text:
relations between topology and organic chemistry
I am a Ph.D student studying  topology and geometry (in mathematics). I find a lot of references giving applications of geometric & combinatorial topology in organic chemistry. For example:
Geometric and Topological Thinking in Organic Chemistry
Graphene, vortices and Graphene, vortices and
the index theorem the index theorem
On borromean links
Is it possible for me to find a career in organic chemistry or organic chemical engeineering after graduation? If I want to do so, could you give me suggestions what background knowledge/textbooks should I learn?
Thanks!

@R. S. is right. If your question was about topology in chemistry and/or those links, it might fit better. But you qquestion is "Is it possible for me to find a career..." which isn't a chemistry question.


text:
How would Dalton's law be affected when there are two ideal gases in a container at different temperatures?

How would Dalton's law be affected when there are two ideal gases in a
  container at different temperatures?

Let the gas with higher temperature be gas A and the gas with lower temperature be gas B. Then heat will be transferred from gas A to gas B due to which kinetic energy of the molecules of gas A will decrease and kinetic energy of molecules of gas B will increase. Hence, molecules of gas B should make more collisions to the walls of the container because of the increased kinetic energy and hence the total pressure exerted by gas B should increase and using the same argument, pressure exerted by gas A should decrease.
Now, we can no longer use P(A) + P(B)= P(T)
; P(A) is the partial pressure of gas A,
P(B) is the partial pressure of gas B
and P(T) is the total pressure exerted by the mixture of gas A and gas B
Am I correct with this logic?

A glass bulb of volume 400 ml is connected to another bulb of volume
  200 mL by means of a tube of negligible volume. The bulbs contain dry
  air and are both at a common temperature and pressure of 293 K and
  1.00 atm. The larger bulb is immersed in steam at 373 K ; the smaller, in melting ice at 273 K. Find the final common pressure.

In the above problem, we cannot simply use PV= nRT to calculate final pressure because we don't know the final common temperature. So, we have to individually calculate the partial pressure of both the gases and then sum them up to get the total pressure. But according to me, as stated above, that should not be correct.
Is my reasoning correct or I am missing out on something? Also, how should I solve using my logic and please tell if there are other ways to go about solving this problem.

According to the problem description, the two bulbs do not reach a common final temperature.  The temperatures in the two bulbs are at 273 and 373 in the final state.  Apparently, the bulbs are isolated from one another.  So even though the pressures of the gas in the two bulbs are equal, the temperatures are not.

When the stopcock is opened, moles of one gas will interact with moles of other gas, and because there is temperature difference, heat will flow and hence kinetic energy of the molecules of those gases will be affected. How will I take into account the pressure and temperature change at the same time?

In the end, when the system equilibrates, all the gas in one bulb will be at 273 and all the gas in the other bulb will be at 373.  We are not concerned with what happens in between the initial state and the final state (in this example).  Our focus is only on the final state.  There will be negligible heat flowing through the tube of "negligible volume" at the end.  You know the total number of moles and you know the two volumes and temperatures.  This is enough information to find the final pressure and the distribution of gas moles between the two bulbs.

The final state is when the stopcock is opened and the gases in individual bulbs interact, and we are required to find the final pressure corresponding to that state. According to you, what is the final temperature when the gases interact.

As I already said, they want you to assume that the final temperatures in the two bulbs are different.  I don't know how to say it any other way.  The temperatures do not equilibrate.

What makes you think when the gases interact they won't change temperature. When two entities come in contact and they are at different temperatures, heat will flow and they will come to equilibrium.  What makes you think they will not?

This is what makes me think that:  "The larger bulb is immersed in steam at 373 K ; the smaller, in melting ice at 273 K. "  The problem statement didn't say anything about next taking the system out of contact with these reservoirs, and then allowing it to thermally re-equilibrate.  So this was my interpretation of the problem statement.  Was my answer of 1.14 atm for the final pressure correct?

See the term : final common pressure? Why would the question maker want  you to find the individual partial pressures of the gases and then sum them up. It doesn't make sense to have connected the bulbs at all with this reasoning. Yes, your answer is correct. And even I got this answer using the same procedure as yours but it is not practically convincing to me.

I think that we have both concluded that the problem statement was confusing and not very well specified.  But, I didn't see anything in the problem statement about partial pressures.

Incidentally, we are dealing here with a single component system (if we treat air as a single component, with MW 29).  In over 50 years of experience in ChE, I have never heard of the term "partial pressure" being used in connection with a single component system.

Well, when two ideal gases are in the same bulb (they will be when the stopcock has been opened) we will talk of partial pressures, at least that's what I have been taught. YunfeiMa correctly pointed out that all the collisions are elastic and that answers the first part of my question. Maybe "we" have got the answer the wrong way because you treated both the gases to be non interacting and a mixture of gases do behave that way in  container. ALl this while you been assuming that both the gases are never in contact, maybe that's your interpretation but I don't agree with you.

Bottomline: I am done with this question. !!

You cannot have two IDEAL gasses in a homogeneous mixture at different temperatures. As the collisions are stipulated to be non-elastic, all the kinetic energy of the particles (temperature) will be instantly transferred and spread out so that all the gas particles follow the Boltzmann distribution.

Maybe not instant exchange of energy but it will be very fast, maybe you should consider the example of a beam of fusion neutrons (14 MeV) in a large tank of high pressure deutrium gas, or maybe C4D10 gas (perdeuterobutane). The neutrons can be considered to be gas (like noble gas atoms) which then hit the other atoms (which are parts of molecules). I am sure that if we have an infinite tank of C4D10 that the neutrons will soon be at thermal energies.

Determine total number of moles of air in bulbs:
$$n=\frac{PV}{RT}=\frac{(1)(0.4+0.2)}{(0.082)(293)}=0.025$$
Consider final state of system.  Let $P_F$ be the final equilibrium pressure, $n_1$ be the final number of moles in the larger bulb, and $n_2$ be the final number of moles in the smaller bulb:
$$n_1=\frac{(P_F)(0.4)}{(0.082)(373)}=0.0131P_F$$
$$n_2=\frac{(P_F)(0.2)}{(0.082)(273)}=0.0089P_F$$
$$n=n_1+n_2=0.022P_F=0.025$$
$$P_F=1.14\tag{atmospheres}$$

We can have two "gases" at different tempertures, neutrons can be considered to be gases.
If we consider a light water reactor then the neutrons formed by fission in the fuel pins are fast neutrons with high energies, they can be considered to be a very hot gas.
If they enter a medium where the temperture is much lower then when they strike atoms in the medium (such as water, graphite or even high pressure hydrogen gas) then exchange of energy occurs between the neutrons and the atoms which make up the medium.
If we choose a medium such as deutrium gas where the chance of capture is low (rather than something like boron trifluoride) then the neutrons will bang around from collision to collision and after a short while the translational energy of the neutrons will be lowered to the same level as gas molecules of at the same temperture.
Based on this observation from the operation of thermal nuclear reactors I strongly hold the view that as soon as you mix a hot gas with a cold gas that exchange of energy between the two gases will occur which results in the cold gas warming and the hot gas cooling to the same temperture.
While it is possible in some reactors to observe a change in the neutron energy spectrum as you change position in the reactor I can not think of an experiment which would allow you to do this with "normal" gases.
I hate to have to say it, but when a chemist spends 20 years of their life thinking about radiaoctivity and nuclear issues they start to see the world through a special lens.

How does this answer the question?

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Behavior of π-conjugated chromophores in the presence of light
In the presence of light, I have read that, structures comprising π conjugated systems like retinol, straighten out from their curled positions, what phenomenon drives this behavior and why does this need to happen? I do understand that for proper π-orbital delocalization in the conjugated system, I would need the p orbitals of all the carbon atoms to be in the same plane, so staying straight would be an assistance, but what I fail to understand is that how does the absorption of light have anything to do with this?

related http://chemistry.stackexchange.com/questions/46567/difference-between-thermal-reactions-and-photochemical-reactions also http://chemistry.stackexchange.com/questions/8910/why-does-the-energy-gap-for-%cf%80-%cf%80-transitions-shrink-with-the-size-of-the-pi-co

but that still doesn't explain the straightening out...!! It does explain the decrease in discretization of molecular energy levels and a red shifted emission spectrum for larger molecules but definitely not why the molecules straighten out in the presence of light of specific energies.

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Does timing matter when adding coconut oil to rice to reduce calories?
I ran in to a rice recipe that can reduce carbs and calories when adding coconut oil. The exact direction is that you 1. boil water 2. add coconut oil 3. cook rice as you normally would.
My question is, does it change the end result (end with less calories) if I add water, coconut oil, and rice at the same time?

I see no way that adding oil (a high-caloric substance) would reduce calories at all.

I'm voting to close this question as off-topic because it is not about chemistry. It is more suited to cooking.SE.

@bon are you able to just migrate it to cooking? I don't know how it's done, but this seems like the appropriate time to do so. They are well suited to tease apart nutritional claims from facts. My 2 cents - the only thing it does is sell more coconut oil.

@Michelle, the question of nutritional calories is more complicated than just chemistry - it depends on the biology of the thing that eats as well. While some people call that chemistry also, it's not a good question here, at least the way you wrote it.

@uhoh Migration can only be done by mods unless migration paths has been set up. They're not.


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How many d-eletrons are in the outer shell of following compounds:
How many d-electrons are in the outer shell of the metals in following compounds:
$\ce{ZnS}$ and $\ce{NiS}$
So sulphur does not has any $\ce{d}$ electrons and $\ce{Zn}$ has $10\ce{d}$ electrons.
But I am pretty sure that this question is more complex to answer like that.
What is considered as the outer shell? as $\ce{Zn}$ has the electrons in the $3\ce{d}$ shell but there is still a filled $4\ce{p}$ shell?
Does it also matter that it is a compound ? ($\ce{ZnS}$ not only $\ce{Zn}$). As I see no reason why they included $\ce{S}$ but I guess it has an important reason.
I am new to transition metals, any help would be really appreciated.

Sulfur does not have any "d" electrons as you said.
Before losing two electrons from the 4s sublevel, Zinc's outer sublevel is 4s (not 4p as you mentioned).  After those two electrons are lost, the outermost sublevel becomes the 3d sublevel which contains 10 electrons.
Yes, it does matter whether the atom forms a compound or not, so we can understand where the electrons are allocated/transferred.
Here is the general rule for how to consider electrons in atoms.  Electrons fill up an atom via the Aufbau principle (image attached).  However, after filling, they arrange themselves in the energy according to the their principle quantum number (1, 2, 3,...) and their sublevel type (s, p, d, f).  For example, 4s fills before 3d in zinc, but after filling, 4s gains a higher energy than 3d.


