Question

The correct answer is C. 21. At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters. So, 120 liters of He gas (helium) at STP would be equivalent to 120/22.4 moles, which is approximately 5.357 moles. The molar mass of helium is approximately 4 grams/mol, so 5.357 moles would be approximately 5.357 x 4 = 21.428 grams.

Answer 1

At STP, 120 liters of helium gas would be approximately 21.428 grams.

Step-by-step explanation:

At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters. So, 120 liters of He gas (helium) at STP would be equivalent to 120/22.4 moles, which is approximately 5.357 moles. The molar mass of helium is approximately 4 grams/mol, so 5.357 moles would be approximately 5.357 x 4 = 21.428 grams.