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Does the Battery Bounce Test actually work?
I heard that a good way to test if a battery is dead or alive is to see if it bounces. Supposedly, dead batteries bounce higher than living ones. Can someone tell me if this is a legitimate claim, and if it is, an explanation would be nice.

Check out what they have to say about this at Skeptics Exchange

Please don't try to summarise your question with tags. For more information, see meta.se and the [help].

Yes.
The battery bounce test kind of works. You can tell whether the battery still has a charge greater than 90%, but bounce hight levels off at about 50% and doesn't go higher. This means that it is not clear whether the battery is completely discharged when you observe a bounce.
Why?
Steingart et al. just recently published a paper on the subject, and through scientific rigor could confirm this claim. The reason they give has  to do with how an alkaline battery is built:

(Source)
As you can see, a large part of the inside is devoted to the zinc anode. During the discharge, a redox reaction takes place to produce the current:
$$ \ce{Zn + 2MnO2 -> ZnO + Mn2O3} $$
The zinc (usually in some gel form at first) is slowly oxidized. As a result of this oxidization, crosslinks between the small zinc granules are formed, which make the whole center more rigid. As such, its "bounce" (or coefficient of restitution) becomes greater, because the downward kinetic energy cannot be fully absorbed by the gel anymore.
If you are interested in more details, check out the article they published. If you don't have access you can read a summary given on the Princeton University News web page, where you also can see an animation of a series of discharged batteries bouncing to different heights.

I have to say, I love how sciencey Steingart et al. sound with their paper title, and how it can be decrypted to just some plain battery bouncing!


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How to figure out which two solutions will emit largest heat when they are mixed in same volumes?
How can I figure out which two solutions will emit the largest heat, when they are mixed in same volumes?
Then  let's assume that there are five pairs of solutions given:
\begin{align}
  \ce{2NaOH   + H2SO4   &-> Na2SO4    + 2H2O} \tag1\\
  \ce{Ba(OH)2 + H2SO4   &-> BaSO4     + 2H2O} \tag2\\
  \ce{2NH4OH  + H2SO4   &-> (NH4)2SO4 + 2H2O} \tag3\\
  \ce{NaOH    + H3CCOOH &-> H3CCOONa  + H2O}  \tag4\\
 \ce{2NaOH    + H2C2O4  &-> Na2C2O4   + 2H2O} \tag5
\end{align}
Which one emits the most heat, when we mix same volumes of given pairs?
These are the balanced reaction equation, that occur when the above solutions are mixed. All of these reactions are acid base reactions. And I know when there is an acid base reaction a salt is formed. So how can I proceed with it?

You should just think about a fact: all chemical processes will correspond to enthalpy changes.
The basic process of acid-base reaction is
$$\ce{H+ + OH- -> H2O}$$
This process will have an enthalpy change of $\pu{-55.8 kJ mol-1}$ at $\pu{298 K}$. (I don’t know why, but in different books or articles this value will be slightly different.) However,

Precipitate formation is usually exothermic, since there are some ion bonds formed.
Weak acid and base usually need some energy, for they are not completely ionized, during the reaction, they will to ionize more since $\ce{H+}$ or $\ce{OH-}$ is consumed. Ionization will be endothermic.

Therefore the answer would be the second reaction, since $\ce{BaSO4}$ is precipitated, which is exothermic, while weak acid and base reactions are endothermic.

(However, there seems to be always some exception, for example, ionization of $\ce{HF}$ will be exothermic. But to discuss this may be out of topic, and to be frankly, I’m not quite sure what its mechanism is.)

Can you figure out whether the reaction is endothermic or exothermic by just looking at the reaction?

This question is too broad and I should say I can't. But you will meet some typical types of reaction and you can expect it. Usually we think it by thinking bonds breaking/forming, equilibrium shifting when heating/cooling, or sometimes but "common senses".

source: Stack Exchange

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How does an electrolyte react to voltages below its electrolysis potential?
An electrolyte can be electrolysed by a voltage higher than the reverse reaction’s cell potential, while conducting a current in the process. But what happens at the electrodes when the applied voltage is below this cell potential, such as below 1.23 V for water or 4.07 V for NaCl? Does the reverse reaction occur simultaneously, or is the electrolyte unable to conduct a current with such a low voltage at all?

Electrolysing water is NOT required for it to conduct electricity. Water ionizes into $\ce{H+}$ and $\ce{OH-}$. Pure water will contain $10^{-7}\ \mathrm{M}$ of either ion at room temperature at equilibrium. This will result in $18\ \mathrm{M\Omega\ cm}$ of resistivity. If and when potentials above the electrolysis potential is applied, the local pH around the electrodes will change and more ions will be created.
$\ce{NaCl}$ dissociates as soon as it is dissolved in water. The $\ce{Na+}$ and $\ce{Cl-}$ ions created in the solution carry current and increase the conductivity.

I think the OP also wants to know what will happen to the water or NaCl outside the window. In short, in water, you'll evolve $\ce{H2}$ or $\ce{O2}$ at the electrodes, or $\ce{Na(s)}$ or $\ce{Cl2}$ with $\ce{NaCl}$ depending on oxidizing or reducing conditions.

What happens when these ions meet the electrode? How does the current continue from the ions to the electrode if the potential is not enough to add/remove electrons from the ions?

The concept appears related to http://chemistry.stackexchange.com/questions/278/how-does-conductivity-work-for-non-redoxed-ions?rq=1 , but neither did answers under that question fully explain the very nature of current in electrolyte and particularly at the metal-electrolyte boundary.

@kadabrium in that case the system will act as a pure capacitor. It will charge until the voltage is fully screened and then the current will drop to zero asymptotically. At DC voltage, there will not be a sustained current without redox reactions

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Periodic cleavage of carbohydrates
Carbohydrate like glucose undergoes periodic cleavage by $\ce{HIO4}$ and forms respective products are five moles of formic acid and one mole of formaldehyde.
Here is my question , what is mechanism and what are the products by fructose with the same? Do we have any specific rules for that?

Yes, there are specific rules regarding the products of periodic acid cleavage of 1,2-diols (Malaprade reaction). As to the mechanism, it is illustrated here. It is similar to the mechanism for KMnO4 oxidation of 1,2-diols. The structures of the products of 1,2-cleavage of carbohydrates is a diagnostic tool used as an aid in determining their structures. The oxidative cleavage of ethylene glycol (1, Fig. 1) affords two equivalents of formaldehyde. Primary alcohols give formaldehyde. Cleavage of the first single bond in glycerol 2 necessarily affords formaldehyde and α-hydroxyacetaldehyde. Aldehydes are readily hydrated in water. As far as periodic acid is concerned, the hydrate is a 1,2-diol. Cleavage of the second C-C bond produces not only another equivalent of formaldehyde but also an equivalent of formic acid. Secondary alcohols produce formic acid via aldehydes. (continued)

From the foregoing information, D-glyceraldehyde (3, Fig. 2) yields formaldehyde from the primary alcohol and 2 equivalents of formic acid, one from the secondary alcohol and one from the aldehyde group. Cleavage of one C-C bond of 1,3-dihydroxyacetone (4) yields formaldehyde and glycolic acid, which in turn provides formaldehyde and CO2. Ketones, and ultimately carboxyl groups, produce CO2. The oxidative transformations are summarized in the red box in Fig. 2.

The expected products from periodic acid cleavage of D-glucose, D-mannitol and D-fructose are predicted below based upon the formulated rules.

can you pls share the link for the mechanism again ... this link is not working ! :)

@visheshjain: The link in the text has been corrected. Either of the first two items under Carbohydrates contains the information.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

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


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Do only outer electron shells take part in forming chemical bonds?
Do only outer electron shells take part in forming chemical bonds? Or could an inner shell create a bond under some conditions?

Let me be the devil's advocate: in a molecule there are no strictly speaking inner or outer shell electrons. Still, it is interesting what you are asking.

I was wondering whether one could get compounds with far greater binding energy than normal

Great Question.
From this article, it seems that extreme conditions of pressure can lead to new type of molecules,bondings and how the electrons participate in the bonding.

Under very high pressures, it appears, electrons in the atom's inner shells can also take part in chemical bonds.
“It breaks our doctrine that the inner-shell electrons never react, never enter the chemistry domain,” says Mao-sheng Miao, a chemist at the University of California, Santa Barbara, and the Beijing Computational Science Research Center in China. Miao's calculations show that under extreme pressures cesium and fluorine atoms can form exotic molecules with inner-shell bonds.
Miao identified two molecules that, at high pressure, would involve cesium's inner electrons as well. To form cesium trifluoride ($\ce{CsF3}$), a cesium atom would share its single valence electron and two inner-shell electrons with three fluorine atoms. Four inner electrons would go into making cesium pentafluoride ($\ce{CsF5}$).
“That forms a very beautiful molecule, like a starfish,” Miao says. He reported his findings in Nature Chemistry. Both the shape of the resulting molecules and the possibility of their formation are “very surprising,” says Nobel Prize–winning chemist Ronald Hoffmann, a professor emeritus at Cornell University.

Found this in the comments about the stability of the exotic molecules at normal pressures

"Since the compound is thermodynamically unstable it would decompose to the more stable form at lower pressures. Since this means that they are going to higher states of enthalpy I would expect that this decomposition would be a endothermic reaction which means that they would get cold during the process."

Any chance of stability when it returns to normal pressure?

Well I think so , but honestly i dont have a single clue . I did not even know this happens till now :P My field is something entirely different from this. Sorry for being not helpful. Also paywall is detering from deeper search :(

JK, I found this "Since the compound is thermodynamically unstable it would decompose to the more stable form at lower pressures. Since this means that they are going to higher states of enthalpy I would expect that this decomposition would be a endothermic reaction which means that they would get cold during the process."  here in comments

@Gowtham For me this web link you have given in your answer is obsolete .  Please provide another link or article's publishing details

@Eka Thanks for noting that , Weird that i could access it for some time ..


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How to identify an emerald green solution with possibly two anions?
The solution is emerald green in colour.
When added with water, it gives a pale blue solution.
When reacted with $\ce{BaCl2}$, it gives a sky blue solution and a white precipitate.
When reacted with $\ce{AgNO3}$, it gives a sky blue solution and a white precipitate.
This does not make any sense to me. The $\ce{BaCl2}$ is used to test sulphates ions and $\ce{AgNO3}$ is used to test halides. How could you have two negative ions?

