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Toxicity of a sheet of lead?
I just went camping and someone brought a game from Australia. There was a wooden box with the top made of a thin, soft sheet of lead with a hole in it. The participants then throw coins at the box, and getting the coin inside gives some points, while landing a coin on top gives a little less. I've been looking online, and I think the game is called "Toad in the hole".
I know that lead is toxic, but I don't know in what way. Is it toxic to touch this lead sheet? And to eat afterwards without washing hands?
Maybe lead is only toxic in some form?

Children used to play with models of soldiers made of lead, so …

@madsskjern How did you know that the sheet was made of lead and not some other metal?

First of all, that is what they told me. And it seemed like lead, it was heavy and soft and had the color of lead. I have bo reason to believe it was not :)

@Jan On its own, that doesn't mean anything. Children may have also received lead poisoning after playing with lead soldiers.

No, it is not toxic to touch solid lead.  Lead poisoning results from

ingestion: paints used to contain lead-based materials, and kids would sometimes eat the peeling paint leading to health problems.
inhalation: if you were grinding or polishing a block of lead and fine particles were generated, you might inhale them if you weren't using the proper protective equipment and health problems could develop.
dermal contact: if you were working with organic lead compounds that were easily absorbed through the skin, serious health problems would result if you weren't using the proper protective equipment.  Gasoline used to contain lead compounds, but they were removed from gasoline in order to prevent problems.

Being around or touching solid lead would not lead to ingestion, inhalation or dermal contact issues.
Here's a link to Wikipedia article on lead, see the section on "Health effects" if you'd like to read more or read about the biochemical mechanism of action.

Thanks. Wikipedia writes: "As lead paint deteriorates, it peels, is pulverized into dust and then enters the body through hand-to-mouth contact or through contaminated food, water or alcohol". Those quantities in the paint are probably very small, but still dangerous. Then imagine throwing coins at a sheet of lead. Is there really no part of the lead sheet that breaks free as lead dust / lead particles? Such particles would also end up in the carpet or atmosphere, like the lead from the paint.

If you did it all day, every day for the rest of your life, there might be a problem.  You were probably a couple of feet away which makes inhalation even less likely.  I just can't see a one-time event like this being significant.  Look at all the folks who pumped leaded gasoline for years and they seem to be OK.

Just to be safe, I'd wash your hands after handling the lead.

@MadsSkjern: I know it's an old comment, but I'd just like to clarify that lead paints used lead compounds as pigments, and thus contained a significant fraction of lead. For example, this page says that "[p]aints made before 1950 contained large amounts of lead. In fact, some paint made in the 1940s contained up to 50% lead by dry weight."

Also note the lead in paints is not pure lead (that would be a dull grey paint) but usually lead oxides that are colourful and provide rust protection, not the same as lead metal.  Water pipes for drinking water were made of lead in many places for long periods and are not the cause of chronic toxicity.  However children's brain development is hampered by exposure to lead COMPOUNDS so best to keep them away from lead paint.

@IlmariKaronen Given how heavy lead is, going by weight might be meant as more of a scare factor. What's the volumetric ratio?

@JAB: Given that the lead in the paint is present as a compound (e.g. an oxide), I'm not sure that's a very meaningful question. You might as well ask how much of the volume of table salt consists of sodium and how much of chlorine. I suppose one could interpret it as "what would be the volume (at STP) of a chunk of pure metallic lead with the same mass as the total lead content of 1 volume unit of dry lead paint?" which would be at least in principle answerable, if one had a figure for the density of the paint (which, alas, I don't have).

@KalleMP: Not sure why you say lead compounds. I believe elemental lead is toxic, never heard different. There may be lead compounds that are way more toxic (like dimethyl mercury is crazily more toxic than elemental mercury) but I would like to see a link that indicates elemental lead is safe.

@Jeff While elemental lead is toxic when it is combined with something in your body it is somewhat hard to get it into your body in this form unless it is a finely divided dust or fume.  You will notice that I did not claim it not to be toxic, I merely indicated that one needs to me concerned with compounds as they are absorbed more easily (skin absorption being the OPs concern here).  -  https://www.osha.gov/pls/oshaweb/owadisp.show_document?p_table=STANDARDS&p_id=10031

If you ingest it, it is toxic and touching is a route to ingesting it. But I would guess that it does not go through skin very well as an element. Tetraethyl lead is absorbed through skin very readily however.

While merely touching lead is probably not directly dangerous, lead, being soft, probably remains on fingers after handling and therefore washing hands before eating, etc. is necessary after such handling.

I'd be interested to see a source for this. "Probably remains on fingers" sounds like speculation.

@Clonkex Not much chance to get response from a guy who's not around for more than five years... He meant some trace amounts - it's not like you're gonna get your hands dirty with it, but better safe then sorry.

@Mithoron Possibly, but I'm commenting on the post more than responding to the author. While I definitely do wash my hands after handling lead (better safe than sorry, as you say), without a source this answer is just speculation and should be a comment rather than an answer. It's not really useful to just repeat common beliefs or assumptions unless you can back it up with evidence to show those beliefs are justified.


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Why do we use helium in balloons?
While I was looking at the periodic table today, I realised that there were gases that were much lighter than helium such as hydrogen. If hydrogen is lighter than helium, why do we insist on using helium in balloons?

For some applications, $\mathrm{H}_2$ is more appropriate. For example, weather balloons often use hydrogen due to cost, with side benefits being lower density and lower diffusion rates across the membrane; flammability is not a major concern.

Since helium is on shortage, fewer and fewer places use it any more for balloons.

Using H2 is needed for balloons too, because the Hydrogen in normal conditions exists only as paired atoms in an Hydrogen molecule. This is important for example for the diffuse rate - where Helium has much higher rate as H2 - exactly because the Hydrogen molecule (H2) is much bigger as one atom He - so the Helium has higher diffuse rate as H2.

As other answers have noted, the only gas lighter than helium is hydrogen, which has some flammability issues that make it more difficult to handle safely than helium.
Also, in practice, hydrogen is not significantly "lighter" than helium.  While the molecular mass (and thus, per the ideal gas law, the density) of hydrogen gas is about half that of helium, what determines the buoyancy of a balloon is the difference between the density of the gas inside the balloon and the air outside.
The density of air at STP is about $\rho_{\ce{air}}=\pu{1.2754 kg m-3}$ , while the densities of hydrogen and helium gas are $\rho_{\ce{H2}}=\pu{0.08988 kg m-3}$ and $\rho_{\ce{He}}=\pu{0.1786 kg m-3}$  respectively.  The buoyant forces of a hydrogen balloon and a helium balloon in air (neglecting the weight of the skin and the pressure difference between the inside and the outside, which both decrease the buoyancy somewhat) are proportional to the density differences  $\rho_{\ce{air}} -\rho_{\ce{H2}}=\pu{1.1855 kg m-3}$  and $\rho_{\ce{air}} -\rho_{\ce{He}}=\pu{1.0968 kg m-3}$.  Thus, helium is only about $7.5\%$ less buoyant in air than hydrogen.
Of course, if the surrounding air were replaced with a lighter gas, the density difference between hydrogen and helium would become more significant.  For example, if you wished to go ballooning on Jupiter, which has an atmosphere consisting mostly of hydrogen and some helium, a helium balloon would simply sink, and even a pure hydrogen balloon (at ambient temperature) would not lift much weight.  Of course, you could always just fill the balloon with ambient Jovian air and heat it up to produce a hot hydrogen balloon (not to be confused with a Rozière balloon, which are used on Earth and have separate chambers for hot air and hydrogen / helium).

Ps. A quick way to approximately obtain this result is to note that a hydrogen molecule consists of two protons (and some electrons, which have negligible mass), and thus has a molecular mass of about $\pu{2 Da}$, while a helium atom has two protons and two neutrons, for a total mass of about $\pu{4 Da}$.
Air, meanwhile, is mostly oxygen and nitrogen: oxygen has a molecular mass of about  $\pu{32 Da}$ (8 protons + 8 neutrons per atom, two atoms per molecule), while nitrogen is close to  $\pu{28 Da}$ (one proton and one neutron per atom less than oxygen).  Thus, the average molecular mass of air should be between $28$ and  $\pu{32 Da}$; in fact, since air is about three quarters nitrogen, it's about  $\pu{29 Da}$, and so the buoyancies of hydrogen and helium in air are proportional to $29 - 2 = 27$ and $29 - 4 = 25$ respectively.  Thus, hydrogen should be about $\frac{(27 - 25)}{25} = \frac{2}{25} = \frac{8}{100} = 8\%$ more buoyant than helium, or, in other words, helium should be about $\frac{2}{27} \approx 7.5\%$ less buoyant than hydrogen.

Pps. To summarize some of the comments below, there are other possible lifting gases as well, but none of them appear to be particularly viable competitors for helium, at least not at today's helium prices.
For example, methane (molecular mass  $\approx \pu{16 Da}$) has about half the buoyancy of hydrogen or helium in the Earth's atmosphere, and is cheap and easily available from natural gas.  However, like hydrogen, it's also flammable, and while it's somewhat less dangerous by some measures (burn speed and flammability range), it's more dangerous by others (total energy content per volume).  In any case, the reduced buoyancy, together with the flammability, is probably enough to sink (pun not intended) methane as a viable alternative to helium.
A much less flammable choice would be water vapor which, with a molecular mass of  $\approx \pu{18 Da}$, is only slightly less buoyant than methane at the same temperature and pressure.  The obvious problem with water is that it's a liquid at ambient temperatures, which means it has to be heated to make it lift anything at all.  This wouldn't be so bad (after all, you get extra lift from the expansion due to heat), except for the fact that it makes any failure in the heating system a potential disaster — whereas a hot air balloon will just gently drift down if the burner fails, a hot steam balloon can experience catastrophic buoyancy loss if the vapor condenses.
Despite these drawbacks, hot steam balloons have certainly been suggested, studied and tried in the past — alas, not always particularly successfully (although, apparently, there have been much more successful attempts as well).  There are various ways in which the condensation issue could potentially be reduced, such as adding extra insulation layers to the balloon envelope, or even surrounding the steam balloon with a more conventional hot air envelope.  So far, however, it seems that steam balloons remain firmly in the realm of nifty but impractical ideas.
Other potential lifting gases, with molecular mass similar to methane and water, include ammonia and neon.  Neon, being a noble gas like helium, would certainly work and be safe, but alas, it's both less buoyant and more expensive than helium.
Ammonia, on the other hand, while much less flammable than methane, is rather toxic and corrosive (not to mention really stinky, which, given its other properties, is probably a good thing).  I don't think I'd like to fly in an ammonia balloon, but apparently, some people do!  It seems that its main advantage (besides being much cheaper than helium) is its relatively low vapor pressure, which makes it easier to store and handle in compressed form.
Thus, at least for some niche applications (mainly hobbyists and some weather balloons, AFAICT), ammonia might actually be the most viable alternative to helium (and hot air) today, with methane / natural gas perhaps coming second.  If helium were to become more scarce and expensive, these low-cost lifting gases (and possibly other alternatives, like helium recovery or even steam balloons) might become more practical.  Then again, so would hydrogen — its safety issues, though well known, are not insurmountable, especially not for things like unmanned weather balloons where the risks are much less.

