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Answer 1 · 2017-05-19T13:58:29

$0.124molN_2$

We'll use the ideal-gas equation for this problem.

First we must convert each unit into the appropriate unit for the equation:

$347mL(1L)/(10^3mL) = 0.347L$

$6680t o r r(1 atm)/(760t o r r) = 8.79 atm$

$27 ^oC + 273 = 300K$

Now, we'll use the ideal-gas equation, and solve it for quantity (in $mol$ ):

$n = (PV)/(RT)$

$n = ((8.79atm)(0.347L))/((0.08206(L-atm)/(mol-K))(300K)) = 0.124 molN_2$