Why can't? The solution get some color, so it's reasonable to surmise there are some cations, and the solution keep electrically neutral. Besides, $\ce {SO4^2-}$ and $\ce {Cl-}$ won't react. However, I'm not sure what ions make those color, so I give a comment instead.

Can you give the procedure of the experiment. I might be some mistakes in it.

What's to say there isn't a mixture in solution?

You can have two different anions if your solution is a mixture of two different metal salts, a chloride and a sulfate. The color change from green to blue upon dilution with water fits well with $\ce{Cu^2+}$ being the corresponding cation. Concentrated solutions of $\ce{CuCl2}$ are green due to the presence of chlorocuprates, like $\ce{[CuCl3]^{-}}$ and $\ce{[CuCl4]^{2-}}$. When water is added, a stepwise ligand exchange takes place, and the pale blue color of $\ce{[Cu(H2O)6]^{2+}}$ starts to appear. The image below shows copper(II) chloride solutions of different concentrations: Green when the chloride concentration is high, and blue when it is lower (Wikipedia: copper chloride).

In summary, the observations (color change, precipitates) indicate that you have an aqueous solution of $\ce{CuCl2}$ and $\ce{CuSO4}$.

Assuming some wording modifications of the problem as stated, I would suggest as a single salt candidate solution: $\ce{[Cu(NH3)5(H2O)]Cl2}$. Other candidates include: $\ce{[Cu(NH3)4(H2O)2]Cl2}$ and $\ce{[Cu(NH3)3(H2O)3]Cl2}$.
The appearance of the cation Cu(NH3)x(H2O)y, where x + y = 6, also changes color with high values of x producing more intense blue coloration.
In the presence of AgNO3, the creation of insoluble white silver chloride:
$\ce{[Cu(NH3)2(H2O)4]Cl2 + 2 AgNO3 -> [Cu(NH3)2(H2O)4](NO3)2 + 2 AgCl(s)}$
Based on the cited reaction (see, for example, this reference):
$\ce{[Cu(H2O)4](2+) + 2 NH3 ⇔ [Cu(NH3)2](2+) + 4 H2O}$
The alkaline nature of the above implies on contact with aqueous barium ions (from BaCl2) the formation of the insoluble white Ba(OH)2.
This is, in my opinion, a good question for grading, given the number of possible reasonable answers.
Note, technically, I am assuming the correct wording of the problem is that 'upon dilution the coloration declines' so as to arrive at a single salt answer. Or, if one is starting with little NH3 in the presence of CuCl2, and one continues to add concentrated aqueous NH3, then coloration increases, with the answer for the starting solution being $\ce{[Cu(NH3)(H2O)5]Cl2}$.

Ammonia will not precipitate barium hydroxide, which is generally considered a strong base.


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Activation energy and temperature
It is said that activation energy does not change with temperature.
But I am unable to figure out why.
If we increase the temperature, the kinetic energy of the molecules will increase and they will need less extra energy and hence lesser activation energy to overcome the threshold energy barrier.
When I surfed the net, at most of the places it has been explained that increasing the temperature increases the no. of molecules having more energy and hence reaction rate is more but activation energy is same. This makes some sense but I still don't get why the activation energy doesn't decrease as the KE of an individual molecule should also increase on increasing the temperature.

Activation energy isn't  "extra" energy, but total.

Try thinking of the term "activation energy" as meaning the height of the barrier between the reactants and the products, rather than the amount of kinetic energy of the molecules in the system. Individual molecules need to have a kinetic energy comparable to the height of the barrier in order to react. When you increase the temperature, it increases the average kinetic energy of the molecules (and hence the number of molecules with sufficient energy to react), but does not change the height of the barrier. So the reactants cross over the barrier more frequently -- hence a faster reaction.

The two distribution plots shown here are for a lower temperature $T_{1}$ and a higher temperature $T_{2}$. The area under each curve represents the total number of molecules whose energies fall within particular range. The shaded regions indicate the number of molecules which are sufficiently energetic to meet the requirements dictated by the two values of $E_{a}$ that are shown.
It is clear from these plots that the fraction of molecules whose kinetic energy exceeds the activation energy increases quite rapidly as the temperature is raised. This the reason that virtually all chemical reactions (and all elementary reactions) are more rapid at higher temperatures.
Activation energy seems to be more of a material or reaction pertinent . Say a reaction has $10J$ as $E_{a}$ to proceed but has $20 J$ kinetic energy - I would say the the additional energy is transferred/carried on by the products after reaction is complete. So having more K.E shouldn't affect the character of a reaction but only its kinetics .


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Density g/cm3 of bleached viscose
I can find in literature what the density is for untreated viscose, but not for viscose that is bleached. Does this change the density? And what is the density for bleached viscose?


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Specific and general acid catalysis
What is the rationalization for calling reactions with rates independent of pH but dependent on the concentration of a particular acid catalyst under "general" acid catalysis, and pH dependent reactions under "specific" acid catalysis ?
It would seem to me that the pH-independent reactions would be better considered to be under "specific" acid catalysis given as they are catalysed by a specific species.

I have wondered about this too. Although I could be wrong it is reasonable to assume (or at least handy to keep as a reminder) that the term "special" refers to the fact that compound is protonated in a very specific state, namely before the actual reaction occurs. General acid catalysis can happen both before the reaction or at any time during the reaction (i.e. in some transition state), whatever is likely in some range around the pH of the acid. Hence it is more general version of the special case mentioned above.

=> specific acid catalysis is specific in so far as protonation occurs "specifically" in the transition state...not totally convinced but this seems plausible.

@J.LS Before the transition state. I agree that it is not totally clear that this is the meaning but it seems very plausible. If anyone knows what the actual reason is, feel free to edit this answer of course. It is provisional.

Hah, of course, it's still the other way round from being intuitive. How is before the transition state a "specific state" ?

@J.LS It is in the sense that the general mechanism can be during the transition state OR before the transition state. In other words SAC is a special case of GAC. That is what I mean. :)

"SAC is a special case of GAC"; that's simply not the case as I understand it; specific acid catalysis is not general acid catalysis because there is a pH dependence across the pH range. If you add some examples/sources to your answer it would really help !


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Why is it that less useful energy can be extracted in explosive reactions?
If you extract energy from a substrate (e.g. glucose) step-wise, a lot more free energy is released that can be used. However, if you were to do this in one single step, very little useful energy could be harnessed. Why is that?

I think there are two relevant points here.
1) explosive reactions can give significant useful energy - e.g. internal combustion engine
2) step-wise processes may go slower and release less heat / light other energy that is not so useful. This is a bit like in thermodynamics you get maximum useful energy from slow reversible processes.


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What are side chain order parameters?
I was reading an article:

We first compared experimental (S 2 exp) and computed (S 2 calc) side-chain order parameters using the simplest approximation of side-chain conformational variability, where motions are only allowed within the rotamer well of the side-chain conformation observed in the crystal structure

Now I am wondering what side-chain order parameters are.

That nomenclature is from the model-free approach from Lipari and Szabo


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Gold recovery under vacuum
Can gold recovery with acid peroxide (ap) method be done under vacuum or left in sealed chamber, as opposed to placement in open air?

Use of peroxides would evolve oxygen, which would need room to expand.

This process is typically done by extracting the gold in a mixture of something like 3 parts concentrated hydrochloric acid to one part 3% hydrogen peroxide. If this is roughly the procedure you are using, which is the common metehod posted all over the internet for recovering gold from used electronics, then the answer is an unambiguous none of the above!
Unless you have experience and training for working with these chemicals in a laboratory setting, then you should not be doing this at all; it is simply not worth it.
Given that, the best answer your question is to use a fume hood, eye protection, gloves, lab coat, etc. The biggest danger you face, other than splashing the solution into your eyes, is inhaling the fumes. As this process produces toxic gases (chlorine as well as the fumes from the reagents themselves) you don't want to use a sealed container or a vacuum system that would just pump the gases out of your pump. You need an appropriate, certified fume hood for working with these materials. Relying on diffusion or wind in the "open air" is just incredibly dangerous and you could permanently damage your lungs in one unlucky breath.
Again, the answer is: none of the above, use a fume hood!


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Why the dilution of solvent doesn't affect the solid phase, in comparison to the liquid phase?

Dilution of the solvent adds new energy states to the liquid, but does
  not affect the solid phase.

The above line appeared on the following page under the section "5: Colligative properties and entropy" and the last paragraph
Actually, I can understand a little bit. According to me, while freezing the solvent freezes separately from the solute( ie If we freeze salt and water solution, the salt will be separated out from ice). Therefore, whatever the presence of the solute maybe, the solid form of solution consists of solid 'ice', which is identical in every case and hence have identical entropy.
Am I going right? Is there an alternative explanation?
I am not clear with the concepts of what actually happens when we freeze a solution. Any Links regarding it or any other related stuff will be welcomed.

Instead of "making the jump" from liquid to gas, the water molecules are trying to stack together into an ice crystal. The salt molecules again get in the way.
Think of making a solid as like stacking shoeboxes, but the shoeboxes are vibrating slightly. The higher the temperature, the more they vibrate. At some temperature (the melting point), the vibration gets so powerful that the whole shoebox stack collapses. Now imagine that instead of a stack of just shoeboxes (pure ice), you start putting golf balls in between some of the shoeboxes. Those are salt molecules. They get in the way of the ordered stack, they mess everything up, they make the stack collapse really easily. There has to be a lot less vibration (lower temperature) to stack the boxes when there are golf balls getting up in everyone's business. So the freezing point is reduced.

Quite a nice analogy. :) But it doesn't answer my question. :/

@SiddharthYadav do you want a thermodynamic explanation for the colligative property ? I can give a free energy explanation. Do you want to know why dilution doesnt affect solid phase or something else ?

I want to know 'why dilution doesn't affect solid phase'...

authors: انس موسى باشا


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Is there a very stinky chemical solution that isn't a health hazard?
An effective way to disperse rioters & looters would be to send an armored vehicle in to spray them with a very stinky liquid. Is there a chemical that stinks but is not  a health risk and that breaks down after a few hours, or could be neutralized with another solution after a few hours?

My vote would be for butyric acid and related short-chain fatty acids with between 3 and ~8 carbons (isobutyric acid, valeric acid, isovaleric acid, caproic acid, etc.)
In pure form these chemicals would be dangerous acids, but in dilute aqueous solution buffered at pH 5 or so, they are not very corrosive or toxic but are still pretty stinky.
Skatole could be another example.  See the Wikipedia article for malodorants for some more ideas.

