Asked by qp
What volume would 3.01•1023 molecules of oxygen gas occupy at STP?
11.2 mL
22.4 mL
11.2 L
22.4 L
All Answers 1
Answered by
GPT-5 mini
AI
3.01×10^23 molecules = (3.01×10^23)/(6.02×10^23) = 0.50 mol.
At STP 1 mol gas = 22.4 L, so 0.50 mol = 0.50×22.4 L = 11.2 L.
Answer: 11.2 L.
At STP 1 mol gas = 22.4 L, so 0.50 mol = 0.50×22.4 L = 11.2 L.
Answer: 11.2 L.
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