56.5/4= 14.125mols
14.125 X 22.5dm3 = 317.8125dm3
1 mole of gas occupies 22.4 L at STP. So 56/22.4 = 2.5 moles. Neon has Atomic Mass of 20 and so 2.5 moles x 20 atomic weight = 50 so the mass is 50g.
p
5
The volume of 5.0 moles of 02 at STP is 100 litres.
1 mole gas = 22.4L 1.5mol C2H4 x 22.4L/mol = 33.6L ethane gas (C2H4)
1 mol of any gas has a volume of 22.4 L at STP
The volume of one mole of gas at a standard temperature and pressure is 22.4 liters. Multiply 22.4 liters by 0.25 moles to get a volume of 5.6 liters.
150 (g O3) / (3*16.0 (g/mol O3)) = 3.125 (mol O3)3.125 (mol O3) * 22.4 (L/mol) = 70.0 Litre Ozone gas
The volume of 5.0 moles of 02 at STP is 100 litres.
0.21 mol
What is the volume of a 24.7 mol gas sample that has a pressure of .999 ATM at 305 K?
The volume is 17 L.
The volume is 22,1 L.
832
One mole has a volume of 22.4l.So the volume is 224ml
At room temperature, neon, which is an inert gas, is far and away less dense than water. Water, as you know, is a liquid at room temperature, and neon is never found in nature on earth as anything but a gas. A link can be found below.
1 mole gas = 22.4L 1.5mol C2H4 x 22.4L/mol = 33.6L ethane gas (C2H4)
0.251 moles neon (6.022 X 1023/1 mole Ne) = 1.51 X 1023 atoms of neon -------------------------------------
The reaction : Mg + O2-----------> MgO The balanced reaction : 2 Mg + O2-----------> 2 MgO that is 2 moles of Mg reacts with one mole of oxygen gas. moles of Mg reacted = 12 g / 24 g mol-1=0.5 mol. Thus the no. of mol of O2 reacted = 0.5 mol /2 =0.25 mol If the gas is at standard temperature and pressure, then Volume of 1 mol of O2 gas = 22.4 dm3 Volume of 0.25 mol of O2 gas = 22.4 x 0.25 dm3= 5.6 dm3 the volume of oxygen that will be used to react completely with 12 g Mg =5.6 dm3 (provided that the gas is at the standard conditions.)
847 m/s