Recent content by Blamo_slamo

Are you trying to do something like: X HCL + W compound1 -> Y compound2 + Z compound3 Perhaps some more information on the chemical equations you're specifically trying to work with would help.

:smile: Thanks a lot tiny-tim!

Would it be reliable, if I just subbed in the bounds of x and y, in r? e.g. r2 = x2 + y2 so as 0 ≤ x ≤ ∞ and 0 ≤ y ≤ ∞; r would then go from 0 ≤ r ≤ ∞ ? and as you explained, we're only looking at the first quadrant, so 0 ≤ θ ≤ π/2 ?

So this region has a peak at x = y = 0, and it's at 1. The region also slopes off to 0, as both x, and y go to ∞ I'm still not entirely sure on this, but technically the original function runs off to infinity, as it gets closer and closer to 0, so would r then be from ∞ --> 1? This is an...

Homework Statement By changing to polar coordinates, evaluate: \int\int e ^(-\sqrt{x^2 + y^2}) dx dy Both integrals go from 0 --> infinity Homework Equations r = \sqrt{x^2 + y^2} x = r cos\theta y = r sin\theta Using the Jacobian to switch to polar coord we get: J = r...

Homework Statement If we were to ignore the interelectronic repulsion in helium, what would be it's ground state energy and wave function? Homework Equations I have created my ground state wave function \psi for 1s: \psi = (1/\sqrt{}\pi)(z/a)3/2(e-zr/a) The operator is the...

First I would like to thank you very much, also I knew they were partials, but I'm a noob to this forum, and didn't know how to put down partials! The way I started to set it up, was with how the book explained to do it, but your way would probably be my normal approach. I did manage to get...

Okay, so my problem lies within taking the second derivative of a change of variable equation. w = f(x,y); x = u + v, y = u - v so far I have the first derivative: dw/dx = (dw/dv)(dv/dx) + (dw/du)(du/dx) = (d/dv + d/du)w Now I'm having problems in finding my second derivative...