Excellent answer. The only thing I have to add is that methane (MW 16)has about half the buoyancy of helium (29-16=13), so if you REALLY don't care about safety you can fill your balloons with that. (Natural gas is a whole lot more readily available than hydrogen.) Hydrogen is more dangerous than methane by some measures (burn speed, flammability limit) but methane is more dangerous by others (total volumetric energy content.)

I'd never heard of Roziere baloons, but they sound interesting.  One thing I've long wondered about would be the practicality of a balloon which had an inner vessel of H2O enclosed within a vessel of air which was heated above 100C.  A lot of energy would be required on the ground to evaporate all the H2O in the inner vessel, but once that was accomplished the H2O would generate more than twice as much lift (assuming vessel temperature 375K and outside ambient ~290K).  Ever heard of such a thing?

@supercat: Googling for "steam balloon" brings up a bunch of links suggesting that it can be done, and apparently has been done at least on a small scale, but it seems to belong in the category of nifty but impractical ideas. I suspect the big problem is the extremely steep sigmoid lift / heat curve near the condensation point, which seems likely to make such a balloon very hard to control safely. Besides, even if you could solve those safety and control issues, I can't think of a good reason to prefer such a steam balloon over a basic helium Rozière.

@IlmariKaronen: Water vapor can be produced using a wide range of resources, while it's almost impossible to procure helium without a source of helium (alpha decay produces helium, but not in useful quantities).  I'm not sure how the cost of energy using H2O would compare with the cost of helium.  Do balloonists have pumps which can re-compress helium for storage when they're done with it?

@supercat: AFAIK, no, but that's probably just because, given current helium prices, it's not worth the trouble. If helium became more scarce and costly, both steam balloons and helium recycling might become more practical in comparison. Then again, so would hydrogen -- while hydrogen balloons (and especially Rozières) have obvious safety issues, they're not insurmountable; and, in any case, the majority of balloons these days are unmanned weather balloons etc., for which the risks are much less.

@supercat - the helium we use comes from alpha decay. We just get to take advantage (for the moment) of accumulation over geologic time scales.

@A.K.: I'm a bit puzzled by your edit to this answer. You seem to have moved a whole bunch of inline URLs to footnotes (and introduced several editing mistakes in the progress, which orthocresol and I have hopefully now fixed) and converted some HTML formulas to MathJax, but I don't see any actual substantive changes. Well, besides converting "AMU" to "Da". Anyway, I'm tempted to just revert it all, because it'd be easier than going over all the changes to make sure there are no more errors, but I thought I'd ask whether you had some specific reason for it first.

It is an excellent and comprehensive answer, I wanted to update it to make the math cleaner and to match current best practices. If there are errors let me know for QA/QI purposes. I was very diligent with the links.

i would like to add an argument that if we really take the example of Jupiter then replacing ballon with any other material will surely help !

Actually, hydrogen is the only gas that is lighter than helium. However, it has a very big disadvantage: It is highly flammable. On the other hand, helium is almost completely inert - this is why it is very much safer to use the latter.
What might happen when you use hydrogen instead of helium was impressively proven by history  when the "Hindenburg" zeppelin exploded on 6 May 1937. There is video footage, that can be seen on youtube.
In some of the comments it was mentioned, that hydrogen alone might not be the cause of the Hindenburg disaster, there were other contributing factors. However, using hydrogen remains dangerous, as this weather balloon experiment shows. In a more scientific setup the burning of a hydrogen balloon is compared to oxygen and a mixture of oxygen and hydrogen. Unfortunately a video of a helium filled balloon is not available, but it basically only ruptures and pops because of the different pressures on the in- and outside.

I can't find a good source now, but I thought in addition to everything else mentioned, hydrogen also diffuses faster than helium meaning the balloon will deflate faster (though this is less of an issue for weather balloons than the 30 cent type party balloons)

While hydrogen is indeed more flammable than helium, and is probably a contributing factor to the Hindenburg disaster, it is not completely certain that hydrogen was the primary cause. (See the article you linked to.)

@200_success Nitpick: Hydrogen is not "more flammable than Helium". Hydrogen is flammable. Helium is not. Helium is completely inert under all situations outside the core of a nuclear explosion or medium-sized stars. Helium (or any other group 18 element) will not oxidize (burn) by any process you'll find on this planet.

@david - I would say that that does actually qualify Hydrogen as more flammable.  It's not hard to be more flammable than something that can't burn.  If it burns at all, it's more flammable.  ---- That said, it's still not a great way to phrase it.

There are likely multiple contributing factors to the Hindenburg disaster (f.e. the paint used had a composition very similar to thermite, a pyrotechnic compound). Hydrogen alone cannot be blamed, although it probably didn't reduce the damage. However, we are unlikely to get a conclusive answer about the accident any more (barring the existence of time travel!)

@david: Just to pick a nit or two: As you move down group 18, the elements pick up a bit of reactivity. Especially when faced with fluorine, things can get a bit wacky, giving rise specifically to neon difluoride, xenon di-/hexa-fluoride and radon difluoride.

If you want to see a comparison of burning H2 and He balloons, Periodic Videos did it.  He is about at 1:50; H2 is before that.  They did a whole separate video on H2 also.  https://www.youtube.com/watch?v=a8FJEiI5e6Q#t=111

Yes, hydrogen is lighter than helium but helium, on the other hand, is an inert gas (very less reactive). Also, hydrogen is highly flammable so that would make it unsafe to play with balloons.

One counter-argument: Helium is essentially a "fossil gas", and there's a limited supply of easy-to-get helium (until we get practical fusion reactors running, at least). Hydrogen, on the other hand, is universally available in $\ce{H2O}$ and needs only a bit of electricity to break it out. Since helium has important industrial uses other than balloons, I expect that we will eventually find it becomes too expensive to throw away on toys.

leaks/diffuses out of the balloon more rapidly than helium does - exactly the opposite is true. Helium leaks more rapidly, because the hydrogen atoms exists only in paired molecules, so "H2" is an much bigger molecule as one "He" atom and therefore the "He" has much-much higher leakage...

Hmm... http://en.wikipedia.org/wiki/Helium says Helium diffuses  3 times after than air but only 65% of Hydrogen gas.  Elsewhere, it's implied that effusion (leak) is primarily driven by mass not size; using 4 for mass of helium and 2 for hydrogen works out to the 65% (1/sqrt(2)) Anyone got a source for @jm666's counterargument?

@Foon aplying the simple Graham's law, the Hydrogen atoms should diffuse faster, but the molecule H2 is bigger than one Helium atom (not by molecular weight). (AFAIK) - going to search some relevant doccu - maybe I'm wrong...

@Foon just found: the atomic radius of the He is 28 (31) pm (by different sources). Now need find the molecular radius of the hydrogen molecule.

For what it's worth, according to Wikipedia, the van der waals radius of He =~ 2.11 A, and H2 is 1.91 A

With hydrogen, you are just one touch away from disaster. A hydrogen balloon goes anywhere near the birthday candles and you end up.
Helium on the other hand is so inert that you can inhale it and all it would do is to make you sound like a chipmunk for a minute or so.

Not completely true. Filling balloons with hydrogen (or methane for that matter) and then making them explode with a flame was a common fun experiment performed in the last lectures before Christmas or carneval. The gist is that while it will burn, it won’t explode unless it’s a $2:1$ hydrogen : oxygen mixture.

From my thinking helium is a stable gas and is a noble gas whereas hydrogen being the lightest gas but is flammable that's why helium is filled in weather balloons.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. This answer is really short and does not offer any new insight to the questions. Please consider expanding it.

@Martin's right. This is sheer opinion, thus is considered better as a comment than answer. Please be more elaborate, cite resources for your claims etc. Otherwise, this may get deleted.

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What happens to the amino acid when it is in a solution with the pH of its isoelectric point? (zwitterions)
I am learning about zwitterions and am confused about the following:

From what I understand, when amino acids are in their solid form, they exist as zwitterions which is why they have such a high melting point and are so soluble in polar solvents.
Then when placed in a solution with pH lower than their isoelectric point, they act as a base and form positive ions, and when placed in a solution with pH above their isoelectric point, they act as acids and form a negatively charged ion and a water molecule.
If the above is correct, what I am confused about is what happens at the isoelectric point. In my textbook it says that the isoelectric point is 'the pH value at which the amino acid exists as a zwitterion', however I thought it also exists as a zwitterion when not in solution. Are they refering just to the case of the amino acid being in solution and ignoring the solid state?
If this is so, I still do not understand how it remains as a zwitterion at the isoelectric point. Why does it not act as both a base and an acid and accept a H+ ion onto the O- atom in the carbonyl group and donate the extra H+ ion from its amine group? Does this have something to do with the fact that then it would no longer have any charges on it so it would not be soluble in the polar solvent?

The accepted definition by IUPAC gold book is: The pH value at which the net electric charge of an elementary entity is zero. pI is a commonly used symbol for this kind-of-quantity. It should be replaced by pH(I) because it is a pH determined under that particular condition.
The solubility of an amino acid is the smallest at its isoelectric point and it easily precipitates.
For question 4 you need to be familiar with dynamic equilibirum. The zwitterion accepts and loses protons all the time, but at the isoelectric point, its concentration is the biggest among all the forms of the amino acid, that is, the species that has a protonated carboxylic group, would easier lose that proton than the zwitterion's carboxylic group would accept a proton.

