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Heats of combustion and stability of rings

Which isomer would have the largest heat of combustion?
A) Propylcyclopropane
B) Ethylcyclobutane
C) Methylcyclopentane
D) Cyclohexane
E) Since they are all isomers, all would have the same heat of combustion.

Can anyone explain why?
If you think about it, without having the data of the heat of combustion of each type of bond, you can't really answer this question. I know that cyclopropane has a higher heat of combustion per bond than cyclobutane. However, that, in my sense of viewing the problem, does not prevent the fact ethylcyclobutane could have a higher heat of combustion as it is a cycloalkane with 4 bonds rather than 3.
It could happen that the three bonds in cyclopropane and the three bonds in the alkyl branch are lower in energy that the four bonds in cyclobutane and the two bonds in the alkyl branch.
For example:
Let's say a cyclopropane C−C bond is 10 J.
A cyclobutane C−C bond is 9.5 J.
A regular alkane bond is 8 J.
For propylcyclopropane:
Total energy is 54 J
For ethylcyclobutane:
Total energy is 54 J
The same! I know this sounds really simplistic and dumb, but I just don't understand how you can assume the answer is A.
How do you analyze this situation? Please help.

Compounds A-D all have the same molecular formula, $\ce{C6H12}$.  We can burn each compound and measure the heat given off (heat of combustion). Since they are isomers, they will each burn according to the same equation
$$\ce{C6H12 + 9O2 -> 6CO2 + 6H2O + heat}$$
Any differences in the heat given off can be used to say that a compound is more stable (it had a lower energy to begin with, so less heat is given off) or less stable (it had a higher energy to begin with, so more heat is given off).
This link provides the heats of combustion for some useful model compounds.  Look at the last column ("Total Strain") in Table I, it shows that cyclopropane is slightly more strained than cyclobutane, while cyclopentane and cyclohexane are both much less strained.  The strain energy (SE) in cyclopropane will not change appreciably when we add a propyl group to the ring, nor will the SE in cyclobutane change appreciably when we add an ethyl group to the ring.
Therefore, since cyclopropane has the most ring strain and since propyl, ethyl and methyl groups don't contain any SE, the correct answer to the question is "A".

Ron, in this case, you are using data to say that the TOTAL  energy of the THREE C-C bonds cyclopropane is higher than the total energy of the FOUR C-C of cyclobutane. If we had only known that each C-C bonds in cyclopropane were higher in energy than the each one in cyclobutane, could we still be able to answer in the way you did?

Sorry for the nitty-gritty details, I'm just the kind of guy who obsesses about that kind of stuff!

yolo123, a) I'm using the data to say that a cyclopropane ring is always more strained than a cyclobutane ring, which is always more strained than a cyclopentane ring (by definition, a cyclohexane ring is strain free).  b) Rather than view the data on a "per bond" basis, it is more commonly viewed on a "per CH2" basis - see the second to last column in the Table, and yes, just using this "per CH2" data we could have answered the question in the same way, e.g. how many CH2's are in each ring.

Sorry for not being clear:
I meant being able to answer this ONLY QUALITATIVELY. Without the data, just by knowing that strain/CH2 is higher in cyclopropane without any quantitative knowledge. After all, in an exam situation for which this question was made for, we do not have data.
I guess we better know that the total strain is also higher in cyclopropane to answer this question.

Yes, just like we learn methyl, ethyl, propyl, we also learn that ring strain decreases from cyclopropane to cyclobutane (but they're close) to cyclopentane to cyclohexane.

The products of combustion are the same so the compound with the highest potential energy (least stable) will have the highest enthalpy of combustion.  All have the same number of carbons.  The cyclopropane has the most ring strain and therefore the highest potential energy.
If you use metacognition you can see what's being asked.  They can only write questions asking "Which has the most ring strain?" so many times.  This question is supposed to involve more thinking but can be distilled down to "Which has the most ring strain?"
A good strategy in answering multiple choice questions is to determine what is the same and what is different about the answer choices.  In this case the number of carbons is the same so the products are the same.  What is different is the number of carbons in the rings and therefore the angle strain is different.

How can you know that 3 cyclopropyl bonds + another C-C alkly bond is more energetic than 4 cyclobutyl bonds?
I know a cyclopropyl bond is more energetic than a cyclobutyl bond.

You can infer it from the intent of the question.  There's no way it could be anything else and be a fair question.  Chemistrywise look at the angles.  60 degrees is a big difference from 90 degrees.  http://classes.yale.edu/chem220/STUDYAIDS/thermo/cycloalkanes/cycloalkanes.html  http://www.sparknotes.com/chemistry/organic2/carbocycles/section1.rhtml

Ok, so, there is nothing certain about the answer :S Are you sure there is no scientific way of proving this with the question without going to the "how to answer test questions realm" and looking up data? By the way, thanks a lot for the help. Really appreciate it.

The scientific way is to look at a heat of formation table.  The thing is, the ring strain tables I found are close enough that I wonder if A really is the answer.  It could be a bad question.

Very confusing.


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What is the Natural abundance of Potassium

The three naturally occurring isotopes of potassium are $\ce{^39K}$ ($38.963707\:\mathrm{amu}$); $\ce{^40K}$ ($39.963999\:\mathrm{amu}$); and $\ce{^41K}$.
The percent natural abundances of $\ce{^39K}$ and $\ce{^41K}$ are $93.2581\%$ and $6.7302\%$ respectively.
a) What is the natural abundance of $\ce{^40K}$?
b) Determine the isotopic mass of $\ce{^41K}$.

So I started to utilize this equation:
$\mathrm{atomic\:mass = (fractional\:abundance\:isotope\:1
\cdot mass) + (fractional\:abundance\:isotope\:2 \cdot mass)}$
But then I realized that I did not have some values. $\ce{^39K}$ is the only isotope which has two values. $\ce{^40K}$ only has the mass weight, however, if I were to multiply $\ce{^40K}$'s mass with ($1-0.932581$) would that help me arrive at an answer?
Also, would that answer be accurate since it only takes into account that there are only to values that add up to 100%.

You can answer a) from the fact that they should add up to 100%. To answer b) you either need to know atomic mass (available from periodic table, if you are allowed to use) or can make e.g an approximation that an added neutron's mass is more or less that same.

All three or just the two? And if so how should I set them up? Thanks in advance.

Sorry, can you rephrase your question? What is three or two? What do you want to set up?

How do I solve for part a in the question? Do I utilize the formula that I wrote with the question?

As I said, one way to answer a) is that you assume the three isotope fractions should add up 100%.

Okay so the set up would be 39.0983u = (fractional abundance isotope-39 [0.932581]x 38.963707u) + (fractional abundance isotope-41 [0.067302] x mass [unknown]) + (fractional abundance isotope k-40 [unknown] x 39.963999)

What should I place in the unknown areas? (0.932581-1) and (1-x)?

A). If you know the percentages of the abundance of $\frac{2}{3}$s of the isotopes then finding the percentage of the $3^{rd}$ is as simple as
$ 100\% - (\%_{\ce{^39K}}+\%_{\ce{^41K}}) = \%_{\ce{^40K}} $
B). Finding the mass of the isotope should be pretty simple. It would be
$(\mathrm{Mass\ of\ proton} \cdot \mathrm{{\#\ of protons}} )+(\mathrm{Mass\ of\ a\ neutron} \cdot \mathrm{{\#\ of neutrons}} ) \approx \mathrm{Mass\ of\ isotope}$
You neglect the weight of electrons because the mass is small relative to the others.

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Regarding preservatives in food
Can sugars like fructose-glucose or high-fructose corn syrup that are found in many types of processed food contribute to shortening a products shelf life?  Don't bacteria love things like sugar so if a product has a lot of fructose-glucose added to it, it will need even more preservative chemicals to compensate for any increase of sugar?

Does the phosphoric acid in POP drinks get rid of bacteria? If so the POP manufacturers wouldn't need preservatives.

Two factors may be at play. (1) While bacteria and molds do indeed grow on sugar, a very high concentration of sugar inhibits growth. So fruit preserves, syrup, honey, molasses etc. do not need to be refrigerated and generally do not require other preservatives to be added. Soft drinks contain less sugar and are more susceptible to spoilage, hence the use of preservatives. I would definitely refrigerate a soft drink after opening it, if not consumed immediately. Acidity also helps inhibit growth (e.g. sweet pickles).
(2) Another factor is that bacteria and molds require water for growth, so foods such as dried fruits, that contain a moderate level of sugar but only a small amount of water, may have a long shelf life.

If the concentration of sugar is high this inhibits bacterial growth. Does this mean Coka Cola and Pepsi and lots of other Pop drinks have a long shelf life? If so why do they need Preservatives?

No, those drinks don't have a high sugar content by microbial standards, even thought they it's high by human dietary standards. In fact, the main microbial effect in those cola's is their acidity (pH=4 IIRC, phosphoric acid)

Does the phosphoric acid content destroy bacteria? If so why do they need preservatives?

Does sugar act like a preservative?

Yes, when it is present at a high concentration (e.g. syrup, honey, fruit preserves, etc.)

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NMR: Most deshielded protons (electron withdrawal)

Of these two molecules, which one would be expected to have the most deshielded signal?
First off, I think the wording of this question is wack because neither of these molecules has only one NMR signal.
In any case it's clear that the question is asking about which molecule has the highest degree of electron withdrawal ...
I chose E, because I was looking at the benzene's signals. This one has an alkyl group, which is a slight inductive donor, thereby shielding the benzene's protons (slightly). This one also has a benzylic carbonyl group, which, according to my teacher, is only a resonance withdrawer and not in any form a resonance donator. Based on this, I chose E as the more de-shielded.
The other group similarly has a C=O resonance withdrawing group, but a resonance donating -OH group. Given that the resonance effect usually trumps the inductive effect, I said that E is still the more deshielded of the two.
I realize I might have now erred in limiting my analysis of shielding to only the hydrogens on the benzene. In E, the methyl's hydrogens are fairly well-shielded still, being connected to a carbon (albeit an sp2 hybridized carbon). On the other hand molecule F has a long linear chain; the hydrogens toward the end of that linear chain will have well-shielded signals! F's linear chain is also longer than E's linear chain ...
Frankly I think this is a horrible question but in any case the answer is apparently "F" ... do you agree?

