woelen

The following reaction occurs, when nitric acid and hydrochloric acid are mixed. This also explains, why aqua regia needs a 1 : 3 molar ratio of HNO3 and HCl. HNO3 + 3HCl --> ONCl + Cl2 + 2H2O The precise reaction is much more complex, the equation above only is a net equation. If you heat a mix of HNO3 and HCl (or KNO3 and HCl), then you'll see that the liquid becomes yellow, or even orange. That yellow/orange color is the color of NOCl. At high concentrations of HCl this is stable. At lower concentrations of HCl it hydrolyses: ONCl + H2O <--> HNO2 + HCl

It might be that in deep space such species exist for a longer time. In the very very dilute high vacuum space, where atoms and ions are very sparse, and where there is no matter to interact with, such ions may exist for a longer time, but on any earthly condition they only exist for milliseconds at best. No, there is no ionic species or whatever species of He or Ne which can be kept in a bottle, not even for minutes. For Ar, I believe there is some known compound, which exists at a few degrees K and can be kept under those conditions in an inert frozen matrix.

The ionic compounds you are referring to only exist as transient species in very dilute gaseous form. By applying high voltages and sparks on a very low pressure mix of helium, neon or argon, people have succeeded to make trace amounts of ions of these gases, which however, only have a fleetingly short period of existence and react immediately with each other and with the environment, forming the normal mono-atomic gas again.

Thanks for making these subforums. Now it is up to the members to fill them with interesting content .

In both cases, you use an emitter follower. For the voltage regulation, you use a fixed resistor (e.g. 120 Ohm) between vref and GND of the LM317 and a variable resistor between the emitter and the vref pin for the LM317. The load them is placed between the emitter and GND. The base is connected to the output pin of the LM317 and the collector to the power supply rail. For the current regulation, you use the same setup, but now without a resistor between the emitter and the vref pin. The load itself must be placed between the emitter and the vref pin, not between the emitter and GND. For a decent current, now you have to use a quite small resistor between vref and GND. For 1 A you need 1.2 Ohm, for 5 A you need 0.24 Ohm.

Keep in mind, I'm not talking about a current limiting thing, I'm talking about a true current source in my previous post. Current limiting is very useful, when you have configured your device as a voltage source. What I described in my previous post is a current source, where the current drawn from the supply is independent from the load.

It does not matter. In aqueous chemistry, nearly always you can exchange NaOH and KOH without problem. Only take into account that the same number of moles needs to be used, so for a 1 : 1 mole ratio of NaOH and KOH, you need a little more KOH than NaOH.

Making a current source with an LM317 can be done, but you are limited to 1A. An LM 317 is intended as a voltage controller. Without external power transistor you can go up to 1A. With a powerful power transistor and a good power supply you can get much more. But, then you can get voltage control. You can also have ccurent control. You do that by placing a resistor between the VR pin and ground. The current then is 1.2/R and can be draw between the VR and VOUT pins (current flowing from VOUT to VR, the cathode hence being the VR-pin). For electrolysis experiments I prefer the simple setup with the resistors and a 12 V DC PSU source with a large elco for ripple voltage reduction. It is more robust and virtually fool-proof. If you really want to use an LM317 for current control, then use a power transistor, otherwise you are severely limited. A 2n3055 is OK. Also use a huge cooling block and good heat-conducting paste if you use a 2n3055. Remember, the 2n3055 is not shortcircuit-protected and can be blown out fairly easily. The LM317 is shortcircuit protected and is not blown up immediately. If you use the LM317 in the standard way, then you have a voltage source and not a current source. Voltage sources are much less suitable for electrolysis than current sources, as explained before already.

Neutron stars are very heavy and here the gravitational pull assures that the neutrons stay together. That is the dominant force for neutron stars, while gravitation does not play a role in nuclei, not even the largest ones.

