text:
What is the order of the following compounds based on polarity?

What is the order of the following compounds from more polar to less polar?

Hydrochloric acid,
Methanol,
Hexane,
Petroleum benzene

I know that $\ce{HCl}$ has the largest polarity. Hexane I guess has the less. I guess the order should be like:
$\ce{HCl}$ > Methanol > Petroleum benzene > Hexane
Is this correct? I'm not sure of the place of petroleum benzene.

While "petroleum ether" confusingly means a mixture of volatile liquid hydrocarbons, "petroleum benzene" doesn't seem to be a specific solvent, and appears to just be benzene. Am I wrong?

I agree. There is no solvent named petroleum benzene. Both hexane and benzene are not polar at all and I do not think it make sense to compare.

@NicolauSakerNeto Well, in my university’s organic lab, petroleum benzine (Petroleum Benzin) and petroleum ether (Petrolether) were confusingly two names for the same thing. Maybe carolina is at some German university and mistranslated Benzin?

@Jan . Im not in German university.but in our language,benzin has only one letter more than benzene,so it is possible there was a typing mistake in the test.and about HCl,it was written Hydrochloric acid,so it is "eq"

The concept of solvent polarity is commonly used to describe solvent effects. For example, the Glossary of Terms Used in Physical Organic Chemistry (IUPAC Recommendations 1994) uses the following general definition (which was originally proposed by Christian Reichardt) of the term polarity:

When applied to solvents, this rather ill-defined term covers their overall solvation capability (solvation power) for solutes (i.e. in chemical equilibria: reactants and products; in reaction rates: reactants and activated complex; in light absorptions: ions or molecules in the ground and excited state), which in turn depends on the action of all possible, nonspecific and specific, intermolecular interactions between solute ions or molecules and solvent molecules, excluding such interactions leading to definite chemical alterations of the ions or molecules of the solute. Occasionally, the term solvent polarity is restricted to nonspecific solute/solvent interactions only (i.e. to van der Waals forces).

However, the polarity of a solvent is not a physical quantity. Therefore, it is difficult to express quantitatively. Several scales exist that involve various physical properties in order to provide a measure of a solvent’s polarity.
In particular, the relative permittivity $\varepsilon_\mathrm{r}$ is often used as a quantitative measure of solvent polarity:
Hexane: $\varepsilon_\mathrm{r} = 1.8865$
Benzene: $\varepsilon_\mathrm{r} = 2.2825$
Methanol: $\varepsilon_\mathrm{r} = 33.0$
Water: $\varepsilon_\mathrm{r} = 80.100$
(values at 20 °C taken from “Laboratory Solvents and Other Liquid Reagents”, in CRC Handbook of Chemistry and Physics, 97th Edition (2016), William M. Haynes, ed., CRC Press/Taylor and Francis, Boca Raton, FL.)
By way of comparison, the empirical $E_\mathrm{T}(30)$ scale is based on the solvatochromism of Reichardt’s dye (Betaine 30):
Hexane: $E_\mathrm{T}(30) = 31.0\ \mathrm{kcal/mol}$
Benzene: $E_\mathrm{T}(30) = 34.3\ \mathrm{kcal/mol}$
Methanol: $E_\mathrm{T}(30) = 55.4\ \mathrm{kcal/mol}$
Water: $E_\mathrm{T}(30) = 63.1\ \mathrm{kcal/mol}$
(values taken from here)

Note that petroleum benzene is nothing but benzene itself. The order must be HCl > Ethanol > hexane > benzene. It is because benzene is a perfectly symmetrical molecule and dipole moment vectors cancel each other completely.
But hexane is not that symmetric and hence there is a dipole moment due to difference in electronegativity of C and H atoms (but it is almost negligible).

However, benzene is capable of dissolving cations much better than hexanes because of it’s polarisability (think $\pi$ system). So I’m pretty sure it could therefore be considered more polar.

Benzene is able to dissolve cations more efficiently because it has more electron density for cations to attack.But that electron density is symmetrical in space so it does not provide dipole moment.Dipole moment is a vector quantity on the other hand electron density is scalar.Dont confuse between them. ;-)

Don't confuse permanent dipole moment and polarity, though ;) 1,4-Dioxane in its trans-configuration is also point symmetric thus has a permanent dipole of $0$ yet is rather polar.

That is what i am trying to say.Polarity is different from polarisability.HCl is most polar because it has got a permanent dipole moment. Theoretically A single oxygen atom is not polar,but will still attract cations, cause it is polarisable

I agree with @Jan. There is only little correlation for a molecule to have dipole moment and being polar. The simplest and most prominent example is carbon dioxide, which is very polar, but has no dipole moment whatsoever. I am pretty certain, that the +/- charge separation is larger in benzene, than it is in hexane. Your ordering is therefore wrong.

@user if you want others to get notifications when you comment, you should use a @ before their username. @Martin polar != polarizable!

I'm going to have to agree with @user247855 on this debate.  A polar molecule is generally referred to as such in a vacuum (i.e. a molecule with a permanent dipole moment).  Molecules can be polarizable (induced polarity) in a different environment, but to say that something like CO2 is very polar is completely incorrect in nearly every general case I can think of.

@user247855 Please be nice.


text:
The effect of air pressure on the melting point

What is the effect of air pressure on the melting point of a substance?
a) No effect
b) The melting point increases when the air pressure increases
c) The melting point decreases when the air pressure increases

The boiling point has inverse relation with vapor pressure of the liquid and positive relation with atmosphere (air) pressure. The melting point has positive relation with vapor pressure of the substance, but what about its relation with air pressure?

Your questions are better, but please use the 'homework' tag

@anshabhi homework questions are allowed, just as long as there is an effort to show working made, see homework policy

@anshabhi, all of this 3 questions were from the test I had today. I just want to see if I wrote the right answer and also I really need to solve my problems.I Could not find the sure answers.now no matter my answers on the exams were right or wrong,I can not change them,but I need to learn

alright.... sorry..!!

For most substances, higher pressure (or air pressure, in your case) will cause the melting temperature to go up. To think about it intuitively, imagine that you have a certain solid. Melting it would increase the volume of that substance because liquids take more space than solids. If you increase the pressure, it becomes harder for that transformation to occur. It's harder to change into a state that needs more volume if you are under pressure! So, you need more energy, hence temperature, to melt.
Notice that this relation is actually reverse in water. Because ice takes more volume than liquid water, it's actually easier to melt it at high pressure and it requires less heat. Try taking an ice cube and apply pressure on it with a knife, see how it liquifies at the contact point.

Usually knifes are made of a metal, which is quite efficient at transferring ambient heat to the ice. So it is not obvious, that it is the pressure which cause the effect. To reduce the possibility of heat transfer contributing to melting in that experiment, the knife could be put in the freezer for a few hours before conducting the experiment.

@kasperd or use a ceramic knife instead

The Melting Point does not go up for water until you apply over 1500psia, in which case it goes down, and then back up, and increases steadily. So for pressures under 1500psia, melting point of water is 0°C. Also water liquid is generally considered essentially incompressible, as with solid form, so I'm not sure why volumetric differences come into play... (Pressure has little affect on these, however vapor is a different story)
Also if the knife thing were true, you would see water phase change back into ice when you removed the force applied. The ice would also uniformly melt, so try doing it at a temperature of 0°C (with a knife at 0°C as well) and see what happens.
Just look up "pressure-temperature phase diagrams" if you don't believe me.

That's incorrect. The melting point of ice decreases immediately as you increase pressure, and you can measure this change for pressure increases far less than 1500 psia.  E.g, at 750 psia, the melting point is -0.4 C.  In addition, the pressure-sensitivity of the melting point is due to the volume difference between ice and water; it does not require that the phases themselves be compressible.   The relative compresibilities of the phases affect how the volume difference changes with pressure.

For any substance to melt, it has to overcome or reduces the interaction forces that keeps the particles together in solid state. As the pressure of substance increases, particles tends to remains compacted, increasing of pressure during melting hindering in melting process, makes it difficult to overcome the strong force of attraction, i.e. more thermal energy is required. That's why the melting point increases as the pressure increase.

That's incorrect.  The melting of water requires that the "interaction forces that keeps the particles together in solid state" be overcome, yet increasing the pressure favors melting (i.e., it reduces the melting temperature).  What matters is that the molar volume of ice is greater than the molar volume of water.

@theorist, how does increasing pressure reduce the melting point? It should increase it, shouldn't it?

@NatashaJ  This phenomenon follows Le Chatelier's principle, which says that the equiibrium will shift to relieve an external stress.  If the external stress is pressure, the equilibrium shifts to favor the smaller phase.  Most substances expand when they melt, so higher pressure favors the solid phase, making it harder to melt, and thus increasing the melting point.  But ice contracts when it melts.  Thus increasing the pressure favors the liquid phase, which means it makes ice easier to melt => the melting point decreases.

If you want an intutitive way to think about this:  As water freezes, it has to push out against the atmosphere.  The higher the pressure, the more work water has to do when it freezes, making it more favorable to stay in the liquid state...which means the melting point decreases.  I'll add that, formally, Le Chatelier's principle follows from the 2nd law of thermodynamics, and can be proven mathematically from the 2nd law.  Some physical chemistry textbooks show this.

Okay, got it!! Thanks for explaining it so well.

According to the phase diagrams of water, the freezing point and melting point is inversely proportional to increase in pressure . About the volume concept , that's actually the concept of Le Chatelier's principle that if you increase pressure on a system, the system will shift in the direction of decreasing volume. When you compare it with the concept of "Ice having greater volume then liquid water', it connects. And I think we can apply the pressure on solid and liquid here because it is the change in pressure around the liquid or solid we are talking about , not the effect of pressure  inside the liquid or solid.

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About pH of an aqueous solution of SO2
Probably we can have an aqueous solution of $\ce{SO2}$ by dissolving it in water, because we would have an equilibrium between $\ce{SO2(g)}$ and $\ce{SO2(aq)}$:
$$\ce{SO2(g) <=> SO2(aq)}$$

How can I compare pH of an aqueous solution of $\ce{SO2}$ with that of pure water qualitatively?
What happens to the pH of an aqueous solution of $\ce{SO2}$ if I aerate it by bubbling air through it?

Is $\ce{SO2}$ an acidic, basic, or neutral oxide? 2. Can $\ce{SO2}$ react with any of the gasses in air, and if so, what is the product?

@  Nicolau Saker Neto: I know that So2 is acidic.And I could assume that I would react with gases, since it is a gas.But how can I relate this?

If $\ce{SO2}$ is an acidic oxide, then which acid does it create when reacting with water, and what is its name? How strong is this acid? Also, notice that this acid is very similar, in formula and name, to another compound which has one extra atom . What is this compound? Could the extra atom come from air? What would the reaction be?

@  Nicolau Saker Neto: When SO2  dissolve in water ,H2SO3  will be the acid that is going to form, it is known as sulphurous acid . And is that extra atom is H or any other?

Think of the most common elements in air, and try adding them to the formula for sulphurous acid. Which do you think is the most likely outcome?