I'm fairly sure that the proton giving the signal with the highest chemical shift (i.e., the most deshielded proton) is in fact the phenolic proton of molecule F. The presence of an EWG in the para- position likely contributes to it having a particularly downfield signal.

Perhaps the question means, in each molecule, which proton is most deshielded?  Draw resonance structures involving the carbonyl and the aromatic ring (E and F) and the hydroxyl group and the aromatic ring (F).  Which aromatic carbon atoms wind up with a positive charge?  Which protons are attached?  These protons are probably the most deshielded.

@ron that is a possibility; I will clarify with my teacher tomorrow. And Greg E. I agree; the oxygen is able to donate electrons to the ring and that leaves the proton quite naked.

@GregE. 's comment is correct.  But if that is the answer they were looking for then, IMO, it is a bad question.  OH signals are all over the place, only with great care and attention to detail can you see them way (10-12 ppm) downfield.

Wouldn't that Oh signal, the one at 10 to 12, be the oH of a COOH?

@Dissenter, aldehydic protons can sometimes be seen in that range (10-12 ppm), as can phenolic protons when the ring is conjugated to strong EWGs, along with those attached to various other less common functional groups (some imines, enols, etc.).

This is the typical result for phenol where traces of water haven't been removed, see http://www.chemicalbook.com/SpectrumEN_108-95-2_1HNMR.htm

I don't believe there is anything really wrong with this question. Yes, both molecules have multiple signals, but one of them contains a signal which corresponds to it being the most deshielded. In fact, one of these, molecule F, contains the most deshielded signal AND the most shielded signal.
"In any case it's clear that the question is asking about which molecule has the highest degree of electron withdrawal ..." not necessarily so. The correlation between electron withdrawing groups and shielding/deshielding is a strong rule of thumb, but can become complicated with aromatic systems and competing electronegativities of substituents. In any case, you need to be looking at parts of the molecule; shielding is quite a localised effect, and as I said above, molecule F actually has the most shielded AND deshielded protons.
In any case, if you consider the electronegativity of oxygen; a proton attached to this will be strongly deshielded. Stick them on an aromatic ring para to a carbonyl group, and you will generate a resonance system like below:

In this case, the phenolic proton is by far the most deshielded of all protons in the molecules shown; I'd expect a chemical shift of over 10ppm. In response to some of the comments about -OH signals being over the place; it is important not to confuse raw chemical shift and averaged chemical shift resulting from chemical exchange with other labile protons, such as the (often observed) exchange of phenol protons with water. In these cases, the resulting chemical shift depends on the relative concentrations of the exchanging species, the exchange rate, and the original chemical shifts of the non-exchanging species.


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Synthesis of synthetic detergent with Friedel Crafts Acylation
I propose the following synthesis.

Oxidation of aldehyde to a carboxylic acid. This sets me up for FC acylation once I ...

Attack the -OH with thionyl chloride ad form the acid chloride.

Reaction with vanilla benzene in the presence of Lewis acid catalyst. Benzene won't go after acid chlorides unless there is some catalyst in the system.

Now that the acid chloride is installed, I can go through the motions of getting rid of the pesky C=O group.

Given that alkyl groups are slightly activating and are ortho/para directing, I can sulfonate with fuming sulfuric acid and expect a mixture of products (including the one desired by the synthesis). Sterics might favor the para product over the ortho product given the sheer size of the SO3 group.

General comments:
I did NOT sulfonate first, as that is an electron-withdrawing group akin to the nitro group. It deactivates the ring and is also a meta-director. I want a para-substituted product. Not a meta-substituted product.
Therefore, how reasonable is my synthesis? Is it correct?
Also would sulfonation before getting rid of the C=O work? Something tells me no; the C=O is able to withdraw electrons from the ring via resonance, and that would make it a meta-director.

Looks good, sulfonation as the last step is the way to go for the reasons you provided.

@ron thanks! Can't wait to ace this test b/c of you!

Dissenter good luck!

Your synthesis looks reasonable, and it is correct to perform sulfonation as the very last step, because the alkanoyl group would be meta-directing for the reason you have provided.
You can simplify your synthesis by reducing the carbonyl group away in one step with the Clemmensen reduction, which works well for aryl-alkyl ketones produced by Friedel-Crafts acylation.

Alternatively, the Wolff-Kishner or Mozingo reaction could be used. Reducing the number of steps in your synthesis would in practice have the benefit of increased overall yield for the final product, as the yield of each single step is usually < 100 %.

When would yield for each step be = 100% then?

@Dissenter When in each step, all educts are converted into the desired product, but this is hard to achieve in practice. The mechanisms of most organic reactions contain a lot of equilibria, and there are also often possible side reactions which lead to by-products.

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Making metal powders using electrolysis - chosing the right electrolyte
Really fine metal powders are good for two things: making paints and pyrotechnics. For this purposes, I can see people usually use thing called ball mill.

It's basically a barell stuffed with heavy balls and the thing to be powdered. You leave it for a while, rotating all the time, and the balls slowly crush your substance into smaller and smaller pieces.
There are many disadvantages to this:

I'd need to construct it
It would be running and making loud mess all the time
soft metals, like aluminium and copper don't even crush into pieces

So I have figured out a different way. More chemical one I'd say. I can dissolve metals in acid by forming salts out of them and then electrolyse those salts. Not so long ago, I complained that instead of copper plating I got copper powder. That's just what I want now. I'm really sorry that powder I created the other day is already flushed down the drain. It could make beautiful green flames.

Now while the general idea is easy, I'm not so sure about choosing the right electrolytes. My theoretical knowledge of electro-chemistry unfortunately sucks. And I'm afraid that every metal will need different electrolyte.
For example I think using any acid for Aluminium could cause something nasty since it reacts with them little bit too quickly. For Iron, I tried hydrochloric acid and then I found out thet $\ce{HCl}$ dissolves iron to form iron chloride.
Are there general rules to figure this out? If not, what sould I consider to find electrolyte individually for these metals:

Aluminium
Iron
Copper
Magnesium

Powdered metals can be bought of course, but I don't want to waste good money for something, that can be easily made from trash (old nails, aluminium foils from chocolate...).

Use sulfuric acid- sulfates don't give significant side reaction. A standard solution of say 30% sulfuric acid should corrode all of those metals.  Chlorides will give some toxic risk with chlorine gas production.

I'm going to try it and post the results right away.

Reaction with aluminium produces gas that is so irritating that I can't tell what does it smell like. It smells like breathing liquid sodium.

So does reaction with iron. My nose really hurts now. And I was sniffing carefully.

Fine copper powder can be prepared using thermal decomposition of copper citrate (can be prepared by mixing dirt cheap citric acid, three equivalents of a base like potassium hydroxide and copper sulphate). I once made pyrophoric (!) copper powder using this technique. There are mentions of preparation of iron powder same way. Anyway, it is much more practical than electrolysis for big batches.
Frankly speaking, fine aluminium flakes are widely employed in various paints and are dirt cheap. They should be available in shops for artists. There is no reason to make them at home (and aluminium reacts with water once cleaned of oxide film)
Magnesium is probably the worst. It is quite costly and active and making fine powder is dangerous. It also reacts with water and alcohols, so making it employing electrolysis is out of question. There are mentions of home smelting of aluminium and magnesium, producing magnalium. Magnalium is brittle and can be easily milled, and it is active enough so fine milling is usually not needed. While fancy, proper pyrotechnics can be made without magnesium with easy, aluminium is usually much more preferable. Magnesium often produce huge volumes of white smoke, a habit aluminium doesn't have.
Regardless of above, I strongly encourage to be cautions when working with metal powders, especially magnesium.

So what do I need? Citric acid? Potassium hydroxide? Copper sulphate?

@TomášZato For copper powder, if you want to make it, you need copper sulphate or chloride and potassium or sodium citrate. If they are unavailable, they may be prepared before use in solution by mixing citric acid and potassium or sodium hydroxide. Precipitated copper citrate should be carefully heated under cotton jam, so gases could exit, but fresh air was blocked. I believe, that for iron similar procedure may be used, except instead of copper salt one should use iron (II) salt, but it is an educated guess, I didn't do it.

After some research, I can't seem to find a way to get aluminium out of aluminium sulfate. Using baking soda, I can turn it into aluminium hydroxide. But what then?

@TomášZato Forget about making aluminium or magnesium at home, using electrolysis or other methods. It is theoretically possible (not in water medium, of course), but requires too much work and involves significant risks.

Permeakra, I tried to perform the sodium citrate method and it doesn't seem to do anything. 1. I mixed NaOH with citric acid. 2. I cooled down the solution and mixed with CuSO4. Result: No precipitate appeared - so there was nothing to heat. I just got blue solution.

@TomášZato First, the amounts of reagents must be right (3 mol NaOH, 1 mol Cictric acid, 1.5 mol CuSO4), and second, the precipitation occurs after some time, not immediately. A deep blue solution is formed first and may be heated to quicken the precipitation. Yeah, it's a bit inconvenient. I heated the mixture to ~50-80 C  for 10-15 min before precipitation occurred,   some sources recommend merely leave the reaction mixture overnight.

Regarding the suggestion in the comments above: Concentrated sulfuric and nitric acids produce $\ce{SO2}$ and $\ce{NO2}$ gases, respectively, as they oxidize metals. This is a common way of dissolving many metals, but you need to do it somewhere where you won't breathe them in.
When you're actually electroplating, you'll want to dilute them with a different electrolyte. Potassium nitrate or citrate or phosphate buffers are common choices. Anything that doesn't complex or form an insoluble salt with the metal will do, though adding things like surfactants may change the morphology of the deposited metal structure. Depending on the metal, you might get precipitation if you raise the pH too high, but you don't want to electroplate in concentrated acid as you'll just redissolve your deposits.

Can I use washing soda?

Probably not a good idea. Many carbonate salts are insoluble (Cu and Fe) or unstable (Al) and it's going to neutralize your acid to make carbon dioxide.


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Preference of certain positions during chlorination of an alkene
I am currently doing the questions:

Write equations for the reaction of chlorine with propene and 2-butene using structural formulas.

