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NH4NO3 produces N2O in itself already, when heated carefully: NH4NO3 ---> N2O + 2H2O A small amount of water can be added to make the reaction somewhat safer. It very easily goes out of control, because it needs heating. Did they in the show heat the ammonium nitrate? The N2O, produced in this experiment is not very pure though. It also contains N2, NO and NO2. Because of the latter two impurities, the N2O produced this way is absolutely unsuitable for inhalation or for use in cream.

How are these holes oriented? I assume that the carbon atoms with 4 holes have all holes oriented towards the end of a tetrahedron. For double-bonded molecules, like H2C=CH2, the 3D structure is not tetrahedral, but it is flat. With these 5-hole carbons, you probably can model molecules like these. Also, can you model molecules like HC≡CH, with a triple bond?

Atheist, thanks very much for your lengthy post. I learn a lot of it. I also perfectly understand that such materials cannot be covered in detail in a single post and using the keywords you supplied I already found out something. I am familiar with the Maxwell equations of electromagnetism, and had never seen your formulation using the d'Alembert's operator, but it is a very powerful representation. Most striking is that both electrical fields and magnetic fields and their interaction now are included in a single equation, which is easier to memorize than Maxwell's equations. I like easy to memorize equations . I have one minor question: I was thinking that □ = (i/c*∂/∂t, ∂/∂x, ∂/∂y, ∂/∂z) (here i is the imaginary number i*i = -1, c is the speed of light in vacuum). Your □ seems to me to be the inner product □•□ (also written as □²). Its value is -(1/c²)*∂²/∂t² + ∂²/∂x² + ∂²/∂y² + ∂²/∂z² Am I right? With these I can derive Maxwell's equations from the formula □²A=j (and not □A=j). I still have to look into the equations for gravity, they look much more complicated to me due to the tensor algebra, but at least I perfectly understand this formulation for electromagnetism now and have a starting point for gravity.

FriedChicken, all of your steps look good and I have the idea that you understand this now. The only missing thing indeed was the omission of -sqrt(2) as the tree pointed out. Just as an exercise for you, try other equations of this type from your textbook or exercise book, but I'm quite confident you can solve this kind of things now. One added comment. You can use the formula for solving quadratic equations also directly for fractional coefficients. Just plug in the values of the coefficients a, b and c as I gave in my first post into the formula for quadratic equations and you'll get the answer. But if that is too cumbersome for you, the multiplication with an integer to get rid of fractional coefficients also is OK.

YT, could you try the KOH-toluene mix again? Add a very small amount of water, such that the KOH is just wetted (e.g. with some damp cotton wadding, you touch the KOH and then add it to the toluene). Try what happens if you allow this to stand for a few hours. At the same time, try this with your NaOH and let this stand as well. If there is a significant difference, then there must indeed be something in the NaOH. Now, it might also be some contamination from your test tube, the rubber/cork stopper, or glass container in which you do the experiment. At least you should be able to track down, which chemical or other thing causes the orange color.

I'm not an expert at all in this area, but if I understand it well from Snail's post, every object has a mass equal to Mt = M0 + E/c², where M0 is the "rest mass", which we in real life simply call "mass" and E/c² is the relativistic mass? And is it Mt, which determines curvature of space-time, instead of M0? In practical/real life of course, Mt is very close to M0, but fundamentally there are differences. And then, what forms of energy E contribute to this? Kinetic energy seems different to me, because we also have Mt = M0 / √(1 - v²/c²) from special relativity. A series expansion, first order in v²/c² gives: Mt ≈ M0 * (1 + ½v²/c²) + O((v²/c²)²) This gives Mt ≈ M0 + Ekin/c² when |v| is much smaller than c. This is close to my initial guess, but not the same . There is some rest term, depending on the fourth power of v/c. And how are things going for massless things? This is really surprising to me. And now I'm doing some math on it, it strikes me . If one of the physics experts could point me in some direction (a good site), or even better, elaborate more on this over here, that would be very nice.