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Activated carbon filter vs. Reverse osmosis filter
I have a 5 month old baby and I'm currently using an activated carbon water filter with the city tapwater supply (I live in Mumbai, India and the water can be highly polluted sometimes with all sorts of things - bacteria and viruses, pesticides and other organic compounds, lead, mercury, arsenic, etc.).
My baby is due to start solids and water soon.  Should I install a reverse osmosis filter instead, because that supposedly eliminates all particles that are larger than a water molecule (yes, the beneficial minerals like calcium, iron, etc. would be gone but I would rather want all the harmful chemicals removed).

Check this website: http://www.kinetico.com/blog/post/ro-vs-carbon-filtration.aspx


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An explosion in a custard factory
I have a vague memory of a question on the BBC panel game QI which started with a long chemical reaction equation on the screen, asking what it might be.
The answer was definitely "an explosion in a custard factory", due in part to high quantities of fine powders in the air, but also some of the chemical constituents of custard powder.
Does anyone know what that equation would be, or how I might re-find it?
How would one go about getting back from that equation to custard?

but also some of the chemical constituents of custard powder

Don't blame it on the custard. As long as there's

dust of a combustible material, preferably with a small particle size
sufficient oxygen in the athmosphere
an igniter, such as a spark or an open flame

there's a good chance for a devastating dust explosion.
Your combustible material can be something seemingly unspectecular, coal dust, saw dust, flour, or custard powder.

UPDATE
The incident in the BBC game most likely took place at a factory in Banbury, UK in 1981. Their traditional custard powder, known as Bird's Custard, is egg-free and mostly consists of cornflour.

how you might use it to tell that the reaction was an explosion in a custard factory

In risk assessment for dust explosions of organic materials, product analysis rarely plays a role; the "equation" isn't interesting and it is usually assumed that all carbon ends up as $\ce{CO2}$.
In order to establish workplace safety, it is more important to determine, how "strong" a possible dust explosion might get.
A measure for the "strongness" of an explosion is the rate by which the pressure (rapidly) increases. This is expressed in the $\mathrm{K_{St}}$ value (in $\mathrm{bar\cdot m \cdot s^{-1}}$), and experimentally determined in standardized tests.
Based on the $\mathrm{K_{St}}$ value, materials of certain properties (particle size, rest humidity) are assigned to the explosion classes (ST 0 to ST 3), and suitable workplace safety measures are taken.

The question was about the reaction equation, how you might use it to tell that the reaction was an explosion in a custard factory. @MaxW's answer appears to have focussed on the high energy content of sugar.

A recipe for custard uses the following ingredients
    2 cups whole milk
    2 eggs (preferably free-range)
    2 egg yolks
    1/3 cup sugar
    1 teaspoon vanilla extract
    Freshly grated or ground nutmeg

In an industrial scale operation the sugar would no doubt be in a large hopper with some sort of auger feed to fill the hopper. The sugar dust can explode due to static electricity in the dust particles. So the reaction is sugar plus oxygen gives a boom!
$$\ce{C12H22O11 + 12O2 -> 12CO2 + 11H2O}$$
Here is an egg free custard powder video.

You could see an industrial scale custard facility also using some kind of non-dairy creamer or something to cut down on volume/cost, and that would just add to the effect.

I'd guess that just about any flammable material when reduced to dust could explode when ignited.

$$\ce{C12H22O11 + 12O2 -> 12CO2 + 11H2O}$$ Sugar can burn and explode? Never knew that before.

@PrittBalagopal Haven't you built any sorbitol rockets as a kid?:)

@andselisk Actually no, although I did hear about them before. I wasn't really the DIY type of guy anyway :)

The reaction given in the QI show was
$$\ce{C6H12O6(s) + 6O2(g) = 6CO2(g) + 6H2O(g)}$$
This basically translates to glucose in solid form with the addition of gaseous oxygen will, with the application of something (I am assuming heat because the show does not say) reacts to form gaseous carbon dioxide and water vapor.
I would like to add that many molecules such as nitrogen and sulfur are excluded from the formula for whatever reason, possibly because they may not have had a role to play. I am not a chemist so I do not know.
What I do know is that in no universe would the chemical formula given lead to the conclusion that it was specifically a custard factory that exploded. It could just as easily have been a chocolate factory or boiled sweet factory or even a marshmallow factory for that matter.
How the contestant rightly guessed a custard factory is beyond my understanding. Besides, unless glucose is used in the production of custard, it is downright impossible to conclude what she did. I assume that they use sugar (sucrose) in the manufacture of custard. Sucrose has the chemical composition of $\ce{C12H22O11}$. And also the presence of lactose (in milk) was not taken into account.

The way I've heard her God-level answer explained is that in UK schools, it was apparently (or perhaps still is) not uncommon to accompany the explanation of the sugar oxidation reaction with an experiment in which powdered custard was mixed with air and lit in order to create a controlled explosion. To me this seems a reasonable explanation why the contestant immediately connected sugar oxidation to custard, perhaps even without remembering that it was a sugar oxidation formula, just remembering the custard part. (after all, the only things we remember from chemistry class are the times when

con't "the teacher made something go bang or on fire. Or both.)"


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What is the preferred IUPAC name of o-toluidine?
Why does o-toluidine have 2-aminomethylbenzene as its IUPAC name?
Why isn't it named as a derivative of amine (like 2-methylaniline)? Aren't amines given a higher priority than alkanes in nomenclature?
See the Priority Chart given on Master Organic Chemistry for reference.

In principle, your rationale is correct.
Generally, the selection of a preferred parent structure is based on the seniority of classes, which gives priority first to characteristic groups expressed as suffixes. Since o-toluidine (note that this name is no longer recommended) contains only one characteristic group $(\ce{{}-NH2})$, it is named as amine.
According to Subsection P-62.2.1.1.1 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the retained name “aniline” is the preferred IUPAC name (PIN) for the functional parent compound. The systematic name “benzenamine” may be used in general nomenclature.

For the retained name “aniline”, substitution is permitted at any position. Substitution of such functional parent compounds is limited to substituent groups having a lower seniority than that denoted by the (explicitly expressed or implied) suffix of the functional parent compound. Anyway, in this case, this limitation is not relevant since o-toluidine contains only one characteristic group and carbon compounds (rings or chains) have a lower seniority than amines.
Therefore, the PIN for o-toluidine is “2-methylaniline”. The systematic name “2-methylbenzenamine” may be used in general nomenclature.

But why is it "2-amino-1-methylbenzene" rather than "1-amino-2-methylbenzene"?

@DGS This appears to be a separate question. The name “2-amino-1-methylbenzene” is unambiguous and describes the correct structure; however, it is not in accordance with IUPAC recommendations for several reasons. Even if we ignore the higher seniority of amines and name the compound as “benzene”, the numbering would be “1-amino-2-methylbenzene” rather than “2-amino-1-methylbenzene” since both names correspond to the locant set “1,2” (all prefixes considered together in a series of increasing numerical order) and “amino” is cited first as a prefix in the name (see Subsection P-14.4, f and g).

That's what I thought. I think it fit in with his question (yes, he specifically mentions aniline in one of his questions), but maybe it should get its own question.

I don't understand your answer,can you explain more?


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How will water move across a membrane that is permeable to only one solute?
If there are two solutions separated by a membrane, which is permeable only to one solute and the solvent, will there be any net movement? Let's assume the two solutes are glucose and sodium ions, with the same molarity, and only the glucose and water can pass through the membrane.

Is the membrane also permeable to the solvent? (water?)

Brownian motion describes there will be some movement. As glucose moves across the membrane, water will follow. I don't understand the nature of this question. Could you clarify? At the moment I think it's more suitable for chemistry.

Let's say the membrane is also permeable to water. James, I am interested if there will be any osmosis, also will there be any diffusion?

There will be net movement of glucose from high concentration to low concentration via diffusion, but not Na (as it cannot pass through).

@another'Homosapien' Let's say that's true. On one side, presumably there will be 1M of Na+ ions, and 0.5M glucose, then on the other side there is 0.5M glucose. If the solutions are in a U-tube, will there be differences in height of solution? Will there be anymore osmosis movement due to the differences in molarity (1.5 to 0.5 total solutes)?

There will be movement, but no net movement as water will keep moving from one side to another. When the side with Na+ becomes more concentrated, water will move that side which will increase concentration in the side not having Na+ by which water will then move that side. Thus there will be no net movement of water.

I am not exactly sure, so I would approach this question mathematically. And there is no point in differentiating osmosis from diffusion since osmosis can be thought of as diffusion of the solvent.

Ya so lets consider it mathematically. Lets assume a beaker of base 10cm x 20cm, of any height. Add 2 litres of water to it, it would come to a height of 10cm. Dividing the beaker equally into two with the semipermeable membrane of your specifications, we'll be left with 1 litre on each side. Now add 1 osmole glucose on one side and 1 osmole NaCl on the other (you can't add only Na+ ions). So now we have the setup, we sit back and watch what happens.

Since we have to predict whats going to happen and we are not able to, I guess it'll be safe to assume that $\ x $ moles of glucose moves to the other side resulting in a change in concentration on both sides. It should be kept in mind that osmosis and diffusion oren't mutually exclusive. Both will happen. As I said, osmosis is just the diffusion of the liquid. So water also move causing not only a change in concentration due to the change in volume distribution but also a change in hydrostatic pressure, a concern that you have pointed out.

Let the change in height be $\ y $ cm. Now the only constraint I can think of is that the pressure should be equal at the base. By pressure i mean the sum of both hydrostatic and oncotic pressure. Oncotic pressure is given by $\ P_o = icRT$ and hydrostatic is by $\ P_h = h\rho g$.
Now I write $\ P_{o1} + P_{h1} = P_{o2} + P_{h2} $
That'd give $\ (h+y)\rho g + \frac{n+x}{(h+y)A}RT = (h-y)\rho g + \frac{n-x}{(h-y)A}RT$
Where $\ h$ is the height 10cm , $\ n$ is the no. of moles which we fixed as 1 mole (or rather, osmole)

So rest all are constants and the only variables left here is $\ y $ and  $\ x $ which we need.
On solving, the equation comes to a cubic in $\ y$ which is actually not so desirable!! So I showed all this since this is the way I would proceed. But I am now struck, so I too want help from here on!! I guess it'd be better posting this on a chemistry forum.


