Asked by Wesley
Two gases HBr and CH4 effuse through a small opening. HBr effuses through the opening at a rate of 4 cm^3 s^-1. At what will the methane gas effuse through the same opening?
Answers
DrBob222
mm = molar mass
(rate HBr/rate CH4) = sqrt (mm CH4/mm HBr)
[(4 cm/s)/(rate CH4)] = sqrt (16/80.9)
Solve for rate CH4. Post your work if you get stuck.
(rate HBr/rate CH4) = sqrt (mm CH4/mm HBr)
[(4 cm/s)/(rate CH4)] = sqrt (16/80.9)
Solve for rate CH4. Post your work if you get stuck.
Wesley
Rate(CH4)= 8.99cm/s
DrBob222
yes