text:
What is the reason why protons and electrons do not collide?
can someone give me an intuitive picture of why electrons don't collide with protons?
I know that electrons move in a sort of cloud, which is our 'orbital', and that they mainly behave like electromagnetic wave because of their very tiny mass. However, how comes they don't collide with the protons? I seriously didn't find a good explanation of these neither by professors I talked to nor by books, so a thousand points from me if you are able to get me into this.
Keep in mind that I am not looking for values, equations, or quantitative explanation of this, for that I have read enough, but for a picture. Think about the subatomic particles as you see them in your mind and describe their relationship.

@bon Thank you sir, I appreciate such a nice welcoming!

They kinda do collide, only they don't seem to care much.

@Ivan Neretin Ok so far it matches my picture, since the probabilistic cloud of the lowest principal quantum number would cover the area of the nucleus as well, but since I am used to think about the acid and base behavior within molecules, I can't understand what is the barrier between electrons and protons..

http://chemistry.stackexchange.com/questions/15727/how-does-the-vacuum-in-atoms-persist

http://chemistry.stackexchange.com/questions/35437/why-dont-electrons-just-crash-into-the-nuclueus

@Mithoron before giving the -1 you should notice that all the answers to those question don't satisfy my question of an intuitive picture of the subatomic particles, which is why I asked on Physics S.E. as well about this, and there as well I didn't get an answer that can lead me to picture this... http://physics.stackexchange.com/questions/257180/atom-particles-relationships

"Collide" is the wrong word: it implies some kind of bouncing solid spheres, which are just not there. There is nothing to collide with. An electron does not just sometimes behave like a wave - it is a wave. Imagine a sea wave running against a lone rock. It will just run around the rock and continue on its way. Sure, once in a great while a wave would bring the rock down. Well, an electron has the chance to react with proton by means of K-capture.

@IvanNeretin this answer is worth the 1000 points. It is a very good picture that perfectly describes it! Indeed the more the rocks and the seawaves the more the power of them on each other, and the chance to get one of the rocks to fall within a wave (forming a neutron), which explains why K-Capture usually happens in bigger atoms. Thank you sir, I would add your answer in a book! If you write it down here I will accept it as answer

They do collide.  But it's not like cars crashing and smashing each other up on the highway.  Their quantum mechanical wavefunctions overlap and the electrons do in face settle in with wavefunctions that are centered on the nucleus.  But the low mass of the electrons, the Heisenberg uncertainty principle (which has its greatest effect on low-mass particles), and the Pauli exclusion principle (electrons can't all pile up in the most stable quantum state)  cause the electron wavefunctions to get diffused out around the nucleus.  Instead of getting smashed up, the electrons basically surround snd embed the atomic nucleus.

This is a good answer. Pretty similar to the one by @Ivan Neretin in the comments, even though his answer was even more accurate onto giving an example. Thank you both guys!

My comment is not more accurate (which, BTW, is why I'd rather not turn it into an answer). If anything, it is more picturesque, but remember: the whole thing about sea waves and rocks is but a metaphor, and those are not intended to draw practical conclusions.

You are welcome.  I woll make sure to +1 Ivan's comment.

@Ivan Neretin yes of course I know that, but it's a good metaphore which gives the idea of that.


text:
Mass of Ice during Phase Change
If 18g of ice at a temperature of -6 ℃ is placed in 250 g of water at 20 ℃, we know that there is enough energy supplied by the water to completely melt the ice and transform it into liquid. My question is: during the phase change of the ice, when the temperature of the ice has reached 0 ℃, is mass from the ice transferred to the surrounding water? In other words, is the mass of the ice changing during phase change or is it constant? By the mass of ice, I mean mass that is still solid.

Have you ever actually seen a piece of ice, and how it melts?

When the temperature of the ice has reached 0°C, additional heat transferred from the liquid water to the ice will not increase the temperature of the ice above 0°C. Instead the transferred heat is consumed as heat of fusion. This lets ice melt, which means that there is a phase change from solid to liquid, which means that a certain mass of the ice is transferred to the liquid water. The mass of the ice consequently decreases during the phase change.

Thank you for your answer. My source of confusion was the fact that ice and water can co-exist at 0 ℃.

source: Stack Exchange

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XlogP of octanol and water itself
According to Pubchem, the XLogP of octanol is 3 and that of water is -0.5. Why aren't they both 0 or 1 which seems to make more sense to me?

These values are consistent with the slight solubility of water in octanol and octanol in water.
For example, the solubility of 1-octanol in water is 0.46 g/L. Assuming the density of a saturated solution of octanol in water is approximately that of water, then the concentration in grams per liter can be converted to moles per liter:
$$\dfrac{0.46\ \mathrm{g\ octanol}}{\mathrm{L\ water}}\times \dfrac{\mathrm{mol\ octanol}}{130.23\ \mathrm{g\ octanol}}=3.53\times 10^{-3}\ \mathrm{M}$$
The molar concentration of pure 1-octanol can be calculated from its density:
$$\dfrac{824 \ \mathrm{g\ octanol}}{\mathrm{L\ octanol}}\times \dfrac{\mathrm{mol\ octanol}}{130.23\ \mathrm{g\ octanol}}=6.33\ \mathrm{M}$$
And the $\log P$ is:
$$\log P = \log \left( \dfrac{\mathrm{[octanol]_{octanol}}}{\mathrm{[octanol]_{water}}} \right) \log \left( \dfrac{6.33\ \mathrm{M}}{3.53\times 10^{-3}\ \mathrm{M}} \right)=3.25 $$
Likewise, if we knew the solubility of water in 1-octanol, we could do the same calculation.

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Chondroitin Sulfate (greater volume in solution than in crystal form)
I've read in a textbook on biophysics that chondroitin sulfate occupies a greater volume in a solution than in crystalline form. Why is that?
Is it because of hydrogen bonds in solution that the volume is larger?

$\hspace{4.5cm}$
In solution, the negative charges on the chondroitin sulfate repel each other and force the molecule into an extended conformation. The polar molecule attracts many water molecules, which increases its molecular volume. In the dehydrated solid, each negative charge is counterbalanced by a positive ion, and the molecule is condensed.


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Are β-sheets formed from one or more polypeptides?
I know that a β-strand forms bonds with another β-strands and so on and thus a β-sheet is formed, but are those strands from the same polypeptide? And does their distance matter? Do they have to be close or it is not necessary? Thanks in advance.

A β-sheet is formed as the result of hydrogen bonding between the β-strands in the secondary structure of a protein. The hydrogen bonding occurs between the carbonyl oxygen and the amide hydrogen of adjacent strands, and can take on either a parallel or an antiparallel arrangement. As can be seen in the diagrams, antiparallel arrangement of the polypeptides results in more stable interactions between the strands, but both arrangements are possible and can be found in nature. Regardless of the type of arrangement, β-sheets can be formed by either interactions between different polypeptides or intramolecular hydrogen bonding within a single polypeptide. In fact, intramolecular hydrogen bonding creating β-sheets is really common. One example of it occurring in β hairpin motifs.

Since the β-sheet is part of the secondary structure of a protein shouldn't it only occur in a single polypeptide?

While you are correct that β-sheets are considered part of the secondary structure,  they generically refer to the shape created by hydrogen bonds between parallel polypeptide chains, meaning that you can have β-sheets that require quaternary structure to emerge.

I think I understand. So when the β-sheet structure is formed from one polypeptide chain, it is considered a part of the secondary stucture, while when more than one polypeptide chains form a beta sheet then it is considered a part of quaternary structure. Thank you very much for the clarification.

No problem, glad it helped.

authors: Αντώνιος Κελεσίδης


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Density Functional Definition
Is it wrong to say that Density Functional means that Electron Density is a function of the orbitals (wave function) of all electrons in 3 dimensions, if so, why ?

I don't really understand your question. The electron density $\varrho (\vec{r})$ is defined to be a function of the many-particle wave function, i.e. $\varrho (\vec{r}) = \langle \Psi | \hat{\varrho} (\vec{r}) | \Psi \rangle$, where $\hat{\varrho} (\vec{r})$ is the density operator, $\vec{r}$ is the position vector, and $\Psi$ is the many-particle wave-function. But that has nothing to do with the definition of a density functional. A density functional in its most general meaning is just a functional of the aforementioned electron density. Maybe you should provide a bit more context.

@Mr.Why The energy is a functional of the wavefunction, i.e. $E=E[\psi]$.

You are right .

Eignvalue is  not as same as function

In general, a functional $F$ is a mapping from an arbitrary set $\mathcal{X}$ of functions to the set of complex numbers $\mathbb{C}$ or the set of real numbers $\mathbb{R}$:
$$
F : \mathcal{X} \mapsto \mathbb{R}.
$$
or
$$
F : \mathcal{X} \mapsto \mathbb{C}.
$$
For example, if you consider $\mathcal{X}$ as the set of polynomials with real coefficients, you can define a functional $F$ as
$$
F[f] = \int_0^1f(x)\,dx
$$
i.e. your functional $F$ takes a polynomial function $f\in\mathcal{X}$ (for example $f(x)=3x+1/2$) as an argument and returns a scalar (2 for $f(x)=3x+1/2$, as you can easily verify).
A density functional is simply a functional $F[f]$ where the argument $f$ is the electron density $\rho(\vec{r})$ (i.e. a density functional is a functional of the electron density). For example Hohenberg and Kohn showed that the energy $\epsilon$ of a quantum system is a functional of the density
$$
\epsilon=E[\rho]
$$
This means that when you plug the electron density of your system $\rho(\vec{r})$ into the energy functional $E[\rho]$ you get a number $\epsilon$, which is the energy of your system. The whole energy functional is not known explicitly, but some of its components are known. For example for the external potential energy we have
$$
V[\rho] = \int v(\vec{r})\rho(\vec{r})d\vec{r}
$$
and for the Coulomb interaction between electrons we have
$$
J[\rho] = \frac{1}{2}\iint \frac{\rho(\vec{r})\rho(\vec{r}')}{|\vec{r}-\vec{r}'|}\,d\vec{r}d\vec{r}'
$$
which are clearly functionals of the electron density.

(i.e. density functional is a functional of the electron density). So cannot it  be said that density functional means that density is it self a functional of some thing else (orbitals=wavefunctions) ?

The density functional is distinct from the density itself.  The density maps from a position vector (3 real numbers) to a single real number $\rho:;\mathbb{R}^3 \rightarrow \mathbb{R}$.  The density functional maps from an entire function to a single real number $F:;\mathcal{X} \rightarrow \mathbb{R}$.
Mapping from a function to something else is what makes something a functional.

Also, the orbitals themselves are fictitious; there are infinitely many choices of orbitals for the same wavefunction but some are more sensible than others.  Also, although the density can be expressed in terms of the wavefunction: $\rho(r) = |\psi(r)|^2$ both side just equal a number and require us to input a position vector so the density is still a function mapping from vectors to numbers, as is the wavefunction: $\rho : \mathbb{R}^3\rightarrow \mathbb{R}$, $\psi : \mathbb{R}^3\rightarrow \mathbb{C}$.

