ce{NO3^-})$, so you have to multiply the number of moles by three. $n(\ce{NO3^-}) = 0.248~\mathrm{mol}$

Now you know the number of moles of nitrate ions and you simply have to multiply with Avogadro's constant.

 $N(\ce{NO3^-}) = n(\ce{NO3^-}) \cdot L = 1.49\cdot10^{23}$

authors: حوراء فلاح عبدالله فلاح


text:
Molar conductivity and concentration
For strong electrolytes, according to Kohlrausch, molar conductivity decreases as the concentration increases.
( i don't know why I'm not allowed to attach a photo here in order to show you the diagram ), but however I'm trying to explain how the diagram looks.
In the diagram we have strong electrolytes like $\ce{HCl,\, H2SO4,\, KCl,\, LaCl3,\, Na2SO4}$ whose molar conductivity decreases as the concentration increases. But, what I don't understand is that in the diagram we have also $\ce{CH3COOH}$ which is a weak electrolyte, and his molar conductivity also is decreasing and even far more rapidly, as the concentration increases.
Can somebody explain this ?

Kohlrausch in 1874 he demonstrated that an electrolyte has a definite and constant amount of electrical resistance. From 1875 to 1879, he examined numerous salt solutions, acids and solutions of other materials. This resulted in the law of the independent migration of ions, that is, each type of migrating ion has a specific electrical resistance no matter what its original molecular combination may have been.So this means that the electrical resistance of the ions in the solution that are migrating, and conductivity is just the inverse of this.
CH3COOH which is a weak electrolyte, so does not dissociate completely. Since it's only the migrating ions that affect it, only the dissociated ions affect conductivity and specific electrical resistance no matter what its original molecular combination may have been.
At lower concentrations, since there are more water molecules for every CH3COOH molecule, more of these dissociate so there are more migrating ions meaning a larger conductivity. The molar conductivity is defined as the solution conductivity normalized by the total ionic concentration. So concentration increases without any significant increase in conductivity meaning the molar conductivity decreases rapidly.


text:
Quality of water for drinking purposes
As I’m reading about water cations and anions, I read that the benefits we gain from water are of course lots, but the most important of them is the neutralisation of acids ($\ce{H+}$) from the anions present in water such as:  $\ce{HCO3-}$ , $\ce{CO3^2-}$ and $\ce{OH-}$.
Does this mean that the higher the concentration of bicarbonates, the better the quality for drinking?
Should we choose the bottle of water with the highest concentration of these anions?
And also, when boiling the water we loose bicarbonates, so is boiled water unhealthier?

You're reading pseudoscience. These are most likely people who write rubbish on the internet trying to manipulate people who don't know better for their own personal gain.

Any bicarbonate in your water will be immediately neutralised by your stomach juices, so unless you are drinking several hundred litres of water, there would be literally no difference between drinking pH 7 water and pH 9 water. None of this has any impact on the pH in your bloodstream, which in any case cannot be messed with - it HAS to be maintained within the range of 7.35-7.45.

I find the term neutralising bicarbonate slightly amusing since the stomach’s pH value is so low that protonating and letting decompose would be a lot more fitting ;)

Finally, there are places in Germany where tap water is slightly acidic due to the ground being acidic and that water isn’t unhealthy per se to the people living there (or others). You can safely drink all German tap water regardless of its pH.

Anyway, this question deserves an answer.

You can try drink so much Pocari Sweat.

Drinking nothing but water saturated with highly soluble carbonates/bicarbonates/hydroxides is one of the ways to die in two days or less. Plus the water will taste terrible.

Does this mean that the higher the concentration of bicarbonates, the better the quality for drinking?

Not necessarily. This is more a subjective question involving what tastes good to you.

Should we choose the bottle of water with the highest concentration of these anions?

Assuming the water is actually monitored for safety/quality, the upper limit on caustic (hydroxide) alkalinity is likely too low to do any damage. And carbonate alkalinity will be limited by pH and equilibrium with atmospheric carbon dioxide. So: If it tastes good to you, then drink it.

And also, when boiling the water we loose bicarbonates, so is boiled water unhealthier?

When you boil water, you are increasing the concentration of dissolved species that do not boil off at that temperature. You can, for example, increase levels of nitrate and nitrite in water by boiling it, which is a bad idea if your water is contaminated by large amounts of nitrate/nitrite. Conversely, you can kill some if not all harmful bacteria in water by boiling it. And I don't think that boiling water results in the loss of bicarbonate in any measurable way, at least not on timescales we're interested in (like cooking).


text:
Why do we need to add FeBr3 for bromination of anisole, but not for bromination of phenol?
I understand that we need to add $\ce{FeBr3}$ while brominating benzene in order to polarize the bromine molecule. However, we do not need to add such a Lewis Acid in the case of phenol, because it is activated by the $\ce{-OH}$ group, and so polarizes the bromine molecule on its own. Anisole is also highly activated, however we still add $\ce{FeBr3}$ in order to brominate it. Why?
(An answer which compares bromination in phenol and anisole will be highly appreciated.)

Can you give a reference to the synthesis, please?

I don't think polarization is the issue. Adding $\ce{FeBr3}$ is just to reduce the $\ce{Br-Br}$ bond strength, speeding up the reaction (attack of the aromatic ring on the bromine is part of the rate determining step, which is slow in the bromination of benzene). It functions as a catalyst being recycled when the $\ce{Br-}$ grabs a $\ce{H+}$ to restore aromaticity in the ring. I'm pretty sure — at least I see no reason to doubt — that the catalyst requirements for the bromination of anisole and the bromination of phenol do not differ much, if at all. That is, no catalyst is required for the bromination of anisole.

