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Light sensitive AgF decomposition reaction
Silver compounds such as silver fluoride are light sensitive (as are all other silver halides) so they will decompose when exposed to light.
I'm stuck with a problem : I want to expose silver(I)fluoride to light to then produce elemental silver and fluorine gas - I'm worried that because of the high reactivity of fluorine silver(I)fluoride will be formed again...
So how could I successfully split silver(I)fluoride by light into elemental silver and fluorine gas without the fluorine reacting again with the silver?

If you want to actually keep the fluorine, you could pull the fluorine gas away with a vacuum and have it pass through a liquid nitrogen trap or two (fluorine boils only 8 K above LN2, however, so you'd need a system with excellent heat exchange). Though liquid fluorine is probably not something I would ever like to see without some serious PPE. It could also damage the pump irreversibly. If for some reason you want to split AgF and not keep the fluorine, it's pretty easy, pretty much any substance will react with fluorine gas preferentially to silver, so you can put some Al foil nearby.

and could teflon be used (or any other fluorine-containing 'polymer') to transport the fluorine gas away? (I suspect that it won't react with fluorine?)

Teflon or other saturated perfluorinated polymers such as FEP are indeed resistant to corrosion by fluorine. I don't know how they behave at cryogenic temperatures, though. Do you have the proper equipment and training to handle fluorine? If not, don't try to isolate it. Many people have died or been severely poisoned in the process.

Actually scratch that, what I suggested won't work. Under low pressure, fluorine's boiling point will definitely be lowered below nitrogen's, so it won't condense at all. I should've realized that sooner. It's probably best if you find some references for this sort of stuff, though I don't know where you might find them.

I think that you understood that I wanted to store the fluorine for a long time - it is just to reuse it in another reaction (so just transporting it to another reaction area to then let it react with something else)

Then if the fluorine is going to be consumed, that will shift the equilibrium in the formation of fluorine gas. You shouldn't even need a pump, just connect a flask with AgF to another flask with the reactants, then begin decomposing the AgF. Are you sure the fluoride decomposes with light, though? It should be the least willing to do so among the halides.

You are right about that- maybe the decomposition could be speeded up with catalysts?

I have no idea about possible catalysts. If you have plenty of AgF you could first do a test by putting Al foil in the other flask, exposing the AgF to light, then checking if the aluminium corrodes. If nothing happens then the photolysis may not be working, or the fluorine may be produced too slowly, or it may be escaping, or it may be reacting with something else present in the system (impurities, glass, etc)

Do you know of a way to predict this theoretically (the decomposition speed of AgF on photolysis)?

No idea. You should be able to get a feel for the spontaneity of the reaction by drawing a thermodynamic cycle and using the Gibbs' free energy state function properties, but I don't know how to obtain kinetic information. The reaction speed may simply be limited by the incident photon flux with sufficient energy to drive the reaction, though, and in that approximation all you have to do to increase speed is find a more powerful source of light of the right wavelengths.

As discussed in the comments, the asker only wanted to create some fluorine in situ for use in another flask, with no desire to store it. Chemistry with fluorine is difficult due to its dangerously high reactivity. Do not attempt anything without the proper safety procedures.
I actually still have doubts whether it is possible to obtain fluorine gas reliably from the exposure of silver fluoride to sufficiently high energy photons, and if there is any simple way to produce photons with the required energy. It is well known that the other silver halides easily suffer photolysis, especially for the heavier halides, but reactions which produce fluorine gas tend to be extremely endergonic, and therefore don't happen except in rather severe situations.
In case it is somehow possible to produce fluorine gas directly by photolysis of silver fluoride, then it is possible to transport the gas to another container where it can react, simply by connecting the flasks with a pipe. Few materials are capable of withstanding exposure to fluorine, however, so the containers must all be made of very well dried gas. Unless the asker has access to a very high power light source, I suspect that the rate at which silver halides decompose is quite slow, which may preclude their efficient usage as halogen generators. This is especially true for silver fluoride, which likely decomposes the slowest, and the fluorine formed in very low concentrations could easily be lost by reaction with impurities or residual water before even reaching the target. I do not know whether there are any catalysts which can be added to increase the photolysis speed of silver fluoride. Presumably some semicondutor with a high bandgap, such as titanium dioxide, could absorb the light and cause photoreduction of the silver/photooxidation of fluoride, but the catalyst would be quickly consumed by the fluorine formed.
If fluorine gas can be formed by silver halide photolysis fast enough, then by simply linking it to another flask with the desired reactants should cause the fluorine to diffuse and react, followed by a continuous equilibrium shift in the production of more fluorine in the flask containing the silver fluoride, and its consumption in the flask with the other reagents.


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Which functional group is -CSSH?
If we replace two oxygen in $\ce{-COOH}$ what is the name of the compound we get?

Does that functional group even exist? If so, do you have an example of a molecule containing it?

Yes the compound exists, it is formed when grignard reagent reacts with thioxide.

Is a dithioic acid.
For example, a benzene with two -CSSH hubstituents in 1, 2 positions is a benzene-1,2-dicarbodithioic acid.

Here's a neat reference containing all sorts of information about carboxylic acid analogues contaning heavier chalcogens.

isn't it just a thioic acid? that thio there means sulfur so why is it a dithioic acid instead of just thioic acid?

@Caters, likely because '-thioic acid' would imply replacement of only one of the oxygens with a sulfur, instead of both.

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Calculating photolysis time of AgF
How can I theoretically predict the time needed to photolyze n moles of $\ce{AgF}$? (with UV) (this is not a homework question)
The photolysis of $\ce{AgF}$ requires UV light - but I don't know which wavelength the UV must be at for the optimal photolysis. Then if the wavelength is known I can calculate the energy of one UV photon of a specific wavelength. After that I can calculate the required amounts of photons required for the reaction
I was thinking of this method:

First I need to find out which wavelength the UV must be at for the optimal photolysis of $\ce{AgF}$ (How?)
Then I need to calculate the amount of energy required for the photolysis of AgF : on wolframalpha I entered this equation: $\ce{2AgF -> 2Ag + F2}$ and it gave me a $\Delta{}H=409.2\:\mathrm{kJ/mol}$

And somehow I must be able to then figure out which lamp I could use and how long I should use it.

Have you read this? I think it will take a very long time until everything is photolyzed.

Unless its initially in very thin film, it will take infinitely long as the silver blocks the ultraviolet and you need separate the fluorine and silver otherwise it just will just react again.

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How does the ambient temperature affect the rate of redox reactions in a salt lake?
Many years ago, I used to work as a gold exploration geologist.  One of the main locations was a salt lake in Western Australia detailed in the document WMC Resources Exploration Successes inN Lake Terrains - Applications of Element Dispersion, Kambalda W.A (Carey and Dusci, 1993).
In the article, and in my research at the time, it is stated that

gold disperses laterally along iron redox fronts that tend to occur higher in the regolith profile

Now the area is arid (up to 50C or 122F in summer), but was not always like this, fossil records suggest far milder conditions occurred in the past.
My question is, how does the ambient temperature affect the rate of redox reaction in a salt lake, such as in the example provided?

The salt lakes that occur in Western Australia are touted by the article Geochemical characteristics of naturally acid and alkaline saline lakes
in southern Western Australia (Bowen and Benison, 2009) are unique environments, in that they possess

an extreme and rare
  range of natural hydrogeochemical conditions. Many WA lake
  waters and groundwaters are hypersaline and acidic.

The article discusses observations of effects due to temperature, especially cycles of dry and wet.  The high ambient temperatures would cause increased evaporation, and hence increase local evapoconcentrations of ions, resulting in a lowering of the local pH, storm/inundation events have the opposite effect.  This leads to further observations by the authors that found that

Field observations of lake pH during storm events
  illustrated that fluids that have been freshened due to meteoric input
  are able to regain acidity rapidly. Part of the acidity may be locally
  sustained by ongoing $\ce{Fe}$ cycling and redox reactions that
  generate $\ce{H+}$ ions. The amount of $\ce{Fe}$ increases significantly in the
  more acid waters, and authigenic $\ce{Fe} mineralization (oxides
  and sulfates, e.g., hematite and jarosite) is abundant throughout
  the acid lake sediments.


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Chromium from electrode not oxidizing in electrolysis
I performed an electrolysis of an aqueous sodium carbonate solution with a platinum electrode and a chromium-vanadium-steel-alloy electrode. The solution turned yellow at the chromium-vanadium-alloy electrode (I suspect that it is sodium metavanadate (NaVO3))
Then I also performed an electrolysis in an aqueous sodium carbonate solution with a platinum electrode and an inox electrode (the inox electrode is composed of iron, chromium,...) . A green compound (insoluble) was formed and when I later added acetic acid, a brown compound (I think Iron(III) acetate ) was formed.
But in both cases there were no chromium compounds formed (at least of what I know) So why didn't the chromium react (why didn't the chromium get oxidized)?
More information:
Voltage was around 13 Volts, the inox electrode was stainless steel (see http://en.wikipedia.org/wiki/Stainless_steel) and has according to wikipedia a minimum of 10.5% chromium content by mass. I read online that a regular chromium-vanadium-steel-alloy has a vanadium  content  of approximately  0.181%  and a chromium  content  approximately  1.00%. (however, I'm not sure about these last two figures) What is true is that there wasn't a lot of yellow sodium metavanadate after an electrolysis of around 20 minutes. ) The electrolysis solution contained sodium carbonate(Na2CO3) and water(H2O) the electrolysis was performed at normal conditions (1atm - 25°C)

First determine what was the actual result what compounds formed - just assuming the result isn't getting anywhere- also the exact condition under which the experiments were done could help- voltage, actual alloys used, chemistry of solutions.

See my edit for more detailed information - I'm also sure that vanadium ions are in the solution because when I reduced the ions from the 2nd liquid (by using zinc) it matched the color changes of vanadium +V,+IV,+III,+II (yellow->blue->green->lilac)

That doesn't prove that there aren't also Chromium ions in solution. The most vibrant coloured ions are the ones you see. I saw that in wikipedia but inox is just a common name finding the actual alloy has given me no success.

the inox-alloy was "18/10 stainless steel"

Some of the ways you aren't at standard condition.
Metal solution so the activity of the metals are at their concentration in the alloys
The solutions have far below 1mol/L of their respective metal ions
The oxygen that you are reducing in basic conditions is probably about 258 micromol per litre-
whereas it should be 1 mol/L. The potential might be too high- its hard to tell the solution resistance is an effect.
If you were at standard condition with low voltage then the first experiment makes perfect sense-
From wikipedia full list of electropotentials
V2+ + 2 e−
    V(s)     −1.13
Cr3+ + 3 e−
    Cr(s)    −0.74
Vanadium is preferred.