I also did the experiment. I added NaOH to toluene, and the solid did not react at all. After 3 hours, there still is colorless toluene and on the bottom there are white granules of NaOH. Next, I added a single drop of water and heated the liquid to 70 C. Now, the NaOH liquefies with the water, and a thin sirupy layer is formed at the bottom. Strong shaking causes formation of small droplets, which quickly settle at the bottom again. The toluene, however, remains colorless and perfectly clear. What sodium hydroxide do you use? I hardly can believe that it is the NaOH which causes this reaction. Are your prills of NaOH white? If so, then they certainly do not contain iron. White NaOH is not very pure, but the only common contaminants are NaCl (remains from the electrolysis production process), Na2CO3, and water. Commercial NaOH may contain as much as 5% of impurities, mostly NaOH and water, and traces of Na2CO3. Other impurities I hardly can imagine, unless added on purpose (such as flakes of Al in some brands of drain opener). Very pure NaOH looks like glass with a matte outside of the granules. So, if your NaOH is white, then I really suspect the toluene is not very pure. The reason that no reaction occurred with your KOH may be because of lower water content, making it totally insoluble. Add a tiny amount of water to your KOH/toluene mix and see what happens. If in that case, you again get orange color, then it really is the toluene, that has impurities.

Your answer is almost correct. You should solve the following equation: (x²-2)(x²-4) = 0 instead of (x-2)(x-4) = 0 Now, show me all steps you take, starting from your original one. So, you first multiply by 4 to get rid of the fractions, then you substitute x² by z, giving you an equation in z. You solve that equation in z, and then you perform a final step to solve the equation in x. Keep in mind that there are 4 different solutions for x. Now you have all steps mentioned. Now you should use them in the right order to solve your problem. So, please write down all steps and let's see if you got the idea. I do not simply post answers here, I really want you to understand how to solve these things...

I assume you know how to solve quadratic equations of the form a*z² + b*z + c = 0 Look at your equation. It can be rewritten as follows: (1/4)*(x²)² - (3/2)*(x²) + 2 = 0 Now take x² = z. I hope this hint helps. Please try to give the solution of the equation over here in this thread. Then I am certainly willing to help you with that solution, or check the correctness of your solution.

Ajb, could you please explain this in more detail? I don't see how light curves space. Of course, light represents a certain amount of energy, propagating through space, but it has no mass. But is it true that energy also contributes to space-time curvature, due to m = E/c²? Or is there another reason for space-time curvature?

The entire idea seems good to me, except the shared userbase. I would prefer to copy all current users with their passwords to the new religion forums, so all currently registered SFN users can login at that site as well. But once the split is made, newcomers who register here are not members at the religion forums and vice versa. If someone really is interested in discussing both science and religion, then they will take the effort to register twice. I think that the religion forums should be an entirely separate site, with its own independent userbase. If this split is not made, then I see no real difference with the current situation. Then you still click the religion link and you get to a forum with religion issues in it. Technically, the solution is very different, from a functional point of view, I hardly see any differences. Having links to each others is OK, and making some advertisement trying to attract people to the other forum also is OK, but please keep them functionally separate. The name theologyforums.net seems OK to me, it is the same type of name as scienceforums.net. Picking yet another domainname different from theologyforums.net or theologyboards.com (you already have reserved two names already) seems a waste of money and resources to me, but of course, I cannot look into your wallet . But on the new forums, make clear that ANY religious person and also agnostic and atheist persons are welcome. I agree with the point of ecoli about the name, but having a clear message on the board will resolve that issue.

Did you pour the toluene over the NaOH, without adding extra water? It seems that there are impurities in the toluene which react with the NaOH. I know of no reaction between NaOH and toluene. I will try with my toluene this evening and report on that. I have "real" toluene, sold as such, and not the stuff from the hardware store. Let's see if this can be reproduced.