Um, as far as I know, FeBr3 is actually used to generate an electrophile. It grabs one Br atom from the Br2 molecule and gives us Br+, which is an electrophile. This make it easier for it to under bromination. Refer to this : http://chemwiki.ucdavis.edu/Organic_Chemistry/Hydrocarbons/Aromatics/Reactions_of_Aromatics/Halogenation_of_Benzene-The_Need_for_a_Catalyst
Thanks for the answer though, it answered my question well. Cheers!

@ag_dhruv I agree, but that is exactly what I said? The $\ce{Br+}$ is not present as a lose species, it would be too unstable. I'm also not that big of a fan of UC Davis website. It contains plenty of incomplete (and erroneous) material. Cheers :)


text:
Which compound reacts faster in the Cannizzaro Reaction?

Which reacts faster in the Cannizzaro Reaction?
  a) $\ce{OHC-C6H4-NO2}$
  b) $\ce{OHC-C6H4-OCH3}$

Obviously, a better hydride releasing group will react faster. Therefore my answer was b, as $\ce{-OCH3}$ shows -I (inductive) effect as well as +R (resonance) effect, as it's in the para position. This increases the electron concentration in the $\ce{C}$ atom of $\ce{CHO}$ and due to the excess electrons, it is a better donor of an electron to the $\ce{H}$ attached to it. Thus, leaving a  $\ce{H-}$.
But the answer given is a. The answer states that $\ce{C}$ can only give electron to $\ce{H}$ when it accepts an electron from the $\ce{O}$ attached to it. For that the $\ce{C}$ should be electorn deficient. Therefore an electron withdrawing group (in this case $\ce{NO2}$, which shows -R effect at ortho and para positions) better releases hydride.
But I don't think that happens. That's because if $\ce{C}$ is electron deficient, then it wouldn't give an electron to $\ce{H}$ (less electronegative) in the first place, even if it takes an electron from oxygen (which is also tough).
Can anyone give a proper explanation to the whole problem?

$\ce{-OCH3}$ usually shows a $+M$ (or $+R$ in your terminology) effect …

The mechanism of the Cannizzaro reaction is illustrated below.

The first step involves attack by the nucleophilic hydroxide ion on the positively polarized carbonyl carbon to form a tetrahedral intermediate.  Once the tetrahedral intermediate is formed, substituents on the aromatic ring can have little resonance interaction with the former carbonyl carbon because it is now $\ce{sp^3}$ hybridized. So we need to look at the step leading to the tetrahedral intermediate - the first step.
However, if we examine some of the resonance structures for the starting carbonyl compounds (where the substituents on the aromatic ring can interact with the carbonyl through resonance) we see that the nitro compound places positive charge adjacent to the already positive polarized carbonyl carbon - a destabilizing situation (remember, destabilizing something means making it higher energy, more reactive).  On the other hand, the methoxy compound places negative charge adjacent to the already positive polarized carbonyl carbon - a stabilizing situation.
The destabilized carbonyl in the nitro compound will be more reactive towards nucleophilic attack.

Im really sorry about asking doubts on closed questions, but i just have a wee bit trouble, you have said that "..nitro compound will be more reactive.." Are you implying that $\ce{H-}$ is easily leaving in the nitro compound?

@SujithSizon No, I'm saying that since the carbonyl in the nitrobenzaldehyde is destabilized, it will form more of the tetrahedral intermediate in the first step (which is an equilibrium).  More tetrahedral intermediate leads to faster reaction.

So far so good, after more of the tetrahedral intermediate is formed (of the nitro compound) what is going to happen, i believe the reaction slows down (for nitro compound) after that? And doesn't more tetrahedral product only mean more final product formed?

@ron whereas the methoxy group though lesser amount of intermediate formed but the H- transfer step happens fast (and in overall could be faster than the latter).

@SujithSizon 1) No, more intermediate can also mean faster rate of reaction (e.g. rate is dependent upon concentration); 2) Even in the H transfer step there is more positive charge on the nitro aldehyde carbonyl than the methoxy case (draw resonance structures)

@ron i agree, but for removing H- shouldn't the carbonyl carbon be more electron rich so that hydrogen can take some extra electrons?

The carbon the H is being removed from is sp3 hybridized, there is no carbonyl and because the C is sp3 there is no resonance interaction between it and the aromatic ring.

@ron in the case of methoxy compound there is resonance interaction between C (from where H is removed) and ring right?

No, that carbon is sp3 hybridized, you can't have resonance with the ring

Looking at the reaction after the fact (always easier!): Two factors favoring compound a are (1) the equilibrium between the aldehyde and its adduct with hydroxide ion will more strongly favor the adduct when it has an electron-withdrawing p-nitro group; (2) once that intermediate is formed, it will be faster to undergo hydride transfer to another molecule of the aldehyde, when the latter has an electron-withdrawing group (nitro).

Why will the equilibrium between the aldehyde and its adduct with hydroxide ion more strongly favor the adduct ?

An inductively electron-withdrawing group (like nitro) on the benzaldehyde will make the aldehyde group more electrophilic, and hence more reactive with hydroxide. On the other hand, the p-methoxy group donates electrons across the ring into the aldehyde group, making it less reactive.

