Recent content by escryan

Oops, just noticed that "Z2eff= 4.129 x 10 23 mol " should actually read "Z2eff= 4.129 x 10 23 mol-1 " I think that the italicized part is what confuses me the most -- what are the units for this portion? I'm going to take a guess that it is currently molecules/mol, but if so, is this always...

Ah.. So subbing in values E=899.4 kJ/mol RH=2.178 x 10-21 kJ n=1 I get Z2eff= 4.129 x 10 23 mol How does one get to the units/value of Z2eff after this?

Homework Statement If ionization energy is 899.4 kJ/mol for Be, what is the effective nuclear charge? Homework Equations Zeff = Z - S E=RH(Z2/n2) ?? E=RH(Zeff2/n2)?? The Attempt at a Solution My attempted solution was subbing into Zeff = Z - S Zeff = 4 - 2 = 2 But I...

Oh wow, that's a really helpful guide! Thanks so much for your help :)! Just one last question, is there any reason as to why the half reaction method is recommended/preferred over the one involving oxidation numbers? I'm just curious because I'd rather use the oxidation numbers one (because...

Ahh.. that's the part that confuses me.. how do you come to find the number of electrons you must add to one side? I've always been under the impression that the electrons added were based sheerly on the oxidation numbers. In this equation, S8 has 0 charge, and an individual S has a 0...

I think that might've been a typo on my part, sorry. I had written on my paper: 2S8 + 48OH- ----> 8S2O32- + 8e- + 24H2O as well as: S8 + 24OH- ----> 4S2O32- + 8e- + 12H2O and neither of the charges worked out to them being equal on either side.. Another typo I noticed (eek.)...

Homework Statement Balance the equation for disproportionation reactions. Homework Equations S8(s) ----> S2-+S2O32- (basic solution) The Attempt at a Solution I attempted into do this with the ion electron method, obtaining an answer: 3S8 + 24OH- ----> 16S2- + 4S2O32- +...

I'm just curious as to how \int (\frac{1}{1+x^{2}}) dx comes to be \tan^{-1} (x) I was able to find the formula on a table of integrals, but I'd just like to know why it works like this, and why we can't use a natural log rule or a substitution method to find this out...

Oh I see now! Thanks so much for your help Dick and sutupidmath!

I don't know how I managed to forget this one, but I did somehow... If there's something like: e^lnx, why is that equal to just x? and same goes for sokmething like: 8^log[SIZE="1"]8x which is just equal to x. I'm just wondering how, algebraically, one could show this to be true.

The chain rule? ... I've never actually seen that before. Haha, I guess that explains a few things! I haven't been taught that yet. Guess I'll go read up on that, and thanks again for your help! I really appreciate it :).

Oh wow, I'm an idiot.. So is the general rule d/dx(sec^n(x)) = nsec^n(x)tan(x) just a combination of the (x^n)' = nx^x-1 and d/dx(secx) = secxtanx? What if I were to be givin d/dx(tan^n(x))... would the answer be like nsec^2(n-1)(x)? nsec^n(x)? ... Thanks so much for your help, by the way :).

Homework Statement Given y=xtanx, find y'' (second derivative) Homework Equations Uh... I'm not even sure if I'm using the right one... d/dx(tanx) = sec^2x The Attempt at a Solution y=xtanx y'= (x)(sec^2(x)) + (tanx)(1) y'= xsec^2(x) + tanx y'' = [(x)(2sec^3(x)) +...