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Is sign convention for delta H and enthapy change of a solution different?
In the Ncert Chemistry Tb, it is given that

$\Delta H$ is negative for exothermic reactions which evolve heat during the reaction and $\Delta H$ is positive for endothermic reactions which absorb heat from the surroundings.

In enthalpy change video lesson which I got access to online, it is given that
for $\ce{NH4NO3(s) -> NH4NO3(aq)}$ since the temperature decreases during the reaction, it is positive enthalpy of solution.
How is it positive? I guess both $\Delta H$ and enthalpy change of a solution is almost change. Since the temp decreases here, it is endothermic, so it should be positive right?
Are both of the above processes different from each other and are the sign conventions vice versa for each other?

Welcome to Chemistry.SE! Please have a look at this tutorial to acquaint yourself with the way math and chemical formulae can be nicely formatted on this site.

The reaction is endothermic, meaning that it takes energy from the surrounding. The "surrounding" in your example is the solvent, so in the dissolution process, energy is taken from the solvent,  thereby decreasing its temperature.


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How does alkaline hydrolysis of peptides occur?
I have been learning about zwitterions and the isoelectric point, and have been trying to apply this to hydrolysis of peptides. From the diagram in my textbook, I can see that when acid hydrolysis occurs, the amino acids formed from the hydrolysis act as bases because the pH is below their isoelectric point, so they accept the $\ce{H+}$ ions forming positively charged ions.
Now when it comes to alkaline hydrolysis, I don't understand why the peptide would not just be hydrolysed, and then act as an acid and donate the extra $\ce{H+}$ ion on its amine group to the $\ce{OH-}$ ions to form negative ions and a $\ce{H2O}$ molecule. Instead, in my text book it shows something like this:
$$\ce{-NH-CH(CH2SH)-C(=O)-NH-CH(CH3)-C(=O)- + NaOH \\-> -NH-CH(CH2SH)-C(=O)-O^{-}Na^{+} + H2N-CH(CH3)-C(=O)-}$$
I'm really sorry if this makes no sense whatsoever...

@Martin Thank you for editing my post. What program are you using for that writing so that I can use it in the future?

No problem at all. For first aid have a look at the [help]. It is basically an implementation of LaTeX as a markup language. These two posts on meta might also be of help: here and here. However, even if you are tempted (everybody is) please do not use markup in the title field, see here for details. Welcome on board!

In the drawing shown below, the first 3 lines show a small peptide and how it will exist under

acidic conditions: the basic amino nitrogen is protonated
neutral conditions
basic conditions: the carboxylic acid group is deprotonated

A peptide contains multiple amide bonds ($\ce{-RCO-NHR'-}$).  When a peptide is hydrolysed under basic conditions, the amide bond is broken.  Note that, as mentioned above, under basic conditions any acid groups in the peptide will exist as very stable (due to resonance delocalization) carboxylate groups.  Line 4 in the drawing shows this in more detail.
The actual mechanism for the hydrolysis of the amide group under basic conditions is pictured in the bottom line of the drawing.  The reaction occurs by a nucleophilic bimolecular substitution reaction ($\ce{S_{N}2}$).  The nucleophilic hydroxide anion attacks the relatively electrophilic amide carbonyl carbon to form a tetrahedral intermediate.  This intermediate can either return back to the amide or go forward to the carboxylic acid (hydrolysed amide).  This same $\ce{S_{N}2}$ reaction can be repeated until all of the amide bonds in the peptide are hydrolysed.

Thank you for your reply! I have a quick question; in the mechanism shown, you have R-CO-NH2.Does it work the same way if you have R-CO-NH-R', as in an amino acid? Then at the end you would have R-COO(minus) + H2N-R' and Na+ attaches to the -COO(minus)...

Yes, that is correct.

The unprotonated amino group ($\ce{R-NH2}$) does not protonate $\ce{OH-}$ because it is an extremely weak acid. Primary and secondary amines have a $pK_a$ of around 35, while the $pK_a$ of hydroxide is about 13-14 in aqueous solution. Deprotonating $\ce{NH2}$ groups requires much stronger bases in non-aqueous medium. Therefore, the carboxylic acid group of the amino acid is deprotonated in aqueous alkaline solution, as shown in the scheme from your textbook.


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The instability of hydrated carbon dioxide or "carbonic acid"
Why is hydrated carbon dioxide - the predominant form of acid that one gets upon dissolution of carbon dioxide in solution - so unstable?
Is the below rationale valid?

Carbon in carbon dioxide has two (empty? no, but still vulnerable to attack) p-orbitals and bears a strong partial positive charge.
Oxygen's lone pair can attack an empty p-orbital and form a formal charge-separated complex with the carbon dioxide.
This is "hydrated" carbon dioxide or "carbonic acid." This form is extremely unstable and subject to disproportionation due to an unfavorable charge and the unfavorable nature of charge separation itself. The change in entropy also favors the products of disproportionation.
However, there exists a pathway to stability - that is - protonation of the oxygen with the negative formal charge by the oxygen bearing the positive formal charge.
This, however, is akin to a forbidden fruit; the ephemeral three-membered ring that would have to be formed exhibits "ring strain" (if you object to this term, can you please elaborate on your objection), and as a result, disproportionation is overwhelmingly favored, especially from an entropic standpoint (reconstitution of carbon dioxide gas is highly entropically favorable).

The situation is described by the following equilibrium
$$\ce{H2O + CO2 <=> HO-CO-OH}$$
In a closed system, using Le Chatelier’s principle you can manipulate the equilibrium in either direction (e.g. pump in carbon dioxide and increase the concentration of carbonic acid, etc.).  However in an open system, as carbon dioxide escapes, the amount of carbonic acid will decrease.
When all is said and done, water and carbon dioxide are two extremely stable molecules.  Since
$$\Delta G = -RT\ln K$$
Their stability is what drives the equilibrium to the left and why carbonic acid appears to be (relatively) so unstable.

I had the same issue in class today; I don't see how we get from water and carbon dioxide to (HO)2CO directly. I think my proposed intermediate explains the instability; it is very hard to actually form (HO)2CO unless under pressure; only 0.4% of the CO2 pumped into water turns into actual carbonic acid or (HO)2CO at STP. The rest is in the charge separated form.

"I don't see how we get from water and carbon dioxide to (HO)2CO directly"  Your scheme is correct, it's just another SN2 reaction of water and a carbonyl.

SN2 but no leaving group?

After water reacts with one C=O, the oxygen in the remaining C=O (let's view the C=O as a two-membered ring) is the leaving group, more specifically, one of the two C=O bonds is the leaving group.  The remaining C-O bond keeps the leaving group connected to the carbon.

Okay, so I just need to generalize the very specific definition of SN2 presented to me in my textbook.

Yes. (but my comment needs to be at least 15 characters)

Apart from the obvious nucleophilic attack a concerted [2+2] cycloaddition reaction can occur. (As for many cases one should not stretch the definition of SN2 too much, it is just a mathematical border case. True SN2 reactions may only happen in gas phase, never in solvent.)

@Martin Three questions: 1) Would you also suggest that a Grignard reagent reacts with $\ce{CO_2}$ via a 2+2 cycloaddition? 2) Would you agree that diethyl carbonate reacts with water by an $\ce{S_{N}2}$ mechanism?  3) If so, where in the series (diethyl carbonate, ethylene carbonate, methylene carbonate [C2H2O3], carbon dioxide} might the mechanism switch to 2+2 and why?

I am not sure, but it is certainly a possibility worth considering. In a Grignard the Mg is only very loosely bound to the C, so a usual nucleophilic attack might be more favourable. 2) I believe the most favourable path will be a nucleophilic attack. But again I cannot be sure. 3) I honestly have no idea. I would consider all pathways and then decide for the energetically most efficient. It will very much depend on the surrounding system. In condensed phase, there will be many factors influencing the actual mechanism. Usually you cannot observe mechanisms, so an educated guess has to do.

This answer looks at the molecular structure characteristics behind the thermodynamics reported by Ron
We are dealing with the reaction
$\ce{CO2 +H2O <=> H2CO3}$
or more accurately
$\ce{O=C=O +H2O <=> (HO)2C=O}$
One might expect the equilibrium to favor the right side through the replacement of a pi bond with a sigma bond and the ability of the hydroxylated compounds to fit better with the hydrogen-bonded water structure. Such bonding features are common drivers of reactions between high oxidation-state nonmetaloxides and water. But the carbon dioxide molecule on the left has two particular features that disfavor such a reaction thermodynamically.
Carbon dioxide polarity: the whole truth
Many a chemistry textbook describes carbon dioxide as nonpolar, but it isn't. It is non-dipolar. Carbon dioxide has two strong opposing dipoles that cancel out as dipoles, but combine to make a powerful quadrupole. This quadrupole is well-sized to be solvated strongly with both the water dipole and the water quadrupole (look along the line from one hydrogen atom to the other, you see a quadrupole). Indeed the water quadrupole, whose dipole components are directed oppositely from the dipole components of the carbon dioxide quadrupole, fits like a hand in a glove when it comes to solvation. The resulting solvation capability is one reason carbon dioxide, despite its "nonpolarity", has fairly good solubility in water in the first place.
The revenge of 4n pi electrons
Carbon dioxide has an especially strong pi bonding stabilized by delocalization; it might be thought of as a linear-molecule counterpart to aromaticity. (In contrast with the usual form of aromaticity, the interaction favors $4n$ electrons; carbon dioxide with all atoms $sp$ hybridized gives $n=2$.) Reaction with water to break one of these pi bonds upsets this delocalization much like benzene resisting addition because of a similarly strong stabilization by delocalization.

Note that waterfree H2CO3 is reportedly kinetically very stable, as  solid or gas, subliming near -55 deg C. A molecule of H2CO3 has reportedly halflife 180 000 years. But there is extremely strong autocatalytic effect of water. //  At ambient temperatures, pure carbonic acid is a stable gas.[5] There are two main methods to produce anhydrous carbonic acid: reaction of hydrogen chloride and potassium bicarbonate at 100 K in methanol and proton irradiation of pure solid carbon dioxide.[2]

@Poutnik all true. But I do not discuss this in the answer because the water-free case seems outside the scope of the question. Aqueous carbonic acid seems to be stable at high pressure and temperature..

