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Washing soda is Na2CO3.10H2O. Simply dissolve it in water and you have an alkaline solution. No heating, driving off something, etc. required.

You can buy it at a supermarket as washing soda. Heating sodium bicarbonate also works. After it has decomposed, dissolve it in a small amount of water. You can also mix NaOH and NaHCO3. As far as I remember, a concentrated solution of Na2CO3 has a pH somewhere near 11, which is quite alkaline already. If you want really strong alkalinity, use NaOH. NaOH however is quite corrosive and exceedingly dangerous for the eyes. Be careful!

If solubility is limited, then the lowest freezing point may be due to the limit in solute. However, this is not always true. Sometimes a phase-system (keyword: eutectics?) is created which has more complex behavior and has a minimum freezing point. Just consider a mix of water and acetic acid. Pure water freezes at 0 C, pure acetic acid freezes at 16 C. Now imagine that you draw a graph with at the x-axis the percentage acetic acid (which ranges from 0 to 100) and at the y-axis the freezing point. This function will have a minimum, quite below 0 C, at 0% its value equals 0 and at 100% its value equals 16. So, you get a 'parabolic-like' shape of the graph. So, in practice, the situation may be more complex.

I actually tried this one and it really is great. Boil some red cabbage for several minutes, such that you get a pruple liquid. At low pH it is beautifully red/pink. At neutrality it is purple. Slightly alkaline makes it blue. More alkaline (e.g. ammonia, dilute sodium carbonate) makes it deep green. Very alkaline (NaOH and very concentrated Na2CO3) makes it yellow. Unfortunately I have no number chart for this, but with some effort you can make your own.

I have another nice variation. Why not try whether the smoke of a joint has a similar effect after exposure to a little copper:D ? You can also try a reefer, leaving out the tabacco. The smell is a little sweeter and spicier already, so it may give a really strong sweet effect. BTW, putting barium salts in your mouth is unwise to my opinion, even with antidote nearby. You also tasted mercury and lead salts? Copper ions are at the border for me, but I personally never tried tasting copper sulfate. I also like to get aquainted with chemicals, so I carefully smell them if they are volatile, but why would I taste them? The only means of getting poisoned is through inhalation or possibly through skin, but in both cases the taste is not needed. I neaver eat when I do experiments and I simply cannot imagine that I ever will get chemicals in my mouth by accident. Inhaling some by accident of getting some on you skin of course is another story...

