woelen

The common form of barium chlorate, when made from aqueous solution is Ba(ClO3)2.H2O. This is the crystalline form. However, on storage, this stuff effloresces, which means that it slowly gives off its water and the crystal structure then is destroyed. So, you get powdered Ba(ClO3)2, which has lost its molecule of water of crystallization. This is a very common effect. It also is known for washing soda, Na2CO3.10H2O. On storage, this material (nice transparent crystals) also becomes powdery, due to loss of water and accompanied destruction of the crystal lattice.

I also did the result with paper tissue also, and that result in something which does not burn better than plain tissue without the nitration: http://woelen.scheikunde.net/science/chem/exps/expshow.cgi?index=281 Apparently, the choice of wadding is important. I use pure snow-white cotton wadding, but I'm not sure how common that is in other countries. I have seen this stuff over here, in Germany and other countries nearby, but maybe in other parts of the world it is less common?

Looks somewhat messy, but I'll give it a try. What strikes me is the short nitration time. It only needs to be in the mix for 2 minutes instead of 20 minutes. I assume this is because there hardly is any water in the paste, while in my liquid with 65% HNO3 there still is some water.

I would not want to work with bare hands with H3PO4, unless it is really dilute. But even with just a 10% solution, longer exposure is bad for your skin. A 1-minute exposure is no problem, but if you are scrubbing for several minutes, then you will hurt yourself. Phosphoric acid is corrosive. The low concentration of this acid in coke has a good reason .

Could you provide a little more detail on what to do with KNO3 instead of HNO3? That would be a nice addition to my webpage. Is it simply adding some KNO3 to H2SO4 and then adding the piece of wadding, or is it more involved? If you provide guidelines for that, then I could try them and add it to my webpage. It makes the experiment accessible to a larger group of people. Another variation may be the use of NaNO3, which is very common at the place where I live, as chili-salpeter.

As I already said in my previous post, my description was somewhat sloppy. If you really want to understand, than you have to resort to complex analysis. This link may be helpful for you: http://clem.mscd.edu/~talmanl/PDFs/APCalculus/OnAnIntegral.pdf Important notice is that ln(|x|) comes in handy, when you do real analysis, but in reality, one can also write ln(x) + C as the integral of 1/x. Also for negative x, this works out OK. E.g. integration of 1/x from -2 to -1 gives ln(-1) - ln(-2) = ln(1) + pi*i - ln(2) - pi*i = -ln(2). In real analysis, you merely take another function (with another C) as the integral of 1/x, in order to get rid of the need to deal with pi*i. These terms with pi*i cancel out. Integration over the singularity indeed is not possible, as Matt pointed out. It can be done, when you choose a path along the singularity over the complex plane, but in that case, you have to take into account which way you take and which branch of the logarithm is used (the log is not uniquely defined, it is a multibranched function: Log(z) = log(z) + 2*k*pi*i). Sometimes, however, integrals through a singularity are regarded as limits of two simple real integrals. In your example: from -1 to -eps, and from +eps to +2 and then you take the limit of the sum of both integrals, while you let eps go to zero. This yields log(2). But I must admit, that this is a sloppy way of working, and not at all strictly correct. The correct way of understanding this is to use complex analysis and integrate along a curve, not through the singularity, but around the singularity. In practical situations, also for real analysis, you do not integrate through this singularity, you either stay at the negative side only, and then it is easiest to work with ln(-x) as primitive, or you stay at the positive side only, and then the primitive of choice is ln(x). Summarizing: In real analysis, you take ln(x) + Cp as primitive for positive x, and you take ln(x) + Cn as primitive for negative x, with Cn and Cp different numbers. Cp usually is taken equal to 0, and Cp is taken equal to pi*i. This is "hidden" by writing ln(|x|) for the entire domain, except 0.