Indeed, many things like butyric acid come from decomposition or decay, vomit (as indicated by Wikipedia), etc. and can be smelled at very low concentration.

Actually, the interesting thing about butyric acid is that washing with alcohol (ethanol) would produce some ethyl butyrate which smells like pineapple.

Already done: See, for example, Skunk which has been widely used by the Israeli Defense Forces.
Per its manufacturer, Skunk is:

A non-lethal ... deterrent [that] consists of an extremely
  foul-smelling liquid, with the viscosity of water, that can be sprayed
  over a large area using a standard water cannon.
The Skunk is not volatile, it's inflammable – in fact, it can
  even be used to put out fires in a pinch – and it poses no
  health hazard. In fact, despite its outrageous smell, the Skunk liquid
  is perfectly safe to consume.


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Specify the formula of the product X_n Y_m

During a chemical reaction, equal moles of two diatomic molecules $\ce{X}$ and $\ce{Y}$ are allowed to react to form one gaseous compound $\ce{X_nY_m}$. After the reaction is complete, it was observed that the density of the product formed is $1.50$ greater than the density of the reactant mixture. Specify the formula of the product $\ce{X_nY_m}$.

Here is my work:
$$\ce{X_m + Y_n -> X_nY_m}$$
Density of product: $D_\mathrm{p}$ = $(m/V)_\mathrm{p}$
Density of reactant: $D_\mathrm{r}$ = $(m/V)_\mathrm{r}$
\begin{align}
D_\mathrm{p} &= 1.5 \cdot D_\mathrm{r} \\
(m/V)_\mathrm{p} &= 1.5 \cdot (m/V)_\mathrm{r} \\
m_\mathrm{p} &= 1.5 \cdot m_\mathrm{r}
\end{align}
I really dont know where to go from there. I would appreciate it if someone could give me step by step how they solved it so I could not just have the answer but also understand how they got to it.
Also, would the formula found be empirical or molecular? How do you know that it is?

possible duplicate of Density of Gas during after reaction greater than reactant mixture

Ok, First of all, MASS is always conserved. So, the density change should only come from the change in volume. Now, V=nRT/P and you are producing one mole of product in the expense of two moles of reactants. So it should be,  vr=1.5*vp
Can you please clarify your question a little bit more. What do you actually want to find? At first you are saying it's diatomic molecule but according to question statement i think you are asked to find the molecular formula.


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How does the atomic structure of crystals determines its shapes?
for example, table salt has an a cubic structure in the atomic level, and they are often cube like in the visible level. I don't see any reason that the atomic crystal structure should reflect the crystal's shape. Can someone explain this to me.

Related: Why does pyrite form cubic crystals?

It’s all about minimizing the energy of the system. Imagine a 2D example:

You can imagine the green circles as the cations ($\ce{Na+}$) and the blue ones as anions ($\ce{Cl-}$). Where is the next chloride you add going to go?

Not a great choice – it doesn’t really balance. The nearest blue ion is already balancing its charge with two other green ions and so can’t help the new ion balance its charge as well.
This is on the grid but still has all the same problems where the new ion doesn’t get its charge nicely balanced because the nearest blue ion is already donating part of its charge to two other ions.
This is the best option because it can share partial charges with two other blue ions. It gets more shared charge and the whole thing ends up being lower energy.

This pattern will keep happening with every addition until you get a macro-scale cube. It is always lower energy to fit new ions into gaps then to branch out and make new structures.
Ions that aren’t part of a complete unit cell will develop surface charges. These surface charges are better at attracting ions that are precipitating onto the surface from solution.


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Dependence of change in entropy on temperature
The thermodynamic definition of entropy is expressed as
$$dS = \frac{dq_{rev}}{T}$$
I understand that it depends on amount of heat transferred because heat transfer can be understood as the degree of dispersion of energy but what does the temperature dependence actually mean?

By "what does [it]... actually mean", do you mean "how can we intuitively understand why there is a temperature dependence"?

yes @ericksonla, I'm looking for an intuitive explanation.

An analogy that I like is one paraphrased from Peter Atkins' great book "The Laws of Thermodynamics - a VSI":

Imagine two rather different places in terms of noise - a silent
  library and a busy train station. Imagine that the temperature is the amount of noise at the given place - a measure of the "state of noise". Let's say that you laugh out loud,
  corresponding to supplying energy as heat.
In the library, where there is little background noise (low $T$) this would cause a large amount of disturbance and
  disorder - corresponding to a large increase in entropy. On the busy
  train station, on the other hand, you laughing would go largely
  unnoticed - there is lots of background noise already (high $T$). The (relative) amount of additional disturbance and
  disorder would be much smaller, corresponding to a smaller increase of
  entropy.

In a very cold system, even a tiny amount of supplied energy in terms of heat would cause a large change in entropy. On the other hand, supplying the same amount of energy to a really hot system - with lots of molecular motion going on - would cause a minuscule change in entropy.
Disclaimer: Note that the analogy and explanation above are two ways of convincing yourself that the entropy change due to added heat is inversely proportional to the system temperature. They do not claim to be correct or physical in any other sense, so read them with a (large) pinch of salt.

that's a good way to think about it but i would also like to hear the physical side of it.

What this equation is saying essentially is that the level of disorder can be modeled as the number of configurations if you have a number of particles and a number of quanta of energy. It's like having three cups and three balls; all the balls can occupy one cup or one ball can occupy one cup and two can occupy another etc. Hence an increase in energy would result in an increase in entropy (more balls and more configurations). Temperature is defined as the average energy of the system. Hence if energy is concentrated (all the balls are in one cup) then the average temperature per particle will be lower and the entropy will be lower which makes sense since the system is more ordered with all the balls in one cup.
I hope this helps answer the question somewhat

Thanks for answering but isnt this the statistical definition of entropy? I believe there might be an alternate explanation that does not involve talking about microstates

If in defining entropy we did not divide the heat by T, heat could flow from a cold body to a hot body with no change in entropy i. e. without breaking the second law.
This does not happen.
Likewise, in order for heat to flow from a hot body to a cold body, the loss of entropy of the hot body has to be smaller than the gain of entropy of the cold body.

I am also often confused and have no clear answer. Consider melting of ice. If we do not allow any physical change (no movement, no pressure change, no volume change), it will not happen. Without volume change when heat is added to gas, entropy cannot increase. We may say the heat was transferred to internal energy...
I think the temperature dependance of entropy is not the nature of entropy Itself but it is caused by the changes of parameters affecting entropy that are affected by temperature such as volume, pressure.. Most thermodynamic equations are  for gases. For liquid and solids the final equations may not be the same as those for gases although all definitions should be the same. Based on statistical definition of entropy, $S=k_\mathrm{B} \ln{W}$, it is obvious that there is no temperature term but it is assumed that particles can freely interchange positions. If particles cannot move ($0\, \pu{K}$), $W=1$ and entropy will be 0 regardless of how many different particles are three. When we calculate entropy changes for melting we just consider $\Delta H$ and T but we do not think about volume changes or anything else. All such changes are already included in the delta H value we use and assumptions. If molecules are not allowed to move, melting will no happen because crystalline state is that molecules cannot move. All reversible processes in thermodynamics allow changes associated With them as long as they are reversible. So, I would prefer not thinking too much about details. Leave them to physicists...

This is a bit of a "chicken-and-egg" problem, as those ultimately dealing with definitions tend to be. The concepts of entropy and temperature were originally developed empirically and were later modified and merged into a coherent theory of thermodynamics. It turns out that temperature can be defined as follows (this follows from the combined 1st and 2nd laws of thermodynamics):
$$T^{-1}=\left( \frac{\partial S}{\partial E} \right)_V$$
$T^{-1}$ describes the sensitivity of the system's entropy to the addition of energy at constant volume.
What the equation means can also be interpreted if we define entropy as follows (modifying a statement in the OP):

entropy can be understood as a measure of the degree of dispersion of energy (the number of possible configurations of the system given the current energy)

When a fixed amount of energy is added to a system more configurations become accessible (invoking concepts from statistical mechanics). At a lower temperature the number of additional configurations - in proportion to the initial number - increases by a greater amount than at a higher temperature. This can be restated as "a higher entropy change results when a very small fixed amount of energy is added to a system at low temperature". This explanation of the difference between the entropy change at low and high temperatures is captured by the silent library/busy train station analogy presented in another answer.


text:
Which has higher priority according to Cahn-Ingold-Prelog priority rules, vinyl or secondary butyl?
According to my textbook, vinyl has higher priority than isopropyl, because an atom which is connected by a double bond can be treated as two identical atoms (one of them is just an imaginary atom).
How about vinyl and secondary butyl? Should the imaginary atom be treated as that nothing else is connected to it? Does secondary butyl have higher priority?

The assignment of CIP priorities has some intricate details when it comes to multiple bonds.
First, the general procedure (Wikipedia):

If there is a tie, we must consider the atoms at distance 2 from the stereocenter — as a list is made for each group of the atoms bonded
  to the one directly attached to the stereocenter. Each list is
  arranged in order of decreasing atomic number. Then the lists are
  compared atom by atom; at the earliest difference, the group
  containing the atom of higher atomic number receives higher priority.
If there is still a tie, each atom in each of the two lists is replaced with a sub-list of the other atoms bonded to it (at distance
  3 from the stereocenter), the sub-lists are arranged in decreasing
  order of atomic number, and the entire structure is again compared
  atom by atom. This process is repeated, each time with atoms one bond
  farther from the stereocenter, until the tie is broken.

Now, concerning double bonds in particular:

If an atom A is double-bonded to an atom B, A is treated as being
  singly bonded to two atoms: B and a ghost atom that has the same
  atomic number as B but is not attached to anything except A. In turn,
  when B is replaced with a list of attached atoms, A itself is excluded
  in accordance with the general principle of not doubling back along a
  bond that has just been followed, but a ghost atom for A is included
  so that the double bond is properly represented from both ends.