So what voltage would you advise to use for the first experiment?

I'm not the person to ask about practical electrolysis. Either repeat it many times starting from low voltage or contact an organisation like NACE (nace.org) if you are in the Americas or  ACA (corrosion.com.au) if you are in Australasia to link you to someone who knows.

I agree as Vanadium is preferred, no other metal would be subject to corrosion until all the preferential metal is consumed.
Research the target of a sacrificial electrode.

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Decomposition products of PET
On Wikipedia, it says that polyethylene terephthalate (PET) decomposes at temperatures above 350°C.
What are these decomposition products? I read further that acetaldehyde is produced, but is that the only decomposition product?
Also, what compounds are produced when PET is burned? ($\ce{+O2->?}$)

Since most PET used for a particular end use requirement the deliberate addition of other plastic additives are added that changes the nature of the toxic degradation products this includes cross contaminates from other polymers added to PET

This is a partial answer to your query.  According to the paper Degradation and Recyclability
of Poly (Ethylene Terephthalate)
 (Venkatachalam, 2012), there are a few degradation products, from the article:

Apart from aldehyde, $\ce{CO}$, $\ce{CO2}$, ethylene, benzene, biphenyl are also identified as
  degradation products

(The article also shows diagrms of the chemical processes of PTE degradation).
an important aspect to note is experiments/observations have

shown that rate of thermal degradation in the closed system is
  about three times higher than that in open system.

In regards to burning of PET, the products can vary according to the page EPB 433- Health and Environmental Effects of Burning Waste
Plastics
 (Saskatchewan Ministry of Environment) stating that

the composition of byproducts of plastic combustion as to their type and concentration depends
  on the combustion temperature and the flame residence time.

According to the Peiyu Plastics page (a reference from Wikipedia), if complete combustion occurs, then the products would be $\ce{CO2}$ and $\ce{H2O}$, incomplete combustion could produce soot and carbon monoxide.
Further reference:
Here is the MSDS for PET
A study of thermal decomposition and combustion products
of disposable polyethylene terephthalate (PET) plastic using
the High Resolution Fourier Transform Infrared
Spectroscopy, Selected Ion flow Tube Mass Spectrometry
and Gas Chromatography Mass Spectrometry  (Sonova et al. 2008).


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Reduction of terephthalic acid to an alcohol (or phenol?)
I want to convert terephthalic acid (TPA) to an alcohol (however I think the correct term would be "a phenol").

I've read that lithium tetrahydridoaluminate ($\ce{LiAlH4}$) can produce an alcohol from a carboxylic acid. And, according to Wikipedia, dicarboxylic acids (such as TPA) react around the same. So does $\ce{LiAlH4}$ reduce TPA to a phenol (or an other product)?

I want to make sure that you understand the different between two commonly confused molecules (their names don't help).
Phenol is an $\ce{-OH}$ group directly attached to an aromatic ring. Because of this connectivity, phenols exhibit different chemical behavior than other alcohols. In particular they are more acidic. Phenol can be abbreviated $\ce{PhOH}$, where the $\ce{Ph}$ represents the $\ce{C6H5}$ group of the aromatic ring. As a class of molecules, phenols can be represented generically as $\ce{ArOH}$, where $\ce{Ar}$ is used not for argon (there are no compounds of argon), but for generic arenes with any substitution.

Benzyl alcohol is a different molecule. In benzyl alcohol, there is a $\ce{CH2}$ group between the benzene ring and the $\ce{OH}$. Benzyl alcohol behaves like other alcohols. Benzyl alcohol is abbreviated $\ce{PhCH2OH}$.

Conversion of benzoic acid to phenol requires loss of a carbon atom.
$$\ce{Ph{\bf C}O2H -> PhOH + {\bf CO}}$$
Lithium aluminum hydride converts carboxylic acids to alcohols, but it does not cause the loss of the carbon atom. Lithium aluminum hydride will reduce benzoic acid to benzyl alcohol. If you want the benzyl alcohol derivative, stop here. If you want the phenol, read on.
$$\ce{PhCO2H} \overset{1. \ \ce{LiAlH4}}{\underset{2. \ \ce{ H2O}}{\ce{->}}}\ce {PhCH2OH}$$
To convert benzoic acid to phenol, we need a way to loose that carbon atom. One way is to oxidize benzyl alcohol to benzaldehyde using perhaps a Swern oxidation. Then benzaldehyde can be converted to phenyl formate using a Baeyer-Villiger Oxidation. The formate could be hydrolyzed in aqueous base.
$$\ce{PhCH2OH \overset{Swern}{->} PhCHO}$$
$$\ce{PhCHO \overset{Baeyer-Villiger}{->} PhOCHO}$$
$$\ce{PhOCHO}\overset{1.\ \ce{NaOH, \  H2O}}{\underset{2. \ \ce{H3O+}}{\ce{->}}} \ce{PhOH}$$
In practice, it would be easier to buy the phenol. In your case, you want hydroquinone.

Thanks for the very good answer - but is it also possible to remove two methanol molecules from the 1,4benzenedimethanol molecule to also yield benzene?

As nice and thoroughly described by Ben Norris, 1,4-(bishydroxymemethyl)benzene is available via reduction of the starting material with $\ce{LiAlH4}$. In order to minimize the amount of alanate used, one might consider to first convert the terephthalic acid to a corresponding ester.
If you're heading for the phenol, which doesn't make a lot of sense since hydroquinone is cheap as dirt, the following "strategy" might work:

brominate terephthalic acid to yield 2,5-dibromo terephthalic acid
heat the dibromo acid in an aqueous solution of sodium hydroxide, cupric bromide and 1,2-diaminoethane. You'd end up with 2,5-dihydroxy terephthalic acid, probably through a copper-catalyzed nucleophilic substitution
decarboxylate the 2,5-dihydroxy terephthalic acid (or its disodium salt) by heating with calcium oxide.

You'll get hydroquinone, and unless your arene was labeled, you won't notice that you haven't replaced the $\ce{COOH}$ with $\ce{OH}$ but actually changed the positions of the substituents.
But all of this is nothing but a mind game ;)

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Converting sulfate ion to sulfur trioxide?
I have a solution of only sulfate ions (no other positive ions belonging to the sulfate ions, because the positive ions are attached to the intermediate in a catalytic reaction).
I want to convert the sulfate ions to sulfur trioxide by using $\ce{V^5+}$ ions...
So what if $\ce{V^5+}$ ions could be present and could gain 2 electrons from one sulfate ion to produce $\ce{V^3+}$ and $\ce{SO3}$ (however I'm not even sure if $\ce{SO3}$ would be formed)? Here the two negatively charged oxygen atoms will each lose 1 electron to the $\ce{V^5+}$ ion, so the oxidation state of the central sulfur atom is still $+6$, but the oxidation states of the 2 oxygen atoms are $0$.
Could this be possible? And if so, would it be $\ce{SO3}$ that is formed, or other sulfur oxides?

Please notice that the oxidation of sulfur in the sulfate ion and and sulfur trioxide it the same; the latter is just the anhydride of the corresponding acid of the first.

And how can they then be converted to each other (sulfate->so3)?

A good analogy to your proposed reaction is the formation of carbon dioxide above a solution of carbonate salts.  Like sulfur trioxide, carbon dioxide is an acid anhydride.  The acid for sulfur trioxide is sulfuric acid, $\ce{H2SO4}$.  The acid for carbon dioxide is carbonic acid, $\ce{H2CO3}$.
Dry carbonate salts like sodium carbonate can be made to release $\ce{CO2}$ by heating to high temperatures.  For example, calcium carbonate decomposes to calcium oxide and $\ce{CO2}$ at around 800 °C to 900 °C.  Since such high temperatures are required, this technique does not work with aqueous solutions at ambient temperatures.   (The water would boil off way before gas is released.)
Thus, sulfur trioxide evaporation from sulfate salts would be even more disfavored at room temperature than carbon dioxide evaporation from carbonate salt solutions, because of sulfur trioxide's affinity for water is much higher than $\ce{CO2}$'s.
Also, the reaction mechanism you have drawn is not a dehydration, but a reduction of the sulfate.  That would not form sulfur trioxide, but a reduced sulfur compound, such as sulfite.  Sulfur dioxide gas evaporation from above sulfite salts is probably more favored than sulfur trioxide evaporation, but that isn't saying much. Either one is still highly disfavored under neutral or alkaline pH.


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Boiling point of Coca-Cola
Inspired in part by the question boiling point of vodka - given that the exact ingredients of Coca-Cola are kept secret, based on what is known, what would a viable prediction of the boiling point of Coca-Cola be?
Would it be significantly different from related products (e.g. Diet Coke, Coke Zero, etc)?

Your best guess for the boiling point of any version of Coke would be 100 C, the boiling point of water.
Diet Coke is mostly water (the flavourings are a very small amount relative to the amount of water). The largest ingredient will be the sweetener but there will be only a fraction of a gram of that. It is unlikely you will notice any deviation from the properties of water.
Standard Coke has quite a lot of sugar in it. A standard can (~300ml) contains about 40g of sugar. To put it another way, the contents are more than 10% sugar by weight and the solution is about 1/3 mol/L of sucrose (other sugars will be slightly different). A standard calculation using the ebullioscopic constant for water suggests the elevation of the boiling point will be barely 0.2 C, so small you'd struggle to measure it without good instruments and a good experimental setup.

+1 Great answer, so essentially, Coca Cola and just about all soft drinks could be considered as sweetened water,in this respect.

@Amaterasu Well, that is indeed what they are.

Atmosferic preasure will produce a greater difference.

Oh pressure produces diffrence so It depends upon pressure, when the bottle is totally closed it has more pressure etc..

Here's an excellent resource that says the difference between 10% sucrose (aka coke) and water is less than a degree Celsius:
http://kitchenscience.sci-toys.com/boiling_freezing_pressure
Why don't you just measure it? All you need is a bottle of coke, a thermometer and a kitchen. There will be some foam when the CO2 comes out, so heat slowly or be prepared for a sticky mess.

I have a bottle of coke and a few kitchen thermometers. Be right back!

Please do post your results if you get any!