Do these really fall in the class of organometal compounds? I regard them merely as salts of the -O-R group, (e.g. -O-CH2-CH3) for ethoxide. The sodium salt can be made by adding sodium to an alcohol and the other metal alkoxides can be derived from that. These ethoxides do not contain direct metal-C bonds.

This is a real classical experiment, but it is always fun to do it. http://woelen.scheikunde.net/science/chem/exps/nitrocotton/index.html This makes nitrated cellulose wadding, which IIRC also is called guncotton. It burns really fast with a brilliant orange flame. Have fun, but please be careful!

How much phosphoric acid is present in coke? I though it is just in the 0.01 ... 0.1% range. I bet, that if I take pure phosphoric acid, diluted to this, that it hardly works. There must be something else in the coke as well, but it could be that the small amount of acid helps or catalyses the action of other active ingredients.

I have some yttrium metal, which is supposed to be quite pure. I dissolved some of this metal in dilute hydrochloric acid of known high quality (no metallic impurities in the acid). The yttrium dissolves quickly with a lot of hydrogen formation, and the resulting solution is light yellow. I suspected an impurity in the yttrium of iron. That would give a yellow color in the dilute acid, due to formation of yellow iron (III) / chloride complex. But on addition of some reagent for iron (III), yellow prussiate of potash, the liquid does not become blue. With iron (III) present, such that it colors the liquid visibly yellow, a thick dark blue precipitate would be formed. Instead, I obtain a light yellow/white precipitate. So, the impurity definitely is not iron. Now I wonder, what is the color of yttrium ions? I cannot find a reliable source. The salt YCl3.6H2O is said to be white, but could the ions in a solution of HCl be yellow? I severely doubt that, I expected the solution to become colorless. So, I still suspect an impurity, but then I ask, what is a common impurity for yttrium metal, which gives a yellow color on dissolving in acid?

I'm a mad scientist and mad scientists make mad avatars. This one actually is from a real mad science experiment, I conducted appr. 18 months ago.

There are two classes of metallic organics. One class is nothing special and can be regarded as salts of organic ions. Examples of this are plentyful, e.g. sodium acetate, calcium citrate, magnesium lactate, copper benzoate, etc. The metal is not really bonded, it is present in conjunction with a negative ion. A little bit closer to a real organic metal compound are the more or less covalently bonded compounds, like "ferric oxalate", basic beryllium acetate, copper citrate. These are complexes of common anions and metal ions. The bond is not purely ionic anymore. Still, this is not specific to metal organics. This also exists in the inorganic domain. Real metal organics are those, where the metal is directly bonded to a carbon atom. A well known example are the Grignard reagents, where Mg is bonded to an aliphatic carbon atom. These compounds also exist for other metals, most notably transition metals. This kind of compounds can be quite stable, but they usually are EXTREMELY air-sensitive, either due to humidity in the air, due to oxygen in the air, or both. This is an area of research, where very unusual compounds are found and the research of these compounds also gives new insights in properties of metals. Unfortunately, experimenting with these organometal compounds is beyond the reach of the hobby-chemist. These are very hard to make and require specialized air-free equipment. The first real organometallic compound, which was recognized as such probably was nickel carbonyl, Ni(CO)4, but it might be that there are even earlier other examples, but I'm not aware of these.

I have 80% acetic acid, and that is not flammable at all. So, even absorbing a relatively small amount of water may render the stuff completely inflammable. Also, the stuff I have does not freeze at all, not even near 0 C.

One can best write: ∫(1/x)dx = log(|x|) + C. This always gives correct results and no surprises. For negative x, this results in log(-x) and for positive x this results in log(x). Also, (improper) integration through the singularity then gives the correct result. E.g. integration from -1 to +2 yields log(|2|) - log(|-1|) = log(2) - log(1) = log(2). This is the expected result, because the infinite areas from -1 to 0 and from 0 to -1 cancel precisely. Integration of 1/x from -2 to -1 yields log(|-1|) - log(|-2|) = -log(2). This also is correct, because the curve is below the x-axis, so we expect a negative value for the integral. This is a somewhat sloppy description. It can be described in a much more formal way, using limits, but I think that this answer is sufficient to understand the problem.

