Please explain each step! A sample of neon gas effuses through a pinhole at a ra

Question

Please explain each step!

A sample of neon gas effuses through a pinhole at a rate of 0.44 mL/s. What will be the rate of effusion of krypton gas through the same pinhole? Graham's law of effusion is given by the equation

r1 r2 A sample of neon gas effuses through a pinhole at a rate of 0.44 mL/s. What will be the rate of effusion of krypton gas through the same pinhole? Graham's law of effusion is given by the equation where r1 and r2 are the rates of effusion of gases 1 and 2, and M1 and M2 are their molar masses, respectively.

Explanation / Answer

r2/0.44 = sqrt(20.18/83.8)

             =0.491

r2=0.22 mL/s.
rate of effusion of krypton gas =0.22 mL/s

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