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Plutonium having more oxidation states than samarium?
Why does plutonium have more oxidation states than samarium?
Electron configuration of Pu: $\ce{[Rn] 5f^6 7s^2}$
Electron configuration of Sm: $\ce{[Xe] 4f^6 6s^2}$
I thought that only the valence electrons affected the oxidation states, so why does plutonium have more oxidation states (6,5,4,3) than samarium (2,3) whilst they both have the same valence electrons (including f-orbitals): ..$\ce{f^6}$..$\ce{s^2}$?
Elements such as scandium (Sc), Yttrium (Y), and Lanthanum (La) all have the same valence electrons (including d-orbitals): ..$\ce{d^1}$..$\ce{s^2}$, but these elements have the same oxidation state : 3 and that doesn't change, even when going down that group and looking at actinium.
Does this have to do with the ionization energy, the energy level, or...?

This is yet another interesting effect of the anomalous compactness of orbitals in the first appearance of each type of subshell ($1s$, $2p$, $3d$, $4f$, $5g$, etc). The solutions to the Schrödinger equation for electron wavefunctions in hydrogen-like atoms are such that these subshells are composed of orbitals with no radial nodes. This means the electrons in these subshells are closer to the nucleus than you might expect, and therefore they are more strongly bound.
The striking chemical similarity of the lanthanides comes from the fact that the $4f$ subshell is both anomalously compact and subjected to a high effective nuclear charge. The electrons end up being held so closely to the nucleus that they effective behave as core electrons, and participate very little in chemical interactions; lanthanides do not form coordination compounds using their $4f$ orbitals, and it is very difficult to go past an oxidation state of +3 in most because few conditions can compensate the large energy required to ionize a fourth electron coupled with the relatively weakly bound compounds they would form.
In the actinides, the $5f$ orbitals do not suffer from the same lack of radial node, and therefore are more chemically available. The first half of the actinides show many compounds with high oxidation numbers and have significant coordination chemistry using $f$ orbitals. Curiously, as you go from curium to berkelium, there is a sudden increase in reluctance to display high oxidation numbers, and the actinides beyond curium tend to behave similarly (though of course exploration of their chemical properties is significantly hindered by their rarity and instability). This is likely because even though the $5f$ orbitals have a radial node, in the second half of the actinides they are subjected to such a high effective nuclear charge that become too strongly bound to display significant participation in chemistry.
This same effect explains why the metals in the first row of transition metals do not display as many stable compounds with high oxidation states compared to the second and third rows, as the $3d$ orbitals are much smaller than $4d$ or $5d$ orbitals.


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Explaining the oxidation states of palladium
Palladium follows an exception to normal electron filling-up rules and so has the electron configuration of $\ce{[Kr]  4d^{10} 5s^0}$
The oxidation states of palladium are $\ce{+II}$ and $\ce{+IV}$.
How can these oxidation states be explained by looking at the special electron configuration of $\ce{Pd}$?

Ions of the transition metals do not often occur "naked" the way one might think of a sodium ion. Of course, sodium ions cannot occur by themselves either, but let's compare the two for a moment.

Sodium nitrate, $\ce{NaNO3}$

When $\ce{NaNO3}$ dissolves in water, the two ions dissociate.
$$\ce{NaNO3 -> Na+(aq) + NO3- (aq)}$$
Both ions are solvated by the solvent. The water molecules (which are polar) surround the ions. The partial charges due to the dipoles in the water molecules stabilize the charges on the ions. However, the interactions between the sodium ions and the water molecules are fleeting. These interactions are not strong enough to be consider "bonds" (there is no covalent electron sharing between water and sodium ion). Thus, we often consider sodium ions in solution (despite their solvation) to be "just" sodium ions.

Palladium nitrate, $\ce{Pd(NO3)2}$

Transition metal ions are different. They tend to form complex ions by coordinating solvent molecules. These dative interactions are far stronger than the ion-dipole interactions in solvated sodium ions.
Palladium nitrate does dissociate, but the palladium ion picks up water molecules that it holds tightly. Thus, we do not worry about the electron configuration of $\ce{Pd^{2+}}$, since it never exists on its own. It is always part of a complex ion or coordination compound, and we care about the electron configuration of that complex ion.
$$\ce{Pd(NO3)2 + nH2O -> Pd(H2O)6^{2+}(aq) + 2NO3- (aq)}$$

But he didn't say solvated in water- Of course the ion in salt going to have some covalent nature.

The OP did not say solvated in water, but then there will always be a counterion.


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Reaction intermediates of MnO2 catalyzed H2O2 decomposition reaction
Manganese dioxide catalyzes the decomposition of hydrogen peroxide to water and oxygen gas.
But what are the intermediates in this catalyzed reaction?

Reaction conditions
As Watts et al. have shown, the decomposition products of this Fenton-like reaction strongly depend on the $\mathrm{pH}$ of the solution.1
If performed in acidic conditions, the reaction generates mostly hydroxy radicals, but no reductants (which would be the hydroperoxide and superoxide anions). If, conversely, the reaction is held in neutral conditions, Watts et al. have shown that the reaction produces significantly more of the aforementioned reductants.
Do et al. have conducted further research into the reaction mechanism at $\mathrm{pH}=7$ and concluded that the observed reaction order may be approximated as a pseudo 1st order reaction, where the ratio $\ce{[H2O2]}/\ce{[#MnO2]}$ is vital to the description of the decomposition rate.2 They have further shown that reactive intermediates, such as superoxide and hydroperoxide anions are generated by the reaction. After slightly modifying the pH towards alkaline conditions, the production rates for the reactive anions increased drastically.
Reaction mechanism
Do et al. have meticulously assembled a table with a proposed reaction pathway:

And finally, I quote one of their closing statements:2

The existence of hydroperoxide/superoxide anion implies that the suggested reaction mechanism could be explained by hydrogen peroxide being decomposed, not only directly on the surface of manganese oxide, but also through a propagation reaction involving intermediates such as hydroperoxide/superoxide anion in solution.

References

Watts, R., Sarasa, J., Loge, F., and Teel, A. Oxidative and Reductive Pathways in Manganese-Catalyzed Fenton’s Reactions. J. Environ. Eng., 131(1), 2005, 158–164, DOI link.
Si-Hyun Do, Bill Batchelor, Hong-Kyun Lee, Sung-Ho Kong. Hydrogen peroxide decomposition on manganese oxide (pyrolusite): Kinetics, intermediates, and mechanism. Chemosphere, Volume 75, Issue 1, March 2009, Pages 8-12, DOI link.

I am adding a more recent study to complement the answer given by tschoppi. This journal addresses the kinetics and mechanism for the decompostion of $\ce{H2O2}$ on transition metal oxide surfaces. However, the paper only considers $\ce{ZrO2}$, $\ce{TiO2}$, $\ce{Y2O3}$. Although  $\ce{MnO2}$ is also a transition metal oxide, $\ce{MnO2}$ may be different enough to lead to different paths or kinetics. The authors even state:

effects such as
  solution pH, type of oxide, temperature, and oxide particle size
  have profound effects on the kinetics and energetics of this type
  of reactions

Thus, take the results of the study with a grain of salt.
I will focus more on the mechanism part of the journal. According to the journal

The kinetic experiments on the decomposition of $\ce{H2O2}$
  together with the experiments on $\ce{HO•}$
  detection show the existence of an adsorption step prior to decomposition. This type of process is also predicted with the DFT calculations. The
  decomposition of $\ce{H2O2}$ follows a similar mechanism for the
  three metal oxides studied. The obtained transition states are
  largely mediated by hydrogen bonding between $\ce {H2O2}$ and surface $\ce{HO}$ groups. Nevertheless, direct interaction between the oxygen atoms of $\ce{H2O2}$ and the metal atoms present in the
  oxide was also observed in the geometries of the transition
  states. The formation of two $\ce{HO}$ radicals as the primary
  product of the decomposition of $\ce{H2O2}$ is confirmed with both the DFT calculations and the experiments. One of these radicals can further abstract a $\ce{H}$ atom initially bound to a surface $\ce{O}$ and form $\ce{H2O}$. The other $\ce{HO}$ radical can adsorb to the surface by forming bonding states with the metal cation.

Reference

Claudio  M.  Lousada,
Adam  Johannes  Johansson,  Tore  Brinck,  and  Mats  Jonsson. Mechanism of $\ce {H2O2}$
Decomposition on Transition Metal Oxide
Surfaces J. Phys. Chem. C, 2010, 116 (17), DOI link.

you said that the statement,
effects such as solution pH, type of oxide, temperature, and oxide particle size have profound effects on the kinetics and energetics of this type of reactions. Makes the paper not that credible.
If you could please state any reason why you said so ,that would be really helpful for me to further understand the problem statement.


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Carcinogens, how do they work?
The easiest carcinogenic thing for me to grasp is radiation, as it directly messes with DNA. Then it seems there are other compounds that simply mimic hormones, but these shouldn't necessarily cause cancer, right? Then there are things that can accelerate mitosis or mess with other processes, which can inadvertently mess with DNA. So my question is, in general, how many different flavors of carcinogens are there? I know there are a lot, but can someone list a good few? Maybe some examples? I'm most interested in how plastics and things like smoke cause cancer. I think a few examples would be best in aiding my understanding.
Thank you!
Edit: I also just realized this is the Chemistry Stack Exchange, I think I am more interested in the chemistry side of this question, but it might need to be migrated.

There are at least two different radiationless pathways to wreak havoc on the DNA:

intercalaction: compounds like benzo[a]yrene or ethidium molecules slide between DNA strands and cause transcription errors
alkylation: compounds like N-Nitroso-N-methylurea may transfer alkyl groups to nucleobases

But the list is by no means complete.


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Are there natural conditions that could enable the formation of noble gas compounds?
Noble gases were considered to be inert until compounds that include them, such as xenon trioxide (as an example) were found.
My question is, what natural conditions allow the formation of noble gas compounds?

Natural? As in compounds you could find on Earth outside of a lab? I find that highly unlikely (other than clathrates), because most noble gas compounds are derived from the fluorides after direct exposure to fluorine, either with heating or light. Molecular fluorine is never found free because it is far too reactive, and it would be promptly consumed by anything else in the vicinity of the noble gas. The noble gas compounds themselves would be highly unstable and prone to decomposition. The only possibility I can think of is in extremely pressurized environments, perhaps in the mantle.

@NicolauSakerNeto yes, natural - I would appreciate any peer reviewed references that detail any research pertaining to the question.