@  Nicolau Saker Neto: Should I add  or subtract that particular element?.Is it HSO3-

You have to add an atom present in large amounts in air to $\ce{H2SO3}$. Can you think of a familiar formula that would result from this? If you don't get it right, no worries, I'll tell you the answer.

Nicolau Saker Neto: I guess it would be H2SO4. But I am not quite sure.

Correct! Okay, so before you bubble air you have a solution of $\ce{H2SO3}$, and after bubbling air, it oxidizes to $\ce{H2SO4}$. Which of these is a stronger acid? What do you expect to happen to the solution pH after bubbling the air?

Btw, how does H2SO3 oxidises, what get reduce here?

Upon adding $\ce{SO2}$ gas to water, it shows its acidic oxide character by hydrolyzing to sulfurous acid:
$$\ce{SO2(g) + H2O(l) <=> H2SO3(aq)}$$
$\ce{H2SO3}$ is a weak diprotic acid in water, with the following acid dissociation equilibria:
$$\ce{H2SO3(aq) <=> H+(aq) + HSO3^{-}(aq)}\ \ \ \ \ \mathrm{K_{a1}=1.4\times 10^{-2}}$$
$$\ce{HSO3{}^{-}(aq) <=> H+(aq) + SO3{}^{2-}(aq)}\ \ \ \ \ \mathrm{K_{a2}=6.3\times 10^{-8}}$$
Thus, addition of $\ce{SO2}$ gas to water will produce an acidic solution.
What would happen if air were bubbled into the solution? Air is a mixture of mostly nitrogen and oxygen. It so happens that $\ce{SO2}$ reacts with oxygen to form $\ce{SO3}$:
$$\ce{2SO2(g) + O2(g) -> 2SO3(g)}$$
Here, the sulfur atom in $\ce{SO2}$ is oxidized (oxidation number goes from +4 to +6) while the oxygen atoms in $\ce{O2}$ are reduced (oxidation number goes from 0 to -2). Now, $\ce{SO3}$ also happens to be an acidic oxide, and it hydrolyzes in water to form sulphuric acid:
$$\ce{SO3(g) + H2O(l) <=> H2SO4(aq)}$$
This time, $\ce{H2SO4}$ is a strong diprotic acid:
$$\ce{H2SO4(aq) <=>> H+(aq) + HSO4^{-}(aq)}\ \ \ \ \ \mathrm{K_{a1} \approx 10^6}$$
$$\ce{HSO4{}^{-}(aq) <=> H+(aq) + SO4{}^{2-}(aq)}\ \ \ \ \ \mathrm{K_{a2}=1\times 10^{-2}}$$
Assuming that not much $\ce{SO2}$ is lost while bubbling air through the sulphurous acid solution, then you're effectively replacing a weak acid with a strong acid, which would cause the pH to decrease.
As a final remark, if the sulphurous acid solution were bubbled with pure nitrogen, then there would be no oxidation of $\ce{SO2}$ into $\ce{SO3}$, and after a while enough $\ce{SO2}$ will have been dragged away for the solution to become a less concentrated sulphurous acid, which would therefore have a higher pH.

If I aerate the sulphurous  acid only with N2 ,why doesn't it convert SO2 to SO3, is it because of not having lower oxidation states for nitrogen, other-than zero?

Nitrogen can reach negative oxidation numbers, but that's irrelevant; without oxygen gas, $\ce{2SO2 + \color{Red}{O2} -> 2SO3}$ can't happen!

Here what happened to the oxygen.You have stated that it is reduced to -2 oxidation state.But where is it?

The oxygen gas was added to sulphur dioxide $\ce{SO2}$ to make sulphur trioxide $\ce{SO3}$. All oxygen atoms in both these compounds have an oxidation number of -2.

Okay, Could you state some other gases behave like this,and can't liquid and solids behave like the way SO2(g) did?

I can't think of another acidic/basic substance off the top of my head which will react and change its acidity/basicity by simply exposing it to air, though it's not really a special feature of sulphur oxides. It doesn't have to be a gas, either. The critical part of this question was the reaction with oxygen.

Okay , so what is the most important thing that I could get form this question?

That is, to a large extent, for you to decide. But some basic points are: (1) sulphur has more than one common oxide; (2) both sulphur oxides are acidic, and hydrolyze in water to create two different acids; (3) one oxide can be converted into the other by simple exposure ti oxygen; (4) atmospheric air is often a source of oxygen for oxidation reactions; (5) sulphuric acid is a much stronger acid than sulphurous acid; (6) comparing two solutions of different acids at the same concentration, the stronger acid will make a more acidic solution.

Let us continue this discussion in chat.

Nicolau Saker Neto:So thank you for the help given to clear me this doubt.

When you dissolve $\ce{SO2}$ in water you get $\ce{H2SO3}$. $\ce{SO2(aq)}$ means $\ce{H2SO3}$ therefor,the pH of the solution will be slightly acidic when compared to that of pure water.
to your other question.when you aerate the solution $\ce{SO2}$ gas right above the solution goes away.The equilibrium is then disturbed.Therefore, $\ce{SO2}$ dissolved in the solution comes out.Because of that pH of the solution increases.

This is incorrect! While bubbling an aqueous solution of a gas does remove some of it from the water, there is another effect to be taken into account in this case.

@ Nicolau Saker Neto:Actually SO2 is dissolved in water.I don't see any other effect to be taken in this case.Please let me know what the other effect is.

@lucyG The other effect is oxidation of $\ce{H2SO3}$ to $\ce{H2SO4}$ by oxygen in the air.

@bon Yes that effect can also be there.But according to the question of  On the way to success don't you think we have to give more priority to the equlibrium that has been made rather than other effects.In my point of view this question inquires about the basic knowledge of Le Chatlier's principle

@lucyG I don't see anything in the question that suggests we should ignore the effect of oxidation.

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What is the order of the reaction with respect to NO?

The following reaction mechanism has been proposed for the reaction of $\ce{NO}$ with $\ce{Br2}$ to form $\ce{NOBr}$
  $$\begin{align}
\ce{NO(g) + Br2(g)&<=>NOBr2(g)}\\
\ce{NOBr2(g) + NO(g) &-> 2NOBr(g)}
\end{align}$$
  If the second step is the rate determining step, what is the order of the reaction with respect to $\ce{NO}$?

Since the 2nd step is the rate determining, shouldn't the order with respect to $\ce{NO}$ be 1? According to my book though, the correct answer is 2. What am I missing here?

The rate of that second reaction will indeed be first order with respect to $\ce{NO}$:
$$
R = R_2 = k_2[\ce{NO}][\ce{NOBr2}].
$$
The problem is, the overall order of the reaction is determined from the rate expression involving only the reactants.  The above rate expression is written in terms of $\ce{NO}$ (reactant) and $\ce{NOBr2}$ (intermediate), not in terms only of reactants.
To convert the reaction rate expression to one involving only reactants, exploit the fast equilibrium of the first reaction,
$$
K_1 = \frac{[\ce{NOBr2}]}{[\ce{NO}][\ce{Br2}]},
$$
to find an expression for $[\ce{NOBr2}]$ in terms only of $K_1$ and the reactants:
$$
[\ce{NOBr2}] = K_1[\ce{NO}][\ce{Br2}].
$$
The expression for $R = R_2$ can thus be rewritten as
$$
R=R_2 = k_2[\ce{NO}]\left(K_1[\ce{NO}][\ce{Br2}]\right) = k_2K_1[\ce{NO}]^2[\ce{Br2}],
$$
which is indeed 2$^{\mathrm{nd}}$ order in $\ce{NO}$.

Even though the 2nd elementary step is the rate determining, you cannot just consider the stoichiometry of a species in one single elementary step to find out the overall order, which is what you are asked to find out in this question. This a complex reaction and thus the order of each reactant will depend on all the elementary steps of the reaction mechanism.
We will use the equilibrium constant method(I actually forgot what this method is called) to find out the rate law for this reaction taking the help of the mechanism that you have mentioned.
The rate law will be $${r}=k\ce{[NO][NOBr2]}$$
Since $\ce{NOBr2}$ is the intermediate we are going to exclude it from the final rate law. (Remember, intermediates should never appear in the final rate law.)
$$K= \frac{\ce{[NOBr2]}}{\ce{[Br2]}{[NO]}}$$
$$\ce{[NOBr2]}=K\ce{[Br2][NO]}$$
Therefore,
$$r=k\ce{[NO]}K\ce{[Br2][NO]}=kK\ce{[NO]^2[Br2]}=k'\ce{[NO]^2[Br2]}$$
Thus you can see that the order with respect to $\ce{NO}$ is 2 and not 1 as you initially though.

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How many times can we recycle one sheet of paper?
Its really interesting to see the process of recycling of paper (as shown in picture below) but at the same time I was thinking like for how many times can one sheet of paper can be recycled ?

Technically, once, since the paper gets shredded and redistributed among many pieces of paper. But more than 90% of the paper can complete each cycle.

Why do you think there is a limit to the number of times a sheet can be recycled?

I doubt that there will be difference in quality which might make that paper quite not useful. I am not sure of that.

Well apart from the fact that the answer depends on the type of the paper that is used. Secondly how much is wasted and also the process used. Below is an article that might be helpful in answering the question

Some industry sources estimate that an ordinary sheet of paper made from cellulose fibers derived from wood can survive only four to six trips through the recycling process. The Environmental Protection Agency puts the figure at five to seven times.
It is not surprising that the rigors of remanufacturing take a toll on the fibers.
Ideally, paper for recycling is separated into types, because paper with long fibers, like white office paper, offers the most flexibility for recycling, while newsprint, with its shorter fibers, is usually reserved for making more newsprint and other low-quality papers.
The paper is shredded and chopped, then subjected to a mixture of chemicals and water and heated as it is repulped. It is centrifuged and screened to remove impurities; de-inked with more chemicals; then sprayed onto a wire screen, drained, dried and squeezed through heated rollers.
With each step, the fibers become shorter, coarser and stiffer, so that eventually, recycled fiber needs to be mixed with virgin fiber to make paper of the desired quality.
According to the American Forest and Paper Association, 63.4 percent of the paper consumed in the United States was recovered for recycling in 2009. C. CLAIBORNE RAY

SOURCES: New York Times - http://www.nytimes.com/2010/12/21/science/21qna.html?_r=0

The fibre length keeps decreasing and the ink load keeps increasing. When no longer suitable for paper the pulp moves to the cardboard recycling stream and eventually leaves through some non-recycled avenue (landfill or burning).

"Energetic recycling" sounds a lot nicer. ;) Even for toilet paper, after the wastewater treatment plant.


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Does 11.2 dm3 of nitrogen consist of 0.5 moles at STP?

Which of the following statements is true?
$\pu{11.2 dm^3}$ of nitrogen at STP-
A: has a mass of $\pu{14 g}$.
B: consists of $\pu{0.5 mol}$ atoms.

My instinct was that the answer is B because any gas has a volume of $22.4$ liters regardless of what the gas is. If $\pu{0.5 mol}$ of nitrogen has a volume of $\pu{11.2 dm^3}$ at STP, then $\pu{1 mol}$ of nitrogen has a volume of $22.4$ liters at STP (if I am correct).
It turned out the answer was A. Why is B not the correct answer?

