Recent content by dorin1993

Thank you so much! :)

Do you mean grouo af all integers - the identity element is 0 and for example 2 +(mod4) 2 = 0 although 2 ≠ 0 (I still trying to understand the modulo)

But the main argument is about ALL abelian group with xx=e

Hi guys, I have quastion about groups: G is abelian group with an identity element "e". If xx=e then x=e. Is it true or false? I was thinking and my feeling is that it's true but I just can't prove it. I started with: (*) ae=ea=a (*) aa^-1 = a^-1 a = e those from the...

Ohh! now I get it. All the time I referred the column as a volume and no thinking at all about the base 1cm^2. Now I realize how easy it is. Thank you so much for halping and explaining!

Ok - so V = 7.47*10^-6 [mole] * 8.314 * 273K / 1.013*10^5 = 0.167 cm^3 again, I get volume - and I need a surface (molec/cm^2) that i can divide in 2.69*10^16 molecules/cm^2 and get DU. But I don't understand why. The concentration of the molecule is given, and it's around the Earth from 15 km...

Sorry, i still don't get it :\ I know the volume and i know how much molecules i have on it. Also i know that the base is 1 cm^2. I even drown it and thinking of it over and over again and I don't know... I keep thinking about the diameter of one molecule and how much is the distans between...

OK - so i have volume of 15km * 1cm * 1cm = 15*10^5 cm^3 then - 3*10^12 [molecules/cm^3] * 15*10^5 [cm^3] = 4.5*10^18 molecules now - how can i calculate the molecule of the base? meaning the number of molecules for cm^2?

This is all the given information... I attach the 2 pages in the book that explain about the dobson unit the question is from this book. http://imageshack.us/a/img824/1039/77622876.jpg http://imageshack.us/a/img29/5307/13079304.jpg

calculate the number of DU assuming that the entire atmospheric O3 column is at a uniform concentration of 3*10^12 molecules/cm^3 between 15 km and 30 km and zero elsewhere. I have no idea how to solve it. I don't know how to convert the 3*10^12 molecules/cm^3 to units of molecules/cm^2 that I...