This is the molar volume at
and at
(At room temperature T= 298 K and p= po the molar volume is 24.5 L/mole)
210.3 moles of H2 are contained in one gallon of H2O
1 mole of any gas occupies 22.4 L at standard temperature and pressure (STP). Therefore, 8.08 L of O2 at STP would contain 8.08/22.4 = 0.36 moles of O2.
At STP, 1 mole of any gas occupies 22.4 L. Therefore, in a 5L sample of argon at STP, there would be 5/22.4 moles of argon, which is approximately 0.223 moles.
See the Related Question "How do you solve Ideal Gas Law problems?" to the left for the answer.
1 mole of gas occupies 22.4 liters at STP. 564/22.4 = 25.18 moles (2 decimal places)
16,8 L of Xe gas at STP is equivalent to 0,754 moles.
210.3 moles of H2 are contained in one gallon of H2O
PV = nRT ⟹ n = PV/RT = 1 * 18.65 / (0.082 * 273.15) = 0.8321 moles.
1 mole of any gas occupies 22.4 L at standard temperature and pressure (STP). Therefore, 8.08 L of O2 at STP would contain 8.08/22.4 = 0.36 moles of O2.
0.25 moles
8,4 liters of nitrous oxide at STP contain 2,65 moles.
Using the ideal gas law, at STP (standard temperature and pressure), 1 mole of gas occupies 22.4 liters. Therefore, a balloon with 560 liters at STP would contain 25 moles of gas (560 liters / 22.4 liters/mole).
At STP, 1 mole of any gas occupies 22.4 L. Therefore, in a 5L sample of argon at STP, there would be 5/22.4 moles of argon, which is approximately 0.223 moles.
The answer is 2,68 moles.
The answer is 0,2675 moles.
See the Related Question "How do you solve Ideal Gas Law problems?" to the left for the answer.
Assuming ideal behaviour, 1 mole of any gas occupies 22.4L at STP. So, moles of 10L = 10/22.4 moles = 0.4464 moles