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Do allenes form rings?
I was wondering whether allenes can form rings; geometrically speaking, this seems like it would be energetically unfavorable due to their 180 degree linear geometry.

Related: A monocyclic 6 carbon ring with 6 double bonds

Strictly speaking, cumulenes with even numbered of C=C bonds (odd numbers of carbons) are tetrahedral, so the bond angle strain not automatically a show-stopper.
That being said, these kinds of systems are not common.

Cyclonona-1,2-diene is the smallest allene in a ring that has been isolated (it's stable at room temperature, but dimerizes at higher temperatures).  Trapping experiments have suggested the possible  intermediacy of cycloocta- , cyclohepta- and cyclohexa-1,2-diene.

Yes they do. Such compounds are even found in nature, e.g. vernonallenolides.

I think OP means "Do allenes react with themselves to form rings," not "Do allenes exist within rings?" I like the post, I just don't think it answers the question's intended meaning. Actually, I may well be wrong. Maybe OP could elaborate.

Well, I guess my main question was whether rings of allenes would be stable enough to exist at all, which was answered by the examples above...However, your question is an interesting one too that I would love to know the answer to.

As small as six-membered rings with allenes can exist, albeit fleetingly. Cyclohexa-1,2-diene has been known since a 1966 report by Georg Wittig (Angew. Chem. Int. Ed. 1966, 5 (9), 846), in which it is prepared by direct elimination of HBr from a vinyl bromide using t-BuOK:

For a more modern take on the matter, see e.g. Nat. Chem. 2018, 10 (9), 953–960 where Neil Garg and coworkers report the generation of azacyclic six-membered cycloallenes. The allene is generated by treatment of a 1,2-silyl triflate with CsF (this strategy will be familiar from benzyne chemistry). The cycloallene intermediates are trapped in various cycloaddition reactions.


text:
How many d-electrons are in the outer shell of the metals in following compounds:
How many d-electrons are in the outer shell of the metals in following compounds:
ZnS
Do I just use Zn^2+ and as I know that in Ions the s orbitals are removed first I count 10 d electrons?
However ZnS makes a metal lattice. Can I use Zn^2+ then?
Help would be really appreciated

Don't ask about the same stuff twice! Don't you know you can edit it?

In zinc sulfide, two s-electrons have been removed from $\ce{Zn(0)}$ to make $\ce{Zn^{2+}}$. Therefore, yes, it does have 10 d-electrons.

In its lattice, zinc is bound to four sulfur atoms which fulfills the 18-electron rule (10 from $\ce{Zn^{2+}}$ and 2 each from four separate $\ce{S^{2-}}$; the 18-electron rule for transition metals is the equivalent of the octet rule for main-block elements). In this case, however, the metal (zinc) still only has 10 d-electrons. The additional electrons are in hybrid, non-d orbitals.


text:
Can I convert from an emission amount (in tonnes per area) to a concentration (in ppbv)?
Hi, I was wondering if it is possible to estimate the concentration of SO2 given the total mass of SO2 emissions?
Europa's EDGAR has a resource called "Global Emissions EDGAR v4.2" (see http://edgar.jrc.ec.europa.eu/overview.php?v=42). A database of SO2 emissions in Metric Tons per Area is provided, but I need concentration instead.
Currently I am determining the volume of the gas by assuming it's at STP, then I am dividing this by the estimated total containing volume, which I tried to estimate as follows.

Total Containing Volume = (11120 m*11120 m)*altitude

or

Total Containing Volume = (apprx. area of 1° latitude by 1° longitude "square")*altitude

I calculated an altitude that makes the calculation match another database of SO2 concentrations that I have. Unfortunately the other database of SO2 concentrations is limited in the area that it covers, so it is not suitable.
Is this estimation attempt misguided or is there a better way to do it? If there is another database that contains global or continental SO2 concentration grids that would also fix the problem.

What you are doing seems reasonable but there are many hidden assumptions in your technique:

The data source you are using actually provides a rate, i.e. metric tons per area per year.
Thus your calculation is also implicitly a rate: the results will be in tons per volume per year.
You are assuming that the only source of atmospheric $\ce{SO2}$ is the emissions quantified by your data source.  It may well be that there are other, natural sources of $\ce{SO2}$, or more likely, natural sinks of $\ce{SO2}$.
If you want to know actual concentrations, you will need to quantify all the other sources and sinks of $\ce{SO2}$, and integrate all sources and sinks over time.


text:
Why is NaBr only a reducing agent? Can you tell if something is a reducing/oxidizing/both agent?
I came across a question today that read "Which of the following can only act as a reducing agent?" The answer was $\ce{NaBr}$, and I knew this from a previous question I had seen before. Is there any way to determine these properties without rote memorization?
(If it helps to use as an example, the other answer choices were $\ce{I2,\;BrCl,\;HIO4}$)

Think about the state of the atoms in the compound and where they can go from there:

$\ce{I2}$

In this molecule, the iodine atoms have $0$ oxidation state, and are easily reduced to the very stable $\ce{I-}$ ions, with $-1$ oxidation state. This means this compound is capable of reacting as an oxidizing agent.

$\ce{BrCl}$

Likewise, in this molecule, the bromine and chlorine atoms have $0$ oxidation state, and are easily reduced to $\ce{Br-}$ and $\ce{Cl-}$ ions, which both have a $-1$ oxidation state. This means this compound is reacts as an oxidizing agent.

$\ce{HIO4}$

The halogen oxyanions ($\ce{F^{(I)}O- ,\ Cl^{(III)}O- ,\ Br^{(V)}O3- ,\ I^{(VII)}O4-}$), and really any oxyanion for that matter, have the marked ability to act as either oxidizing or reducing agents. These anions can accept electrons on the central atom in exchange for bonds to oxygen, acting as an oxidizing agent. They can also also accept additional oxygen bonds to the central atom, acting as reducing agents. Of course there is a limit to how reduced and how oxidized an oxyanion can become, but what I am trying to get across is that as a whole these species can act either as an oxidizing or reducing agent.

$\ce{NaBr}$

Here the ions exist as $\ce{Na+}$ and $\ce{Br-}$, which have $+1$ and $-1$ oxidation states respectively. Here, $\ce{Br-}$ is unable to accept any more electrons and can only donate them to form $\ce{Br2}$, making it an effective as a reducing agent. It can also be oxidized from $\ce{Br-}$ in the presence of $\ce{O3}$ to $\ce{Br^{(V)}O3-}$, and is still acting as a reducing agent.

Na+ may be reduced to metal by Cs and some other reducers.

Good luck with that.

Haha, but do you have proof it can't happen with molten salt and metal? Overpotential or sth?

I'll cede you that in terms of electrochemistry, $\ce{Na+}$ could be reduced, but it's going to be very hard to do and won't act as an oxidizing agent in most normal situations.

In the presence of something reasonably oxidizing, it could form $\ce{Br2}$ and reduce whatever oxidizes it. It could also oxidize to $\ce{NaBrO_{x}}$.

Those other things are not going to easily oxidize because that would put a formal positive charge on a highly electronegative atom. In the case of $\ce{HIO4}$, it would increase the charge on iodine from $+7$. This would be very hard.


text:
Reaction Penetrating Metal Sheet
I'm writing a novel set 28 years in the future in which the protagonist attempts to manipulate a communications satellite by directing a swarm of nanorobots (or possibly micro-robots, depending on plausibility).
An initial wave of nanorobots is tasked with creating a path, such as a tunnel, through the satellite's casing so that a second wave can enter.
Inspired by real-life nanotechnology that delivers chemotherapy drugs only to leukemia cells, I thought to have each nanorobot carry a tiny amount of super acid, each depositing its payload onto the same spot of the metallic hull of the satellite. My thinking is that the vacuum of space would carry off the product of metal + acid, leaving fresh metal for the next microrobot to attack.
Is it plausible to erode a narrow path through metal in this manner? What acids would be well-suited to the task? How would I go about calculating the order-of-magnitude number of nano-robots needed to complete a path through a given thickness of (let's say) aluminum?

I'm not sure you'd need a super acid. Also, how would they contain that? I think maybe just a strong acid/oxidizer would work.

I'm not sure of the specific mechanics of containing a super acid, though in practice highly potent chemotherapy drugs are already contained in nano-scale containers (see linked article). I did come across nano-thermite (https://en.wikipedia.org/wiki/Nano-thermite) The article speaks of macro-scale use but it seems it might work as well if delivered at small scale to a tiny target.

What would the protagonist prefer to do: a) turn the satellite into debris, b) only kill its function, or c) reuse it for his own purposes?

There could be problems with making them impervious  to aggressive payload. Chemistry aside it looks like terribly problematic way to try to manipulate a satellite. For example how would you delver them to target in such precise manner?

@KlausWarzecha: The idea is to create a small entry in the hole, then a different type of nanoprobe will enter through that hole to reprogram the satellite's encryption system.

@Mithoron: The nanoprobes are deployed from a stealthy ship in a nearby orbit. I currently plan to have them ride a laser from that ship targeted at the satellite (idea based on http://www.space.com/32026-photon-propulsion-mars-three-days.html). Part of the drama of the scene is whether the ship is "stealthy enough."

I knew that you would decide for the most challenging option: tinkering in geostationary orbit ;)

Depending on the material that the casing is made out of, you could have a mercury or gallium payload.  They both are capable of significantly weakening the structures of some metals (such as aluminum).  See the following video of gallium weakening a coke can.  Mercury can also weaken the structure, especially in conjunction with gallium as seen here.  Mercury will form an amalgam with aluminum and significantly weaken it, as will gallium.  You might conceivably have nanobots bring a payload of the two metals to weaken the structure, especially since they are not nearly as reactive with other things.
Unfortunately, this would only work if the structure you are trying to damage is primarily aluminum, but might give a starting place for non-acidic solutions.

Problem is it's supposed to be in orbit - there would be probably problem with any liquid as they don't stay this way in void, also @EricJ.