So I know that propene is $\ce{H3C-CH=CH2}$ and chlorine is $\ce{Cl2}$.
So:

From this I observe that naturally, the double bond is removed in the presence of an addition reaction and that the chlorine atom bonds to the $\ce{CH}$ pair and the other $\ce{Cl}$ atom does not bond to any of them but attaches itself at the end.

Why did the second $\ce{Cl}$ pair not attach itself to $\ce{CH2}$? Does this mean that atoms can only bond to $\ce{CH}$ pair and nothing more than that, e.g. not $\ce{CH2}$ or $\ce{CH3}$, etc.

Second question for the reaction of 2-butene with Chlorine. I know that 2-butene is $\ce{H3C-CH=CH-CH3}$

I noticed, like what happened above, the chlorine atom bonds to only the $\ce{CH}$ pair. Here however there are two $\ce{CH}$ pairs so the Chlorine doesn't need to attach itself at the end of the chain, unlike the first picture.

Are my observations correct? My textbook has no mention of this.

With respect to (1), I imagine that Cl bonding in those locations is energetically favorable given that there are no C-H bonds being broken.  Of course this does NOT mean that atoms can only bond to the CH pair and nothing else but then we'd have to consider the C-H bond rupture which is an entirely separate process.

In your first example, propene plus chlorine, one of the Cl atoms bonded to a "CH" and the other, that you described as bonding to the end, bonded to a "CH2," so it's not just CH groups getting all of the chlorines.  Instead, it's the carbons at the ends of the double bond, however many hydrogens they might be attached to.
I think part of your misunderstanding lies in the way that you're writing/imagining the alkene.  The propene molecule has a shape, and the geometry of the molecule plays a major role in its reactions.  The double bonded section is more or less flat, and the CH3 is spread out in space.  The chlorine atoms attach to the ends of the double bond, wherever that may be.  The mechanism for this attachment is beyond the scope of your question, but if you're curious, you can find more information here.  At the very least, know that the chlorines do not attach to the carbons at the same time; there is an intermediate compound formed.

(If my drawing isn't clear, you can try building a model of the molecule.)
It would require the breaking of a C-H or C-C sigma bond in order for the chlorine to get to the CH3 carbon, and that's unlikely.  As I mentioned above, the C=C section is planar, so these carbons are open on the sides for attack from a halogen.  A bond still has to break for a new C-Cl bond to form, but in this case it is the weaker C-C pi bond that breaks.  With easy physical access and a lower energy barrier, the reaction goes forward.

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Ionic Equilibrium
Calculate the pH of a solution which contains 100mL of 0.1M $\ce{HCl}$ and 9.9mL of 1.0M $\ce{NaOH}$.
While calculating the concentration of $\ce{H+}$ ions, why shouldn't we consider the H+ ion concentration from water?

Check the Homework Policy for specific information regarding asking homework, or homework-type questions.  Please try to narrow your question to a specific concept and show your reasoning thus far.

You can consider the hydrogen ion concentration from the ionization of water. However, that quantity is a million fold less than the amount of hydrogen ion coming from $\ce{HCl}$ and can be considered insignificant.
Incidentally, in this problem you could ask the same question about hydroxide from the ionization of water, and you'd find that the added hydroxide is ten-million fold greater than from the ionization of water.


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Simple and cheap anti-graffiti layer based non-paint solvent
I have this issue with a group of people intentionally damaging a bench. So far, they would paint curses over the bench with permanent marker. However after I removed the curses with ethanol, they damaged the bench physically and threw the trash all around (we used to have a trash can there, with sign "Please do not throw trash around" - they threw this trash can away too).
I would like to keep this question non-personal though, so my question is:
How can I protect a nitro-based paint layer from other paints?
The professional solutions are way too expensive, while the thing I want to do is simple: I'm looking for non-nitro based material and solvent that will form transparent layer over the original paint. In case of damage by nitro-based paints, I could wash the layer under them and then apply it again.
The theory is following: you apply the protective layer over your paint. When somebody applies any paint over it, you've got 3 layers.

It does not matter what the top layer is, as long as it does not dissolve our protection. After this, we apply solvent for the protective layer. Assuming there are microscopic holes in the top paint, the protective layer will dissolve and the top layer will go off with it.

The layer must not be obvious.
I was considering these:

Ethanol-based transparent paint (too expensive, suprisingly)
Wax (parafin) based layer (I don't know how to dissolve wax, it's crystalic so not transparent)

I think you need a motion detector hooked up to the quick release on a crane that is precariously holding a piano 200 feet above the bench.

The "professional" solutions here in the US are often fluorinated paints. My understanding is that the graffiti literally slides off and people give up after a few tries.
Given the costs, my suggestion would probably be a thick layer of polyurethane. These are often applied to seal wood floors and can be transparent or stained. If you apply multiple layers (say 2-3 on a floor) the surface is very smooth and slick. It's hard for paint or other stains to stick.

I realized after reading the linked Wikipedia article that polyurethane is actually mentioned as an anti-graffiti coating. Definitely it's easy to remove and reapply the polyurethane coatings if needed.


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Electrolysis not running, maybe dependant on distance of electrodes?
So I created a DC power supply of 15 V (special transformator connected to the net) and I have a graphite anode and stainless steel cathode. When I put them in a beaker next to each other in water, the electrolysis takes place as I see gas bubbles. However, when i put them in another container that consists of two beakers connected with a tube to allow liquid to flow through but to separate the gases, the reaction doesn't run while it is the same setup more or less. How come? It is true that the tube connecting the beakers is relatively thin and the distance is also quite large (10 cm or sth), so does this lead to a higher required overpotential? I really don't know why it doesn't work in the larger container. Any help is appreciated!

Adding a small amount of salt (or acid/base) will increase the conductivity of the solution and improve the rate of electrolysis.  This can generate side products, like small amounts of chlorine gas, as well, just FYI.

Don't use stainless steel in electrolysis! Carbon electrodes are quite fine.

Yes indeed, I have read before stainless steel releases toxic chromates, but I only use stainless steel for the cathode as it is being reduced, not oxidized.

Solution resistance in a two electrode cell can make a big difference as it is totally uncompensated. You may be applying 15 V between the electrodes, but the redox reaction only occurs at a very close proximity to the surface (<10 nm), so it's the interfacial potential we're concerned with, which can be significantly different than the applied voltage if the solution is highly resistive. A tight constriction in the cell can certainly increase the resistance and you should make sure you're using an electrolyte as plain water is very resistive.

Resistance is proportional to length and reverse proportional to сross section of the conductor. When you increase distance by 20 cm from 2-3 cm (10 times resistance) AND indtroduce ion bridge about 1 or less cm in diameter (probably also 10+ times resistance, likely more), you effectively increase the resistance by several orders of magnitude, proportionally reducing the current and the speed of the reaction.
In industrial-grade applications when speed of reaction is of essence, like eletrolitic cutting, the distance between electrodes may be as small as several millimeters. Strong electrolite flow is also used.


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Synthesis of benzophenone
Would his be a valid synthesis? I'm just unsure about the oxidative clevage step. I know that there is a benzylic hydrogen. So should oxidative clevage work here? What happens to the fragment that is lopped off? Is it also oxidized?

Your synthesis will work in theory, but it can be shortened. I think ron's method below is very nice, but it might be possible to do it all in one step with phosgene ($\ce{COCl2}$) in the presence of $\ce{AlCl3}$.

@GregE.  That's a better approach then what I suggested, especially using something like triphosgene instead of phosgene.  Why not post it as an answer?  I'd vote for it.

@ron, thank you. I'll expand it into a fuller answer a bit later when I have more time. Thanks also for the reminder about triphosgene; its existence had completely slipped my mind.

There is another nice synthesis, you can hydrolyze dicholodiphenylmethane.  The latter can be synthesized from (a double Freidel-Crafts) benzene and carbon tetrachloride.. And carbon tet is a bit less toxic to work with than phosgene.

Your basic strategy to involve a Friedel-Crafts acylation is correct, but your route is a bit circuitous.  You could acylate benzene directly to produce benzaldehyde in one step using carbon monoxide and hydrochloric acid to generate formyl chloride in situ.

While it is "just another" Friedel-Crafts acylation, it has it's own name, the Gatterman-Koch reaction, after its discoverers.  Once in hand, you could oxidize benzaldehyde to benzoic acid.
An alternate approach would be to 1) brominate benzene to form bromobenzene, 2) form the Grignard reagent with magnesium and 3) react it with carbon dioxide to produce benzoic acid.
In either case, with benzoic acid in hand, you can procede as you've indicated, on to benzophenone.

I disagree with Ron on this one. First off, he got the Gatterman reaction wrong. The reaction he showed does not work, since the formyl chloride $\ce{(HC=OCl)}$ breaks down under the conditions required for Friedel–Crafts to occur, giving you $\ce{HCl}$ and $\ce{CO}$.
The actual Gatterman reaction would work, since it uses a $\ce{C=N}$ instead of a $\ce{C=O}$ for the reaction.

Simply oxidize this compound (benzaldehyde) with something like LiAlH4 (edit: this is a reducing agent, use KMnO4 in aqueous base or Chromic Acid) to get to benzoic acid, then continue with your thionyl chloride and then the second Friedel–Crafts reaction with another benzene.
Ron's second suggestion of using a Grignard to synthesize benzoic acid is also good, just note that you need to add an acid (even just water is a strong enough acid) to remove the Grignard reagent afterwards. Here's an image of the reaction he is referring to, except they used $\ce{HCl}$ to remove the Grignard.

The Gattermann-Koch reaction is not the same as the Gattermann reaction, they are two distinct reactions, see the link in my answer for clarification.  The G-K reaction was specifically designed to generate formyl chloride in situ so that aromatic substrates could be formylated.  The reason why the G-K approach is preferred over the Gatterman is that the latter route involves the use of $\ce{HCN}$.  Also, I'm not sure what you mean by "oxidize this compound (Benzaldehyde) with something like $\ce{LiAlH4}$."  $\ce{LiAlH4}$ is a reducing agent.

Well done @ron. Two equiv. of phenyl Grignard added to 1 equiv. ethyl formate gives benzhydrol, readily converted to benzophenone by a myriad of oxidants.