I can give an example to make you know, one of the condition. Let us understand as to why $He_2$ molecule is not formed. This can be easily understood by calculating the attractive and repulsive forces when two helium atoms approach each other. Each helium atom has two electrons in its 1s-orbital. There are two attractive forces between the nucleus and the electrons of each atom. It has been found that when two helium atoms approach each other, new four attractive forces and five new repulsive forces come into play. Therefore, repulsive predominates and the potential energy of the system...

...increases which leads to unstability. Thus, $He_2$ represents unstable state and a chemical bond is not formed between helium atoms.

I bet that electrical discharges in evacuated metal tubes, filled with noble gases and halogenes don't count as natural conditions, although the compounds formed were happily emitting in the uv :D

This is just thinking out loud, but could this type of reaction occur with lightning? @KlausWarzecha

@Amaterasu Honestly, I really don't know whether a situation like in an excimer laser might happen in the atmosphere under the extreme conditions of a lightning. I wouldn't even dare to speculate on that.

Here is a video about xenon in sand (theory) http://www.youtube.com/watch?v=fZ7_CyLpwPc

As noticed in the comments, there are probably no natural conditions that would allow formation of noble gas compounds; although it should be noted that definition of 'natural' conditions should be specified. Lightning bolts surely carry enough energy to force such a reaction, but it is impossible for them to meet right mixture of gases, due to scarcity of both noble gases and super-strong oxidants and abundance of other substrates; as noted in comments (again), substrate able to create a compound with noble gas will create it with everything else first, and noble gas last.


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Converting 1,4-benzenedimethanol to benzene and methanol?
I have the molecule 1,4-benzenedimethanol:

How could I produce two methanol molecules and one benzene molecule? (Is there a catalytic reaction method available for this?)
I'm guessing that hydrogen ions could be used to split off the methanol molecules on both sides?

Oxidize to the dialdehyde (Jones reagent), decarbonylate with rhodium (do it at the bench before building the pilot plant), reduce the CO to methanol.  Why would you want to do that?

I will not ask why you would want to do that.
You're right that the benzylic alkohol can be deprotonated. But that won't help you for the intended reaction.
Regarding the possible mechanisms, I suggest to do it on paper.
Draw the starting material and split the bonds. Draw the fragments that result from splitting the $\ce{Ph\bond{-}CH2OH}$ bond. Do this first for the two heterolytical pathways possible in theory. Inspect the fragments and ask yourself, how stable these might be in general, and in the presence of other molecules of the starting material.
Are they likely? ;)
Compare with the following experiment on paper: Which reactions are possible with your starting material in the presence of strong acids? (Assuming that their corresponding bases are lousy nucleophiles.) Yes, it has nothing to do with the reaction you'd like to perform. But it works ;)
Now let's inspect the homolytical case.
To cut the "lecture" short, it has indeed been observed that benzylic alcohol furnishes benzene upon sonication (PDF). The yields however are only in the ppm range and I'm not aware of a comparable study on the title compound.

Edit:
Concerning the question in your last comment, have a look at the prices for methanol and benzene at the chemical supplier of your choice:
Methanol (z.S.)         2,5 l   14.00 €
Benzene  (reinst)       1,0 l   32.90 €
1,4-Benzenedimethanol   50 g  294.30 €
Let's just say that the production of methanol and benzene the way you suggested is not a promising concept.

thanks :) (My organic chemistry knowledge is not very good) but would it be possible to create other organic molecules, derived from this molecule that then can be eventually reacted to form methanol and another compound?


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How well (or poorly) does natural leather insulate?
What is the general thermal conductivity of animal skin leather?
I understand there are many different animal's skins (cow, sheep, buffalo, deer, pig), but in general, does natural leather insulate well (compared to, for scale, an esky for food insulation), or not?

Interestingley enough, there are papers entirely devoted to the subject of measuring the properties of living skin. The paper linked gives a value of $7 \times 10^{-4}~\mathrm{\frac{cal}{cm\, s\, °C}}$ for human skin, or in everyday units used nowadays: $0.29 ~\mathrm{\frac{W}{m\, K}}$. This is only for the dermis, the part of skin that then gets transformed into leather.
Now, this website has a comprehensive list of many different thermal conductivity values, and dry leather clocks in at $0.14 ~\mathrm{\frac{W}{m\, K}}$, which is about 2 times less the conductivity than that of living skin. What a difference the tanning process can make!
Given the similarity  observed in the review paper between man, beef and porcine tissue, I think it is fair to say that the value for leather will not change greatly across species.

I feel that my answer is somehow insufficient, since it's gotten so many upvotes but hasn't been accepted. Is there something that's missing, @foregon?

sorry about that, your answer was fabulous I think it was just me not finishing looking into the matter and I tend to wait until I definitively find the answer until I can ark something answered (if it is not clear) but upon  cursory look of yours again it is indeed sufficient, thank you for the superb research and links!!

I never worked in that field and thus had to google it myself ;)
Apart from the general observation that leather is indeed a good heat insulator, there seems to be an article in the Journal of the AQEIC (Asociación Química Española de la Industria del Cuero) addressing the question quantitatively.


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How to demonstrate the safety of wearing leather shoes in the chemistry lab?
Quite often (from my own experience anyway), one of the safety rules in a school (high school or university) laboratory is that the students must wear leather shoes.
I like to demonstrate some of the safety rules, so my question is, what is an effective and safe experiment to demonstrate the safety of wearing leather shoes while experimenting in the chemical lab?
*This is in comparison to say sneakers or canvas shoes

Tangential I know, but a rule like this may not be up-holdable in the case of, e.g. vegan lab users.

@Aesin interesting point, I must admit, I have never encountered that

Most institutions I know of (in the US) would have trouble enforcing the leather part, because of 1) price and 2) moral/ethical objections by certain groups of the population.

@BenNorris in some places, such as here in Australia, it is enforced rules.  I am just interested in the scientific experiment above.

Interestingly, the European standard EN 13832-2:2006 for rating footwear based on chemical resistance specifically excludes footwear with leather outsoles -- presumably because leather's actual resistance to materials varies significantly.

The simplest thing to do is to sacrifice two pairs of shoes, one with leather uppers and the other with canvas/cloth/whatever. Put something inside the shoes to represent "feet". Do something nasty to each pair, like pour concentrated acid over them (using secondary containment to avoid spills and splashes!). Then, extract your indicator and compare (assuming the different in damage to the shoes was not enough). You can also try lighting them on fire (though canvas shoes probably have fire-retardant compounds added), or seeing how long it takes water to soak through (leather will win here).
And, whatever you do, record it to video so that next time you only need to play the video and not buy more shoes. If this is for your students, I would make sure you appear in the video to increase "authenticity".

love it! A very simple but extremely effective method!  Now to choose what kind of nastiness I will do to them...

Use directly shoes could be not so practical and reproducible. I will try to find first to figure out which experiment could be more convincing.  So why don't take many little piece of leather, cotton and synthetic fiber and make some general test (maybe useful even for clothes). These are what is passing through my mind.
Wettability and permeability
One of the  most important thing is that substance drop over the shoes don't get in touch with the skin. So you could measure the contact angle of polar and non polar liquids over different "shoes textile" and find which has the lower one.
For permeability there are also some simple instrument you can buy.
Thermal conduction and inflammability
Even if this is more difficult to perform you could try to make your own system with a little hot plate and some thermocouples to see which materials do not conduct heat or is less flammable. One simple test could be the Clothes iron test (this in fact doesn't exist) you can palace a piece of different shoes textile and place it beneath the clothes iron measure the time at which it burns.
Acid and bases resistance
Why don't you put a little piece of textile over a piece of paper with an acid-base indicator and then put a known amount of acid or base over it if you do it in a glass table you can monitor when the acid reach the other side watching when the paper change color.
Penetrability
Pieces of broken glass could cut the textile so you could perform some test for determinate which textile is more hard to penetrate.


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How to calculate the frequency of cyclohexane conformational interconversion?
My book mentioned that the energy barrier for cyclohexane to reach the half-chair conformation is $50.6\ \mathrm{kJ\ mol^{-1}}$. It says that from this value, it was calculated that cyclohexane undergoes $10^5$ flips per second.
How did they calculate this? We can assume that all the conformational states possible will be of the same energy as the chair conformation (so they all encounter the same energy barrier).
I'm more interested in the principles behind the calculation – why the equation works – than just seeing the calculation done.

Look up application of the rate of reaction equation given barrier height and energy available at room temp, $kT$ (Boltzmann's constant).
$k = A \mathrm e^{ - \frac{E_\mathrm a}{RT} }$
http://research.cm.utexas.edu/nbauld/teach/cyhexconf.jpeg
Twist-boat leads to mathematically perfect geometric chirality,
Symmetry: Culture and Science 19(4) 307-316 (2008).


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Unexpected reaction during acid hydrolysis of cellulose
Recently I tried to hydrolyze some cellulose:
I added some acid to it ($\ce{H2SO4}$) then noticed nothing. Even after a few days there was no visible change and after doing some tests for glucose, those were negative too.
So at the end - expecting nothing would happen - I added some sodium hydroxide ($\ce{NaOH}$) to the solution to neutralize the acid and then the most unexpected thing happened, the solution began to warm up and there was an immediate color change (turning from white (from the cellulose) to yellow/brownish) I suspect that color came from the dehydration of the glucose  that was produced at that moment due to instantaneous hydrolysis of cellulose. I finally diluted the solution with more water and the reaction stopped
So somehow there was no interaction between the sulfuric acid and the cellulose, but when sodium hydroxide was added, the solution did more than just neutralize the acid!
Does anyone have any idea why this is?

We miss some data to make a good hypothesis.

The concentration of the $\ce{H_2SO_4}$ solution.
The temperatures.

These are the main two factors that affect cellulose hydrolysis. Usually are used temperature above 200 °C and pressure above 25 MPa. For me its strange that the exothermic reaction between SA and $\ce{NaOH}$ could lead to an increase of temperature such that.  However some studies show that the reaction can occur at lower temperature with concentrated acid.
Another possible solution is the formation of alkali cellulose.

I used 50 grams of cellulose, then I added 55 ml H2SO4 solution (99%) and it was conducted at normal room temperature and normal air pressure And in the end I added about 10 sodium hydroxide pellets (1 pellet being around 1cm long)

@LievenB was very concentrate maybe you have done a partial hydrolysis using the heat provided by the reaction. Maybe the crystalline structure was broken by the neutralization reaction. I don't know...

and so the yellowish/brown colour is then because of further hydrolysis (by H2SO4) of the produced glucose?

