Draw the cyclic structure, then draw in double bonds where necessary to keep the octet rule for carbon. Should end up with 4 bonds to one of the Nitrogens, which gives a + charge. Now if you draw the resonance forms you'll see that the double bonds (and + charge) can bounce around throughout...
Why do people always go to this defense?
Let's compare the two situations in lesser terms.
Iraq = 5 year old with a plastic baseball bat
Iran = 10 year old with a hammer
N. Korea = 15 year old with a shotgun
You know -Iraq- wants you dead (hypothetically, not trying to justify the...
Don't know if it's been posted, but here is a [PLAIN]mms://wmscnn.stream.aol.com/cnn/world/2005/07/22/eyewitness.london.shooting.affl.ws.wmv[/URL] with some eye witness acounts.
http://www.fda.gov/ola/2003/canadian0612.html
Right now people only have themselves to blame if they purchase Canadian (or from god knows where) drugs and they are faulty.
Another thing is that most drug companies aren't making a huge amount of profit from their drugs. If the price was...
Remote helicopter and a buddy would make sense...
I say they really test mister Yahweh by blindfolding him and taking him to a remote location unknown to him (while making sure nobody is following) then have him perform his rain dance. Can't be too hard to tell if someone is following you in...
I was not confused by the algebra at all. Really what I wanted to find out was that my answer, which I obtained from the very equation for P2 I posted above, was correctly stated as a pressure. The basic algebra of the problem made sense to me, however my instructor circled the units on my...
Yes, this was the heart of my question. We use Troutons rule to determine the pressure at the vaporization temperature, yet to do this we must take the exponential of the equation. So, for instance:
{ln(\frac{{P}_{2}}{{P}_{1}}) = -\frac{{\Delta}_{vap}\overline{H}}{R}(\frac{1}{T}_2...
Ok so i was doing a problem involving finding the pressure of mercury at its boiling point (630.05K) using Troutons rule and the final answer seems a bit strange to me.
Integration of the Clausius-Clapeyron Equation:
{ln(\frac{{P}_{2}}{{P}_{1}}) =...