When you have a single positive point charge and an electron 'orbiting' around the point charge, then the electron can only be in discrete energy levels. In an atom, the nucleus can be regarded as a single point charge and the error introduced with this is VERY small. The nucleus is very small, compared to the total atom size. Having said that, the motion of an electron around a nucleus can be described by means of a partial differential equation. By assuming certain forms of the solution, the problem is reduced to an eigenvalue problem, which has a discrete set of solutions. Such a solution is fully specified by four numbers: 1) The principal quantum number, which tells how far the electron is 'orbiting' around the nucleus. I.e. this tells something about the size of the orbital. 2) The angular quantum number, which tells something about the shape of the orbital. 3) The magnetic quantum number, which tells something about the orientation of the shape, described by number (2). 4) The electron spin. The principal quantum number can be any positive integer and this number corresponds to the shell. Principal number 1 corresponds to shell 1, etc. Shell 1 only supports one spherical orbital. Shell 2 supports a shperical orbital and a lobe-shaped orbital. Shell 3 supports a spherical orbital, a lobe-shaped orbital and a kind of cloves-shaped orbital. Shell 4 supports a spherical orbital, a lobe-shaped orbital, a kind of cloves-shaped orbital and an even more complex form of orbitals. Shell 5 .... Now comes into play a lot of geometry. A sphere is symmetric about every plane which goes through its center and spheres of different orientation cannot be distinguished from each other, if they are totally featureless. So, for spherical orbitals, only one exists. So, each shell has precisely one spherical orbital. The lobe-shaped orbital can have three orientations. If you fix one of them, then two perpendicular orbitals can be constructed, which are independent. So, three lobe-shaped orbitals can exist inside a single shell. For the cloves-shaped orbital there can be 5 different orientations, and for the even more complex one, there can be 7 different orientations. A spherical orbital is called s-orbital. A lobe-shaped orbital is called p-orbital. A cloves-shaped orbital is called d-orbital. The complex shaped orbital, allowed by shell 4 is called f-orbital. In shell 5 there can also be a g-orbital, which is amazingly complex. Now, summarizing: Shell 1: s Shell 2: s + 3p Shell 3: s + 3p + 5d Shell 4: s + 3p + 5d + 7f Shell 5: s + 3p + 5d + 7f + 9g Inside an orbital, there can be at most two electrons of opposite spin. This is due to the Pauli exclusion principle. Google this for more info. Based on the above info, the shells can accomodate at max the following number of electrons: 1: 2 2: 8 3: 18 4: 32 5: 50 Now, let's look at a real atom, the element sodium: Sodium is element 11, so there are 11 electrons. Two electrons go in shell 1, 8 electrons go in shell 2 and 1 electron goes in shell 3. First s orbitals are filled, then p orbitals. So, we have configuration 1s2 2s2 2p6 3s1 In fact, 2p6 usually is written as 2px2 2py2 2pz2, indicating that there are three p-orbitals, perpendicular to each other, each filled with two electrons. Now for calcium, which is element number 20. The first two electrons go in shell 1. The next 8 electrons go in shell 2. The next 8 electrons after that go in shell 3. And now we have a 'problem'. The next two electrons do not go in shell 3's d orbital, but in shell 4's s-orbital. So for calcium we have: Ca: 1s2 2s2 2p6 3s2 3p6 4s2 When we go to element 21 (scandium), then the next electron is added to the d-orbital of shell 3: Sc: 1s2 2s2 2p6 3s2 3p6 3d1 4s2 This is where the term 'transition metal' is coming from. While going from scandium to zinc, the 3d orbitals are filled and after zinc, the 4p orbitals are filled. There are two anomalies in the transition series, but for the time being it is best to skip that detail . Why is the 4s orbital preferred over the 3d orbitals? This is something which only can be understood when quantummechanically, the energy levels in the atoms are computed and it appears that first filling the 4s orbitals is more favorable energetically. With the next row of transition metals, a similar thing happens. First the 5s orbital is filled for Rb and Sr, and then the 4d orbitals are fileld for Y to Cd, after Cd the 5p orbitals are filled. So, here we see another strange thing. The 5p orbitals are preferred over the 4f orbitals. The last 6 elements of row 5 fill all three 5p orbitals. In row 6, first the 6s orbital is filled for Cs and Ba, then the 4f orbitals are filled (these are the lanthanide elements), then the 5d orbitals are filled (these are the third row of transition elements) and finally, the 6p orbitals are filled, ending at Rn. In row 7 again something like this happens. First the 7s orbital is filled, then the 5f orbitals and then it stops... we have entered the region where there are no more stable elements. In theory, here we would have completion of 6d orbitals and then completion of 7p orbitals. The g-orbitals are purely hypothetical, because these would only come up in a very long 8-th row of the periodic table. With this theory in mind, you can now perfectly understand the shape of the periodic table. This is really beautiful. From underlying quantummechanics and orbital theory, the periodic table can be understood very nicely. I hope that digesting this long post is not too much for you .

Having a good understanding of orbitals allows you to predict chemical properties of compounds, the spatial geometry of compounds and it also is very helpful in understanding (and predicting) the formation of complexes. A forum like this is not a really good place to explain all this (that would result in a post of many many pages), but I suggest you to grab the book of Linus Pauling. It contains a good explanation of basic quantum mechanics and application of this on orbitals. Once you grasp that concept, you also can understand much more about orbital hybridization, which helps very much in understanding many (organic) reactions.

As long as it is not yet dry, you can use white spirit, but of course the fabric must withstand the white spirit. Dried paint is very hard to remove and then I think it is cheaper to buy new clothes.

Well, a word of caution must be given at this place, however. High concentration H2O2 is not explosive, but even the slightest contamination can make it decompose almost explosively. When decomposition starts somewhere in the very concentrated liquid, then the heat evolved at that place causes rapid decomposition in the surrounding liquid, which in turn produces heat, etc. So, in practice, having H2O2 around with concentrations above let's say 35% by weight is quite dangerous. Accidental contamination with a piece of hair, some dust or a small amount of some transition metal-salt may result in nasty and violent decomposition. Even 30% H2O2 can be quite nasty with this.

CO-ligands prefer low-oxidation state metal centers. This is unlikely to form a complex with Co(2+). O-ligands prefer high-oxidation state metal centers, think of Mn(VII) or Cr(VI) in the permanganate ion and chromate ion. I(-) is intermediate, it indeed forms a complex with Co(2+). I've tried this and Co(2+) forms a deep green complex with I(-), but it requires some heating and a really high concentration of iodide. As for redox reactions and for acids, there also is a table for ligands. They can be ordered from affinity for low-oxidation state metal centers to very high oxidation metal centers. With this table at hand, you can make reasonable predictions on which ligands form a complex with a certain metal. This table, however, is not as predictive as a table for redox reactions or for acids. Hard lewis bases prefer hard lewis acids and soft lewis bases prefer soft lewis acids. The ligands are the lewis bases and the metal centers are the lewis acids. Metals can be classified as hard, intermediate or soft lewis acids and so can ligands be classified. By using these tables you can predict to some extent, whether a complex is stable or not. If you google for the four words hard lewis base complex then you'll find a lot of documents, with explanations about this stuff. A nice one is the following: http://www.meta-synthesis.com/webbook/43_hsab/HSAB.html