Just summarizing it all: Try to approach a current source as much as possible. That gives much more consistent results with your electrolysis. In order to approach a current source, it is MUCH better to use the 2.8A 12 volts line than the 3.8A 5 volts line. Take for granted the lower maximum amperage. In fact, 2.8 A is quite a lot already for electrolysis and if you want to use that current, then you need a GOOD anode, otherwise it will pulverise in minutes. Why is 12 V better than 5 V? Suppose you do an electrolysis. The current as function of voltage, applied over the electrolysis cell is highly non-linear. Suppose for a certain electrolyte, you have the following function I(V), current as function of voltage: I(0.5) = 0.0 A I(1.0) = 0.0 A I(1.5) = 0.0 A I(2.0) = 0.0 A I(2.5) = 0.1 A I(2.8) = 0.4 A I(2.9) = 1.0 A I(3.0) = 2.0 A I(3.1) = 3.5 A I(3.2) = <almost like short circuit> A characteristic like I sketched above is very common. Now you also understand, why having a voltage source has such inconsistent results. If you adjust your source at 3 V, then the current depends very sensitively on the applied voltage. A little fluctuation in your output voltage, due to imperfections in your circuitry gives a huge fluctuation in your current. This makes it very hard to obtain consistent results. If you use a current source of 1 A, then the voltage will be 2.9 V. You'll also see that in the range from 0.1A to 3.5A there only is around 0.5 V voltage change. Now, suppose you have a 5V power supply and you want to select a certain current as function of resistance, then you need to determine the voltage over the cell. Over the entire range, it may fluctuate between 2.5 V and 3.1 V. The voltage over the resistor is 5 - 3.1 to 5 - 2.5 volts, hence it is between 1.9 and 2.5 volts in this example. You see that the variation is quite strong in relative sense (over 25%). So, selecting a certain current is not that easy. Now, if you use a 12 V power supply, then the voltage over the resistor is between 8.9 and 9.5 volts. Now, this voltage is constant withing 6%. The absolute variation is the same, but the relative variation is smaller. If you use a resistor to control the current, then you have quite an accurate (more than enough of accuracy) for electrolysis. If the characteristics of your electrolyte change somewhat, due to heating, change of chemical composition, etc, then with a voltage source you'll observe a huge change of current, while with a current source, you'll only notice a small change of voltage and if you use a 12 V power supply, then your current hardly will change, because the total voltage drop over the resistor network hardly changes. ---------------------------------------------------------- Now the question, why not take a single 10 Ohm resistor in series with 12 Volts. In the power supply scheme, I posted, a 10 Watt resistor is used, which is connected to the case and this is sufficiently cooled. For the other ten resistors I suggest to buy lower rated 5 Watt resistors (these are a lot cheaper). If you use a single 10 Ohm resistor and you have a short circuit, then the power dissipation will be V*V/R = 12*12/10 = 14.4 Watt. This may burn out a resistor, rated for 5 Watt. ------------------------------------------------------------ Parallel circuits. Why 10 pieces of 10 Ohm resistors? With suitable combinations of series and parallel circuits all resistances, mentioned in my previous long post can be reached. As an example, I show you how to compute the resistance of the network RR//RRR//RR. RR is a series of 2 resistors, having a total resistance of 20 Ohms (series --> addition of resistances). RRR is a series of 3 resistors, having a total resistance of 30 Ohms. The total resistance of this network is 20//30//20 How do you evaluate this? 20//30//20 = ((20*30)/50)//20 = (600/50)//20 = 12//20 = (12*20)/32 = 240/32 = 7.5 Ohm Like with addition and multiplication, the order of evaluation of // operators does not matter: (a//b)//c = a//(b//c) = (a//c)//b. How do you build a network RR//RRR//RR? You take 7 resistors. Two of them you connect in series, three of them you connect in series and the final two are connected in series. This yields 3 series resistors. These three series resistors in turn are placed in parallel. ------------------------------------------------------------------------- Finally, this theory only holds when the used power supply allows the selected current to be delivered. From a flimsy cell-phone charger you'll never obtain 2.5 A or so, but a PC power supply certainly can deliver that kind of currents.

The solution, I have proposed, gives you the opportunity to use any current between let's say 100 mA and 2.5 A. It also has the advantage of current control, instead of voltage control. This is a great added advantage. Besides that, it also is a VERY cheap and very simple solution if you already have a working PSU from an old PC. The basic idea behind my R-networks is to allow a safe and reliable way of controlling current from a 12 V power supply (which every PSU from a PC has). The reason that you have current control is that 12 V is much more than 3 V, which is common in electrolysis. Of course, you can buy a regulated and adjustable power supply, but that is more expensive. My idea is the idea of Raivo in his last post, but worked out in more detail with a lot of different preset currents and the most important difference being the fact that I use 12 V instead of 5 V. With 5 V the current regulation is much worse, because the electrolysis voltage is quite large, compared to the 5V of the power supply.