Sure, I have mentioned it as a side note. This On the Surprising Kinetic Stability of Carbonic Acid (H2CO3) may be worthy to read too, assuming with the access.


text:
Volatile and non-volatile solutes in solution
What is the difference between having a volatile versus a non-volatile solute at the boiling point of the solution?
I have been getting mixed answers online concerning the differences outlined above. This is what I have gathered so far:

A non-volatile solute does not produce vapour at the boiling point of the solution.
A volatile solute does produce vapour at the boiling point of the solution.

Can someone confirm these definitions?

Nonvolatile solutes have negligible vapor pressures when doing problems in chem texts.  Volatile solutes must be accounted for.

A substance with higher vapour pressure vaporizes more readily than a substance with a lower vapour pressure. In case of a volatile solute this has a high vapour pressure and hence produces vapour. While in the case of a non-volatile solute due to its lower vapour pressure it does not produce vapour. The vapor pressure of a substance is the pressure at which its gas phase is in equilibrium with its condensed phases (liquid or solid).


text:
What is the highest relative abundance for diatomic bromine?
I was introduced to this thing today, and saw a graph showing the detection of ions with different relative abundances.
Just as an example here, $\ce{Br2}$ is shown a detection chart. And at $160~\mathrm{m/z}$ the ion has the highest relative abundance, but why? Since the $\mathrm{m/z}$ ratio is just the mass, for $\mathrm{z}$ is always $+1$, the one which is supposed to have the highest peak should be at $162$?

See https://chemistry.stackexchange.com/questions/86158/why-is-the-probability-for-the-signal-at-160-in-the-bromine-mass-spectrum-twice?rq=1 for a similar answer

In $\ce{Br2}$, the possible combinations of isotopes are:

$\ce{^{79}Br-^{79}Br} : \mathrm{m/z}~158$
$\ce{^{79}Br-^{81}Br} : \mathrm{m/z}~160$
$\ce{^{81}Br-^{79}Br} : \mathrm{m/z}~160$
$\ce{^{81}Br-^{81}Br} : \mathrm{m/z}~162$

Bromine's isotopic distribution is essentially a $50:50$ ratio of mass $79$ and mass $81$. That means that there's a $25\%$ chance of $\mathrm{m/z}~158$ $(0.5 \cdot 0.5 = 0.25)$ arising from $\ce{^{79}Br-^{79}Br}$. There's a $50\%$ chance of $\mathrm{m/z}~160$ since there are two combinations ($\ce{^{79}Br-^{81}Br}$ and $\ce{^{81}Br-^{79}Br}$) that give that mass $(0.5\cdot0.5 + 0.5\cdot0.5 = 0.25)$. There's a $25\%$ chance of $\mathrm{m/z}~162$  $(0.5\cdot 0.5 = 0.25)$ arising from $\ce{^{81}Br-^{81}Br}$.

Since they are all ions, how could ions be combined with ions to make new ions? And why can some large ions fragment?

jerepierre's answer is correct, you should see a statistical distribution of 3 different peaks, but also keep in mind that the m/z is not always the same as the mass. Depending on the molecule you're looking at and the ionization technique, the charge can be much larger than 1. Electrospray ionization, commonly used for ionizing biomolecules, routinely produces multiply charged ions. +20 charge is not uncommon. Additionally, many types of ionization cause fragmentation of molecules. For electron impact ionization, it's often the case that a peak for the full molecule is not observed at all.


text:
Why is magnesium hydroxide not hygroscopic?
I asked this question today and my professor was unable to provide any rationale.
I know that sodium hydroxide is hygroscopic, but why not magnesium hydroxide (at least to the same extent as $\ce{NaOH}$)? Could it be that $\ce{Mg(OH)2}$ has a higher lattice energy; the hydroxide is more closely associated with the magnesium ion and less able to react with carbon dioxide?

Not an answer to the question, bet also a partial refutation of what has been said: aluminum is highly hydroscopic, mainly because of its high affinity for oxygen. Yet, magnesium seems to avoid the reactions of both sodium and aluminum.

Try looking at Inorganic Chemistry in Aqueous Solution by Jack Barrett around page 41 on google books or elsewhere.
Entropy of hydration of Mg2+ is very negative compare to Na+.
According to Barrett, more highly charged ions restrict the movement of water molecules of hydration more.  Also, the smaller the radius of the ion, the more restricted the movement of the hydrating waters is.  Both the number of waters whose movement is restricted, and the degree to which they are restricted, is affected by charge and radius.
A complete analysis would consider both the entropy and enthalpy contributions.


text:
Why is the water column empty in ICE table
In the ICE table pure solids and liquids have a value of one in the equilibrium expression but when we create an ICE table, why is the $\ce{H_{2}O}$ column empty? Should it not have a value of one?

Presumably water is the solvent for this reaction, which means that it will be in extreme excess and the amount will not change to a measurable extent.


text:
What is the molecular mechanism of ammonia's toxicity?
How does it damage proteins/lipids/DNA? There seems to be surprisingly little literature on its toxicity.

If you have access to the article, you can read about excess activation of NMDA receptors in the brain here.

Ammonia is reportedly not toxic in bacteria even at molar conentrations (PMID 16604417).
However, yeast, plants, and animals (presumably, all eukaryotes) are unable to tolerate high intracellular ammonia concentrations.  The best explanation for this phenomena I've seen is a hand-waving assertion that intracellular ammonia taxes overtaxes the cell's ability to maintain membrane gradients (PMID 7764110).  However, that isn't a particularly detailed or satisfying response.
I agree with your assertion that there's a paucity of literature on the subject - we probably don't know the answer yet!

Ammonia will harm you like any other base. Although it's not directly caustic, it will cause inflammation and general "corrosion" of whatever membranes it encounters, such as those in your lungs when you get a good drag of ammonia gas. Upon contact with your body it becomes NH4OH and the hydroxide hydrolyzes amides in proteins.

Here are some articles about the neurotoxicity of ammonia.
Identifying the direct effects of ammonia on the brain
Bosoi, C.R.; Rose, C.F. Metab Brain Dis 2009, 24, 95-102

Ammonia in solution is composed of a gas (NH3) and an ionic (NH4 +) component which are both capable of crossing plasma membranes through diffusion, channels and transport mechanisms and as a result have a direct effect on pH. Furthermore, NH4 + has similar properties as K+ and, therefore, competes with K+ on K+ transporters and channels resulting in a direct effect on membrane potential. Ammonia is also a product as well as a substrate for many different biochemical reactions and consequently, an increase in brain ammonia accompanies disturbances in cerebral metabolism. These direct effects of elevated ammonia concentrations on the brain will lead to a cascade of secondary effects and encephalopathy.

Molecular mechanism of acute ammonia toxicity: role of NMDA receptors
Monfort, P.; Kosenko, E.; Erceg, S.; Canales, J.-J.; Felipo, V Neurochemistry International 2002, 41, 95-102

(1) depletion of brain ATP, which, in turn, leads to release of glutamate; (2) activation of calcineurin and dephosphorylation and activation of Na+/K+-ATPase in brain, thus increasing ATP consumption; (3) impairment of mitochondrial function and calcium homeostasis at different levels, thus decreasing ATP synthesis; (4) activation of calpain that degrades the microtubule-associated protein MAP-2, thus altering the microtubular network; (5) increased formation of nitric oxide (NO) formation, which, in turn, reduces the activity of glutamine synthetase, thus reducing the elimination of ammonia in brain.

Oxidative and nitrosative stress in ammonia neurotoxicity
Skowrońska, M.; Albrecht, J. Neurochemistry International 2013, 65, 731-737

An increase of brain ammonia in experimental animals or treatment of cultured astrocytes with ammonia generates reactive oxygen and nitrogen species in the target tissues, leading to oxidative/nitrosative stress (ONS). In cultured astrocytes, ammonia-induced ONS is invariably associated with the increase of the astrocytic cell volume. Interrelated mechanisms underlying this response include increased nitric oxide (NO) synthesis which is partly coupled to the activation of NMDA receptors and increased generation of reactive oxygen species by NADPH oxidase. ONS and astrocytic swelling are further augmented by excessive synthesis of glutamine (Gln) which impairs mitochondrial function following its accumulation in there and degradation back to ammonia (“the Trojan horse” hypothesis). Ammonia also induces ONS in other cell types of the CNS: neurons, microglia and the brain capillary endothelial cells (BCEC). ONS in microglia contributes to the central inflammatory response, while its metabolic and pathophysiological consequences in the BCEC evolve to the vasogenic brain edema associated with HE. Ammonia-induced ONS results in the oxidation of mRNA and nitration/nitrosylation of proteins which impact intracellular metabolism and potentiate the neurotoxic effects.

Ammonium salts are routinely used in protein purification.  It does not damage proteins.

That's a salt. I think the question here is about $\ce{NH3}$ itself.

only above pH 9 will it be mostly in the free base form

Indeed. I was commenting that the OP probably wanted more discussion of the free base, which isn't really indicated in your answer.

One process I can think of is that it's toxic through alkalosis and a high enough dose can overwhelm homeostasis.


text:
Does hydroxytyrosol have the same neurotoxicity potential as dopamine?
It has the catechol group, and it's the oxidation of the catechol group that makes dopamine produce reactive oxygen species (ROS).
Hydroxytyrosol isn't that different from dopamine (the distal N being replaced with an OH). I'm not sure if that really affects anything. Yet - it's often sold as an antioxidant.
Couldn't it have the same ROS potential that dopamine and other catecholamines have?

I have expanded your post with some links. I hope with ROS you meant what I inserted. Please check.

Okay thanks! Yup - what you inserted was correct.

while hydroxytyrosol, ( lets call it HT because I can't keep spelling that ) is a catechol, it lacks the amine of dopamine. Dopamine gets metabolized to an aldehyde, this aldehyde can form schiff bases with amines on proteins. The catechol can also react with thiols on proteins. So in addition to dopamine's oxidative damage potential, it can crosslink proteins. HT doesn't have the amine, so it probably doesn't get metabolized in the same way.
Here's a paper with more information on dopamine metabolism and toxicity. Look up other papers by that group for more.


text:
What has stronger antioxidant potential: BHT or BHA?
Butyl Hydroxy Toluene (BHT) has two tert-butyl groups that can stabilize a free radical. Butyl Hydroxy Anisol (BHA) doesn't, though it has the 4-substituted methoxy group (I'm not sure what that does).
What is the stronger antioxidant?

I have added the full names and links, please check if this is correct or if you would have referred to something else.

