Question

A mixture of 7g of nitrogen and 8g of oxygen at stp occupies a volume of

Answer 1

Answer:

11200 ml

Explanation:

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Answer 2

A mixture of 7g of nitrogen and 8g of oxygen at STP occupies a volume of 11200 mL.

• At STP 1 mil of any gas = 22400 mL
• Which is equal to 22.4 L
• Molecular mass of N2 = W/M = 7/28 = 0.25
• Similarly, Molecular mass of O2 = W/M = 8/32 = 1/4 = 0.25
• By adding the above values we will get 0.5 moles
• We know that 1 mole of a gas = 22400 mL
• Therefore, for 0.5 mole of gas mixture,

              =0.5 * 22400 = 11,200 mL.