can anyone please solve this...
Attachments:
Answers
Answered by
10
3.6g of oxygen contains 3.6/32 = 0.1125 moles
0.1125 moles corresponds to 0.1125*22.4L of oxygen gas at STP = 2.52L
Thus 2.52 L of gas is adsorbed on 1.2g of metal powder,
thus on 1g of the metal powder oxygen adsorbed is 2.52/1.2 = 2.1 L/g
0.1125 moles corresponds to 0.1125*22.4L of oxygen gas at STP = 2.52L
Thus 2.52 L of gas is adsorbed on 1.2g of metal powder,
thus on 1g of the metal powder oxygen adsorbed is 2.52/1.2 = 2.1 L/g
Similar questions