@user55119  Nice, only 3 steps and certainly more convenient to carry out.  Take a look at Greg E.'s suggested synthesis up above - only 1 step!

Oh yeah, thanks Ron, I messed that up with LiAlH4, it's quite obviously a reducing agent. KMnO4 in aqueous base or Chromic Acid would do the trick


text:
Completing inorganic reactions and writing spectator ions

Complete the following reactions and write the spectators for each if any
A. cobalt(II) chloride (aq) and cesium phosphate (aq)

I am not quite sure if I have the correct answer to this equation, but I have $\ce{ CoCl2(aq) + Cs(PO4)(aq)}$ (for the equation setup) and for spectator ions $\ce{Co^2+ (aq) + 2 Cl- (aq) + Cs^+ (aq) + (PO4) (aq) -> CsCl(aq) + Co(PO4)2(s)}$ and so for the spectators I have $\ce{Cs^+(aq)\ and\ 2Cl- (aq)}$, Is this correct?

B. sodium phosphate (aq) and calcium chloride (aq)

For this equation I have $\ce{6Na^+\ and\ 6Cl-}$ as spectator ions, $\ce{Ca3(PO4)2(s)}$ (as a solid)

For B, you have correctly identified calcium phosphate as the precipitate, as it is the least soluble of all possible compounds. In the case of A, cobalt phosphate precipitates from the solution containing cobalt(II) and phosphate salts.
$$\ce{3Ca^{2+} + 2PO4^{3-} ->~ Ca3(PO4)2\downarrow}$$
$$\ce{3Co^{2+} + 2PO4^{3-} ->~ Co3(PO4)2\downarrow}$$


text:
t-butylbenzene in highly acidic medium
Why would the following mechanism be valid for the conversion of t-butylbenzene to benzene in highly acidic medium followed by an aqueous workup?
The main problem I have with this mechanism is the formation of the secondary doubly allylic carbocation. Why not go for the tertiary, inductively and doubly allylic stabilized carbocation?
But I guess the problem with going down this path it becomes hard to see how the t-butyl group might leave.

As you can see, the book's mechanism is entirely dependent on having the secondary allylic carbocation.

All 6 positions on the benzene ring are being protonated (the frequency of protonation at the different sites is related to the stability of the carbocation generated) in acidic media, so all of the ions you suggested are being formed.  Protonation at 5 of these positions can only regenerate starting material when one of the two protons at the initial site of protonation is removed (an identity reaction, we could make this pathway visible by running the reaction in deuterio-acid + $\ce{D2O}$).  Protonation of the benzene ring at the carbon bearing the substituent provides a choice.  Either the proton can again be removed or the stable t-butyl cation can be ejected.  All of these things are happening in the reaction, but only removal of the t-butyl group is visible.

I think the problem you are having stems from locally optimizing the energy, but disregarding the global energy well that you get in the end.
The mechanism your book gives is valid because it explains how this reaction could take place along a minimal-energy pathway. Since the reaction is observed, it somehow has to proceed from tert-butylbenzene to the products benzene and tert-butanol.
I know that this is a slight tongue-in-cheeky kind of argument, but in the end it is a valid one.
If you protonate elsewhere (and we haven't discussed the protonation kinetics involved here1 but they might be important) you will not end up cleaving the right $\ce{C-C}$ bond (or any bond at all). The tert-butyl cation is quite stable, and having stable reaction products means that the reaction is more happy to proceed. Since the protonation is likely to be a fast equilibrium reaction, it will happen at many sites on the aryl group.

1 Frankly I have no clue about those, so maybe someone can help me out here...

I agree with what you're saying; I'm perhaps trying too much to stabilize the carbocation when, from a thermodynamic standpoint, the stability of the intermediate carbocation isn't worth a damn. Kinetically it might matter, but thermodynamically, we're going from a (slightly) activated benzene to a plain old benzene! Plus I'd expect the carbocation to be protonated somewhere along the path.

Another way of thinking about tschoppi's opening statement: the mechanistic step that you propose is reasonable but not productive. It may happen under the reaction conditions but cannot proceed to the product, and the resulting intermediate an only revert to the starting material. You are seeing it here, but keep this in mind when you get to carbonyl chemistry (assuming you haven't covered it already).


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How can we recreate old pharaos' perfumes?
There is ongoing research where efforts were launched to find out the original constituents of a perfume for the female pharaoh Hatshepsut:

(Source)
In a perfume flask that was found in her tomb, there are apparently some remnants of the (assumed once-liquid) perfume that she used to wear. The perfume flask was sealed, so there was probably a very low amount of gas exchange taking place during the millennia that the flask took to wait for being discovered with its treasure.
However, there should still be a lot of reactions that can happen in this timescale, and since most of the reactions that just happen by themselves tend to lower energies, at some point different organic molecules would lead to the same product (e.g. acetic acid, or similar).
It seems to me, that then the only reasonable approach is to make a list of possible constituents and then start to play with the amounts thereof and model the decay reaction pathways. Then you could check if your theory predicts the experimental results.
Is this really how you find out what an ancient perfume was made of?
References:
Here, this, this one and that (talking about Tutankhamun, but the same principles would apply).

The analysis won't be done before lunch, but neither will it be an extremely difficult process.  Certainly some of the researchers will have a fragrance industry background  and access to necessary equipment like  a tandem gas chromatograph - mass spectrometer.  Also, Pharoah's perfumes were undoubtedly less complex than todays, and all oils used in perfumes are available as "known" compounds for comparison.  Additionally, all fragrances have been "fingerprinted" making identification of the compound in a sample fairly routine and straightforward.  By "fingerprinted" I mean that if a certain pattern of compounds can be detected than they will suggest the presence of a certain oil.  For example, look at lavender oil, there are about 30 chemical components in it - just like a fingerprint.
Perhaps the base solvent has evaporated through the wall of the container and oxygen has flowed into the container.  Certainly some of the ingredients will have degraded, but the container won't be full of acetic acid.  Some amount of an alcohol may have oxidized to the corresponding ketone, aldehyde or carboxylic acid, perhaps some amount of a conjugated diene may have oxidized to the corresponding dienol; but that shouldn't make the process that much more difficult.

It's definitely not a pure chemistry problem which would be a fair bit harder. Given the location, time period, and the social class of its owner, one can probably narrow down the list of possible components considerably.

And indeed, there are people who specialize in archeological chemistry (i.e., analytical chemistry applied to archeological tasks). There are even books on the subject.


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Quantifying CO2 generated/released by the soft drink industry and breweries
I need some help in developing a method by which to calculate how many tons of "evil" $\ce{CO2}$ are routinely being released into the atmosphere SOLELY due to our beverage addictions, notably beer and soft drinks.  I realize that not all manufacturers and products are consistent in their carbonation levels, and that an average probably has to be applied industry-wide (unless individual companies are required to record and publish such data) per unit measure or monthly/annual volume output.
Is this the proper forum to ask such a question?  If not, suggestions on where else I go?
On a previous post I got part of the answer, relating to an individual 12 oz can, or 1 liter, but I'm wondering if there is an easier way in which I can go about plugging weekly, monthly or annual volumes of beverages into formulas which will yield the results I'm looking for, notably a discussable unit of measurement in tons or ?  I'm willing to do all the numbers work once I'm pointed in the right direction.
I'm not sure if anybody's ever thought about this before, because I've never heard it discussed.  But if, according to some environmentalists, we should all watch our carbon footprint (and I'm not debating that here), then it seems like the most logical place to start would be with our hundreds of millions of daily beverage choices?  Isn't it?

This might answer your question. also check Carbonated water and Coke servings. The more important question may be  where does the CO2 come from. They probably don't burn gas to carbonate your soda/beer. edit: maybe they do

If you're familiar with the xkcd What If blog, the author addressed a similar question here. It leads you through the thought process required to answer your question.  The answer to your question winds up being, "Not much at all," in case you're wondering.  1 ppm of atmospheric CO2 equals 450 quadrillion cans of soda, by the author's reckoning.

So if I grow a bushel of wheat (consuming atmospheric CO2) and brew beer with it, the carbon dioxide generated is somehow more evil than if I make bread with the wheat, eat it, and breathe out carbon dioxide?

@JasonPatterson I'd say that's just about an answer if you add in some of the details.

A mistake in my comment - it was 450 quadrillion cans to equal the amount of CO2 that humans have added to the atmosphere, 130ppm = 450 quadrillion cans, not 1ppm.  I misread the linked article.

The xkcd What If? blog, by Randall Munroe, addressed a similar question.  He states the result better than I can.

For most of the history of civilization, there were about 270 parts
  per million of carbon dioxide in the atmosphere. In the last hundred
  years, industrial activity has pushed that number up to 400 parts per
  million.
One "part per million" of CO2 weighs about 7.8 billion tons.
  A can of soda contains in the neighborhood of 2.2 grams of CO2, so you
  would need about 450 quadrillion cans of soda. That's enough to cover
  the Earth's land with ten layers of cans.

Randall Munroe - xkcd What If? #88:Soda Sequestration. http://what-if.xkcd.com/88/

For further perspective, each breath of a healthy adult male involves about 500mL of gas exchange.  The air is, on average, 0.04% CO2 prior to inhaling.  Exhaled air contains between 4 and 5 percent carbon dioxide, I'll call it 4.04% for a nice, round 100-fold addition.  The mass of 500mL of air is about 0.6g, and though it will be slightly more dense with the added carbon dioxide, I'll assume that it remains at that mass for simplicity's sake.
So 4% of the 0.6g is CO2 that was added to the atmosphere by a single adult male breath.  That's 0.024g.  About 90 breaths add as much CO2 to the atmosphere as the carbonation in a can of Coke.  Where does the additional carbon in those breaths come from?  The Coca Cola itself, of course.
The sugar in the can of Coke (39g) contains about 16g of carbon atoms.  After being metabolized, that turns into 57.2g of carbon dioxide.  Not considering any of the energy required to harvest, process, or transport the sugar, that's still 26 times a much carbon dioxide going into the atmosphere from the calories in a can of Coke as opposed to its carbonation.
To put it more succinctly, the CO2 output from the carbonation in Coke could be completely offset by reducing overall caloric intake by 6 calories per can.