You really have to put a lot of effort in making ammonium nitrate detonate. The average starter in home chemistry will not be capable of doing that, so indeed, no need to worry about accidental explosion by simply bending the crystals. If it were really a sensitive explosive, then it definitely would not be sold to farmers as a very common fertilizer.

I think for me it is safer to use perchlorate, instead of BF4(-) or PF6(-). I don't have such fluoro-compounds anyways, but even if I had them, I fear their toxicity much more than the possible explosion risk of perchlorate (I work with just small pinches of chems, for safety, but also for keeping cost low). ================================================================ I did the experiments with colored gasses. First unexpected thing that I observed is that it is important to have a small amount of water present. E.g. when NaCl is added to a notrosyl/sulphuric acid solution, then only HCl is produced. If a small amount of water is present (e.g. wet NaCl), then a nice orange/yellow gas is produced. This gas is ONCl. I did the same experiment by adding NaBr (also wet). This produced a chocolate/brown very dark compound in solution, and a dark brown gas. This gas is very heavy and can be poured out of the test tube and see really nicely see it falling downwards. This is an indication of large molecular weight. I think that the brown compound is ONBr. It definitely is not bromine, because that does not have a chocolate/brown color, that is much more red. With dry NaBr, I get HCl and orange/red/brown fumes of Br2. The color of these are quite different from the color of the chocolate/brown compound. A small whiff of this dark brown gas (done by moving the hand along the open end of the test tube towards my nose) has a weak smell, totally different from the smell of bromine (once you have smelt bromine, you recognize it for ever). Most funny result with NaBr is that even on strong dilution, this dark brown compounds still exists. As a counter test, I did an experiment by adding NaNO2 to a dilute solution of H2SO4 and KBr. This results in formation of the dark brown compound, even in water! Under these conditions, the color is dark brown/green, very different from the color of bromine in water. With NaSCN, a lot of a colorless gas is produced and the liquid starts foaming a lot, and an orange/yellow solid is produced. The liquid itself turns grey/blue and turbid, but that color is transient. In contact with air, the colorless gas turns brown, so it at least contains NO, it may even be pure NO. The orange/yellow solid is insoluble in any aqueous solvent, either acid or base. Only with strong scrubbing, it could be removed from the test tube. Finally, I did experiments, hoping to get blue ON-CN. This gas, however, is not formed. The solid NaCN is covered by a dark brown/black layer and the solid hardly further reacts. There is some formation of gas. I did not dare to take a small whiff of this. When a small amount of water is added, then I just get copious amounts of brown gas, indicating quick decomposition of the nitrosyl to NO + NO2. This brown gas could contain some HCN as well, I have been overly cautious while doing this experiment and did not smell anything.

It might also be that the crystals form an intercalation compound, with many water molecules trapped into the lattice. If the amount of trapped water is relatively large, then the crystal lattice can be deformed, with watermolecules moving from one hole to another hole. Because these watermolecules need some space, the crystal then is permanently deformed and does not snap back, when the force is taken away. It is quite an interesting phenomenon though. I have a bag of 25 kilo of KAS fertilizer and I'll see if I can do something like this with that fertilizer (it also contains chalk, magnesia and other insoluble stuff, so I need some filtering).

Phrased in a more simple way: You can regard the 6 electrons, which would be used for making the three additional bonds, as being distributed over all 6 C-atoms. So, the structure of benzene is that all 6 C-atoms are connected to each other with a single bond, there also is a single bond with the H-atoms, and then 6 electrons remain (one from each C-atom). Each of these 6 electrons contributes to a bond, connecting all 6 C atoms. Plastically, think of each of these 6 electrons, giving each other hands. Because all 6 electrons contribute to all 6 C atoms in the same way, there is no difference in bond length, and all C atoms are equivalent. Schematically, this usually is depicted by means of a ring in the hexagon for the benzene-ring, instead of drawing three double bonds. This delocalization in fact is very common, and many molecules, but also ions, contain delocalized electrons. Another nice example is the nitrate ion, NO3(-). This sometimes is presented as (-)-O-N(=O)2, with two O-atoms bonded to the N with a double bond, and one O-atom bonded with a single bond, and the remaining electron attached to the single-bonded O-atom. This structure is not correct, however. It would make one O-atom very different from the others, but in reality, in NO3(-), all O-atoms are equivalent. The real structure is that of three O-atoms, all bound to the nitrogen atom with a single bond, and then there are 6 electrons remaining, which together are distributed over the entire ion. So, each O-atom has 1/3 of the -1 charge, and each O-atom can be regarded as being bonded for 5/3.