@Mithoron to contain any liquid until use, you just need to maintain its vapor pressure. For Gallium that's extremely low, and mercury only reaches 1 mm of mercury (1 Torr) at 400 K, dropping to a few mTorr at 300K. Once exposed to space, un-reacted/amalgamed mercury would slowly evaporate, but the gallium should stick around quite a while. cf https://www.powerstream.com/vapor-pressure.htm for plots of vapor pressure vs temperature.


text:
What do DNA crystals look like?
I had about 5 ug DNA suspended in water, which I dried thoroughly in a vacufuge.
I obtained a whitish crystal (very similar to table salt, although the structure seems uneven and it's hard to tell) at the bottom of my wells, apparently of about 30 ul volume (I am not counting the dead space in between bits of crystals).
Is this crystal DNA? Or is it some sort of salt that was bound to the DNA? It was my understanding that DNA is very hard to precipitate without using ethanol (I didn't) but it seems I precipitated something anyway.
The DNA came from a prep (a Qiagen kit), so it's highly pure - the A260-A280 ratio is about 1.8.


text:
Acidity of non-transition cations
My general chemistry textbook discusses how cations behave as (Brønsted-Lowry) acids in water. There's a picture of a $1.0~\mathrm M$ solution of Calcium nitrate with a pH indicator indicating a pH of around 6.9. How is it that $\ce{Ca^{2+}}$ ions behave as acids at all?
The book does explains how transition metals behave as acids: they form coordination spheres with water, and one ligand water molecule loses a proton to a free water molecule. The hydroxide is bound to the transition metal, thus not affecting pH.
However, to my knowledge, non-transition metals, such as $\ce{Ca^{2+}}$ or $\ce{Al^{3+}}$ do not form coordination compounds. Even if they can solvate some hydroxide ions, isn't the bond much weaker and therefore the hydroxide not effectively removed from solution like by a transition metal?
What would be the acid equilibrium expression for these cations?

Calcium actually does exactly what you're describing transition metal cations to do (binds $\ce{H2O}$ and then releases $\ce{H^+}$). Aluminum does it even better, as does magnesium. This is exactly why using $\ce{MgSO4}$ to dry solutions with highly acid-sensitive compounds is not recommended.

So why is a solution of $Ca(OH)_2$ basic? It dissolves to form $OH^-$ ions, but the subsequent $H_3O^+$ released by the solvated calcium ion counteracts that to some degree.

Calcium hydroxide forms a lot more hydroxide ions than the amount that get "lost" to binding with the calcium ions.  So there are plenty of hydroxide ions left over to make limewater basic.  Ditto for magnesium hydroxide solutions, albeit with much smaller solubility.


text:
Nucleophilicity/basicity in protic and aprotic solvents

In polar aprotic solvent, why is $\ce{F-}$ a stronger nucleophile than $\ce{I-}$?
What I think: If I talk about their basicity, $\ce{F-}$ would be more basic due to instability from its high electron density. Is it applied same when talking about nucleophiliicty?
Why is an atom with high polarizability a strong nucleophile?
What I heard: an increased atom size means a high polarizability, and a high polarizability means a high nucleophilicty because the atom can give its weakly-held $e^-$ easier.
What I think: above sentence makes sense to me. But in another point of view, isn't it also right that the atom with properties like that would be stable, making it poor nucleophile?
When talking about nucleophilicity, why does the polarizability matter in a polar protic solvent, but not in a polar aprotic solvent?
What I heard: when deciding the nucleophilicty of an atom in a polar protic solvent, there are two factors that come into play: solvation and polarizability.
What I think: in a polar protic solvent, since the effect of solvation is significant, the polarizability shouldn't mean so much. Rather, its effect should be significant in polar aprotic solvent since there's no hydrogen bonding. Is this wrong?

Why is an atom with high polarizability a strong nucleophile?
An atom with a higher polarizability will share his electrons more easier since they are less attracted by the nucleus. The more the electrons are near the nucleus, the more it's hard the share them with an electrophile. $\mathrm{I^-}$ is thus more nucleophile than $\mathrm{F^-}$.
As you'll se below, the solvent can affect the nucleophilicity and reverse what I just said.
In polar aprotic solvent, why is $\mathrm{F^-}$ a stronger nucleophile than $\mathrm{I^-}$?
When talking about nucleophilicity, why does the polarizability matter in a polar protic solvent, but not in a polar aprotic solvent?
First, basicity and nucleophilicity are not the same at all. For instance, a strong base can be non-nucleophile, such as LDA for instance.
Then, let's talk about the polarizability of the $\mathrm{F^-}$ and $\mathrm{I^-}$ anions. The fluoride has a small polarizability: its charge is dense, very localized. It's called an hard anion. Iodide has a great polarizability with its dispersed charge. It's called a soft anion.
Using a protic polar solvent, the anions will be well solvated, even better if they are hard anions because the hydrogen bondings will be more efficient. The anion $\mathrm{F^-}$ is thus more solvated than $\mathrm{I^-}$ and less reactive.
In the case of an aprotic polar solvent, the anion is not so well solvated and the fluoride will be more reactive.


text:
The stability of the cupric and cuprous ions
My teacher told me that $\ce{Cu^2+}$ is more stable than  $\ce{Cu+}$ in solution, but why? I think $\ce{Cu+}$ is $\ce{[Ar] 3d^10 }$ ,$\ce{Cu^2+}$ is $\ce{[Ar] 3d^9}$, why is $\ce{Cu^2+}$ more stable then?

possible duplicate - http://chemistry.stackexchange.com/questions/42855/why-cu-is-unstable-in-aqueous-medium

Hund's rules, which are where we get the idea that half filled and completely filled orbitals are more stable than partially filled orbitals, are specifically about the ground states of atoms. Hund's rules do not apply to ions in solution which are complexed with ligands (like water).
What you fail to consider is that $\ce{Cu+}$ is larger than $\ce{Cu^{2+}}$ but with only half the charge, so its charge density is much lower. As a result, so its enthalpy of hydration for $\ce{Cu+}$ is much less, meaning it is much less stabilized by the formation of the metal-ligand bonds. As a result, in aqueous conditions $\ce{Cu+}$ tends to disproportionate to $\ce{Cu}$ and $\ce{Cu^{2+}}$.


text:
Do the tetramminecopper(II) and hexathiocyanatoferrate(III) complexes obey the 18-electron rule?
Can anyone tell me whether $\ce{[Cu(NH3)4]^2+}$ and $\ce{[Fe(SCN)6]^3-}$ obey the 18-electron rule, and if not, why? I found $\ce{[Cu(NH3)4]^2+}$ and  $\ce{[Fe(SCN)6]^3-}$ have 17 electrons, was I wrong?

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This is an unsatisfactory answer so I will just comment: there is absolutely no reason why those compounds should obey the 18e rule.

@orthocresol if there is no reason,why we need 18-electron rule

Because there are other compounds that do obey it.

@orthocresol such as?

Try Google or Wikipedia. They're pretty good.

I found $\ce{[Cu(NH3)4]^2+}$ and $\ce{[Fe(SCN)6]^3-}$ have 17 electrons, was I wrong?

You are correct for the iron(III) complex, which has an uneven electron count and is a $\mathrm{d^5}$ system. You are almost correct for copper(II), but that complex is actually $\ce{[Cu(NH3)4(H2O)2]^2+}$ where the two aqua-ligands are further away from the metal centre. In any case, it is a $\mathrm{d^9}$ system and has 21 electrons.
No system in which the central metal has an uneven electron number can fulfill the 18-electron rule unless there are radical ligands around which happens rather rarely.
So what is the benefit of the rule? Well, complexes involving 18 electrons — that would be ten d-electrons, two s-electrons and six p-electrons in extremely simplified terms — are often noted to be more stable than others with respect to decomposition. So iron forms a pentacarbonyliron(0) compound ($\mathrm{d^8}$) with ten electrons coming from the five carbon monoxide ligands. This compound is rather stable with respect to oxidation and decomposition while a hexacarbonyliron(0) or a tetracarbonyliron(0) are not. But you should take this 18-electron rule with a much larger grain of salt then the octet rule for main group elements.

Only for some complexing agents do we typically expect the 18-electron rule to hold.  The best candidates are very "soft-base" ligands with strong pi-electron interactions, like CO.  Ammonia and thiocyanate do not do this.


text:
How does cooling increase the selectivity of a reaction producing the kinetic vs the thermodynamic product?
I know that cooling gives a lower ratio of molecules with sufficient energy to overcome the activation barrier, but I'm not sure how this selects for a kinetic or thermodynamic product. I also know that the kinetic product is favoured if there is not enough energy supplied to reach equilibrium, and the opposite for the thermodynamic product. Can cooling select for either product in different circumstances?

First, there is no rule that "the" kinetic product will be different than the thermodynamic product.  It depends on the particular reaction you are interested in.
Thus, I think you are really asking "in reactions where the kinetic product is different than the thermodynamic product, what does cooling select for?"
Let's consider two reactions, the first with a reactant R to a thermodynamic product T and a second, separate reaction where R goes to the kinetic product K.  The rate constants for these reactions are $k_1$ and $k_2$, respectively.
$$\ce{R ->[k_1] T}$$
$$\ce{R ->[k_2] K}$$
Let's suppose both reactions follow classical transition state theory, which is a very good approximation to many everyday reactions and is a great starting place even for the weird cases that do vary.
TST says (if you assume that the $\kappa$ transmission coefficient is 1):
$k_1 = \frac{k_b T}{h} e^{-\Delta G^{\ddagger}_1/RT
}$
$k_2 = \frac{k_b T}{h} e^{-\Delta G^{\ddagger}_2/RT
}$
The "kinetic" product is the one with the higher rate constant; this is the definition of kinetic.  Doing some algebra,
$$\ln{\frac{k_2}{k_1}} = -\frac{1}{RT}\left(\Delta G^{\ddagger}_2 - \Delta G^{\ddagger}_1 \right)$$
...we see that for $k_2$ to be higher than $k_1$, i.e. for $\ln{\frac{k_2}{k_1}}$ to be greater than 0, then $\Delta G^{\ddagger}_2$ must be smaller than $\Delta G^{\ddagger}_1$ (note the negative sign).
The effect of temperature is still in the equation:
$$\ln{\frac{k_2}{k_1}} = -\frac{1}{RT}\left(\Delta G^{\ddagger}_2 - \Delta G^{\ddagger}_1 \right)$$
The assumption of TST is that $\Delta G^{\ddagger}$ is not a function of temperature.  As temperature is lowered, then the $RT$ divisor in the equation gets smaller, which means that the difference $\left(\Delta G^{\ddagger}_2 - \Delta G^{\ddagger}_1 \right)$ affects the ratio of rate constants more.  Thus, under the assumptions we have made, cooling will favor the kinetic product more. As temperature is raised, eventually $RT$ becomes much greater than $\left(\Delta G^{\ddagger}_2 - \Delta G^{\ddagger}_1 \right)$, so selectivity for the kinetic product vanishes because $\ln{\frac{k_2}{k_1}} \rightarrow 0$, i.e. $\frac{k_2}{k_1} \rightarrow 1$.
You might ask, how can the ratio go to one; doesn't that mean that we have assumed the equilibrium constant between T and K is also 1?  The answer is no, we haven't assumed that, but we have made a number of other assumptions that mean we we won't be able to predict the net rates of T and K formation.  The most important one is that we have neglected the reverse reactions $\ce{T -> R}$ and $\ce{K -> R}$; introducing them would give the same qualitative effect of cooling the temperature but would complicate the math considerably (but would let us wind up with net rates of T and K instead of the gross single-directional rates we have calcualted here.  Second we have assumed that $\kappa = 1$, which might not always be true.


text:
Making o-bromoaniline from aniline
How can I make o-bromoaniline from aniline (as a major product)?
I know how to make p-bromoaniline from aniline, by acetylation followed by treatment with $\ce{Br2}$/$\ce{CH3COOH}$ and hydrolysis. But in this process the ortho isomer (if formed) will be the minor product.

