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Yep, Na2SO3 is even better than H2O2, good that this one is mentioned as well. It is called sodium sulfite. At shops, which sell stuff for making wine, you can also buy sodium metabisulfite, Na2S2O5, or potassium metabisulfite, K2S2O5. These also are suitable, but they are somewhat more smelly. This smell, however, does not stick to everything and quickly disappears. At low concentration, I do not find that smell (SO2) unpleasant. At high concentration it becomes pungent though.

Inside a single molecule of methane there is approximately 1/6.02E23 mol of carbon and there is approximately 4/6.02E23 mol of hydrogen. But I have serious doubt that this answers your question. If I am right, then please rephrase your question and be more clear.

In engineering and electronics, the use of complex numbers also is very widespread. In electronics, you can model linear capacitors and resistors as frequency dependent linear resistors: resistor with resistor R: [math]Z_R = R[/math] capacitor with capacitance C: [math]Z_C = 1/i \omega C[/math] inductor with inductance L: [math]Z_L = i \omega L[/math] Here Z is the generalized impedance (complex resistance) and [math]\omega[/math] is the frequency of the electrical signal across the component (2*pi*f). You can do calculations with these complex imedances, as if you were computing currents and voltages in ordinary resistor networks. These computations then yield output voltages or currents, which e.g. can be written as [math]V_{out} = H(\omega) V_{in}[/math] Here, H(ω) is some freqency dependent complex number. The absolute value of H equals the gain (or attenuation) of the signal at freqency ω, while the argument of H(ω) equals the phase shift between output and input at frequency ω. Another important application of complex numbers in engineering is in the determination of the poles (and zeroes) of a linear system. The location of poles in the complex plane gives a lot of information about a system's stability and usefulness. Poles in the right half plane (with positive real part) indicate that the system is unstable.

The funny thing about simulating such large systems is that the solutions are chaotic. The qualitative behavior of the systems (e.g. spectral properties of positions of function of time, shape of orbits) can be reproduced very well with such simulation software, even by using simple first order approximations for the next time step. The precise quantitative behavior as function of time (exact position of all objects at a certain point in time) is amazingly hard to reproduce. Luckily, when one wants to study the behavior of such systems, then the qualitative behavior is sufficient. The fact that precise quantitative behavior is very hard to reproduce tells something about the real system. If such a system exists in reality, then the tiniest perturbation by means of an external (not modelled) force also screws up the precise quantitative behavior, while hardly affecting the qualitative behavior. What I told here is not relevant for simulating relatively simple systems with a few tens of objects (e.g. our solar system with the planets and the heaviest asteroids and several moons), but it becomes important for simulating systems of 1000's or even 1000000's of objects. I did experiments with this (I have written a program, using DASSL and later DASPK) and have found that this behavior in fact is quite common for many natural phenomena.

Why doesn't it exist? That is an interesting question. There definitely is a pattern among all transition metals. All first row transition metals are most stable in lower oxidation states, while the second and third row transition metals are more stable in their higher oxidation states. E.g. V - Nb - Ta. Vanadium has extensive aqueous chemistry for +2, +3, +4 and +5 oxidation state, +5 is actually a fairly strong oxidizer. Nb only has fleeting existance in the +4 oxidation state in aqueous chemistry, +5 is by far the most stable oxidation state, and Ta is more or less limited to the +5 oxidation state, and only under very specific anhydrous conditions it exists in lower oxidation states. The effect is more marked with Cr - Mo - W. Cr is strongly oxidizing in the +6 oxidation state, while for Mo and W this is the most stable oxidation state. With Mn - Tc - Re this effect is even stronger. Permanganate is one of the strongest common oxidizers, while pertechnate and perrhenate are not oxidizing. I do have some ammonium perrhenate, and that chemical is rather inert. Only strong reductors, like sodium borohydride and metallic zinc in acid (nascent hydrogen) are sufficiently strong reductors to reduce the ReO4(-) ion to a lower oxidation state. The triad Fe - Ru - Os is the turn-over point. Fe(VIII) would be so strongly oxidizing, that it cannot exist anymore in normal environments. Molecules or ions with this would be so unstable that they fall apart through internal oxidation/reduction. Even iron(VI) already is very strongly oxidizing. On the other hand, RuO4 is only a moderately strong oxidizer and the same is true for OsO4. Ru(VI) is quite stable in alkaline environment, Ru(VII) is quite stable in acidic environments. For the further transition metals, a similar thing is true, but less pronounced. Usually the first row allows oxidation states +2 (and sometimes +3) and the higher rows then allow +3 and sometimes +4. Platinum also allows +6, but that is an extremely strong oxidizer, e.g. PtF6 is capable of oxidizing oxygen. I think that really understanding this pattern requires quantum mechanics. What I did is just describing observations, QM computations, however, reveal these properties.