Electron density, is defined as the probability density of finding any of the $n$ electrons with arbitrary spin at some point $\vec{r}_{1}$ in space,
$$
    \newcommand{\el}{_\mathrm{el}}
    \newcommand{\dif}{\mathrm{d}}
    \rho(\vec{r}_{1})
    :=
    n \sum_{m_{s1}} \sum_{m_{s2}} \dotsi \sum_{m_{sn}} \iint \dotsi \int
    | ψ\el(\vec{q}_{1}, \vec{q}_{2}, \dotsc, \vec{q}_{n}) |^{2}
    \dif \vec{r}_{2} \dif \vec{r}_{3} \dotsb \dif \vec{r}_{n} \, .
$$
From the first Hohenberg–Kohn theorem, it is known that electronic energy is a functional of electron density,
$$
    E\el = E\el[\rho(\vec{r}_{1})] \, .
$$
i.e. electronic energy $E\el$ is a function that takes another function, namely, electron density $\rho(\vec{r}_{1})$, as its input argument and returns a scalar value (real number). So, density functional is a functional of electron density that returns (possibly approximate) electronic energy or a part of it, if $E\el$ is subdivided into parts.


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How to transform an aliphatic primary amine to the corresponding nitro compound?
Could you suggest a reagent or procedure to transform an aliphatic primary amine to the corresponding nitro compound.
I know this is a bit backwards of a request because the usual procedure is the reduction of nitro groups to their corresponding amine, but there is a need of its electron-withdrawing properties for a subsequent condensation to form the intended intermediate. If this approach could be solidified, then it would make things a lot easier for this recent graduate, as the starting material is a lot less expensive.
I have found references to ozonolysis, but am unsure of it, as I have no previous experience with these sorts of reactions. I have also seen in abstracts where it is discussed formation of the diazonium compound, which was the used to form the nitro compound. There are a couple of others I have come across, but I am not sure of their efficacy or plausibility.

Using silica gel, primary amines can be oxidized to their corresponding nitro compounds using ozone ($60$-$70 \%$ yield). In this paper, dry silica gel (dried for $24\ \mathrm{h}$ at $450^\circ \mathrm{C}$) was mixed with an amine. The silica gel (ca. $\mathrm{30\ \mathrm{g}}$) containing the amine ($0.1$-$0.2\ \mathrm{wt/wt \%}$) was cooled to $-78 ^\circ \mathrm{C}$, and a stream of $3 \%$ ozone in oxygen was passed through it.
Aliphatic amines and aromatic primary amines are rapidly and effectively  oxidized to nitro compounds by dimethyldioxirane ($80$-$90 \%$ yield). Dimethyldioxirane is prepared by reaction of OXONE ($\ce{2KHSO5 \cdot KHSO4 \cdot K2SO4}$) with buffered aqueous acetone. In a typical reaction, $0.7\ \mathrm{mmol}$ of amine in $5\ \mathrm{mL}$ acetone is treated with $95\ \mathrm{mL}$ of dimethyldioxirane in acetone ($0.05\ \mathrm{M}$). The solution is kept at room temperature for $30\ \mathrm{min}$ with the exclusion of light.
Oxidation of amines to nitro compounds in high yield has also been carried out using oxyacids like peracetic acid, peroxytrifluoroacetic acid, and meta-chlroperoxybenzoic acid, though the anhydrous forms of these acids are hazardous and difficult to use.

Ono, N. The Nitro Group In Organic Synthesis; Wiley-VCH: New York, 2001.

Thanks for the reply. I will take a look into these methods.

The peroxy acid oxidation was another method I came across in my original query but was curious about the conditions.
Thank you for the reference! I started to put in the request for that reference and realized it was link to a version available on the web. Nice!

@JeffreySchultz Happy to help!

I am also going to attempt the second reaction you have provided. The 80-90% yield reported is somewhat of a confidence booster. Beyond that it will be whatever is easier or what further advice I receive on the matter. We have a few guys that might be able to provide some guidance on the ozone reaction.

These are the 2 methods I could think of to convert primary amine to corresponding nitro compounds.
The first method:
I am not so sure about this method. I think the yield would be low.
Second method:http://onlinelibrary.wiley.com/doi/10.1002/(SICI)1099-0690(199804)1998:4%3C679::AID-EJOC679%3E3.0.CO;2-W/abstract
Here is an abstract from the paper:- Primary aliphatic amines are oxidized with tert-butyl hydroperoxide to the corresponding nitro compounds in 50−98% yield using Zr(Ot-Bu)4 as the catalyst.

Thanks for the reply. I will take a look into these methods.

Happy to help others learn :)

This seems to be a very attractive route. It's sad that copper seems to have been lost in the shadow of palladium. I believe this will be a very good starting point for my attempts as everything is on-hand. I started looking into diazonium chemistry more and have found some very interesting reaction possibilities, however, most of them fall outside the scope of the current task. Introduction of the cyano group with CuCN seems most interesting, but it looks to be falling in popularity due to cyanide toxicity even with its negligible water solubility.

As for yields, they will be determined, and if there is issue, it at least justifies the more tedious procedures, in my mind at least.

Diazonium reactions are very neat. It helps with a lot of conversions. And yes, palladium does shadow copper, maybe because of its versatility.

Wouldn't addition of $\ce{HBF4}$ yield a fluoride?

No, adding $HBF_4$ to diazonium chloride gives diazonium fluoroborate, which on further heating gives a fluoride. But on heating with $NaNO_2+Cu$ it gives the corresponding nitro compound.


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Does any compound exist with multiple liquid phases?
I've seen some phase diagrams and I found really interesting, how many exotic solid phases of most elements exist on higher pressure and temperature.
But I've seen always only a single liquid phase and also only a single gaseous phase.
The second is understable, but the first not. I think, if there is an unconventional potential energy field between the molecules, maybe it is not impossible.
Does any chemical compounds with multiple liquid phases exist?
If not, why not?

Extension: Well, superfluid helium is one of them. Maybe I don't formulate enough well, but I think on a distinct phase border, like between ice-I and water. As I know, the superfluid helium is a mix of its superfluid and normal fluid phases.

Well, superfluid Helium is one of them. Maybe I don't formulate enough well, but I think on a distinct phase border, like between ice-I and water. As I know, the superfluid Helium is a mix of its superfluid and normal fluid phases.

related http://chemistry.stackexchange.com/questions/49510/why-so-few-fluid-phases

I suspect helium may be the only one. At sufficiently low temperatures, superfluid $\ce{^3 He}$ phase separates from $\ce{^4 He}$, producing two distinct superfluid phases, presumably with an almost invisble meniscus between them. Not sure whether this satisfies your requirements.

@NicolauSakerNeto Both of He3 and He4 can be superfluid.

There's no reason why chemical substances with multiple liquid phases shouldn't exist.
In fact there is evidence for the existence of a least a second form of (surprise) liquid water. If you search the Web for liquid-liquid phase transitions you will find more of these, e.g. cerium.

You may want to describe the second liquid phase of water in a little more detail, or else readers might mistake it for "polywater" which turned out to be a miss.


text:
Heat Transfer during Phase Change
When ice is melting, we know that the temperature stays constant at 0 ℃. If we were to place ice at 0 ℃ in a sufficient amount of water at ambient temperature, would the temperature of the water itself also stay constant during the time the ice is melting, or is it only the temperature of the ice that stays constant? Does the actual cooling of the water occur after or during phase change?

If temperature of water stayed constant, where in the world the heat would be coming from?

But constant temperature does not imply that there is no temperature difference (between the ice and water). There is still a temperature difference, but my question is whether this difference remains constant or not.

See, there is a substantial amount of heat transferred from water to ice. In effect, you are asking whether water can give off heat, while staying unchanged at constant temperature. I wish it would; then we would not need heaters in the winter.

No, as the ice is melting, the water is cooling down. Ice melts in water because the water molecules in solution collide with those in the ice, giving them energy. In exchange, the water molecules in solution slow down a little during the collision. While the Ice does not change temperature as it is melting, the energy from the water is going into breaking the crystal structure of the ice, turning it into water. This process consumes energy, and the energy to fuel this process comes from the water in solution. Another factor to consider is conservation of energy - if energy is being put into breaking the ice, where else could that energy come from but from the water?


text:
How does a lower pH change the alcohol concentration of wine?
I have been told it that when the pH of wine is lowered the alcohol concentration will decrease because of Le Chatelier's Principle. Which states; if a  equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. How can I best expand on this?

Possible duplicate of How does a lower pH affect the alcohol concentration of wine?

This is a slightly broad question, but ill try to assist some. Google has a large assortment of articles describing pH, Le Chatelier's Principle, and EtOH.
Im not sure if you mean altering the pH during the fermentation process or altering the pH of the final product? It also makes a difference on the types of yeast you use. Some yeasts can handle different conditions and produce a different amount of alcohol.
Here is a great response from last year on the Stack Exchange, regarding fermentation without the addition of commercial yeast, using the change in pH as a catalyst for fermentation.
With the production of wine that uses commercial yeast for fermentation (which is most commercial wines), altering the pH too far outside of it's "designated" range will cause it to produce less alcohol. Some connoisseur winemakers will alter the pH slightly sometimes, varying on the types of grapes, to alter the terpene profile present. Different variables, equals different systems of equilibrium.

For my chemistry EEI I decreased the pH of my varied fruit wine just before fermentation. I have quite a lot of information about Le chatterer's principle but am still unsure of how it causes the alcohol concentration to decrease if pH is increased?


text:
Why does K2ZnO2 + H2SO4 yield K2SO4 + ZnSO4?
$\ce{K2ZnO2 + 2H2SO4->ZnSO4 + K2SO4 + 2H2O}$
How does the ZnO2 ion disassemble here? Does it perform as a basic oxide, since we know that a basic oxide + an acid --> a salt + H2O?
I'm not sure, because a basic oxide does not carry a charge, and usually sports only a single oxygen atom.
When trying to write down this reaction on my own, I wrote it as
$\ce{K2ZnO2 + 2H2SO4->H2ZnO2 + K2SO4}$
It was a surprise for me when I looked up the answer.

Zinc oxide (or hydroxide) is amphoteric and behaves as such. With strong acids, it acts like a base, and vice versa.

@IvanNeretin - even if it is in the ionic form, and not as your 'average oxide' without charge?

You seem to already know what happens if zinc is in the anionic form. Look again at the reaction you wrote. This is right. Only it does not stop there.

@IvanNeretin - thank you! Maybe in the ionic form the oxide is even more basic. Or maybe there's no difference.

Of course the zincate anion is much more basic than zinc hydroxide. If you add a very weak acid (or a limited amount of a strong acid), you will end up with $\ce{Zn(OH)2}$.


text:
Na2CO3 + FeBr3 + H2O: why do we get Fe(OH)3?
$$\ce{3Na2CO3 + 2FeBr3 + 3H2O -> 6NaBr + 2Fe(OH)3 + 3CO2}$$
Why $\ce{Fe(OH)3}$? Where do these $\ce{-OH}$ groups come from?
This is from a series of reactions in a Russian sample exam question.
P.S.
Following Ivan't hints, I've gone through the process of reaction, and I get some water in the right-hand side of the equation. I wonder why the answer has no water in the right-hand side.

From the reaction of the carbonate with water.

Carbonic acid is a weak acid, $\ce{Fe(OH)3}$ is a weak base, so their salt hydrolyzes completely.

@IvanNeretin - but isn't $\ce{Fe(OH)3}$ a salt? And I've just read that $\ce{Na2CO3}$  should only hydrolyze as far as $\ce{NaHCO3}$ without heating.

You might have to review your definition of salt. As for $\ce{Na2CO3}$, it is a salt of a weak acid and a strong base, therefore it hydrolyzes only partially.

@IvanNeretin - oops, indeed, it's a base. So it is $\ce{Fe2(CO3)3}$ that instantly and completely hydrolizes. Now I see. Thank you!

