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Everything posted by hermanntrude

  1. if you're having trouble with it crystallising too soon, just add bit of extra water, and make sure all your glassware is clean and free of cracks and dents
  2. Heavy water is usually D2O. D stands for deuterium, which is an isotope of hydrogen which has a neutron AND a proton in its nucleus (unlike normal hydrogen, sometimes called protium, which has only a proton). There is another, even heavier isotope, with two neutrons and a proton, called tritium. I've never heard of water being made from tritium but i expect it can be done and probably has been. Of course you can also get combinations... things like DOH, TOD etc. Heavy water (D2O) is interesting from several viewpoints, and wikipedia has a lot of info on it. One interesting fact is that ice-cubes made from D2O will sink in normal water.
  3. As a guess i'd say it can, but H2O2 is usually in water and P4 is very hydrophobic, so it might not wet enough. However, I wouldn't reccomend trying it! P4 can be soooo nasty
  4. personally, when I've been put on oxygen in hospital it's always made me feel awesome. Not an objective statement by a long way, but it's certainly true that the body responds differently even to small changes in concentration
  5. I knew a fellow who wrote a thesis in archaeology. it was 4 volumes. yuck.
  6. Congratulations in getting this far, ajb! It's easy from now. Generally they'll ask you to change a couple of minor things and perhaps write an extra paragraph somewhere. Don't fret! In my viva they told me I hadn't written enough about molecular wires and that I should write an extra chapter on them. I told them it wasn't relevant but i'd do it if that was what they wanted for me to get a PhD. They said yes it was. So I wrote a short paragraph and labelled it as a chapter. They passed it. It's all just hoop-jumping. PS my thesis was 89 pages, double spaced. Length being proportional to chance of passing is a fallacy.
  7. The decomposition of H2O2 can be catalysed by many things. Perhaps one component of your buffer solution is acting as a catalyst?
  8. do you squeeze it or does it work simply from the heat of your fingers? if it's heat then I doubt the gas inside is just air, or the liquid just water.
  9. first, figure out how much solvent was used at 100°C to dissolve the acetanilide. Then figure out how much acetanilide that quantity of solvent will dissolve at 25°C, subtract one from the other, then compare that quantity with your 8.3g to find the percent yield.
  10. nope. Doesn't work like that. Gonelli, you'll always get hydrogen but you might get chlorine instead of oxygen or mixed with it, depending on the concentration of chloride ions and the voltage used. You will NEVER get sodium this way. link: http://www.scienceforums.net/forum/showthread.php?t=40177
  11. adianadiadi, this is a science forum. You can't just make a statement without backing it up. Give us some examples... what hybridisation is used? how and when is a pentacoordinate carbon possible?
  12. I teach chemistry at a college in Canada, and I also have family members at high-school and I have noticed (more than once) that advanced science students in Canada are severly understimulated. All I can say is try to have patience, jump through the stupid-hoops they want you to jump through, pay attention to every detail (because sometimes you WILL learn something), and look forward to the days when things get harder. Meanwhile, work on your study skills. I often encounter students who have gone through high-school with little or no effort and really start to struggle when they get to the point where they have to work hard to keep up. Perhaps you could study something you don't find easy, just as an exercise in study-skills?
  13. I would not do it unless at least a few million had had it done and the benefits shown to outweigh the negative effects. I would also be wary about potential issues with reproductivity. Modified species tend to have trouble reproducing.
  14. the only things a liquid's boiling point depend on are what the liquid is and the pressure. the only relevant equation i can think of is the clausius-clapeyron equation, which relates the vapour pressure of a liquid at one temperature to the vapour pressure at another. it can be used to find (for instance) the normal boiling point (the boiling point at 1atm). a liquid's boiling point is when the vapour pressure is equal to the external pressure. the only other information you'd need is the heat of vapourisation.
  15. cant think of any right now. many reactions are catalysed by acid or base... I guess ester formation might be an example... also some reactions are "catalysed" by light.
  16. hydrogen peroxide decomposition. google "elephant toothpaste". you can also see an example in my youtube video at http://www.youtube.com/hermanntrude it's in the video on decomposition reactions.