You haven't considered that once the tetrahedral intermediate is formed, the nitro group will also compete for the hydride.
When I realized the Cannizzaro reaction with o-nitrobenzaldehyde, the only product I obtained (in high yield) was the o-aminobenzoic acid (i.e. an intramolecular Cannizzaro).


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Why does an element's chemical properties rely only on its valence electrons and not on anything else?
I understand that elements use their valence electrons for their reactions and whatnot, and that the whole idea of the electron-dot structure is all about valence electrons. But why can't the other electrons play a part in it?

They play a part but it's smaller, related http://chemistry.stackexchange.com/questions/27827/do-core-electrons-have-molecular-orbitals

It seems that in extreme conditions this rule can be broken: http://www.scientificamerican.com/article/chemical-bonds-inner-shell-electrons/

@Mithoron If you can formulate an answer based off that article in layman's terms and explain why that is in only extreme cases, that would be awesome!

Technically, other electrons could play a part in determining chemistry. However, there is a significant energy barrier between the p-orbitals of one period and the s-orbital of the following period.
Consider copper. The characteristic deep blue colour of $\ce{[Cu(NH3)4(H2O)2]^2+}$ derives from the absorption of photons whose energy corresponds to the difference of the $\mathrm{3d}_{z^2}$ and $\mathrm{3d}_{x^2 - y^2}$ orbitals. This is spin-allowed because copper(II) is a d⁹ system. The wavelength corresponds to approximately $610\:\mathrm{nm}$.
On the other hand, the energy between a $\ce{2p}$- and a $\ce{3s}$-orbital can be measured using x-ray transition — it corresponds to a photon of a wavelength of approximately $1\:\mathrm{nm}$. This is about six-hundred times the energy difference compared to inside the d-orbitals.
Of course, this is a slightly unfair comparison. The energies between $\ce{3d}$ and $\ce{4s}$ are noticeably larger than inside of the $\ce{3d}$ orbitals, and the energy difference between $\ce{4s}$ and $\ce{4p}$ should again be well larger — but they will remain significantly smaller than energy differences across shells. The former are typically visible or UV, the latter are typically x-ray.

As the valence electrons are the furthest from the nucleus, they are more loosely held by the nucleus. So, they can readily be lost, gained or shared to form ionic or covalent bonds. The inner electrons are tightly held by the nucleus. So, they need higher energy to participate in the formation of chemical bonds.

Yes, but why can't the other electrons also play a part?


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Is the relative natural abundance of isotopes of an element the same everywhere?
Wherever you find potassium, the isotopes are present in a set percentage that exists the same everywhere in nature, but why is that? Does it have something to do with how the element is produced?

Define 'in nature'.  Small isotopic differences can be used in archeology to determine where minerals and gemstones came from to track prehistoric trading routes.  The natural abundance of even stable isotopes will likely be different in different solar systems.

There are plenty of natural processes that alter the ratio of isotopes so the answer in general will be no. An example, evaporation of water alters the ratio of both deuterium and oxygen isotopes and this can lead to significant changes in the ratios depending on the source and history of the water you are examining.

"presnt" -> "present"

The relative natural abundance of isotopes is not the same everywhere.  Depending upon what you mean by "everywhere", there are two cases to consider.

Extraterrestial

Dust from before the sun was formed (stardust, presolar grains) has a very different elemental and isotopic composition than that found on earth.  Depending where a star is in its life cycle different elements and isotopes will be created and destroyed, so it is not surprising that dust from different stars has a very different composition than what is found on earth. As Wikipedia notes,

In these materials, deviations from "natural abundance" are sometimes
  measured in factors of 100

Terrestial

When our solar system was forming, elements and isotopes were not uniformly distributed - close to uniform, but not exactly uniform.  This is due to processes like diffusion in which a mass-dependent fractionation occurs.  A heavier isotope won't travel as far as a lighter isotope in a given amount of time, so the resulting distribution of the two isotopes will not be exactly equal.  While these mass-dependent processes account for most of the isotopic variation in our solar system, mass-independent processes can play a smaller role.  It is also suggested that catastrophic events in nearby stars may have also influenced local elemental and isotopic homogeneity.  These "original" inhomogeneities will be further altered by a continuation of the mass-dependent (diffusion, bond-breaking and making, etc.) and -independent processes. Natural radioactive decay processes can also lead to changes in local isotope ratios.  For example, $\ce{^238U}$ ultimately decays to $\ce{^206Pb}$, so it wouldn't be surprising to find an altered lead isotope ratio (e.g. enriched in $\ce{^206Pb}$) around large uranium deposits.
This paper presents a study of the natural isotopic variation found on earth.  Here is a table from the paper that gives an idea of the range in variation.  As expected it is much smaller than the extraterrestial variation, but certainly not insignificant.  For boron and copper, the isotopic variation is on the order of parts per hundred, but more typically isotopic variation is on the order of parts per thousand.

And of course, humans modify these ratios as well - the easiest visible example being nuclear power plants. Maybe not something you'd call natural, but it seems that there have been natural nuclear reactors on Earth in the past, working much the same way (though with vastly lower intensity).

Yes, good point.

Short answer: it's not the same everywhere on earth.
Ron's answer is excellent, but the main point is that isotopic ratios vary from place to place and over time.
We can use these geographic variations with isotope analysis to track locations and dates of things.

