If these atoms are loose unities, not bonded in a molecule, it would be about 22 dm3 at 0oC and standard pressure (STP). If the atoms would have formed gas molecules, e.g. CO2 or SF6, the answer would be different, because 6.02 x 1023 molecules always take in a volume of about 22 dm3 at STP. But with CO2 you would have then 18 x 1023 atoms because there are 3 atoms in one molecule.
The volume of 10.9 mol of helium at STP is 50 litres.
The volume of any gas at STP (standard temperature and pressure) is 22.4 L/mol. The molar mass of helium is 4.0026 g/mol. So, 84.6 grams of helium would be 84.6/4.0026 = 21.1 mol. Therefore, the volume of 84.6 grams of helium at STP would be 21.1 mol * 22.4 L/mol = 472.64 L.
It would be 7 x 22.4 litre ie 156.8 litre.
Because at STP, Chloroform is liquid and Helium is in gaseous state. When something is in a gaseous state, it occupies a larger space than the liquid. I thought however, that chloroform would occupy less than that
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The volume of 10.9 mol of helium at STP is 50 litres.
The volume of any gas at STP (standard temperature and pressure) is 22.4 L/mol. The molar mass of helium is 4.0026 g/mol. So, 84.6 grams of helium would be 84.6/4.0026 = 21.1 mol. Therefore, the volume of 84.6 grams of helium at STP would be 21.1 mol * 22.4 L/mol = 472.64 L.
It would be 7 x 22.4 litre ie 156.8 litre.
The volume is 0.887 L.
The mass of 43,7 L of helium at STP is 7.8 g.
Because at STP, Chloroform is liquid and Helium is in gaseous state. When something is in a gaseous state, it occupies a larger space than the liquid. I thought however, that chloroform would occupy less than that
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Helium exists as a gas at STP
Helium is a gas at STP.
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.
Gas at STP.