I would say, it is somewhat inbetween. The sample I have is quite pure basic copper carbonate (see link, two links above) and this is blue/green, I would call this color cyan. Compare the color of basic copper carbonate with the color of copper sulfate or copper nitrate. Then you see that the latter two are really blue.

It will last, provided your diodes are sufficiently large. So, you need to take good power diodes. Suppose your electrolysis takes 5A maximum, then take some headroom, make 10 A of that for headroom, so buy your diodes, such that they can stand 10A of current. I have been thinking about this and an even better approach would be to buy a bunch of 10 Ohm power resistors and just 2 silicon power diodes. With that setup you have very high flexibility at low cost. Buy 10 resistors, each 10 Ohm and each allowing between 5 and 10 Watts of power dissipation. Now you can implement a reasonable approximation of a current source with your resistors. Electrolysis voltages are around 3 volts. Some cells require 4 volts, others less than 3 volts, but 3 volts is a fairly average voltage. Fluctuation around that is small. Now some theory. Because the voltage of your voltage source is 12 Volts, and the voltage of your cell is around 3 Volts +/- 1 volt, the voltage accross your resistor network is 9 volts +/- 1 volt. So, the current remains constant within let's say 11 to 12% of your selected current. Selecting currents can be done with resistor networks. Remember: Never put a single 10 Ohm resistor in series with an electrolysis cell. That may destroy your resistor with the rating I have provided you! The current you select for electrolysis is 9 / Reff. You select Reff with your resistors by means of series and parrallel circuiting of resistors. I'll try to explain to you how this can be done. I introduce a mathematical operator // with the property x//y = (x*y)/(x+y). Examples: 2//3 = 6/5 = 1.2; 3//3 = 9/6 = 1.5 Now, I also need to introduce a notation for networks. I use notation RRR for a series of three resistors (30 Ohm), RRRRR for a series of 5 resistors (50 Ohm). I use notation RRR//RRRR for a series of three resistors, and a series of four resistors, with both series put in parallel. The resistance of such a network equals 30 // 40, which is appr. 17 Ohms. You never may have a single R in one branch of your parallel network. That can lead to destruction of that resistor. So don't use networks like R or R//R or RR//R. Now, some useful Reff values: RRRRR --> 50 Ohm RRR --> 30 Ohm RR --> 20 Ohm RRR//RRR --> 15 Ohm RR//RRR --> 12 Ohm RR//RR --> 10 Ohm RR//RR//RRRR --> 8 Ohm RR//RR//RRR --> 7.5 Ohm RR//RR//RR --> 6.7 Ohm RR//RR//RR//RR --> 5 Ohm RR//RR//RR//RR//RR --> 4 Ohm With electrolysis experiments, you'll most frequently use between 10 Ohm and 20 Ohm networks. For lower resistance you need a good and thick graphite anode, otherwise it will be pulverized in minutes, so going with networks, lower than 10 Ohm should only be done with high quality heavy electrodes. Now, what can you do with the diodes? These can be used for fine-adjustments. Suppose you want to electrolyse bromide, to make bromate and you see some bubbling at the anode. Then oxygen is formed and you know that the voltage is too high. You can decrease your current in that way by selecting a network with higher resistance, but you can also put a single or even two diodes in series with your resistor network. In fact you make a new network ReffD or ReffDD, with Reff your original resistor network. Never ever put a diode in parallal to your network, e.g. RR//RR//D or RR//RR//DD WILL destroy your diodes and most likely also your power supply. What is perfectly safe is a network like this: RRD//RR or RRDD//RR, as long as in a single branch there are at least two resistors, you are safe. It may seem a little awkward when you start with this, but when you have some experience, then you'll get a quick feeling for this. When your solutions are very dilute, then the current source approximation does not work anymore, then the voltage accross the electrolysis electrodes will rise to much higher values than 3 Volts and in that case you have a low current and high voltage and you mostly produce oxygen at the anode. For easy working, make nice plastic boxes, one for each resistor, with heavy duty connectors for both sides of the resistor and thick short wires (e.g. loudspeaker cables) with contra-connectors at each end, such that you easily can plug your networks together. I'm going to make such a thing as well, that is why I showed you the page for the PC power supply project. I have an old power supply left and with $20 I can buy all resistors, connectors and plastic boxes to make it robust and easy to use. I already have an LM317 circuit, but for electrolysis I have discovered that current control is much better, and then the resistor networks are superior.