@LievenB yes this for me is quite probable, search in youtube glucose and sulphuric acid and you will see a lot of yellow and black. However I never read about hydrolysis with 99% concentrated acid, you should use a more diluted solution in my opinion...

ok - thanks! I'm planning on doing some more experiments with the hydrolysis of cellulose by using sulfuric acid so I'll definitely use your advice!

Take all the precautions! Never add water to sulphuric acid but viceversa and avoid strong exothermic reaction without the right protection! Good luck!

Chromophoric bi-product of alkali hydrolysis on cellulose is called oxycellulose. It is this which will have produced the colour - like you get on old clothes. This acid will have made the cotton stiffer/brittle, but the colour will come from the alkali.

From the comments:

I used 50 grams of cellulose, then I added 55 ml $\ce{H2SO4}$ solution (99%) and it was conducted at normal room temperature and normal air pressure, and in the end I added about 10 sodium hydroxide pellets (1 pellet being around 1cm long)

So unless I am missing something, in an attempt to hydrolyze cellulose, no water was added to cellulose powder.  The exo hydrolysis of cellulose is
$$\ce{(C6H10O5)_{n} + H2O -> C6H12O6 + (C6H10O5)_{n-1}}$$
Since water is a reactant, adding it is probably required to make the reaction work!
These are some hypotheses to consider:

Mostly anhydrous sulfuric acid and cellulose probably did not react, although perhaps some solvolysis to form sulfated glucose monomers may have happened.
When NaOH was added, water was formed nearly instantly by reaction with the acid, and the liberated heat strongly heated the reaction.
It doesn't sound like you added enough NaOH to neutralize the acid completely.    The reactions you observed probably happened in acidic conditions.

I think the strong heat and water formation that happened led to simultaneous hydrolysis and dehydration of the cellulose, leading to a complex mixture of hydroxymethylfurfural and further polymerized/carbonized derivatives.
I'm a little surprised that the 99% sulfuric acid you used did not carbonize the cellulose eventually even without water.  It's a very powerful dehydrating agent.$%edit$

Aside from the exotherm... Repeat the experiment with warming.  Glucose can air-oxidize, removing the aldehyde with which glucose tests react,  In fact, you can get the peracid from aldehyde free radical air oxidation, ruining reductive glucose assays toward any ring-opened glucose present.  Glucose slow hydrolytic release in a vessel exposed to air can be tested by doing it in seltzer as the reaction medium, then close the jar or cap cap after fizzing subsides, thus purging solution and containment with carbon dioxide.

I don't understand quite well what you are saying, could you please clarify? (I haven't learned yet about peracids,....)

R-C(=O)-H + •O-O• = R-C(=O)• + •O-O-H = R-C(=O)-OOH Oxygen is a triplet diradical.  It abstracts the aldehyde hydrogen, flips around, and bonds.  A peracid has a carbonyl bonded to -OOH not to -OH.  Peresters are with peroxides R-00H not with alcohols R-OOH.  The Alpha effect makes peroxide anions strong nucleophiles.   Absence of steric hindrance promotes fast kinetics.


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Extraction of starch from soup
I have a solution containing many different organic compounds (originally from leek and potato soup) and I would like to extract the starch from it. However, starch is insoluble in water. How could I go about extracting the starch?

You can't simply filter it and take the feed because all the vegetables fiber remain in the filter with your starch. If you want to make a starch with a decent level of purity you should start directly with the potatoes: grating the potatoes, putting them in hot water (preferably salted: this make more effective the separation from proteins) for a bit of time. Then take the filtered solution (not the residual) make the water evaporate now what you get is the starch with the potato protein, amino acids and minerals (what is called fruit water). There is this video if you want to see.

If you want to have a more pure starch also you can try to precipitate the proteins warming the solution and adding an acid. If you are at home acetic acid could be enough.
Eventually you can cool down the solution this make starch precipitate so you can remove the water with a filter and take the precipitate.
If you really want to isolate starch there are more sophisticated methods see here and here for references.

Doesn't dissolve in water?  Filter it.  Celite is your friend.

If the soup was cooked the starch is gelatinized - hydrated goo.  Centrifuge it.

thanks! I didn't know that there was a filter-option, but i'll use it!

For me this is not the way to do it, if you filter  a soup with celite what you have are for the most part the vegetables fibers with some starch in it...

@LievenB I think this is simply a soup without water...


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Why does the 3rd electron shell start filling up with scandium?
The electron configuration of calcium is 2, 8, 8, 2, where up to that point each shell, asides from the first shell counts up to 8 - why then does scandium have an electron configuration of 2, 8, 9, 2?
What causes the 3rd shell to start filling up, rather than the fourth?

Some part of the answer is contained in these posts. This essay on the matter is interesting, but it feels a bit like an a posteriori qualitative argument. This article goes a bit more deeply into the matters of the essay. I haven't properly processed all this information myself, even though I've read most of the material before.

Relevent post http://chemistry.stackexchange.com/questions/6921/why-does-ruthenium-only-have-one-electron-in-its-5s-orbital-in-the-neutral-oxida/6928#6928 (first paragraph)

This post might help you: https://chemistry.stackexchange.com/questions/73325/while-filling-electrons-we-follow-aufbau-principle-but-not-while-removing-why

The relative energies of the electronic subshells have been calculated for atoms in the vicinity of $Z=20$ (J. Chem. Educ., 1994, 71 (6), 469), and the result is surprising:

Looking at this graph, by all means the electronic configuration of scandium should in fact be $\ce{1s^2 2s^2 2p^6 3s^2 3p^6}$ $\color{blue}{\ce{3d^3}}$ in order to minimize orbital energies! The graph does not show the energy of the $\ce{4p}$ subshell, but it would lie somewhat above both curves shown. However, there is an important effect not being considered here, which is the destabilizing interelectronic repulsions. Electrons in a same subshell tend to repel each other more (intrasubshell repulsion) than electrons in different subshells (intersubshell repulsion). It turns out (and no one can really qualitatively explain why, see below) the repulsion is such that in an atom with $Z=21$, the configuration $\ce{[Ar] 4s^2 3d^1}$ is lower in energy than other potential candidates, such as $\ce{[Ar] 3d^3}$ or $\ce{[Ar] 4s^1 3d^2}$. Theoretically, the odd configurations $\ce{[Ar] 4s^2 4p^1}$ or $\ce{[Ar] 4s^1 3d^1 4p^1}$ would have lower interelectron repulsion energies than the observed ground state, but in these cases the lowered interelectron repulsions do not compensate the requirement of populating the higher-energy $\ce{4p}$ subshell. The comparison of several possible electron configurations and the joint minimization of orbital energy and electron repulsion energies does not seem to be something that can be done without resorting to heavy calculations or direct experimental validation.
In addition to some material I linked in comments, this article from Eric Scerri's blog (a chemist who focuses on aspects of periodicity, including electronic distribution) states, regarding the electronic configuration of scandium compared to calcium and to the $\ce{Sc^3+}$ ion (emphasis mine):

This amounts to saying that all three of the final electrons enter $\ce{3d}$ but two of them are repelled into an energetically less favourable orbital, the $\ce{4s}$, because the overall result is more advantageous for the atom as a whole.  But this is not something that can be predicted.  Why is it 2 electrons, rather than one or even none?  In cases like chromium and copper just one electron is pushed into the $\ce{4s}$ orbital.  In an analogous case from the second transition series, the palladium atom, the competition occurs between the $\ce{5s}$ and $\ce{4d}$ orbitals. In this case none of the electrons are pushed up into the $\ce{5s}$ orbital and the resulting configuration has an outer shell of $\ce{[Kr] 4d^10}$.
None of this can be predicted in simple terms from a rule of thumb and so it seems almost worth masking this fact by claiming that the overall configuration can be predicted, at least as far as the cases in which two electrons are pushed up into the relevant s orbital.  To those who like to present a rather triumphal image of science it is too much to admit that we cannot make these predictions. The use of the sloppy Aufbau seems to avoid this problem since it gives the correct overall configuration and hardly anybody smells a rat.

So unfortunately, it seems that even though most of the effects which combine to result in the observed electronic configurations are known, there is no qualitative way to predict where the configurations are going to mismatch with the Aufbau principle or the energy levels of the orbitals. I have read the statement that the Aufbau principle is most decidedly wrong for practically every atom with respect to the placement of the orbital energy levels, but incredibly it happens to predict the configuration of the valence shell for most atoms.

The electronic configuration model is anyway a limited model, in the sense that the different ROHF orbital energies could provide the same total energy (Table 1 in http://scitation.aip.org/content/aip/journal/jcp/125/20/10.1063/1.2393223). One may fix a form of orbital energy in ROHF then discuss on top of it, nevertheleess each explanation is based on the choice of orbital energies. Historically, many plot of orbital energy are based on a rather primitive treatment http://prola.aps.org/abstract/PR/v99/i2/p510_1, for a historical survey, see Levine's quantum chemistry 5th edi, p313-315

The official electronic configuration is extrapolated from experimental spectroscopy data, but the first-principle (without recursion from experimental data) theoretical model is not unique (ROHF, UHF, simplied HF with Thomas-Fermi-Dirac potential, KS-DFT(sometimes)), which complicated the case. If one wants to explain the experimental extrapolated electronic configuration, perhaps it is a problem to go through the mental setup of the argument which is used for the extrapolation (somehow discussed in Levine's book mentioned above).

@user26143 I appreciate your contributions in discussions, as they are always very insightful and deal with a facet of chemistry I am woefully unfamiliar with. I shall be sure to read what I can from your links.

Could you correct me if I'm wrong here? In summary, the Aufbau principle predicts the correct configurations, but not always the order of filling or the energy order. In the example of Z=21, Scandium, (not considering the interelectronic repulsions) the energy levels contrast with the predicted energy levels; 3d<4s. The electronic repulsions that arise when electrons are filled change it a bit; the first electron finds 3d to be at a lower energy, but the next two prefer 4s.

(continued)But the damage is already done; the Aufbau principle is misused to predict that 4s is always at a lower energy than 3d, which is not the case here. This is the flaw. Have I understood it correctly?

@HarryHolmes I would weaken that initial statement to something like "the Aufbau principle usually predicts the correct configurations of light neutral atoms, sometimes predicts the configurations of ions and heavy neutral atoms, and rarely predicts the correct order of subshell energy levels in any case, especially the relative energies of the inner shells". For example, consider that via the Aufbau principle, 4s is always "filled before" 3d, but for every neutral element past Sc (Z=21), the 3d subshell has a lower energy than 4s.

