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Practically speaking, you can make in the order of magnitude of 100 ml of H2 gas within an hour or so if you use 10% H2SO4 and a voltage of 5 to 10 volts. At the same time you'll make 50 ml of oxygen. The reaction is not that fast, but fast enough to do some nice experiments. I once did the experiment by taking a tub of appr. 1 liter content, I took two small 100 ml glass bottles and put them upside down, completely filled with the liquid and keeping the electrodes completely under the bottles. At the cathode I simply used copper wire. The anode must be constructed, such that absolutely no copper touches the liquid, only non-conductive stuff like wire-isolation and the carbon rod may touch the liquid. Btw, what do you intend to do with the H2 and O2? You can make some detonation gas with that, but if you do so, only make ml quantities and never light more than just a few ml of mixed gas at the same time, if you want to keep your ears operational.

I do not agree. If you use NaCl, you will get quite a lot of chlorine if the conditions are chosen right. If you use a carbon anode, then the amount of chlorine you get is only a little less (on a volume by volume basis) than the amount of hydrogen you get at the cathode. You loose some chlorine, because it dissolves in water somewhat and you'll also get some oxygen.

In fact, the redox potentials for I2+2e-->2I(-) and for the copper reactions are fairly close to each other. What you get is an equlibrium reaction: 2Cu(2+) + 2I(-) (+ 2I(-)) <----> 2Cu(+) + I2 (+ 2I(-)), where the quantities at both sides of the arrow are appreciable. The two I(-) ions between parentheses also are present, because we have 4 iodide ions for two copper (II) ions, so I include them in the equation. These play an important role in the rest of my 'story'. The compound CuI, however, is highly insoluble. So, if some Cu(+) is formed in the equilibrium reaction, then it precipitates as solid CuI with the half of the iodide ions (given in the equation between parentheses) and in that way it is taken out of the equilibrium system. Because the CuI is taken out of the equilibrium system, the reaction is driven to one side, being the side with Cu(+) and I2.

You are lucky to have such a supplier over there.... Over here we have no reasonable choice in many cases.

In order to do electrolysis you need a DC voltage supply. Using an AC-voltage does not work or gives you a mix of hydrogen and oxygen at both electrodes (DANGER!!). Besides that, you need to add some electrolyte to the water, otherwise the reaction will be extremely slow. If you want hydrogen and oxygen, then the best thing you can do is add some dilute sulphuric acid or nitric acid to the water. Probably it also works with sodium sulfate. It also works with sodium hydroxide as electrolyte, but this is very bad for the skin, much worse than dilute sulphuric acid or nitric acid. So, if you use sodium hydroxide as a electrolyte, then be even more careful than with the dilute acids. A nice thing of using sodium hydroxide, however, is that you can use a simple copper wire electrode, also at the anode. It still gives oxygen at large quantities and hardly corrodes. This is quite remarkable and I have no real explanation for this. If you perform electrolysis in a copper sulfate solution, then metallic copper will be formed at the cathode, regardless of the cathode material used. At the anode you will get oxygen (and acid) with graphite anode and copper will dissolve when you use a copper anode.

It is ridiculously sensitive. I have made some of this some years ago by adding a drop of concentrated H2SO4 to a few crystals of KMnO4. What I saw is formation of a purple vapor (which is Mn2O7) and the most remarkable was that every few seconds I saw a little speck of light on the dark brown/green drop of H2SO4/KMnO4. These little specks of light were due to dust particles from the air, touching the drop and being oxidized at once, as soon as they touch the surface of the liquid. While I was watching this beautiful phenomenon it suddenly did BENG... a lot of little brown droplets of H2SO4 and MnO2 were sprayed around and I had many little dark spots on my clothes. So, if you experiment with Mn2O7, be really careful and only use milligram quantities.!