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@IvanNeretin - I've gone through the calculations, and I get some water in the right-hand side of the equation, because $\ce{H2Co3}$ breaks down into carbon dioxide and water. I wonder why there's no water in the equation provided in the answers section..

The water on the right cancels out with (part of) that on the left.

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

I think it gives iron(III) oxide-hydroxide, sodium bromide and carbon dioxide. see this:-http://chemiday.com/en/reaction/3-1-0-5170

You have certainly already recognized $\ce{Na2CO3}$ as the base in your equation.
Did you also consider the following:
When $\ce{FeBr3}$ is dissolved in water, the hexaaqua complex $\ce{[Fe(H2O)6]^3+}$ is formed. This complex is fairly acidic:
$$\ce{[Fe(H2O)6]^3+ + H2O<=> [Fe(H2O)5OH]^2+ + H3O+}$$
Ignoring ligand exchange processes, you might state that the $\ce{OH}$ groups come from the deprotonation of the water ligands of your initial complex.

I see. $\ce{Na2CO3}$ creates a basic environment, because CO3 binds protons tightly and leaves a lot of OH(-) groups floating about. These groups deprotonate three of the H2O molecules from the $\ce{[Fe(H2O)6]^3+}$ complex, and we get $\ce{Fe(OH)3}$. Now it's much more clear, thank you!

Draw a lewis diagram of the molecules to give you visual clues. From a glance it looks like the halides that are connected to the iron take their electrons leave and then are captured in solution to the sodium molecules. (this wouldnt be such a great bond since its ionic and the sodium and bromine ions would really act more like spectators).
From thee the CO3 attacted to the Na would decompose to CO2 and the extra O would attach itself to the iron and and somewhere before or after that instant those oxygens would strip a hydrogen from the water.


text:
Organic Chemistry Exercise Book
An Organic Chemistry Book that has only exercises? Better with solutions so to see if the answer is correct. Even more better if there is a PDF version.
The exercises I am looking for are exercises about Hydrocarbons and their reactions, Isomerism exercises, haloalkanes and their reactions, alcohol, eters and thiols, ammines, Fisher's projections and the absolute configuration, sugars, lipids, and proteins.
Thanks a lot!

@Martin-マーチン Hello sir! You might want to check it again since it only lists books that have both theory and exercises, and I already use 'Introduction to Organic Chemistry' by Brown for that. I was looking for a book containing exercises only as I expressed in my question.

Well the general problem here is, that book recommendations are somewhat always biased and probably highly opinionated. You might want to be more specific on the topic: organic chemistry is far too broad. You should include examples of what you are looking for specifically.

@Martin-マーチン I am looking for a book that has all organic chemistry exercises, but exercises only, since I use PDF books and it is very unconfortable to scroll between 200 pages to find different kind of exercises because they are related to the topic they cover, so I think my question was broad because it covers a broad bunch of exercises, simply that sir.

Yes, and that is exactly the problem. The question is too broad. Please narrow it down. Otherwise it will probably be closed or it will not gather much attention in the first place and will remain unanswered. Take the [tour] to learn more about which questions are on-topic here, please also check the [help]. You can always add more information to your post by [edit]ing it.

@Martin-マーチン I understand that, but how can I make it another question if this is what I am looking for?

For example, you could narrow it down to retrosynthesis or reaction mechanisms, or a few functional groups like carbonyl chemistry, organo-catalysis, etc. - OC is quite broad. You will also probably have already some experience with the type of question you are looking for, so you can  include a worked example. Focus on a few things first and hope for answers, if that does not help, come back and ask a new question.

Geez, go to your library, ask librarian or check shelves. Borrow book...


text:
Is the equation of an aldehyde to a carboxylic acid in alkaline conditions correct?
The equation is:
$$\ce{RCHO +2OH- -> RCOOH + 2e- + H2O}$$
If it isn't correct, then please explain.

What will happen to a carboxylic acid in the presence of hydroxide?

@bon I was trying to work out the half equation of it and combine it with the half equation of Cu2+ to copper oxide(1).

Ok but what will happen to the acid in the presence of hydroxide?

@bon Become water?

@bon So I started off with RCHO + H2O = RCOOH + 2H+  +  e- .  Then I added hydroxide ions onto both sides and then simplified the water molecules.

Yes but you have still missed something. Hydroxide is a strong base so what will happen when you add it to the carboxylic acid.

@bon  I think I understand now. So you get a carboxylate ion (RCOO-). and , therefore the half-equation would be : RCHO +3OH- = RCOO- +2H2O +2e-. Is that correct, Also, if I wanted to retain the RCOOH, what would I also have to do.

Protonate the carboxylate by flooding the reaction mixture with acid after the oxidation has finished.

What is it supposed to be? Half of Cannizzaro reaction? Or what?

This is almost correct. The only thing you have missed is that hydroxide is a strong base, so when you have a carboxylic acid in a reaction mixture with hydroxide it will be deprotonated. Therefore the overall half equation would be: $$\ce{RCHO + 3OH- -> RCOO- + 2H2O + 2e-}$$
If you want to get the carboyxlic acid you would have to flood the reaction mixture with acid after the oxidation has finished, in order to protonate the carboxylate anion.


text:
Why doesn't iron(III) iodide exist?
$$\ce{Fe2O3 + 6HI -> 2FeI2 + I2 + 3H2O}$$
Why don't we get $\ce{FeI3}$? After all, iron's oxidation state is $+3$ in the reagent.
Should one just memorize that up to bromine, it's $\ce{FeX3}$, and below it's $\ce{FeX2}$?

Because the iodide will reduce $\ce{Fe^3+}$ to $\ce{Fe^2+}$.

@bon - the iodide (-3) ion will give an electron to Fe(+3)? Because their electronegativity difference is low?

There is no $\ce{I^{3-}}$ ion.

@bon - I see. So the I(-1) ion will give it's electron to Fe(3+), even though I(-1) is more electronegative.

Look at the redox potentials for iodide, bromide and chloride here.

To quote wikipedia "Ferric iodide, a black solid, is not stable in ordinary conditions, but can be prepared through the reaction of iron pentacarbonyl with iodine and carbon monoxide in the presence of hexane and light at the temperature of −20 °C, with oxygen and water excluded"  Housecroft and Sharpe or Greenwood and Earnshaw (don't have them to hand) syas similar. So it does exist, but is fairly unstable, both thermodynamically (as covered above) AND kinetically.

What's the color of FeI2? Wikipedia says it's whitish, while other site (don't remember the name) says it's dark...

@Zenix Varied chemist had noted varied colors of iron(II) iodide. To quote from this paper: "The color of ferrous iodide, therefore, is not by any means settled, the balance of evidence in the chemical literature being in favor of grey or white"

Aww man, "this paper" is from 1899! We have not solved the color of ferrous iodide in the 120+ years since?

The standard reduction potentials for the following half reactions can be found here.
$$
\begin{align}
\ce{Fe^3+(aq) + e- &-> Fe^2+(aq)} &\quad E^\circ &= \pu{+0.77 V} \\
\ce{I2(s) + 2 e- &-> 2 I-(aq)} &\quad E^\circ &= \pu{+0.54V} \\
\ce{Br2(l) + 2 e- &-> 2 Br-(aq)} &\quad E^\circ &= \pu{+1.07V} \\
\ce{Cl2(g) + 2 e- &-> 2 Cl-(aq)} &\quad E^\circ &= \pu{+1.36V}
\end{align}
$$
You can see from this that only iodide is a strong enough reducing agent to reduce $\ce{Fe^3+}$ to $\ce{Fe^2+}$ at standard conditions. Even with non-standard concentrations it will be very difficult to get bromide to do the reduction because the difference in electrode potential is large.
The trend in electrode potentials for the halogens can be explained in terms of the increasing electronegativity going from iodine to chlorine which increases the first electron affinity. It just so happens that the crossover point with the iron reduction is between iodine and bromine.

As it is pointed out in other answers and in literature, it is indeed thermodynamically unstable and its reaction synthesis has unfavorable pathways. But, is it "non-existent"? Not quite. There was a possibility of its existence indicated by the fact that hydrated ferric oxide dissolves in hydriodic acid, yielding a brown solution. Its thermodynamic limitation was known in aqueous medium, so its synthesis was thought to be carried in non-aqueous medium. It was the year 1989 that the synthesis was somehow achieved. Diiodotetracarbonyl(II) iron was photochemically reacted with iodine in n-hexane medium to yield the product:
$$\ce{2(OC)4FeI2 + I2 ->[h\nu][C6H14] 2FeI3 + 8CO  }$$
However, there were some limitation to this reaction:

The photochemical conversion was limited due to photochemically induced decomposition:

$$\ce{2FeI3 ->[h\nu] 2FeI2 + I2}$$

It is metastable and hence attainment of pure iron(III) iodide was quite difficult
The iodine concentration was needed to be precise. More iodide led to formation of complex:

$$\ce{FeI3 + I- -> FeI4-}$$
Ref.: Ferric iodide as a nonexistent compound, K. B. Yoon and J. K. Kochi
Inorganic Chemistry 1990 29 (4), 869-874, DOI: 10.1021/ic00329a058

One idea: It may not he iron(III) iodide as such that is unstable, but the combination of the separate ions which are prone to react in the usual way. The difference is that when iron is bonded with iodine there is a lot of covalent character, which in my answer shows up as the molecule becoming a soft acid. Were we to prevent the ions from forming and keep away hard bases (like water or THF) that would upset the iron-iodine bonds, we might better sustain $\ce{FeI3}$.

Iron(III) iodide as a binary salt is highly unstable/transitory, but stable complexes are known with appropriate ligands.
Pohl et al. [1] first synthesized such a complex, $\ce{FeI3(SC(N(CH3)2)2)}$, actually oxidizing iron(II) iodide with elemental iodine in the presence of a carefully controlled amount of $\ce{((CH3)2N)2CS}$. The iron(III) iodide, despite the high oxidation state of iron, acts as a soft acid, coordinating to sulfur instead of nitrogen (picture from Ref. [1]):

Barnes et al. [2] report that reaction of iron metal with trimethylarsine diiodide affords the complex $\ce{FeI3(As(CH3)3)2}$, a trigonal bipyramidal complex with a structure similar to that of lighter halide complexes (picture from Ref. [2]).

In a variation, Bernstein an Herbstein [3] describe the salt ferrocenium triiodide, $\ce{[Fe(C5H5)2]+I3^-}$, in which iron(II) incorporated into ferrocene reduces elemental iodine and the iodide ion then complexes with additional iodine. Thereby iron(III) is combined with the triiodide ion. The authors even indicate that the triiodide ion is symmetrical, a condition alkali metal triiodides fo not achieve.
These reactions, together with the oxidation of the iron(II) iodide-carbonyl complex reported in Nilay Ghosh's answer, share some common characteristics:

The iodine comes from nonionic sources, even where the reactant has it in the -1 oxidation state as in Ref. [2].

The environment avoids water and other hard bases, which would presumably bind to the iron (III) and displace iodide ions. The iron(III) iodide, which has significant covalent character and would be a softer acid than ionic $\ce{Fe^{3+}}$, is instead bound to soft bases.