Please explain (a) how a tert.-butyl substituent stabilizes a radical intermediate and (b) how BHT, BHA (or TBHQ) manage to act as antioxidants.

yes - the full names and links are correct - thanks! regarding the second point - I'll try to add that in later (in a rush now)

That's a good question. In literature there are some contrasting opinions about the antioxidant power of BHA vs BHT. TBHQ (2-tert-butyl hydroquinone) is also used as antioxidant, although maybe is less common.
There is a paper from 1997 in which the antioxidant activity of BHA, BHT and TBHQ are examined with Miller's test (M. Karamac and R. Amarowicz, Grasas y Aceites, 1997, 48 (2), 83-86), and in this test BHA and BHT gave similar results, although in the initial phase of the experiment BHA seemed to have a higher antioxidant activity than BHT.
In the paper precedent studies are also reported. Quoting from the paper:
Higher antioxidative activity of BHA than of BHT and lower for TBHQ was
demonstrated in thin layer model system of methyl linoleate on cellulose (J. Am. Oil Chem. Soc. 52,174-178). Berner et al. (J. Am. Oil Chem. Soc. 51, 292-296) reported antioxidative index in lard emulsion 30-fold higher for BHA than for TBHQ. Yet, in active oxygen method, antioxidant index for TBHQ was 3-fold higher than for BHA and BHT. Higher antioxidative activity of TBHQ compared with BHA and BHT in model studies with soybean oil and chicken fat in emulsion versus dry oil tests was reported by Cort et al. (J. Am. Oil Chem. Soc. 52,174-178). Relative effectiveness of antioxidants-linoleic acid monolayer on silica gel for TBHQ was almost the same as for BHT, while for BHA almost 3-fold higher (J. Food Sci. 42, 1533-1535).


text:
Why are d-d electronic transitions forbidden and weakly absorbing? Why do they occur at all?
I am unable to understand why d-d electronic transitions are forbidden? Why are they weakly absorbing and apart from that, why do they occur at all?

Welcome to Chemistry.SE! Please have a look at this tutorial to acquaint yourself with the way math and chemical formulae can be nicely formatted on this site. As for your question: For most of it you can find an answer in this Wikipedia article.

In order for an electronic transition to be allowed (occur with strong intensity), certain "selection rules" must be obeyed.  You may already be familiar with the rule that the electron spin quantum number cannot change during a transition. It is this rule that forbids singlet-to-triplet absorption or emission (emission follows the same rules as absorption) and consequently makes phosphorescent emission so much weaker than fluorescent emission.

$\ce{\Delta S = 0}$ the Spin rule

Another rule, often referred to as the "LaPorte" rule or the "Orbital" rule, states that in molecules with a center of symmetry (centrosymmetric molecules), transitions within a subshell are forbidden when $\ce{\Delta l =0}$.

$\ce{\Delta l = \pm 1}$ the LaPorte rule

The orbital angular momentum quantum number "l" describes the type of orbital (s, p, d, f) within the subshell.  In centrosymmetric molecules, the various s, p, d, f orbitals cannot mix within a subshell.  For example, the 4d orbital cannot mix with the 4p orbital in a centrosymmetric molecule, such mixing being symmetry forbidden.  Octahedral complexes can be centrosymmetric.  In such a case 3p->3p, 3d->3d transitions would be forbidden by LaPorte's rule.  Tetrahedral molecules do not have a center of symmetry and p-d orbital mixing is allowed, so in the case of tetrahedral molecules 3p->3p and 4d->4d transitions may appear stronger because a small amount of another orbital may be mixed into the p or d orbital thereby removing the violation of LaPorte's rule.
If we can do something to remove the centrosymmetric nature of an octahedral complex, then we "relax" LaPortes rule and weak transitions can be observed.  For example, if a molecular vibration removes the molecular center of symmetry, then a d->d transition can occur if light is absorbed at that instant.  Things such as spin-orbit coupling, vibronic coupling, or ligands that can mix with the metal's d orbitals would all allow relaxation of LaPorte's rule and thereby permit weak transitions to be observed.

An important point is that only spin-orbit coupling allows singlet and triplet states to mix and so overcomes the 'forbiddenness' of the $\Delta S$=0 selection rule. This coupling is enhanced by paramagnetic species and heavy  atoms in a molecule or solvent. Also note that phosphorescence always has a longer lifetime than fluorescence whatever its quantum yield.


text:
Can carbon dating tell how long a species would survive?
Carbon ($\ce{^{14}C}$) is used to determine how long a particular species has lived or how old it is. Can we use the same isotope to derive how long a species may survive? (My understanding is still quite at infant level on half-life concept)
Is it prohibited or is it impossible? If not possible, why?

No, it can't be used for that. Read the mechanism of radiocarbon dating.

Radiocarbon, often abbreviated as $\ce{^{14} C}$, is constantly being created in the atmosphere by the interaction of cosmic rays with atmospheric nitrogen. The resulting radiocarbon combines with atmospheric oxygen to form radioactive carbon dioxide. This is then incorporated into plants by photosynthesis, and animals acquire $\ce{^{14}C}$ by eating the plants. When the animal or plant dies, it stops exchanging carbon with its environment, and from that point the amount of $\ce{^{14} C}$ it contains begins to reduce as the $\ce{^{14}C}$ undergoes radioactive decay. Measuring the amount of $\ce{^{14}C}$ in a sample from a dead plant or animal such as piece of old wood or a fragment of bone provides information that can be used to calculate when the animal or plant died. The oldest dates that can be reliably measured by radiocarbon dating are around 50,000 years ago, though special preparation methods occasionally permit dating of older samples.

I have seen that link. I am also referring to other articles. And I don't mean to be rude, but please take a look at (A) how to write a good answer in Stack Exchange by help. (B) how to write a good answer in SE by Meta. I am stating this, mainly 1. Chemistry is still in beta 2. I have seen Stack Exchange community reviewing questions and answers. It's obvious you seem to have a knowledge in Chemistry. So please share with the community when answering.

Alright I am glad you at least added some text...

I dimly remember a paper contributing to the controversy if nerve cells in the brain continue to divide in adulthood.
The experiment relied on the 14-C spike (I think it was carbon and not tritium) caused by above-ground nuclear testing.  The authors extracted DNA (which we all know is synthesized during cell division) from brains preserved before the Sixties.  It turns out that no excess 14-C was incorporated.
So, yes, in its own way, radiocarbon dating can be used to gauge the age of a living being.

Your answer says yes, in its own way, radiocarbon dating can be used to gauge the age of a living being but the above answer and some other materials specify that carbon dating can determine dead species...or is this another angle of looking at the same thing?


text:
Which of the following statements is true about the trend down group 7?

Which of the following statements is true about the trend down group 7 (halogens) with increase in atomic number?

The reactivity increases.
The atoms get smaller
The boiling points decrease
The melting points increase
The reactivity decreases

For me D and E must be right but only one can be right.

Just from a test taking point of view, the answer is likely either A or E, unless you are convinced there is no trend in reactivity...

Related: Why does color of halogen deepens and the boiling point decreases down the group?

The answer is D, the melting points increase. This is absolutely true (source for values):
\begin{array}{lrr}
\text{Halogen} & \text{Melting point}/^\circ\mathrm{C}& \text{Boiling point}/^\circ\mathrm{C}\\\hline
\text{fluorine}   & -220  & -188  \\
\text{chlorine}  & -101  & -35   \\
\text{bromine}  & -7.2  & 58.8  \\
\text{iodine}   & 114   & 184   \\
\text{astatine}  & 302   & 337   \\\hline
\end{array}
And the boiling point increases, too, so answer C is definitely wrong.
Another obviously wrong answer is B, the atoms get smaller. With increasing main quantum number the atoms obviously have to become bigger since the main electron density moves further away from the nucleus. In numbers:
\begin{array}{lrr}
\text{Halogen} & \text{Covalent radius}/\mathrm{pm}& \text{Ionic radius }\ce{(X^{-})}/\mathrm{pm}\\\hline
\text{fluorine} &  71  & 133  \\
\text{chlorine}   &  99  & 181  \\
\text{bromine}  & 114  & 196  \\
\text{iodine}   & 133  & 220  \\
\text{astatine}  & 150  &      \\\hline
\end{array}
For answers A and E it cannot be unambiguously answered, because reactivity is not a well defined concept, see "What is reactivity really, and can it be quantified?"  for example. The IUPAC goldbook states, that reactivity is a kinetic property. It goes on to describe, that it can only be absolutely used in a given context or in reference to another system. However, it is also often used in expressing general trends, which is the case when looking at this question.
Given this context, you can find the following portion on UC Davis ChemWiki:

Reactivity of Elements: decreases down the group
  The reactivities of the halogens decrease down the group ( At < I < Br < Cl < F). This is due to the fact that atomic radius increases in size with an increase of electronic energy levels. This lessens the attraction for valence electrons of other atoms, decreasing reactivity. This decrease also occurs because electronegativity decreases down a group; therefore, there is less electron "pulling." In addition, there is a decrease in oxidizing ability down the group.

In this context, answer A is wrong. However, I would not consider E to be strictly true either. Have a look at the electron affinities, and you will find, that fluorine behaves anomalous. This was also discussed in "Why does chlorine have a higher electron affinity than fluorine?"
However the general statement will usually be given as: "Electron affinity decreases down the group."
\begin{array}{lc}
\text{Halogen} & \text{Electron Affinity}/\mathrm{kJ\cdot mol^{-1}}\\\hline
\text{fluorine}  &  -328.0  \\
\text{chlorine} &  -349.0  \\
\text{bromine}   &  -324.6  \\
\text{iodine}    &  -295.2  \\
\text{astatine}  &  -270.1  \\\hline
\end{array}
The same trend can be observed with bond enthalpies of the $\ce{X2}$ series (source):

This trend is broken for the hydrogen halides $\ce{HX {(g)}}$.

Your assumption is therefore correct, maybe you have already done the same analysis. However, we can only definitely state and proof, that answer D must be correct, while answer E might be correct depending on the reference system.

Your second table is missing the ionic radius for astatine :/

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


text:
Which is the most intuitive to perform Dynamic Monte Carlo calculations?
I'm trying to find a (free is possible) code to perform Dynamic Monte Carlo calculations over chemistry systems. Which one is the most intuitive one?

What do you mean by "Dynamic" Monte Carlo? Dynamic generally suggests motion, e.g. molecular dynamics simulations. MC does not simulate motion, per se, but rather samples configurational space in a random way.

