First we need the moles of krypton gas. Monoatomic gas.
0.405 grams Kr (1 mole Kr/83.80 grams)
= 0.004833 moles krypton gas
======================
Now the ideal gas equation.
PV = nRT
(1 atm)(X volume) = (0.004833 moles Kr)(0.08206 L*atm/mol*K)(298.15 K)
= 0.118 Liters of krypton gas
--------------------------------------
first, convert the 0.416g into moles. o.416/83.80 = 0.0050mol second, since there is 1 mol for 22.4L at STP, you write: 0.0050mol x (22.4L / 1mol) and your answer: 0.1112L
1 mole of gas particles at STP (Standard Temperature and Pressure) occupies a volume of 22.4 liters.
1 mole occupies 22.4 liters. 0.5 moles occupies 11.2 liters at STP.
PV=nRT 32 gram O2 = 1 mole O2 (1atm)(V) = (1 mole)(.0821)(273) V = 22.4 L
Krypton is a colorless, odorless gas at room temperature.
first, convert the 0.416g into moles. o.416/83.80 = 0.0050mol second, since there is 1 mol for 22.4L at STP, you write: 0.0050mol x (22.4L / 1mol) and your answer: 0.1112L
This volume is 79,79 litres.
The volume is 64,8 L.
The volume of ammonia is 19,5 L.
The volume is approx. 15,35 litres.
0.00922 g of H2 gas will occupy approximately 0.100 L at STP
At STP conditions, 11g of SO2 will occupy a volume of approximately 5.6 liters.
1 mole of gas particles at STP (Standard Temperature and Pressure) occupies a volume of 22.4 liters.
The volume is 22,1 L.
At STP (standard temperature and pressure), one mole of any gas occupies a volume of 22.4 liters. This is known as the molar volume of a gas at STP.
A 0.50 mole sample of helium will occupy a volume of 11.2 liters under standard temperature and pressure (STP) conditions, which are 0 degrees Celsius (273.15 K) and 1 atmosphere pressure. At STP, one mole of any gas occupies a volume of 22.4 liters.
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, a volume of 22.4 liters will be occupied by 1 mole of Cl2 gas at STP.