You could dump 3% H2O2 over it. H2O2 and bleach react with each other to make oxygen, water and salt. But if you spilled a lot of bleach, then you may need quite a lot of H2O2. I would just try to clean up everything and then after cleanup wipe all of the floor with 3% H2O2 to get rid of the smell.

I have the book "Chemistry of the Elements" and this books says that no compounds of iron(VIII) have been isolated and confirmed by independent groups. So, any compounds like FeO4 or K2FeO5 are very doubtful. I think that no iron(VIII) compounds exist, and certainly not at room temperature. ================================================== For layout: Use [ ce]...[ /ce] sections: [ ce]K_{2}FeO_{4}[ /ce] ---> [ce]K_{2}FeO_{4}[/ce] [ ce]CrO_{4}^{2-}[ /ce] ---> [ce]CrO_{4}^{2-}[/ce] You need to skip the spaces after the [ of course. See also here: http://www.scienceforums.net/forum/showthread.php?t=10188

I don't have crystals, but a very fine powder. A picture of that is not interesting, it just is a fluffy off-white powder. Today I did some further tests with it, and despite the off-white color (most likely due to tiny traces of I2 in it), it is remarkably pure. I cannot find any traces of iodide in it (tested with Pb(2+), no yellow precipitate of PbI2), it also seems to be free of periodate, because I do not obtain a precipitate with copper (II) sulfate. Copper (II) gives a green/yellow precipitate with periodate, but I could not obtain such a precipitate with my sample.

YT, no ClO2 is formed in this reaction. What is formed is Cl2, H2O and Na-acetate. 2CH3COOH + NaClO + NaCl ---> Cl2 + 2CH3COONa + H2O @konishewah: The Cl2 in turn reacts with Fe. That reaction you should write down yourself. The reaction between Cl2 and Fe is independent from the first reaction. You cannot write a single reaction equation with all chems in it in an unambiguous way, the solution space of all possible reaction equations is 2-dimensional, so it is best to write down two independent equations, which together describe what happens.

Bleach is not only NaClO, but also NaCl. With vinegar, this results in formation of HOCl, which in turn reacts with chloride ions and excess vinegar to form Cl2, water, and sodium acetate. The Cl2 in turn reacts with the iron to give iron (III) chloride. With this info you should be able to write reaction equations. Show us some of your work and we will help you further.

I have done the synthesis of KIO3 from KI. I dissolved approximately 10 grams of KI in a little over 10 ml of water, giving a total of appr. 20 ml of liquid. First I did the electrolysis with 2 A current, using a platinum anode and a titanium cathode. After 5 hours, the yield was only 1 gram. Pathetic! The liquid become really hot, almost boiling. The next day, I added approximately 50 mg of K2Cr2O7, dissolved in 1 ml of water, and continued the experiment. With this K2Cr2O7 added, the result was amazing. After 2 hours, I already had a nice thick crop of KIO3. The liquid again becomes VERY hot. I have had this running for 4 hours total, and that has given me a nice amount of KIO3 (appr. 7 grams). I rinsed 5 times with 15 ml of cold distilled water, the 5th time, the rinsing liquid is not yellow anymore. The powder is very light beige, almost white, but not 100% purely white. I did a test with 1 M H2SO4. A small spatula full, added to 1 M H2SO4 does not result in a brown color, the solution becomes colorless. This means that the iodate is free of iodide, and that even without recrystallization. Just a 5 times rinse! I am very content with this synth and I'll do further experiments with this KIO3, most notably the Briggs-Rausch reaction, mentioned by Encipher.

