D'Nalor

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 cause ammonia gas, which would make it distinctive, the article didn't state the states of the matter. following a link from the page, there's another test that would work. thanks for that.

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 everything listed in my textbook, and so makes a good carrier ion) Other thing, for the cations, I need to add Hydroxide ions (OH- to the solution (to remove copper, iron, zinc, etc.), but later I need to flame test for sodium, potassium, barium, stromium, etc. basicly, all the ions that actually dissolve in water when they're with hydroxide. my textbook doesn't actually show lots of the tests that you could use, so if there are any easy ones that would work, I'd like to know them.

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...

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 learning, nor with wanting to learn, is there?

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 existence of. after all, 3-1=2 So I'd say the question given by swansont would have the same answer as this question. Base chances.doc

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 should take the remaining casket. This always seems quite odd to me. It seems to work on the principle that at the beggining, he has a 2/3 chance of being wrong. but once one casket is opened, that should reduce the chances of being wrong to 1/2, shouldn't it? that there were originally three shouldn't make any difference, should it? Am I right, or is the book?

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!

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 one that is very common or easy.

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)

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.

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. having said that, I think that there are probably a few too many laws at the moment. Some are very odd. Sticking to just the biblical 10 commandments or "do unto others as you would have done unto you" should be all the law we need. After all, most of them are just sensible and common sense. I know that I'm not really in much of a position to post here, because I know nothing of psycology, but nullnaught did say it was for everybody, and this is sort of a different view compared to the others here.

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?

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!

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 that quater at the front, but that seems unnesscessary at the moment. There we are, I can do the question now. thanks! I looked back over the rule, and, stupid stupid, I forgot that [math]\cos(x)=\cos(-x)[/math] That would be why I thought that it wouldn't work. sorry...

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 right? from [imath] \sin \alpha \sin \beta = \frac{1}{2} (\cos(\alpha-\beta)-\cos(\alpha+\beta)) [/imath] [imath] \sin \beta \sin \alpha = \frac{1}{2} (\cos(\beta-\alpha)-\cos(\beta+\alpha)) [/imath] and as [math]\sin \alpha \sin \beta = \sin \beta \sin \alpha[/math] but I don't think this is right... [math]\frac{1}{2} (\cos(\beta-\alpha)-\cos(\beta+\alpha)) = \frac{1}{2} (\cos(\alpha-\beta)-\cos(\alpha+\beta))[/math] Just so I can show an example of [math]sin(\frac{\pi}{3}) = \frac{\sqrt {3}}{2}[/math] [math]sin(\frac{\pi}{6}) = \frac{1}{2}[/math] [math]sin(\frac{\pi}{2}) = 1[/math] [math]cos(\frac{\pi}{3}) = \frac{1}{2}[/math] [math]cos(\frac{\pi}{6}) = \frac{\sqrt {3}}{2}[/math] [math]cos(\frac{\pi}{2}) = 0[/math] [math]sin(\frac{\pi}{3}) = \frac{\sqrt {3}}{2}[/math] so [math]\sin (\frac{\pi}{3}) \sin (\frac{\pi}{6}) = \frac{\sqrt 3}{2} \times \frac{1}{2} = \frac{\sqrt 3}{4} [/math] which does not equal [math] \cos ((\frac{\pi}{3}) - (\frac{\pi}{6})) - \cos((\frac{\pi}{3}) + (\frac{\pi}{6}))[/math] which equals [math] \cos(\frac{\pi}{6}) - \cos(\frac{\pi}{2}) = \frac{\sqrt {3}}{2} - 0[/math] ending with [math] \frac{\sqrt {3}}{2}[/math] So either I've done something wrong in there, or the rule is wrong. They should equal the same thing, so I think the rule is wrong. Phew, that was a lot of work to type that many equations through the system...

ok then. Formula is to integrate is [math]\int{sin(2x)cos(3x)sin(4x)dx}[/math]

Quick question, how do I integrate this? ~sin(2x)cos(3x)sin(4x)dx I can't figure out how to do it. I've managed to get an answer using cos(nx)=(z^n+z^-n)/2 and sin(nx)=(z^n-z^-n)/2i, but somehow I get the feeling that isn't how I'm supposed to do it... I am familiar with inegration by parts, ~F(x)=~G(u)u' [and another one like the previous that I can't remember how to write], double angle rules, and naturally 1=sin(x)^2+cos(x)^2 I still can't figure out how to do it though, given that we're supposed to be able to do it using only those. I think I remember reading somewhere a rule that I think was sin(a+b)=sin(a)-sin(b), but I don't think it will really help in this one. Please help me. note: ~ represents integration.