Did you consider different sequences? Remember that aromatic nitro compounds can be reduced to the corresponding amines ;-)

@Klaus Warzecha But even if I consider nitro benzene, the ortho isomer will be the minor product. Would the yield be satisfactory ?

Remember how the nitro group directs. If you start with nitrobenzene, chances are good that the bromination will almost exclusively yield the meta-substituted product. But what if you try to nitrate bromobenzene?

Didn't you consider the way through sulfanilic acid, the para-position will be protected. In this case both substituents will direct the electrophilic substitution in ortho- position with respect to amino group. Then remove sulfonic acid group by desulfonation and you will obtain your ortho-bromoaniline.

Wont it exist in zwitter ion form and loose its directing influence?

I think that you probably cannot avoid separating ortho from para isomers (and possibly disubstitution products) after using a method like the one you suggest or a method of blocking the para- position.
Another method that is probably not taught to undergraduates (?) is the Directed ortho metallation (DoM) or Snieckus chemistry. The image below has been copied from that link.

As outlined in a comment direct bromination of nitrobenzene with $\ce{Br2}$ and $\ce{FeBr3}$ is likely to give the meta-substituted product.
If you want to avoid time-consuming and expensive procedures, your options are:

Change the order of events. Nitrate bromobenzene under mild(er) conditions, or
Find conditions, under which bromination of nitrobenzene furnishes the ortho-product

For the latter, have a look at Chem. Commun., 1996, 2679-2680
(DOI). The authors state:

We have observed that monosubstituted benzene derivatives
  react with a mixture of dimethyl sulfoxide and aqueous
  hydrobromic acid […] to yield regiospecifically only the
  monobromo derivatives. It is interesting to note that electron-
  withdrawing groups such as $\ce{NO2}$, $\ce{CHO}$ yield ortho-bromo
  derivatives, while electron-donating groups such as $\ce{NH2}$, $\ce{OH}$ or $\ce{Me}$ give only para-bromo derivatives.


text:
Will there ever be a "last element"?
Recently the IUPAC added four new elements to the periodic table and it seems the race for discovering new elements is just getting faster. I was wondering if a "last element" exists; an element with so great an atomic number that it would be impossible to make anything larger.
There should probably be a bound placed on the maximum number of protons in an atom by quantum theory. Because, elements with extraordinarily large an atomic number should become so unstable that it would impossible for them to come into existence(in theory).
I do not know quantum theory so I am not sure; but is there such a possibility? Or the number of elements created is just limited by our current technology and the energy of our particle colliders meaning that this race will never end?

@Martin-マーチン You are basically saying I asked an already answered question and that too on the wrong site? Well, at least, thanks for being polite.

No that is not what I am saying. I am saying that questions regarding that topic in general have been asked before, here (it is on topic), but also on physics. I regret I am not a specialist on that, so I couldn't give you related information that you can dig in until someone else finds the time to answer. My comment was in no way critique. However, this question still might be on interest to you.

@Martin-マーチン that feels like a direct duplicate to me :/

@AbdurRahman Please have a look at the island of stability and you might find a good starting point for further investigations.


text:
Calculating rate constant of intersystem crossing from Jablonski diagram
I am to calculate the rate constant for $T_1 \to S_0$ intersystem crossing, with reference to the Jablonski diagram attached below. I know this should not be difficult, but for some reason I cannot put the number together to get the correct result.
I have that
$$
\Phi_\text{ISC} = \frac{k_\text{ISC}}{\sum_i k_i}
$$
where the denominator represents a sum of all decay pathways for the $T_1$ state. However, how can this sum be calculated if we do not know the rate constants to begin with?

The lifetime of the excited state is the reciprocal of the sum of rate constants which you are trying to find:
$$\begin{align}
\tau(T_1) = \frac{1}{\sum_i k_i(T_1)} &= \pu{6 \times 10^-3 s} \\
\sum_i k_i(T_1) &= \frac{1}{\pu{6 \times 10^-3 s}} \\
&= \pu{167 s^-1}
\end{align}$$
The two lines going from the $T_1 \to S_0$ state represent intersystem crossing (ISC) and phosphorescence. Squiggly lines represent nonradiative transitions, of which ISC is one, so the quantum yield for ISC is 10%. (Straight lines represent radiative transitions, in this case phosphorescence.)
Therefore:
$$\begin{align}
k_\text{ISC} &= \Phi_\text{ISC} \cdot \sum_i k_i(T_1) \\
&= 0.1 \cdot \pu{167 s^-1} \\
&= \pu{16.7 s^-1}
\end{align}$$

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


text:
Double Bond or Aldehyde priority?
Pent-3-enal. In the structure of pent-3-enal, the third carbon atom has aldehyde group or double bond?  Also is the priority same for all other functional groups?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), locants are placed
immediately in front of the part of the name to which they refer.

P-14.3.2 Position of locants
Locants (numerals and/or letters) are placed immediately before that part of the name to which they relate, except in the case of the traditional contracted names when locants are
placed at the front of their names.

Therefore, the locant ‘3’ in ‘pent-3-enal’ refers to the ending ‘ene’ (not to the suffix ‘al’).
The class name ‘aldehyde’ refers to compounds containing the group $\ce{{}-CH=O}$
attached to another carbon atom. In systematic substitutive nomenclature, the suffix ‘al’ is used for $\ce{{}-(C)HO}$ (see Subsection P-66.6.1). Thus, the aldehyde carbon atom is always a terminal carbon atom.
The characteristic group cited as suffix (here: ‘-al’) receives the lowest locant possible (here: ‘1’). However, terminal locants are omitted in preferred IUPAC names (PINs) for such aldehydes.

P-14.3.4.1 Terminal locants are not cited in names for mono- and dicarboxylic acids derived from acyclic hydrocarbons and their corresponding acyl halides, amides, hydrazides, nitriles, aldehydes, amidines, amidrazones, hydrazidines, and amidoximes, when unsubstituted or substituted on carbon atoms.

Therefore, the PIN for the example given in the question is ‘pent-3-enal’ (not ‘pent-3-en-1-al’).

The third carbon has the double bond, not the aldehyde.  Aldehydes exist on terminal carbons, so the aldehyde group is on the first carbon atom.  The aldehyde takes priority over the alkene (double bond).  See the attached structure.

The number 3 indicates the position of the double bond in this case and since there is no number before the -al suffix it means the aldehyde must be present at a terminal position i.e 1.
Always when there is a presence of a functional group, it is given preference over other unsaturations and other substituents and lowest locant number is assigned to it.


text:
What is the significance of "magic numbers" of atoms in nanoparticles?
In gold nano particles, as the synthesis method improves to form smaller and smaller nanoparticles, they tend to congregate in magic numbers.
I want to know if these magic numbers occur in other metals besides gold, and whether they contribute to stability, reactivity, catalysis, increased surface area, and their lattice structure? In short, what are the main differences between magic and non-magic nanoparticles?

https://en.wikipedia.org/wiki/Superatom and https://en.wikipedia.org/wiki/Cluster_chemistry

As indicated in the comment above, "magic numbers" (i.e., atom counts with much higher stability) have been known in atomic clusters of many types for a long time.
Each magic number has a specific geometry. Some are non-spherical, so it's hard to make generalizations about surface area or lattice structure.
A very famous non-gold example is buckminsterfullerene ($\ce{C60}$). Multiple groups had seen the signature of 720 Daltons in mass spectra of soot for years before the official discovery. There are other magic numbers for $\ce{C_n}$ clusters.
In nanoparticles and nanoclusters, this is an active area of research, but the main difference is stability. The so-called magic-numbers are simply more stable. (Case in point, the thiol-termated gold nanoparticles were created by heating up the samples until only the most stable ones survived.)
While I'm not an expert, I suspect the increased stability likely means less reactivity, lower catalytic activity (i.e., the species is more stable, less reactive so less likely to activate).
I'll give a comparison with Hückel's 4n+2 $\pi$ electron rule. We know that benzene is aromatic and has increased stability relative to the 5-membered and 7-membered conjugated hydrocarbon rings. It's less reactive and thus requires specific reaction chemistry.


text:
If chemical reactions occur because everything tends to a lower energy state, why is there such a thing as equilibrium?
It just doesn't make any sense, if a reaction occurs because the reactants go into a lower energy state, then there should be no such thing as equilibrium. right? Because that would mean that the products are going back into their higher energy states..... Is it just some sort of random chance that they keep changing form?

Everything tends to a lower energy state, that's the reason spontaneous reaction occur without absorbing any energy. But when you  supply energy to some substances they will show some molecular activity.

So there is equilibrium because the products re-absorb the energy? But why? why don't they just release that energy in some other form besides enthalpy?

@bluebellae I'm afraid your comment is bad, not the question.