B looks like a trick answer. It would be 0.5 moles of N2 molecules but not 0.5 moles of nitrogen atoms.

So you thought B was correct, and consequently A was not correct, but did you stop to independently check whether A was correct or not?

Checkout WolframAlpha

Subtle point.  In the 1980s the standard pressure part of STP was changed from 1 atmosphere = 101325 Pa to 1 bar = 100000 Pa.  With this reduced standard pressure the corresponding molar volume has expanded to 22.71 L.  But it seems to have been overlooked, even 30+ years on, by most texts.

You are right that 0.5 mol of $\ce{N2}$ is 11.2 L, but read the question carefully. Option B says that 11.2 L of $\ce{N2}$ contains 0.5 mol of atoms, which is wrong, it contains 11.2 L of $\ce{N2}$ contains 0.5 mol of molecules. It seems to be a trick answer. The first option is correct as 0.5 mol of $\ce{N2}$ has a mass of 14 grams.

The correct answer for this question is A.
Prior to start explaining the answer for your question, I would like to show you a mistake you have done.
You have said

"any gas has a volume of 22.4 litres regardless of what the gas is."

This true only if the gas exist in standard temperature and pressure or (STP). And note that per one mole of any gas at STP condition occupy a volume of 22.4 litres or $22.4\ \mathrm{dm^3}$. Simply we call this as molar volume.
Here is how you can derive it.
Subject $V$ in the formula $pV=nRT$, and then you can substitute values. Where $T=273.15 \ \mathrm{K}$ (value of standard temperature) and $p=101325 \ \mathrm{Pa}$ (value of standard pressure) and $R=8.314\ \mathrm{J\ mol^{-1}\ K^{-1}}$
$$V_\mathrm{m}=\frac{V}{n}=\frac{RT}{p}$$
You have asked why not the answer B is correct?
It is not correct. Here is why:
In one mole of any compound you have $6.022\times10^{23}$ atoms
$1\ \mathrm{mol} \Rightarrow 6.022 \times 10^{23}$ atoms.
$6.022 \times 10^{23}$ is the Avogadro constant. (If you have no idea about it check this.)
So here in this question you have found number of moles of nitrogen molecule, it means $\ce{N2(g)}$ as $0.5\ \mathrm{mol}$. Which means you have $0.5\times2\ \mathrm{mol}$ of atoms. So as I said before if you have one mole mean you already get $6.022 \times 10^{23}$ atoms of $\ce{N}$ atom. So you have $6.022 \times 10^{23}$ atoms or $6.022 \times 10^{23}$ nitrogen mol atoms. So now you can clearly see why answer B is incorrect. I hope my answer is helpful enough to solve your doubt.

The answer is not correct, in 0.5 mol of N2 you would have 6.022e23 nitrogen atoms and 1 mole of N atoms (if that is even a thing).

@orthocresol: Hey, as far as I know, in one mole of any compound you have 6.022e23, so how can you say " in 0.5 mol of N2 you would have 6.022e23 nitrogen atoms and 1 mole of N atoms "

@Onthewaytosuccess one molecule of diatomic nitrogen has 2 nitrogen atoms

@orthocresol: It means you are saying for 1 mol  of nitrogen atom we have  6.022e23 atoms?

@Onthewaytosuccess yes, that is right. That is assuming that you could even have one mole of nitrogen atoms (which would, as far as I know, immediately combine to form diatomic nitrogen).

@ orthocresol: Oops, till now I have the idea in the wrong way. Because I thought , it is possible only for one mole of compound!


text:
Quantum mechanical model of atom and the quantum numbers
I studied Bohr's model of atom and then the drawbacks of it and then quantum mechanical model of atom. Now quantum model is according to uncertainty principal and dual nature of matter and it says we can not talk about fixed orbits (uncertainty principle) and can only talk of  an area of maximum probability of finding an electron known as "orbital". Now when I'm studying principal quantum number, it says it is the "shell number". So how come we are talking about fixed shells again and returning to what Bohr said?

The Bohr model is completely wrong and really shouldn't be used for anything other than a historical account of how Quantum Mechanics came about.  It did okay for explaining the atomic emission line spectrum for the Hydrogen atom but that is about it.  Related: http://chemistry.stackexchange.com/questions/31639/what-do-the-quantum-numbers-actually-signify

There's no fixed in the quantum theory! Unless, if you consider the energies as fixed. FWIW, I think if you study a little bit more, you simply wouldn't have this question, as it rises from a misunderstanding about what shells are.

So how come we are talking about fixed shells again and returning to
  what Bohr said?

Nope, we are not returning to the old quantum theory. We do not talk about orbits anymore, we rather have orbitals. Note that it is important to bear in mind that orbitals also aren't real, i.e. this description is still approximate, but it is a much better way of thinking about atoms (and molecules) than Bohr model.
Now, to understand the concept of orbitals you have to learn the basics of quantum chemistry first, namely, the Hartree-Fock method. Until then just realize that:

We do use hydrogen-like one-electron wave functions (orbitals, or, spin-orbitals, to be more precise) to describe the state of each and every electron in any atom.
These hydrogen-like orbitals are described by a set of quantum numbers: the principal quantum number, the azimuthal quantum number, the magnetic quantum number, and the spin quantum number.
The set of orbitals belonging to a given principal quantum number $n$  constitutes what we call a shell in the orbital model.

When you say that orbitals aren't real, do you mean to say that orbitals are not an exact description of the system? One might argue that (not in the complex number sense) wave functions are a very realistic aspect of any quantum system.

Also, the hydrogenic orbitals are exact in one-electron atoms like H or He+.

@Ian, the reason to say that orbitals aren't real is that the electronic wave function for a many-electron system could not be written as a product (antisymmetrized product, to be more precise) of one-electron wave functions (say, hydrogenic orbitals). As a result, one-electron wave functions (orbitals) arise in the description of a many-electron only as a part of an approximation (the HF one). It is in this sense they aren't real.

@Ian, well, strictly speaking, the hydrogenic orbitals are not exact even for one-electron atoms. They are exact only in the non-relativistic approximation, which is, well, an approximation. :D So even in this case one can argue that the hydrogenic orbitals aren't real, though, in this case it is a sort of nitpicking.

We are not talking about fixed shells. Rather, we are observing that for any given atomic orbital you can find an orbital of very similar shape but of a higher energy. Examples:

The simplest orbital is the 1s orbital of hydrogen. We can find an orbital of higher energy that has the same shape (namely: spherical symmetry): The 2s orbital.
There is an orbital called $3\mathrm{d}_{z^2}$ which is rather important in transition metal chemistry. Its ‘shape’ consists of two lobes of identical phase with a nodal plane in the middle plus a ‘ring’ of the opposite phase around the middle. Again, an orbital of similar shape but higher energy can be found ($4\mathrm{d}_{z^2}$).

This fitted in so nicely with the previous models, that they were called ‘same orbitals of different shells’. They’re still orbitals, so wave functions, but we grouped them into shells.

As Ian correctly mentioned in a comment, when dealing with a one-electron system (such as the basic hydrogen atom) all orbitals of the same shell will have identical energies. (That’s just what Schrödinger’s maths tells us.) Another reason to still use the word shell without referring to it’s meaning of the fixed Bohr orbits.

Also, we organize the orbitals by energy. In one-electron atoms like hydrogen, any orbitals with the same principal quantum number will have the same energy. This fact is not quite true for multi-electron atoms, but there is a sufficiently reasonable direct mapping from the orbitals in a one-electon atom to the "orbitals" in a Hartree-Fock multi-electron atom such that it still makes sense to assign orbitals with principal quantum numbers and to organize them into groups by principal quantum number.


text:
Claisen condensation

Aren't two last steps superfluous? I just do not see how the fourth and last forms differ.

The deprotonation step is essentially irreversible due to the acidity of the proton. Up until that point, the steps are reversible and favor the left hand side of the equilibrium. The deprotonation is often called the driving force of the reaction because when the reaction manages to beat the odds and reach that step of the reaction it is prevented from returning to the starting material by this "irreversible" deprotonation. Acidic workup yields the product. If deprotonation could not occur, only the starting material would be recovered.

The Claisen reaction occurs in a strong base $(\ce{RO-})$, which is a conjugate base of an alcohol $(\text{p}K_a \approx 16-18)$; while for the $\beta$-dicarbonyl compound, $\text{p}K_a \approx 10$. The fourth molecule cannot help but be deprotonated in the strongly basic reaction mixture to produce the fifth  molecule. Acid needs to be added to reform the desired product.

You are correct that there is no difference between the last fourth and last molecules shown.
However, the thermodynamic driving force behind the Claisen condensation is the formation of the highly stable enolate of the $\beta$ ketoester, or in your case $\beta$ diketone.

In order to form this stable enolate the original ester must have at least two $\alpha$ hydrogens so that there is at least one left at the end to deprotonate. Therefore the Claisen condensation will not work with esters having only one $\alpha$ hydrogen (and obviously not with esters with no $\alpha$ hydrogens).
Once you have completed the reaction to yield the enolate product, it can be acidified as shown in your reaction scheme to yield the neutral molecule.

The 4th and the last structures are same but the last two steps are not superfluous, because the 4th structure is consumed right after it is generate and will not persist while after addition of acid, the last structure persist.
Please pay attention to the condition of the last step (acid). All steps except the last step happened consequently once you mixed two starting materials with base. The last step happens after addition of acid.
Under basic condition (alkoxides), the beta-keto ester generated in step 3 will be deprotonated immediately to give the second 5th enolate structure which will be protonated after addition of proton source (acid in this case).


text:
Are the p orbitals of the biradical dioxo compound in the HOMO perpendicular to the plane?
I tried to draw the frontier molecular orbitals of the following biradical structure:

(A) At the top I've drawn the HOMO and LUMO, because I know that the reaction coordinate belongs to the $\ce{C_{$2v$}}$ point group. These two orbitals of the HOMO (green) lies in the O-C-O plane. In each $p$ orbital I would locate one electron (triplett state is favoured because of Hunds rule). Oxygen has six  valence electrons, so I must bring four electrons in orbitals of each oxygen. I can place two electrons in a $p$ orbital perpendicular to the greenish depicted orbital. Then I would place the other two electrons in a $sp$ hybrid orbital.
(B) In the second possibility (at the right at the bottom) I also used a $sp$ hybridization for the oxygen atom. The two radicals are placed in the $p$ orbital perpendicular to the O-C-O plane. I guess the structure is less favoured than the first possibility (with the two radical in the O-C-O plane) because there are unfavoured filled-filled interactions between pink-pink and green-green while above there is onyl a pink-pink destabilizing interaction.
(C) As a third possibility I've used a $sp^2$ hybridization of the oxygen. The radicals are now perpendicular to the plane. Without exact calculation of the hybrid orbitals I would predict that 4 electrons in $sp^2$ hybrid orbitals plus one electron in a $p$ orbital is lower in energy as case (A) (3 electrons in $p$ orbital and two electrons in a $sp$ hybrid orbital). However the green lobes (both two electrons and hence repulsive interaction) here in the case (C) in the $sp^2$ hybrid orbitals seems to come closer (120°) than the two electrons in the orthogonal $p$ orbitals (also filled-filled) in case (A).
Are these reasons why the HOMO is best described by case (A)>(C)>(B)?
Probably this question has similiarties to the hybridization in a previous question (please note I'm a novice concerning hybridization)
Why allene cannot be described with an allyl system?