I have the imrpression that the OP has erroneously described the problem. I have been looking into this problem and as it is formulated, it only can be tackled by means of iterative numerical methods for root finding and then an answer, like raptor has given pops up, but I suspect this is not the answer the OP is expecting.

Currently, truly simulated environments are built in computer systems. Of course, these do not simulate at the level of the chemistry of the environment, these are more behavioral studies, but they nevertheless are attempts to see effects of evolution of "life" over long periods of time. In a computer system, time can go much faster than in a real world.

I have to work a lot with younger people (20 .. 25 years old). The good side is that they are very independent, and willing to arrange their own affairs. This is a good thing in consultancy. But the bad side sometimes is that they think that all achievements of today are as common as that the sun rises every morning. Grandpa is visiting his grand child Mike. Mike wants to show a new computer program to his grandpa. He turns on the computer and then .... the computer does not start up. Mike is totally upset: "The PC does not work!!!! How is this possible?". Grandpa answers: "Mike, everytime when you start your PC, I think it is a miracle that this thing works again.". This is the attitude which I notice with many younger people. Yes, it is a miracle that all these things are working, all over again...

Just to keep things simple. The graph is not an exponential, because it is about a chemical reaction and its rate, as Tartaglia already pointed out. Reaction rates usually are polynomial order in the concentration of reactants. The reaction between thiosulfate and acid is quite complicated and its kinetics are not that easy. The graph is inverse quadratic for very low concentration and apparently it is reciprocal at higher concentration. The relation will be something like t = A/x² for low concentration x (then A is approximately 0.15). For higher concentration it will be something like t = B/x. The transition between both is smooth and the best approximation probably is to use a function t = C/xⁿ, with n somewhere between 1 and 2, and C suitably chosen. But keep in mind, such an expression only is based on mathematics and does not reflect the properties of the chemical/physical process, underlying this set of data points. Hence, such an expression also is useful for predicting times in the interval of [0.02 .... 0.1] mol/l, maybe a little outside this interval, but not too much.

I would like to hear about the people who have read some of Rebiu's posts.

With this method you will almost take all iodine out of the seaweed, so in terms of efficiency, it is good. But the absolute amount is very small. From such a small quantity of seaweed you only will obtain a few tens of mg, maybe 100 mg but certainly not more. That depends on the percentage of iodine in the seaweed, but I think that will be at most of few 0.1%'s by mass of iodine.

Your quote mentions adding bromide. That seems pointless to me. It still does not release any iodine. I can imagine adding bromine, but adding bromide is not of any use. What I understood is that seaweed is heated (burnt and calcined) such that only inorganic matter remains. This tuff is dissolved in water and all insoluble matter is filtered from the solution. Then, one can bubble chlorine through the solution. This releases the iodine. But adding too much chlorine is not good, that results in formation of iodate. I personally would first determine how much iodide is present and use that for the determination of the amount of oxidizer for releasing all iodine.

Inorganic chemistry also is a good place for this. Making phosphoric acid from burning red P actually does work quite well, but you should assure a very good supply of air. If insufficient air is used, then lower oxides of phosphorous are produced and you also get phosporous acid, H3PO3. The reaction is somewhat cumbersome though. Collecting the smoke and leading it through water is a lot of a hassle. Also, you should be very careful not to burn too much red P at once. It is VERY flammable and burns very violently, with a LOT of smoke. Be careful! The reaction of P2O5 (better: P4O10) with water is very violent, but the smoke of burning red P is consisting of small particles, spread over a large volume, and then the voilent reaction is not that problematic. But on the other hand, isn't it a pity to burn valuable red P to make such a dull compound like H3PO4? Can't you buy chalk-remover for coffee machines? Many brands simply are dilute H3PO4. Over here in the Netherlands, a common chalk remover for coffee machines simply is 8% H3PO4, the rest is water. By boiling away the water, one could make fairly pure H3PO4. Go and look at hardware stores for chalk removers, maybe the situation is the same as for us and you also can buy dilute H3PO4 somewhere.