If you restrict yourself to using the Aufbau principle solely to explain the valence shell of neutral atoms, then the principle looks nicer because you're unintentionally (or intentionally!) throwing away most of the discrepancies. However, to my understanding, restricting the Aufbau principle to describe only the valence configuration was not an original part of the procedure when it was first proposed, nor do most people obey the restriction.

The more fundamental error is to think that energy levels of electrons are set in stone (like holes in a wall) as the number of protons and electrons changes, which is what the Aufbau diagram implicitly does.

So energy's not fixed like in the arrow diagram, but multiple factors cause it to vary. Got it, thanks.

The electron configuration for scandium is:
$$\ce{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1}$$
The $\ce{4s}$ and $\ce{3d}$ orbitals are relatively close in energy, so there are a few anomalies in the order of filling.  A fair question might be, why isn't the scandium electron configuration $\ce{[Ar] 3d^3 4s^0}$.  There is a special stability associated with completely filled orbitals, so the $\ce{3d^1 4s^2}$ configuration is of lower energy than the alternative $\ce{3d^3 4s^0}$.

I don't find this answer very satisfying. To say something is the way it is because the total energy of the system is lower that way, while entirely correct, is too shallow; you could answer hundreds of completely different questions with the same idea. What is really interesting is why should the energy be minimized in that specific configuration. Is there a reasonable theoretical justification for why the $(n-1)d$ orbitals fill after the $ns$ orbital, without necessarily having to go into lengthy and complex computer-assisted calculations?

@NicolauSakerNeto yes, what you have described is precisely what I am after - the 'why' of this phenomena intrigues me

@ron This would better be explained in terms of electron spin pairing energy.  Your answer would be very nice to have this concept incorporated.

Perhaps I am missing something.  I have read and reread the comments several times.  I didn't say "something is the way it is because the total energy of the system is lower" and just left it like that.  The points I tried to make are: 1) the 3d and 4s orbitals are close in energy, so small factors can tip the energetic balance favoring the formation of one electronic configuration over the other (3d1 4s2 vs. 3d3 4s0 example I gave), and 2) that full shells have a special stability and lower energy (inert gas configuration), hence the 3d1 4s2 configuration is energetically favored over

the 3d3 4s0 configuration.  If questions remain, please ask them in a clear, specific way and I will try to answer them.

I hope I didn't come across as insulting, as that was certainly not my intention. I'm fine with the statement that the orbital energies for 3d and 4s are close, hence more subtle effects would need to be taken into account. But then you write about the "special stability" without going into it more, so it sounded like a bit of a handwave to justify that the $3d^14s^2$ has a lower energy than $3d^34s^0$ and hence the former is observed instead of the latter. I was hoping I could stir someone into giving a more rigorous, possibly semi-quantitative answer.

No, no offense taken.  You're right about me "glossing over" the "special stability".  I did that because what more can be said than, empirical observation shows us that the inert gas configuration (e.g. completely filled orbitals) has "special stability."

What is being looked at, is how to know or predict what an electronic configuration is going to be.
When some speak of filling up shells with electrons, one has to consider what that means. Does it mean a kind of cartoonish idea of take a particular element whose electronic configuration we know eg. Carbon atomic number 6, electronic configuration 2,4.  Add a proton(making Nitrogen), and then add an electron, where will that electron go. And it goes in the last shell making Nitrogen 2,5.
Another way that might be meant by filling up is if you take the neutral atom, remove all  or some of the electrons and start adding them in then what order would they fill up.
For example neutral scandium has 21 electrons, if you remove 3 electrons, to give 18 electrons which is Argon's configuration, so that goes from neutral scandium to Sc 3+. Then if you were to add electrons, where would the 19th, 20th and 21st electron go.  Or, just for one electron, take a neutral atom, e.g. neutral Scandium,  remove an electron, so we are looking at the 1+ cation, (maybe assuming that the electronic configuration is now that of the previous element, or maybe checking to find out what is the electronic configuration of that 1+ cation), then add the electron and consider where it's going to go.
Or for a simpler example, taking Carbon, imagine it has one less electron so has configuration 2,3 (Boron's configuration), then add an electron, and it becomes 2,4.
There's the idea that each shell is identified by a number n. The second goes further out than the first. The third goes further out than the second etc.
It'd be wrong/inaccurate to say that shells are filled inward to outwards, they aren't necessary.(Though the first and second shells do follow this).  It'd be wrong/inaccurate to say that a shell fills to completion before the next shell fills, they aren't necessarily. (Though the first and second shells follow that). And it'd be wrong to think that the third shell has a max of 8, it doesn't.
Some basic material is excellent.
https://www.middleschoolchemistry.com/lessonplans/chapter4/lesson3
"When the third energy level has 8 electrons, the next 2 electrons go into the fourth energy level."
and
"The third energy level can actually hold up to 18 electrons, so it is not really filled when it has 8 electrons in it. But when the third level contains 8 electrons, the next 2 electrons go into the fourth level. Then, believe it or not, 10 more electrons continue to fill up the rest of the third level."
And you see the electronic configurations in that format here https://ptable.com/?lang=en#Properties e.g. Zinc 2,8,18,2
A view that explains a bit further than just looking at shells, is to use a model involving "subshells" rather than shells, and to say they are filled in order of energy level.  That speaks of the Aufbau principle and order of energy levels and subshells, the madelung rule  aka n+l rule.  At 16-19 you might learn about it.
The electronic configurations for neutral atoms are shown in abbreviated form here https://sciencenotes.org/list-of-electron-configurations-of-elements/
That rule, the n+l rule, or madelung rule. Some might call it the afbau rule (though calling it afbau rule might be a bit ambiguous 'cos by that some just mean the idea of filling up in order of increasing energy level, whatever that order is).
The n+1 rule gives us the following order of subshells.  "1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, and 7p". Those are all listed here https://chem.libretexts.org/Courses/Valley_City_State_University/Chem_115/Chapter_2%3A_Atomic_Structure/2.4_Electron_Configurations
n and l are quantum numbers.. n is the principal quantum number.  Each subshell s,p,d,f respectively, have l values of 0,1,2,3 respectively. For The 4s orbital ,  n=4 and l=0, thus n+l=4+0=4. The 3d orbital has n=3 and l=2, thus n+l=3+2=5. So  the n+l rule  has 4s below 3d.  4s lower energy than 3d.  The n+l rule always comes out with that order. It's a rule that helps to find the electronic configuration of neutral atoms.
When it comes to the transition metals, scandium onwards, technically, the order is a bit different. (i.e. not the n+l rule). Here, 3d is before 4s.  (which helps to explain the electronic configuration of some of the ions but doesn't help to explain scandium's electronic configuration, that would take some further explanation).
The n+1 rule gives an easy "explanation" of scandium's electronic configuration. If we assume as the rule says that 4s is below 3d.
Also an explanation behind why 4s is below 3d, is looking at a radial probability distribution graph for the orbitals, "Why 4s before 3d" by mommachem https://www.youtube.com/watch?v=5hzv5KQ4LoM     4s is said to penetrate the nucleus more than 3d.
The n+l rule also answers a more foundational question to the one you ask. Of why is Potassium [Ar]4s1
You might have thought that the third shell has a max of 8 electrons, because some basic books wrongly state that(and are wrong in doing so).  Infact a shell n, has a max of 2n^2 electrons. So the third shell has a max of 18 electrons.
But because 4s is less than 3d , (in energy level),  for (neutral)  Potassium and Calcium, the fourth shell is next to fill after 3p.
An s subshell can take a max of 2 electrons, a p subshell can take a max of 6 electrons, and a d subshell can take a max of 10 electrons.
Notice that before the third shell is maxed out. Specifically, after it has 8 electrons, (3s and 3p), then, the fourth shell fills up a bit(4s), getting two electrons, and then the third shell continues to fill(3d).
So that explains(or at least justifies!) why/how potassium, calcium and scandium have the electronic configurations they do.
And if you look at a periodic table with blocks drawn in, you see that.  (e.g. looking at the fourth row  left to right. 4s then 3d then 3p).

As you can see, an inner shell can have electrons added to it. And so that's where many books will put aside a simple view of just the shells, and from scandium onwards will tend to put away the format of electronic configurations of K,L,M,N like 2,8,9,2 And they'll go to showing electronic configurations as e.g. 1s2 2s2 2p6 3s2 4s1 e.t.c. And that model is in a way, simpler for elements from scandium onwards, and even Potassium onwards, in the sense that it's clear which subshell is going to fill up next.
When using the madelung rule(n+l rule), for getting the electronic configuration of neutral atoms, which is what it's for, then there are 21 exceptions.   No exceptions in the first three rows. Two exceptions in the fourth row - Chromium and Copper.  (Notice that Scandium is not an exception). And the rest of the exceptions are in the fifth sixth and seventh rows.  No exceptions in the s and p blocks aka the main blocks.