I think you are asking what does the HOMO and LUMO of the 1,3-dioxabiradical look like.
Let's start by using cyclopropane as a model compound.  In cyclopropane the $\ce{C-H}$ bonds are approximately  $\ce{sp^{2.46}}$ hybridized and the $\ce{C-C}$ bond is approximately $\ce{sp^{3.74}}$ hybridized (see here).  That's a lot of p-character in the $\ce{C-C}$ sigma bond. Dioxirane is likely similarly hybridized.  As soon as the 1,3-biradical forms (as soon as we break the $\ce{O-O}$ bond), the HOMO and LUMO will look like what is shown in the "initial" drawing below.  If the biradical lives long enough to undergo rotation about the $\ce{C-O}$ bond, then the HOMO and LUMO will look like what is shown in the "rotated" drawing.

These drawings just point out the conformational dependence of the HOMO and LUMO.  Note that the two lone pairs on each oxygen have not entered the discussion.  They are orthogonal to, and have no effect upon, the HOMO and LUMO orbitals.
Imagine bringing two oxygen atoms closer to each other.  If we push them closer and the orbitals containing the odd electrons are pointing towards each other as in the "initial" geometry, we will form a sigma bond and we know what the HOMO and LUMO look like for a sigma bond (just like what is pictured in the "initial" geometry, only the orbitals are closer together).  If we push them closer and the odd electrons are pointing parallel to each other as in the "rotated" geometry, we will form a pi bond and we know what the HOMO and LUMO look like for a pi bond (just like what is pictured in the "rotated" geometry, only the orbitals are closer together). In the 1,3-biradical the carbon atom holds the 2 oxygen atoms together, they are closer than they would be at infinity, but further apart then they would be if there was a bond between them.  This tells us that the HOMO-LUMO energy separation is greater than it would be at infinity (0), but less than if they were closer together.
We could refine our model by noting that, according to Bent's rule, we would expect the orbitals containing the oxygen lone pairs to have a bit more s-character that the carbon $\ce{C-H}$ orbital in cyclopropane.  Therefore, in dioxirane and the corresponding 1,3-dioxabiradical, we would expect both the $\ce{O-O}$ bond and the orbital containing the single electron in the biradical to have even more p-character than $\ce{sp^{3.74}}$.
Again, however, to re-emphasize an earlier point, the lone pairs on oxygen have no bearing on our discussion.  That said, if we would like to speculate on the hybridization of the orbitals containing the lone pairs, we may do so.  Most texts tell us that the oxygen lone pairs in water, ethers and alcohols are in (roughly) $\ce{sp^3}$ orbitals.  However data from photoelectron spectroscopy experiments seems to suggest that one lone pair is in (roughly) a p-orbital while the other lone pair is in an $\ce{sp}$ orbital.

IR data shows also that the C-H bond in cyclopropane has $sp^2$ character. I tried to figure out the C-C hybridization without your formula (just with the "valence orbitals"). 3 C gives totally 3 s + 9 p orbitals. Now I subtract 3x(s+2p) for the 9 $sp^2$ orbitals and I obtain 3 p orbitals. In the lecture the cyclopropane has been described with Walsh orbitals and three perpendicular p orbitals (Möbius set with smaller $\beta$ than the Hückel set of the Walsh orbitals pointing to the center of the cycle).

Frankly I think this is not the truth, because I expect a part of the $sp^2$ hybridization coming from 3 H s orbitals therefore instead of 3 perpendicular p orbitals I get 3 sp orbitals for the C-C bond.  Different webpages treat the Möbius banana bonds as p orbitals. Möbius sp orbitals would propably give more overlap than just p orbitals, yes? Are these sp orbitals also perpendicular to the Walsh orbitals as the p orbitals would do?

A few comments: 1) both Coulson's theorem and pKa data suggest that the cyclopropane $\ce{C-H}$ bond is neither $\ce{sp^2}$ nor $\ce{sp^3}$, but rather something in between, 2) you're analysis is off because there are only 6 $\ce{C-H}$ bonds in cyclopropane, not 9, 3) "I expect a part of the sp2 hybridization coming from 3 H s orbitals" , the substituent (H in this case)  electronegativity will affect the carbon hybridization.  The $\ce{C-C}$ bonds have much more p character than sp;  p (90° bond angle) will provide for much better overlap in cyclopropane than sp (180° bond angle)

Google "Coulson's theorem" and read about it if you have time.

IR: 3080$cm^-1$, H-C-H bond angle 121° (reference: from lecture notes).  2) I used 3x3 $sp^2$ orbitals if I assumed one $sp^2$ pointing to the middle of the cycle. If we really can do so (as first approximation for 3 $sp^2$ orbitals at each carbon) then it would be only 3 sp orbitals left over. 3) I'm not familiar with hybrid orbitals: from Google I know that sp hybrid orbitals have a shape (I know orbitals are not real :) ) like p orbitals. What do you mean with 180°?

google cyclopropane bond angle, most values are 114-116° (example); 2) OK, now I see what you're doing, but again, the C-H bonds are more like sp^2.5; 3) if the C-C bonds are sp hybridized (they are not) you would need 2 sp orbitals on each carbon and they would point 180° from each other, impossible in cyclopropane

Shouldn't the biradical have a triplet ground state and therefore two (degenerate) HSOMOs and possibly two (degenerate) LUMOs?  Or If you'd like to argue a biradicaloid, where the ground state invokes an excited configuration (strongly correlated electrons)? At least the wave function would have to be treated with more than one configuration and the hybridisation concept is very likely to break down. Even in UHF treatment you have two determinants.


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Chemistry of cereal in milk
In a nearly finished bowl of oat cereal, why do the pieces of cereal stick together? I figure that since they are composed of starches it could be hydrogen bonding. Is this true or is there some other reason?

Surface tension, not hydrogen bonding, is the primary phenomenon which causes cereal rings such as Cheerios to adhere together while floating in milk.
The total linear perimeter of the 3-phase cereal-air-milk interface is minimized when pieces cling together.
Here are some links:

http://www.livescience.com/9350-cereal-science-floating-objects-stick.html
http://phys.org/news/2010-09-well-known-effect-breakfast-cereal-physicists.html
https://en.wikipedia.org/wiki/Cheerios_effect


text:
Why does the Gibbs free energy only correspond to non-expansion work?
It has been defined as the energy available for work other than expansion work. Why can't it be used for expansion work

The association of the Gibbs free energy with “additional”, or “non-expansion” work, is simply a mathematical result of its definition: $G = H - TS = U + pV - TS$.
From the First Law, for a closed system, we have $\mathrm{d}U = \delta q + \delta w$.
For a reversible change, the Second Law tells us that $\mathrm{d}S = \delta q_\text{rev}/T$; so $\delta q_\text{rev} = T\,\mathrm{d}S$.
We can split $\mathrm{d}w$ into two parts, expansion work and additional work: expansion work is given by the formula $\mathrm{d}w_{\text{exp}} = -p\,\mathrm{d}V$, and for now we can just label additional work as $\mathrm{d}w_{\text{add}}$. In this context, additional simply means anything that isn’t expansion work. Putting all this together, we have
$$\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V + \delta w_{\text{add}}$$
and from the definition of the Gibbs free energy in the first paragraph,
$$\begin{align}
\mathrm{d}G &= \mathrm{d}(U + pV - TS)\\
&= \mathrm{d}U + (p\,\mathrm{d}V + V\,\mathrm{d}p) - (T\,\mathrm{d}S + S\,\mathrm{d}T)\\
&= T\,\mathrm{d}S - p\,\mathrm{d}V + \delta w_{\text{add}} + p\,\mathrm{d}V + V\,\mathrm{d}p - T\,\mathrm{d}S - S\,\mathrm{d}T\\
&= V\,\mathrm{d}p - S\,\mathrm{d}T + \delta w_{\text{add}}
\end{align}$$
Now, if a closed system is kept at constant pressure and temperature, $\mathrm{d}p = 0$ and $\mathrm{d}T = 0$. Therefore $\mathrm{d}G = \delta w_{\text{add}}$.
The simplest application of this interpretation of $G$ is in electrochemistry, where one could move $|\mathrm{d}n|$ moles of electrons through a certain potential difference $E$; the “additional work” done in this case is given by $-FE\,|\mathrm{d}n|$ (you could consult a physics text for details). This gives us $\mathrm{d}G = -FE\,|\mathrm{d}n|$, and (after several more steps) the familiar equation $\Delta_\mathrm{r} G = -zFE$.

While writing the expression for U, why dq has been replaced by dq(rev), does that mean Gibbs free energy will be maximum non-expansion work only for a reversible process?


text:
Have P-acyl phosphines ever been synthesized?
I've done some amount of searching and can't seem to find any evidence in the literature that there exists a phosphine analogue of carboxyl amides. I'm looking for compounds of the form $ \ce{ R(C=O)PR_{2}} $; basically a moeity similar to a primary or secondy amide but with a phosphorous atom instead of a nitrogen.
I would also wonder what the proper name for these functionalities would be. Taking a shot in the dark I might guess "phosphomide", "acyl phosphine", or "acyl phosphinide".
I would like to specifically disqualify Acyl-phosphine oxides or any compounds where the acyl-connected phosphorous is also connected to an oxygen.

When I saw this question, I realized I couldn't remember ever seeing an acylphosphine, but I was curious and went looking. I found that they're actually quite easy to make. One common thing I saw was the use of trimethylsilylphosphines with acyl chlorides to form acylphosphines.
The easier method, however, was simply to react a substituted phosphine in the presence of triethylamine with an acyl chloride. The analogous reaction can be done with amines to form amides. In this paper, they describe the synthesis of P-acetyldiphenylphosphine using diphenylphosphine, triethylamine and acetyl chloride. They have some other examples in there, but this suggests that these compounds are typically fairly trivial to synthesize.
One more note: In my searches, I saw "acylphosphine," "acyl phosphide," and "phosphomide" as named. The paper I linked uses the term "phosphomide."


text:
Can a precipitate dissolve upon addition of a diluted acid?

Is it true that precipitates dissolve upon addition of diluted acids?
If a precipitates is dissolve , does it disappear from solution , or  any thing else  happened out there?

I  am just curious about this. Because I have heard that precipitates can be dissolved if you  add a diluted acid. But why it would happen? And is it valid for any diluted or strong acid ?.Do all the precipitates dissolve upon addition of any acid ?.I would like to know more about this.

Another example of a precipitate dissolving in acid is the reaction $\ce{AgNO3 + HCl -> AgCl (v) + HNO3}$. At first, the silver chloride precipitates out as a white powder (if kept away from direct light). Adding more acid, though, forms ligands with the additional chlorine ions:  $\ce{AgCl(s) + Cl- (aq) -> AgCl2- (aq)}$. See also Physics Forums.