I have constructed a table of these atom lists for the vinyl, isopropyl and sec-butyl rests:
$$\small
\begin{array}{llll}
\hline
\text{Dist} & \text{Vinyl} & \text{Isopropyl} & sec\text{-Butyl} \\
\hline
1 & \text{C} & \text{C} & \text{C} \\
2 & \text{C} & \text{C} & \text{C} \\
  & \text{(C)} & \text{C} & \text{C} \\
  & \text{H} & \text{H} & \text{H} \\
3 & \text{(C)} & \text{H} & \text{C} \\
  & \text{H} & \text{H} & \text{H} \\
  & \text{H} & \text{H} & \text{H} \\
  & & \text{H} & \text{H} \\
  & & \vdots & \vdots \\
\hline
\end{array}
$$
The sec-butyl rest has thus the highest priority of the three, followed by vinyl and then isopropyl.
The ghost atom at distance three is the ghost carbon at distance one, as seen from the ghost carbon that was doubly bonded to it. If the distance 4 was to be considered, no more ghost atoms would appear, as the ghost atom at position 1 does not have any more ghost atoms bonded to it (atom 1 however is bonded to the ghost atom at position 2).

The quoted procedure from Wikipedia is actually incorrect. The rules get very messy beyond a "distance" of 2 (or what the 2013 Blue Book calls "Sphere II"). In Sphere III, it is not sufficient to simply list all the (up to 9) possible atoms in this sphere. One must first sort the three atoms in Sphere II by applying the CIP rules to Spheres III and beyond, and then the hierarchy in Sphere I (i.e. the original substituents) can be determined by comparing the highest-priority groups in Sphere II. In this case, the wrong procedure happens to give the right answer.

Great correction! If only we could reassign the "accepted" status...

The accepted answer is incorrect
There are some subtleties in the determination of stereochemistry in this question which need to be carefully explained with reference to the official rules, as laid out in Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book).
The gist of it is similar to that provided in other sources, e.g. Wikipedia, as cited in the accepted answer. However, many secondary sources do not fully capture the complexity of the Blue Book, and consequently end up providing simplified "rules" that do not really exist. Thus, although the actual ordering given in the accepted answer, i.e. sec-butyl > vinyl > isopropyl, is correct, the means by which this ordering is reached is not fully correct.
The actual rules are quoted verbatim at the bottom of this answer, and the following is an explanation tailored to the question at hand. We will first talk about sec-butyl and isopropyl as the rules are easier to follow, then we will move on to the vinyl substituent.

sec-butyl vs isopropyl
First, we must construct what is called a digraph for each of the three substituents, which refers to the hierarchical "tree" of atoms where each (non-terminal) atom is bonded to three other atoms that are one bond away. For sBu and iPr this is easy.

The Roman numerals I, II, III, IV refer to the "spheres" that are talked about in the Blue Book, which denote the distance from the stereogenic centre. Generally, this "distance" refers to the number of bonds: so Sphere I is one bond away from the stereogenic centre, and so on.
In order to rank the substituents we must go from Sphere I upwards.

In Sphere I, no decision is possible because C = C.
In Sphere II, no decision is possible because CCH = CCH.

When we get to Sphere III, it is not sufficient to simply list all six elements and declare that because CHHHHH > HHHHHH, sec-butyl has a greater priority over isopropyl. We need to prioritise our exploration: that is to say, we only search in one of the two branches going from Sphere II to Sphere III (labelled (a) or (b) in the diagram).
In order to make any progress for the original question, we therefore need to choose which of the two branches we preferentially explore. That is to say, we need to figure out which of the two branches has a higher priority. We do this by recursively applying the CIP rules to each of the two branches but starting this time from Sphere II (which is where the branches originate). Consider first the two branches labelled (a) and (b) in the sec-butyl digraph.

In Sphere II, no decision is possible because C = C.
In Sphere III, we can rank (a) > (b) because CHH > HHH.

Thus in sec-butyl the preferred branch to explore is (a) and we can reject (b). In isopropyl, we find that both branches (a') and (b') are equivalent, thus it does not matter which we pick.
Going back to the original issue of sec-butyl vs isopropyl, in Sphere III we do not consider all six items, but only the three items that result from the preferred branch that we selected. Therefore:

In Sphere III, we can rank sec-butyl > isopropyl because CHH [branch (a) of sec-butyl] > HHH [branch (a') of isopropyl].

This seems like a long-winded way of reaching the same conclusion, but it is the correct way.

The vinyl digraph
When double bonds are present, duplicate atoms are added to the digraph on each end of the double bond (P-92.1.4.2). These duplicate atoms are "duplicate representations of the atoms at the other end of the double bond". However, only the doubly bonded atoms themselves are not duplicated: the atoms or groups that are bonded to them are not also duplicated. Instead, each duplicate atom is considered to "carry three phantom atoms of atomic number zero".
Thus, the complete digraph for the vinyl group is as follows.

To the left of the actual digraph we can see the duplicate carbon atoms, denoted (C), explicitly drawn out. In the digraph these are connected to phantom atoms of atomic number 0, which are denoted as 0.

Vinyl vs isopropyl
When we compare vinyl against isopropyl, we note that:

In Sphere I, no decision is possible because C = C.
In Sphere II, no decision is possible because CCH (isopropyl) = C(C)H (vinyl). Note that duplicate atoms (C) do not have any inherently lower priority compared to normal, genuine C atoms.

At this point, we realise that we need to perform the branch tiebreaker for the vinyl group. It is quite clear here that for the vinyl group, branch (a") > branch (b") because (C)HH > 000. Thus, returning to the vinyl vs isopropyl question,

In Sphere III, we have that (C)HH [branch (a") of vinyl] > HHH [branch (a') of isopropyl]; therefore vinyl > isopropyl.

Vinyl vs sec-butyl
Again, in Spheres I and II, no decision can be reached. For Sphere III, we therefore need to compare branch (a") of vinyl with branch (a) of sec-butyl. Unfortunately, again, no decision can be reached here because CHH (sec-butyl) = (C)HH (vinyl)!
In order to go on to Sphere IV(!), we need to first rank the three atoms in each of the Sphere III trios, in order to prioritise our exploration. Clearly, this must be accomplished by choosing the carbon atoms in each trio.
So, in Sphere IV we have HHH (sec-butyl) > 000 (vinyl), and therefore the overall ranking is given by sec-butyl > vinyl > isopropyl.

The rules, verbatim
The Ranking Rules quoted below describe how to use the Sequence Rules to explore a digraph (i.e. the order in which to proceed, and the direction in which to explore). On the other hand, the Sequence Rules explain how to assign CIP priority between two branches of a digraph. There are five Sequence Rules which are too long to be fully quoted here, but a summary is available in P-92.1.3, and examples are explored in P-92.2 to P-92.6. For the purposes of this question, the only relevant one is Sequence Rule 1(a), which states that higher atomic number precedes lower atomic number.
Note that P-92.2.1.1.3 provides a worked example of a digraph that is rather similar to the above (where Spheres III and IV must be explored to fully assign the ranking).

P-92.1.5 Exploration of a hierarchical digraph
Digraphs are constructed to show the ranking of atoms according to the toopological distance i.e. number of bonds, from the core of the stereogenic unit (i.e., center) and their evaluation by the Sequence Rules.
(a) Atoms lie in sphere and atoms of equal distance from the core of the stereogenic unit are in the same sphere; spheres are identified as I, II, III, and IV, as shown in Fig. 1. The first sphere is occupied by the 'proximal atoms', 'p' and 'p′'. Those in the sphere II are numbered '1,2,3' and '1′,2′,3′'. Those in sphere III are numbered '11', '12', '13', '21, '22', '23', ... '11′', '12′', '13′', ... and so on for each further sphere. Indicated branches may not be present in all molecules.
(b) Atoms in the nth sphere have precedence over those in the (n + 1)th sphere (Ranking Rule 1).
(c) The ranking of each atom in the nth sphere depends in the first place on the ranking of atoms of the same branch in [the] (n − 1)th sphere, and then the application of the Sequence Rules to it; the smaller the number, the higher the relative ranking (Ranking Rule 2).
(d) Those atoms in the nth sphere which are of equal rank with respect to those in the (n − 1)th sphere in the same branch are ranked by means of the Sequence Rules, first by the exhaustive application of Sequence Rule 1; if no decision is reached, Sequence Rule 2 is exhaustively applied, and so on.

Can you give a more general idea when to prioritise CIP assignment through branching? Is it when comparing more than 3 atoms or is it something else.? Thanks.

It's wrong. Vinyl loses at position 2 of distance 2 against both isopropyl and sec-butyl and therefore gets priority 3. sec-butyl beats isopropyl at the first position of distance 3 and therefore gets priority 1 thus isopropyl gets priority 2.
Phantom atoms are less worth than real atoms but more worth than real atoms with lower atomic number.

Note that what you call "phantom" atoms (and the accepted answer calls "ghost" atoms) is actually called a duplicate atom according to the 2013 Blue Book; these duplicate atoms are each considered to be "carrying three phantom atoms of atomic number zero" (P-92.1.4.2). However, contrary to your last sentence, there is no rule in the Blue Book that says that a duplicate atom has an inherently lower priority compared to a genuine atom. Instead, one must move on to the next "sphere", where vinyl beats iPr as stated in the accepted answer.


text:
PAPS , is it a carrier or not
In sulfate conjugation , is it considered as a carrier or help to make sulfonation if I got a question and I have to choose ( enzyme helps to make the reaction or is a carrier to transfer the sulfate ) I know isn't an enzyme but it does help in the reaction , that what I understand

https://en.wikipedia.org/wiki/3%27-Phosphoadenosine-5%27-phosphosulfate, https://en.wikipedia.org/wiki/Sulfotransferase Btw bould you improve grammar?


text:
Name for analog of salt hydrate with non-water solvents?
Salts can bind with some water to form hydrates. For example, we can call $\ce{CoCl_2\cdot 6H_2O}$ cobalt chloride hexahydrate. There is no reason that water should be the only solvent that can bind with salts though. Dimethyl sufloxide is pretty polar and can dissolve some salts, so I might hypothesize that we could make salt crystal that has some bound into it, say $\ce{CoCl_2\cdot 6DMSO}$. I might call that cobalt chloride hexadimethyl sulfoxate, but that is totally made up. Is there a name for that sort of thing?

Have you tried using the term solvates? For example, my top Google search result for "salt solvates" was this article.

@NicolauSakerNeto Excellent! That is exactly what I was looking for, thank you!