Ryan, good to see you here again. And indeed, as YT said, hopefully your stay in the hospital was not due to a chem experiment! Anyway, I hope to see your contributions again, and also nice new experiments and ideas. EDIT : Your reply to YT just crossed my post

If you carefully bend them and then release them, do they go back to their original shape, or do they keep their bent shape?

I also want to add to Borek's explanation, that some gases also are far from ideal, even at low pressure, due to chemical equilibria. Most common example is 2NO2 <---> N2O4 This equilibrium strongly affects the properties of NO2/N2O4 gas and this gas is far from ideal at any concentration. However, for the majority of the gasses, this is not the case and then at sufficiently low pressure (better: sufficiently low relative partial volume of the molecules) they behave like ideal gasses.

I don't know of any of them, and where I live, pentane also is not something you can buy OTC. Technically, it would be perfectly possible, but I see no reason why one would like to use an expansive chemical like pure pentane for such a thing, while a much cheaper chemical mix like 40 .. 60 ligroin also works.

This is a structural isomer. In one molecule, there are 2 C-atoms between the double bonds, and in the other molecule there are 3 C atoms between the double bonds. So, these molecules really have different structures. Next time, use the [ code][ /code] tags for the drawing and insert data between these tags. Then, the molecules look as follows: O ll / \ l l \_// O ll / \ ll l \__/

Yes, the gas nitrosyl cyanide is blue. There are more blue gases, the better known gas CF3NO also is blue, see http://www.google.com/search?q=trifluoronitrosomethane+blue. And undoubtedly there are more blue gasses, but they will be less common. In my home lab I have made gasses with yellow, brown, red, orange, green and purple color, but up to now no blue gas, so I'm definitely going to try this reaction next weekend.

I also posted this question on a Dutch forum. There I also got the answer of formation of nitrosyl sulfate. They told me, this is a salt, and not a covalent compound. It has NO(+) cations and SO4(2-) anions, or HSO4(-) anions, in the excess H2SO4. Nitrosyl ions are interesting and as I understood, it seems to be possible to make pure nitrosyl compounds from this solution, by adding the appropriate anion. I'm going to experiment with this, by adding NaCl (could make pure ONCl, an orange gas), adding NaClO4 (could make ON-ClO4, a colorless liquid) and NaCN (could make a blue gas, ON-CN!). I'll keep you updated.

I think phosphorous acid would be an easier base for esterification (analogon: nitrite also is easier than nitrate). But you will get another ester. Also keep in mind that phosphorous acid is two-basic, its structure is H-P(=O)(OH)2, and only the H-atoms, attached to the O-atom can be split off.

When you add solid NaNO2 to an acid at room temperature like dilute H2SO4, HCl or HNO3, then the stuff heavily bubbles and you get a big brown plume of NO2. This is well known: NO2(-) + H(+) ---> HNO2 2HNO2 <---> H2O + NO + NO2 If I, however, add NaNO2 to concentrated H2SO4 (98%), then no gas is produced. Even if the stuff is heated gently, hardly any gas is produced and the solid completely dissolves. The liquid becomes colorless. When I add this liquid to water, then it starts bubbling heavily and I get the well-known brown plume of NO and NO2. My question is: What happens with the nitrite ion in sulphuric acid? Why don't I get NO2 + NO? Is some interesting sulphur/nitrogen/oxygen compound formed?