Then if ligands is formed upon addition of excess dil. acid does the precipitate dissolve?

Yes; see the references.

I would like to answer your second question first:
When a precipitate dissolves, it basically reacts to form some other soluble compound.
For your first question:
In many cases, a precipitate is soluble in a dilute acid solution. For example:
$$\ce{BaSO3 + 2HCl (dilute) -> BaCl2 (soluble) + SO2 (evolve) + H2O}$$
However is not necessarily true that a precipitate will dissolve in a dilute acidic solution. For example, calcium carbonate on reacting with dilute $\ce{H2SO4}$ evolves $\ce{CO2}$:
$$\ce{CaCO3 + H2SO4 -> CaSO4 (insoluble) + H2O + CO2 (evolve)}$$
It depends on the hydration enthalpy of the (would be) product after dissolution of a precipitate, whether the precipitate will be soluble or nor. In these examples, hydration enthalpy of $\ce{BaCl2}$ is −8.8 kJ/mol and that of $\ce{CaSO4}$ is −73 kJ/mol.

How can I relate hydration enthalpy to this?


text:
Trans effect on square planar complex
We've learnt about the trans effect but I have a question regarding how this plays a role with water in the equation.
trans-effect:
$$\ce{H2O < NH3 < Cl-}$$
Based on this, given cis-$\ce{[Pt(NH3)2Cl2]}$, if you were to add water to form a dicationic complex, would a substituion reaction occur between all the substituents, and then $\ce{Cl-}$ reacts again to form the trans-complex?
Since water has a weaker trans effect than both $\ce{NH3}$ AND $\ce{Cl-}$, I can't imagine that a cis-complex would form as any cis-combination of $\ce{NH3}$ and $\ce{Cl-}$ has a stronger trans effect than any combination of $\ce{H2O / Cl- / NH3}$.
Are there any other kinetic / electronegativity factors that I am missing?

Just to clarify.. You're asking if you replace both $\ce{Cl-}$ ligands with two $\ce{H2O}$ ligands, what would you get? I indicate the chloro ligand here because you said "dictation". I'm just trying to get the sequence of reactions (i.e., first one, then two $\ce{H2O}$ to get $\ce{[Pt(H2O)2(NH3)2]}$ and then react with $\ce{Cl}$

It's really not clear to me what sequence of reactions is intended in your question, making it very hard to answer. If you list each reaction, I can probably help.

Related: http://chemistry.stackexchange.com/a/18409/5017


text:
Proper use of vacuum pump during vacuum distillation
I am preparing to do a vacuum distillation and am unsure exactly how to manage the vacuum. The vacuum pump I have is a two-stage rotary vane pump which is capable of 250 microns or lower. It has a triple inlet, one 1/4", 3/8" and 1/2".
My glassware is typical distillation glassware including a Graham condenser and a vacuum takeoff joint that connects between the condenser and the receiver.
My concern is that if I apply too much vacuum that it will cause a glass implosion. Is this not a concern and I can just operate the vacuum pump freely, or do I need to take special steps to limit the degree of vacuum?
I watched a YouTube video of a guy doing a vacuum distillation and I noticed that he did not use a capillary bubbler. Under what circumstances can the bubbler be dispensed with?
Another concern is the need for a manometer. I have read that I need to use a closed-end manometer (very expensive) to measure the pressure in a vacuum distillation. Why can't I use a normal vacuum gauge? Also, if I am doing an azeotropic separation such as a water-ethanol separation, is it critical I monitor the pressure? I am assuming it is.

@Jan I do not have enough reputation to create new tags. There is no distillation tag. There is no vacuum tag even. Pretty lame that a forum supposedly about chemistry has no tag for distillation.

There is an [tag:experimental-chemistry] which would have fitted your question a lot better.

Currently, the tag [tag:distillation] currently only refers to this question. We just graduated not so long ago, so it is not that unusual that we do not yet have every tag that we might need as a community. There are 44 questions to which this tag could probably be applied, so it looks like a tag we might need. However, we do not need duplicate tags. If all questions tagged distillation also share another tag, then we may not need the new tag.

@Ben Norris: I suppose your reasoning is technically correct, but I do think distillation is an essential chemistry tag, and I was surprised to hear we did not have this tag, which is why I added it.

@trb456 I think you were right to add it, since there are 44 questions with distillation in the title. I was worried that all distillation questions might also be tagged purification, but that does not seem to be so.

I very much disagree, that distillation needs its own tag, so I removed it.

@Martin-マーチン: Why do you very much disagree? Distillation is an essential chemistry tag, in my view. What is an acceptable substitute, which must be a larger category than distillation? Purification? That seems too general given how important distillation is. And p.s. I voted for you as moderator, so I highly respect you, but this call seems wrong.

@trb456 Thank you for your trust. I saw that you have found the meta post, where I admit that I was wrong here. I should have left a more elaborate comment here in the first place. I am rolling back my edit here, so that the tag gets recreated. If you have the time, it would be nice, if you could write a short tag wiki.

@Martin-マーチンI just added the distillation tag back. Perhaps you or someone you know could approve? ;-)

This pump is way overkill for a vacuum distillation. The problem is not glass implosion necessarily (round bottom jointed glassware is designed to be put under vacuum, less clear on the condenser), but rather that you may not be able to condense your product given the low vacuum. Vacuum distillation is typically done with an aspirator that gets down to 25 Torr (25,000 microns) or so. At too low a pressure, you may need a very cold receiving vessel and condenser, and depending upon the solvent dry ice slush may not even be enough. You need to determine the boiling point of your solvent at low pressures to figure this out. Your pump is for Schlenk lines and similar applications.
I don't see a reason not to use a vacuum gauge (and you are not supposed to use mercury anymore in any application), though again, depending upon the setup, it is possible to harm the gauge. You should try to monitor pressure as part of the experimental process.
Not using the capillary bubbler seems foolish, as bumping is a major problem with vacuum distillations. A capillary bubbler is a very thin tube (like when you draw glass tubing after melting it) that goes into one joint of the stillpot (drawn end in the reaction mixture). When vacuum is applied, air is forced through the tube, but because the drawn end is so narrow it only allows small bubbles into the mixture. These take the place of boiling chips which will not form bubbles as all gases are extracted from the chips via the vacuum.

Right, I have been figuring that out. I either have to get a "dry" vacuum pump with a regulator, or else use a vacuum regulator with my existing pump. The problem there is that a 3 CFM vacuum regulator is pretty expensive.

How about a water aspirator? Cheap!

Just looked up the 3 CFM, and unfortunately this is probably what you need for your overkill pump. You need a weaker pump! I don't know where you are working, but if you are in an educational or professional lab, do you have access to house vacuum? That tends to be low enough for vacuum distillation.

@ShakaBoom: If this answer is sufficient, could I ask you to accept it? Thanks in advance!

@trb456: Could you clarify what a capillary bubbler is and how it should be configured in a vacuum distillation apparatus?  (Or address the same question here?)  I can't find any component by that name, though I know some bump countermeasure is required for vacuum distillation!

@feetwet: Please see edited answer!

Oof, I definitely don't agree that aspirator is nearly exhaustive for vacuum distillation use. Ever distill DMPU? Heck, I had to reflux HMPA under high vac in undergrad.

If you can, use a digital vacuum guage. I never use a bubbler as it can draw air in and introduce oxygen into your material as a result.  Not ideal. You could try and use nitrogen through the capillary,  but trying to manage the flow balanced with the vacuum drawing it in is a pain. Plant vessels never use bubblers and they distil under vacuum just fine.

Apparently continuous stirring is an adequate substitute for a bubbler.


text:
test tube cracked when added glycerin to potassium permanganate
I am not expert in chemistry.I learned about combustion in reaction when added glycerine to potassium permagnet.I tried it at home, I took potassium permagnet in test tube and when I added glycerine to it test tube just CRACKED as combustion started.I want to know why, because it lab we some time directly heat test tube, it does not cracked so why this time..
I am not sure I should continue on sulfuric acid and potassium parmagnet test in a TEST TUBE after seeing this affect...
Any Suggestion..

I am not sure I should continue on sulfuric acid and potassium parmagnet test in a TEST TUBE after seeing this affect...

In case of doubt you should not. Anyway, I recommend to avoid doing it with any significant amount. It may and will spread around and even if not it produces manganese (IV) oxide smoke that settles down nearby and is quite hard to wash out.

I want to know why, because it lab we some time directly heat test tube, it does not cracked so why this time..

Potassium permanganate is a strong oxidizer, glycerol is a reducer. Mixed together they quickly start to react. The exact nature of the products is debatable, but the mixture self-ignites if warm. The thermal shock is enough to crack glass. This is a common weakness of glass and ceramics.
Furthermore, in case of bad luck it may produce enough gases to spread the mix around. Consequently, safety precautions must be always employed (incombustible table, safety goggles etc). A movie how it may go http://www.youtube.com/watch?v=6F4cuQkKt2U . The reaction has some quirks, specifically if the glycerol is cold, it may not start immediately, so a rather well mixing is achievable. Now, once it warms up to room temperature, it suddenly bursts in flame, potentially delivering itself to nearby surfaces, including your face.


text:
Maximum observed optical rotation through a polarimeter
Is there a maximum optical rotation of plane polarized light that can be observed using a polarimeter?
I understand that in a polarimeter, light passes through a filter that converts it into plane polarized light. This type of light only oscillates in one plane. When it is passed through an optically active solution, the plane is rotated due to interactions between the molecules. To measure this rotation, we use another a filter and try to orient it so that light passing through it appears brightest. The difference in angular orientations of the two filters is read as the observed angular rotation.
Here is an image I've taken from Solomon and Fryhle's Organic Chemistry textbook:

So is there a maximum value for this rotation? Surely after a 90 degrees rotation, the angle would be the same as another angle is we'd rotated in the opposite direction. For example, I could say that a +100 degrees rotation is the same as a -10 degrees rotation.
Could the angle be greater than 90 degrees?
The question might seem simple, but it's difficult for me to take into consideration multiple rotations of the plane polarized light before it passes through the second filter. Are the multiple rotations even possible?