These features suggest that $\ce{FeI3}$, as such, is not all that unstable; rather it is the combination of $\ce{Fe^{3+}}$ and $\ce{I^-}$ ions that doesn't hold up. If the iron(III) iodide were to be formed in a process that avoids the use or generation of the separate ions, it would more likely be sustained.
Compare these results with cerium(IV) chloride.
References

Siegfried Pohl, Ulrich Bierbach, Wolfgang Saak; "FeI3SC(NMe2)2, a Neutral Thiourea Complex of Iron(III) Iodide", Angewandte Chemie International Edition in English (1989) 28 (6), 776-777. https://doi.org/10.1002/anie.198907761

Nicholas A. Barnes, Stephen M.Godfrey, Nicholas Ho, Charles A.McAuliffe, Robin G.Pritchard; "Facile synthesis of a rare example of an iron(III) iodide complex, [FeI3(AsMe3)2], from the reaction of Me3AsI2 with unactivated iron powder", Polyhedron (2013) 55, 67-72. https://doi.org/10.1016/j.poly.2013.02.066

Tsur Bernstein and F. Herbstein (1968). "The crystal structure of ferricinium triiodide, (C5H5)2FeI3". Acta Crystallographica Section B Structural Crystallography and Crystal Chemistry. 24. 1640-1645. 10.1107/S0567740868004784.

A side note: same holds with cobalt(III) and chloride; cobalt(III) chloride does not exist per se, but hexaamminecobalt(III) chloride does (and is stable enough to be sold commercially).

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text:
Is there are any universal methods or software for reaction rate modelling?
In software like DWSIM are used databanks which contains enthalpy and entropy at 298 K and more other information about chemicals. This software can create a steady-state system, where showed state of matter, temperature of reaction, etc.
But I can't find any software for calculation of reaction rate. I know, that chemical kinetics is using experimentally measured time of reaction and concentration of matter. So what should we do, if we want to know an estimated time of reaction?

Like Aspen etc? https://en.wikipedia.org/wiki/List_of_chemical_process_simulators

@Mithoron, Aspren looks like software for thermodynamical modelling. Or may be you did mean that there is available to find time of reaction? I watched tutorial on Aspren. Looks like it's same as DWSIM.

@Mithoron, yes! Aspren is one which supports kinetic calculations. Thanks.
Yeah, and may be you know any freeware alternatives to it?

Most modern process simulators, including DWSIM, feature reactor models which can calculate and display reaction rates. For PFR and CSTR models you must also consider the reactor volume to estimate the reaction time.


text:
How do we get NaClO3 in the reaction between Cl2 and NaOH?
From problem no. 4080 in a Russian example exam set:
$$\ce{6NaOH + 3Cl2 -> NaClO3 + 5NaCl + 3H2O}$$
How do we get that $\ce{NaClO3}$? I can't seem to guess. I found this explanation on YouTube, but there the products are different:
$$\ce{Cl2 + 2NaOH -> NaCl + H2O + NaClO}$$
In both reactions, there is twice as much sodium chloride as there is chlorine. How do the two additional oxygens connect to the chlorine?
P.S. A related question - thanks, Nilay Gosh.

Why, this is a typical disproportionation reaction. Many non-metals would do something of this sort. Related: http://chemistry.stackexchange.com/questions/41182/products-of-electrolysis-of-nacl/41190#41190

@IvanNeretin - thank you! What I don't get is how we get a whole of 3 oxygens there, on the chlorine. I'll reread the basics about electrolysis and will try to wrap my head around this.

This is not related to electrolysis (chlorine that goes into the reaction may or may not be produced by electrolysis; that's not important). As for the particular oxidation state of chlorine, that's something to be memorized rather than deduced.

also see this:- http://chemistry.stackexchange.com/questions/50333/reaction-of-chlorine-with-hot-cold-solution-of-hydroxide

If you substitute the spectator ion, the answer in the question Nilay linked is pretty much the solution to your problem. It depends on the reaction conditions.

I find that when I'm faced when problems having to do with why the products are formed I make lewis structures.
if you make a lewis diagram of the $$\ce{NaClO3}$$
First off, you will see that a single Cl will be bonded to three oxygen molecules whom all carry full valance shells which were striped from the hydroxide ions in the reactants. (Remember that in solution the Na and OH will be around as free spectators.)
From the structure you will see the reason why there was a formation of the three waters. Since there were three oxygen stripped from the hydroxides and then recombination with more OH to make a reaction like this:
$$\ce{OH- + H+ -> H2O}$$
Also notice that on the products side for every molecule of NaClO3 made there will be a molecule for NaCl made. From there you can speculate that when the Cl-Cl bond breaks three oxygen atoms and a sodium atom will surround one of the Cl while another is tied to a Na molecule simultaneously.

Thank you for the answer, and welcome to the SE! There, however, actually will be $\ce{5 NaCl}$ molecules per one sodium chlrorate

There are quite a few issues with charge balance here. In none of the examples hydroxide will act as a spectator.


text:
What is the correct way to verify a structure's geometry, for example for benzene?
My goal is to compare a calculated bond length with experimental data. Benzene is obviously common enough to expect sufficient experimental data to be available for it on the internet.
I have found two sources for the C-C bond length in benzene, and the numbers differ very slightly. This source says 140 pm for benzene, while this source says 139 pm for benzene. Both claim their number comes from x-ray diffraction.
Is the discrepancy possibly attributable to a difference in solvent? Is a difference of 1 pm even significant? What is a good way to know for sure which quantity is correct?

I would worry when confronted with a 10 pm difference. 1 pm may pretty well be just a rounding error.

It may very well also be within the error margin of the experimental setup.

Most likely 139pm and 140pm are the same given the experimental uncertainty of the technique. If your sources are good they will quote an error bar around their estimates and you could check this directly.

Notice calculated bond length is typically in gas phase at 0 K without vibration averaging ($r_e$),  x-ray diffraction may be for crystal with finite temperature and vibrational effects.

I understand. So I won't quibble over the 1 pm then, sounds good. I'd star an answer for helping but can't seem to do that in comments? Thanks guys!

I must point that in case you need a reference point for calculation involving isolated molecule without solvation model, you should go for data on gas phase, which implays gas difraction, like, for example, in this work http://www.sciencedirect.com/science/article/pii/S0022286006006958

A deviation within the low picometres is nothing to worry about, there are many reasons for this.
Primary literature, like peer-reviewed journals, will always publish an analysis of the obtained values along with any error bars, i.e. how accurate the measurement is. When you have disagreeing values it is worthwhile to have a look at these, too.
Secondary literature, like encyclopaedias, will usually only quote values and might not include all details due to their general reference nature.
Quite often if you have different values, the conditions of the measurement simply were not the same. Also keep in mind, that within the recent years a lot of progress has been made for spectroscopic measurements. Hence you can find values from older publications that are significantly different from more modern values.
For example Tamagawa et al. reported $r_g(\ce{C-C}) = 139.9 \pm 0.1~\mathrm{pm}$.[1] They used a method, which combines diffraction and spectroscopic data. This article was published in 1976 and it was introduced with the following paragraph. I think that pretty much summarises the predicament of modern structural analysis.

It is often found that the structural data of the most basic molecules are not very up-to-date because such molecules were studied in the early stages of the development of experimental methods for structure determination. Benzene is one such molecule.

And since then many more studies on benzene have probably been published. Chemspider.com lists (currently) seven data sets[2] and I am confident this is just the tip of the iceberg.
In a quite recent publication by Budzianowski and Katrusiak the crystal structure of benzene has been studied at various pressures. They pressure froze the molecule(s) in situ in phase I.[3]
We learn from this, that there is more than one phase in which benzene crystallises. And a little more digging leads us to a publication by Piermarini et al. dealing with the crystal structure of benzene II at 25 kbar.[4] But I will not go into that and rather focus on the former publication.
This paper illustrates nicely that there cannot be only one true value for the compound. I will reproduce a part of table 2 from the publication here.
$$
\text{Bond lengths (}\mathring{\mathrm{A}}\text{) and angles (}^\circ\text{) for benzene in phase I.}\\
\begin{array}{llll}
             & 0.3~\mathrm{GPa} & 0.7~\mathrm{GPa} & 1.1~\mathrm{GPa} \\\hline
\mathrm{C1—C2}     & 1.40 (4)   & 1.397 (8)        & 1.420 (16)       \\
\mathrm{C2—C3}     & 1.38 (8)   & 1.384 (10)       & 1.41 (3)         \\
\mathrm{C3—C1^i}   & 1.38 (5)   & 1.354 (10)       & 1.32 (2)         \\\hline
\end{array}\\
\text{Symmetry code: (i)} 1-x; -y; -z.
$$
As you can see even in a single crystal structures, the bond lengths are not equal. This is obviously due to packing effects and intermolecular interactions. When we talk about a single value for the carbon carbon bond length in benzene, then we assume an idealised, even rigid, molecule.
And there are more ways to obtain the geometry of a molecule: IR, microwave, Raman to name a few.[5] Nowadays we can employ computational chemistry and try to predict bond lengths, but there is a catch to that, too. These calculations take place at absolute zero and no pressure and we have a couple of other approximations in the mix. You can have a look at the Computational Chemistry Comparison and Benchmark DataBase, which lists values from 138.3 to 144.3 pm for the carbon carbon bond.[6] For example, my own calculation on the MP2/def2-SVP level of theory gives me a carbon carbon bond length of 140.0 pm. But that is a value that corresponds to a stationary point on the potential energy surface. We all know that this molecule has thirty vibrational modes, which are not yet accounted for. Below is just one, one that I find particularly funny.

TL;DR Don't worry about deviations of about 1 pm. When there are larger differences, look at the original publications about the conditions and trustworthiness of their measurements. Always consider the error margins on your own experiments when you compare values. It also depends on how accurate you need your value to be before you go and dig deep into publications from 50 years ago. (For what it is worth, you could round 139 pm to 140 pm.)
Notes and References

Tamagawa, K.; Iijima, T.; Kimura, M. Molecular structure of benzene. J. Mol. Struct. 1976, 30 (2), 243–253. DOI: 10.1016/0022-2860(76)87003-2.
CSID:236, http://www.chemspider.com/Chemical-Structure.236.html (accessed November 18, 2018)
Budzianowski, A.; Katrusiak, A. Pressure-frozen benzene I revisited. Acta Crystallogr Sect B 2006, 62 (1), 94–101. DOI: 10.1107/S010876810503747X.
Piermarini, G. J.; Mighell, A. D.; Weir, C. E.; Block, S. Crystal Structure of Benzene II at 25 Kilobars. Science 1969, 165 (3899), 1250–1255. DOI: 10.1126/science.165.3899.1250.
Wikipedia has a longer list: Molecular Geometry.
Unfortunately I am unable to generate a permalink from the database: http://cccbdb.nist.gov/ I welcome you to play around a little by yourself.
However, I'll include the summary from the geometry comparison between the C1–C2 bond length:

distance is atom 1 (C) to atom 2 (C)
  Experimental bond length is 1.397 Å     Calculated bond lengths range from 1.383 to 1.443 Å
  The bar charts show differences (theory - experiment) and the range is −0.046 to 0.046 Å

This is just a logical explanation but this might or might not be wrong.
Due do resonance in benzene if we look at canonical forms, at a single moment there are alternate single and double carbon carbon bonds..due to rapid interchange between these positions the average(or experimentally calculated) value of carbon-carbon bond in benzene  will oscillate between $\mbox{139}$–${\mbox{140 pm}}$.