Looking at your setup, I expect you made basic copper carbonate. I have a picture of this stuff on my website. Compare it with the color of your precipitate. http://woelen.scheikunde.net/science/chem/compounds/copper_carbonate_basic.html If it is purely blue, then you probably have copper hydroxide, but in the presence of bicarbonate I would expect the basic carbonate. ----------------------------------------------------- Many metal ions, especially many transition metals, but also bismuth, lead and antimony, have a strong tendency to hydrolyse. The main mechanism is that the metal ion has water molecules coordinated around it and that these water molecules loose H(+) ions. For copper: [Cu(H2O)4](2+) <--> [Cu(H2O)3(OH)](+) + H(+) For some salts this goes even further, with a second or a third water molecule loosing H(+). When such solutions are evaporated, they tend to loose acid and a basic salts remains in solution, which finally precipitates. When the counter ion itself already is basic, such as carbonate, then the hydrolysis reaction is favored even more. Stating it very simplistically, when carbonate and copper (II) ions are brought together, then the copper (II) ions tend to hydrolyse more strongly and then there almost only is [Cu(H2O)3(OH)](+), which with carbonate gives rise to formation of precipitates like [Cu(OH)]2CO3.xH2O on precipitation. On heating, the water is lost and the carbonate and hydroxide are retained. This explanation is somewhat simple (reality is more complex), but it certainly explains the basic idea behind the formation of basic carbonates. Because of the alkalinity of carbonate, most transition metals only form basic carbonates. The exceptions are FeCO3, MnCO3 and CrCO3. All other ions form basic carbonates, and some ions do not form carbonates at all, such as Fe(3+), Cr(3+). With these, hydrolysis is so strong, that in the presence of carbonate they simply precipitate as Fe(OH)3 and Cr(OH)3.

Many battery chargers are not simply voltage sources. They contain quite a lot of intelligent sensing electronics, which determines what voltage the battery to be charged has and how much current has to be emitted. When no battery is present, then probably no current is applied at all. Have a look at this: http://www.hw.cz/data_ic/lm317.pdf It is the LM317 datasheet, which also contains some circuits you can build around this chip. If you are not confident enough on building your own electronics, then another fairly inexpensive alternative is the simple PC power supply, with some wiring. http://web2.murraystate.edu/andy.batts/ps/powersupply.htm You can adjust voltage by buying a set of power diodes. Each diode takes 0.6 volts. If you put them in series, then you can make many voltages in steps of 0.6 volts. If you also can get two germanium diodes, then you can also adjust voltage in steps of 0.2 volts. That should be perfectly OK and provide you good flexibility in your electrolysis experiments.

Xeluc, you are quite close with your description: tH4T'5 k3Wl M4N But, just to get serious. This kind of persons is a real threat for the hobby chemistry. When accidents happen, when things go wrong, then companies may become even more careful when selling chemicals. Last summer a Dutch boy blew off his hands with acetone peroxide and since then we have a discussion on whether H2O2 should be forbidden or not for the general public. Fortunately the discussion fades away and things do not change, but when such things happen too often, then indeed common chemicals may become much harder to obtain.

Now I first need to make new bromine. I used up my last small amount of bromine in an experiment with selenium. It is good that all bromine is gone again. It is nasty stuff on storage, as jdurg stated already. If I make bromine again, then I'll try to see what happens if I add it to molten sodium metal. Cutting sodium metal under a thin layer of bromine does not seem a very wise thing to me. Being with my nose over all these fumes and having the risk that the sodium/bromine mix bites at me... no, not this time . Right now, I've had a lot of fun with just 1 ml of bromine . My final experiment with the selenium also was very nice. Adding selenium to bromine results in a violent reaction (while sulphur only reacts very slowly in the course of a day). The reaction product is a rust-like solid powder, SeBr4. With a little heating, driving off all excess bromine, a nice dry powder of SeBr4 can be obtained and then it really looks like rust, brown with an orange hue. When SeBr4 is added to water, then it dissolves, leaving a colorless solution with H2SeO3 and HBr in it. When some Na2SO3 is added, again amorphous red selenium is produced. This is not the way to make red selenium, but it is very interesting to have made a strange compound like SeBr4 and seeing this as a pure compound. It really is a pity that bromine is so bad on storage, otherwise I would make much more. You only need NaBr, electrolysis and a little acid. ----------------------------------------------------------------- Yet, a nice suggestion, which does not require isolation of liquid bromine. Make a solution of NaBr or KBr and add some dilute hydrochloric acid. Add some bleach, such that the liquid becomes deep red and red vapors are above the liquid. Carefully tap a small amount of red phosphorous into the red vapor. The phosphorous ignites. @Jdurg: A similar experiment can be done with chlorine gas. Take a very small pinch of your red P (just 10 mg or so, not more) and tap this in a bottle with some chlorine gas in it. Quite spectacular. The chlorine does not need to be pure. This experiment is also on my website: http://woelen.scheikunde.net/science/chem/exps/P+Br2+Cl2/index.html .

