1.

If 4g of O_(2) diffuse through a very narrow hole, how much H_(2) would have diffused under identical conditions ?

Answer»

1g
`1//4g`
`16G`
64g

Solution :`( 4g//t)/(X//t) = SQRT((2)/( 32))`
( x is the AMOUNT of `H_(2)` diffused and t is the time )
`( 4)/( x) = sqrt((1)/( 16)) = ( 1)/( 4)`


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