The corresponding rule in the Red Book (‘Nomenclature of Inorganic Chemistry – IUPAC recommendations 2005’) reads as follows:

The names of the individual components of such a generalized addition compound are each constructed by using an appropriate nomenclature system, whether compositional, substitutive or additive. The overall name of the compound is then formed by connecting the names of the components by ‘em’ dashes; the proportions of the components are indicated after the name by a stoichiometric descriptor consisting of arabic numerals separated by a solidus or solidi. The descriptor, in parentheses, is separated from the compound name by a space.

$\ce{CaCl2.8NH3}$ calcium chloride—ammonia (1/8)
$\ce{AlCl3.4EtOH}$ aluminium chloride—ethanol (1/4)
$\ce{BiCl3.3PCl5}$ bismuth(III) chloride—phosphorus(V) chloride (1/3)
$\ce{2Na2CO3.3H2O2}$ sodium carbonate—hydrogen peroxide (2/3)

@Loong Emotionally unsatisfying but you can't argue with IUPAC (unless, I guess, you're in those sort of circles). Unfortunately, that naming convention is utterly unhelpful when it comes to googling these compounds.


text:
Azathioprine CIF data file rendering problem in Accelrys
I have been working on a CIF data file based on experimental data [1] (free full text isn't available, but the page 2 with the experiment details can be found in the free preview provided by Springer).
Unfortunately, I don't know an awful lot about crystallography parameters (although I have been consulting the CIF core dictionary). The CIF file I linked to was created using EnCIFer & VESTA (I mostly used VESTA to get a handle for the text formatting of CIF files). The non-coordinate data mentioned in this paper includes:
\begin{array}{ll}
\hline
Z & 4 \\
\text{Formula} & \ce{C9H7N7O2S} \\
M_r & 277.3 \\
\text{Coordinate system} & \text{monoclinic} \\
\beta & 105.99^\circ \\
a & \pu{4.488(2) Å} \\
b & \pu{31.886(4) Å} \\
c & \pu{8.067(2) Å} \\
V & \pu{1149.5 Å^3} \\
D_m & \pu{1.59 mg mm^{-3}}\\
D_c & \pu{1.60 mg mm^{-3}} \\
\lambda & \pu{0.7107 Å} \\
\text{Space group} & P2_1/c \\
\text{Crystal dimensions} & 0.7\times 0.1 \times 0.7\, \pu{mm} \\
\hline
\end{array}
The problem is that Accelrys DS Visualizer so far renders my CIF file as:

Notice how some of the hydrogen atoms are disconnected from the main chunk of the azathioprine molecule.

Acharya, K. R. Proc. Indian Acad. Sci. (Chem. Sci.) 1984, 93 (2), 183–187 DOI: 10.1007/BF02842141.

The coordinates of the hydrogen atoms of the methyl group in question are there (H16a, H16b, H16c).
The problem is a rendering problem of the Accelrys visualizer.
Other tools don't have a problem with this CIF file. The image below is a medium-quality PovRay rendering generated through PLATON.

Indeed. I highly recommend the free CCSD Mercury program, which renders like this: http://i.imgur.com/OXOXO0f.png


text:
What is the reason that tertiary alcohol pKa is higher than primary alcohol pKa?
A tertiary alcohol has a $\mathrm{p}K_\mathrm{a}$ of 18 while a primary alcohol has a $\mathrm{p}K_\mathrm{a}$ of 16.
Shouldn't the tertiary alcohol be more acidic since its conjugate base is more stabilized due to induction?

Which tertiary and primary alcohol are you referring to? Please give some references for your data.

I meant (CH3)3COH for tertiary and CH3CH2OH for primary.

It's other way around with induction - tert-butyl group is electron donating stronger than ethyl.

Here is a nice explanation.  Considering polarizability and solvation effects you can explain the solution pKa data, as well as why it reverses in the gas phase.

Methanol is a weaker acid than t-BuOH in the gas phase. http://pubs.acs.org/doi/pdf/10.1021/ja00723a029

The short answer is that the steric hindrance of the tert butyl alcohol makes it hard to add an protonate the hydroxy group.

Based on this lecture from the vanderbilt university. The answer is due to steric hindrance:

The hydroxy group of the tert butyl alcohol is intramolecular stabilized by hydrogen bonds (2nd image), between the methyl hydrogens and the oxygen. In the other hand, i.e. the hydroxy group of methanol is intermolecular stabilized with another methanol molecules ("Behave more like a water") (3th image). Therefore, the equilibrium $\ce{ROH}\rightleftharpoons\ce{RO^-}+\ce{H^+}$ is more displaced to the right with the methanol than with the tert-butil alcohol.

The protonation of ter butyl alcohol is slower than methanol (Figure 4th and 5th), for the same reason: steric hindrance.

The conjugate bases of tertiary alcohols are LESS stabilized due to induction!
Because carbon groups are donating groups they push their electrons towards the negatively charged oxygen making it even more negative and therefore making it more reactive. For the conjugate bases of less substituted alcohols there are less carbon groups pushing their electrons into the already negatively charged oxygen which makes it less reactive. Now when comparing something like methanol and water their pKa's are very similar! Water has a pKa of 15.7 while methanol has a pKa of 16. When you look at the conjugate base of methanol the one carbon group is pushing its electrons towards the oxygen making the oxygen slightly more negative and therefore more reactive. The conjugate base of water does not have any donating groups pushing their electrons into the oxygen to make it more negative and is less reactive.


text:
Debye-Huckel theory with non-polar solvents
I want to calculate the activity coefficients of mixed solvent salt solutions. I am seeing very strange behavior when I try calculating the activity coefficient of salts in non-polar solvents using Debye-Huckel theory though and its messing up my down-stream calculations.
In a simple example, let's compare the activity of NaCl in hexane and water:
Inputs: (hexane / water)

Density: 654 / 1000 kg/m^3
Dielectric constant: 1.89 / 80.4
Temperature: 293 K
Molarity = 0.1 mol/L

Calculations:

$A = consts \frac {\sqrt {\rho}}  {(\epsilon * T)^{1.5}} = $ 230 / 1
$I = \frac12 \sum_{ions} cz^2 =$ 0.1 / 0.1
$ln(\gamma) = -Az^2\sqrt I = $ -73 / -0.32
$\gamma = 10^{-32}$ / 0.72

In equilibrium the salt activity should be the same in each phase, $\gamma_{aq}x_{aq} = \gamma_{org} x_{org}$, which would imply that the salt should all be flying over to the organic phase. That clearly makes no sense.
Why does this give such totally wrong behavior in a limiting case?

I've never applied Debye-Huckel theory outside of aqueous solutions, but does the ionic strength equation really apply directly to hexane in the same way as water? I would imagine the actual concentration to be taken into account would be the equilibrium concentration of the dissolved ions in the hexane phase; in reality, practically all the salt would precipitate out of the hexane, so $\rm{ln(\gamma )}$ would be very close to zero.

@NicolauSakerNeto I agree. In theory, the solubility is derivable from activity so I was really hoping for a robust way to (roughly) predict solubility without any pre-sorting. It would be nice if the limiting case worked but you may be right that the ionic strength equation just doesn't make sense.


text:
How can you tell if a sugar is D/L in its chair form?
I'm having trouble determining if sugars are D/L in their chair forms. My textbook has generously provided this image of D-Glucose.
So I know that from the fischer, when I put everything into a chair view, the things on the left point up, and the things on the right point down.
However, what about the bottom $\ce{CH2OH}$ group. Which way does it point, and does it determine D/L in a ring? I'm imagining "rotating" all the groups 1 position over clockwise (does it matter if its clockwise/counterclockwise? How do you know which way to spin it?) such that the hydroxyl is now where the $\ce{CH2OH}$ was, H is on the right and $\ce{CH2OH}$ is on the left.
Does that sound right? I can vaguely remember this from high school, but it's become very clouded.

D-glucose stays D-glucose regardless of whether it is in chair (hemiacetal) form, open chain form, or any other minor form. These different forms (if you could isolate them) would likely have different specific rotation values (interaction with plane polarized light). There are two chair forms-one with an axial OH and the other equatorial (probably favored)--but both are forms of D-glucose.


text:
What are the correct resonance structures of nitrous oxide?
There was a question in one of my exam papers to draw the resonance structures for $\ce{N2O}$.
These are the ones I drew, but they were not awarded marks:

The ones in the marking scheme are as follows:

Can anyone please explain why mine aren't correct?

Related: http://antoine.frostburg.edu/chem/senese/101/inorganic/faq/laughing-gas.shtml

Firstly, neither of the resonance structures that you drew for your test are possible because they both violate the octet rule. For the structure on the left, the leftmost $\ce{N}$ is in control of only 6 electrons, and will not exist in this form. The structure on the right is not possible because central $\ce{N}$ is participating in 5 bonds, which $\ce{N}$ cannot do. The maximum number of bonds that $\ce{N}$ can participate in is 3 (covalent bonding - like $\ce{NH3}$) or 4 (coordinate bonding - like $\ce{NH4+}$).
As long as you satisfy the octet rule (or at least as best as you can - here are some exceptions) and you wish to check the stability of a molecule that you have drawn, simply use formula for formal charge. This is given by the equation:
$$\mathrm{FC} = \mathrm{V} - (\mathrm{N_B} + \frac{\mathrm{B}}{2})$$
Wher $\mathrm{FC}$ is the formal charge, $\mathrm{V}$ is the number of valence electrons that the atom under consideration usually has, $\mathrm{N_B}$ is the number of non-bonded electrons, and $\mathrm{B}$ is the number of electrons shared in a covalent bond on the atom. If a molecule is stable, the sum of the formal charges of each substituent atom should be 0. If you were to test this on each of the resonance structures that the test provides as answers, you would see that this holds true.

'If a molecule is stable, the sum of the formal charges of each substituent atom should be 0' — I'm not even going to try and list the number of ways in which that sentence is wrong. Even if we don't understand stable as a thermodynamic term, it still is wrong for molecular ions.

@Jan A molecular ion is not a molecule, it is a charged molecular entity and a molecule is by definition neutral. Hence the only flaw that is present in that sentence is, that it is redundant, as a molecule is also by definition always at least meta stable.