I expect barium benzoate to have a low solubility. If copper benzoate has a low solubility, then the barium salt almost certainly also has low solubility. So, I would expect the precipitate to be barium benzoate, contaminated with mainly unreacted benzoic acid and unreacted barium carbonate. The problem of this synthesis is that both reactants are not soluble in water very well and the expected product also is not soluble in water very well. Having a really good mix therefore is difficult. ===================================== You could try the following to see whether barium benzoate really has low solubility: Dissolve a pinch of BaCO3/BaS in a small amount of conc. HCl and evaporate to dryness. This will leave reasonably pure BaCl2 with some BaSO4 and S contamination. Dissolve the solid and let turbidity settle. The clear liquid above it is an almost pure solution of BaCl2. Add the clear liquid to a solution of sodium benzoate. This probably will show a thick white precipitate of barium benzoate.

Just to prove that perchloric acid is not very reactive when below 72%, try the following for comparison (YT probably has the chems): Add a pinch of calcium hypochlorite or 1 ml of hypochlorite bleach to some conc. HCl: You'll see formation of chlorine gas. Note the stench. This reaction is immediate. Do this outside and do this with no larger amounts, than mentioned above. Add a pinch of KClO3 to conc. HCl: The liquid will turn yellow and you get a deep yellow gas. This is a mix of ClO2 and Cl2. Quite dangerous, but in the small amount used here, this reaction in fact is quite fun. It demonstrates that chlorate is less reactive than hypochlorite. ClO2 is not nearly as toxic as Cl2, but it is dangerous. It may spontaneously explode, but if this experiment is performed with a clean test tube and the gas is allowed to simply diffuse into the air, then you do not need to worry about explosion. Finally, add a pinch of KClO4 to conc. HCl. Nothing happens. Bring the solution to a boil. The KClO4 dissolves and still nothing happens! Hardly any Cl2 gas, only the stench of HCl vapor. I even boiled 60% HClO4 with solid KI and even then only a tiny amount of the iodide is oxidized to I2, really remarkable. So, from these three experiments you can see that perchlorate in aqueous solution is as energetic as a dead dog and even NaCl is more interesting in aqueous solution than KClO4. Only at very high concentration of acid (72+% of HClO4 or at highly elevated temperatures, red hot, the perchlorate ion shows its real nature). What jdurg mentions about perchloric acid vapor, soaked into wood and other organics, yes that is a real risk. This, however, will not occur with 60% solutions. The vapor pressure of HClO4 is very low. It forms an azeotrope at 72% which boils at 190 C or something like that.

This can perfectly be explained. A very common contamination of barium carbonate (and also of strontium carbonate) is barium sulfide (or strontium sulfide). This contamination can be up to 6% by weight. So, what you actually received is a mix of BaCO3 and BaS. When you add this to nitric acid, then the BaCO3 decomposes as follows: BaCO3 + 2HNO3 --> Ba(2+) + 2NO3(-) + H2O + CO2 (this does not surprise you ). The BaS does the following: BaS + 2HNO3 ---> Ba(2+) + 2NO3(-) + H2S (smell!!) Part of the H2S is oxidized by HNO3 and both sulphur and sulfate are formed: 3H2S + 2HNO3 ---> 3S + 4H2O + 2NO 3H2S + 8HNO3 ---> 3H2SO4 + 4H2O + 8NO The sulphur precipitates as a very fine white or very pale yellow precipitate. The H2SO4 forms the highly insoluble BaSO4 with Ba(2+) in solution. With HCl you also get H2S, but this is not oxidized. However, H2S is also oxidized by oxygen from the air and this explains why even in HCl you get a very small amount of precipitate. So, if you want to make Ba(NO3)2 from the barium carbonate, you dissolve it in excess amount of dilute HNO3 and let the precipitate settle. What remains is a very pure solution of Ba(NO3)2 in excess dilute HNO3. By decanting the clear liquid and heating this, the dry Ba(NO3)2 can be obtained. If you want really pure Ba(NO3)2, then add a few drops of 10% H2O2 and swirl, after the initial fizzling stops, but before you let the white precipitate settle. This destroys any small remains of dissolved H2S and converts this to sulphur and water. This H2O2 will be destroyed on heating to dryness. Any remains of H2S will lead to sulphurous contamination of your Ba(NO3)2, making it impossible to make a clear solution with this, which is not nice.