The stable isotopic ratios of drinking water are a function of location, and the geology that the water percolates through. $\ce{^87Sr}$, $\ce{^88Sr}$ and oxygen isotope variations are different all over the world. These differences in isotopic ratio are then biologically 'set' in our hair as it grows and it has therefore become possible to identify recent geographic histories by the analysis of hair strands. For example, it could be possible to identify whether a terrorist suspect had recently been to a particular location from hair analysis.

Similarly, explosives and other compounds are tested for isotopic ratios to determine their origin (e.g., if the material at a crime scene matches the isotopes in a suspect's house).


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Why there is no change in water level when salt is added?
Let us say we have one glass of water and after that when we add one or two spoon of salt then we notice that salt dissolves in it but when we measure the water level we found that there is no difference between initial reading and final reading. So what's happening over there ? Why water level doesn't changed when salt is added ?

A useful term for exploring this phenomenon is the partial molar volume of mixture. Some background can be found here.

Expanding on NSN's point, cases where volume doesn't change at all when a component is added to a mixture point to cases where the partial molar volume of that component is zero.  As the answers say, sometimes adding material can decrease the total volume of a mixture, so partial molar volume can sometimes be negative.

The same can happen with sand!  (At least to some extant.)

This is an example of electrostriction - see, for example, this paper for a full explanation.
In short, the solvent molecules become more ordered in the vicinity of dissolved (charged) ions and less ordered as we go further away from the dissolved (charged) ions. The increase of solvent density near the ions is offset by the decrease of bulk density in the solvent as distance from the ions increases. This results in an overall net reduction in the observed volume of the solution.
From the Discussion section of the manuscript:

The interaction of the electrostatic field of an
  ion with water tends to align the dipolar water
  molecules in the direction of the field. In this way
  the field tends to disrupt hydrogen bonded
  structures in liquid water, and to compress the
  water molecules surrounding an ion. These
  electrostatic effect give rise to a shrinkage of the
  water.

Is it ever possible that dissolving a solute in a solvent will cause the level of the solvent to fall? @ToddMinehardt

@TheCrypticCat - absolutely, and that's something that Michael D. M. Dryden addresses in his answer (which I should have included in mine!).

when measuring concentration in molarity, do we only take the volume of the solvent before adding the solute? do we not care about any volume change which might occur upon adding the solute?

@ibuprofen - For molarity, you are concerned with the volume of the solution after you have added the solvent to the solute. Therefore, the volume change is implicitly included as you correctly note (weigh the solute, add solvent to desired volume - done).

What happens for sugar?

Actually, sodium chloride added to water will decrease the volume of the solution, up to around 2% for a saturated solution. Even when fully saturated, that's not a big change, so you may not have been able to observe it without something with a narrow neck like a volumetric flask. If you start with a large volume and carefully mark the level, you should be able to see it.
Todd's answer explains why this happens with water, but it should also be noted that deviations from the ideal can happen in both directions (increasing or decreasing the volume) and changes in volume also occur when mixing two miscible liquids together, sometimes in a rather complex way. A good example is a binary mixture of acetone and chloroform (there's a good figure for this in the Engel and Reid physical chemistry textbook, but I can't find a decent one online). Starting from a pure acetone mixture, as we add chloroform, the volume is more than the sum of the two component volumes, but once we get above 20% chloroform or so, the volume drops below the sum of the two pure component volumes. Why this happens is a bit complicated, but it has to do with different competing factors affecting how compact the local structure around the molecules is.

When sodium chloride dissolves in water to make a saturated solution there is a 2.5 per cent reduction in volume. One would never notice that in a beaker. Even in an ordinary flask it would be barely perceptible. However, if a volumetric flask is available from the chemistry laboratory, the volume change will be noticeable in the narrow neck. It is essential to remove all air bubbles from the salt that is to be dissolved. Therefore, it must be thoroughly wetted at the start of the demonstration.
The solubility of salt does not change much with temperature, so there is little profit in using hot water.
The salt should be in small crystals and not in rocks or very fine powder.


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Is pigment grade titanium dioxide likely to contain any impurities that are toxic?
I have a bag of titanium dioxide of pigment quality, unknown source. I am curious to know if the manufacturing process of titanium dioxide makes it likely that it contains any toxic by products / impurities.
Would it for example be safe to use pigment grade titanium dioxide to manufacture sunblock?

According to the MSDS for titanium dioxide, the substance itself is suspected of causing cancer. See, for example, this link.
My opinion is that, in general, it's not a good idea to use reagents from unknown sources for any purpose, especially something like sunblock.


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Wavelengths of the visible spectrum
Why is data about wavelengths of different colors and the visible spectrum in general so different in different sources? On Wikipedia, the numbers differ by up to $\pm\mathrm{30~nm}$.

It may all be because there's no distinguishable "line" between for example blue and violet. I imagine this is like saying all compounds with species with electronegativity difference of more than 1.9 are ionic; which is very crude, at times leads to wrong results and confusion.

Many different light wavelengths can produce the same perception of color. Besides, the color perception is not purely physical process, but rather  psychophysical and physiological part can differ from one person to another.

And I seriously doubt that this question is about chemistry: color perception is physical and biological process. There is some chemistry out there, of course, but I don't feel the question touches chemical aspects in any way.

@Wildcat  Chemistry enters in the absorption spectra of the cone receptors -- there's gotta be electronic excitation in there somewhere.

@Brian, I knew someone gonna say something like this. But using that logic I can copy Biology.SE site almost entirely and paste its content here, since ultimately all biological processes are rooted in chemical (and physical) processes. But we don't do so, we distinguish the scientific disciplines by focusing on the most important aspects and abstracting out non-essentials. And in this question the actual chemical way receptors work is non-essential. And besides, electronic excitation is a physical process.