See http://chemistry.stackexchange.com/questions/4/gibbs-free-energy-minimum-or-zero

http://chemistry.stackexchange.com/questions/44919/in-equilibrium-if-bonds-are-constantly-being-created-and-broken-where-does-the and http://chemistry.stackexchange.com/questions/7260/does-minimizing-any-type-of-energy-always-predict-a-state-of-equilibrium are related

Going to the lowest free energy state is not the same thing as minimizing potential energy.  In classical thermodynamics, the idea that equilibrium can't exist because it isn't the lowest-energy state could be true, but only at zero kelvin.


text:
Orbital and state correlation diagram for ring-closure of pentadienyl to a cyclic allyl
This is an exam question. I am to construct the orbital and state correlation diagram for one of the reaction below (disrotation or conratation). Also, I should explain whether the reaction is thermally or photochemically allowed.

I know the procedure to these kinds of problems:

Draw (Hückel) π orbitals of the reactant
Draw relevant orbitals of the product. Here: π and newly formed σ orbitals
Identify the symmetry element that is preserved through the reaction (common for reactant and product)
Assign symmetry labels to orbitals
Correlate orbitals of same symmetry

I know what the molecular orbitals of the reactant looks like, but I am unsure about those of the product. Here's what I have

Using the vertical reflection plane perpendicular the screen, I get the MO's of both reactant and product (in the order from bottom to top) to have the following symmetries: S, A, S, A, S. However, the orbital correlation diagram will be very boring if this is correct, and the reaction will be allowed both thermally and photochemically.
So are the product orbitals correct? I just assumed that an in-phase and out-of-phase combination of the π orbitals on C1 and C5 lead to the desired σ orbitals. Or perhaps my ordering is incorrect?
Final orbital correlation diagram:

Concerning the state correlation diagram, I look at the electron configurations of the states of interest, which I take to be $S_0$, $S_1$, and $S_2$. I find the symmetry of each state by multiplying the symmetry of each electron in every MO. However, should I just draw straight lines between each state? The corresponding states are of the same symmetry, so no avoided crossings occur. I don't see whether the reaction is thermally or photochemically allowed from the state correlation diagram. Or did I do something wrong?
Electron configurations:

Non-finished state correlation diagram:

Your assigned symmetries are correct, however the second cyclopentyl MO from the bottom should be the highest in energy since it is an antibonding sigma orbital. You've drawn it like a pi orbital, but is a sigma orbital.

Obviously, that was a stupid mistake! Thank you for pointing it out. I am confused about the disroratory and conrotatory pathways. As for which of the pathways that are photochemically allowed, it is fairly easy to see that only the disrotatory pathway is possible in the ground state; that's the only way to gain a bonding $\sigma$ orbital. Thus, the conrotatory ring-closure is photochemically allowed, which we also see from the LUMO orbitals. Is that an acceptable assessment, would you think?

I have been thinking about this, and the question specifies clearly the pentadiene anion. Hence, there are five electrons occupying the Hückel orbitals, so my figure is incorrect. Fixing this, I find that the $S_0$ state of the reactants becomes the $S_1$ state of the product, and vice versa, indicating that the photoinduced reaction is allowed, while the thermal reaction is not.

Also, what is the $S_1$ occupations? Is the HOMO excited to the LUMO, or does the HOMO become fully occupied, and the HOMO-1 becomes singly occupied? Both possibilities are a singlet state. And then, what about the $S_2$ state - which transition leads from $S_1$ to $S_2$?


text:
Mass of salt required to make a 1.2 g/ml solution
I would like to make an NaCl and water solution with a density of 1.2 g/ml. How would you calculate the grams of NaCl required to make a 1000 ml solution?
I have used density of mixture = (mass NaCl + mass H2O)/1000 ml and solved for the mass of NaCl. However, the resulting mass does not give a density of 1.2 g/ml in practice (and in fact does not dissolve) so is there another way of calculating this? (at room temperature (~21 °C))

The density also depends on the temperature..

To answer in detail Jan, I didn't make stranged calculus, now I answered more than a year ago I can't tell you if my calculus are the same, but here is the logic.

For any solution (aqueous) we have,
$$m_{solution}=\rho_{solution}\times V \tag1$$
We also have,
$$\rho_{solution}=d\times \rho_{water} \tag2$$
where d is the relative density of the solution and then, merging both $(1)$ and $(2)$,
$$m_{solution}=d \times \rho_{water}\times V \tag3$$
The mass of solute in the solution is a percentage of the masse of the whole solution then, because,
$$n_{solute}=\frac{m_{solute}}{M_{solute}} \tag4$$
We have,
$$n_{solute}=\frac{p\times d \times \rho_{water}\times V}{100\times M_{solute}}\tag5$$
Then find my previous result with $C\times V=n_{solute}$.

Now I think everyone is able to goes from molar to mass concentration then I can stop here. If not ask me.

The answer is: as much as water can dissolve.
The density of saturated sodium chloride in water at 21°C is $1.20~\pu{g cm^{-3}}$.
So you may take roughly $1000~\pu{ml}$ water of 21°C and put in sodium chloride such that not all of it can be dissolved (it will take some 400 grams). From the resulting solution‘s supernatant take $1000~\pu{ml}$.

table of densities at 60 F (15.5 C).  http://www.brinemaker.com/TechnicalDocument/Sodium-Chloride-Table.pdf

How do you do if in your laboratory water is at 25°C ? :-)

At 25°C and normal pressure there exists no such thing as sodium chloride solution with a density of 1.2000 g/ml (as requested). You had to live with a density of about 1.1978 g/ml which is 1.2 g/cm when rounded to two significant figures. ;-)


text:
Crystal structure vs melting point
Calcium oxide has a melting point of $\ce{2700^\circ C}$ and sodium chloride has a melting point of $\pu{801^\circ C}$.
If they have the same crystal structure and ions are about the same distance in each crystal, why is there such a big difference?

Calcium oxide is double charged, meaning the force required to separate ion pairs is much larger than Sodium chloride.
Melting an ionic solid requires you give an ion sufficient energy to become free of the ionic bonding of its neighbors.  In a simplified view, each ion has 6 neighbors in bulk (both are fcc, I believe) or 5 at surface face, 3 on corner and 4 on edge.  You need to add at least 3-5 times as much thermal energy to liberate the ions in calcium oxide as sodium chloride due to the doubling of charge from each neighbor.


text:
Is there an algorithm for drawing resonance structures?
Question
What is an instance of a procedure for writing resonance structures for any given compound's name or structural formula?
I would prefer any of these formats to represent the algorithm:

Numbered list
Flow diagram
Decision tree
Pseudocode

Any format that is human-readable and has ordinated the steps is probably okay.
Purpose
For myself, I would like to have such an algorithm because I prefer to systematically solve problems. I may also wish to write a computer program to perform this task in the future, so understanding the logic in more detail would help me create the decision tree.
For others that do not understand how to draw resonance structures, I hope the responses to this question are useful for their learning. I've seen a lot of people struggle with this topic.
Demonstration of Effort
I have been combing through my organic chemistry textbook (Organic Chemistry, Seventh Edition, L.G. Wade, Jr., Whitman College, Pearson, Copyright 2010) for rules to drawing resonance structures.
Here are the general principles I have found:

Each resonance structure has the property of being a valid Lewis structure
Electrons can be delocalized
Nuclei cannot be delocalized (ie don't move them)
Prefer placing pairs of electrons to instead of single electrons
Never separate pairs of electrons
Maximize the number of octets acheived among the atoms that obey the octet rule
Maximize the number of bonds in the structure
Prefer to place negative charges on electronegative atoms
Prefer to produce structures with minimized charge separation
Prefer structures with a negative formal charge instead of positive formal charge

Never separate pairs of electrons?

A problem here is that your general 'principals' are guidelines, rather than rules.  This is clear in your last 3, when you use the term 'prefer'.  At best, these rules will give you the 'best' structure, that is the one most descriptive of the molecule as a single picture.  However the purpose of drawing resonance structures is to make several, and apply value/weights to them to get an average view of how the molecule will behave.  I understand this makes the topic maddeningly frustrating for some.

@ringo The text could be wrong, or I misunderstood. It states "Most stable compounds have no unpaired electrons, and all the electrons must remain paired in all the resonance forms". In any case, as long as a proposed algorithm demonstrably works whenever tested, we're in good shape.

@Lighthart I agree, some or all of the points above are guidelines. The implication of your statement about the purpose invokes a thought that there are a subset of structures that are useful for prediction of molecular interaction, as well as a subset that are expected as answers on tests or assignments. For both of those subsets, I don't always find that the combination of resonance structures I draw are in the subset I want to draw resonance structures for.

Sadly, as a former chemistry professor, I find 'experience' is the best help here.  For example, don't consider open shell systems unless you are talking about a carbocation explicitly, or aldehydes or ketones.  That is something which becomes apparent after awhile studying, but not from the start.  From the student's perspective, there are a myriad of these 'exceptions' and chemistry becomes a mess of trivia.  Really, what we want you to learn is how to judge the importance of various factors, and then apply the most important ones to the problems.  I understand it is frustrating.


text:
Why does rdkit's `AllChem.ReactionFromSmarts.RunReactants` return a tuple of tuples?
In rdkit, the function AllChem.ReactionFromSmarts.RunReactants returns a tuple of tuples. Does anyone know what the two dimensions are for? From the (admittedly few) examples I've seen, the inner tuples are always just one molecule.
The docstring says,
RunReactants( (ChemicalReaction)arg1, (tuple)arg2) -> object :
apply the reaction to a sequence of reactant molecules and
return the products as a tuple of tuples

example:
from rdkit import rdBase, Chem
from rdkit.Chem import AllChem, Draw
rxn = AllChem.ReactionFromSmarts('[c:1][c:2][c:3]>>[c:1][c:2]([F])[c:3]')
m = Chem.MolFromSmiles('c1ccccc1')
ps = rxn.RunReactants((m,))

((<rdkit.Chem.rdchem.Mol at 0x10be03818>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef92f0>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef9a28>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef99d0>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef9768>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef9920>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef96b8>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef98c8>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef95b0>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef9710>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef9558>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef9ad8>,))

The reason I ask is that I want to put the products into a pandas dataframe (already accomplished) and label the columns and indices according to the intended scheme.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I know that rdkit is used extensively in cheminformatics, but this question is not about chemistry itself and not a question how to apply the specific software to a given problem. I would think this is off-topic for our site.

Thank you @Martin, I see your point. I've posted the question to rdkit-discuss@lists.sourceforge.net

From a tutorial I wrote on SMARTS reactions in rdkit:

The output is a tuple of tuples. The inner tuples are there because even reactions that take only a single input molecule can result in multiple output molecules (e.g. hydrolysis). The outer tuples are there because rdkit applies your reaction incidence of the appropriate motifs in the input molecule, even if it occurs more than once.