This is wrong. Take a look at the answer I just posted to address this misconception.

also http://chemistry.stackexchange.com/questions/50005/does-benzene-structure-stand-for-a-single-resonance-form-or-the-whole-molecule

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text:
What is resonance, and are resonance structures real?
My teacher told me about resonance and explained it as different structures which are flipping back and forth and that we only observe a sort of average structure. How does this work? Why do the different structures not exist on their own?

related Does benzene structure stand for a single resonance form or the whole molecule?

You could get more information from https://en.wikipedia.org/wiki/Resonance_(chemistry).

This answer is intended to clear up some misconceptions about resonance which have come up many times on this site.
Resonance is a part of valence bond theory which is used to describe delocalised electron systems in terms of contributing structures, each only involving 2-centre-2-electron bonds. It is a concept that is very often taught badly and misinterpreted by students. The usual explanation is that it is as if the molecule is flipping back and forth between different structures very rapidly and that what is observed is an average of these structures. This is wrong!  (There are molecules that do this (e.g bullvalene), but the rapidly interconverting structures are not called resonance forms or resonance structures.)
Individual resonance structures do not exist on their own. They are not in some sort of rapid equilibrium. There is only a single structure for a molecule such as benzene, which can be described by resonance. The difference between an equilibrium situation and a resonance situation can be seen on a potential energy diagram.

This diagram shows two possible structures of the 2-norbornyl cation. Structure (a) shows the single delocalised structure, described by resonance whereas structures (b) show the equilibrium option, with the delocalised structure (a) as a transition state. The key point is that resonance hybrids are a single potential energy minimum, whereas equilibrating structures are two energy minima separated by a barrier. In 2013 an X-ray diffraction structure was finally obtained and the correct structure was shown to be (a).
Resonance describes delocalised bonding in terms of contributing structures that give some of their character to the single overall structure. These structures do not have to be equally weighted in their contribution. For example, amides can be described by the following resonance structures:

The left structure is the major contributor but the right structure also contributes and so the structure of an amide has some double bond character in the C-N bond (ie. the bond order is >1) and less double bond character in the C-O bond (bond order <2).
The alternative to valence bond theory and the resonance description of molecules is molecular orbital theory. This explains delocalised bonding as electrons occupying molecular orbitals which extend over more than two atoms.

Since it came up, here is a nice question about fluoro bullvalene. It should also be noted, that MOT and VBT are congruent theories, at their theoretical limit (exact treatment) they become identical.

@Martin-マーチン What if we measure the position of an electron in a molecule? There will be some uncertainty, but what if we are able to find some position(small finite volume of space in some part of the molecule) of the electron, won't it suggest that the "electron is whizzing around the molecule"? (I've not yet studied quantum mechanics; I'm ready to delete this comment if it's rubbish)

@ApoorvPotnis That is not possible. Electrons behave like particles and waves, it is that duality, among others, that make your suggestion impossible.

@Martin-マーチン But electrons also behave like particles, don't they? And can't a particle be only in one position when observed? 'Position' meaning the text in brackets in the previous comment.

@ApoorvPotnis Yes, let's say you measure the position of the electron. Now what? Resonance forms do not exist; that does not mean that electrons in the resonance hybrid just "sit" happily at one point in space forever. An electron always has a wavefunction and as such you can theoretically always measure a position or momentum. That is not relevant to the question, which is whether the system is described by 1 wavefunction (i.e. figure (a), a resonance hybrid) or ≥2 wavefunctions (i.e. figure (b) above, rapid interconversion, not resonance) ...

... You seem to think that an electron will have zero momentum in the first case and nonzero momentum in the other, which is not true.

@orthocresol I think that I should wait until I learn quantum mechanics fully. Only then, I'll have a clear understanding.

Bon has a very nice example of motion potentially restricted because of a barrier, and although it turns out that this is not the case for the norbornyl cation, there are simpler examples that do show inversion through a barrier, such as ammonia inversion or cyclopentene ring puckering. The two structures either side of the barrier would be not be called resonances any more that one would cis or trans isomers, where the barrier is substantial. The time to move back and forth across the barrier can be measured spectroscopically; in the case of $\ce{NH3}$ inversion this is only a few picoseconds.
The theoretical idea of resonance is only necessary to perform an accurate calculation in the valence bond method. In benzene, for example, the three double bonds can be drawn in different ways by rotating them around the ring. But there are other 'resonance' structures that could be added, for example a biradical structure such as the following:

and so on.  The more of these we add the better the calculation should become. So the resonance is just a way of figuring out what structure to add into our calculation. We could end up with one electron on each carbon, or +/- charges here and there etc. All very high energy structures no doubt, but it may still be necessary for them to be included in a calculation although their contribution may be small. However, there is no suggestion of any conversion between them; they merely form a 'basis set' from which to perform the calculation.

Is biradical structure a valid structure ?. cause there are lots of university websites stating that no of unpaired electron will not change in a valid resonance structure.

Resonating structures are totally imaginary structures as they are only used to explain the properties of a single molecule. There are no concrete experiments to prove the real existence of resonating structures. But the resonance hybrid has real existence.

This doesn't really add anything to what has already been said.

Nevertheless, it answers the main question, so no flagging or deleting.

This is not correct. First of all the concept is called resonance and the contributors are resonance structures; it doesn't have anything to do with resonating. Secondly, these structures are as real as a resonance hybrid. The viewpoint from VBT favors resonance, while MOT favors the hybrid (or delocalised structure). Both descriptions are correct and within the same framework at their theoretical limit identical.


text:
How does a fire start?
I know that fire in a few words is the exothermic reaction of a fuel with an oxidizing agent, but I can't fully understand what exactly happens to piece of wood when it is ignited. How do molecules start producing a flame? In other words, what is the chemistry behind the production of flame?

Fire is about physics rather than chemistry. Sure, there is a chemical dimension to it, but it is quite trivial: $\ce{C + O2->CO2}$ (that is, if we burn coal; wood is somewhat more complicated, but whatever).

related http://chemistry.stackexchange.com/questions/1254/what-are-the-chemical-reactions-behind-fire does not describe start of fire unfortunately

http://chemistry.stackexchange.com/questions/44526/why-does-shaking-a-match-put-the-fire-out is useful here

and http://chemistry.stackexchange.com/questions/6954/why-does-a-wood-fire-create-benzene

This is a classic: https://www.youtube.com/watch?v=ITpDrdtGAmo (Always copy Feynman's answer)

https://www.youtube.com/watch?v=B0E4PX3e3RE It feels somewhat weird to answer my own question but I think this video describes exactly what I wanted. As it supports, when heat is applied to a piece of wood, some bonds of the molecules that make up wood, break and thus different compounds are formed. These compounds are not held back by some force and so they are released in the air. When these compounds meet atmospheric oxygen, under heat (=energy), they burn and thus more heat is released along with carbon dioxide and water. That stage can be described as ignition. Finally, this produced heat is able to preserve the fire.

Answering your own question is good. There is nothing wrong with it on Stack Exchange.

The only bit you are missing is that, for a fire to start, the reaction has to be self-sustaining when the applied heat/flame is removed. In other words the energy released from the initial reaction has to be large enough that the reaction continues to happen when the initial heat source is removed.


text:
Lowering energy in molecular orbital mixing
When drawing molecular orbital diagrams for mixing two orbitals we always end up with one orbital which is lower in energy than the initial orbitals and another which is higher in energy. The principle extends to mixing of more than two orbitals with the total number of orbitals always being conserved.
Often we will use arguments involving mixing of molecular orbitals to form MOs of a slightly lower energy to explain phenomena such as hyperconjugation with diagrams such as this.

In this situation, when you have one filled and one empty orbital, the electrons always end up in a lower energy than they started.
What stops you from mixing these orbitals again to produce even lower energy orbitals, even if the energy difference is very small? I understand that symmetry might have something to do with it but I am not well versed in the more mathematical elements of MO theory such as symmetry groups. Or is my initial assumption that mixing two orbitals always results in a lower energy wrong?

You start with two functions $f_1$ and $f_2$, and you are trying to mix them to get new functions $g_1 = c_{11}f_1 + c_{12}f_2$ and $g_2 = c_{21}f_1 + c_{22}f_2$, in order to minimise the energy of $g_1$. The problem is that there are constraints on the constants $c_{ij}$, namely $c_{i1}^2 + c_{i2}^2 = 1$ and $c_{1i}^2 + c_{2i}^2 = 1$ ($i = 1,2)$. These mean that there is a fixed value for which the energy is minimised and this is the closest approximation to the ground state energy that you can get (by the variational principle) using the basis set that you have

In less mathematical terms, one could look at it as the strength of mixing decreasing with the size of the energy gap between the two orbitals. The larger the energy gap, the weaker the mixing. This means that the total mixing you can get is asymptotic and will eventually reach an end (the most stable configuration). Also, symmetry has nothing to do here, symmetry only dictates which wavefunctions can mix with which but not how much they mix.

I object. Symmetry has very much to do with this. Being different eigenfunctions of the same problem, our resulting functions are orthogonal. They can't interact (that is, "mix").

@Ivan That is true but I don't think that was what was meant by symmetry, symmetry was referring to the irreducible representations under which the orbitals transform.

@orthocresol Well, take two H atoms so they produce $\sigma$ and $\sigma^*$ orbitals, and voila, these two belong to different irreducible representations, so (in a way) they are orthogonal "because of symmetry".

@Ivan take LiH and you get two orbitals which are both $\sigma$ symmetry. What you said is of course right, you get two eigenstates which are orthogonal and that is the "point where the mixing stops", but not because of them being in different irreps

Well, yeah, come to think of it, you are right. I guess the cases where symmetry actually kicks in (like in my example) are a relative minority.

I think one or both of you should write answers ;)

The orbital $\psi$ is obtained as a mixture of the orbitals {$\phi_i$}
$$\psi= c_1 \phi_1 + c_2 \phi_2$$
choosing the {$c_i$} through energy minimization.
Once obtained $\psi$, if you try to mix it with the initial {$\phi_i$} to obtain a new function, lets say $\mu$, you'll have:
$$\mu = c_3 \psi + c_4 \phi_1 + c_5 \phi_2$$
(notice that the coefficients don't need to be the same)
or equally
$$\mu = c_3 (c_1 \phi_1 + c_2 \phi_2) + c_4 \phi_1 + c_5 \phi_2$$
$$\mu = c_3 c_1 \phi_1 + c_3c_2 \phi_2 + c_4 \phi_1 + c_5 \phi_2$$
$$\mu = (c_3 c_1 + c_4) \phi_1 + (c_3c_2+c_5) \phi_2$$
that has the same functional form that
$$\psi= c_1 \phi_1 + c_2 \phi_2$$
so as the criteria is the energy minimization it should be no surprising that we can not get better energy that making
$(c_3 c_1 + c_4) = c_1$ and $(c_3c_2+c_5)=c_2$.
The idea behind this is that you want to obtain the "better" function $\psi$, but you are restricted to a set of functions of the form: $c_1 \phi_1 + c_2 \phi_2$.  You impose this restriction because it only requires to obtain two coefficients, that is far simpler than obtaining an arbitrary function (defined with infinite images).
One restriction is that the equation $c_i \phi_i = c_k \phi_k$  can only be true if $c_i = c_k = 0$.
You can expand your search to a larger set of functions, for example: $c_1 \phi_1 + c_2 \phi_2 + c_3 \phi_3$, but it requires that the set still be linearly independent, that is, you can not get any of the functions with a linear combination of the others, (i.e. it is not possible to get $c_1 \phi_1  =  c_2 \phi_2 + c_3 \phi_3$ if at least one $c_i$ is different from 0). If not, as in the discussion above, the space of function will contain the same functions so the inclusion of the new function is in some sense redundant or useful.
You can add infinite linearly independent functions, but, it does not mean that the energy will be always get lower.


text:
Balancing the redox disproportionation of chlorine by half-reactions
I am having some difficulty balancing the following reaction using the half-reaction method:
$$\ce{Cl2(g) -> Cl-(aq) + ClO3-(aq)}$$
My attempt at the problem:
By determining oxidation numbers it is possible to see what is being oxidised and what is being reduced:
$$\ce{\overset{0}{Cl}_2(g) -> \overset{-1}{Cl}^-(aq) + \overset{+5}{Cl}\overset{-2}{O}_3^-(aq)}$$
From this I determined the following two half-reactions:
Reduction Half: $\ce{Cl2(g) + 2 e- -> 2 Cl-(aq)}$
Oxidation Half: $\ce{Cl2(g) + 3 H2O (l) -> 2 ClO3-(aq) + 6 H+ + 4 e-}$

Net ionic equation: $\ce{6 Cl2(g) + 6 H2O(l) -> 8 Cl-(aq) + 4 ClO3-(aq) + 12 H+}$
The answer to this problem is however given as

$$\ce{2 Cl2(g) + 6 H2O(l) -> 2 ClO3-(aq) + 12 H+ + 10 Cl-(aq)}$$

I have tried this problem several times over and cannot work out how this answer was reached.