To append Ringo's good answer and to add some more insight into the bonding situation, I performed a calculation on the DF-BP86/def2-SVP level of theory. Since this is a linear molecule, there are symmetry restrictions. Its point group is $C_{\infty\mathrm{v}}$, which means, that there are degenerated orbitals. As we will see, these correspond to $\pi$ bonds. Here is an image of the valence orbital scheme:

The orbitals 1-3 are core orbitals (1s) at the respective elements. MO 4 corresponds to a $\ce{O-N}~\sigma$ bond. It is noteworthy, that this bond is delocalised and also describes part of the $\ce{N-N}~\sigma$ bond. Analogously this holds for MO 5, but the other way around. In MO 6 we can see the presence of the lone pair at oxygen. MOs 7 and 8 describe two delocalised $\pi$ bonds, bonding with respect to all atoms in the compound. MO 9 has mainly lone pair character and is located at the terminal nitrogen. Tho HOMOs are also $\pi$ orbitals, bonding with respect to $\ce{N-N}$ and anti-bonding with respect to $\ce{O-N}$. This somewhat implies lone pair character for these orbitals, too. Please note, that there is no orbital with lone pair character at the central nitrogen.
I further analysed the wave function with the natural bond orbital theory. This method localises the orbitals (and electron density) and compares it to the ideal electron density of a given Lewis structure. High percentages mean that the configuration agrees well with the Lewis concept.

I included the structures you have suggested. You can see, that while configuration 1 is not an ideal Lewis structure, it is still agrees to a certain point with the Lewis concept. If someone would do a more rigorous approach with Valence Bond theory, I would expect this structure to have (at least a minor) contribution to the overall bonding situation. The basic flaw with this structure is the electron sextet at the terminal nitrogen, as Ringo already mentioned. The positive aspect of this structure is, that it does not need formal charges. I would not consider this structure wrong.
The second structure, however, is not correct. This can only exist in an excited state. Nitrogen cannot have more than eight electrons binding around it.
The third structure is the ideal structure, i.e. the one with the highest agreement. I would expect it to have the highest contribution to the overall bonding.
Structure 3 accounts for delocalisation effects and is one of the major contributions to the overall resonance stabilisation of the molecule. Basically you can see the character of this in the MOs 7 and 8 above.
The last structure (5) tries the same thing, but for my taste, this is a little bit far fetched. It still agrees somewhat with the Lewis-concept, but like configuration 1 I would only expect a smaller contribution.

Nitrogen cannot expand its octet (it cannot accommodate more than eight valence electrons). In the resonance structures that you have drawn, both the nitrogen and oxygen on the ends are stable (having no charge) but nitrogen in the middle has five bonds hence violates the above rule. So, the maximum number of bonds for nitrogen in the middle is four. That is why the middle nitrogen atom (in the correct answer)
has four bonds but with a charge.
Also, an expanded octet is generally observed in elements which are located below the second period.

"Expanded octet is used for elements which are located below period 2 cause they have d-orbitals to expand." Um, what?! How does Iodine expand octet then?

There always some expection

How does Bromine and Chlorine expand octet then? The point is, this statement is wrong! Wrong I say!

@MARamezani if we consider the concept of expanding octet to be true (which I disagree with) then Simon-Nail-It's example is correct, because iodine is in the 5th period and 5 > 2, thus lower. Same for bromine (4th) and chlorine (3rd).

@Jan then maybe, either me or you (or even Simon) have misunderstood what is meant by lower. If lower indicates the number, then the statement isn't true, which is what I had in mind. And when I asked about iodine, Simon said it's an exception, so he must be understanding the statement like me. Of course, if lower points to the relative position in the periodic table, then the statement is still incomplete, if not wrong. Expanding octet (if exists) happens because of the closeness of d orbitals in energy to s and p. (Which is a flawed concept I agree) Not due to simply vacant d orbitals. [...]

[...] Heck, even H has vacant d orbitals, but it takes a considerable amount of energy to make electrons reach there. So, all in all, this statement is misleading...if not totally based on a wrong premise.

@MARamezani: Yes, I can agree with that; the statement being incomplete. I hardly ever address elements based on the number of their period (or even group) but rather according to their relative positions in the periodic table … apparantly usage of that is different around the world.


text:
Between ethene and propene which is more acidic and why?
Between ethene and propene, which is more acidic and why? If they are they are the same then why so?
 Their $\mathrm{p}K_{\mathrm{a}}$ is the same $\mathrm{p}K_{\mathrm{a}}$ as ~44.

related http://chemistry.stackexchange.com/questions/18805/abstraction-of-a-proton-in-cycloheptatriene

I dont think that quite answers my question.

The difference between there acidity arises only due to the difference in inductive effect[or the electron donating capacity of the alkyl group] of the alkyl group attached to the $\ce{C=C}$ in both the cases.
By this site there exists a difference of total one unit
For more info on Inductive Effect
Relative acidity of a compound is checked by removing the acidic hydrogen and then comparing the stability of the anions.The more stable the anion is the faster it is formed and hence the acid is stronger.
The anions formed are $\ce{H3C-CH=CH-}$ and  $\ce{H2C=CH-}$ . In the former one the anion or the negative charge on the anion is destabilized by the electron donating ability or the inductive effect of $\ce{CH3-}$, hence it is less stable and less rapidly formed and hence less acidic.

Why $\ce{H3C-}$ has electron donating effect?

In an alkyl group, for example, in methyl group carbon is linked to three hydrogen atoms. Carbon is more electronegative than hydrogen. So carbon pulls bonded electrons nearer to carbon and carbon acquires a slight extra negative charge (delta-) and correspondingly hydrogen lose hold over bonded electrons and acquires a slight positive charge (indicated by delta +). Thus carbon get a little more electronic charge which the carbon will push towards the group to which it is linked.When this electron rich carbon centre (methyl group) attached with other compounds it behaves as electron donor.
And thus methyl group acts as electron donating group.
Be in mind, the electronegativity difference between carbon and hydrogen is very less.
$\ce{H3C-CH=CH-}$ is less stable due to presence of above factor.
You can also use conjugate base method or charge holding capacity method to compare the acidity of compound to give the same result as above.


text:
Smoke alarm ion neutralisation
Smoke alarms are a necessary part of many modern day buildings, especially in laboratories and of course, the family home.
According to the World Nuclear Association page Smoke Detectors and Americium (2014), a major part of the operation of a smoke detector is when the Americium-241 emits alpha particles collide with atmospheric oxygen and nitrogen, ionising them - which form part of an electric current between two electrodes.
When there is smoke (from the article) - bolding mine:

When smoke enters the space between the electrodes, the smoke particles attach to the charged ions, neutralizing them. This causes the number of ions present – and therefore the electric current – to fall, which sets off an alarm.

What is the chemical process that causes smoke to neutralise the charged oxygen and nitrogen ions in smoke alarms?

From this paper, found the mechanism for neutralisation,

Smoke particle charge interaction
When smoke particles enter in the ionization chamber, they are
  subjected to the electric field generated by electrodes but also to the
  one generated by the charges. Close to the charges, the electric field is
  highly divergent resulting in dielectrophoretic force acting on smoke
  particles. The force tends to bind the smoke particles to the charges.
  When smoke particles are considered as isotropic neutral dielectric
  bodies linearly and homogeneously polarisable, dielectrophoretic
  force F can be expressed as
$ F = \frac{1}{2} \alpha v \overrightarrow{grad} |E^{2} |$
where $\alpha$ depends on smoke particle polarizability, $v$ is the smoke
  particle volume and $E$ is the electric field produced by the charges
  [11]. Coefficient $\alpha$ is typically of the order of $\epsilon_{0}/3$ [12] where $\epsilon_{0}$ is the
  dielectric constant of air. Volume $v$ is approximately $4 \mu m^{3} $ for a 1-$\mu m $
  diameter smoke particle.
  Amplitude of electric field $E$ produced by charges is given by
  coulomb’s law:
$ E = \frac{1}{4\pi \epsilon_{0}} \frac{q}{r^{2}}$
in which $r$ is the distance between the charge and the smoke particle.
  Due to the divergence of the field, the dielectrophoretic force
  decrease as $\frac{1}{r^{5}}$. If a smoke particle is close to a charge then they can bind. If the smoke particle is relatively far from the charge, the time
  of interaction may be relatively long and it is necessary that the
  charge drifts sufficiently slowly in the chamber to optimize the
  interaction. Indeed if the charge passes too rapidly there will be not
  enough time for the smoke particle to bind and thus nothing would be
  detected. In fact the capture radius reduces with the electric field and
  is of the order of 10 $\mu m $ for $E= 1  V/mm$.
When the smoke particles attach to the free charges due to
  electrophoresis force, the charges become heavy bodies then the
  charge mobility $\mu$ decreases resulting in a drop of the electric current
  measured between the electrodes of the ionization chamber.
  Even if ionic detectors are very efficient, according to new
  standards they are no longer authorised in almost all of the developed
  countries because of the radioactive sources they include for
  generating the ions. Such radioactive sources are indeed difficult to
  collect and to recycle.


text:
Molecular orbitals of CH2 an SLACs
I was wondering about molecular orbitals of methylene different states 1B1 3B1 1A1
and how are they connected to symmetry point group lower case 1a1, etc orbitals what those numbers and alphabets mean?
It would be great if someone can derive those orbitals from symmetry and show me which states 1B1, 3B1 and 1A1 are referred to and what does up script, subscripts and alphabets stand for and connect them to lower case 1a1 ,etc.

Related: http://chemistry.stackexchange.com/questions/16305/symmetry-labels-for-orbitals

The first question to always ask when you are trying to build an MO diagram is what point group does my molecule fall into?  In this case, the $\ce{CH2}$ molecule has C2V symmetry (like the water example shown above).  Each of the atomic orbitals (AO's) on carbon (left) and the combination of $1s$ orbitals on the hydrogen atoms (right) has a symmetry label that corresponds to one of the irreducible representations of the point group.
You can determine by symmetry label for the AO's by operating on the orbital with each of the symmetry operators in the C2V group (E, C2, σ, σ').  For example, the 2py orbital is in the plane of the nuclei and so it is symmetric with respect to E and σ' but anti-symmetric with respect to C2 and σ and has b2 symmetry from the C2V character table.
For the hydrogens, you have to consider the combination of both 1s orbitals at the same time.  The a1 combination is when the phase of the wavefunction is the same and the b2 combination is when the phase is opposite (i.e. H1 is positive and H2 is negative or vice versa).
AO's that have the same symmetry can combine to form MO's with that symmetry label.  In the above example, the 2s and 2pz orbitals on oxygen have a1 symmetry, and they combine with the a1 combination of hydrogen 1s orbitals to form three MO's with a1 symmetry.  The 2py (b2 symmetry) orbital combines with the b2 combination of the hydrogen atoms to form two MO's. Finally, the 2px orbital remains unchanged because there are no other SLAC's with b1 symmetry.
The MO's are qualitatively ordered based on the number of nodes that are found when you combine the SLAC's.  The numbers in front of the MO's differentiate the orbitals with the same symmetry.


text:
sp, sp2 and sp3 hybrid orbitals are actual solutions of Schrodinger's equation or only a heuristic?
According to Wikipedia's article about orbital hybridisation:

[...] today it is considered an effective heuristic for rationalising the
structures of organic compounds. [...] Hybrid orbitals are assumed to be mixtures of atomic orbitals, superimposed on each other in various proportions.