Also, at AP level like 16-19, or A level in UK like 16-18,  you might also be expected to know electronic configurations for ions of transition metals.  Removing electrons and knowing the electronic configurations of various cations.
And that's where another rule comes into play. For working out electronic configuration of transition metal ions.  That is that electrons come into 4s first, and go out of 4s first.
Some will explain  that "into 4s first and out of 4s first", by saying that, assuming n+l order of energy levels with 4s below 3d, that for transition metals,  electrons go in based on order of energy level, but they go out based on  which shell is further out. So the fourth shell is further out, hence they leave 4s first. And 4s is lower in energy so they enter 4s first.
There is another explanation.. Which is that according to HF calculations, when it comes to neutral potassium and calcium, 4s is below 3d(so, like the n+l rule).  But for transition metals scandium onwards, 3d is below 4s.  So that then explains very well why electrons leave 4s first, i.e. leaves 4s before leaving 3d,  it's because 4s is higher in energy than 3d.  (So we wouldn't be using the n+l rule there).
Here is a graph based on HF calculations, that shows that , for Potassium and calcium 4s is below 3d. But for Scandium onwards, 3d is below 4s. So it's not just an explanation, it's well grounded.
https://pubs.acs.org/doi/abs/10.1021/ed071p469
Transition Metals and the Aufbau Principle
L. G. Vanquickenborne, K. Pierloot, and D. Devoghel

That then brings up another question, which is if 3d is below 4s in neutral transition metal atoms, scandium onwards, as HF calculations show, then why is  Scandium's electronic configuration [Ar]4s2 3d1. Why isn't Scandium's electronic configuration [Ar] 3d3. And the explanation given, is that if we take Sc 3+  So scandium but with all fourth row electrons removed, so 3d and 4s electrons removed, to give Sc3+. Scandium has 21 electrons. Sc3+ has 18 electrons.   And then we add  three electrons, then the 19th electron goes into 3d. And the next two electrons initially go into 3d but then jump into 4s, because it's more stable so more preferable.  This is explained nicely on Dr Eric Scerri's blog in these two articles     here and here and on RSC here.  And also explained by Jim Clark, an educator who has written some articles explaining that   here  and  here
There is an article by a chemist, Geoffrey Neuss, here that tries to differ with Dr Scerri, and while Neuss makes some points that I think are fair and could be incorporated into Dr Scerri's explanation, such as that an ion doesn't necessarily have the same order of energy levels as a neutral atom (and perhaps shouldn't be assumed to be).. and that an ion's energy levels would be a bit different to a neutral atom. e.g. even if the order is the same, the distance between energy levels could differ. And Neuss also makes the good point, that when we see an electronic configuration lists 3d before 4s, that's just because it's listed in order of principal quantum number, it's spectroscopist notation and one shouldn't conclude from that, as some students might, that it is making a point about the order of 3d and 4s.
Neuss might not have been aware that the basis for Dr Scerri saying that 3d is below 4s, is HF calculations, and very strong.  Neuss uses NIST data that looks at electron excitations and tries to conclude that neutral Scandium has 4s < 3d in energy level, like the n+1 rule has it. But that's flawed for Neuss to do that, because when a photon is fired in to cause an excitation, the energy levels are changed quite a bit.. And it's a different set of rules to work out what transition is going to happen, called "selection rules". So e.g. even looking at something simple like Hydrogen and the excitations of that, you can't get any list of orbital energy order.  So i'd definitely go with Dr Scerri's explanation.
Orbitals fill up low energy first, then to higher energy.  But since an electron could go into one orbital then jump to another, (as explained by Dr Scerri), we can't really always predict electronic configurations.
Jim's chemguide article mentions Vanadium. We can use NIST data to see the ground configurations of neutral atoms and of ions.     For Vanadium  V+ is [Ar]3d4  and V is  [Ar]3d3 4s2
So we see that to get from V+ to V, when an electron is added to V+, one of the electrons jumps out of 3d into 4s. Or to look at it the other way, when an electron is removed from V to make it V+, an electron leaves 4s but then another electron jumps out of 4s and into 3d.  The rule that electrons leave 4s first, would not predict V+ 'cos it didn't account for an electron jumping from one orbital into another!
There might still be some further advanced issues re the discussion of ordering of 3d and 4s,  mentioned in the accepted answer here  Why do 3d orbitals have lesser energy than 4s orbitals in transition metals?
At a very high level (and not relevant to the energy of orbitals subject), they'd say electrons are not particles, some say they are waves, or something with properties of each. And also, at a high level  see this ACS paper, https://pubs.acs.org/doi/pdf/10.1021/ed200673w  they say electrons don't "occupy" orbitals, they might even be spread out. And orbitals are just visual representations mapping to some Mathematics, not direct representations showing shapes of regions of space. But for the purposes of considering electronic configurations, people are fine with the concept *as a model*, of orbitals as regions and of electrons in orbitals.

This is Dr. Eric Scerri. I would be interested in hearing why Neuss' version is considered to be superior to mine, as claimed above. Eric Scerri
www.ericscerri.com


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Extract acid from a solution
I have a solution containing glucose and some sulfuric acid. I want to distill out the sulfuric acid to then reuse it for other purposes. The problem is the presence of glucose in the solution. I'm afraid that, when increasing the temperature, the sulfuric acid will dehydrate glucose to carbon and other compounds - something that I don't want to happen. Would distillation cause such a reaction?
And in case it would - are there any other methods to extract the sulfuric acid?

I don't know your equipment, but the boiling point of sulphuric acid is > 300 °C.

Is there a specific reason you want to recover that sulphuric acid? Ultimately it'd be much simpler, cheaper, and less energy intensive to simply discard it (after proper neutralization) and use new sulphuric acid.

I used the sulfuric acid for hydrolysis of starch and as it is a catalyst, it would be better to recover it? Or is that more expensive than using new sulfuric acid?

Sulphuric acid is one of the cheapest chemicals of the world. Don't recycle.

As above, your labor all by itself (even if you are grad student) is worth more than the sulfuric acid.  As a pedagogic exercise, run the solution down a strong anion exchange resin washed OH-.  Only glucose comes out.  Recover the sulfuric acid by displacing it with phosphoric acid.


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Photoelectric effect and electron loss
I've read that when a certain amount of UV light is shone on a metal surface, electrons are ejected (the photoelectric effect).
Are these electrons from the metal atoms themselves? and in case they are - what happens to the metal atoms seen as they have lost an electron? (Are they chemically different?)

You may find http://en.wikipedia.org/wiki/Photoelectric_effect and especially http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html illuminating.

It is NOT the amount of light (number of photons), it is the energy of each photon (its frequency).  When the energy of a photon exceeds the vacuum work function of the surface, an electron is unbound.  There obtains a charged surface and a free electron.  Unless the surface is part of a circuit (e.g., grounded or connected to the cathode), the next electron out must overcome the vacuum work function plus the electrical potential.
The electron is emitted from the conduction band of the bulk metal surface.  The positive charge is delocalized across the surface if it is of constant radius of curvature.   Surface electric field divergence is inversely proportional to radius of curvature.  Unless the surface is very smooth you can get cold cathode emission or electrical breakdown of the encapsulating medium.


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Biochemistry for a mathematics student?
I am a second year mathematics student. I am mostly interested in statistics. Right now my curriculum doesn't offer me much choice (all second year courses are mandatory) but next semester I will be able to select the courses I want. Among the ones that are offered is a course called Biochemistry. It is actually a second year class for those how study chemistry and chemical engineering but I guess there must be applications of statistics in this field since it is also offered to third year math students.
My problem is that I am a math student: I have no knowledge or whatsoever in chemistry or biology. Granted that I had some in high school but back then I just learned everything by heart in order to pass.
My questions are the following:

With no knowledge of chemistry, is it realistic to undertake a biochemistry class?
The course description lists the following prerequisites: "Basic knowledge in general and organic chemistry". How important are these subjects in biochemistry? What are some good references (books) for self-learning these subjects?
Is Stryer's Biochemistry a good book for self-learning?
Are biochemistry and biostatistics related at all?
Are there many applications of statistics in biochemistry?

I have already searched for some information on the internet but I'd like to have some specific answers.
If it helps here is the information page of this class.
Finally I hope this kind of question is accepted here, if not please point me to a better place. Thanks in advance!

EPFL? Home of Michael Grätzel! Nice! Anyway, I can't give you a real advice, I'm "just an organic chemist". Does the faculty have a "Fachschaft" with students whom you could ask?

I will definitely try to find someone that took this class but I fear that there are not many math students that took it (it is optional after all).

Talk to the prof or to somebody who has taken it. I cannot imagine the value of memorizing amino acid structures, buffers' pH and capacity, or where the acetate carbon atoms are in cholesterol.  However, genomics is heavily statistical.  You can be of considerable value there.

It appears that you will need at least 2 semesters of O chem. I recommend Loudon for O chem. Take a look at it and if it makes sense to you, you should be ok.

It also looks like an intro class. I think that you may be ok without knowing all the o chem. However, you will have to put a significant amount of effort in filling in the gaps. It shouldnt be as bad as in mathematics but nevertheless be cautious.

I'd also suggest that you read up beforehand on pH and the related calculations and get a glimpse of physical chemistry (thermodynamics: reaction enthalpy, free energy, equilibria; kinetics: order of reactions, reaction speed, Michaelis-Menten kinetics) Maths behind that is easy, but you should be familiar with notations and useful as well as widely used approximations (I emphasize the widely used: as a maths student you may solve directly without approximation - but nevertheless approximations are employed in practice and if you don't know that it can be very confusing)

As chemist, I like the Stryer. As chemometrician I can tell you that there is a considerable need for statisticians in (bio)chemistry. Including but not limited to genomics.

With no knowledge of chemistry, is it realistic to undertake a biochemistry class?

I would say that taking a biochemistry course without any background in chemistry or biology would be very difficult, but it would be possible.

The course description lists the following prerequisites: "Basic knowledge in general and organic chemistry". How important are these subjects in biochemistry?

The info page link that you gave would not work for me, so I can't tell how in-depth this course would be, but I know that the undergraduate biochemistry courses I have experience with expect you to be familiar with (at a minimum):
Chemistry

Bonding theory
Molecular structure
Nomenclature of organic molecules
Polarity and its relation to hydrophobicity/hydrophilicity
Basic organic reaction chemistry (particularly Lewis acid/base and functional group chemistry)
Basic reaction kinetics
Basic thermodynamics

Biology

Macromolecules (proteins, carbohydrates, fats)
Cell structures
Basic anatomy/physiology

All of these are fundamental to just about every concept in biochemistry, and they will probably expect you to already be familiar with if not good at them.

What are some good references (books) for self-learning these subjects? Is Stryer's Biochemistry a good book for self-learning?

That I can't answer. I can tell you that Voet & Voet's "Biochemistry" is a good biochem book, but I am not sure if it would be suitable for self-learning. I don't have experience with Stryer's book. Voet & Voet also have a "Fundamentals of Biochemistry" book that might be better - but again, I don't have any experience with that one and can't compare to Stryer. In my personal experience, I have found that reading as many different books as possible is the best way to self-study, but your results may vary.
I can give you some suggestions for general chem and organic books:
General Chemistry

Brown, Lemay, and Bursten's "Chemistry: The Central Science"
Tro's "Chemistry: A Molecular Approach"

These are both very easy to read and clearly explain all of the fundamental concepts, with lots of example problems worked out in detail.
Organic Chemistry

Wade's "Organic Chemistry"

This is the only organic chem book I have much experience with, but I remember it being very easy to understand.
For biology textbooks, I'm afraid I can't be much help.

Are biochemistry and biostatistics related at all?

Only in the sense that biostatistics is statistical methods applied to problems found in biology, and many problems found in biology are related to biochemistry. You would certainly want to learn some basic biochemistry if you were going to work on bioinformatics problems, for example.

Are there many applications of statistics in biochemistry?

