1 mole of any gas at STP is 22.4L, so 22.4L of nitrogen gas is 1 mole of nitrogen gas.

16,8 L of Xe gas at STP is equivalent to 0,754 moles.

1 MOLE of gas occupies 22.4 Liters at STP so a 4 liter flask conatins 4/22.4 = 0.1786 moles

PV = nRT ⟹ n = PV/RT = 1 * 18.65 / (0.082 * 273.15) = 0.8321 moles.

1 mole occupies 22.4 liters. So 2/22.4 = 0.0893 moles

O.8 moles of 122.4 L of neon at STP.

8,4 liters of nitrous oxide at STP contain 2,65 moles.

See the Related Question "How do you solve Ideal Gas Law problems?" to the left for the answer.

22.4 liters will have 1 mole of Helium at STP. So, 6 liters will have 0.23 moles

At STP/NTP, 10dm3 of oxygen contains 0.45 moles

1,75/21,4 = 0,0818 moles

The answer is 0,2675 moles.

The answer is 2,68 moles.

Assuming ideal behaviour, 1 mole of any gas occupies 22.4L at STP. So, moles of 10L = 10/22.4 moles = 0.4464 moles

The volume of 5.0 moles of 02 at STP is 100 litres.

The answer is 0,305 moles at 1 at and 25 0C.