For preparation of some acids, H3PO4 is the preferred acid. HBr and HI cannot be made in a pure state from bromide/iodide salts and H2SO4. With bromide, some Br2 and SO2 come over as well, but the majority still is HBr. With iodide, the method of adding H2SO4 is totally useless. The concentrated acid oxidizes the iodide to iodine, and the acid itself is converted to a dirty mix of SO2, S and even H2S. With H3PO4, which is non-oxidizing, even at high concentration, one can obtain quite pure HBr and even HI.

I think that you are right and that the amide (peptide) group is a naming convention, specific for the -C(=O)-N(H)- group. With other acids, such groups exist, but IIRC then they are not called amide.

BH3 This is a non-existent molecule. There exists a molecule B2H6, which has a very special structure. I write it as H2B(-H-)2BH2. The left two H-atoms and right two H-atoms are bonded to the B-atom using a normal bond, sharing two electrons (one from the B-atom and one from the H-atom). The two H-atoms in the middle, however, are bonded by means of the so-called 2e3c bond, 2 electron 3 centered. For each of the H-atoms, one electron of a B-atom, and its own electron are shared by both B-atoms and the H-atom. Unfortunately I cannot draw over here, but the following is an attempt to draw the structure of B2H6. It does not resemble C2H6 at all, it uses a TOTALLY different bonding structure. The =|= stands for a type of bond, with 4 electrons in total (kind of double bond), but with the H-atoms associated with it as well. Google for "2 electron 3 centered" bond for more details. H H \ H / B =|= B / H \ H H Have a look at this page: http://en.wikipedia.org/wiki/Diborane PCl3 P has 5 electrons in its outer shell. Three of them are shared with Cl, 2 remain as free electron pair. The two electrons repel the bonds to the Cl-atoms, so the molecule will have the three Cl-atoms on one side, with the Cl-atoms spanning a triangle with equal sides. I expect this molecule to have a distorted tetrahedral shape. SiH4 Si has 4 electrons in its outer shell. All of them are single bonded to H-atoms. This molecule will be symmetric as much as possible and will form a tetrahedron, with the Si-atom inside, and the H-atoms on the vertexes. H2S S has 6 electrons in its outer shell. Two of them are used for bonding with H, the other two free electron pairs repel the bonds with the H-atoms. This molecule will have the shape of the letter V, but with a larger H-S-H angle. The two free electron pairs will be above and below the plane, in which the H-S-H atoms are.

It can only be ignored in aqueous solution. In non-aqueous solution the water, formed in such reactions can be quite disturbing. As an example, I tried to make sodium nitronate from nitromethane and sodium hydroxide in methanol solvent. This did not work, because the nitronate ion is destroyed quickly by the water, formed in this reaction. Only the use of sodium methoxide as base does the job without problem.

Ryan, I have to disagree with your last example of reaction with magnesium hydroxide and hydrochloric acid. It is the hydroxide and oxonium ions which do react, the Mg(2+) and Cl(-) ions are just spectator ions. OH(-) + H3O(+) --> 2H2O (+ a lot of heat) When you add hydrochloric acid to solid Mg(OH)2, then you obtain Mg(2+) and Cl(-) in solution and some water is formed. On evaporation you obtain MgCl2.

There already is a thread on this subject. http://www.scienceforums.net/forum/showthread.php?t=15636

I indeed intend to use my Pt-electrodes for making KIO3. My previous sample is brown/yellow and has a fairly strong smell of iodine. I hope to get a better result this time. I'm not sure whether this is due to the use of carbon rods, but using platinum electrodes always is very neat, the result is really clean.