Dynamic or Kinetic Monte Carlo is a technique to perform out-of-equilibrium calculations using the basic ideas proposed by the "usual" MC approach.

Got it. I'm familiar with Kinetic MC, just never heard it referred to as Dynamic MC before.

In chemistry, dynamic Monte Carlo (DMC) is a Monte Carlo method, also called Kinetic Monte Carlo for modeling the dynamic behaviors of molecules  http://en.wikipedia.org/wiki/Dynamic_Monte_Carlo_method

If you want to perform Kinetic Monte Carlo for a chemical reaction system, you can use ZACROS. I am not sure what your system is? You should follow some good text book like "Understanding Molecular Simulation" to identify the type of your problem and appropriate software for that. There are plenty of freely available code on the internet but they all are for some specific purpose, So basically it is your responsibility to choose the best available software that will suit your need.

Welcome to chemistry.se! Thank you for your insight. I edited in some links, so people have an easier way of finding things. This is always very much appreciated. (In case of ZACROS you might find a japanese touch screen company.) Please make sure, that what I inserted is correct. Thank you.

You are welcome :) you have hyperlinked the right website.


text:
How to verify the accuracy of a pH meter
One suggestion I received is that I prepare a solution of a known activity and probe the solution with the pH meter. The problem I see with this is that any error in solution preparation can lead us to conclude that the pH meter is faulty.
Is there any other way to test the accuracy of a pH meter? We need to make extremely accurate measurements (we are studying the activities of ionic solutes).
Also, what solution has an activity of exactly 1.0?

Your school should have a purchased buffer solutions that have been calibrated against a NIST standard.  If they don't, have them purchase several in order to span the entire pH range, here's one from Sigma-Aldrich.  For your calibration, pick a standard solution that is close to the range you plan to work in.  Here's a couple of links with some more "tips" on how best to actually perform the calibration.

tips #1
tips #2

You don't suggest just making a solution?

No, not for "extremely accurate measurements."

There are usually standard buffer solutions available at pH 4 and 10 used for calibrating pH meters. I assume the manufacturer takes great care to verify the accuracy of these solutions.
Here is a list of standard solutions and their pH. If you trust your scale more than your pH meter, you could make some of these buffers from dry reagents and see how your meter reads them.

This is a good question. If you calibrate a meter and then put the electrode back into the 4 or 10 buffer (i.e. not the 7 buffer) the meter reads exactly what the meter thinks the pH of the buffer is. For example, with an operational buffer labeled pH 4.00 ± 0.02 pH at 25 °C the meter thinks the pH is 4.01 and, at 25 °C it will read 4.01 but not for long. Soon the electrode will begin to drift and you will get readings like 4.03, 4.02.4.01, 4.00. 3.99 etc. The error (relative to 4.01) may exhibit a trend over time or may stay centered around 4.01. Whatever it does, the rms error over some reasonable period of time (such as an hour or 2) is a measure of the meter's accuracy. It says nothing about the accuracy of the buffer itself and thus you cannot talk about the accuracy of the pH reading based on the accuracy of the meter alone. To estimate the accuracy of the reading you can rms the accuracy of the meter with the accuracy of the buffer. Thus if the buffer is ±0.02 (you can also buy ±0.01) and the meter rms error measured 0.03 the assumed accuracy of readings is sqrt( (0.02)^2 + (0.03)^2 ) = 0.036


text:
Why does steel wool catch fire when rubbed with a battery?
I was watching an episode of Man vs Wild where steel wool was rubbed with phone battery. To my surprise, the wool caught fire. What is the reason for this?

Steel wool is a mat of very fine steel fibres. Because of their small size, it's easy to heat them up quickly to the point of burning and their relatively large surface area to volume ratio lets them oxidize quickly. If you try it with a flame, you'll find that it's quite easy to burn steel wool. When you short the terminals a battery across steel wool, a lot of current passes producing heat (think about incandescent light bulbs), and especially if you touch the battery such that only a few fibres are touching, enough heat is produced to cause them to burn.

Why is it easy to heat them because of their small size

Less material to heat. Think about heating a large bar of steel versus a tiny wire. While steel doesn't have spectacular thermal conductivity, it's enough that it takes a much hotter flame to heat a bar to the point of burning. A small wire doesn't have much material to transfer the heat to. The other thing is a bigger piece needs more oxygen to burn than can be readily obtained from the air by diffusion. This is why cutting torches are designed to give a blast of extra oxygen to the hot metal.

The small size means higher electrical resistance which means more heat from a current flowing through.

Which is why I indicated that only touching a few fibres at a time helps, but I suspect it's not the only factor. I'm not sure whether oxygen transport or heat capacity would be more important, but the iron fibres don't have to get that much thicker before it doesn't work with a normal flame, so it can't be just current. Also, assuming a fixed voltage, the point of maximum heat production in the wire is going to be when the external resistance matches the internal resistance of the battery. Beyond that, increasing resistance decreases joule heating.

What effect, if any, does coating the steel wool in soap have - could you start a fire with a Brillo pad or would you have to rinse out all the soap and dry it first?
Also, how would rust effect the reaction - could you do this with a highly-used piece of steel wool?


text:
Basicities of meta-nitrophenoxide and para-nitrophenoxide
The negative oxygen is the site of protonation. Size and EN are the same for both oxygens. We are mainly looking at resonance (and possibly inductive) effects here. We want the least resonance withdrawal of electron density from the oxygen. And if possible, the least inductive withdrawal of electron density from the oxygen (although this is not a major issue; resonance trumps induction). Recall that the nitro group cannot withdraw electron density via resonance effects at the meta position. Therefore meta-nitrophenoxide should be more basic than para¬-nitrophenoxide.
Is my reasoning correct?

Yes, it is. These resonance structures describe basicity in para-nitrophenoxide.


text:
Van der Waals equation at high temperature and high molar volume

At high temperatures the Van Der Waals equation has the form
  $$P\approx\frac{RT}{V_\mathrm{M}-b}$$
What would happen with the equation if in addition to high T, molar volume of the gas ($V_\mathrm{m}$) will get very large too?

The only sense I was able to make out of this was that the pressure would be reduced, but if $V_\mathrm{m}$ stands for the volume occupied by a mole of a substance, than wouldn't a larger $V_\mathrm{M}$ mean less available void space, less volume, and thus a greater pressure?

First, let's look at this from a purely mathematical standpoint. This isn't going to be rigorous at all, but I'm happy to clarify anything if asked.
If we assume we have 1 mole, and we say that $V_\mathrm{m}$ and $T$ become very large, then the question becomes what happens to $P$ (our only remaining variable)? Well as $T$ and $V_\mathrm{m}$ become very large with respect to (w.r.t) our excluded volume parameter, $b$, (also written as $T$, $V_\mathrm{m} >> b$), the -b term in the denominator becomes negligible and we approach the conventional ideal gas law:
$$
P \approx \frac{R T}{V_{\mathrm{m}} - b} \implies P \approx \frac{R T}{V_\mathrm{m}} = \frac{n R T}{V}
$$
If this is not clear, try plugging in some numbers to the equation and calculating what $P$ would be for small $P$ and $T$ (same order of magnitude as $b$) vs. large $P$ and $T$ (100 times larger than $b$). I will let you determine if this leads to greater or less pressure.
Now let's switch gears to a chemistry point of view. What Johannes van der Waals did, is add this little $b$ parameter in the denominator of the ideal gas law. This parameter is a correction to account for the actual volume that each mole of particles can occupy. You can think of it equivalently as the actual volume each particle can occupy, since a mole is just a really big number of particles.
This $b$ correction is taking into account the fact atoms can repel each other at a relatively large distance. If we approximate these atoms to be hard sphere bouncing off of each other, the radii of these hard sphere is what we call the van der Waals radii - the distance at which atoms "bounce" off of each other if they come into contact. So the  parameter $b$ in the denominator is derived purely from the van der Waals radii of the atoms in question. So in the case of large $T$ and $V_\mathrm{m}$, we find that this correction is no longer necessary to make a reasonable approximation, because its presence makes such a small difference.
I recommend looking at the Wikipedia page for more information.


text:
Acidities of amines and carboxylic acids

I have been told that the first of the two acids is the stronger acid. According to this person,deprotonation for both occur at the nitrogen.The resulting conjugate bases include,respectively a neutral amine group,and a negatively charged amine group.Because negatively charged nitrogens are highly unstable from an acid/base standpoint (consider lithium diisopropyl amide)... the first must be the stronger acid (because of its more stable conjugate base).
The problem I have with this line of reasoning is that it's based on a faulty premise - that the nitrogen is the site of deprotonation.I see a much better site of deprotonation for both - and that is the carboxylic acid site. A negative charge here will not only be resonance stabilized but also inductively stabilized by oxygen.Plus, the Ka of ammonium ion is 5.6*10^-10, while the Ka of acetic acid is 1.8*10^-5.In other words - orders of magnitude of difference. I can't see why something that is structurally analogous to ammonium ion (much less ammonia) would be deprotonated preferentially over a carboxylic acid.

This question is nonsensical.  The acidity of compound two cannot be determined because it does not exist in most environments.  It actually exists as a zwitterion, that is, $$H_3^+N+CH(CH_3)COO^-$$.  Compare that to compound 1.

@Aditya It's not necessary to add to the situation.

The first compound is more acidic. The ammonium functionality is inductively electron withdrawing, stabilizing the conjugate base (carboxylic acid).
This answer is assuming that we are talking about the drawn structures exactly. User137 is correct that it's not possible to have the second compound, because the amine is a strong enough base to deprotonate the carboxylic acid, and it exists as the zwitterion. However, assuming that we could hold the amine as neutral, it would be less inductively electron withdrawing than the cationic ammonium.

The extent to which the zwitterion exists is very low.

Compound 1's conjugate base IS compound 2. The positive NH3 loses a proton to become the neutral NH2, but deprotonation of the NH2 group is extremely unlikely. Compound 2 will lose a proton at the carboxylic acid group first, so your friend is wrong to say deprotonation will occur at the nitrogen in both.
This will likely form a zwitterion, where the nitrogen is NH3+ and the carboxylic acid is COO-. This compound will have 2 pKas, one for the carboxylic acid, 1 for the amine. The carboxylic acid will be the lower of the 2 pKas. So if we start at very low pH and add base, the compound will start like compound 1. As we approach the pKa for the carboxylic acid, it will deprotonate to form the zwitterion. Then, as we approach the pKa of the NH3+, it will also deprotonate, forming a compound with an NH2 and COO-.