!!!!!!!!!!!!! Now THAT is an impressive name!!!! I'm very glad I didn't try to name that myself, that would have been horible. I would be quite intrested in actually knowing how to name something like that. The (E) at the front is something wierd though. Do you have any idea as to what it means? Obviously, the person that designed that program did a good job. probably why it normally costs $2000. Back soon with more evil molecules, that program's going to get a workout...

Right, although noone's even looked at the thing(easter holidays for everyone, I think), I realised that It isn't actaully a very good organic substance. Too many things that don't make it a good example of a good organic molecule. I have revised it, and here it is, made anew.

Alright then. how about this one? It's much more complicated, so I hope you appreciate it. I am actually really interested as to how you'd even begin to go about naming it. Heh, I'd love to write up a whole bunch of these molecules on the board in class and name them, That'd be hilarious. now that i think about it, this is a slightly boring molecule. next time I'll put in some metals as well maybe.

Thanks for clearing that up. how about this one? It's not quite as big or complicated as some of the other ones, but it would be interesting to name anyway. There was an even more complicated one, but I can't manage to draw it in ASCII, so I might try to scan it from my hand drawn ones.

Ok, I've thought some more about these, and now i'm wondering: how do you name these groups? ~ is once again a place holder, and =- is a triple bond CH3-C-CH3 ~~~|| CH2-C-CH3 ~~~| CH=-C-CH-C-=CH ~~~~~| CH2=C=CH-CH-CH=C=CH2 ~~~~~~~~| CH ||| C | That's all the simple ones for now. I'll put in some really complex ones if I can't figure out how to name them from this. I've Given up looking for signs of the previous ones existance, they obviously don't exist at the moment, or aren't on the internet. Please help me name them

I found references to all the molecules... but they were all from this forum, with the exception of 6-isopropylundecane. even that however was actually 4-ethyl-6-isopropylundecane. from just the quick glance down the page, it is used in acrylic paint for some purpose. it wasn't what i wanted, so i didn't really bother looking too far into it. apparently then, those molecules exist only to puzzle students like me, who are the only people who'd really care. Ah well. maybe this thread will inspire someone out there.

Nope, wasn't a homework excersise; I merely did it out of interest. We are familiar with the longest chain rule, it's just that we haven't heard of an isopropyl group. 2nd and 3rd make sense, I hadn't known that you could assign those groups groups of their own(at least with the names). 4th one makes sense as well now, thanks very much. It doesn't answer my qusetion of: Do they exist, and what are they used for if they do? thanks for the names though. Now that I think of it, would what would a groups like these be called(just assume that they're on a large chain. ignore the ~s, they're just placeholders for the ASCII pics, because it didn't work without them): 1.~~~CH3-CH2-CH-CH2-CH3 ~~~~~~~~~~~| 2.~~~~CH3-CH-CH2-CH2-CH3 ~~~~~~~~~| I know these might seem like easy things for a year 12 student to be asking, but I'm in australia, a country not renowned for its fantastic education system. I'm even in Queensland, the state that is well known in australia for having the worst education system. That's a good enough excuse for such a pathetic question, isn't it?

I've been doing organic chemistry in my year 12 chemistry class, and I wanted to know how I'd go about naming certain chemicals. My teacher and I went through some of the molecules that i'd drawn, and found the only names we could think of for them. but, I'd like to know how they really are named, if they actually exist, and (if they do exist), what they are used for. the list is attached, so if you can help me, that would be appriciated. Hard-to-name molecules.doc