If the density of oxygen atSTP is 1,429 g/L the mass of 180 L is 257,22 g.
If the mole of oxygen (O2) is 15,999 g the number of moles is 16,077.
0.25 moles
16,8 L of Xe gas at STP is equivalent to 0,754 moles.
8,4 liters of nitrous oxide at STP contain 2,65 moles.
The answer is 0,2675 moles.
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
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.
No. of moles = mass/relitive molecular mass in this case = 10/16 = 0.625 so that's 0.625 of a mole and a mole of anything contains 6.022 x 1023 atoms = 3.76 x 1023 atoms in 10g of oxygen.
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.
0.25 moles
16,8 L of Xe gas at STP is equivalent to 0,754 moles.
8,4 liters of nitrous oxide at STP contain 2,65 moles.
Ideal gas law.PV = nRT(1.00 atmosphere)(30 L) = n(0.08206 L*atm/mol*K)(298.15 K)n(moles O2) = 30/24.466= 1.2 moles oxygen gas================
The molar volume of a gas at standard temperature and pressure (STP) is 22.4 L/mol. Therefore, the volume of 2 moles of oxygen gas at STP would be 2 moles * 22.4 L/mol = 44.8 L.
The amount of oxygen is 0,067 moles.
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.