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