Update:
Silly mistake, oxidation reaction should be: $\ce{Cl2 + 6H2O -> 2 ClO3- + 12 H+ + 10 e-}$ hence solving the mystery of why my answer is wrong.

It is an arithmetical problem, watch out over the mathematical operations you did.

Your oxidation half-reaction (and, consequently, your tentative answer) are not balanced in oxygen. You should have known something's amiss when you saw that $4e^-$ part. We oxidize from 0 to +5, that is, by 5 steps; the number of electrons should be a multiple of 5.

You have an error half-way through your oxidation half-reaction which probably multiplies itself. You correctly determined the oxidation states so it should be simple to spot it. But for reference, here they are:
$$\begin{align}&\overset{\pm 0}{\ce{Cl2}}/\overset{\mathrm{+V}}{\ce{Cl}}\ce{O3-}\\[0.9em]
\ce{Cl2 \phantom{\ce{{} + 6 H2O }} &-> 2 ClO3- + 10 e-}\tag{Ox1}\\
\ce{Cl2 \phantom{\ce{{} + 6 H2O }} &-> 2 ClO3- + 10 e- + 12 H+}\tag{Ox2}\\
\ce{Cl2 + 6 H2O &-> 2 ClO3- + 10 e- + 12 H+}\tag{Ox3}
\end{align}$$
Thenceforth it should be clear to see that you need five reductions per oxidation, thus the redox equation is:
$$\begin{align}\ce{6 Cl2 + 6 H2O &-> 2 ClO3 + 10 Cl- + 12 H+}\tag{Redox1}\\
\ce{3 Cl2 + 3 H2O &-> ClO3 + 5 Cl- + 6 H+}\tag{Redox2}\end{align}$$
(The second version is the first but divided by two; you only have even coefficients.)


text:
Role of acetate ion in formation of hydrogen sulfide ion from thioacetamide
We are using TAA to form metal sulfides. What is the effect of sodium acetate on sulfide ion concentration?
The procedures instruct me to add sodium acetate and to heat the test tube with the metal ion if no precipitate forms after the initial addition of TAA.
From the equilibrium given below, it seems that addition of sodium acetate would actually reduce sulfide ion concentration by driving the equilibrium to the left! That would be antithetical to sulfide salt formation.
However, this cannot be the case; the instructions clearly tell me to add sodium acetate if no precipitate has formed; the instructions are clearly suggesting that sodium acetate helps a precipitate form.
Thinking about it some more, could it be that the addition of sodium acetate basicifies the system by forming hydroxide ion, thereby driving the equilibirum to the right? This seems to make sense, because the addition of sodium acetate is to be accompanied by heating, and wouldn't heating drive the formation of acetic acid and hydroxide ion from acetate ion? And acetic acid is rather volatile, so it'll just leave the solution and not interfere by making the solution too acidic. Am I correct in my thinking? How can I predict the effect of heat on an acid-base reaction? I know heating ammonium chloride causes the ammonium ion to lose its proton. Will heating acetate ion in the presence of water cause it to gain a proton?

You're simply conducting basic hydrolysis, with higher temp. accelerating that.

You are correct.
Adding sodium acetate to an acidic solution forms a buffer and thus decreases the acidity of the solution without making it basic.
$$\ce{CH3COO- + H3O+ <=> CH3COOH + H2O}$$
This helps to precipitate sulfides $\ce{S^{2-}}$ without forming basic precipitates like hydroxides that might disturb the analysis.
$$\ce{S^{2-} + H3O+ <=> HS- + H2O}$$
$$\ce{HS- + H3O+ <=> H2S + H2O}$$
Heating the test tube just speeds up the hydrolysis of thioacetamide. To keep $\ce{H2S}$ in the system this is sometimes done in a hermetically sealed glass tube.


text:
Which single organic molecule has strong absorbance throughout the visible range?
Is there a single organic molecule that has strong absorbance throughout the visible range? For example a black dye that is based on a single molecule?

Black dye is normally based on polymeric conjugated systems.

pentacene is a bluish-black powder, suggesting that it does have some adsorbance throughout the visible range... though primarily in red part.

Does graphite count as organic?

@bon doesn't graphite's absorption depend on the van der waals gap of the graphene sheets?

Maybe - I don't know much about graphite's absorption properties but it seems pretty black to me.

related http://chemistry.stackexchange.com/questions/23608/is-it-possible-to-make-broad-spectrum-light-absorbing-paint http://chemistry.stackexchange.com/questions/34007/a-compound-that-absorbs-all-visible-light

Quite a few dyestuffs will have strong absorption across the visible range of wavelengths, though few will have an entirely even absorbance. This won't matter much is they are concentrated enough and absorb most of the light at each wavelength.
Acid Black 1, for example, shows strong absorbance from the UV to just under 700 nm, which is most of the visible. Absorbance is stronger in the red but overall absorbance is large enough that it is sometimes used in black inks. See this paper for a sample absorbance spectrum:

There are quite a few other dyes with strong absorbance at most visible wavelengths and whether they appear black depends on the concentration.

Short answer: no, not yet
This question has been the focus of intense research for dye-sensitized solar cells and other organic conjugated solar cells. Such a dye would absorb a high fraction of photons and likely improve device efficiency.
Some metal complexes have been found as strong absorbers for the DSSC, starting with the so-called "black dye" ($\ce{[Ru(4,4',4''terpy-(COOH)3)(NCS)3]}$) ~1999.
Suffice to say that I have not seen anything to match the ideal "equal black" profile you mention that's also an organic molecule.
As mentioned in the other answer, there are many organic dyes that are strong absorbers and in high concentration will appear black. This is one reason that "black" pens or markers or inks will become purple or blue or green or mixtures of colors under chromatography.


text:
Can you have a meta stereocenter?
Imagine a compound such as 1,3-dibromo-1,2,3-trichloropropane. Since the first and third carbons are attached to four different groups, they are both chiral centers. Assuming that both of these stereocenters are mirror images of one another (as in (1$S$-3$R$)-1,3-dibromo-1,2,3-trichloropropane), would the second carbon also be considered a stereocenter? If so, how would you differentiate the two enantiomers around that carbon using IUPAC nomenclature? Also, would such a carbon be optically active?

Related: IUPAC name for 1,2,3-trichlorocyclopropane?

Thanks! That addresses the nomenclature part, but would such a compound be optically active, as the two substituents on either side of the central carbon are different conformations despite being a meso compound?

It's also addressed in the question: "Note that the stereodescriptors “r” and “s” describing a pseudoasymmetric stereogenic unit are invariant on reflection in a mirror" Hence the mirror image is superimposable on itself and the molecule is optically inactive. For example if you look at the trichloropropane example given, the molecule clearly possesses a plane of symmetry

Yes, carbon-2 is a stereocenter, and by (arbitrarily) giving the R conformation higher priority than the S, we can assign its stereochemistry using s/r notation:
$\hspace{2cm}$
Pseudochirality simply means that a molecule has two substituents of a stereocenter are mirror images of each other, and the other substituents are either achiral or also mirror images of each other. Despite having multiple stereocenters, pseudochiral molecules are ahciral (optically inactive) because the mirror images of both the s and r conformations are identical.

Pseudo-pseudo-chirality should also exist, right? If so, how do you designate it?


text:
Sodium percarbonate and sulfuric acid reaction
If an excess sulfuric acid is added to a detergent containing sodium percarbonate, would it be possible for the excess acid to react with hydrogen peroxide?
I believe that the hydrogen peroxide will be released from the reaction with sulfuric acid given by this reaction
$$\ce{2Na2CO3·3H2O2 + 3H2SO4 -> 3H2O2 + 2NaSO4 + 2CO2 + 2H2O}$$
So if an excess acid is added will it react with hydrogen peroxide to yield peroxymonosulfuric acid?


text:
How to separate and analyse a sample of cations of alkaline and earth alkaline cations and ammonium?
I am teaching a section of qualitative analysis and I am trying to get a better grasp on the assignments the students are doing. These are the cations they may be presented with and must identify:

$\ce{Na+}$
$\ce{K+ }$
$\ce{NH4+}$
$\ce{Ca^2+ }$
$\ce{Al^3+ }$
$\ce{Mg^2+}$

Now, this lab hasn't changed for the last 20 years, and there is only one way the students perform the separations, which happens to be the same method found in an old lab report that is easily found with Google. The method is:

Add excess 3 M ammonia to the sample.
If a gel-like precipitate forms, then $\ce{Mg^2+ }$ and/or $\ce{Al^3+ }$ ion is present. They have been converted into their respective hydrated hydroxides. Separate the precipitate from the supernatant liquid.
Since hydrated $\ce{Mg^2+  }$ hydroxide is not readily amphoteric, while hydrated $\ce{Al^3+ }$ hydroxide is, addition of excess  $\ce{NaOH}$ should cause the an $\ce{Al^3+ }$ precipitate to dissolve. If the precipitate from the previous step does dissolve with excess $\ce{NaOH}$, then $\ce{Al^3+ }$ ion is present in the original sample. If the precipitate does not dissolve, then $\ce{Mg^2+  }$ ion is present.
Time for the supernatant analysis (if there is a supernatant). If there is no ppt, then we can immediately rule out the presence of $\ce{Mg^2+}$ and $\ce{Al^3+ }$. Either way, the solution, non-precipitate portion can only contain $\ce{Na+}$, $\ce{K+ }$, $\ce{NH4+}$, and/or $\ce{Ca^2+ }$ since none of these form insoluble hydroxides at the concentration of ammonia used. Yes, calcium hydroxide is insoluble, but it doesn't form with 3 M $\ce{H3N}$; I've done the $Q_\mathrm{sp}$ vs. $K_\mathrm{sp}$ calculation.
Add sodium carbonate to the liquid portion. If a solid forms, the solid must be calcium carbonate, and therefore $\ce{Ca^2+ }$ was present in the original sample.
If no solid forms, or even if a solid forms, separate the solid from the liquid and perform further tests on the liquid to determine the presence of $\ce{Na+}$, $\ce{K+ }$, and $\ce{H4N+}$. A yellow flame tests tells us $\ce{Na+}$ is present and $\ce{K+}$ may or may not be present. A purple flame test tells us $\ce{Na+}$ is absent.
Heat the remaining solid sample (not any precipitate, but the original sample itself) to ascertain the presence of $\ce{NH4+}$. Heating a sample that contains $\ce{NH4+}$ will result in the release of white smoke. After heating the solid sample until smoking ceases or until one is certain the sample will not start generating white smoke, perform a sodium coboltinitrite test if necessary to determine whether $\ce{K+ }$ is present.