The xkcd information is interesting, but doesn't really answer the question, which is about how much CO2 is produced by the industry. For that, we need to understand a key difference between fermented and non-fermented carbonated beverages.
Non-fermented carbonated beverages are carbonated using CO2 provided by industrial gas companies who capture this CO2 from waste exhaust of other industries and clean it up. For example, hydrogen production by steam-reforming of methane produces high CO2 concentration gas as waste, as does the ethanol fuel fermentation industry. So the non-fermented beverage producers aren't generating CO2. They're just delaying its release into the environment from other processes.
The fermented beverages are quite different, in that CO2 is a product of the conversion of sugar to ethanol by the yeast (or in a few cases bacteria). Most of this CO2 leaves the tank and doesn't remain in the beverage as carbonation, so the amount of carbonation in the beverage isn't the number to focus on. Instead, we can estimate the CO2 generated by assuming that most of the fermentation operates anaerobically at a typical product ratio of 1 mole CO2 per mole ethanol.
According to the WHO, Americans drink about 5 liters or so of pure alcohol per capita per year, which we can round up to 100 moles (5.9 L) for convenience. Assuming roughly 350 million people in the US, we have total consumption as 35 B moles of CO2. 35 B moles of CO2 is about 1.5 million metric tons of CO2 produced during fermentation of the alcoholic beverages that are consumed in one year by Americans.
For comparison, a typical car produces about 4.5 metric tons of CO2 per year as exhaust, according to the EPA, so the beverage fermentation industry is equivalent to about 350,000 cars.

The answers so far leave out any Co2 generated by the electricity consumed in refrigeration of the beverages (which in a retail outlet may be as long a a week or more) and any Co2 produced by the electricity used in the manufacturing process. World wide this gets a bit tricky! I couldn't begin to know how to work that out.   There's also Co2 emitted in the transport of the ingredients to the manufacturing plant and of the finished product to market as well. Also I am not sure that Co2 being used for soft drink is always the bi product of another already existing project, and perhaps not even that frequently, in 2016 I think this was still developing technology (see)https://www.abc.net.au/science/articles/2001/11/29/405835.htm An article from Time Magazine in 2008 if skim read seems to suggest this technology is in wide use, however if read more closely, it doesn't actually. http://content.time.com/time/specials/2007/article/0,28804,1730759_1731383_1731989,00.html

Dicobalt, eh?$%$

To clarify - the cited articles describe the potential for capture of CO2 from combustion, which is a low grade stream currently not reused. The major sources of industrial CO2 are steam methane reforming and ammonia fertilizer production. In both cases, the CO2 is a byproduct whose value is not enough to justify running the process if the main product (H2 or NH3) is not sold. Ethanol fermentation is a third, smaller scale source. There is no large-scale process that produces CO2 as the main product for sale.


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Methods for Determining Partial Charges
I want to run classical molecular dynamics simulations of a periodically replicated surface (rutile $\ce{TiO2}$ with grooves). In order to do so, I first need to solve for the partial charges residing on each atom of the surface. What electronic structure methods are available for determining the partial charges on a surface?
I am vaguely familiar with Mulliken population analysis, but my understanding is that this method is not the most accurate. What other methods are commonly used, and how accurate are each of these methods?
Finally, what are some good (and free) packages for performing such calculations? I've used SIESTA in the past, is that well-suited for determining partial charges of a surface?

There are lots, and I mean lots of methods and software programs to produce partial charges. See Wikipedia for a small (incomplete list)
Let's start with the basics. The idea of a partial atomic charge, while useful for concept, can not be defined uniquely. Quantum chemical methods (whether wavefunction or DFT) produce some sort of electron density. Different schemes divide up that electron density in different ways.
I like Cramer and Truhlar's categories:

Class I: Some sort of intuitive or empirical approach (i.e., non-QM). These include methods based on experimental dipole moments and atomic electronegativities, including Gasteiger-Marsili partial charges and other "electronegativity equalization methods" (EEM).
Class II: Partitioning using wave functions / orbital schemes (including Mulliken charges while very efficient to compute, are basis set dependent.
Class III: Partitioning the electron density or electrostatic potential (e.g., Hirshfeld, density-fitting schemes, etc.
Class IV: Semiempirical mapping from Class II or Class III (preferred) to match experimental data like dipole moments.

Almost any modern quantum package will include several schemes to fit partial charges. My personal preference is for something like Hirshfeld charges (i.e., from the electron density, so not basis-set dependent), or electrostatic potential fitting schemes like Merz-Kollman or CHelpG.
I think the latter (electrostatic potential fitting) will be better generally for molecular dynamics, since you're attempting to produce point charges that will best represent the quantum mechanical electrostatic potential.

I'd be remiss if I didn't mention my colleague Ken Jordan who studies the electronic structure of molecules (water, $\ce{CO2}$, etc.) adsorbed onto $\ce{TiO2}$ and has found dispersion-corrected DFT methods to be crucial for many properties.

Relevant article by Cramer and Truhlar: http://link.springer.com/article/10.1007/BF00117280

@gncs I feel like they have much more recent review articles. I'll find one later.

Kinda obvious, but still.
First, you have to decide, what face you are working with. $\ce{TiO2}$ may exhibit several different faces, you'll have to consult literature to find out which are the most frequent.
Second, you have to consult literature to find which surface species exist on the surface in your conditions. Generally, oxides tend to adsorb water forming hydroxide groups and some other small molecules.
After that you have to build a model slab of the surface. Generally, two-three layers of atoms close to surface are 'relaxed' and two-three 'deeper' ones are fixed in positions, modelling unperturbed crystal structure. However, ideally you should perform several computations, gradually increasing amount of layers in slab until the characteristic you considering stabilize.
It is generally recomended to employ periodic approach using plane-wave sets, possibly in DFT method, as it greately reduces memory requirements and amount of atoms (and electrons) considered.  Given you care to use the charges to model electrostatic potential, Mulliken and Natural charges go to garbage. Ideally you should use full-scale electrostatic potential. However, if it is not an option, analog to Gaussian Pop=CHelpG should be used, that fit atomic charges to produce  electrostatic potential closest to one produced for given electronic distribution. Not aware of other programs implementing similar schemes, so you have to find one on your own.


text:
What are derived names?
Q)The derived name of (CH3)4C is
A)Tetramethylmethane
B)2,2-Dimethylpropane
C)Neopentane
D)None of these
The answer was A.
What is a derived name?Never heard of it in any book.
Thanks

There are (were) three main nomenclature systems used for naming organic compounds:

Common Name (or trivial name). Under this convention this molecule is neopentane. For alkanes, this naming system is based on the positioning of methyl substituents, and gives rise to the iso-, sec-, neo- namiong prefixes.
Derived Name. Who uses this method these days???? This system identifies the central carbon atom which has the most carbon groups attached, and then names the groups as a prefix, and ends the name with 'methane'. The formula are often written with the central carbon unit in square brackets. You may still see derived names on MSDS/SDS under the 'Other Names' section. Under this system this molecule is tetramethylmethane. I can think of more important things to teach you in class if this is what you are learning.
IUPAC Name
This is the accepted naming convention in use today, which allows for retained names, which usually have their origins in the trivial naming system. Its naming system is based on a systematic framework and is used universally. In older texts (pre 1970s), it was known as the Geneva system or congress. Under this convention, this molecule is 2,2-dimethylpropane.

I can't even find any decent references for you at this point, but will update if I find some.

Derived names were not taught but it appeared on this test from nowhere and I marked the IUPAC name :P


text:
Nomenclature in Other languages
Even one letter can make a huge difference in Chemistry Nomenclature. So I am looking for credible resources that is specifically published for explaining Nomenclature rules, examples in Spanish. Any suggestions?

By quick google search I found this web page where there are few titles in Spanish. For instance, there is a book entitled "Nomenclatura de la Química Orgánica (IUPAC)". Is it what you are looking for?

@Wildcat I was trying out that link before and after you have pointed out. However from U.Barcilona the pdf or the web page isn't loading. I looked for other avenues - no luck as of now. I am thinking a Spanish Chemistry book for Nomenclature might include the required info. However it's vital to get accurate nomenclature as of IUPAC though...

I do not speak any word spanish, but this might look promising: http://books.google.co.jp/books?id=Q_Z8RrSUuu8C&dq=Nomenclatura+de+la+Qu%C3%ADmica+Org%C3%A1nica+(IUPAC).&hl=de&source=gbs_navlinks_s You also might want to check your local organisation for more information: http://www.rseq.org/

That's indeed quite helpful, although it only covers the organic set.


text:
What do the stereochemical labels D and − mean in tartaric acid?

Why is the isomer of tartaric acid above labelled as ᴅ and (−)?

it rotates plane-polarized light clockwise
it rotates plane-polarized light anticlockwise
it is an equimolar mixture of the enantiomorphs
its configuration is related to that of ᴅ-glyceraldehyde

If I am not wrong, the ᴅ in ᴅ-glyceraldehyde refers to (+)-glyceraldehyde, and likewise ʟ-glyceraldehyde is (−)-glyceraldehyde. The (+) and (−) labels indicate that these isomers are dextrorotatory and levorotatory, respectively.
So in ᴅ-(−)-tartaric acid, what does the ᴅ mean, if not dextro, and what does the (−) mean, if not levo (the minus sign indicating anticlockwise rotation of plane-polarized light)? How can one molecule be both?

Wikipedia has a useful section on this topic.

As far as I remember, small d always means the same as (+), not small cap ᴅ. So, a dextrorotatory compound is prefixed with either (+) or d while a levorotatory compound is prefixed with either (−) or l.
Small caps ᴅ and ʟ prefixes have different meaning: if an enantiomer can in principle be synthesized from naturally occurring (+)-glyceraldehyde without affecting its configuration, it is considered the ᴅ-form. Otherwise, it is considered the the ʟ-form. And, say, the ᴅ-form of a molecule is not necessarily a dextrorotatory one.
So, the answers to your question are options 2 (the (−) sign indicates that it rotates plane-polarized light anticlockwise), and option 4 (the ᴅ label indicates that it is related to ᴅ-glyceraldehyde).