The examples in the tutorial illustrate both cases where a single application of the reaction rule creates multiple molecules (the inner tuple(s) would have length > 1), and also cases where a reaction rule can be applied more than once to a molecule, which would lead to the outer tuples having length > 1.

Thank you for answering, and thanks for the tutorial as well Curt!


text:
Are empty orbitals degenerate?
This is apparently a true statement:
"In a hydrogen atom, the 2s and 2p subshells have the same energy."
Why is this the case? Why would they be of the same energy? Wouldn't the 2p be slightly higher in energy than the 2s?

While addressing a different question, this should answer your question pretty precisely.

Incidentally, the answer to your title is "no" for the general case. The hydrogen atom is a very specific case as mentioned by @Philipp.

My chemistry teacher Mr Ong Chiau Jin answers:

Why are all the orbitals that have the same principal number in
  Hydrogen degenerate?
In hydrogen, all orbitals with the same principal quantum number 'n'
  (1,2,3...) are degenerate, regardless of the orbital angular momentum
  quantum number 'l' (0,1...n-1 or s,p,d,). However, in atoms with more
  than one electron, orbitals with different values of l for a given
  value for n are not degenerate. Why is this? Surely the radial
  distribution functions are similar for hydrogen (in that there's still
  penetration of orbitals and so on). Or is it that orbitals with
  different values of l are degenerate for a value of n greater than
  would be occupied in that particular atom's ground state?
The answer is right there in your question. It's only the interaction
  of multiple electrons in an atom like He, Li, Be, etc. that makes the
  different angular momenta wave functions differ in energy. Consider
  this. For the one electron system, why should a p or d orbital differ
  in energy from an s? What makes them differ? In the multi-electron
  case, the p orbitals have different spatial extent, different angular
  components, so the electron density caused by an electron in that
  orbital will interact differently with the other electrons. In other
  words, you need to have more than one electron for the "shape" of the
  p and d and f orbitals to matter to the other electrons. In the H
  atom, there's only one electron, so there's no electron-electron
  repulsion to differentiate the s, p, and d orbitals.

This may not be the perfect answer but it does capture the essence of the perfect answer, which is the idea of multi-electron interactions. Hope this helps.

Do you think you could possibly find an authoritative source for this? My chemistry teacher Mr Ong Chiau Jin answers does not a reference make, even though the information is/may be factually accurate.

@NotEvans. No, I don't think I have come across this piece of information in the literature I have read before and hence, I do not have any authorative sources to cite. Do let me know if you have come across any such source.

If you go through the trouble of setting up the Hamiltonian for a single electron in the vicinity of a point-sized, positively charged particle (our approximation of the proton) — i.e. that for the hydrogen atom — and then solve the Schrödinger equation, you will realise that the energy of said electron depends only on the principal quantum number (typically labelled $n$).
This is true for all hydrogen-like atoms, i.e. also for $\ce{He+, Li^2+},\dots$ (Note, however, that the energies of a helium-cation’s orbitals are naturally different from those of a hydrogen atom’s.)
It is not until you add additional electrons to the species that their potential induces a difference between the energies of s and p subshells of a given shell.


text:
Electrolytic conductivity vs metallic conductivity in a galvanic cell
I'm analyzing the current output of a galvanic cell with aluminum electrode in aluminum sulfate solution and platinum electrode in acetic acid solution.
What I am confused about is whether molar conductivity will account for current resistance in the cell even though the electrons are being transferred through the metallic electrodes. Or is it just the metallic conductivity that accounts for current resistance?
If molar conductivity is also one of the reasons for current resistance in the cell, how is it?

The resistance you'll find in your system basically corresponds to that of the solution (plus some charge-transfer resistance if your kinetics are somewhat sluggish).
Remember that for current to circulate, you need a closed loop; so the movement of electrons from anode to cathode is only part of the big picture. An equal ammount of current will be transported in the solution by ions approaching/leaving the electrode's surface. So, if you want to minimize your resistance effects, two intuitive approaches come to mind: i) make your ions move faster and/or ii) have a lot of ions moving at the same time. That means that the diffusion coefficient and the concentration of your ions (hence, the molar conductivity) will have direct effect on the solution resistance.
Since metals/alloys are, normally, excellent electronic conductors when compared to electrolytes (which, in turn, have ions as charge carriers), the overall cell resistance is pretty much due to the latter.


text:
What will form in the hydrolysis of urea in urine?
I just wanna know what is the end product of hydrolysis of urea in urine. I know that the hydrolysis of urea will form ammonia and a subsequent rise in pH but will the ammonia react with the water from urine to form ammonium and hydroxide?

The hydrolysis is the first of two reactions that occurs in aqueous urea. This reaction forms one equivalent of ammonia and one equivalent of carbamic acid. Carbamic acid rapidly decomposes to form gaseous carbon dioxide and another equivalent of ammonia:

The ammonia then simply participates in an equilibrium where it reacts with water to form ammonium hydroxide, or so it is commonly described. There is actually a fair amount of debate on this topic, but it typically written like this: $$\ce{NH3 + H2O <<=> NH4OH}$$

thanks for the answer! So basically you're saying that both of them would exist at the same time?

Yes, in aqueous ammonia, you have both ammonia and ammonium hydroxide, but primarily ammonia.


text:
Product formed under ethoxide/ethanol and heating?

What would be the major product of the following reaction sequence?

I believe that the first step will be the formation of a radical at the first carbon of the ethyl group because benzylic radicals are quite stable. However, I'm not really sure what happens after that. The answer key has 'I' as being the correct answer.

You are correct in your prediction for this first reaction:

The second is simply an $\mathrm E2$ β-elimination reaction, which forms product I:

With heat, would any IV formed just eliminate to the same product? Or do you think they're just doing the main product? Or would IV just not be very likely with a hard nucleophile like that?

I don't think it would very easy for IV to form in basic conditions. In acidic conditions that could easily form, however.


text:
How to compare acidity in the following aniline derivatives?

Let the simplest one be a), ortho be b), meta be c) and para be d).
I have confusion between ortho and meta. How to compare their acidity?

Acidity or basicity?

The ortho is tricky, the rest is covered by Hammet relationship, https://en.wikipedia.org/wiki/Hammett_equation

Taking the case of ortho and meta in which you have confusion:
The acidity of aniline will increase if there is a electron donating group in the compound to stabilise the conjugate positive acid formed.
Since OMe is shows +R and -I effect, it is a net electron donating group as +R effect generally dominates over -I.So in case of ortho it is able to show electron donating nature due to +R effect which makes it more basic than the meta isomer which shows electron withdrawing effect due to only -I effect as +R effect will not affect from meta position.

I just got more confused. Can you please help me compare the basicity? Should not ortho be the most basic out of the four due to distance factor?

Distance factor plays role only in I effect not in R effect.Para is the strongest although both o and  p show +R because  the ortho since located closer to nh2 has more - I effect which withdraws more electron and hence stabilises the conjugate base lesser than para which has very kess - I effect due to distance factor.

Oh. I am new here. Done!

According to source the pKa values (of the conjugate acid) are:
compound            °C    pKa
Aniline             25    4.87
2-Methoxyaniline    25    4.53
3-Methoxyaniline    25    4.20
4-Methoxyaniline    25    5.36


text:
How does a catalyst accelerate both forward/reverse reactions simultaneously?
A catalyst is known to speed up both forward/backward reactions of a reversible reaction. But how does this work, because the mechanism for the forward and backward reactions could be different right? I mean the adsorbed species for the forward and backward reactions when conducted independently will be different right? Is that because all the steps are exactly reversible?

The idea is that the activation energies is lowered for both the forward and backward reactions.

Though the activation energy for the backward reaction is higher than the activation energy for the forward reaction, it is nevertheless lowered.
To answer your second question, reactions aren't always reversible, but a lot of the time they are, and are subject to the principle of microscopic reversibility. That means when reactions are performed as a series of steps, at equilibrium, each individual reaction occurs in such a way that the forward and reverse rates of reaction are equal, and these equilibriums can be affected to drive the reaction one way or the other.

how does this assumption of microscopic reversibility hold good in reality?can I assume  a catalyst that accelerates the forward reaction can also be used to catalyze the baxckward reaction always(given thermodynamic limitations of course)?sometimes catalyst chemically decompose in presence of some reatants and it is this decomposed state that actively catalyzes the reaction. But when you start with the products and do the reverse reaction I am not sure if the same exact decomposed states woud be formed

It's very important in synthetic organic chemistry. It simplifies what you need to know (if you know the forward reaction you know the backward reaction as well), it can help you to predict side products, and it can make undoing a reaction easier by understanding what changes need to be made to affect the equilibrium one way or the other.

so any catalyst that accelerates forward reaction(in heterogeneous gas-solid reactions) will be equally effective if I need to accelerate the reverse reaction as well? Is this true under all circumstances, just for confirmation?

Yes. This doesn't change yield at all, but the rates of reaction are effected equally.

The equilibrium constant for the reaction is equal to the ratio of the forward rate constant to the reverse rate constant.  In order for the equilibrium constant to be independent of the catalytic effect, the forward and reverse rates must speed up equally.

