A sample of neon effuses from a container in 72 seconds. the same amount of an unknown noble gas requires 147 seconds. you may want to reference (pages 442 - 444) section 10.9 while completing this problem. part a identify the second gas.

The second gas is identified as follows

by Graham law formula

let the unknown gas be represented by letter y

=time of effusion of Neon / time of effusion of y = sqrt (molar mass of neon/molar mass of y)

= 72 sec / 147 sec = sqrt (20.18 g/mol / y g/mol)

square the both side to remove the square root sign

72^2/147^2 = 20.18 g/mol/y g/mol

=0.24 = 20.18g/mol/y g/mol

multiply both side by y g/mol

= 0.24 y g/mol = 20.18g/mol

divide both side by 0.24

y = 84 g/mol

y is therefore Krypton since it is the one with a molar mass of 84 g/mol