It is remarkable how the availability of chems varies over different countries in the world. Where I live, HCl and HNO3 are fairly easy to find, H2SO4 is more difficult already. What I understand is that HNO3 is really difficult to obtain in the USA. So, it is quite well understandable that there are places where HCl is hard to find. For me, for instance, obtaining KClO4 is very hard, while I2 is not that difficult. In the USA it is the other way around, because of fear of meth-making. Over here, laws are VERY strict on pyrotechnics, so a chem as KClO4, which hardly has any uses besides pyro, is hard to obtain. Congratulations . In fact, not only the quantity counts, but also quality. But the quality of your posts also is worth a congratulation .

If the power supply is disconnected, then the production of gas stops immediately, both at the anode and at the cathode. If the explanation were formation of H2, due to some acidic compounds formed, then the production of H2 would continue for at least several seconds or minutes, because this acidic stuff would be spread around throughout the liquid. In fact, I believe that the liquid is more alkaline instead of acidic, because of the fact that flocculent Mg(OH)2 is formed as well (not immediately, but after a few minutes, when hydroxide formed at the cathode has moved towards the anode region as well). A similar problem I have with the heat formation. If heat is the cause of formation of H2, then I also would expect the formation of H2 to continue for a while after disconnecting the power supply. Any heat, produced, will take some time before it has spread sufficiently through the liquid.

I did an experiment in which I electroysed a plain table salt (NaCl) solution with different metal electrodes. Among the experiments there was an experiment with magnesium ribbon electrodes (both anode and cathode). When a sufficiently high voltage is applied, then a gas is evolved at the cathode (hydrogen gas) and the anode dissolves slowly and falls apart (formation of MgCl2). But, I also get a gas at the anode (approximately 1/3 of the amount at the cathode) and the really funny thing is, this gas is hydrogen! I collected gas from both electrodes and both can be confirmed to be hydrogen gas (they burn with kind of whoosh sound). The result of this experiment really startles me. I never expected the formation of H2-gas at the anode. I simply expected to dissolve the magnesium metal to form some MgCl2.

YT, isn't that a little overreacted ? I have 60% HClO4 and I'm using it fairly comfortably. Nice stuff to play with, but unfortunately it is quite expensive. No, HClO4 is not that bad, in fact even less worse than HNO3 of the same concentration. Have a look at this: http://www.gfschemicals.com/technicallibrary/perchloricacid.pdf This link takes away a lot of the myth around HClO4. I own some 60% HClO4 and in fact it is favorite for me, because it is totally unreactive (other than being a strong acid). It does not form any complexes with metal ions and it does not act as an oxidizer, other than the simple H(+) ions do. This makes the acid very valuable for me, because I do a lot of experiments in coordination chemistry and then sometimes sulphuric acid and certainly hydrochloric acid are a real pain, because of complex formation. Nitric acid also is quite problematic, because of its oxidizing properties at higher concentrations. What remains is HCl04. HClO4 only becomes really dangerous, in combination with other strong oxidizing acids and at very high concentrations (more than 80% or something like that).

My best experience is concentrated HCl, cheap and simple. It works well, because of its strong acid content, combined with the good complexing capabilities of chloride ion. Sulphuric acid does not nearly work as well. Concentrated nitric acid also works great in many cases, but why use such a precious chem, while dirt cheap and easily available HCl also does the job. If nothing of these works, then you a mix of HCl, HNO3 and some HClO4 certainly will work, but only as a last resource, it uses up very precious chems.

Xeluc, you are very right with your explanation about spectator ions. They just are there, looking to all the interesting things the Cl(-) and Cu(2+) ions are doing . The reaction indeed works with CuCl2 only, but as this is a scarce resource you can better use some NaCl/CuCl2. The latter also works and NaCl is cheap and available without limits. From my experiments I have the impression that the reaction rate is proportional to [Cu(2+)]*[Cl(-)]. So, having a low concentration of Cu(2+) and a very high concentration of Cl(-) may be as effective as a medium concentration of both, that it why I suggested you to use a very concentrated solution of NaCl with some solid CuCl2 added. I encourage you to experiment a little with both chems and see what is optimal. Beware, this reaction also is quite exothermic.

This will be quite hard in the dry way. A very tedious and long way would be to go over the wet way, dissolving it in acid, purifying a calcium salt by fractional crystallizing, precipitating CaCO3 and then heating again. A better and cheaper way probably is to go to a drugstore and buy some pure calcium carbonate and heat that.