A sample of gas mixture from a neon sign contains 4.95x10^-2 mol Ne and 3.07x10^-2 mol...

A sample of gas mixture from a neon sign contains 4.95x10^-2 mol Ne and 3.07x10^-2 mol Kr . What are the mass percentages of Ne and Kr in the gas mixture?

Mass percentages of Ne = %

Mass percentages of Kr = %

Solutions

Expert Solution

mass of Ne = no of moles * gram atomic mass

= 4.95*10^-2 * 20.1797 = 0.9988g

mass of Kr = no of moles * gram atomic mass

= 3.07*10^-2 *83.798 = 2.573g

Mass percentages of Ne % = 0.9988*100/0.9988+2.573

= 0.9988*100/3.5718 = 27.96%

Mass percentages of Kr% = 2.573*100/0.9988+2.573

= 2.573*100/3.5718 = 72.04%

queen_honey_blossom answered 1 year ago

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