On the same ground, I remove quantum-chemistry tag from the question: I don't feel quantum chemistry is in any way essential for the matter. Of course, one can tag almost any question on this site with the quantum-chemistry label, since its ultimately where the roots of all chemistry are, but, again, we don't do so.

I'm voting to close this question as off-topic because I don't feel that it is about chemistry. There is much-much more physics & biology out there.

And if you are a fan of Wittgenstein, lots of philosophy in it as well...

@Wildcat Glad to be the one not to "disappoint" you.  :-) Concur it's fuzzy. I certainly wouldn't fuss if it did get migrated to Bio.SE. Or, Neuro.SE or whatever's out there.

Given that chemistry significantly applies spectroscopy, I would say that this question is on-topic and rightfully belongs here.  Just because I have a dictionary sitting on my desk doesn't negate the contributions I can get from using Google when looking up a word.

There is a lot of distracting comment here. Distracting because there is a simple and not-entirely-unrelated-to-chemistry answer. Physiology and philosophy are irrelevant. The reason there are different definitions is because the colour scale is continuous and any division into simple colours is arbitrary and different people's definitions will differ. The specific wavelength of the bright lines in the emission spectrum of sodium, for example, is unambiguous, but whether we classify them as orange or yellow is arbitrary.

The comments on the question discuss this may not be about chemistry.
One could argue that colors are chemistry related because
they are associated with chemical substances: like minerals, chlorophyl, rust, dyes etc.
So colors seem to be properties of substances.
And that's fundamentally wrong.
Color is a function of the visual system of the viewer!
If that sounds surprising, think of color blind people, which see different colors.
Also, there are not just us - there are other animals, with different visual spectrum, partly including what we call UV.
Colors are not well defined
Actually, there are more than enough ways, even standards, to define which part of the rainbow is green, exactly - people working with visual media like print design can tell details.
And there are way more details than anyone could expect.
So, without a common definition of the colors, it's not the wavelengths that vary in different sources - it's the colors themselves.

To illustrate that color is more complex when looking closely, below you see a diagram defining color names. It is part of a formal standard which is in heavy use:

From https://en.wikipedia.org/wiki/CIE_1931_color_space


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Solubility in homogenous mixtures of solvents
Alcohol and water are miscible.  I am coming across some chemicals that are soluble in alcohol and insoluble in water.  If I know solubility in two solvents separately, is solubility in a mixture of the two always straightforward, or are there confounding variables?
E.g., if some chemical will dissolve to 100g/L alcohol at 25°C, then will its solubility in a 50/50 mixture by volume of water and alcohol at 25°C always be 50g/L?
Does this generalize to all miscible solvents?

Short answer "no, it doesn't generalize." See the link above for more - solubility in mixed solvents is very complicated.

@Mithoron: Not a duplicate -- in this question I am saying that I know the solubility in both of the solvents.  However I can see that one of those answers (http://chemistry.stackexchange.com/a/32535/5354) seems to answer this question also.

Hi feetwet - do you think you have an answer to your question ? I have an example that may help if not.

@Beerhunter: Sounds like the answer is "no," but examples and explanations would be appreciated!

You can read the IUPAC publication, its a bit hard it depend of your studies level. http://pac.iupac.org/publications/pac/pdf/1990/pdf/6211x2069.pdf
EDIT : Read it also it's a bit better http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2525813/

The complication comes from solubility of the solvents with each other whilst simultaneously having some solvation power of a solute. I have made a compound before that was so water soluble that it caused sodium chloride to precipitate out. A real example was of an organic compound that had 1%w/v solubility in ethanol but around 6.5%w/v in toluene. In a 50:50 mixture,  the compound had solubility of 13%w/v. A completely non-linear result! The solubility was even higher in 90:10 toluene-ethanol. So, the complex nature of three way interactions mean it would be very difficult to predict these solubility graphs just using two data points of the two pure solvents.
If we asked the same question of boiling points, we would point straightaway to azeotropes. So, analogously, other properties exhibit this behaviour. However,  in this instance the nature of the solute as a third component makes it too complicated to predict without having the wealth of property data available for all components. If this is known and you have the software, you may have a chance.


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Magnesium air-fuel cell or battery?
I'm a bit confused.
I've got this "toy" ...
with the help of this site I was abled to find out that the reaction stops with a lack of air. Link to the old question.
The battery/fuel cell or however you call it consists out of following components:

magnesium plate
isolator which consists out of something like paper, the surface feels more like cotton
the black plate is called "black cathode" in the manual, my hypothesis is that it might be rubbered graphite or a polymer, because it has a very low electrical resistance.
the electrolyte is a $\ce{NaCl}$ dispension- I think it just need to lower $\ce{H2O}$'s electrical resistance
plastic stuff is only the case of the "cell"

I did some research about this and the results are very confusing to me.
In the manual they call it a fuel cell.
My chemistry teacher says that in his hypothesis it is impossible to be a fuel cell and has to be a battery. -Because of the great answer to my last question I doubt it or at least I'm too confused to agree.
My research results:
►Battery:

news about a battery for cars
an article about the upper link and about the toy

►Fuel cell:

Wikipedia-MAFT
a kind of datasheet of a "battery" which is in development

►My hypothesis for beeing called fuel cell
→ because fuel cells require according to their definition a constant source of fuel, it might be called fuel cell because of the fact that the "thing" requires air to react.
►My hypothesis for beeing called battery
→ oxygen as fuel?- some people might see it different and call it battery because of that (then there's no constant fuel source →battery) .
►My idea about the chemical equation is:
$\ce{Mg + H2O → Mg(OH)2 + H2}$  (but I doubt it, because I can't imagine that it is responsible to sell a toy which emitts $\ce{H2}$ - or what else happens with it?-maybe the black cathode has to do something with it?)
If you've got a better idea about the chemical equation- let me know.
I would like to know if there's a mistake in my thoughts- and if it's actually a battery or a fuel cell.(the main problem).
And if you've got a better idea about the black cathode component's material- let me know please.
please don't mark it as duplicate to my older question-there's a difference because I added information about the theories and reasearched a lot about it and found the reasons/arguments I've posted here out- the last question was doing it by its components.

What you have is classified as a Mg-air battery. You can read more about it here : https://en.wikipedia.org/wiki/Category:Metal-air_batteries
These are a class of primary(i.e. not rechargeable) batteries, where the anodic reaction is a metal being oxidised and the cathodic reaction is oxygen reduction.
The reactions are:
$\ce{Mg → Mg^{2+} + 2 e^{-}}$ on the anode
$\ce{O2 + 4 e^{-} + 4 H+ → 2 H2O }$ on the cathode, if the reduction goes to completion. Most of the time, it is only reduced to the peroxide or the superoxide which is considered incomplete reduction.
For it to be classified as a fuel cell, the fuel for the anodic reaction would need be brought from the outside(source: anectodal, personal thought).

from where do you get the H+  ?

From water dissociation. The potential and the kinetics of the reaction is pH dependent. As I mentioned, in most practical applications, the reaction only proceeds to superoxide which doesn't involve H+

so from the OH-  or H3O+ ?

H3O+ if you will.

It is a fuel cell because the magnesium anode is consumed by the oxidation reaction and is then replaced by a new anode. This example being a toy is a very simplified version where the full scale fuelcell has a continuous roll of magnesium foil, once the surface is oxidized the roll slowly advances until it is entirely expended and then replaced to start again. If the container were not excess able for cartridge replacement it would be disposable and would be considered a primary cell.

I would say more :
$$\ce{O2 + 2H2O + 4e- → 4OH-}$$
$$\ce{2Mg + 3OH- → 2 Mg2+ + 4e-}$$
(2.93V global potential so 6,46 kWh/kg theoritical energy density)
It is a potential fuel cell by adding magnesium pellets continuously with oxygen from the air.
Great research anyway and good scientific spirit, I'm now looking for information about KIST car but can't find. Maybe it has a link with the fact that the autonomy is twice compared with today's electric cars?

Please use mathjax to format mathematical expressions and chemical equations. To learn more about mathjax, refer to How can I format math and chemistry expressions.


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Deciding the major product in this Wurtz reaction
Today I was told:

When 2-chlorobutane is reacted with 2-chloropropane in the presence of sodium and dry ether the following products are formed:

2,3-dimethylpentane
3,4-dimethylhexane
2,3-dimethylbutane

3,4-dimethylhexane, being most symmetric, is the major product.

I'm not satisfied with the reason provided. Can anyone please explain the fact in a better manner?

What I'm unable to understand is that both 3,4-dimethylhexane and 2,3-dimethylbutane and equally symmetric since both have a plane of symmetry, don't they? Then how are we using symmetry to compare the yield?

What part of it didn't you understand? You may get better answers if you exactly point that out.

"2,3-dimethylhexane, being most symmetric, is the major product." 2,3-dimethylhexane is not an expected product, do you mean 2,3-dimethylbutane?  Were the starting materials present in equimolar amounts?

@ron I think OP meant 2,3-dimethylpentane, which would statistically be the major product if the starting materials are present in equimolar amounts. (That has nothing to do with symmetry... so I don't know what OP was told.)

@ron the OP had it right then did an edit.

Sorry for the confusion here, it's actually 3,4 and not 2,3. :)

I've edited the question, @M.A.Ramezani

I think that there should be a 'statistical product distribution', if I'm correct the products should be in the ratios 2:1:1.
You get to this conclusion as follows, call 2-chlorobutane A and 2-chloropropane B.
A can react with A to give AA, and
A can react with B to give AB, while
B can react with A to give BA, while
B can react with B to give BB.
When you realise that AB and BA are the same molecule you get AA:2AB:BB; that is, 1:2:1.

+1 Then, is symmetry responsible for anything here?

Not as far as I can see.


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How does the catalyst in a polyester resin or epoxy work?
I would like to understand how the polyester resin or epoxy catalyst works and to what degree it must be mixed with the base material in order to be effective.
Typical product instructions advise to thoroughly mix the catalyst for a specified period of time.
How thorough does this really have to be in order to have the material cure properly? Is the mixing done just to initiate a thermosetting behavior and does that just continue to propagate heat to the rest of the material to complete the curing of the entire mass? If this is the case, wouldn't the base material cure if it were heated sufficiently without the presence of a catalyst?
Or is something else happening when the catalyst begins the cure? And how does mixing effect the quality of the cure?

Are you sure that epoxy relies on a catalyst?  I thought that at least sometimes, the materials mixed are two solutions of two different monomers which can undergo a condensation polymerization with each other but not with themselves.