Do you think the Cl can oxidize the Al? The Cl is present as Cl(-) ions (of course you know that ) and will not be transformed to Cl atoms or Cl2 molecules. The sodium ions also remain present as ions. The Cl and Na will not be available as elements at the temperatures involved in the microwave. I think that the temperature needs to be thousands of degrees centigrade, before free sodium atoms and chlorine atoms become avaiable. I can imagine formation of an aluminate, but the mechanistic pathway to this certainly will not be through chlorine and sodium in the zero-oxidation state. If sodium aluminate is formed, then the only reasonable pathway could be formation of Al2O3, which in turn reacts with NaCl, forming NaAlO2 and AlCl3. In this pathway, the sodium and chlorine remain at oxidation states +1 and -1 all the time. This reaction may occur at very high temperatures when the NaCl melts, but I do not expect this to happen in the microwave experiment.

Most likely it is Al2O3. If you heat Al-foil in a flame, then you'll also see it shrink and crumble when it becomes sufficiently hot. The very hot Al-metal is oxidized by oxygen from the air to Al2O3.

Thats' strange, I can buy it at several places in concentrations up to 52% and I even was lucky to find some 58% HNO3. It is not as common as hydrochloric acid, but it certainly is not that hard to obtain (at quite some drugstores and also from chemical supply houses, who sell to private persons). Is it really so difficult to obtain chemicals in the USA? I sometimes order chems from the USA or Canada (only solids, no liquids like HNO3 of course ) taking the shipping costs for granted. I thought the USA is the walhalla for citizen chemists or has this changed due to 'war on drugs' and 'war on terrorists'?

Cesium is VERY reactive and will ignite (explode?) on contact with air and water. Whether your 0.5 gram sample may do harm depends on where it is stored. A spark from a cigarette can be harmless, but it can also initiate a fire, which destroys complete forests, so think twice about the place where you store your cesium. If you store it in a thick-walled metal can with metal screw cap, in all wrappers you already have, then I would not worry about it. Even if it would explode, I would not expect it to rupture the metal container, because it is just 0.5 grams.

Nickel does dissolve in nitric acid, but the concentration must be carefully adjusted. I have made nickel nitrate from the Dutch coin 'dubbeltje', which is made of 99+ nickel metal (with mainly tin as an impurity). I found that in HNO3 (60%) the nickel did not dissolve, in 40% HNO3 it dissolves vigorously, in 20% HNO3 it did not dissolve. Quite an excess amount of HNO3 is needed in order to dissolve all metal, because at lower concentrations of HNO3, the metal does not dissolve anymore. Dissolving the metal can best be done by putting it in conc. HNO3 (60% or something like that) and dripping in water slowly and swirling. Making the acid luke-warm helps a lot. When the reaction starts, a LOT of brown NO2 is formed and the liquid becomes quite hot. Be careful and do this experiment outside or in a very good fumehood!

I also heard this story from a Dutch guy, but I can hardly believe it. I actually ordered some chems from this site (once in Oct. 2004 and a few weeks ago). Both times, the parcel I received really was from the UK, with Great Britain stamps and postmarks. I indeed think that kno3.com is not the most 'legal' company, but the story that it is an Arizona-based anti-meth agency simply seems false to me. I think that the original story is from a member of the scimad forums, who had a really bad dream .

Many oxides are remarkably inert, once they are calcined. I have some praseodymium oxide Pr2O3 (lightgreen powder), which does not dissolve in hot conc. HCl, not even in hot conc. H2SO4, nor in HNO3. Dilute acids also do not work! I also have Cr2O3, TiO2, and SnO2 and these oxides behave similarly, no dissolving at all. I also have calcined Fe2O3 and Co3O4 and these oxides only dissolve at a mg per day rate in concentrated HCl. Really annoying... I'm afread Mendelejev's oxides also will be VERY inert. For making the salts of the lanthanides I have switched to the metals as starting point (currently Pr and Er, but I also want to experiment with Dy and Sm), the oxides simply are too inert to be useful. If you want a really beautiful fluorescence, then use the dye fluorescein or its sodium salt, sodium fluorescienate. The solutions of these compounds are orange, but even on a cloudy day, the liquids give an amazing intense lime-green light. If you put the liquids in a ray of sunlight, then the entire beaker emits a lot of green light. Imagine what your UV-LED's will do with this!