This is something I discovered after an attempt at a more mathematically rigorous investigation of how to interpret polarimetry results. It's not that hard, but I don't think you'll ever find a similar analysis in a book because in practice the details are pretty much unnecessary for experimental purposes (though fun mathematically!).
Before we begin, it's very important to distinguish two types of optical rotation. The first is the real optical rotation, $\alpha_{real}$. This is the twist which a substance actually causes on the plane of polarized light as it travels through the optically active solution, and therefore is the rotation which is part of the specific rotation equation, shown further below. The numerical value of this angle can be any finite real number, in degrees or radians. There is no constraint on how much an optically active solution is "allowed" to twist a plane of polarized light; the fact that you can't directly see how many complete twists the plane of polarized light made is entirely an experimental limitation.
This experimental limitation is what gives rise to the other type of rotation: the observed or apparent rotation, $\alpha_{obs}$. This is an angle which is also real number, but which is limited between $0$ and $180$ degrees or $0$ and $\pi$ radians (or you could say between $-90$ and $90$ degrees or $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ radians, it makes no conceptual difference). The observed rotation is what you measure in a polarimeter, and in principle it cannot be directly put into the specific rotation equation.
Why is $\alpha_{obs}$ allowed only in the range $[0, 180]$ degrees? Because you can't observe a difference between an angle $\alpha_{obs}$ and $\alpha_{obs} + 180\times n$ (where $n$ is a positive or a negative integer). Look at a figure such as this (source) and try to convince yourself of this. If it helps, remember that two identical polarizers on top of eachother with an angle of $0^\circ$ let light through completely, but rotating one will block more and more light until they are perpendicular ($90^\circ$) and no light at all comes through, at which point increasing the angle will brighten the light until it reaches a maximum again at $180^\circ$ before once again starting to dim. This means that, in polarimetry the observed brightening or dimming by rotations of plane-polarized light are cyclic with period $180^\circ$, not $90^\circ$ as you suggest.
Turning the previous sentence into a simple equation, we obtain:
$$\alpha_{real} = \alpha_{obs} + 180\times n \ \ \ , \ \ n \in \mathbb{Z} $$
(The angles in this answer will always be treated in degrees, though for simplicity, in the equations I shall forget about the $^\circ$ unit and treat angles as pure numbers*).
Now we use the equation for optical rotation to find the specific rotation $[\alpha]^T_\lambda$, which characterizes a substance when measured at a temperature $T$ with a monochromatic light source of wavelength $\lambda$:
$$\alpha_{real} = [\alpha]^T_\lambda \times l \times c$$
, where $l$ is the optical path and $c$ is the concentration of the (enantiomerically pure) optically active substance. Substituting that into the previous equation:
$$[\alpha]^T_\lambda = \frac{\alpha_{obs} + 180 n}{lc}$$
So for each individual measurement of $\alpha_{obs}$ made, there are infinite solutions for $[\alpha]^T_\lambda$. This comes as no surprise.
But if we take several measurements, then we can find a value of $[\alpha]^T_\lambda$ as a unique solution to the equation, right? Speaking in strictly mathematical terms, no. Any finite amount of measurements, even with perfect accuracy and precision, will still yield infinite possible values of $[\alpha]^T_\lambda$. In fact, there are infinite levorotatory and dextrorotatory solutions, so in principle you can't even determine the direction of rotation by this method! See the graph below:

EDIT: In my previous graph, I incorrectly equated the values of $\theta$ with $\alpha_{real}$. The actual identity comes from $\mathrm{tan}\ \theta = \frac{d\alpha_{obs}}{dc} = \frac{d}{dc}(\alpha_{real}+180n)=\frac{d\alpha_{real}}{dc}=\frac{d}{dc}(lc[\alpha]^T_\lambda)=l[\alpha]^T_\lambda$, from which $\theta = \mathrm{arctan}\ (l[\alpha]^T_\lambda)$. This however does not change the theoretical development of the answer.
Note that since the measurements are cyclic with period $180^\circ$, straight lines wrap around. The two black points are two measurements of $\alpha_{obs}$ at two different concentrations. There is an "obvious" red line connecting them, but it need not be the correct one. There are infinite lines which cut both points; I have represented two more perfectly valid solutions with the green and blue lines. Each line which cuts all points represents a possible solution, with an inclination given by the angle $\theta = \mathrm{arctan}\ (l[\alpha]^T_\lambda)$ (see derivation above). From this, $[\alpha]^T_\lambda = \frac{\mathrm{tan}\ \theta}{l}$. Interestingly, the red and blue lines are represent dextrorotatory solutions to the problem, while the green line indicates a levorotatory solution for the same compound. There are infinite more solutions, characterized by steeper lines which wrap around more times. See the comments for information about how they come into play.
However, in practice, materials only have a finite and usually relatively small ability to perform optical rotation. Most of the more commonly encountered substances have a value of $[\alpha]^T_\lambda$ which tops out at a few hundreds. Some extreme examples are helicenes, for which specific optical rotations can reach as high as $\approx 10000^\circ$ in [13]helicene. After performing enough measurements (especially if you do them intelligently, by not measuring at different concentrations which are related to eachother by simple fractions), then you basically reach a situation where you can find an infinite family of solutions such as $[\alpha]^T_\lambda = -534 + 462785\times n$, in which case the true specific rotation is probably levorotatory with a value of just $-534^\circ$.
As a small additional comment, usually the filters in an optical polarimeter are rotated until the least amount of light comes through as the human eye is better at detecting small differences in light levels in darker conditions, which will make for a more accurate measurement of the observed rotation.

Originally I had answered this question by bringing up modular arithmetic, but while it's interesting it can possibly make the answer harder to understand at first. I have relocated my previous argument here.
The first equation shown is equivalent to:
$$\alpha_{obs} = \alpha_{real}\ \ (\mathrm{mod\  180})$$
The $\mathrm{mod\  180}$ part comes from modular arithmetic, which may sound scary and new, but it's just normal mathematics, with the small quirk that every time a number appears out of the range $[0, 180]$, you just add or subtract $180$ as many times as necessary for the answer to appear between $0$ and $180$.

* In reality some of the $\alpha_{real}$ and $\alpha_{obs}$ variables in certain equations above would have to be divided by $^\circ$, so that all equalities are dimensionally valid. Equivalently, the modulus could be expressed with the unit as $\mathrm{mod\  180^\circ}$, but I'm not sure if the modulus is allowed to have physical dimensions, even if I don't see any issue with it.

Suppose I change the concentration by 10%, the result will also have several equations, but only 1 from each situation will fit the data. Therefore, if a compound has a specific rotation of -200°, then that is the rotation for that compound, not 160°.

@LDC3 There will always be an infinite amount of lines passing through all points, though almost all lines will be extremely steep, represented by very high specific rotation constants. I have tried to illustrate this by adding a third point to my graph. Obviously it is collinear with the red line, but not with the blue or green line. However, among the previous infinite solutions, there was a very steep orange line I had not drawn, which goes through both the initial two points and the new third point. Picking other solutions is always possible.

Your comment is insightful though; making at least two measurements in close concentrations is precisely the best way to eliminate large swathes of physically plausible solutions, tempered only by possible measurement inaccuracies.

Errata: As mentioned in the edit section of my answer, please note that the graph linked in the comments also incorrectly labelled the angle $\theta$ with $\alpha_{real}$. However, once again, this does not alter the previous argumentation.

Just for clarification: Your argument does not invalidate the determination of purity by optical activity, does it? If the literature measures $+11.9$ and I get $+5.95$, and I know from the mechanism that racemisation is impossible (too many stereocentres) I still know that my compound is $50:%~m/m$, don’t I? Same for e.e. measurements.

@Jan I hadn't considered a case with mixtures of enantiomers just to keep the argument as simple as possible, but I think all these calculations readily extend to enantiomerically impure mixtures (so long as there is only one pair of enantiomers, of course), with the added detail that the concentration being measured is the concentration solely of the excess enantiomer. Some other measurement must be performed to relate this excess concentration to the total concentration of both enantiomers.

@Jan That said, if you want to determine purity and enantiomeric excess, you generally start off already knowing what the specific rotation is, so very little of this involved analysis is really necessary!

Ah, but one doesn’t know if the literature value is the one one could have guessed or not — the exact point of the discussion.

Could you expand on "Most of the more commonly encountered substances have a value of [α]Tλ which tops out at a few hundreds", Is this due to the limit in how concentrated we can make an optically active substance?, can't we then instead compare the pathlength so that we don't reach this limit? or would there be a limit also there?

This was aimed to @NicolauSakerNeto , sorry didn't specify

@AndreaRowlatt The specific rotation is an intensive property of a material (i.e. doesn't change with amount), which measures its "innate strength" to twist polarised light. Why most materials have relatively small values of specific rotation is a deep question, for which I have no good answer. It just seems that, in our system of units, "maximal twistiness" seems to be around $\mathrm{100,000°\ mol^{-1}dm^{-1}}$ for relatively small molecules, and since "random" molecular structures aren't expected to have high specific rotations, most compounds are closer to zero.

@NicolauSakerNeto If I understand correctly then specific rotation is the strength to "twist" polarized light in a substances' given concentration. But would the pathlength need to be fixed to make this comparison? Would there be a similar limit with the pathlength so that the relationship of: α=[α]Tλ×l×c   only works for a limited path length (l), where 0< l < d  and d is the limit specific to a substance?

@AndreaRowlatt First I must clarify a typo in my previous comment; the unit of measurement for the specific rotation should have been  $\mathrm{°\ M^{−1}\  dm^{−1}}$, or equivalently but more obscurely, $\mathrm{°\ dm^2\ mol^{−1}}$ (recall that $\mathrm{1\ L=1\ dm^3}$). Thus, by definition, the specific rotation is "normalized" with respect to both concentration and path length; doubling the path length makes the same difference to the specific rotation as doubling the concentration (that is to say, none). It is only the real rotation (measured purely in degrees) which changes.


text:
Ph of acid would be one?
I am interested in the chemical lsd or "acid". I wanted to know if lsd is acid in its pure form would that mean it is water?

No! Don't get fooled by the slang term.
LSD is the abbreviation for lysergic acid diethylamide. However, the parent lysergic acid (see structure) is indeed a carboxylic acid.


text:
Pseudo-asymmetry-like situation in alkenes
We have pseudo-asymmetric carbons in optical isomerism which dictates that if 2 groups on the carbon atom are different in terms in configuration then the carbon is chiral. In alkenes can the same thing happen with respect to geometrical isomerism? For example consider
If both the alkenes on the left were E or both Z, then certainly there wouldn't be any chance of geometrical isomerism. But as in this case if the configurations are different ,i.e, E and Z, can the alkene be contended for geometrical isomerism ,i.e., whether the compound is a Z or E isomer in terms of the middle alkene (to which a methyl group is attached)?
While giving the answer to this question, it would be extremely helpful if you cite the source of your information for me to look it up.

Related: How to determine priorities for this compound according to the Cahn-Ingold-Prelog rules?

@Loong: It's not a question of finding the priorities but rather would even one be applicable in this case.

The second answer to that question answers your question exactly though. If one is E and one is Z, the Z substituent would have higher priority.


text:
How are point group character tables typeset correctly?
I want to know how to typeset character tables of point groups correctly. Consider the following two extremes using the $C_{\text{2v}}$ (or is it $\text{C}_{\text{2v}}$?) point group as an example.
Firstly, one where all the symbols (except for the symbols denoting vertical) are cursive (your default LaTeX math mode behaviour):

Second, a table where all the symbols are in roman font.

Which is correct? Or is it a mixture?

In this book about group theory written for mathematicians and physicists they use upright letters for the irreducible representations and "group names", e.g. $\mathrm{B}{3\mathrm{u}}$ and $\mathrm{D}{6\mathrm{h}}$, while the symmetry elements are typeset in italics but subscripts that do not correspond to variables get an upright font, e.g. $\sigma_{\mathrm{d}}$ and $C_{4z}$.