Dextrorotatory and levorotatory are a set of two terms that designate the direction polarised light is rotated to by a specific compound. Chiral compounds can be identified as being chiral due to their optical activity only. The optical activity is a physically measurable parameter, determined in polarimeters and calculated to a specific optical activity by dividing by concentration and length of the cuvette. Rotation to the right is indicated by a $(+)$ sign, while rotation to the left is indicated by a $(-)$.
The names dextro- and levorotatory derive from the Latin words dexter (right, as in right hand) and laevus (left, as in left hand).

The D/L-nomenclature is entirely different from the above. While the prefixes D and L also derive from the same Latin words, they are used in a different way. They key to D/L-nomenclature is a Fischer projection. If a chiral molecule is drawn in such a (linear) projection, and the group of the highest priority (typically the group with the highest oxidation state) is at the top, a compound is defined as L if the non-hydrogen atom on the lowest asymmetric carbon of the chain points to the left. Likewise, if it points to the right, a compound is termed D.

From left to right: D-Glyceraldehyde, L-tartaric acid, D-tartaric acid.

It is very important to know that there is no connection between D/L and the direction of rotation of polarised light! Likewise, one cannot deduce the direction of optical activity from knowing whether a compound is (R) or (S). D/L and (R)/(S) can be determined on paper or via crystal structures while $(+)/(-)$ can only be determined experimentally in a polarimeter or similar experiments.
That D-glyceraldehyde is dextrorotatory is a mere (but maybe lucky) coincidence.

Therefore, the correct answers to your given question are (2) (it rotates to the left; $(-)$) and (4) (it relates back to D-Glyceraldehyde).


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Relation in the sign (+/-) for gyromagnetic ratio and magnetic susceptibility
Is there a relation (or formula) between the sign of the magnetic susceptibility and the gyromagnetic ratio of a nucleus in NMR?

I should preface this with the caveat that I'm a chemist and not a physicist, so it's possible that I haven't understood your question correctly.
Magnetic susceptibility, $\chi$, is an electronic property, not a nuclear property, so one could simply write the question off as not making sense.  That being said, let's look a little closer as I believe we can demonstrate an answer to this question.
First, magnetic susceptibilities are typically divided into diamagnetic susceptibilities (which are negative and arise from paired electrons), paramagnetic susceptibilities (which are positive and arise from unpaired electrons), and then a few others (ferromagnetic, ferrimagnetic, etc. which arise from the cooperative behavior of unpaired electrons).
Second, the shell theory of the nucleus provides a nice picture when discussing the NMR properties of different isotopes as we can picture protons and neutrons existing in discrete nuclear energy levels and being paired or unpaired, much like chemists picture electrons in atomic or molecular orbitals with their discrete energy levels.  One of the neat results from this approach is that any isotope with an even number of protons and and even number of neutrons will have all of the protons paired up and all of the neutrons paired up, resulting in a total nuclear spin = 0.  Thus $\ce{^4He}$, $\ce{^12C}$, and $\ce{^16O}$ are all expected to be spin 0.  On the other hand, an isotope with an odd number of protons will have an unpaired proton; and an odd number of neutrons gives rise to an unpaired neutron.  Having a single unpaired nucleon gives rise to a half-integer nuclear spin (1/2, 3/2, 5/2, etc.) where the unpaired nucleon provides a unit of 1/2 to the angular momentum and any remaining integer spin comes from the nuclear equivalent of orbital angular momentum ($l=0,1,2,$ etc.)  Having both an unpaired proton and neutron gives you an integer-spin nucleus (1, 2, 3, 4, etc.) where 1 unit of angular momentum comes from the two unpaired nucleons and the rest comes from the shell theory equivalent of orbital angular momentum.
OK, so why I am boring you with all of this detail?  Well, nuclei with unpaired nucleons are our nuclear analogues to atoms or molecules with unpaired electrons.  Thus, all NMR active isotopes (spin > 0) can be considered to exhibit nuclear paramagnetism.  It's the presence of the unpaired nucleon that gives rise to a non-zero spin angular momentum.  On the other hand, all NMR inactive isotopes (spin = 0) would be expected to exhibit nuclear diamagnetism.
Now, at this point I'm venturing beyond what I actually know, as I have never seen any values reported for nuclear magnetic susceptibility.  However, one would expect that, analogous to the electronic case, nuclear paramagnetism should have a positive susceptibility and nuclear diamagnetism should have a negative susceptibility.  In other words, the only nuclei that should have negative nuclear magnetic susceptibilities have a spin of 0 and no gyromagnetic ratio.  Any nucleus that is NMR active should have a positive nuclear magnetic susceptibility; but, some NMR active isotopes have positive gyromagnetic ratios and others have negative gyromagnetic ratios.
So the answer to your question (if I've understood it correctly) would be NO, there is no formula relating the sign of the gyromagnetic ratio to the magnetic susceptibility.
In the case of normal (electronic) magnetic susceptibility, the nucleus plays no role (or at least such a small role that it's completely negligible) and in the case of the extremely weak nuclear magnetic susceptibility, all NMR active nuclei have positive susceptibilities regardless of the sign of their gyromagnetic ratios and all NMR inactive nuclei have negative susceptibilities and no gyromagnetic ratios.
As a footnote that you may or may not find interesting.  Picturing integer-spin nuclei as the nuclear equivalent of di-radicals gives a nice rationalization for why they are so rare (only ~8 stable isotopes with integer spin vs. ~112 stable isotopes with half-integer spin vs. I don't know how many isotopes with spin = 0).  Just like molecules, the di-radical isotopes are quite unstable, with some exceptions (e.g., $\ce{^2H}$ is spin=1 and has an unpaired proton and neutron, but is a stable isotope; similarly, molecular oxygen is probably the most abundant di-radical (note: you need M.O. theory to see why it's a di-radical, but you can easily show that liquid $\ce{O2}$ is paramagentic - go do a youtube search) and is relatively stable, at least compared to many other di-radicals).

Your last paragraph makes me want to shout ‘Oh nitrogen, why do you have to be special! D=’

I also indented to point out that we can not yet predict gyromagnetic ratios from the known nuclear structure (neither the sign nor the magnitude of the gyromagnetic ratio).  I believe that we are able to predict the values of the nuclear spin (I vaguely recall that the shell theory is able to predict the spin angular momentum for excited nuclear states, which is something neat to think about).  However, my understanding of nuclear structure pretty much ends with the shell theory (and a rather superficial understanding at that).


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Relation between thermodynamic reversible process and reversible reaction
I know it seems to be a weird question. But for long I have been thinking whether there is any relation between thermodynamic reversible process and reversible reaction. Do they have any connection and if so how?

To complement other answers, I provided here definitions extracted from official IUPAC sources:
reversible process
A definition comes via the concept of entropy (Ref. 1):

Quantity the change in which is equal to the heat brought to the system in a reversible process at constant temperature divided by that temperature. Entropy is zero for an ideally ordered crystal at $\pu{0 K}$. In statistical thermodynamics
$$S=k \ln W$$
where $k$ is the Boltzmann constant and $W$ the number of possible arrangements of the system.

The second sentence is a formulation of the third law, while the last is a definition of entropy viewed from the context of statistical thermodynamics. The first sentence describes the thermodynamic entropy. Coupled with the second law expressed as an inequality relation between heat and entropy,
$$\Delta S \ge \frac{q}{T} \tag{const. T}$$
it provides a functional definition of a reversible process (it is a process that maximizes $q$).
reversible reaction
The definition provided by IUPAC is related to microscopic reversibility:

In a reversible reaction, the mechanism in one direction is exactly the reverse of the mechanism in the other direction. This does not apply to reactions that begin with a photochemical excitation.

Such a definition implies thermodynamic reversibility.
References

IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.

Reversible reaction
According to Merriam Webster, a reversible reaction is:

a reaction that takes place in either direction according to conditions (as the formation of hydriodic acid by union of hydrogen and iodine or its decomposition into these elements)

For this type of reaction, you would use a double-harpoon to write the chemical equation:
$$\ce{H2(g) + I2(g) <=> 2HI(g)}$$
and would be able to write an equilibrium constant expression
$$K = \frac{[\ce{HI}]^2}{[\ce{H2}][\ce{I2}]}$$
with [] denoting activity or fugacity. If you start with pure reactants, the reaction would go forward (decrease in reactants, increase in products). If you start with pure products, the reaction would go backwards (increase in reactants, decrease in products). Once the reaction has reached equilibrium, concentrations would no longer change.
In any case, reactions in both directions would occur at the particular level. When the reaction is at equilibrium, you just won't know at the macroscopic level and would have to disturb the equilibrium to see macroscopic changes again.
Reversible process
Again according to Merriam Webster, a reversible process is:

an ideal process or series of changes of a system which is in complete equilibrium at each stage such that when the process is reversed each of the changes both internal and external is reversed but with the amount of transferred energy unaltered

Applied to chemical reactions, a reversible process is one where the Gibbs energy is constant throughout (or where the entropy of the universe is not increased by it). We would not expect any macroscopic changes (the process is at equilibrium), which why this is an ideal process (not real). The closest real process is one where the system is near equilibrium, and the increase in entropy of the universe is minimal.
What does IUPAC say?
The IUPAC Gold book makes reference to a paper (DOI: https://doi.org/10.1351/pac199466051077) offering a glossary of terms used in physical organic chemistry. In their definition of chemical equilibrium, it says:

While the definition of reversible process matches the one I cite above, and they use the same notation for the equilibrium reaction. However, the connection between reversible process and chemical equilibrium surprises me. When a reaction approaches equilibrium, the Gibbs energy does change, and it will not return to the initial state unless work is done on the system. Maybe the confusion regarding reversible process vs. reaction stems in part from this.