As I remember from posts a few months ago, you came from/attended a church in your home-town? Then try and follow Severians advice, that is a really good place to get in contact with people, and no one will force you precisely to believe what they believe. I also am in a similar boat like Severian, and we do all kinds of nice things as well, such as going out in the city with small groups of people, going to a restaurant, or just visit each other at home, discussing things and doing games (I most like the board games like Risk and so on ). Besides that, I want to add, if you are going to college, aren't there any student's clubs? At the place where I studied, there were student clubs of all kinds of people. General clubs, just for being together, clubs, specialized in doing games with each other, all kinds of religious clubs (name it and it was there). You do not have to achieve all kinds of things in those clubs, they differ from sports clubs. Being there is what they want, being together with other people. Many people come from a remote town and hardly know anyone, so these clubs are very good meeting places. In my time, these clubs also did lots of fun things together and during those years I met a lot of people and many became friends. And yes, also many of those friends, I do not have contact with them anymore, but a few friends remained, even some, who live 2500 km from us and we meet almost every year. Do not fear to loose contacts over time. You can't keep up with all of them, life will learn you anyway, especially if you all move away, finding jobs all over the country. Enjoy the contacts you create now, and the really good ones remain. Finally, I want to tell you that every person is here on earth with a purpose. You also have a purpose, please don't forget that. You should know better .

This whole discussion reminds me of my own studies. I studied electrical engineering and I had to learn a lot about the physics and the chemistry of making integrated circuits. It even went so far that I had to be able to compute 3D potential fields in semiconductor devices and I also had to know all kinds of chemistry of silicon in more detail than I wanted at that time. Right now, I'm not using that knowledge anymore, but yet, I think it is good that I had to study it. It gives basic understanding and now I do understand much better how electronic circuits work, even if these are built with discrete components. I have some foundation of knowledge and understanding. In a similar way I'm quite sure that people who study medicine also need to have a foundation of knowledge and understanding. This understanding may help them to take wise decision about situations, which they are not trained for. This understanding also helps being creative and make you better at your work.

This test looks quite hefty to me, at least for people who have to study it all over again. I have acquired my knowledge slowly over the years, so it is hard for me to estimate the time needed to study this, but a book, covering all these subjects will be a thick one with 1000+ pages. That will give you an indication of the time needed to study it all. It will be quite some effort. The list of things you need to study, however, is quite detailed. So, if you have a textbook (or a few textbooks), which cover all these things and you have studied them well, and understand them, then you could step into the test with confidence. The only part of the test-description which is not sufficiently detailed is the experimental chemistry part. Common apparatus, used in basic chemistry are the following: - test tubes, beakers, erlenmeyers, drippers, etc. for mixing and observing reactants - heating apparatus (bunsen burner, sand bath, oil bath) - distillation equipment for separation and purification of chemicals, used mainly in syntheses. Quantitative equipment: burettes for titration. You definitely have to understand how titrations work and how stoichiometry of reactions can be determined by means of titration, or how titration for known reactions can be used to measure concentration. Some examples: iodometric titration, acid/base indicator used in titration. Read something about volumetric glassware and how it is used. Read about measurement of heat of reactions: calorimetry. Also read about the problems in this and how systematic errors easily creep into measurements with this, e.g. by capacity of heat of the apparatus itself. Then of course there are the more advanced things, such as gas chromatography, resonance spectra, UV-VIS measurements, absorption spectra. These, however, require costly and large equipment, which the average mad scientist doesn't have in his basement ;-). I hope this explanation helped a little. But be prepared, if you have to start from scratch without significant chemistry knowledge, then a lot of effort is needed.