Answer to Question #321077 in General Chemistry for zzz

Question #321077

F2(g) (M=38.00 g/mol) effuses through a pinhole at a rate of 0.01287 mol/s. At what rate does N2O4(g) (M=92.02 g/mol) effuse at the same pinhole?


1
Expert's answer
2022-03-31T10:19:02-0400

vAvB=MBMA\frac{v_A}{v_B}=\frac{\sqrt{M_B}}{\sqrt{M_A}}

vN2O4=vF2MF2MN2O4v_{N_2O_4}=v_{F_2}\sqrt{M_{F_2}M_{N_2O_4}}

vN2O4=0.0128738x92.02=0.761mole/sv_{N_2O_4}=0.01287 \sqrt{38x92.02}=0.761 mole/s




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