I found the solution. The old kernel module for the nvidia driver was loaded before the new module was loaded. This is built-in default behavior of Ubuntu. I remove the kernel module again, at the end of the /etc/init.d/rc.local file. The effect of this is, that when the X-server starts for the first time, it sees that no kernel module for the nvidia video adapter is loaded, and then it loads the kernel module, but now for the new driver (the load library path is correct at that late point in the boot/startup process). I don't think it is the 'correct' solution for this problem (first loading an old kernel module, then unloading it again), but I don't care. It works and that is what counts For the rest, Ubuntu 6.10 is quite a nice distro, it looks good and certainly does not look over-engineered like other Linux distros, such as Fedora Core. But again, I still think Linux is not suitable for the masses as a desktop system. I really like Linux as a server platform (I have worked with it in that way for many years, and my website also is running on Linux, on a tiny device, called NSLU2, only somewhat bigger than two cigarette boxes), but for desktop work, again I needed many hours of tweaking before it works. I needed to compile a kernel module for my network adapter (Attansic L1 10/100/1000 Mbit/s), I had lots of trouble getting my video adapter working, and now I'm still struggling with my audio device (an Analog Devices AD1986 High-Definition integrated audio chipset). It works flawlessly under Windows XP, but it works with terrible clicking and popping under Ubuntu.

All together I think these developments will not be good for the home chemist. I purchase many chems from the UK, simply because I cannot buy them over here. If the Uk obtains the same rules as the Netherlands for selling chemicals, then things will become more difficult, not only for people from the UK, but also for me.

YT, I wish it were true what you are saying here, but unfortunately things are not that good. Only 1/6 becomes iodate, the rest becomes iodide: 3 I2 + 6 NaOH ---> 5 NaI + NaIO3 + 3 H2O To my opinion it is cheaper to use KI as a starting point, I2 also is an expensive reagent. Using KIO3 also results in formation of less soluble KIO3, which separates from solution, if a concentrated solution of KI is used as a starting point. I plan to do such an electrolysis with KI next weekend. I'll dissolve 20 grams of KI in as little as possible of water (I expect this will be 30 ml or so) and then proceed as with my KBr synth. I'll let you know how things are going if I've done the experiment. In the past I already made some KIO3, but that material is beige/brown, and I want a nice white chemical, so I'll retry again, now with more experience from my KBrO3 synthesis.

Yes, unfortunately Gilded is right. But things are changing in the Netherlands. In Amsterdam, last week, 33 sex-houses were closed. Most cities are banning the so-called "tippel-zones", places where people can sell themselves for quick sex. The Netherlands are becoming less liberal, and things are changing quickly. The rules for pyrotechnics also will change. All EU-countries will grow towards each other and have the same rules. For the Netherlands this means that rules will be relaxed somewhat, but for the UK and Sweden things will be tightened. For many other countries things will not change very much.

I did the experiment with KBrO3/malonic acid/H2SO4, without manganese (II) sulfate added. To my surprise it did not react at all. The solution remained colorless and no gas was evolved. Next, I heated the mix. When it was near boiling, a very faint red color could be observed, not like the color of bromine, but more like a purplish red. But it was very faint. Also, there was some slow bubbling. Finally, while the liquid was still hot, I added a pinch of Ce(SO4)2.4H2O. At once, I obtained a fantastic oscillating reaction. Vigorous bubbling and really fast pulsing. It pulsed bright yellow two times per second or so, for several tens of seconds. It was really amazing to see this, especially because the pulsations were so fast and so clear. So, with cerium (IV) you get beautiful oscillation, better than I expected.

To my opinion it is quite insane to use more than a few grams of flash powder. I do not advice against making such powders, but I really urge anyone who intends to do so to carefully study the chemistry and the risks of it. You would not be the first one to loose limbs (or more). Another very nasty risk is the risk of self-ignition on storage. Some powders (most notably sulphur-based) can ignite after weeks of storage. Also certain metal-compositions (most notably with Mg) can ignite spontaneously, especially if they become a little humid.

The green material indeed most likely is copper carbonate, or some basic copper carbonate/sulfate. The black material probably is CuO.