That's interesting. I only thought there was a single crystalline allotrope. The crystalline piece I have is a very nice sample, but I also have some very fine black powder, which I did not bother about putting on my website. This powder was sold to me as high-purity silicon (99.999%) to be used for chemical experiments and as a solid state reductor. It fairly easily dissolves in NaOH-solution, giving hydrogen gas and a colorless solution. Is that stuff the amorphous form?

nitrogen-triiodide monoammine Ammonia adducts are named, using the word "ammine". Amine (with a single m) denotes a -NH2 group in organics.

Don't mess around with what you call ammonium triiodide. First, that is NOT the name of that compound, ammonium triiodide is as explosive as a dead dog. You made a different compound. Second, please no further discussion about that XXX stuff you made. K3wls are reading this as well and we should not give them noughty and dangerous ideas. If a k3wl blows off his hands (or more), then this forum may cease to exist if it becomes clear that that info is from here.

You cannot make anhydrous ferric chloride from the hydrous salt. This is the case for MANY hydrated salts, most notably those of iron (III), aluminium and chromium (III), but also for almost all nitrates. All these salts decompose on heating, loosing acid and leaving behind an oxide, hydroxide or basic salt (which is mixed hydroxide/oxide and the original ion). A few chlorides can be obtained in the anhydrous state from their hydrated salts, e.g. CuCl2 and CoCl2. This problem with dehydration also is the reason, why the stuff you buy at electronics stores for etching PCB's is not FeCl3, but FeCl3.6H2O. Making anhydrous FeCl3 can only be done by heating the hydrated salt carefully in a stream of dry HCl or by burning iron in dry Cl2-gas. Another nice example of such an elusive anhydrous salt is Cu(NO3)2. From aqueous solution you get Cu(NO3)2.3H2O, a nice blue crystalline solid (I actually have some of that salt lying around and it looks nice). When this is heated, HNO3 is formed and CuO remains. The HNO3 is decomposed further at the high temperatures involved, giving water, oxygen and nitrogen oxides. People, however, have succeeded in making anhydrous Cu(NO3)2, which is a volatile blue compound, it is not a salt. It was prepared by dissolving Cu-metal in liquid NO2 and carefully allowing the NO2 to evaporate. This gives a solvated salt Cu(NO3)2.2NO2, which on heating at 80 degrees C slowly looses its NO2 without decomposition of the Cu(NO3)2. You see, not something to be done at home.

No, I don't think so. In the meantime I have done some studies on this subject in my book "Chemistry of the Elements" by Earnshaw and Greenwood. This book states that sodium metal can be stored in liquid bromine indefinitely, as long as no moisture is present. I was surprised to read about this, but my observations perfectly confirm this. A very similar thing exists for many other reactions. Another example on my website is the reaction between iodine and magnesium and as far as I remember, Jdurg's experiment with sodium and iodine also required some water to start the reaction (or heating). My experiment: http://woelen.scheikunde.net/science/chem/exps/mg+iodine/index.html Jdurgs experiment: http://www.chemicalforums.com/index.php?op=articles;id=22 In my experiment I also said, that the piece of sodium was freshly cut. I cut a small piece from a larger piece, rinsed it in low boiling ligroin (40 - 60 degrees C), took out the piece of sodium, allowing the ligroin to evaporate (just takes seconds) and then put it in the bromine. So, it was not heavily oxidized (at least not at one side).