If you look at the two tubes you will probably see that one component is a diamine and the other is a di- or tri-terminated oxirane / epoxide.  Both are required for polymerisation as the manufacturers intended. Water may in theory react with the epoxy but it might be a lot slower and not give you a hardened adhesive.

The degree of mixing is very important in "two part" epoxy binder systems for coatings and adhesives.
In most cases the two liquids are quite viscous and the liquids themselves are better described as suspensions or colloidal dispersions rather than solutions. Unlike say two simple aqueous solutions when combined our two prepolymer liquids will not readily homogenize in a reasonable amount of time unless they are mechanically mixed in some way.  Even if the liquids do combine very slowly they will often gel or vitrify by the molecules at the interfaces reacting before the liquids homogenize.
In many epoxy systems we have, as Beerhunter stated above, some oligomer or polymer with epoxy functionality in one part and some liquid with amine (or sometimes hydroxyl) functionality in the other. Every amine needs to "find" an oxirane ring to attack or the curing process simply can't occur. The curing process works by nucleiphillic attack from amines to open epoxied rings.
The presence of leftover unreacted amine in cured epoxy materials often results in a phenomenon known as amine blushing or amine blooming. leftover amines or polyamides migrate to the surface and may in some cases react with air to form carbamates. This can be disastrous for desired materials properties like tack, adhesion (for the epoxy as a substrate), or cohesive strength of the bulk material.
In summary you can't build a robust three dimensional polymer network without thoroughly mixing because all the reactants can't react. Always mix as thoroughly as possible to get complete conversion.
My experience with polyesters is limited so I will not comment on that.

Excellent response, for a chemist kind of mind, a layman need to have a little interpretation .. what he basically said id that the hardener (part b) affects the resin (part a) chemically. the manufacture has simplified the mixing by changing the basic concentrations so that ...generally part A & B are the same amounts.

I have been using epoxies for decades and I can tell you mixing is the most important step. I use ziplock bags to mix the parts in . one corner a the other b. and you can knead the two together in the bag then icing


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Calculating the mass of acceptable carbon monoxide in a room

The acceptable concentration of $\ce{CO}$ in the air is $10\:\mathrm{mg/m}^3$. In
  a room that is $19\:\mathrm{m\times4.0\:m\times25\:m}$, what is the acceptable mass in
  kilograms of $\ce{CO}$?

The room has a volume of $1900\:\mathrm{m}^3$. Since there are $10\:\mathrm{mg}$ of $\ce{CO}$ per $\mathrm{m}^3$, there are $19000\:\mathrm{mg}$ of $\ce{CO}$ in this room. That’s $19\:\mathrm{kg}$.
But the book says the answer is $1.9 \times 10^{-2}\:\mathrm{kg}$. How come?

You have to be careful with your conversions - you are missing a step in your conversions.
You have 19000 mg
As 1 mg = 10-3 g  and
1 g = 10-3 kg
So:
19000 mg = 19 g = 1.9 x 10-2 kg


text:
How does Ammonia and water mixture works in an absorption refrigeration?
As per one of the questions in this site $$\ce{NH3 + H2O -> NH4 + OH}$$
However, per another perspective in absorption refrigeration

The mixture is placed in the same container. Then when it is heated the mixture is separated and $\ce{NH3}$ changes phase, becomes a gas which moves further to cooling, while water is trapped down in the reservoir.

Which one is correct?

The difference between the absorption refrigerator and the more common refrigerator is that there is no compressor in the absorption refrigerator.  The Absorption refrigerator still needs a way to convert the evaporated refrigerant back to a liquid state.  Instead of compressing it in one step, the evaporated refrigerant is absorbed by a liquid.  The liquid is heated and the refrigerant evaporates from the liquid.  The refrigerant is cooled in a heat exchanger and condensed.  It is now a liquid and ready to go back through the cycle.
The equation you have written is a reaction between ammonia and water.  This reaction will occur in an absorption refrigerator that uses water as the absorber and ammonia as the refrigerant.  The equilibrium constant of the reaction is low so it is not accounted for in a discussion of how an absorption refrigerator works.


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Free Web-Based Orbital Solver to look into the Walsh orbitals of a perturbed cyclopropane
There exists a free online orbital calculator.
When I draw cyclopropane it plots three molecular orbitals, but unfortunately it doesn't use the Walsh orbitals.
Are there any free online tools which can solve such the problem?
(It would be very nice if it would also be an android app. Please let me know if you know one.)

The section below is for interested chemists:
The reason why I want to do that:
Cyclopropane can, to a good approximation, be described with the following orbitals:

The question was how the molecule orbitals behave if we add an electron withdrawing group (a $\pi$ acceptor because some people call it the "CXZ model")

The solution says:

In the substituted case all energy levels are lying more down in energy.

I don't agree because I only see overlap with the Walsh Möbius set of $sp$ orbitals in the following sketch:

The Hückel set (a simpler way to put it: the energy lowest orbital in the scheme of cyclorpropane; used above) gives no overlap with the $p$ orbitals of an electron withdrawing group, e.g. X=COOH.
As I see from the orbitals' interaction, I believe the solution is wrong, I don't agree that all orbitals are lowered.
EDIT:

Sorry for the mistake in the sketch, there is of course only one of the two degenerate Möbius orbitals lowered.
I guess COOH not only has a $\pi$ acceptor effect but also a $\sigma$ acceptor effect which could lower all orbitals.