What other valid separations are there? Is this the easiest method, or are there easier ones?
What about initially adding $\ce{NaOH}$?

Add $\ce{NaOH}$. If a white precipitate forms, the sample contains $\ce{Ca^2+}$ (the white ppt is calcium hydroxide). If a gel-like ppt forms, then the sample contains either $\ce{Al^3+}$ or $\ce{Mg^2+}$. Perform the necessary tests described above to determine the nature of the gel-like ppt.
The liquid part of the solution must now only contain $\ce{Na+ }$ or $\ce{K+ }$. Ammonium ion has been converted to ammonia thanks to the hydroxide ion. $\ce{Na+}$ or $\ce{K+ }$ can easily be tested for using a flame test. The sodium coboltinitrite test can also be employed to detect $\ce{K+ }$ if the color of $\ce{K+ }$ has been masked by the color of Na+ in the flame test.

This method seems a bit faster. But are there any pitfalls?

Dude, you gotta get with the program in re: typesetting those formulae...

related http://chemistry.stackexchange.com/questions/31431/differentiation-between-zn2-al3-mg2-in-solution

You want more than Mart's answer? o.O

@Mithoron well I wanted to reward his answer and to see if anyone else wanted to chime in :).

In Qualitative analysis, cations are arranged in groups. For each group, there are a specific set of reactions that determines the cation. If result are negative, we proceed to next group.
\begin{array}{c|c}
\mathbf{Group} & \mathbf{Cation} \\\hline
\ce{Zero} & \mathrm{\ce{NH4^+}}\\
\ce{I} & \mathrm{\ce{Pb^2+}} \\
\ce{IIA} & \mathrm{\ce{Pb^2+{,Cu^2+}}}\\
\ce{IIB} & \mathrm{\ce{As^3+}}\\
\ce{III} & \mathrm{\ce{Fe^3+ {,Al^3+}}}\\
\ce{IV} & \mathrm{\ce{Co^2+ {,Ni^2+} {,Mn^2+} {,Zn^2+}}}\\
\ce{V} & \mathrm{\ce{Ba^2+ {,Sr^2+}, {,Ca^2+}}}\\
\ce{VI} & \mathrm{\ce{Na+ {,K+} {,Mg^2+}}}\\
\end{array}
Let's start with $\ce{NH4+}$. A simple test is Nessler's reagent test ($\ce{K2[HgI4]}$) which gives brown ppt.
$$\ce{2K2[HgI4] + NH4OH -> NH2.HgO.HgI↓+ 7I- + 2H2O}$$
Then we proceed to $\ce{Al^3+}$. A test called Lake test is performed which gives blue ppt. on a colorless solution.
$$\ce{Al(OH)3 + 3HCl -> AlCl3 + 3H2O}$$
$$\ce{AlCl3 + 3NH4OH -> 3NH4Cl + Al(OH)3↓}$$
Otherwise perform the cobalt nitrate test as described in Martin's answer.
Then proceed to  $\ce{Ca^2+}$. For this, add ammonium oxalate. Explanation for this test is given in aventurin's answer.
For the alkali metal ion, there are indeed specific chemical test but the best test for determining them is the Flame test as described in Jan's answer.

Magnesium

Ammonium phosphate test (described in Martin's answer).

Sodium

Potassium pyroantimonate test (Milky ppt.)
$$\ce{2NaCl + K2H2Sb2O7 -> 2KCl + Na2H2Sb2O7↓}$$

Potassium

Sodium cobaltinitrate test (yellow ppt.)
$$\ce{3KCl + Na3[Co(NO2)6] -> 3NaCl + K3[Co(NO2)6]↓}$$
Or Picric acid test (yellow ppt.)
$$\ce{KCl + C6H6(NO2)3OH -> HCl + C6H2(NO2)3OK↓}$$

Here is the updated table.

In a qualitative analysis, the fast way is always the wrong way. There are already so many issues with the "long" version that I am surprised that this is remotely successful.
The cations of groups 1-3 are always the most difficult to separate and during my time as a teaching assistant were by far the biggest source of error.
The proper course of analysis is

dissolve sample (completely)
precipitate ion
dissolve precipitate
specific reaction
repeat

The reason for separating the cations in the first place is, that they often act in similar ways and the reactions are not very specific. Certain cations should be tested multiple times, too.
All the reactions you are performing are equilibria and very dependent on the overall concentration. None of them form a precipitate quantitatively. Sometimes you might have problems recognising precipitates in the first place.
You need to perform double blind tests for all used chemicals, since already small contaminations may lead to false positives.
The main issue with your proposed strategy is that you will not precipitate aluminium with this strategy, since this is extremely pH dependent.
$$\ce{Al^3+ + 4OH- <=> Al(OH)3 + OH- <=> [Al(OH)4]- }$$
Without a buffered solution at around pH 9 and ageing the solution, there should not form a precipitate. I learned this by adding ammonium sulfide first and raising the pH slowly with ammonia. (This will co-precipitate cobalt, nickel, manganese, zinc, and iron [as sulfides], and chromium [as hydroxide]. This might be useful if you have to analyse for more ions later on.)
If a precipitate forms you can dissolve it with sodium hydroxide and use cobalt(II) nitrate to form Cobalt blue.
You can also test the presence of aluminium with morin as it forms a fluorescent complex with it. Note that $\ce{Mg(OH)2}$ is slightly fluorescent, too.
I cannot remember precipitating magnesium with that strategy in any sufficient amount. I would prefer precipitating it as an ammonium phosphate, which you can try to (re-)crystallise as struvite and look at them with a microscope.
$$\ce{Mg^2+ + NH4+ + PO4^3- <=> MgNH4PO4 v ->C[H2O] MgNH4PO4.6H2O}$$
If you want to precipitate out aluminium and magnesium (and almost every other heavier element) you can use oxine, $\ce{C9H7NO}$ (abbr. $\ce{Hox}$), solution.
$$\ce{ M^{$x+$} + $x$\,Hox <=> [M(ox)_{$x$}] + $x$\,H+ }$$
Then precipitating calcium is straight forward with carbonate solution. Other earth alkaline metals will precipitate, too. You can dissolve this precipitate with acids or you can use the solid to perform flame tests. Since there can only be calcium, strontium, or barium, this should be conclusive. (Of course you could precipitate barium sulfate.)
The main pitfall with your proposed strategy is, that you will not quantitatively precipitate calcium hydroxide and it is as mentioned above quite concentration dependent. The solubility constant is not too small, i.e. $K_\mathrm{sp}=5.5\times10^{-6}$.
If you have precipitated everything so far, then the last thing left over are ammonium, potassium, sodium(, and lithium). The biggest problem is, you cannot mask them properly and you cannot precipitate them. You can use flame tests to determine their presence, but as long as you have sodium in it you might not see anything else. You should use a prism, or a spectrometer to determine the characteristic spectral lines. Apparently phones can do it now, too. Also see publiclab.org... makes you think why separate at all.
You will need to analyse ammonium from the sample. You can add sodium hydroxide, which releases ammonia. You can detect the presence of ammonia in the sample by holding a wetted universal indicator paper over the sample. Additionally, you should be able to smell it.
In other words ...
... be patient with your analysis as there are more pitfalls, false negatives, false positives, contaminations. Don't use short cuts, except atomic emission spectroscopy.

Flame test, flame test, flame test!
I assume that the lab has access to a simple optical spectroscope that will actually help you discern the lines different cation give in the flame test. If it doesn’t have one … Seriously how, because the technique is over a hundred years old.

Sodium, of course, gives a very recogniseable yellow double line.
Lithium (which is not present — why?) would give red lines where nothing else would.
Potassium gives a faint red line at the very edge of the spectrum. It can also be detected by looking at the flame through a blue-coloured glass (it was called Kobaltglas, cobalt glass, in our lab) and noting a different flame colour.
Calcium gives ‘traffic lights’, a red, yellow and green line approximately at equal distances of each other.

Ammonium
This is easily detected by just adding 6 M $\ce{NaOH}$ (or whatever the highest non-concentrated solution your lab may have) to the sample, putting a watchglas above with a bit of wet indicator paper stuck to the watchglass’ bottom side. If ammonium is present, the indicator paper will quickly get coloured blue (or whatever the colour of ‘alkaline’ is for your indicator). Make sure that no students ‘accidentally’ added ammonia to the sodium hydroxide (i.e. perform a blind experiment without sample).
The remaining
Which leaves magnesium and aluminium. I personally didn’t have magnesium in my ion lottery class, so I don’t really know how to cope with it. However, your method seems reasonable.
A secondary test for working with aluminium may be either the morin fluorescence (which is disturbed by sodium) or alizarin S precipitation (red). The latter worked like a charm in my ion lottery.
I’ll add a link to this beautiful graph from Professor Klüfers of the LMU Munich which shows the prevalent aluminium species at different pH values. The notation is as follows: $\ce{Al}$ is $\ce{[Al(H2O)6]^3+}$, and an added $\ce{H_{$-n$}}$ shows that this number of protons is missing. Hence, $\ce{AlH_{-3}}$ is equal to $\ce{Al(OH)3}$. Since I do not own the graph and am unsure about its copyright status, I prefer to link rather than upload here.

One additional suggestion for the identification of calcium is the precipitation as calcium oxalate (weddellite).
$$\ce{Ca^{2+} + C2O4^{2-} + 2 H2O-> CaC2O4 \cdot 2 H2O}$$
Adding a solution of ammonium oxalate to the weakly ammoniacal or weakly acidic (buffered acetic adid) solution will form a white crystalline precipitate. The reaction is very sensitive and specific if only $\ce{NH4+}$, $\ce{Na+}$, $\ce{K+}$, $\ce{Mg^{2+}}$, $\ce{Ca^{2+}}$, and $\ce{Al^{3+}}$ can be present.


text:
Making sense of J coupling data for Hydronium
I found J coupling data for hydronium using DFT and GIAO but I need help making sense of the data. Can someone with help?

Without knowing the atom numbers, it's gonna be pretty hard to know what's going on. My (educated) guess is that atom #4 is the oxygen, so it's coupled equally to the 3 other hydrogens, which are each equally coupled to each other.

The oxygen is atom #4.

Can you tell me what the negative values mean?

All couplings may be positive or negative, and the relative sign can provide useful structural insights. It arises from the relative energies of the mixed spin states, most commonly influenced by Fermi contact mechanism. 1JOH are typcially negative; 1JCH are typically positive for instance. The sign of coupling does not change the appearance of a first order spectrum. If you need further clarification, you should consider asking another question.

We are using the J coupling data to justify the trends we see for the spin echo experiment. We used varying pHs (all acidic) and found that T1 (spin-lattice) relaxation was lowered. I'm still an undergrad but I do know that adding a proton is going to affect the magnetic moment of the molecule, I just need to justify it. All the couplings for water were negative whereas the only the hydronium oxygen couplings were negative. Do you think you can give me any insight as to why this is? Even pointing me in the right direction would be greatly appreciated.