One of the implications of the second law of thermodynamics is that, if a catalyst increases the rate of the forward reaction by a factor of, say, 745492, it obligatorily increases the rate of the reverse reaction by the same factor, 745492, despite the fact that the two reactions - forward and reverse - may be entirely different (e.g. the diffusion factor is crucial for one but not important for the other) and accordingly require entirely different catalytic strategies. The absurd implication is usually referred to as "Catalysts do not shift chemical equilibrium".
A paradigmatic example of the inability of catalysts to accelerate both - forward and reverse - reactions to the same extent are diffusion-controlled reactions:
Wikipedia: "Diffusion-controlled (or diffusion-limited) reactions are reactions that occur so quickly that the reaction rate is the rate of transport of the reactants through the reaction medium (usually a solution). As quickly as the reactants encounter each other, they react. The process of chemical reaction can be considered as involving the diffusion of reactants until they encounter each other in the right stoichiometry and form an activated complex which can form the product species. The observed rate of chemical reactions is, generally speaking, the rate of the slowest or "rate determining" step. In diffusion controlled reactions the formation of products from the activated complex is much faster than the diffusion of reactants and thus the rate is governed by collision frequency."
Can a catalyst accelerate a diffusion-controlled reaction? Obviously not - catalysts do not accelerate diffusion. On the other hand, nothing prevents the catalyst from accelerating the reverse reaction, which means shifting the equilibrium and is tantamount to violating the second law of thermodynamics:
Consider the association-dissociation reaction
A + B <-> C
which is in equilibrium. Let us assume that the forward reaction
A + B -> C
is diffusion-controlled. The reverse
C -> A + B
is not diffusion-controlled, obviously.
We add a catalyst, e.g. a macroscopic catalytic surface, and it starts splitting C so efficiently that the rate of the reverse (dissociation) reaction increases by a factor of, say, 745492. If the second law of thermodynamics is obeyed, the catalyst must increase the rate of the forward (association) reaction by exactly the same factor, 745492, which is impossible - the catalyst cannot accelerate the forward reaction at all!
Scientists have always known that catalysts do shift chemical equilibirum:
https://www.facebook.com/ParadigmEnergy/posts/249600938581128
 "For 50 years scientists have seen in experiments that some monomers and dimers split apart and rejoin at different rates on different surfaces. The eureka moment came when we recognized that by placing two different surfaces close together in a way that effectively eliminates the gas cloud, the energy balance would be different on each of the two surfaces. One surface would have more molecules breaking apart, cooling it, while the other surface would have more molecules joining back together, warming it."
The second-law-violating effect is presented by Wikipedia as a fact:
Wikipedia: "Epicatalysis is a newly identified class of gas-surface heterogeneous catalysis in which specific gas-surface reactions shift gas phase species concentrations away from those normally associated with gas-phase equilibrium. [...] A traditional catalyst adheres to three general principles, namely: 1) it speeds up a chemical reaction; 2) it participates in, but is not consumed by, the reaction; and 3) it does not change the chemical equilibrium of the reaction. Epicatalysts overcome the third principle..."
Wikipedia: "Consider a dimeric gas (A2) that is susceptible to endothermic dissociation or exothermic recombination (A2 <-> 2A). The gas is housed between two surfaces (S1 and S2), whose chemical reactivities are distinct with respect to the gas. Specifically, let S1 preferentially dissociate dimer A2 and desorb monomer A, while S2 preferentially recombines monomers A and desorbs dimer A2. [...]
http://upload.wikimedia.org/wikipedia/commons/c/ce/NatureSLTD-Fig1c.jpg
In 2014 Duncan's temperature paradox was experimentally realized, utilizing hydrogen dissociation on high-temperature transition metals (tungsten and rhenium). Ironically, these experiments support the predictions of the paradox and provide laboratory evidence for second law breakdown."

The point missing so far in the answers is that a catalyst causes the reaction to occur by a different mechanism than it would normally take. (This is why its actually very hard to find good catalysts)
The catalyst lowers the activation energy but its not the same barrier as without the catalyst. So in that sense it does not 'lower the barrier' as commonly stated.
As the reactant and products are still at the same relative energy (no matter if catalysed or not) but the reaction occurs by a lower activation barrier then both forward and reverse rate constants are speeded up.


text:
Are there ion combinations which can't be separated using the ion exchange resin method?
I am currently investigating the uses of ion exchange resins (both cation and anion resins). I have not come across any ion combinations that cannot be separated using this method, are there any exceptions or does this method work for all ion combinations? If there are ions that can not be separated using the resin, what methods would work to separate them?

Ion Exchange resins are not selective.  You are essentially just statistically overpowering the situation by providing many more times the concentration of the ion you want compared to the ion you are 'extracting'.
There are some situations where chelation is so strong you can't exchange because the other ion is less stable in said position, say for example in crown ethers.
So the answer to your question, is no, there are not specific ions that can't be exchanged as a general rule, but there are ion pair combinations which won't exchange.


text:
Is styrene more or less reactive than benzene towards electrophiles?
Is styrene more activated or more deactivated than benzene towards electrophilic aromatic substitution? The C=C double bond doesn't seem to have any particular electron-withdrawing or -donating properties, so I'm not sure what to make of it.

I guess for many agents the double bond is likely to be attacked first.

Styrene (vinylbenzene) undergoes electrophilic aromatic substitution much faster than benzene. Although the vinyl group seems to be withdrawing, it offers +M effect in the carbocation intermediate. Resonance forms of the intermediates to explain these results. and the products are found to be primarily ortho- and para-substituted styrenes. The products are found to be primarily ortho- and para-substituted styrenes.

Do you have a reference to or example of EAS where the ring reacts rather than the double bond?


text:
Why does sodium chloride not conduct as much as aluminum chloride in water?
I understand how $\ce{NaCl}$ and $\ce{AlCl3}$ dissolves in water but I'm wondering why one is more conductive than the other. Is it their extra ions to conduct the electricity?
I know that in $\ce{NaCl}$ the water molecules 'steals' the Sodium or the Chloride depending on which end becomes closest to the material due to the Dipole-Dipole attraction. However, this reaction alone cannot break down the ionic lattice alone so there is a separate endothermic reaction. But I still can't understand the science behind why one conducts more than the other.

Your question might really benefit from adding more details, e.g. what do they conduct and in what conditions. Also, welcome to Chem.SE.

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

If the anions are identical, it must be due to the cations. What happens when you dissolve both salts in water?

@KlausWarzecha No, it isn't in this case.

@Kelsey ; clarify molar concentration of the solutions at the very least and if any other compounds are dissolved in the solutions

@permeakra Did I miss something? Under otherwise similar conditions, I would attribute the effect to the higher ion mobility of protons (hopping mechanism). $\ce{AlCl3}$ gives an acidic solution.

@KlausWarzecha You worded it in a very misleading manner.

@KlausWarzecha I don't think a high school student is going to be familiar with the Grotthuss diffusion mechanism. Heck, the people who wrote OP's textbook may not even know of it....

@permeakra Might be. An author tends to read what he has meant. The only misunderstanding I can think of is that somebody might mix both salts in the same beaker and add water. That's definitely not what I meant :D

@Brian You're probably right and I was just overdoing the problem. Counting ions might be good enough here :)

I'm guessing OP is asking about same concentration of the salt, not ions.

With all the discussions, we missed to tell you one important thing: Welcome to Chemistry.SE!

Make sure you note that aluminum chloride is actually $\ce{AlCl3}$.

Okay, so the reason that is important is because of the van't Hoff factor. Is the question more complicated than this or are the other folks here maybe going a little more in-depth than required?


text:
Correlation between functional group and viscosity of an organic compound?
I'm looking to explore the correlation between the type of functional group on an organic molecule and the viscosity of the compound. I know that viscosity is affected by intermolecular forces (Ion-induced dipoles, Ion-dipole, Hydrogen bonding, van der Waals), and that the type of forces present depends on the functional group. I'm aiming to explore viscosities of compounds with the following groups:

Alkenyl
Alkynyl
Hydroxyl
Ether
Aldehyde
Carbonyl
Carboxyl
Amide
Amine
Nitrile
Phenyl

How do the (absolute) viscosities of compounds with each of these groups (assuming single-functional group compounds) differ, and why (in terms of intermolecular forces present)? Are there any databases with values for viscosities that I could use?
Thank you for the help.

There are empirical methods that can give you the viscosity from the structure. For lower temperatures, Orrick-Erbar equation, and for higher temperatures, Letsou-Stiel equation is what you are looking for. Orrick-Erbar maybe more suitable for you, because it is a group contribution method, and by looking at the values assigned to functional groups, you can develop an intuition about how each group affects viscosity. Letsou-Stiel uses acentricity factor, because at high temperatures, secondary bonds become less important.
As for database, NIST may be useful. Or if not, look at some supplier webpage (I use SigmaAldrich for this).

This is a great chance for you to do some searching through the literature and draw conclusions on your own. Make a list of molecules containing each functional group, group them by trends, and compare their viscosities. (alkyl: ethane, alkenyl: ethylene, alkynyl: acetylene, etc... to see the effect of adding pi bonds, for example). Most, if not all of this data is freely available through the CRC Handbook of Chemistry and Physics.
Once you've searched and found the trends on your own, you should be able to tell from a structural standpoint why these trends exist. To measure viscosity, liquids are sheared between two solid surfaces, and the viscosity is proportional to the force required to move the surfaces. At the molecular scale, parts of different molecules must move past each other under shear, and the "ease" of molecule sliding past one another is what determines viscosity. One can imagine that molecules with stronger intermolecular forces will have higher viscosities, since those forces must be overcome when molecules slide past one another. For example, consider acetone and acetic acid, which have viscosities of 0.306 and 1.056 mPa-s at 25 C, respectively. Acetic acid has much stronger intermolecular forces because of the dipoles present (so much that it forms dimers in the gas phase), and as a result has a higher viscosity.


text:
How did chemists in the 18th century learn that magnesium gain mass when it's burned?
Crucial to disprove the phlogiston hypothesis was the observation that magnesium gains mass when burns.
But how was this observation made in the 18th century? Is there any way this property of magnesium would be obvious to a layman today?

Presumably by weighing magnesium before and after it was burnt? It's not that difficult to do.

Per Wikipedia "The metal itself was first produced by Sir Humphry Davy in England in 1808." So burning the metal would have taken place in the early 19th century. https://en.wikipedia.org/wiki/Magnesium#History "Phlogiston remained the dominant theory until the 1780s when Antoine-Laurent Lavoisier showed that combustion requires a gas that has mass (oxygen) and could be measured by means of weighing closed vessels." https://en.wikipedia.org/wiki/Phlogiston_theory#Challenge_and_demise So by the time metallic magnesium was isolated the phlogiston hypothesis was dead.

The phlogiston hypothesis is about a substance, phlogiston, which was supposed to exist within matter that be consumed or released when it is combusted.  If this were true, then when magnesium is combusted the mass would decrease.
To someone who believed this to be true, it would come as a great surprise to find that if they combusted 100g of magnesium the mass would increase by about 65%.  This would be easy to measure with a set of balances, even with much smaller masses of a few grams.
A layman might not come across many day-to-day examples of a combustion reaction with a mass increase, since most things that burn release a lot of mass as gases, and that effect is what phlogiston was intended to explain.  One possibility, aside from burning metal, is the oxidation of metal in air.  Someone could weigh an iron nail or other object, allow it to rust, and weigh again to show it has gained mass.  This could easily be demonstrated by any merchant with a set of scales, but it's not obvious the rust is an oxide related to the combustion products.

Unless of course phlogiston had negative mass, which was an initial hypothesis to explain these observations.