Which one is the stronger acid? From what you're saying you are saying that compound 2 is the stronger right? @user137

I believe compound 2 is stronger, since compound 1 exists at low pH. Compound 2 would be hard to make because the NH2 only exists at pH well above where COOH converts to COO-. So if you had compound 2 in a very high pH solution, it would immediately give up that proton.

Are you really saying that compound 2 $\ce{H2N(CHMe)COOH}$ is more likely to release a proton than compound 1 $\ce{H3+N(CHMe)COOH}$? Also the amount of zwitterion in compound 2 is very low, but non negligible.

@Martin My reasoning is that compound 1 can be easily made by putting it in low pH solution, using the pKa's for alanine provided by Aditya's answer, the pH must be below 2.33. Since increasing the pH will cause the COOH to deprotonate first, and NH3 holds on to its proton until about pH 9.71, NH3 must have higher affinity for its proton. If you could get compound 2 in a high pH solution, the COOH would most likely lose the proton very quickly. And these are all equilibriums, with small amounts of each protonation state, but compound 2 would be present in very small amounts.

Some answers suggest 1 and others suggest 2. Which is the right one?

These are two possible forms of Alanine(Ala||A), an amino acid with hydrophobic side chain. Here are the exact $pK_A$ values:
$$-COOH:2.33\\-NH_3^+:9.71$$
Compund two is stronger as you anticipated.Cheers!


text:
Acidity of ester and benzoic acid

Can the oxygen in an ester participate in resonance donation? I think so, although the donation is slight. The hybridization is correct. However, resonance withdrawal by the ring via the oxygen doesn't seem possible as my picture depicts. Therefore, because of resonance donation from the oxygen in the conjugate base of the ester, would that make benzoic acid the stronger acid? It seems that the resonance donation would destabilize the conjugate base of the ester.

Can the oxygen in an ester participate in resonance donation?

Sure, structure III in the following diagram is a valid resonance structure.

However, resonance withdrawal by the ring via the oxygen doesn't seem
  possible

No, that's possible too.  It might not be a major contributor, but structure V is a valid resonance structure.

would that make benzoic acid the stronger acid?

I think you're asking which is the stronger acid, benzoic acid or the half-ester of carbonic acid. Resonance structures like III and VII will stabilize both the acid and the corresponding carboxylate anions.  One molecule doesn't appear to have a more resonance stabilized starting acid or a more resonance stabilized carboxylate anion than the other.  So resonance arguments don't seem to suggest that one molecule should be more acidic than the other.  This conclusion, that resonance factors do not control the acidity in these two compounds, can also be arrived at by noting that the starting acids and particularly the carboxylate anions are analogous to the allyl MO system which has a node on the central carbon atom in the HOMO ($\ce{\Psi 2}$). The node effectively  disconnects the substituent (-OR or phenyl) from the MO, meaning that the substituent cannot interact via resonance with the MO.
Inductively one can argue that the electron withdrawing oxygen in the half-ester would destabilize the already positively polarized carbonyl carbon in the starting acid and then also provide some inductive stabilization of the electron-rich carboxylate anion, more so than a phenyl group would.  So perhaps, based mainly on inductive arguments, one could argue that the carbonic acid half-ester would be more acidic than benzoic acid.
I couldn't find any pKa values for carbonate half-esters, but we can compare the pKa of

carbonic acid 3.6

to

benzoic acid 4.2

The values are close, as expected - even in the right direction, but that's probably just luck.

So the phenyl cannot interact with the carboxylate via resonance? Why then are there resonance pictures suggesting this?

Resonance theory suggests that a substituent can interact with the carboxylate carbon via resonance.  MO theory refines this and suggests that a substituent at C2 in the allyl system is effectively disconnected from the HOMO and therefore the substituent should not have a significant influence on the MO.  Note that we have a 1,3-dioxa-allyl system, this perturbation may place a coefficient at C2 - I don't know. If the coefficient is significant, than the substituent could have a greater effect in the 1,3-dioxa-allyl system.

I think one reason why you did not find a pKa for the carbonate half ester is, that it is very unstable and releases carbon dioxide readily. Also phenolates are quite good leaving groups. $$\ce{Ph-O-CO2H ->[\Delta T] Ph-OH + CO2 ^}$$ (Water would also be a quite nice catalyst for this reaction.)

@ron how can you tell whether there is a node in the wave function or not?

@Dissenter  The primary method would be to actually calculate the coefficients for each atomic orbital in each MO; if you get a zero coefficient, then there is a node.  For simple systems like ethylene, allyl, benzene, etc. you can find the coefficients in most books on the subject of Molecular Orbital theory, or on the internet, or just remember them from past usage.

@ron if the phenyl is disconnected from the substituent because of a node, then why is resonance structure V "valid"? Wouldn't that just be showing something that can't happen?

When I wrote, "The node effectively disconnects the substituent (-OR or phenyl) from the MO", I was speaking of the corresponding carboxylate anion - where the dioxallyl system exists. V is not a carboxylate anion.


text:
Baking soda bath to deacidify newsprint?
I have some oil-pastel sketches on newsprint I would like to treat for framing (and longevity). My plan is to rinse them in a deacidifying bath, then use wheat paste to wet-mount them to heavyweight pH-neutral 100% rag paper. And finally, to frame them with UV-protective glass recommended for art.
My question is this: Is there any reason not to use baking soda for the deacidifying bath?
I have experimented with it, and it seems to work great. I used a garden-soil pH test solution before and after, and sure enough, the treated paper shows no sign of acidity. The wheat paste bond seems just as good with the treated newsprint as with the untreated newsprint. The oil pastel colors seem unaffected (not surprising, oil pastel being very water-resistant).
Is there anything about sodium bicarbonate that might, over time, cause a visible effect on the paper?

As the sodium bicarbonate solution dries, it gives off carbon dioxide and water, leaving a residue of sodium carbonate. Sodium carbonate is hygroscopic, meaning it will attract moisture from the air. Moisture can cause paper to buckle and mildew, and that is good a reason not to use a baking soda bath to deacidify newsprint for archival purposes.

I thank @Michael for the reference to the Wikipedia article on acid-free paper. Somehow, in all the reading I've done about deacidifying newsprint, I missed that article. Reading it and then looking up sodium carbonate is what gave me the answer.

It seems that adding bicarbonate is how acid-free paper is made, though more modern paper is neutralized by the use of chalk as filler. The wiki page says that magnesium or calcium bicarbonate are commonly used, but that may be just due to cost rather than any specific issues with using sodium bicarbonate. Source

Aha! The Wikipedia page on acid-free paper says that as the magnesium or calcium bicarbonate deacidifying solution dries it leaves a carbonate residue. Presumably that means that using baking soda would leave sodium carbonate? Sodium carbonate is apparently hygroscopic, which could cause paper to buckle or mildew. And that's a very good reason not to use baking soda to deacidify paper!

Presumably that means that using baking soda would leave sodium carbonate? || Yes, it would. In case you really care, consider ammonium bicarbonate or aqueous ammonia. But it STINKS.

You can always try it on something you don't care about and see if it's actually a problem, or there's the option of just rinsing well with DI water to remove any remaining bicarbonate after you've neutralized the acid.

I am hoping not to have to subject the wet newsprint to too much manipulation -- the sheets are large and rather fragile while soaking wet. I am going to try one of the other bicarbonates next, maybe calcium.


text:
Influence of Temperature on Biochemical Reactions
Short question
How (quantitatively speaking) does temperature influences rate of decay of proteins? I am looking for some general number/function, the influence of temperature on an average protein.
Long question
For computer modeling purposes, I am looking for some referenced quantitative measurements of the effect(s) of temperature on the dynamic of biochemical reactions. In particular, my question is:

How does temperature influences rate of protein decay?

I am looking for a function (eventually expressing the michaelis-menten constant as a function of the temperature) that I can plug in my algorithms to incorporate the influence of various temperatures on the process I want to simulate. The model under simulation is an eukaryote such as Saccharomyces telluris (a yeast) for example living at some common range of temperature [5-35 °C]. I am not thinking to any particular protein rather I am looking for some approximative effect of temperature on the decay rate on an "average protein". I welcome any kind of expression of how temperature influences the rate of decay. Some values using the Erying equation, Van't Hoff equation or a $Q10$ (measured in some reasonable range [5°C - 35°C]) for examples would also fit my needs. I would be very surprise that there is not enough literature on the subject in order to get some average impact of temperature on protein decay rate but as a biologist (and not a biochemist) I can't wrap my head around this problem and can't find any satisfying article. Thanks for your precious help!
I am not looking for...
I am NOT looking for a theoretical explanation of how temperature influences these reactions (I have basic understanding of the importance of the activation energy as displayed on a Maxwell-Boltzman distribution and of the michaelis-menten equation). I am just looking for some empirical observations of how temperature influences proteins decay rate. I am not looking for any accurate value but only for some estimation that I can plug into my algorithms.
Update
Reading the comments I realize that my question is too broad in the sense that there might have too much variance in how different proteins respond to temperature in order to draw in general tendency. If this is so, someone willing to answer may reduce the question to transcription factor, to General Transcription Factor (GTF) or even to TFIIA.
What about this article for example? I am not really able to extract from it the info I am looking for but it might be in here.

This question is extremely broad if you want to go beyond general arguments about activation energy. If you want specific functions, data, then you have to pick a specific system and go find literature about it. There is no one magic formula.

@Greg thanks for your comment. I reduced my post to one question and described the organism and the protein I am interested in. Hope that makes this question easier to answer

I am not thinking to any particular protein rather I am looking for some approximative effect of temperature on the decay rate on an "average protein". || I'm afraid you won't find any. It really depends on the protein in question. For example, some enzymes successfully works at 90+ celsium, while others break down at ~30 celsium. I suppose, this have to deal with folding of proteins: a folded protein is significantly more stable, than linear polypeptide chain, but the folding breaks down at highter temperatures.