I read this as indicating 2JH-H = 9.6Hz and 1JO-H = -163.8Hz. That's a large O-H coupling.
For water, experimental values have been reported as 2JH-H = -7.3Hz and 1JO-H = -80.6Hz 1, and calculated values vary, but within an order of magnitude. Gas phase studies for 1JO-H also provide values around -80Hz. I can't explain the large discrepancy here; perhaps there is a significant change in bond angle for hydronium? Or perhaps a poorly refined calculation set?

I went to the solvation tab in Guassview 5.0 and set our model to be IEFPCM with water as our solvent. We wanted to mimic our chemical environment Could that have any anything to do with the discrepancy? . Also, we are limited on resources and could only use an augmented triple zeta basis set.


text:
Is the weakening of O-H bond due to -I effect of sugar groups?
In DNA molecule after formation of phosphodiester bond with two nucleosides the O-H bond of phosphate group becomes highly weak and the H+ is lost. This makes the DNA acidic but what causes the weakening of O-H bond? Is it due to the -I effect of sugar groups?

? Free phosphoric acid dissociates anyway... you don't need a sugar there to make it dissociate

The PO42- group is not due to phosphoric acid. It is derived from dNTP/NTP.

which is in turn derived from phosphoric acid


text:
Why does the reaction of sodium nitrate, aluminium, and sodium hydroxide yield ammonia?
From Russian test problem 4301:
$$\ce{3NaNO3 + 8Al + 5NaOH + 18H2O -> 8Na[Al(OH)4] + 3NH3(g)}$$
How does ammonia evolve here? Is it that we get hydrogen gas evolving in the reaction between Al and NaOH and this gas reacts with the NO3 anion?
Is there a logical way to deduce this or should one just memorize that "Al and Zn reduce $\ce{NO3-}$ to ammonia in basic solutions"?
I learned that the reaction is called "a nitrate test using Devarda's alloy", but there is no specific description of the process on Wikipedia. All I learned is that the reason is certainly not the freshly-minted hydrogen:

Nascent hydrogen was supposed to be responsible for the reduction of arsenate or nitrate in arsine or ammonia respectively. Nowadays, isotopic evidence[8] has closed the nascent hydrogen debate, presently considered to be a Gedanken artifact of romanticism.

As a side question, why is NaOH needed? My thought is that it activates the aluminium by breaking through the oxide layer.

Related question: testing for nitrate ions with NaNO3

In the test problem linked you are not expected to know this is the case, but to guess from other data with more common knowledge. And no, neither guessing, nor remembering such corner-cases is expected and/or possible. Some should be remembered if they are part of your field of work, and that's it.

@permeakra - thank you! So I was told on a Russian chem forum. Still I'm curious to know how exactly that ammonia evolves. ^_^ P.S. I'm finding that part of the USE exam the hardest yet. Too many reactions for rote memorization.

You should use reduction potential to deduce whether it will yield ammonia or higher nitrogen oxidation states compound. But in this case, Al and Zn are quite strong reductive metal, so both can strongly reduce nitrate ion to ammonia.

That's true, but the reduction potential alone is not enough. Thermodynamically, nitrate can oxidize pretty many things (including nearly any metal), but kinetics would sometimes forbid that.

Does this answer your question? Testing for nitrate ions with NaNO3 in Devarda's test

@NisargBhavsar I really do not think the other question answers this at all, that's basically just the same reaction equation given here in the first paragraph.

Assuming a redox reaction between metallic aluminium and nitrate, I'd predict that the aluminium gets attracted to the pi-system of nitrate(which contains oxygen, which aluminium has an extremely high affinity for) and donates its electrons to the antibonding N-O orbitals, splitting them and leaving off with the oxygen that it loves. What is left behind is the nitride, which immediately abstracts three protons to become ammonia.

@KanghunKim probably you don't actually get nitride ion. As 9xygen is stripped off and electron pairs beconlme available on the nitrogen they protonate. So you probably get various intermediates where nitrogen is bonded to hydrogen and oxygen, until the oxygen is gone and the bonding with hydrogen is maximized. Thus, in basic solution, NH3.

Maybe ask in /chempros on Reddit.

For electrochemical reduction of nitrate to ammonium on a catalyst, there are intermediates for almost every oxidation state from +5 to -3: https://i.stack.imgur.com/L1izy.png


text:
How to derive van 't Hoff equation for Henry's law constant?
While studying about Henry's law, I encountered this equation:
$$\frac{\mathrm d\ln H}{\mathrm d(1/T)}= \frac{-\Delta_\text{sol}H }{R},$$
where $\displaystyle\mathrm d(1/T) = -\frac{\mathrm dT}{T^2}$ and the $\displaystyle\frac{\Delta H}{RT^2}$ reminds me of some kind of $\Delta G$ equation.
I didn't know how to derive it, so I searched about it and got nothing. I tried to substitute $H$ (Henry's law constant) by $\gamma\cdot f_i$ where $\gamma$ is the activity coefficient at infinite dilution and $f_i$ is the fugacity of pure species.
But still I have no idea what to do to obtain the formula that I'm familiar with, so I welcome any idea about how to derive or how to reverse engineer the formula.

Assuming Henry's coefficient $\mathcal{H}_i$ is defined by the following relation:
$$
\mathcal{H}_i = \gamma_i^\inf \cdot p_i^\mathrm{sat}
$$
Where $\gamma_i^\inf$ is the activity coefficient for an infinite diluted solution and $p_i^\mathrm{sat}$ is the vapor pressure of pure component $i$.
One can try to extrapolate Henry's coefficient at another temperature by assessing the following ratio (this is a common approach in Chemistry, not always valid):
$$
\frac{\mathcal{H}_i(T_1)}{\mathcal{H}_i(T_0)} \approx \frac{\gamma_i^\inf(T_1) \cdot p_i^\mathrm{sat}(T_1)}{\gamma_i^\inf(T_0) \cdot p_i^\mathrm{sat}(T_0)}
$$
The RHS maybe rearranged in order to have the well know form of ratio of  mixed equilibrium constants $K_i$ considering that bulk concentration are close and ratio of activities also simplifies (over a small range of temperature).
$$
\frac{\mathcal{H}_i(T_1)}{\mathcal{H}_i(T_0)} \approx \frac{\frac{\gamma_i(T_1)\cdot p_i(T_1)}{\eta_i(T_1)\cdot x_i(T_1)}}{\frac{\gamma_i(T_0)\cdot p_i(T_0)}{\eta_i(T_0)\cdot x_i(T_0)}} =\frac{K_i(T_0)}{K_i(T_0)}
$$
That is, we are stating that ratio of Henry's constants approximates to the ratio of equilibrium constant because we assume that only ratio of pressure is significant and other ratios tend to unity. This is, off course, not always acceptable, but in some restrained conditions it might hold.
Where the concerned reaction is the converse of  gas solubilization:
$$
\ce{X_{i,(aq)} <=> X_{i,(\mathrm{g})}}{\quad \Delta_\mathrm{R}H = -\Delta_\mathrm{sol}H}
$$
And its mixed equilibrium constant:
$$
K_i = \frac{\gamma_i\cdot p_i}{\eta_i \cdot x_i}
$$
Then one may apply van 't Hoff relation using enthalpy $\Delta_\mathrm{sol}H$ of the concerned reaction, from its integrated form, it comes:
$$
\frac{\mathcal{H}_i(T_1)}{\mathcal{H}_i(T_0)} \approx \exp\left[\frac{\Delta_\mathrm{sol}H}{R}\left(\frac{1}{T_1}-\frac{1}{T_0}\right)\right]
$$
This formulae is generally a good approximation around $\pu{20°C}$ over a small range of temperature. Do not forget that $\Delta_\mathrm{sol}H$ is not a constant but rather a function of temperature.
If you wish to get your definition, you just have to differentiate the last relation with respect to temperature:
$$
\frac{\mathrm{d}\ln(\mathcal{H}_i)}{\mathrm{d}T} \approx \frac{\Delta_\mathrm{R}H}{RT^2} = -\frac{\Delta_\mathrm{sol}H}{RT^2}
$$
Or equivalently:
$$
\frac{\mathrm{d}\ln(\mathcal{H}_i)}{\mathrm{d}\left(\frac{1}{T}\right)} \approx -\frac{\Delta_\mathrm{R}H}{R} = \frac{\Delta_\mathrm{sol}H}{R}
$$

Can you please explain how you changed the right side of second equation?

assuming  that ratio of $x's$ and ratio of activity coefficients are 1 right? so what's the formula used for  $K$ here  ?

How the equation for $Ki$ has obtained ? I know $K$ is the ratio of $fi$ to $fi$ (standard condition)، I understand  the numerator,But Can't understand the denominator.

This is the expression of a mixed (polyphasic) equilibrium constant. $x_i$ stands for the bulk dissolved gas concentration at equilibrium and $\eta_i$ is its activity coefficient. It is just the ratio of product/reactant as you always write them for aqueous solutions.

If you feel uncomfortable with this mixed constant, try convert it into $K_c$ or $K_p$. You will get some $RT$ terms that simply because of the ration of constants.

Oh so $\eta_i$ is an activity coefficient,it's okay,Thank you.


text:
How to extract caffeine from coffee
To extract caffeine from coffee, I tried reacting it with baking soda to make tannins more soluble. I then put in salt to push caffeine out of solution. I saturated it with salt and then mixed the solids with acetone to get the caffeine out. There is also some salt in the acetone. Washes with water should remove the salt and tannin salts leaving behind the caffeine(not pure). Would this procedure work? Is there another procedure that would be more efficient and not require DCM.

What is the question? The text says nothing about DCM and doesn't actually ask a question.

There is a lot of stuff in coffee. If you try to selectively wash away the non-caffeine stuff you will probably fail. You would do better to try extracting it into a non-water-soluble solvent (DCM might work).

Thank you. Are there any household solvents that are not soluble in water

In the organic practical we used black tea: less contamination and higher caffeine content. It is a nightmare though. I think it was about 10 gram tea for 100 milligram caffeine.

Ethyl acetate is in fingernail polish remover (along with acetone).  It is not water soluble, although the presence of acetone might cosolubilize it.  Adding salt to the water should effect a partition.  Also starting fluid is diethyl ether, which is water immiscible.

Is there a method of purification for that impure solid mush.

Supercritical CO2 is used but I doubt you find it in house :)

The popular procedure to experimentally isolate caffeine is sublimation.
Slowly heat a teaspoon of coffee or tea from 150°C to 250°C. Let the sublimated caffeine precipitate at a watch glass cooled with ice. The resublimated caffeine forms white needles that can be identified under a microscope.
The sublimation temperature of caffeine is 178°C.

Sorry, are we talking liquid coffee here? How exactly are you heating it to 250C?

Solid coffee, preferably powder.

I did this in a practical lesson in chemistry. We filtered the coffee (1.2um filter paper), concentrated it 10x using distillation and then extracted the caffeine into an equal volume of chloroform in a separatory funnel. After evaporating the solvent, we were left with fairly white powder.
It was probably impure, containing lots of other alkaloids.

Do you think i could use isopropyl alcohol instead of chloroform

Not really. Although the caffeine would dissolve in it ok, you would have the problem that it would also mix well with the water, so the separatory funnel would not work. The whole idea of using one is to get two separate layers, so that you can drain off and discard the water layer, leaving the stuff you want in the chloroform. Hexane is an alternative, though.

Oh :( I was looking for a home-and-teenager-friendly alternative.