Indeed, aluminium is remarkably inert in sulphuric and nitric acids. This has to do with the nature of the process for breakdown of the oxide layer. Both the sulphuric acid and nitric acid have to eat throughthe oxide layer by means of the equation you mention: Al2O3+3H2SO4=>2Al2(SO4)3+3H20 This process is very slow. Many oxides are remarkably inert, especially when they have been heated. The sulfate ion and nitrate ion of these acids do not coordinate to the aluminium ion. With hydrochloric acid, the situation is quite different. The oxide coating now is quickly eaten away, due to the coordinating properties of the chloride ion. A complex ion AlCl4(-) is formed and apparently, there is a fast mechanistic pathway for formation of this ion from the oxide and hydrochloric acid. Just try it. Add some Al-foil to 30% HCl or to your battery acid, to which quite a lot of table salt is added. You'll see that within a few tens of seconds the reaction with the aluminium becomes very vigorous. Things become even more striking, when a combination of copper (II) ions and chloride ions is present. Then the acid is not even needed anymore. The CuCl4(2-) ion in a concentrated solution of a chloride, with some copper (II) added is capable of instant destruction of the oxide layer. If you have copper sulfate, then try dissolving some of this in your battery acid and also add quite some table salt. You get a green solution. You can also dissolve some copper sulfate in 30% HCl if you wish. If you add the Al-foil to this, then you get a very violent reaction (be careful!) in which a lot of H2 is produced, a lot of steam and a small amount of copper metal. But even in a plain solution of table salt, to which some copper sulfate is added, the Al-foil already dissolves quite vigorously.

Not only bases, also a lot of acids, such as SO2(OH)2, PO(OH)3, HPO(OH)2 (sulphuric acid, phosphoric acid and phosphorus acid). As a rule of thumb, inorganic compounds of the form X(OH)n or XOm(OH)n are basic if X is in a low oxidation state (1, 2 and sometimes 3), amphoteric if X has an intermediate oxidation state (4, 3, sometimes 2) and acidic if X has a high oxidation state (>= 4). There is also NH2OH, which is slightly basic, but not because of its OH group, but in a similar way as NH3 is basic, by accepting a proton, forming NH3OH(+).

Besides the radioactivity, uranium also is very toxic from a chemical point of view, both as an element in the form of very fine powder, and in its compounds. Uranium toxicity is comparable to lead and mercury toxicity.

What I tried to point out in previous posts is that I do not expect the reaction to work. The lithium most likely will be oxidized by the KNO3, not resulting in formation of K metal and LiNO3, but the nitrate ion will oxidize the Li with great violence. If you have molten KNO3 and you throw in some reducible material then you get quite a violent reaction. Remember, KNO3 is a main component of black powder, which acts as oxidizer. As Li is even a stronger reductor, than C and/or S, expect the reaction between Li and KNO3 to be even more violent than the reaction between KNO3 and S+C, as in black powder.

Indeed, nitrate does not react with hydroxide, but... what is the relation with the topic on making K-metal from Li and KNO3?

I do not expect the reaction to work anyway, regardless of the precise redox potentials. In order to have a reaction, the reactants need to be liquid, but at those elevated temperatures I expect the nitrate ion to act as a strong oxidizer towards the alkali metals. Finely powdered Li + finely powdered KNO3 probably would be a potent pyrotechnic mixture .

Indeed, you just made Cl2. Then you only observed a faint green color. If you make ClO2 then you get an intensely colored yellow gas. You certainly won't overlook it . See the pictures at my website: http://81.207.88.128/science/chem/exps/clo2/index.html Making a strong base from a strong acid? That is cool . OK, no kidding, the reaction is as follows: HClO + Cl(-) + H(+) --> Cl2 + H2O Written without ions (simpler but less accurate description): HClO + HCl --> Cl2 + H2O The HClO is from ClO(-) from the bleach and acid. HClO is a very weak acid, so it hardly is dissociated in its ions. In some countries it is used as herbicide, I'm not sure whether this is the case in your country. As an alternative you can use KClO3, but the reaction is somewhat slower in that case, but the final result is the same. KClO3 can be prepared with some effort at home from table salt and low-sodium salt (mostly KCl), see http://www.wfvisser.dds.nl/EN/kclox_EN.html