  17. I've learned something new. I had no idea there were actual examples of hybridisation between orbitals of different principle quantum numbers. Can you give me some links or diagrams? i'd love to see what the hybrid orbitals look like.
  18. in all hybridisation examples ive ever seen, the atomic orbitals which were hybridised all had the same value for n. If that were the case here, we'd need some 2d orbitals. These don't exist so my statement remains valid.
  19. seriously. You come to a science forum and ask us not to be pedantic??? really???
  20. it depends on exactly which kind of percent you're using. the most common percentage used in solutions is weight/volume, which means you'd put 15g in every 100 mL of solution (in other words, dissolve 15g in the minimum quantity possible and then top up to 100mL after it's dissolved) other types are volume/volume (usually for liquids dissolved in liquids), weight/weight (often used in temperature-sensitive situations... volume changes with temperature, whereas weight doesnt) and mole percent (percentage based on the numbers of molecules of solute and solvent). I'd guess you're almost certainly gonna want weight/volume, which means you'd need 150g for every liter of solution (not every liter of water, though... the volume of water will be different to the volume of solution. Just dissolve it then top up to 1L.)
  21. I can see your point, though, and i'm only really trying to simplify... the transition state essentially has 3 normal bonds and two half-bonds, and would have a trigonal bipyramidal shape, but that requires an sp3d hybridisation, but that's not possible in carbon, since it doesn't have the unused d orbitals available. I suspect the answer to your question would become an essay quite easily, given half a chance and a lot more knowledge than I posess
  22. the point is that it's planar, and sp2 hybridisation would explain that. I guess you'd have to assume that both the leaving group and the attacking group are not (entirely) bound at that stage.
  23. In the past i've had yields anywhere between 120%( water in my product making it seem heavier than it is) and 0.5% (actually not a bad yield for that particular reaction, which has literally hundreds of possible by-products. If you're sure about your calculations, that's what you got.
  24. OK here goes: the conversion factor on a simple metric conversion: convert 2.34 mm to m: The conversion factor between mm and m is 1000. There are 1000 mm in a m. We can represent that as an equation: 1000mm = 1m Or we could represent it as a fraction (conversion factor): [math]\frac{1000mm}{1m}[/math] The conversion factor is useful because technically it's equal to 1, so we can multiply any number by it and not change its overall value. We can also turn it upside down and not change its value. So we can multiply our 2.34 mm by our conversion factor and the only thing that will change is the units. the actual value won't have changed, although the number and the unit will. 2.34mm x[math]\frac{1m}{1000mm}[/math] = 0.00234m Notice we had to decide which way up we wanted the conversion factor. the rule for that is "wanted over given" in this case meters over mm. What about converting volumes? It's a bit more confusiong here but let me show you how it goes: convert 1L (1dm^3) to m^3: Now we can use the conversion factor again but we have to be careful either we can use the conversion factor for dm^3 to m^3 if we absolutely sure we know what it is, OR we can go back to basics and use the conversion factor between dm and m. if we do that, though, we must be sure to cube that conversion factor: [math]1dm^3 \times (\frac{1m}{10dm})^3=1dm \times \frac{1m^3}{1000dm^3} = 0.001m^3[/math] To apply this to your original question, the slight problem is that the units are difficult, and not easy to bring back to their original definitions in terms of simple distances, but remember that these are volumes, and the metric prefixes are already applied, so we don't need to get too deep into conversion factors this time. All you need to do is realise that the conversion factor between microliters and milliliters is 1000microliters = 1 milliliter. Then 100 micoliters becomes 0.1 ml, and you're given the information for cc to ml: [math]100\mu L \times \frac{1mL}{1000\mu L} \times \frac{1cc}{1mL} = 0.1cc[/math] Ideally, you'd cancel out the units as you wrote this but I haven't found a way of showing that. Notice that all the units you don't want appear both above a line and below one, so they cancel out, and you're left with a unit you DO want. The math looks awfully long and drawn out but it's actually what you ought to be doing every time you do a conversion from one unit to another, even when the conversion factor is 1 to 1 (like cc and mL).
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