I agree with both the MW and the IUPAC definitions, but perhaps you can help me checking if I am getting this right.
If a reactive system involving a reaction that is reversible in the chemical sense (i.e. it "goes both ways" around its equilibrium (i.e. the  activities ( ~ concentrations) of its components can match the value of the equilibrium constant $\mathrm{K(T)}$ ), then as a whole there are changes within the system that satisfy $\Delta{S_{system}} = 0$ (i.e. reversible in a thermodynamic sense) because the reversible reactions within have molar changes of entropy + $\Delta S$ (for the forward reaction, typically non-zero) and $- \Delta S$ (for the backward reactions, with a minus sign in front), with forward and backward reactions at the same rate (dynamic equilibrium).
In a previous reply, I attempted the following wording, which perhaps is more detailed but does not make explicit the distinction between entropy of reaction and entropy of the system:
With $\Delta G = \Delta H - T\Delta S$ for a reversible reaction, where $\Delta G$ is the change in Gibbs energy, $\Delta H$ the enthalpy of reaction and $\Delta S$ the change in entropy (and $T$ and $P$ constant), I think what is going on then for a system where a dynamic equilibrium is established between backward and forward reaction, which could qualify as a reversible process, once the equilibrium is reached. $T$ sets the value of the equilibrium constant and hence $\Delta G$ for these conditions. $\Delta H$ and $\Delta S$ typically are considered to be fairly constant (you could see how $\Delta S$ may in fact be dependent on temperature though, just through different vibration energies of different bonds).
In the special case where $\Delta G = 0$ (when equilibrium constant $K = 1$), with $\Delta G = \Delta H - T\Delta S$, one has to assume that the $\Delta S$ is fully reversible, which I think assumes faultless conversion of $T\Delta S$ to $\Delta H$ and vice-versa, between reacting species.


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Effects Guiding Electrophilic Aromatic Substitution

This is a question which I come across in an examination. I know the answer and also know the solution. Well when I first saw this problem I didn't know how to approach this problem and I think I will still not be able to predict the correct answer of a question which will be based on this question's type . I am not asking that this is "this" reaction and it must proceed like" this " ,but what I want is how will you approach this problem . What will you think on first look of the problem. What all things you will observe from question and also do write what you think the answer must be . Also BEWARE THIS HAS MULTIPLE ANSWERS CORRECT. I am not telling you the answer because it would be easier to explain that answer . Also, please those who have seen this problem earlier. Please resist yourself from posting answers.

The first thing I notice is the hydroxyl group, a very powerful activating group and an ortho-para director.  The hydroxyl group is such a strong activator that if you take a solution of phenol and drip bromine into it, the phenol will react with the bromine so rapidly that the bromine color will be immediately removed from the solution as 2,4,6-tribromophenol is formed.
The next thing I notice is the very bulky t-butyl group, also an activator, but much milder than hydroxyl and also a mild ortho-para director.
Next I notice that the two substituents are arranged meta to one another.  There is one position in between the two groups that will be sterically blocked (double-ortho effect); the position on the other side of the t-butyl group will also be sterically blocked, but not as bad.  Also the directing effect of the hydroxyl group will overwhelm the directing effect of the t-butyl group.
Finally, in the halogen series size increases from chlorine to iodine.
In the reaction with iodine, the position ortho to the hydroxyl group and para to the t-butyl group is the only position that is electronically activated and sterically accessible to the bulky iodine.
In the reaction with smaller bromine, the position ortho to the hydroxyl and para to the t-butyl and the position para to the hydroxyl and ortho to the t-butyl are electronically activated by both groups and sterically accessible.  The "double-ortho" position is still sterically blocked.
With the smaller chlorine atom, all 3 positions - even the "double ortho" - position are sterically accessible to the smaller chlorine.
Based on this reasoning, I would select answers A, B, and C.  I would probably also include D because the results are (weakly) reinforced by the electronic effect of the t-butyl group.

I agree, both answers should be graded correct ie (A, B, C) and (A, B, C, D)

@ron do you know a name of book of organic chemistry that deals with such things I mean not of these question type but which gives such good comparision between one molecule interacting while other not . Also I wanted to know How do you immediately catch this problem I mean this is because you have done a lot of practice and have seen many situations in organic chemistry that make you ponder or the question was that obvious

@DSinghvi  Sorry, I don't know of a particular book that approaches problems this way. And yes, just practice and experience have taught me this approach.


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Conical intersections - phase
Can anyone explain why at a conical intersection the wavefunction changes sign? My understanding is that it is a test to see if the crossing IS indeed a conical intersection or just a coincidence of potential energy surfaces, which do not exhibit such phase changes upon periodic cycling. If this is correct ... why? Further what can we gain from knowing that?

A rather quick googling shows up that there is some Longuet-Higgins theorem, which states almost what you are looking for. For instance, the abstract of this paper says:

It is proved that if the wave function of a given electronic state
  changes sign when transported adiabatically round a loop in nuclear
  configuration space, then the state must become degenerate with
  another one at some point within the loop.

In the paper itself it is further stated that this in turn implies that

[the corresponding] potential energy surface intersects one of another electronic state

and that the theorem

makes it possible to diagnose the presence of an intersection between potential energy surfaces from the behaviour of the electronic wave function as it is transported adiabatically round a closed loop remote from the intersection

which answers both your questions, if I understand them correctly.


text:
Why do atoms "want" to have a full outer shell?
Okay, so I know that this is about filling the orbitals of the atom, and I understand that. What I don't understand is why? For example, an Oxygen atom has 8 protons and 8 electrons spinning around it. Oxygen wants to have 8 valence electrons so it takes 2 electrons from other elements when bonding. But, since Oxygen has 8 protons and now has 10 electrons around it, it has a charge of -2. I don't understand how having a charge is more stable than having no charge at all, shouldn't the most stable version of an element be its neutral state? Why do the atoms want to look like the next noble gas?

Have a look here.

Thanks, I looked into it. But it explains why the number of electrons in a specific shell's full state is what it is, not why it's more stable than the neutral state of the atom, which is what I'm asking

I've wondered about this issue also. The answers that I've found generally use the rules of quantum chemistry as an explanation. But while these rules are consistent they don't provide the kind of explanation I and others would find more satisfying. As a starting point, it seems to me that there has to be some relationship between a full outer shell and the ability to construct a low potential energy arrangement. Using the case of a diatomic like H2 for simplicity, a low potential energy arrangement exists a the particular distance at which there is neither attracting nor repelling force.

So the question becomes: Why does the creation of a full outer shell for a pair of hydrogen atoms produce a low potential energy arrangement... "bonding distance"... at which there is zero force?

Related: Why are atoms with eight electrons in the outer shell extremely stable?

You are attaching too much importance to Lewis structures. The 8-electron rule and Lewis structures which are derived from it are only rough guidelines for working out the electronic structure of a compound in very broad strokes. Often these broad strokes are accurate enough to make some meaningful statements about molecular properties but it does not accurately describe the true electron or charge distribution in a compound. Take water for example. As you say, the 8-electron rule would predict that the hydrogen atoms each transfer one electron to the oxygen molecule which would then carry a charge of -2. But experiments show a different result: here you can find a paper that determined

that a charge of approximately $0.5 e$ is transferred from each
  hydrogen atom to the oxygen

which would amount to a partial charge of $-1 e$ on oxygen and here you can find a similar claim:

The charge distribution depends significantly on the atomic geometry and the method for its calculation but is likely to be about
  $-0.7e$ on the O-atom (with the equal but opposite positive charge equally divided between the H-atoms) for the isolated molecule.

And there not being a $-2e$ charge on oxygen is quite understandable if you look at the electron affinities of oxygen (e.g. here): You see that the first electron affinity is $-142 \, \mathrm{kJ}/\mathrm{mol}^{-1}$ and thus energy is gained by the process of transferring a free electron to a neutral oxygen atom. But the second electron affinity is $+844 \, \mathrm{kJ}/\mathrm{mol}^{-1}$, so another electron transfer is very costly. The quite extreme situation of a -2-charged oxygen described by the 8-electron rule is very rarely found in real compounds. Even in ionic compounds like $\ce{NaCl}$ the electron is not 100% completely transfered from the electropositive to the electronegative element.

Essentially, its a result of energetics.  Energy generally goes down as we approach the 8 e- valence space.

Its not a problem about being charged, its about the symmetry. The quantum states of those valence orbitals are stable, in that there is little interaction with the other electrons. Since there are already filled orbitals, electrons will want to bind to complete the shell because as it happens extra binding energy is available. With a full shell maximum binding occurs.


text:
13C NMR peak location in butanone

The above is the correct $\ce{^{13}C}$-$\mathrm{NMR}$ spectrum of butanone.
In the $\ce{^{13}C}$-$\mathrm{NMR}$ spectrum of butanone, I figure that the peak locations of the first $(\pu{27.3 ppm})$ and third $(\pu{35.2 ppm})$ carbons should be switched. Shouldn't the third carbon be shifted upfield by the terminal methyl group and therefore have a "smaller $(\pu{ppm})$" than the first carbon?

No, substitution by an alkyl group produces a small downfield shift in both proton and carbon nmr. Therefore, in a typical hydrocarbon a secondary carbon will be downfield of a primary carbon.  The assignments in your figure are correct.
A possible explanation for the downfield shift of a carbon atom when a proton is replaced with another carbon atom is as follows.  Hydrogen is more electropositive than carbon, therefore, in a C-H bond a significant amount of the electron density in the bond resides around the carbon.  When we replace the electropositive hydrogen with a relatively more electronegative carbon, then there is less electron density around the central carbon atom.  This reduction in electron density deshields the carbon nucleus and shifts it downfield.  Every time another hydrogen is replaced by carbon a further downfield shift occurs.

Thanks! I guess I had always thought about alkyl groups as being more shielded than the rest, but I just needed to think about the fundamental electronegativity behind shielding/deshielding.

I agree with the Ron's notation that a substitution by an alkyl group produces a small downfield shift in both proton and carbon NMR. My intention here to educate OP that he/her can learn some mathematical calculations, which can predict expected chemical shifts of most of simple compounds such as 2-butanone. If OP is studing spectroscopy, my recommendation is Ref.1, which gave pretty good amount of mathematical calculations. For instance, following is the predicted $\ce{^{13}C}$ chemical shifts for 2-butanone:
The difference between butane and 2-butanone is the carbonyl group at second carbon. According to Ref.1 (page 269):

Replacement of $\ce{CH2}$ of alkanes by $\ce{C=O}$ causes a downfield shift at  $\alpha$-$\ce{C}$ $(\pu{10-14 ppm})$ and an upfield shift at the $\beta$-$\ce{C}$ (several $\pu{ppm}$ in acyclic compounds).

