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About D'Nalor

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  1. Anions to test for: S2-, OH-, SO32-, SO42-, NO3-, Cl- and CO32- Cations to test for: Iron (II), Iron (III), Copper (II), Lead (II), Stromium, Calcium, Sodium, Pottassium, Aluminium (III), Barium, Silver (I), Zinc (II) and magnesium. I have no idea how much we'll be allowed, but I should immagine as much as we need. Not sure of the concentrations, but they should be about .1 M. If it's any guide, about 5 drops(not very acurate, I know) is usually enought to get a distinctive reaction. Devarda's alloy... sounds as if it would be usefull. I'm assuming that the reaction would caus
  2. I've got a chemistry practical exam, and in preparation for it, we have to devise tests for certain anions/cations. I need a bit of help for the Nitrate Ion (NO3-. is there any test that you can perform on nitrate to get a distinctive reaction? precipitation reactions are out, any Cations that might cause a precipitate to form would be unavilable in a school laboratory. is there any easy test? otherwise, I might just have to hope it isn't on the exam. I'm hoping to use the acetate ion when possible too so it won't interfere with any other things i'm testing for(because it's soluble with
  3. Heh, that was irritating. I realised I'd forgoten to put up an example. Then, When I put up the example, I realised that That You can't help me If I don't show you what i'm doing wrong. And then I managed to solve every example I could think of, and all the ones in my book. Isn't it typical? You spend ages trying to figure out how to do something, and then It works first time when you show it to someone else. sorry guys...
  4. Hi, Just need a bit of help putting things into conjugate normal form(for those who have forgoten, here's an example: [math](x+y+z)(x+y'+z')(X'+y+z')[/math]) I can manage to put things into the disjunctive n.f. easily, but for some reason am unable to change it to Conj. n.f. Can I just use the duallity principle on the disj. n.f.? I think I can do it by adding things like z*z', but I only seem to be able to get it to the disj. n.f. Yes, I know you're probably sick of me now, asking all these questions and not putting anything else back in to the system, but there's nothing wrong with le
  5. Ok, sorry it's taken me a while to get back to this, but here we go... I've provided a spreadsheet showing why I think the answer is wrong. I have put it in more detail that I probably need to, but I want to know the flaw in my reasoning, and the more detail I provide, the easier it is to find. I haven't managed to find a way to show that you are 2x more likely to find it in the second case. Looking at what swansont said, I would have thought that the main info that changes the situation would be that you have one case less than you began with, which you could basically ignore the existenc
  6. Here is a question I have always wondered how the answer is correct, because I can't manage to make it seem right. A chap at the fair is tring to find a prize that is hidden in one of three caskets. He chooses a casket, whereupon the man running the stall opens one of the other cases, which is empty. He then asks the chap if he wants to stick with the casket he chose originally, or switch to the remaining one. which casket will give the chap the greatest odds of winning? The answer in the back of the books states that if he switches, his chances of winning are doubled. so the chap
  7. Oops, silly me, I got the first part but not the second... I really should have tried just doing that second part as well. It flows quite nicely after that. oh well, at least it was the only one that I couldn't get. thanks for that. nothing else I need to ask at the moment, and it will be unlikely that I'll need to again on this topic for quite a while. Thanks again for your help!
  8. I spoted that you could just differentiate it, the only trouble was that I was trying to get it into the form of the rule(which obviously you can't[although in retrospect i should have noticed that]). I really should have noticed that I could have just done [math] \frac{d}{dx}arsin\left(\frac{x-a}{x+a}\right)=\frac{\frac{1}{x+a}-\frac{x-a}{(x+a)^{2}}}{\sqrt{1-\frac{(x-a)^{2}}{(x+a)^{2}}}} [/math] Which would have been much easier than what I was trying to do... now at least i understand how I was supposed to have done it. There might have been some identity, but It mightn't have been on
  9. Umm... how? only rule I know for this is [math]\int \frac{dx}{\sqrt{a \times a - x \times x}} = \arcsin(\frac{x}{a})+c[/math] This is the only rule For deriving that I can think of, and the teacher has said that it is possible to do it using this rule. but how? (Couldn't figure out how to do powers using system)
  10. Do you think that study is cheating? after all, you are supposed to know the all stuff before the exam. If you need to study, you don't know it well enough, and if you don't know it well enough, you're going to forget it. there isn't really much point in learning the stuff if you just forget it afterwards, so I think study is sort of cheating. maybe it is merely legalised cheating. of course, I don't study myself and get high marks, so maybe that's just me descriminating against dumb people.
  11. Here's an alternative view that I hold. Free will Is basically being able do do what you want to do without any concequences. of course, this would be slightly complicated if we had complete free will, because this would make laws redundant, which could be rather awkward. someone could murder you and your family without paying any penalty(except your friends might take vengance) under complete free will. there are plenty of people who'd do that too(i know that i have occasionally). So, while having no free will would make us zombielike, we can't have an overdose. all we need is a balance
  12. Just finished my exam. Guess what? that question wasn't in there. what a suprise, and waste of time. I think I did well though. there was one hard question that I couldn't get though, and it would be nice if someone could show me how to do it: show that [math]\int \frac{\sqrt{a}}{(x+a)\sqrt{x}} dx = arcsin(\frac{x-a}{x+a})[/math] hence find [math]\int \frac{dx}{(4x+3)\sqrt{x}}[/math] I found most of the other questions very easy, but this one was just ridiculous. help please?
  13. That's fine. It happens. Have an exam on this subject tomorrow, and thanks to you lot helping me with this, I think I should do quite well. It was the only thing that I wasn't quite sure of. Thanks!
  14. heh heh... oops... I did forget about that one didn't I... Yes, that does work. sorry... Heh, I even forgot about the Cap'n's other rule... that's why I still couldn't get it right. I should just sub those other values in to the continued formula. continueing from there... [math] \frac{1}{2}(\int cos(-2x)cos(3x)dx -\int cos(6x)cos(3x)dx) [/math] [math]\frac{1}{2}(\int (\frac{1}{2}(\cos(-5x)-\cos(x)))dx - \int (\frac{1}{2}(\cos(3x)-\cos(9x)))[/math] [math]\frac{1}{4}(\frac{-1}{5}\sin(-5x)-\sin(x)-\frac{1}{3}\sin(3x)+\frac{1}{9}\sin(9x))+c[/math] and I could remove th
  15. Urm... I'm affraid you lost me there. Is there some other rule in there I don't know about? the quoted rule doesn't use sin and cos at the same time. Did you mean to bracket it like this? [math]\int[\sin(2x)\sin(4x)]\cos(3x)dx[/math] That would allow you to make that substitution. Certainly from what you put down, the answer is right. Using the corectly bracketed version, that would make it... [math]\int[cos(-2x)-cos(6x)]\cos(3x)dx[/math] [math]\int cos(-2x)cos(3x)dx -\int cos(6x)cos(3x)dx[/math] ...which doesn't make it too much easier... Are you sure that rule is
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