stupid question
You can use the ideal gas law to solve this problem. First, convert 0.30 g of Cl2 to moles. Then use the molar volume of gas at STP (22.4 L/mol) to determine the volume of Cl2 gas needed. Convert this volume to milliliters (1 L = 1000 mL) to find the answer.
This volume is 79,79 litres.
The volume is approx. 15,35 litres.
The volume is 64,8 L.
The volume of ammonia is 19,5 L.
You can use the ideal gas law to solve this problem. First, convert 0.30 g of Cl2 to moles. Then use the molar volume of gas at STP (22.4 L/mol) to determine the volume of Cl2 gas needed. Convert this volume to milliliters (1 L = 1000 mL) to find the answer.
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.
what is the volume of a balloon containing 50.0 moles of O2 gas at a pressure of 15.0 atm at 28 degrees
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.
At standard temperature and pressure (STP), one mole of any ideal gas occupies 22.4 liters. Therefore, the volume of 1.9 moles of chlorine gas (Cl2) can be calculated as follows: 1.9 moles × 22.4 L/mole = 42.56 L. Rounding to the nearest option, the volume of 1.9 moles of Cl2 at STP is approximately 43 L.