Following diagram would show the actual $\ce{^{13}C}$ chemical shifts butane (Table II; page 260, Ref.1) and calculated $\ce{^{13}C}$ chemical shifts of 2-butanone accordingly:

According to the calculations, $\ce{^{13}C}$ chemical shift of $\ce{C}$1 of 2-butanone (a carbon $\alpha$ to carbonyl carbon) is approximately $13.4 + 14 = \pu{27.4 ppm}$, actual value of which is $\pu{27.3 ppm}$. Similarly, the calculated $\ce{^{13}C}$ chemical shift of $\ce{C}$3 of 2-butanone, which is also $\alpha$ to the carbonyl group, is approximately $25.2 + 14 = \pu{34.2 ppm}$. The actual value given by OP is $\pu{35.2 ppm}$. This prove that the original assignment is not by mistake. they are the actual values of 2-butanone.

References:

Robert M. Silverstein, G. Clayton Bassler, and Terence C. Morrill, In Spectrometric Identification of Organic Compounds; Forth Edition; John Wiley & Sons: New York, NY, 1981 (ISBN: 0-471-02990-4).


text:
Show the vibrational frequency of fluorine molecule anion is much lower than that of fluorine molecule
Basically I need to draw the molecular orbital for $\ce{F2}$ and then answer a bunch of questions about it. I have drawn it correctly, as far as I know, but I don't know how to use it to show that the vibrational frequency of $\ce{F2-}$ is lower than $\ce{F2}$.
I worked out the bond order for $\ce{F2-}= 0.5$  and for $\ce{F2}= 1$. Is this what you use to find it out, or is it something to do with the HOMO-LUMO?

Want to bust out some electronic structure theory? raises and lowers eyebrows rapidly

My answer is intended to be some sort of interactive guide to the solution. So, when you have looked at my hints so far and they aren't enough to get you to the solution simply notify me, tell me what points are unclear and I will add something to my answer that should get you back on track.

You should definitely post the MO diagram of $\ce{F2}$ that you worked out. When trying to solve such a problem you should try to analyse the question as thoroughly as possible to make sure that you collect all the information available to you beforehand. Especially important here is that the question already tells you that the vibrational frequency of $\ce{F2^{-}}$ is lower than that of $\ce{F2}$. So the first question you should ask yourself is: Under what circumstances is a vibrational frequency lowered and which relation do I know that links the vibrational frequency to some variable that is connected to the MO diagram of $\ce{F2}$? I give you a hint: You will need the equation of a harmonic oscillator
\begin{equation}
 \omega = \sqrt{\frac{k}{m}}
\end{equation}
where $\omega = 2 \pi f$ is the angular frequency, $f$ is the frequency, $k$ is the force constant (in your case of the bond and thus its a measure of the bond strength) and $m$ is the reduced mass of the oscillating system. (The harmonic oscillator is one of those simple model systems that can help you quite often when you want a starting point for the investigation of much more complicated ones.) Now ask yourself: Which of the variables $k$ or $m$ might have a relation to the MO diagram and how will they be related?


text:
Electrochemical potentials
I was carrying out an experiment into the effect of temperature changes on the standard potential of a cell.
$$\ce{Zn(s) +2Fe^{3+}->Zn^{2+} +2Fe^{2+}}$$
I used equi-molar concentrations of $\ce{Fe(CN)_6^{4-}}$and $\ce{Fe(CN)_6^{3-}}$. Thus the Nernst equation
$$\mathrm{E^{\theta}_{cell}=E_{cell}-{\frac{RT}{nF}}*ln1}$$ reduces to $$\mathrm{E^{\theta}_{cell}=E_{cell}}$$
We plotted our $\mathrm{E^{\theta}_{cell}}$ values versus T and noticed that our $\mathrm{E^{\theta}_{cell}}$ values were actually increasing with temperature.
I would have expected that these values go down as the concentration of the reactant falls.
I asked my demonstrator and sourced information from different web resources, but can't access a definitive answer.
Any hints even?

@JohnSow use E^\circ for $E^\circ$

My demonstrator told me that he expected the entropy change to be negative as well. I would have that with a larger atom $$Fe^{2+}$$, there would be less solvation and thus a positive change in entropy?

A few thoughts:
For starters, $\ce{Zn^{2+}}$ is also part of the reaction quotient, so the log term in the Nernst equation will not necessarily be zero if the hexacyanoferrate concentrations are equal, depending on the $\ce{Zn^{2+}}$ concentration:
$$E = E°+\frac{RT}{zF}\ln\frac{[\ce{Zn^2+}][\ce{Fe^2+}]^2}{[\ce{Fe^3+}]^2}$$
Depending on whether the reaction quotient is greater than or less than 1, the potential can have either a positive or negative temperature dependance.
Second, the Nernst equation uses activities not concentrations so depending on the ionic strength and other factors, the activities of the ions in solution may not be what you think. This deviation is greater for multiply charged ions and may be significant for concentrations over 1 mM.
Finally, since the hexacyanoferrate species are both soluble, I'm assuming you're using some kind of reference electrode for the second electrode. This is also temperature dependant in the same way as the rest of the cell.

use \ln for $\ln$


text:
Can I use piranha solution to clean copper out of impurities?
I have some old pieces of wire. Since I want to use copper in homemade fireworks, I'm trying to gather it from the wires. I could buy it, however harvesting is much more fun.
Sulphuric acid doesn't dissolve copper in normal conditions, even when it's concentrated. However adding $\ce{H2O2}$ causes vigorous reaction during which copper sulphide is created.

On the solution surface, remains of impurities are floating. I'm afraid, however, that most of impurities were dissolved. Were they?
I don't need my $\ce{CuSO4}$ chemically clean, but too much impurities could increase the risk of toxic gases during electrolysis.

The difficulty here is that you've almost surely got excess peroxide and sulfuric acid in your solution, and it's going to be a challenge to eliminate them in a safe way in your kitchen/garage without resorting to something more expensive than just buying the chemical you want to make or producing a toxic byproduct.  If I were going to do this, I would just buy a couple of pounds of copper sulfate.  It's sold at home improvement/hardware stores as root killer in high purity form for very little money.

This way I could just buy copper dust as well, since that's what I need. No, I'm really trying to purify it.


text:
Prevent molten glass from cracking when cooling down
My mother works in a ceramics workshop. Besides giving lessons she creates various ceramic dishes, cups and plates. She often uses the effect of colored glass melting in the product and then leaving beautiful patterns at the bottom.
However it always cracks a lot as cooling down. Sometimes, it looks good when it's cracked, but other times she's trying to make it look like a gem - and cracked gem is not very gemmy.
So I was curious if there is something non toxic, insoluble in water and alcohol, that would melt at the similar temperature as glass and would prevent it from cracking in the cool down phase.
The glass doesn't have to be temperature resistant in the final products. We just want to avoid the cracks

This might be better asked in an art exchange. (Zoinks! There's no Art Exchange on here!?!) In any case, the way that glassmakers usually avoid this problem is through the use of an annealing oven.  The forces involved in it cracking are very large; I wish you the best of luck in finding the sort of product you're after, but I'm doubtful about its existence.  I'll be delighted if I'm proven wrong though.

At first I would suggest using glass with low coefficient of thermal expansion, e.g. http://en.wikipedia.org/wiki/Borosilicate_glass
And then of course annealing, as sugested by @JasonPatterson, see http://en.wikipedia.org/wiki/Annealing_%28glass%29

You might not need a specialized annealing oven if you can control the kiln correctly. Just cool it down VERY slowly, maybe over a 24 hour period. This would require the kiln to allow some sort of temperature gradient and I don't know anything about kilns so I don't know if yours will do that. If you can adjust the temperature without turning off or opening the kiln you can probably do this manually. Every few hours reduce the temperature by a bit until it's off. This could require staying up night and playing with how quickly you cool it down until you find the right conditions.

@JasonPatterson Though there is Arts & Crafts with dedicated tags «ceramics», «glass», and «material-selection».

@Buttonwood And slowly but surely things change for the better.  :-)  Looks like that opened a year and a half after my comment, on April 26 of 2016.  The update is sincerely appreciated.

The surface cracks due to a mis-match in the temperatures inside and outside the ceramic. Things contract as they cool. So, the cooler outer surface is contracting against the hotter inner surface, which is still not contracting. The outer surface wants to be closer together than the inner material, but the inner material forces the outer material to stay apart. When the ceramic on the outside transitions to a solid (at what is called its "glass temperature"), it can no longer flow like a liquid. The only way it can continue to stay apart due to the forces from the inside is to "fall apart" ... i.e. to crack.
Your mother may be able to coat the pottery with a different glaze that forms a glass at a lower temperature than the one she has now. Also, the suggestions to cool VERY SLOWLY are directly on target. The cooling should happen in a place that has no air flow, ideally in a closed box. Absent an annealing oven, a different option may be to put the pots in an conventional NON-CONVECTION oven that is held at a temperature just below the glass temperature of the glaze being used. When the entire pot reaches that temperature, it can be removed to cool further in a box at room temperature, again with no air flow.


text:
Can functional groups exist by themselves
Just started to study organic compounds and I am curious about functional groups. I want to know if functional groups can exist as molecules themselves.

By definition, a functional group is a non-carbon atom (or group of atoms) attached to the carbon compound. If the carbon compound isn't there, you wouldn't have an organic molecule.

Sorry but some functional groups may contain carbon atom within it. For example keto group, carboxy group etc

As molecules? The reason a functional group is a group is because it is attached to a molecule. There is no point of saying that a functional group which exists by themselves. Even if miraculously it does, it's high instability would make it highly reactive. I guess you've just started organic, so you'll get to know about all this.

@Rohin97 why are you critcal about the question. If I knew that functional groups were highly unstable if they existed by themselves I wouldn't have asked the question. But I am not surprised, most comments and answers from individuals on stack-exchange's websites are usually critical, prideful people. But not everyone: thanks Cheran for your answer.

A functional group on its own essentially has at least one non-bonded electron, thus making it a free radical, this makes them highly reactive molecules, making them react with other substances (including themselves) and can thus only exist in very low concentrations in inert medium/vacuum.
However, there are molecules such as $\ce{NO}$, $\ce{NO2}$ which do have an odd electron in their molecular state, can exist in reasonably large quantities, and can be functional groups in organic chemistry, these are in some sense functional groups that exist as molecules.