So are $\mathrm{sp}$, $\mathrm{sp^2}$, $\mathrm{sp^3}$ hybrid orbitals 'real'? Are they solutions of Schrodinger equation? Are they a superposition of atomic orbitals?

The classic solutions to the orbitals for hydrogen atoms (s,p,d,... Legendre polynomials)  form a complete basis set.  Any wave function can be written using linear combinations of them. As for 'are they real', various STM and other surface imaging techniques show them to be amazingly accurate in some circumstances.

@JonCuster, I was actually asking about the hybrid orbitals sp, sp2 and sp3. In Modern valence band theory it says that the molecular orbitals are LCAO and not superposition of only s and p.

From the wikipedia link it says , LCAO is a quantum superposition, which again will lead you to a dead end . Reality and Quantum mechanics just dont go well together, yet.

As far as I remember, hybrid orbitals are not real. I don't exatly remember were I read it. Maybe in Atkins or in Class Notes.

Obligatory reminder that in general "orbitals" are not real - they are the solutions to mathematical models of the behavior of electrons. The electrons are real. Thus, the hybridized orbitals are no less real than molecular orbitals or atomic orbitals. They are just the solutions to a different set of equations modelling electron behavior. The better question would be Are hybridized orbitals a useful model?

The hybrids $sp, sp^2, sp^3$, are not solutions of the Schrödinger equation in a polyelectronic atom. The atomic orbitals $s$ and $p$ are solutions, but their linear combinations are not, because their energy level are different. Not very much different I know. But if the energy levels of $s$ and of $p$ were equal, their linear combination would be solution of the Schrödinger equation. And this may only happen in the Hydrogen atom.

As soon as you move to systems with more than one electron them all orbitals are approximations (i.e. not real). Hybridized orbitals are no more "heuristic" than atomic orbitals.
The transformation between hybrid and atomic orbitals can be mathematically, though not uniquely, defined (e.g. as the minimization of inter-orbital electrostatic repulsion) and the transformation does not affect the total electron density.
See for example http://molecularmodelingbasics.blogspot.dk/2010/03/canonical-and-localized-molecular.html and http://molecularmodelingbasics.blogspot.dk/2010/04/double-bonds-banana-and-otherwise.html

so minimization of LCAO energy is not an asymptotic approximation of actual ("real") orbitals?

No, because there are no actual ("real") orbitals

I'm not sure what you mean. It sounds like you're rejecting a core assumption: hydrogen AOs obtained from solving Schrodinger equation represent actual orbitals of the electron of hydrogen.

No, for hydrogen what you are calling AOs are actually wavefunctions, from which you can compute the densities which is are physical observables.  It's when you start using these 1-electron functions to describe multi-electron wavefunctions that they cease to be "real".

My understanding is the following: You can only get exact analytical solution for the Schrödinger equation for a) a single electron and b) a single proton - nucleus. That is, for the hydrogen atom with a single electron. The solution for this atom exactly describes the wavefuntion for the electron in any of the orbitals (the base one, or a excited one).
When you have more than an electron, and/or more than a nucleus, then you cannot get exact analytical solutions for the Schrödinger equation.
(I am not sure, if you have a multiproton nucleus with a single electron, e.g. He+, if you can get exact solutions; I believe so).
As for atoms with multiple electrons, you can consider that the efect of other electrons shields the charge in the nucleus, and then you can approximate the orbitals in this case to the single electron case.
When you have multiple atoms forming a molecule, and there are shared electrons positioned around two or more of their nuclei, you cannot get an exact analytical solution - even if you still had a single electron, as in H2+. But it was found that by adding or substracting the atomic orbitals, you get a reasonable approximation for the molecular orbitals that form covalent bonds. Eg. By combining the s orbitals of two H atoms, you get a close approximate solution of the molecular orbital of the H2 molecule. (Note that in these H2+ or H2 examples, we are not "using" hybridization yet.)
When you consider covalent molecules formed by atoms with electrons in different types of orbitals, e.g. s and p, then again, it was found that a reasonable approximation for the solution of the molecular orbitals was obtained by a) combining the wavefunction of the different orbitals in each atom (the ones with similar energy, symmetry, etc.) to calculate a "hybrid orbital" and b) combining the hybrid orbitals of each atom to get an approximation of the wavefunction of the molecular orbits.


text:
What kind of math should I expect in first year chemistry?
The course calendar for my university describes Chem 101 and 102 as "math intensive". However, it doesn't say what kind of math is involved.
What kind of math should I expect in first year chemistry?
Would it be helpful to have first year calculus under my belt first?

Welcome to chem.SE! Also, which country? It could differ you know.

@MARamezani - Thanks for the welcome. Re: which country? Canada. I think that the curriculum here is the same as it is the US. The MCAT website categorizes the tested materials according to classes that teach it; they don't distinguish between American and Canadian schools. Likewise, the Canadian medical schools that I've looked into don't distinguish American credits for their prerequisite classes from Canadian ones, but for other countries, the prerequisite credits vary.

A quick search gives me Calculus and Linear algebra from here

Let's answer three ways:

Specific expected math  Important mathematical skills will include algebra (including understanding variables and constants conceptually), interpreting graphs, and logorithms. There is very little chance you will need to know calculus in first year chemistry. If you have specific concerns, like those described in the comments, your advisor or department assistant should be able to directly address those.
First year chem philosophy  A good goal for first year chemistry is to understand fundamental chemistry concepts so that you can apply them in future courses and work. To do this, you need to know enough math that it doesn't prohibit you from understanding the concepts.
Continuing Education This one is just my opinion and I'm sure many will disagree. If you want to work in chemistry, take more math and physics than the department requires. It will give you a more thorough and complete understanding of chemistry. It will give you an edge in the future in solving interesting problems. This can be done in parallel however- the department won't expect that you're doing this.

Bonus: As many people before me have suggested, go to 6-8 classes on syllabus day even if you're only signed up and intending to take 4. Its always better to know your options and theres no risk to going and seeing whats available. In this case, you'll learn if you've got the background the professor expects for this course and can compare it with other options.

There is relatively little mathematics required for a typical first year chemistry course beyond what most will have studied at school.
A solid understanding of algebra, trigonometry and diffrentiation/integration is necessary. More advanced topics which may be useful include partial derivatives, solving simple first and second order differential equations, solving partial differential equations, group theory, Fourier transforms.
Appreciation of group theory in particular is important for a good understanding of spectroscopy and quantum mechanics. A good (but difficult!) introduction to this topic with suitable examples is Chemical Applications of Group Theory by F.A. Cotton. For the other mathematics you might find Mathematics for Engineers and Scientists by Alan Jeffrey useful, although there are lots of other texts which cover this sort of material well.

To clarify, you're not suggesting that spectroscopy and quantum mechanics will be covered from theoretical perspective in first year chemistry are you? I would expect that at most students would be exposed to spectroscopy as a block box that helps you in the lab.  Also, can you give an example of problem a student would need to solve first year that requires differentiation/integration? I can't think of one.

Many of my physical chemistry problems in first year (at a European university) in kinetics or thermodynamics or chemical physics used calculus. Maybe the pedagogy is different at other universities but I struggle to think how you could teach any substantial physical chemistry without it.

@J.LS IRC physical chemistry is taught as its own course at my (and I presume, at most north american) schools. At least, from what you're saying, I hope that's the case. Word on the street is that there's reason for students to fear integrals and derivatives. Learning those as their own concept along with chemistry would be an intimidating proposal.

Most of the first year physical chemistry we read (chemistry major) included lots of calculus, specifically things like euler chain law, partial diffrential equations, various operators like laplacion. For both multi-component thermodynamics as well as quantum chemistry depends on it. For chemistry minor courses thermodynamics was bit relaxed but still considerable amount of calculus was there in quantum chemistry.


text:
Old sample of sodium tetraphenylborate—why it is red?
I have a really old sample of sodium tetraphenylborate. It should be white, but it is red. Does anyone know what was formed and how it influences precipitation of cations in water?

Hm. How was the sample stored? It might give us an idea of what was reacting with it.

Yes, more details would be useful.

@BreakingBioinformatics because it was stored about ~30 years in closed bootle, water, oxygen and carbon dioxide could somehow get in. Tetraphenylborate ion does not looks stable in acidic medium, so oxidation of boron atom may begin. These are only my thoughts about what is going on here.

Haven't found information about tetraphenylborate ion and triphenylborane oxidation products  named boroxines. These could be precipitated out of  tetraphenylborate ion containing solution with aluminium ion containing solution (aluminium(III) chloride) most likely due to its attraction to oxygen. Feel free to research ;)

Was the bottle glass, plastic, or metal?

Glass bottle with plastic covering which was additionally covered with isolating tape (or adhesive tape, i do not know exact English equivalent).

I work for a company that's produced STPB for 27+ years and I've seen the material decompose\turn colors in glass and plastic. Material shouldn't be stored in metal due to the corrosive nature.

According to Decomposition of Sodium Tetraphenylborate, which relates to the Savanna River site for nuclear weapons material production,

Stable aqueous NaTPB solutions are a transparent,
  flesh-toned color. During the initial stages of decomposition (or
  instability),the solution turns a golden color. As the quantity
  of TPB- that has decomposed increases, the solution turns dark
  red, then brown, and eventually black.

In the presence of oxygen "phenol and phenylboronic acid
are produced as well as benzene, biphenyl, and other benzene ring
derivatives".


text:
Failed synthesis of aspirin due to excess sulfuric acid
Last week, I tried synthesizing acetylsalicylic acid - the reaction is shown below - using $\ce{H2SO4}$ as a catalyst. However, as the title suggests the synthesis failed as I used too much $\ce{H2SO4}$ - approximately four times more than the prescribed volume. Needless to say, I had to redo the synthesis.