Notations and conventions used for the description of symmetry in rigid molecules are established in Notations and conventions in molecular spectroscopy: Part 2. Symmetry notation (IUPAC Recommendations 1997):

Symbols for symmetry operators are printed italic, with sub- and super-scripts that are upright except for the variables $k$ and $n$ which are replaced by numbers for a specific operator.

For example:
$E$ – identity operator
$\sigma$ – reflection operator for reflection across a general plane
$\sigma_\mathrm{h}$ – reflection operator for reflection across a horizontal plane
$\sigma_\mathrm{v}$ – reflection operator for reflection across a vertical plane
$C_n{}^k$ – $n$-fold rotation operator for $k$ successive rotations through an angle of $2\pi/n$ about an $n$-fold rotation axis, where $n = 2, 3, \ldots$; $k = 1, 2, \ldots, (n-1)$
$S_n{}^k$ – $n$-fold rotation-reflection operator for $k$ successive rotation-reflections about an $n$-fold rotation-reflection axis for a rotation through an angle of $2\pi/n$ followed
by a reflection in a plane perpendicular to the axis, where $n = 2, 3, \ldots$; $k = 1, 2, \ldots, (n-1)$
$i$ – inversion operator through the centre of symmetry
Likewise,

Symbols for symmetry groups are printed italic with upright subscripts, following common usage.

For example:
$C_n$, $S_{2n}$, $D_n$, $D_{n\mathrm{h}}$, $D_{n\mathrm{d}}$, $C_{n\mathrm{v}}$, $C_{n\mathrm{h}}$, $T$, $T_\mathrm{h}$, $T_\mathrm{d}$, $O$, $O_\mathrm{h}$, $I_\mathrm{h}$, $C_{\infty\mathrm{v}}$, $D_{\infty\mathrm{h}}$
However,

Symbols for irreducible representations of point groups, which are usually called symmetry species in spectroscopy, are printed upright.

For example:
$\mathrm{A}$, $\mathrm{B}$, $\mathrm{E}$, $\mathrm{A}_1$, $\mathrm{A}_2$

The example table given in the question should thus be typeset as follows:


text:
How can I figure out whether a reaction is endothermic or exothermic without its enthalpy value?
I already know the way to determine whether if the given reaction is exothermic or endothermic by the enthalpy values. But is it possible without enthalpy values, and just by looking at the reaction?
So as far as I know there are two types of chemical reactions, namely one way reactions and reversible reactions.
So how can I determine this factor in one way reactions and reversible reactions?
I guess that for reversible reactions we can use equilibrium constants such as $K_p$, $K_c$, but I am not quite sure whether it is correct. On the other hand, I feel that there is no such way to determine what I want in one way reactions. So how can I decide it?

Hello everyone , I want to know whether the English that I have used  here  is  good enough or is it horrible?Because I am not a native Englishman.If so please show me that mistakes also, while you  helping me to come out of this question.

I don't think there is a way. The concept of enthalpy came to be for a reason.

You can make an educated guess based on the relative strengths of the bonds broken vs bonds formed.

@On the way, your English is quite good. It might need some polishing, but then what are the editors for? So no worries. :) Also, regarding your question, if one knows about the bond strengths in both reactants and products (for example they'd say an F-F bond is way weaker than say, an H-H bond) , they could decide the sign of enthalpy, but it would be very very crude and it could lead to wrong results. Unfortunately, I have nothing else related in my mind right now.

@M.A.Ramezani I agree that using bond strengths would lead to a very crude estimate, but if you only want the sign of $\Delta H$ and not the magnitude, for 99% of reactions it should suffice. The 1% would be when your estimated $\Delta H$ is close to 0. Regarding the question itself, I'm just nitpicking here but all reactions are reversible except for reactions where a product escapes from the system. As such, you can theoretically use le Chatelier's principle to determine whether the reaction is exothermic or endothermic based on the change in equilibrium position when temperature changes.

@M.A.Ramezani:  Then how about comparing equilibrium constants at different temperatures? (by the way thank for check-up you did for my question)

After thinking about it I think I have exaggerated quite a bit. If you only had access to a table of bond energies you would only be able to estimate enthalpies for gas-phase reactions; for any other reactions involving solid/liquid/aqueous reactants you would have to consider lots of other factors such as intermolecular forces, lattice energies, enthalpies of hydration etc... After lots of experience in chemistry it might be possible to identify whether a reaction is exo/endothermic at first glance, e.g. thermal decomposition of CaCO3 is endothermic. But in the absence of data it's not easy.

@ orthocresol: How about using equilibrium constants?

If you know the signs of the Gibbs' energy change and entropy change you could use Δ G = Δ H - T Δ S. For instance, if you know that the reaction is spontaneous (Δ G is negative) and the products have a lower entropy than the reactants (Δ S is negative), then you know that the reaction must be exothermic because Δ H must be negative.

@ Derek: Okay so , is it possible to find if a reversible reaction is exothermic or endothermic by Kp and Kc  value.?

How about going into the lab, mixing the reactants, and observing whether they react, and if so, does the flask warm up.

You can use the value of the equilibrium constant to determine Δ G. If K is 1 then Δ G is 0, if K is greater than 1 then Δ G is negative, and if K is less than 1 then Δ G is positive. This is because Δ G = -RT ln K. Δ G can then be used to find Δ H, per my previous comment.

It means , there is no way to compare Kp and Kc  in-order to have a clue about the type of reaction  thermodynamically.(either exothermic or endothermic)?

Then how about comparing two Kp values at two different temperatures ?

First of all, the equilibrium constant $K$ in general doesn't tell too much about the enthalpy change of the reaction as it is related to the (Gibbs) free energy change $\Delta G = \Delta H - T\Delta S$. It is true that in some simple cases (such as gas phase reactions) the  entropy change can be approximated by looking at the stoichiometry of the reaction but in general it's usually more complicated.
Regarding your distinction between one-way and reversible reactions I think you may as well assume all reactions are reversible, but some have incredibly huge $K$ so the equilibrium is shifted totally to one side of the reaction - this is the result of a very negative $\Delta G$.
In case of a "reversible" reaction you can always look at the change of the equilibrium constant with temperature. The constant is
$K = \exp\left(-\frac{\Delta G}{RT}\right)=\exp\left(-\frac{\Delta H}{RT}\right)\exp\left(\frac{\Delta S}{R}\right)$
so from the temperature dependence you can deduce the sign of $\Delta H$. Naturally, this is assuming that $\Delta H$ and $\Delta S$ doesn't depend on the temperature in the temperature range you're looking at.

The figure shows the change of equilibrium constant with temperature (arbitrary units) for an exothermic (yellow line) and an endothermic (blue line) reaction.
Just by looking at the reaction I don't think you can tell the enthalpy change. Normally you have bonds breaking on the reactant side and others forming on the product side and normally there is a delicate balance between the two. People have come up with bond strength values (which could be determined from known $\Delta H$ values) and you can use those in your reaction.

How about comparing equilibrium constants  at two different temperatures ?

But can't we compare  Kp ,Kc for  different temperatures , and then come to a conclusion?

Yes, you are right. I have modified the post so it makes much more sense.

If you would give few examples that would be great.

Then how about comparing two Kp values at two different temperatures ?

If I compare Kp value at two different temperatures would I be able to find whether the reaction is exothermic or endothermic?

Well, yes. Look at the figure I added to my answer, it show how the equilibrium constant changes with temperature for exothermic or endothermic reactions.

Hey albapa , is this valid for Kp and Kc?

You need to look at the temperature dependence of Kp and Kc. I guess Kp has an extra temperature term so it's slightly more complicated.

Anyhow, can I use this theory to determine what I want? regardless of Kp or Kc?

If the enthalpies of formation for all reactants and products are known then you can compute the standard enthalpy change ($\Delta H^\circ$).
If they are not known you can estimate $\Delta H^\circ$ using bond energies as described in this video.
Finally if you happen to know that the reaction is spontaneous ($\Delta G^\circ < 0$) and if the standard entropy change is (likely) negative (e.g. if the number of particles decrease) then it must be an exothermic reaction.

not sure what you mean by "one way" and "reversible" reactions

@ Jan Jensen:I used one way to indicate the reactions which have one solid forward arrow, while reversible for ones which are in equilibrium.

$\ce{A -> B}$ means that at equilibrium there is more $\ce{A}$ than $\ce{B}$, while $\ce{A} \rightleftarrows \ce{B}$ suggests that there is roughly an equal amount of  $\ce{A}$ and $\ce{B}$.  However, both usually imply equilibrium $\ce{A <=> B}$.  Either way, thus does not affect how you would estimate endo/exothermicity.


text:
Is this sydnone aromatic?
Is the following molecule aromatic?

It does seem like it will be aromatic because it is similar to furan. I am confused because of the positive charge on the $\ce{N}$.

The pyridinium cation is aromatic and also has a positive charge on nitrogen. So, positive charges by themselves do not preclude aromaticity.

The molecule in your question is a sydnone. For those molecules you can find evidence that suggests that they are not aromatic (see here) even though Huckel's rules may tell you otherwise.

Yes it's rather tricky and good question. This mesomeric structure suggests aromaticity, but there are many other...

As correctly pointed out in the comments, the compound you present belongs to the class of sydnones. According to Hückel's rule, it should be aromatic, as the molecule is planar and contains $4n+2$ π-electrons.
However, calculations performed in 1998[1] that mainly looked at chemical shifts and inferred aromaticity from the magnitude of observed diamagnetic shielding, show that sydnones in fact are not fully delocalized, i.e. they are not aromatic. To quote from the linked publication:

Mesoionic compounds  are  planar  five-membered  heterocyclic  betaines
  with at least one side chain whose α-atom is also in the ring plane and with dipole moments of the order of 5 D. Electrons are delocalized over two regions separated by what are essentially single bonds. One region, which includes the α-atom of the side chain, is associated with the HOMO and negative π-charge whereas the other is associated with the LUMO and positive π-charge.

The publication is also riddled with images that depict syndones like this:

It ends with another quote from Gilchrist's book "Heterocyclic Chemistry", 2nd edition, 1992, p. 329:

It does not therefore seem useful or appropriate to classify sydnones and other mesoionic compounds as aromatic, despite the existence of a cyclic array of p-orbitals.

Reference

Simas, A. M.; Miller, J.; de Athayade Filho, P. F. Are mesoionic compounds aromatic? Can. J. Chem. 1998, 76 (6), 869–872. DOI: 10.1139/v98-065.

