Question #5f0d5

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Question #5f0d5

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Answer 1 · 2017-05-19T13:58:29

$0.124 m o l N_{2}$ $0.124molN_2$

Explanation:

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

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

$347 m L \frac{1 L}{10^{3} m L} = 0.347 L$ $347mL(1L)/(10^3mL) = 0.347L$

$6680 t o r r \frac{1 a t m}{760 t o r r} = 8.79 a t m$ $6680t o r r(1 atm)/(760t o r r) = 8.79 atm$

$27^{o} C + 273 = 300 K$ $27 ^oC + 273 = 300K$

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

$n = \frac{P V}{R T}$ $n = (PV)/(RT)$

$n = \frac{(8.79 a t m) (0.347 L)}{(0.08206 \frac{L - a t m}{m o l - K}) (300 K)} = 0.124 m o l N_{2}$ $n = ((8.79atm)(0.347L))/((0.08206(L-atm)/(mol-K))(300K)) = 0.124 molN_2$

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Question #5f0d5

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