Don't know if this help but you can compute the canonical MOs (at the RHF/STO-3G//PM3 level of theory) using http://molcalc.org/

Related: http://chemistry.stackexchange.com/a/10655/4945

@JanJensen I tried your page, however I like to get a graphical interface similar as in my SHMO solver, so I don't have to manually plot the energy levels. If there exist a freeware for Windows, which can easily do that (or an Android app) I would also be interested in.


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How to perform solvent extraction of a solid
Suppose I have a solid compound from which I want to extract a target chemical I know is soluble in acetone.  So I crush and mix the compound in acetone, dissolving the target and suspending the detritus.

What are methods for filtering the detritus from the solution? I have frequently seen coffee filters used for this purpose.  Are commercial coffee filters considered "lab-grade?"  Or, suppose that the target chemical has some affinity for cellulose, what alternatives exist for suspension filtration?
Now, having the target chemical in solution I assume it is easiest to extract it by evaporating the acetone.  Is there a process for this that doesn't leave any of the target stuck to the glass or whatever was containing the solution?

What are methods for filtering the detritus from the solution? I have frequently seen coffee filters used for this purpose. Are commercial coffee filters considered "lab-grade?" Or, suppose that the target chemical has some affinity for cellulose, what alternatives exist for suspension filtration?

The method that comes to mind for this task is a Soxhlet extraction. Commercial coffee filters are not "lab-grade." Soxhlet extraction thimbles are available in cellulose.

Now, having the target chemical in solution I assume it is easiest to extract it by evaporating the acetone. Is there a process for this that doesn't leave any of the target stuck to the glass or whatever was containing the solution?

Yes - rotary evaporation, in conjunction with Soxhlet extraction.

I don't follow the second answer: How does rotary evaporation avoid adhesion problems?

The rotation of the evaporation flask ensures that a minimal amount of solute remains on the wall of the flask. As you probably suspect, it's not perfect: you will still have to wash down the sides of the flask with the solvent of your choice, pour it out, and dry it on a watch glass. In my opinion, it's the best way to quickly concentrate an extraction without inducing thermal decomposition/reaction.

A typical undergraduate experiment is the extraction of the triglyceride of myristic acid from nutmeg using Soxhlet apparatus.


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Why is lithium bromide used in absorption refrigerators?
Lithium bromide refrigerator
Why is lithium bromide used in absorption refrigerators?  It is more expensive than sodium chloride.  The wikipedia article on lithium bromide says it is extremely hygroscopic.  However, it is less soluble in water than NaCl.  I am thinking that a salt capable of generating a more concentrated solution would work better in the absorption refrigerator as it would increase the propensity of water to leave the gas phase and enter the solution.  Am I wrong about that?  Is the affinity of water for lithium bromide so high that it is the most important factor?

Your summary is correct; $\ce{NaCl}$ is more soluble than $\ce{LiCl}$ or $\ce{CaCl2}$, for that matter, but both $\ce{LiCl}$ or $\ce{CaCl2}$ are far  more hygroscopic. The explanation of hygroscopicity involves various mechanisms, including adsorption (surface forces) and absorption (requiring transport to the interior of the substance) due to hydrogen bonding, coordinate covalent bonding etc.
So these are distinct, though related, properties: solubility (limit on how much solute the solvent holds) and hygroscopicity (how avidly the solute attaches to the initial molecules of solvent).
For that reason, $\ce{CaCl2}$ is used to absorb water in damp closets, though it will eventually deliquesce into a sodden mess.
BTW, Isaac Asimov wrote of a compound so hygroscopic that it dissolved before it hit water: The Endochronic Properties of Resublimated Thiotimoline, often cited.

Calcium chloride does itself dissolve before hitting water! Wait long enough, and the sodden mess will be a clean solution. That's slow because the liquid has a much smaller surface.


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Can all species that undergo aldol reactions also undergo Cannizzaro reactions?
Can all species that undergo aldol reactions also undergo Cannizzaro reactions? Ater all, the nucleophile can always attack the carbonyl group.

No. The Cannizzaro reaction only works for aldehydes with no $\alpha$-protons, benzaldehyde being a classic example.
For a start, ketones and any other species that can form stable enolates, such as esters or carboxylic acids, can undergo the aldol reaction or similar reactions like the Claisen condensation. However, the Cannizzaro reaction is limited to aldehydes because only they can disproportionate to form alcohols and carboxylic acids.

Secondly, the Cannizzaro reaction involves strongly basic conditions, typically using sodium or potassium hydroxide. Under these conditions an aldehyde with $\alpha$-protons will undergo deprotonation to form an enolate much more readily than nucleophilic attack on the the acyl carbon by a hydroxide ion.


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How does sulfur has atomicity 8
How does sulfur has atomicity 8, when its valence electrons are 6 like oxygen- ie even 2 sulfur atoms can share a pair of electrons and achieve the octet rule, why should it involve 8 atoms?

You may want to clarify the question you're asking. Sulfur exists in multiple allotropes.

Octasulfur is just one of few different allotropic forms of sulfur. It happens to be the most stable form at room temperature, but different conditions favor different forms. Thus, for instance, sulfur vapor usually contains $\ce{S, S2, S4, S6, S8}$, and the higher the temperature, the lower is the amount of $\ce{S8}$ molecules since they break up into smaller allotropes of sulfur.
So, while two sulfur atoms can share a pair of electrons by forming a diatomic disulfur molecule with a double bond, so can also eight atoms, but by forming a puckered ring structure with few single bonds. Then it is basically the question of thermodynamic and kinetic stability, which form would dominate others at some particular conditions.