One of Paul Schleyer's students, Matthew Duvall, has studied the aromaticity (or lack thereof) of mesoionic compounds, such as sydnones and münchnones, using various techniques.[1] In his thesis, Duvall directly refutes the article linked in tschoppi's answer (emphasis mine):

In contrast to the major arguments of Simas et al. (the endocyclic C–O bond with little or no π character and the fact that mesoionic compounds had well-separated regions of negative and positive charges separated by what were essentially single bonds), sydnones and münchnones are aromatic and are comparable to the aromaticity of benzene, furan, and pyrrole by magnetic (NICS(0)πzz) and energetic (ASE) criteria.
$$\begin{array}{cccc}
\hline \text{Molecule} & \text{NICS}(0)_{\pi zz}\text{ / ppm} & \text{ASE / }\pu{kcal mol-1} & \text{Point group} \\ \hline
\text{Benzene} & -36.3 & 31.0 & D_\mathrm{6h} \\
\text{Pyrrole} & -33.5 & 16.7 & C_\mathrm{2v} \\
\text{Furan} & -28.9 & 12.5 & C_\mathrm{2v} \\
\text{Sydnone} & -23.7 & 27.0 & C_\mathrm{s} \\
\text{Münchnone} & -20.6 & 28.2 & C_\mathrm{s} \\ \hline
\end{array}$$

Negative nucleus-independent chemical shift (NICS)[2] values correspond to magnetic shielding in the middle of the ring, which you would expect to see in an aromatic compound with ring currents (this is strikingly illustrated in [18]annulene). On the other hand, aromatic stabilisation energies (ASEs)[3] refer to the difference in energy between the compound and a non-aromatic isomer, although the exact details are somewhat more involved. For example, these were the reactions used to determine the ASEs of the benzene and sydnone rings. The value for benzene must be corrected in order to account for the different number of individual butadiene units in syn or anti conformations.

References

Duvall, M. B. The Aromaticity of Mesoionic Compounds: Sydnones and Münchnones. M.S. Thesis, University of Georgia, Athens, GA, 2013 (website, PDF).

Chen, Z.; Wannere, C. S.; Corminboeuf, C.; Puchta, R.; Schleyer, P. v. R. Nucleus-Independent Chemical Shifts (NICS) as an Aromaticity Criterion. Chem. Rev. 2005, 105 (10), 3842–3888. DOI: 10.1021/cr030088%2B.

Schleyer, P. v. R.; Pühlhofer, F. Recommendations for the Evaluation of Aromatic Stabilization Energies. Org. Lett. 2002, 4 (17), 2873–2876. DOI: 10.1021/ol0261332.

Work from Schleyer and his group is always incredible. I also want to add that sydnones undergo very rapid Diels-Alder reactions, another evidence that it's not a (very) aromatic system. A lot of work by the Houk group on this afaik.


text:
Can Polyethylene Dissolve on 50-60 °C
I know that polyethylene's melting point is between 115–135 °C, however I would like to know if can it start dissolving between 50-60 °C?
In a real example, if I put warm food (straight from the oven, I think it can be max. 50-60 °C) into a re-closable polyethylene plastic bag and close it, will any chemical reaction affect the food (from the polyethylene's side)? Or is polyethylene safe at that temperature, and if the bag is warm it doesn't mean that any chemical reaction is started? I didn't experienced any alteration, but I'm not an expert and don't know much about chemistry yet, so I would appreciate some guidance.

There may be better answers forthcoming, but a chemical does not need to melt (i.e. be in the liquid phase) to be reactive. For example, your food very likely has some liquid component, and that could be enough to react with some solid. Now polyethylene is used all over the place and is very nonreactive in many non-rigorous situations (like food service), so you are probably just fine. But I wanted to make the more general point that you don't have to be a liquid or gas to be reactive.

This should be a comment.

@LDC3: No, it is a perfectly fine answer. It may not be the best answer, as I noted, but it is an OK answer.

Am I right that you are pointing that if something (polyethylene in this case) doesn't melt or smokes it doesn't means that there is no chemical reaction (releasing any type of gas or something)?  As an instance if I put ketchup/soups into a heated plastic bag it may be reactive  because it's liquid  and it can be mixed up with the molecules we get during the chemical reaction - despite the change is not apparent -, and it won't if I put something that contains much less or not contains water like dried spices etc.. ?

I'm saying that lots of chemical reactions are possible besides liquid-liquid. Now PE is quite chemically resistant, to the point that it is used for lots of food containers, even for hot food.  So I'm guessing you have no issues. But the premise of your question seems flawed.

Though polyethylene (polythene in the UK) food bags and containers are quite safe to use even with hot food, oils, fats and other materials do penetrate polyethylene. For example, lycopene in tomato sauce or paste is absorbed so that the container is stained orange in a few days at room temperature, or in minutes when microwaved.
Whether this is true solution of the lycopene and/or oil carrier into the polyethylene, or penetration of microscopic pores, I do not know. This patent claims "staining may occur because the relatively non-polar nature of these commodity resin materials allows greater amounts of diffusion of non-polar substances (such as oils), which may contain dispersed pigments (e.g., lycopene), into the resin."

In addition to the answeres from @trb456 and DrMoishe, please be careful if you talk about PE as a homogene material.
There a quite a lot different "types" of PE, mostly categorised by chain lenght and side arms.
LLD-PE (linear low density Polyethylene) can start to melt at temperatures as low as 45°C. Containers for food normaly have a stamp on the bottom to indicate if they can be used for microwave ovens or hot food/liquids. Same should be there for bags sold for storing food. They should have a lable if you can use them for hot food (at least in EU and US).
From a reactivity standpoint, PE is mostly unreactive due to being composed of saturated hydrocarbon chains. Therefore, you can safely eat normal PE from a chemical health perspective. As long as you can't cut or otherweise harm yourself with it, it should leave your body in a state similiar to how you've eaten it (excluded are nanosized particles, which are still in discussion). Health risks from (mostly cheap) PEs normally come from additives like colours and softeners.

So if it's a normal PE without any additives it can't release toxic gases inside the plastic bag when heated, like chlorine and benzene gas? (Of course, I'm talking about normal use and not burning it) Or if it does the amount of these chemicals won't be more harmful than smoking a tobacco?


text:
Is the Arrhenius Equation only valid for reactants in the gas state?
I couldn't find a source that explicitly say this, but given the gas constant R is used, is the Arrhenius equation only valid when all reactants are gases? Do they have to be ideal gases?
If the above is true, is there a similar equation for reactions where at least one reactant is in liquid/solid form?

No, it has no such limits, obviously R being named "gas constant" is misleading.

$R$ is called the gas constant because it was first determined in relation to gases. However, $R$ is related to two other constants that have wide usage outside the gas world: Boltzmann's constant $k_B$ and Avogadro's number $N_A$: $$R=N_A \cdot k_B$$

@BenNorris  From what I understand the exponential is a probability function, is that the same regardless of phase? If the equation has no phase restriction then I am curious about the theoretical reasoning behind using the gas constant R as opposed to some other constant (i.e. Boltzmann constant)?

@Yandle - calling the 'gas' constant such is a misnomer. The constant has applications all over thermodynamics at the molar scale. It just happened to be discovered during the study of gases. We choose not to use $k_B$ very often because it is more convenient for small numbers of particles.

@BenNorris My textbook describes the $e^{\frac{-E_a}{RT}}$ as an energy factor that expresses the frequency of collisions that occur with an energy above threshold for reaction ($E_a$). Am I correct to interpret this expression as valid regardless of what phase or combination of phases the reactants are composed of?

As Ben Norris explained, $R$ can be just more convenient so that one can deal with quantities per mole.
I.e. one could write either $e^{(-E_a/RT)}$ or $e^{(-E_a/K_BT)}$, where in the first instance one would use $E_a$ in $\ce{J~mol^{-1}}$, in the second instance just in $\ce{J}$. The term in the exponent should of course be dimensionless.
Since the numbers in $\ce{kJ~mol^{-1}}$ are much more familiar for chemists, one would often use the first one to avoid having to use very small numbers.
The crux is that the Boltzmann factor in the Arrhenius expression is related to the chance for particles having enough energy at a given temperature, $T$, to surmount the activation barrier.

So is the exponential (to my understanding is expresses the probability of a collision resulting in successful reaction) in the Arrhenius equation valid for all phases and not just gas? Is there an explanation/proof as to why this is the case?

Indeed, but not exactly. The exponential relates to the probability of having sufficient energy to surmount the barrier upon collision. Whether or not, given sufficient energy, the collision leads to reaction depends on other factors. E.g. in gas phase theory, the reaction probability depends on the collision cross section and a 'geometrical' factor depending on whether the reaction partners are oriented in the right way to react. (For reactions involving hydrogen atoms and/or reactions at low T, also collisions with insufficient E can lead to reactions: tunneling).


text:
Is this product description wrong and a simple battery and not a fuel cell?
I won this "toy" at a science fair...
Now I have to make a short presentation about it in school.
On the package description they describe its energy source as "fuel cell"...
It consists out of a metal plate(magnesium), a black metal plate with a black coating and a paper-cotton-like isolator which is between those plates.
Through putting salt water on the isolator, there's a voltage difference of 1.76V between the black and the magnesium plate.
After a couple of seconds on load the voltage stays constant at 1.31V.
My theory is that it's only a simple battery, because there's no continious source of fuel which should be required for naming it fuel cell (according to wikipedia).
Is my thought right?
My idea about the black coat is, that it is a rubberd graphite or "coal" cathode, rubberd because coal or graphite would fall apart quickly. It is described as air-cathode in the package description. My idea is that the black coal or graphite coat is used to provide a larger amount of air to the reaction...
The magnesium plate gets damaged through this reaction, the black thing not.
Which material is the black coat?
and what is the full reaction?
a picture of the components

Your analysis seems correct, i.e. it is only a fuel cell if the magnesium is being oxidized by the air. You can prove if that is the case, and do so as part of your classroom demonstration, by excluding any air, and therefore oxygen, from the fuel cell.

Use distilled water and fairly pure $\ce{NaCl}$ (rather than table salt) to make the salt solution.
Boil the salt water for a minute or so to remove air bubbles (some oxygen from air dissolves into water).
Put the cell with electrolyte in a plastic bag sealed around the wires and a narrow tube such as a plastic coffee stirrer or soda straw that you'll use later to admit air, carefully squeezing out air bubbles. Seal both bag and tube with plasticine modeling clay or soft wax.
Observe the open circuit voltage of the cell.
Then run it for a while with a load (e.g. a 75 mA 1.5 V lamp) or a 100 ohm resistor, which will draw 1.3 mA), depending on the current output of the cell. You would expect some voltage drop over time, either due to cell polarization or due to any residual oxygen being used up.
To prove that this is a true fuel cell oxidizing the magnesium using air, bubble some air into the plastic bag through the (unplugged) tube. *If the voltage goes back up, then drops down until yet more air is added, you've shown that oxygen is being used and it's a true fuel cell.

Please accept my compliments for your thoughtful question; do post your findings here!

I accept your compliments thanking you very much, I'll make your  experiment and will add it to this question... after that you'll get your laurels

Hey,
I found out that it stops working without oxygen. The voltage goes back up after fresh air can contact the cell.
Could you took a look at this link and tell me if this might be mine?
https://en.wikipedia.org/wiki/Magnesium-air_fuel_cell
I think it is, because it would fit perfectly on mine. (Magnesium plate, "unknown" cathode.... If I got it right, this seems to be a fuel cell, right?

Sure looks like you proved the point! Thanks for going to the trouble of testing it and updating your post.  Hope you get extra credit for the thoughtful analysis, as well as the classroom presentation.