There are - particularly in areas like design of experiments, genomics, proteomics, and drug screening, and other various assays.


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Conversion of potassium dichromate solution to a less toxic solution
I have a solution of potassium dichromate ($\ce{K2Cr2O7}$) and I want to convert it to a less toxic solution. What would be the best way to do this?
I was thinking of using the dissolved potassium dichromate as an oxidizer in the oxidation of alcohols to organic acids, to eventually produce the chromium(+III) ion (what is less toxic)
Or are there any other, easier methods?

Or are there any other, easier methods?

The cheapest and safest one is probably to use ethyl alcohol and some acid (acetic acid should work).

Ethyl and any other alkohol is a bad idea. Reaction of chromate with alcohols is slow. Try FeII as Kwak recommends.

That looks like one way to reduce the dichromate ion which is known to be toxic. MSDS sheets indicate that potassium dichromate has category 1B carcinogenicity.
Another way I believe is to just start from an aldehyde and convert it into a carboxylic acid:
$$\ce{3RCHO + 2Cr_2O_7^- + BH^+ -> 3RCOOH + 2Cr^{3+} + 4H_2O} $$
Alternatively, one can put dichromate in acidic solution with iron:
$$\ce{Cr2O7^{2-} + 6Fe^{2+} + 14 H^+ -> 2 Cr^{3+} + 6 Fe^{3+} + 7H_2O }$$
Dichromate oxidizes $ \ce{Fe^{2+}}$ ions and produces $\ce{Cr^{3+}}$ and $\ce{Fe^{3+}}$ ions.

thanks! I'll use the iron oxidation method, however : would an acid such as acetic acid be sufficient for this redox reaction?

@LievenB Hi, I believe so. As long as you have enough acetic acid to form a ratio of moles of 14:6:1, that would be sufficient for this redox reaction.


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What is a product of water ionization?
I've read that radioactive compounds are dangerous because they emit electromagnetic waves (gamma-radiation etc.) that are dangerous for living organisms because they can ionize atoms in molecules.
So if one water molecule is ionized by using radiation, would you then get:
$\ce{H2O -> H^+ + OH^-}$
or would you get free radicals : $\ce{H2O -> H^{.} + OH^{.}}$?

Water, when hit by radioactive particles, will most likely suffer ionization (or radiolysis). The ejected electron very quickly leaves the vicinity of the affected molecules (in the case of exposure to gamma or beta radiation) or a very chemically stable neutral helium atom is formed (in the case of exposure to alpha radiation), so you obtain short-lived species which will tend to react with most other substances nearby, even if neutral. The electron, after being slowed down considerably, becomes hydrated and can participate in reactions itself. This article has several examples of reactions triggered by radiolysis, and also separates them by their relevant timescales.
Some further examples of reactions involving radiolysis of water via beta particles and gamma rays are (from the very well-cited article "Critical review of rate constants for reactions of hydrated electrons, hydrogen atoms and hydroxyl radicals (⋅OH/⋅O−) in aqueous solution"):

Of course, not only are the decay particles damaging, but so are the entities produced by their interaction with matter in the vicinity. If the nuclear decay particle doesn't directly ionize organic matter, reactive species such as $\ce{OH^{.}}$ or $\ce{H^{.}}$ could then react with biomolecules, degrading cells and tissues.
This Wikipedia link has some details on the radiolysis of water. Interestingly, it seems that dissolved hydrogen in water can efficiently suppress the formation of unusually reactive species, presumably as the side reactions eventually cause the formation of hydroxyl anions (if electrons are present in excess) or hydronium cations (if positive charges are present in excess). This image (source) shows a collection of possible reactions in sequence. If you search for articles on water radiolysis, there are many other results, such as these two. It's interesting to note that the effect of radiation on water still doesn't seem to be entirely understood, in part due to our incomplete knowledge of the behaviour of hydrated electrons.

Here's an extra bit of background on radiation and its biological effects (at least for the most common types. There are many ways nuclei can decay).
Depending how far the radiation source is from you, gamma radiation is either the least or the most dangerous type of nuclear decay. If the radiation is coming from well outside your body, then the danger scale goes alpha < beta < gamma. If you happen to have a radioactive nuclide in your pocket, the scale may well become alpha < gamma < beta. However, if the radiation is coming from inside your body (avoid ingesting glowing substances), then the danger scale goes gamma < beta < alpha. Why does radiation behave like that? There are two main effects in play: penetration capacity and ionization capacity.
In collision theory, different types of projectile and target can be characterized by something called an collision cross-section (or absorption cross-section, in nuclear physics), which basically gives you a probability of how likely it is for the projectile to hit a certain target. If the combination of projectile and target have a high collision cross-section, they they are very likely to interact, and so almost any projectile will be stopped by even a thin target. If the collision cross-section is low, then the projectile has a low chance of hitting the target, and therefore will tend to pass through it entirely. To compensate, you need to make a very thick target so there are more chances for the projectile and target to interact.
The different types of radioactive decay have different penetrating capacities. All of matter as we know it is composed of atoms, which contain well-defined oppositely charged regions (the nuclei and the electron shells), and so we can expect them to interact strongly with other charged particles. Alpha and beta(-minus) decays emit particles with a charge of $+2e$ and $-e$, respectively ($+e$ for beta-plus decay). We then expect these particles to interact strongly with matter, being stopped easily. Indeed, most alpha particles are efficiently stopped even by air. Beta particles have half the charge of an alpha particle and are going quite a bit faster (having approximately 1/7000th the mass), so they manage to go further before interacting. A few centimetres of of metal shielding or a concrete wall are usually enough to stop them. However, gamma rays have no charge, so they interact comparatively weakly with matter. These highly energetic photons will go though dozens of centimetres of solid matter. To maximize the odds that they will hit something and be stopped, you want to use a shield made of a very dense substance (which would pack a lot of possible targets in a small space). This is why lead is the substance of choice for radioactive shielding; its high density combined with low cost allows the fabrication of relatively compact shields which are capable of stopping even gamma radiation, when around 10 cm thick.
Now, for the mechanism of how particles from radioactive decay are stopped. For uncharged particles, the absorption cross-section is low because interactions only happen when the uncharged particle comes very close to a target particle. For gamma rays (very high energy photons), when there is an interaction, it is usually due to Compton scattering, where an energetic photon interacts with an electron, giving it a large amount of kinetic energy, while the photon is re-emitted with a smaller amount of energy. The target electron (not necessarily a valence electron) often gains enough energy to be completely removed from the atom, causing ionization.
For collisions between charged particles (both same and opposite charges), interactions happen much more easily, as charges interact strongly even at comparatively much larger distances. Beta-minus particles (very fast electrons) can more efficiently repel electrons and shove away from atoms, creating cations. Alpha particles (bare helium-4 nuclei) are interesting because their positive charge means they can tear electrons from any atom or ion. In fact, it's defensible that alpha particles are the strongest Lewis acid one can commonly find; they are effectively free protons with twice the charge.
Notice that penetration capacity and ionization capacity are inversely correlated; if a radioactive decay particle is strongly ionizing, it interacts well with matter and is stopped in short order. If it is weakly ionizing, one of the main interaction mechanisms is suppressed, and the decay particle can go further before it is stopped. In order to not damage you, you want the nuclear decay products to not dump their energy in your body, i.e. you want them to pass right through you. Here, it is clear that alpha particles are the most dangerous, followed by beta and gamma radiation. However, alpha particles interact so strongly with matter that they literally need to be released closer than your skin to dump energy into living tissue, so you need to ingest, breathe or inject alpha emitters. When this happens, they are extremely deadly. It is a very rare occurrence, however, since radiation usually comes from far outside. In that case, gamma radiation has a larger chance of reaching you, so it is said to be the most dangerous.

You can get the free radicals upon direct UV irradiation of water at wavelength shorter than $\lambda$ = 190 nm.


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Single Displacement Reactions: A + AX =?
This seems like a stupid question, but how would I find the product of a single displacement reaction like the following?
$$\ce{Pb + Pb(NO3)2 -> \ ?}$$
Normally, a single displacement reaction follows the path of $\ce{A + BX -> AX + B}$. However, what will the equation look like when A and B are the same element?

Try this link. Maybe that helps some. (Don't know if it's valuable advice, though...)

If Pb(I) were stable, 2PbNO3.  Nitrate itself might be a sufficient  oxidant to prevent that.  Cu(0) + CuCl2 gives 2CuCl when refluxed in acetonitrile (soft Lewis base ligand).  A blue solution with red metal dust turns opaque brown-green, then goes clear and colorless.

Well, nitpicking involved, there could be the isotopic statistical equilibrium: $\ce{^{206}Pb(s) + ^{208}Pb(NO3)2(aq) <=> ^{208}Pb(s) + ^{206}Pb(NO3)2(aq)}$

Nothing. All you would have is just a mixture of white powder and lead metal.
In general, lead (II) nitrate is synthesized by:
$$\ce{PbO (s) + 2 HNO3 (aq) -> Pb(NO3)2 (aq) + H2O (l)}$$
Lead (II) nitrate is soluble in water.
Alternatively, you could do:
$$\ce{PbCO3(s) + 2HNO3(aq) -> Pb(NO3)2(aq) + H2O(l) + CO2(g)} $$
Single displacement reactions are a kind of redox reaction. For single displacement reactions to occur:
1.) A and B must either be different metals OR
$$ \ce{2AgNO3(aq) + Zn(s) -> 2Ag(s) + Zn(NO3)2(aq)}$$
2.) A and B must be a halogen
$$\ce{Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g)}$$
This is necessary as for the redox reaction to occur, you need two differing metals on the reactivity series in order for the single replacement to work, or alternatively, use compounds that contain halogens.
Even so, in combining two compounds with metals or halogens, if the element is less reactive than the element in the compound, the single replacement reaction will not proceed.
Some examples are:
$$\ce{Ag (s) + Cu(NO3)2 (aq) -> } \text{No reaction}$$
$$ \ce{I2 + 2KBr ->} \text{No reaction} $$

If you consider reaction of type A + AX ->, of course they can be achieved. However, they won't be considered displacement reaction but rather disproportionation reaction. A handful of interhalogen compound synthesis proceeds through this reaction format (if I consider X to be the more electronegative halogen than A).
$$\ce{Cl2 + ClF3 ->[300 °C] 3ClF}$$
$$\ce{Br2 + BrF3 -> 3BrF}$$
$$\ce{I2 + IF3 ->[-78 °C][CCl3F] 3IF}$$