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...