The volume of one mole of oxygen can be estimated by the ideal gas law. In this case, you will use V = nRT/P, where n is the moles of gas, R is the ideal gas constant, T is the temperature in kelvin, P is the system pressure.
1 mole of any gas occupies 22.4 liters at standard temperature and pressure (STP). Therefore, 68.5 liters of oxygen gas at STP would be 68.5/22.4 = 3.06 moles of oxygen gas.
One mole of any gas at STP occupies 22.4 liters. Therefore, one mole of oxygen gas at STP also occupies 22.4 liters.
At standard temperature and pressure, one mole of gas takes up 22.4 liters. So the amount of gas necessary to occupy 2 liters is:2 L ÷ 22.4 mole/L = 0.08929 molesOne mole of oxygen gas (O2) weighs 32 grams per mole, so:0.08929 moles * 32 g/mole = 2.857 grams of O2The density of liquid oxygen is 1.141 g/cm³, and so the volume is:2.857 grams ÷ 1.141 g/cm3 = 2.50 cm3 = 2.50 mLIn other words, oxygen expands by a factor of 800 going from liquid to gas!See the Related Questions link to the left for more information on solving Ideal Gas Law problems of this type.
At room temperature and pressure, 1 mole of ideal gas occupies 22.4 liters. Therefore, 8.00 moles of oxygen will occupy 8.00 x 22.4 = 179.2 liters.
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, 15 liters of oxygen at STP would be equivalent to 15/22.4 = 0.67 moles.
1 mole of any gas occupies 22.4 liters at standard temperature and pressure (STP). Therefore, 68.5 liters of oxygen gas at STP would be 68.5/22.4 = 3.06 moles of oxygen gas.
One mole of any gas at STP occupies 22.4 liters. Therefore, one mole of oxygen gas at STP also occupies 22.4 liters.
The volume is 64,8 L.
1 mole = 22.414 liters So, 3.5 mole = 78.45 liters
At STP, one mole of any gas occupies 22.4 liters. This is called molar volume. 113.97 liters ÷ (22.4 L/mol) = 5.09 moles Then convert moles to molecules (1 mole = 6.02 × 1023 molecules) 5.09 moles × (6.02 × 1023 molecules/mol) = 3.06 × 1024 molecules
One mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters. Therefore, one mole of helium would also occupy 22.4 liters at STP.
At standard temperature and pressure, one mole of gas takes up 22.4 liters. So the amount of gas necessary to occupy 2 liters is:2 L ÷ 22.4 mole/L = 0.08929 molesOne mole of oxygen gas (O2) weighs 32 grams per mole, so:0.08929 moles * 32 g/mole = 2.857 grams of O2The density of liquid oxygen is 1.141 g/cm³, and so the volume is:2.857 grams ÷ 1.141 g/cm3 = 2.50 cm3 = 2.50 mLIn other words, oxygen expands by a factor of 800 going from liquid to gas!See the Related Questions link to the left for more information on solving Ideal Gas Law problems of this type.
At room temperature and pressure, 1 mole of ideal gas occupies 22.4 liters. Therefore, 8.00 moles of oxygen will occupy 8.00 x 22.4 = 179.2 liters.
1 mole occupies 22.414 liters So, 1.84 moles will occupy 41.242 liters
1 mole (or 4 g of He) occupies 22.414 liters. So, 2.3 mole occupies 2.3 x 22.414 liters = 51.5522 liters
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, 15 liters of oxygen at STP would be equivalent to 15/22.4 = 0.67 moles.
At standard temperature and pressure (STP), one mole of any ideal gas occupies 22.4 liters. To find the number of moles of ammonia gas (NH₃) required to fill a volume of 50 liters, you can use the formula: moles = volume (liters) / volume per mole (liters/mole). Therefore, the calculation is 50 liters / 22.4 liters/mole = approximately 2.24 moles of NH₃ are needed.