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September 22, 2014

September 22, 2014

Posted by **bubble** on Friday, January 15, 2010 at 2:16am.

by GC 1/lambda = R/ni^2–R/nf^2

y = mx + b (standard form) of a linear equation)

x = (y-b)/m

Let,

y = 1/lambda

m = -R

x = 1/nf^2

b = R/ni^2

1/nf^2 = [(1/lamda)-(R/ni^2)]/(-R)

Solve for nf

i still dont get how you get rid of y and x to get m and b in the end

- chemistry -
**UCI CHEM1B**, Friday, January 15, 2010 at 5:39amAnswer is (m/b)^1/2

- chemistry -
**zot!**, Saturday, January 16, 2010 at 9:23pmI don't know the answer to this question but I put the answer given above is wrong if you are doing the uci webwork from chem 1LB

- Chem1LB -
**Alex Le**, Sunday, January 17, 2010 at 2:06amdude its (-M/B)^(1/2)

- chemistry -
**bowisha**, Sunday, January 17, 2010 at 5:22pmim doing uci webwork and thats right

(-M/B)^(1/2)

- chemistry -
**jolie**, Sunday, January 17, 2010 at 8:08pmAlex Le and bowisha

you're all probably stupid in school

thats y u want to let people down idiots

answer is correct as mb to the 1/2

not negative

this is the Right answer

- chemistry -
**wow**, Sunday, January 17, 2010 at 10:36pmI'm doing the webwork for uci, not sure what the answer is but it's not (m/b)^-1/2. thanks people....

- chemistry -
**wow..**, Monday, January 18, 2010 at 2:28amit's negative

- chemistry -
**Antonio**, Monday, January 18, 2010 at 4:23pmyo it is (-m/b)^(1/2) i just did it and submitted.

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