Ok, so do you think I could ask the same question for the General Transcription Factor (GTF)? If not I may even reduce my question to TFIIA. What do you think? (I updated my post in reaction to your comments) Thanks

@Remi.b With such specific, but deeply specialized question you should go to scholar google com and search for something like 'activity decay of ' or similar.

well... that's the point I can't find anything. I might be too ignorant in the relevant field in order to really judge whether an article is of interest to me based on its title, I don't know but I personally keep failing to find the answer I need. That's why I came here on ChemistryBeta to ask you guys for some help.

I don't think you're going to find any non-specialists able to help here. As @permeakra mentioned, my feeling about "decay" here is going to depend on denaturation - when does this protein unfold or mis-fold. That depends entirely on the protein. I'd e-mail other people who do research (esp. folding studies) on that protein.

What about this article. I don't know sufficiently about Gibbs free energy and stuff to understand it but the article might give this answer I am looking for. What do you guys think? Rethinking about my issue I realize that I probably missed two keywords in my article research: Degradation and stability curve.

@Remi.b As a rule of thumb I would say, that the protein is active at least up to temperature that kills the cell. Other than that your only hope is that someone made measures for your specific protein. As for the article, it describes empirical correlation of stabilities in homologous proteins i.e. 1)it requires statistics on homologous proteins to be of any use and 2) it may be used to predict, but not guarantee.

@permeakra Does your rule of thumb imply that the rate of degradation is zero under ~50°C and after ~50°C all proteins would suddenly be non-functional.

@Remi.b Nope, it means that the degradation rate is slow enough to natural protein synthesis to overcome it. Protein renewal rate is easier to find, even though it depends on specific protein, and it ranges from about a hour to first tens of days for rats, for example. Still a too broad range, but at least it is something. But it does imply, that some human proteins may become non-functional somewhere around 43-50 celsium.

@permeakra Thanks a lot for all your comments. In reaction to your last comment I asked this question.

@permeakra imho, not even 'degradation' is sufficiently defined. Degradation could refer to (a) amide bonds being hydrolysed, the protein degrading, (b) biological protein degradation processes, usually ubiquitination followed by degradation, (c) denaturation due to change of environment, followed either by ubiquitination-degradation or unspecific degradation. Luckily, (a) can be disregarded at temperatures you mentioned/pH-values anywhere near to in vivo. If intracellular protein levels are kept constant, it's either a physicochemical decay rate equal to the renewal rate or a biochemical one.

This question might be better suited for Bio.SE, since proteins don't compare well to dead-chemistry polymers.

Since this was already asked and answered on Bio, I am going to close this one.


text:
Safe explosion demonstrations
As an end of high school treat, my class is trying to convince our chemistry teacher to make an explosion as a demonstration. What sort of reactions are beautifully explosive but safe enough to do in a school lab environment, when performed by a teacher who knows what they're doing?

From our friends at Periodic Videos, the following come to mind:
Gun cotton, more precisely known as nitrocellulose, which is a very simple nitration that you could show prior to the pyrotechnics.
Sodium and Potassium are also good, though you may not always get an explosion, depending upon the available hydrogen above the metal.  Which suggests...
Hydrogen! This is the classic explosion in oxygen. In fact, on a small scale, the electrolysis of water into hydrogen and oxygen can be used to show that the mixture of these gases is very explosive. But keep it small!
All of these experiments avoid outright explosives, which seems prudent in a high school setting.

In my high school chem class we filled plastic pipette bulbs with different ratios of H2 and O2 made by H2O electrolysis, then jammed them on a partially disassembled kitchen lighter. When we pulled the trigger on the lighter, it would spark and ignite the gases, blowing the bulb down the hall. Measured the distance each ratio of gas would produce and determined ideal ratio of H2 to O2. So these demonstrations can be more than just "HEY LOOK AT THIS KICKASS FIREBALL!"

Another variant of the hydrogen plus oxygen demo is to pass the gas mixture into a soap bubble solution. The bubbles rise in the air and can be ignited with a taper or handheld propane lighter. Start out with pure hydrogen, which burns gracefully with a whoosh, the add more and more oxygen until, when you are at the stoichiometric ratio, the bubble will detonate with a loud report.

Nitrogen triiodide is pretty spectacular and not dangerous so long as you keep the batch small. I remember my high school chem teacher had a batch left in the back of the classroom to go off randomly to scare us all. It's basically a matter of mixing aqueous ammonia and iodine crystals, leaving them to react a couple hours, then gravity filter. The result is pretty stable so long as it's damp, but explodes with a cloud of purple smoke if even slightly disturbed when dry. Even a feather or something touching is likely to set it off. As I mentioned above, keep the batch small. It doesn't take much to give a good show.

Contact Explosive - Detonating Nitrogen Triiodide, The Royal Institution

I am giving you the link where you can find many interesting explosion experiments.but let me tell you these are little dangerous too.so be careful while performing these.
http://youtu.be/sfW7KztrA8I

Explosions usually happen too fast to be overly fascinating. Explosions with true detonation are quite dangerous.
There is quite a set  of flash-powders that produce a cloud of flame and smoke. They are not true explosives as they do not detonate. The easiest one is $\ce{1Zn +1S}$ equimolar mix, it produces bright green-blue burst. I also found funny $\ce{3CuO +2Al}$ mix - it burns with a report, producing yellowish flame and black smoke, and covering nearby surfaces with yellow layer of copper, and $\ce{1Bi(OH)2NO3 +5Mg}$ - in small doses (and I mean really small) it burns with a report, bright (may be even blinding) yellow flash and mushroom cloud of bright yellow smoke. Bismuth oxynitrate is quite easy to prepare via hydrolysis of commercially available bismuth trinitrate. $\ce{2Fe +KNO3}$ mix burns without explosion, producing fountain of yellow sparks.  Generally, however, mixes with metal powders must be handled with great care as the mixes may be sensitive, burn without oxygen and produce toxic fumes. Metal powder are available via specialized pyro- or chemical stores and aluminium powder is available also via painting store. However, in reciepes above the finest powder is required.
There are very few explosives that are easy to make. If you have time, you can go with synthesis of barium styphnate (lead is much more toxic, and explosion products will contain lead oxide smoke. Barium is toxic too, but to much, much lesser extent, so it is reasonable safe). It IS sensitive, and will explode if left on heating plate, but is quite easy to produce (styphnic acid is produced by nitration of resorcinol in two stage process with $\ce{NaNO2 + HCl}$ first and $\ce{HNO3}$ second, barium styphate is easily produced from barium chloride, styphic acid and a base, though don't do it without finding proper procedure), so no need to store it. I tested it and found... funny, if not very spectacular.
Here is a funny video for you http://youtu.be/7gnC0mXE0Vk


text:
Are there any solvents that dissolve talcum powder?
I understand that talcum powder is insoluble to water, would alcohol or something be able to dissolve it?

Talcum powder is made from talc, a silicate mineral with formula unit: $\ce{Mg3Si4O10(OH)2}$. Silicates tend to be pretty much insoluble in water, but it is possible to dissolve them in strong acids. I don't know about talc specifically, but many silicates can be dissolved in concentrated nitric acid. Hydrofluoric acid is also commonly used for dissolving minerals for chemical analysis. Both are very very dangerous substances that really shouldn't be used without proper training and safety equipment. I don't think there's really anything accessible for a home experimenter, unfortunately.

would dissolving talc in strong acids then make the properties of talc change? I understand that acids can be solvents but i would think acids are very strong and are mostly oxidizing agents (especially nitric acid)

Well it won't be talc anymore. Dissolving anything changes its properties to some extent. Talc is only talc because it has the composition it does in the atomic arrangement it does. In order to dissolve, the talc must lose its solid structure. In the case of using an acid, you're right that an oxidation reaction does sometimes cause dissolution, especially with things like elemental metals. When used for digestion, HF tends to pull out Si as $\ce{SiF4}$. In acidic conditions, I found a reference to this dissolution reaction for talc: $\ce{Mg3Si4O10(OH)2 +6H+ + 4H2O -> 3Mg^{2+} + 4H4SiO4}$

I consider that there is no solvent that can dissolve talcum powder, indeed if you consider "solving" as having your material into solution but maintaining the crystalline structure, only some inorganic compounds (most of them transition metal complexes) are soluble even in water or in other organic solvents (like alcohol).
Quoting wikipedia, we find:

Talc is not soluble in water, but it is slightly soluble in dilute mineral acids.

In any case, solving it will keep intact its structure. This is naturally true for the case of silicates, which are characteristics by their insolubility.


text:
Is the colour of leaves due to d-d transitions in chlorophyll?
While reading a chapter on transition metals, I came to know that d-Block elements have colour compounds due to d-d transitions.  A question came to my mind as to whether there is any d-d kind of transitions in plants as they also show characteristics colours.
I know the green colours of leaves is due to presence of chlorophyll and also there are coloured pigments in plants.
I may be wrong in my thought, but I really want to clarify my doubt.

Let's rephrase your hypothesis and test it:

The green colour of leaves results from a d-d transition of the central atom of metal complexes known as chlorophylls.

Green leaves contain two types of chlorophyll, Chl a and Chl b, both have $\ce{Mg}$ as the central atom and show two distinct bands in the UV-VIS absorption spectrum.
The stronger ($\epsilon \approx 100000\, \mathrm{cm}^{-1}\cdot \mathrm{M}^{-1}$) band in the 400 nm (blue) region is known as the Soret band, the weaker  $\epsilon \approx 30000\, \mathrm{cm}^{-1}\cdot \mathrm{M}^{-1}$) in the 660 nm (red) region is known as the Q band.
Now imagine that Chl a is treated with weak $\ce{HCl}$. In the resulting product, known as Pheophytin a, the central atom ($\ce{Mg^{2+}}$) is replaced with two protons.
In the UV-VIS absorption spectrum, however, we would still observe the two bands. They are slightly shifted and show somewhat different extinction coefficients, but the pattern is fundamentally the same.
We may conclude that our initial hypothesis is wrong. If it is not the absorption of the central atom in our complex, the ligand must be responsible.
Our new working hypothesis therefore is:
The colour of green leaves results from the absorption of the polycyclic pheophytin ligand and its extended $\pi$ system.
Let's simplify the Pheophytin a molecule: Removal of the E ring and replacing all the fancy substituents with ethyl groups leads to octaethylporphyrin as the core structure.

In the UV-VIS spectrum, we would still be able to see the strong Soret band, which corresponds to a $\pi,\pi^*$ transition from the ground state to the second excited singlet state ($S_0 \rightarrow S_2$).