But now I try to understand why the synthesis failed - I know that catalysts are used in small amounts, but I thought that much of it would only lead to equilibrium state being reached faster.
I think that maybe the acidity of the reaction environment could've somehow explained why the synthesis didn't take place. But I don't understand why - shouldn't more protons in the environment lead to faster protonation of the carbonyl oxygen of acetic anhydride? Could you please explain it to me?

You want to protonate the carbonyl group on the anhydride without protonating the phenolic hydroxy group of salicyclic acid, because a protonated hydroxy group is a terrible nucleophile. Adding a large excess of aqueous acid would also hydrolyze the anhydride...

@J.LS Thanks for answer. I'm not sure if I understand correctly - there are two competitive processes - protonation of the phenolic OH group and of the carbonyl group of acetic anhydride. The concentration of protons affect the equilibrium between the two (leaning on the salicyclic acid side when concentrations are high). And that's what happened? What do you mean by hydrolization of anhydrate? That the protons and OH group of acetic anhydride formed water and the reaction do not occur environment with water?

@Studen - You will just protonate both. The anhydrate will undergo an acid catalyzed reaction with water to give two equivalents of acetic acid.

@J.LS Because I added too much H2SO4 most of the anhydrate reacted with water forming acetic acid. It may be a stupid question, but where did the extra oxygen come from (to make two acectic acids from one anhydrate I need another oxygen - I get it from water, but the oxygen in water comes from? air?). Is protonation of the two compounds independent of the concentration of H2SO4? Because I thought that they are both protonated always.

Wait, wait, how do you know you have to redo the synthesis? It's not that obvious.

The reaction of the anhydrate is via nucleophilic attack of water to the carbonyl carbon, try drawing a mechanism. The oxygen is the same one as the one in the attacking water molecule !

@Mithoron I know, because aspirin did not crystallize in the next step, even after cooling it down (the solution with added water).

@J.LS Thanks for answering! This I understand, but the synthesis took place in water free conditions. So where did it come from? Is it from the protonation of anhydrate?

I assume you were using aqueous sulfuric acid ?

Concentrated sulfuric acid.

If you were using reagent quality anhydrous sulfuric acid then you probably wouldn't get any hydrolysis of the anhydride.

But I don't understand why - shouldn't more protons in the environment lead to faster protonation of the carbonyl oxygen of acetic anhydride? Could you please explain it to me?

The short answer is "side reactions".  Without more information about your exact experimental conditions, or even better, data about the obtained yields of the desired product and of byproducts, it isn't possible to explain conclusively what went wrong, but here are some ideas.

Acetyl sulfate formation.  In the comments you you implied that you used anhydrous sulfuric acid.  Under those conditions, acetic anhydride can undergo solvolysis to yield acetyl sulfate and water.  $$\ce{CH3COOOCCH3 + 2H2SO4 <=> 2 CH3COOSO3H + H2O} $$  Acetyl sulfate might be a less active acetyl transfer reagent than acetic anhydride, or the water formed may hydrolyze some of the acetic anhydride or even the formed acetyl sulfate.
Dehydration/sulfation of the starting material instead of acetyl transfer.  If you really did use anhydrous sulfuric acid, it is a powerful dehydrating reagent.  It could be possible that you are sulfating the alcohol or even the acid group (or both?) of the substrate, forming either o-sulfoxybenzoic acid (sulfation at $\ce{-OH}$ group) or sulfosalicylic acid (sulfation at $\ce{-COOH}$ group).

The heat of mixing of concentrated sulfuric acid with water is about $-74~\mathrm{kJ~mol^{-1}}$, judging from Figure 34b of a NIST treatise on the subject.  The heat of reaction for hydrolysis of acetic anhydride is only $-41~\mathrm{kJ~mol^{-1}}$.  Thus under anhydrous conditions, I think sulfated products will be more thermodynamically favored than acetylated products.  Sulfated products will be far more soluble in water or sulfuric acid than would be acetylated products.
I would think that using more water in the reaction would help.  The reason is that the heat of mixing of sulfuric acid is a very strong function of water concentration, but the heat of hydrolysis of acetic anhydride is not.  So raising the concentration of water a little bit will disfavor sulfation significantly by lowering the heat of mixing of sulfuric acid, but will leave the heat of hydrolysis of acetic anhydride the same.  It's worth noting that this argument is purely thermodynamic, not kinetic.  But I don't think sulfation reactions are kinetically hindered under ambient conditions, so although it isn't always true, I think thermodynamic arguments are valid in this case.


text:
Why is a tin can that appears like galvanised steel rusting?
I bought a canned pineapple and finished the content. The part of the can that is in contact with the liquid appears like galvanized steel (crystalline surface or spangles sort of appearance), while the part that are not looks pretty shiny like the outside of the can.
I then washed the can and added water to it, and heated the bottom of the can with a butane torch. After the water was heated up, I removed the torch and left it out for a while.
When I came back later, I emptied the water but noticed that the bottom rim of can had quite a bit of rusting. After inspecting it closely, I found that there are spots of (concentrated) rust on the cylindrical wall of the can as well. There are also light patches of rust on the inside bottom of the can. The seam on the cylindrical wall that is in contact with water is also badly rusted.
I thought galvanised steel only rusts when the layer of galvanisation wears out.
So, is the tin made with galvanised steel?
If yes, why are there concentrated spots of rust on the wall and other rusting?
If not, why does the inside of the can look like galvanised steel?
And BTW, why does the emptied and washed can smell "metallish"?

First of all, any 'tin' can you are likely to find is actually made of steel. Tin is too expensive. Also, tin does not 'rust', although it oxidizes. Your rust is iron oxide. Galvanized steel is steel with a thin zinc coating, likely hot-dip galvanization. What you describe (crystalline surface or spangles sort of appearance) does sound like galvanization, although I am not aware of any steel cans that are galvanized.
I thought galvanised steel only rust when the layer of galvanisation wears out. This is true. If your can is indeed galvanized, then it has likely chipped off, possibly on the rims of the can where the lid it crimped or pressed together, or by cutting the can open. Strong acids will dissolve zinc  Also interesting to note: sodium chloride (salt) will dissolve zinc over time by converting the galvanization to zinc chloride, but that's not applicable here.
So, is the tin made with galvanised steel?
  Well, that does sound like what you were describing, but galvanized steel is not fit for foodstuffs. Zinc is mildly toxic and ingesting it can make you ill. I like feetwet's suggestion; that universally the inside of metal food cans are lined, by actually an epoxy that is sprayed on. This is almost certainly to be the case. Perhaps this makes sense: if your can was indeed galvanized, then the rational by the manufacturer was that the lining will keep the zinc separated and the food safe. Sounds legit. It might be good to be aware that the type of epoxy being used here is the one that uses as its plasticizer Bisphenol A and Bisphenol S (BPA/BPS). Ill leave it up to you to decide if that should bother you or not.
If yes, why are there concentrated spots of rust on the wall and other rusting?
Well, if there is rust, then we can deduce that water must have be coming in contact with the iron in the steel, so whatever was protecting the steel has some failure.
And BTW, why does the emptied and washed can smell "metallish"?
While zinc does have a metallic smell to it, iron oxide's is much stronger. Since you visually identified rust, this is what you are smelling.

Thanks. Yep I am aware that the can isn't made of tin. Just that it is commonly called 'tin'. What other material could give a galvanised steel kind of appearance?

@dennis97519 I believe zinc is unique in having this property, and I was able to find the correct term for it, 'spangle'. Zinc exhibits visible spangle. Other metals will exhibit spangle, but they are likely too small to see. Two examples would be silicon electrical steel, which predictably has silicon in it, and something called malleable iron.  Malleable iron has a low carbon content. Its something about the annealing process that cause the carbon to aggregate into tiny spheres of graphite.

I see, thanks. Also, isn't the zinc acting as sacrificial anode so that even if it is scratched off at particular spots the steel will still be protected and not rust?

@dennis97519 Correct

Most modern steel cans for food are given a hermetic plastic coating on the interior.  Before that steel cans were tin-plated.  (I believe all-tin cans haven't been used in over a century.)
Depending on how hot you got the can when testing it you could have melted or vaporized the plastic coating, exposing the unprotected steel.  Also, if the can is dented then the coating can be compromised and the steel will rust where exposed.
Finally, when you open the can you are cutting through the hermetic coating, exposing the steel at least on the edges where you cut, which is why opened cans quickly develop a "metallish" rust smell.

Believe it or not, the causative agent is the Ascorbic acid (presence in pineapple juice), which is known to cause pit corrosion (described here) with stainless steel! Here is a source "Effect of ascorbic acid on the pitting resistance of 316L stainless steel in synthetic tap water", to quote:
"Above the 10^-4 M of A.A concentration, A.A generates soluble chelate rather than absorbs on the steel surface and it causes passive film deterioration and severe pitting corrosion."
And per the question: "why are there concentrated spots of rust on the wall", which supports not normal rusting, but pit corrosion, which I have experienced. Interestingly, I actually did a related thread on it in another forum when my cutlery was unexpectedly (and seemingly, unexplainedly) attacked. Others reported similar experiences, with my experience starting with mango juice exposure to stainless steel steak knives left unwashed in a sink.
Apparently, in the presence of moisture, oxygen, iron, Ascorbic acid  (a source of H+ and a chelate) and tap water minerals (electrolyte), an electrochemical based pit corrosion reaction can proceed with time.

The tin is on the outside surface of a tin can . The inside may be zinc coated ( as apparently yours was ) ,tin coated ,or coated with an organic, eg. lacquer. These protective coatings have limits to life depending on the contents ( salts , pH , and the particular acid ) and any damage - scratches, dents.

Food cans are usually tin coated steel. They may or may not have an organic coating, and many fruit cans leave exposed tin to reduce browning of the fruit with age. The surface phenomena that you see that looks like galvanizing is the organic acids in the fruit attacking the tin, which forms tin oxide and has a grey modeled appearance. It was just as shiny on the inside as you see on the outside when the can was first filled.
The tin coating is only resistant to corrosion in the absence of oxygen; once you opened the can, the tin is able to oxidize and leave exposed steel, subsequently allowing rust to form.