The volume of 10.9 mol of helium at STP is 50 litres.
what is the volume of a balloon containing 50.0 moles of O2 gas at a pressure of 15.0 atm at 28 degrees
1 mol of any gas has a volume of 22.4 L at STP
C5H12 + 8 O2 --> 5 CO2 + 6 H2O (1 mol O2)(6 mol H2O/8 mol O2) = 0.75 mol H2O
0,019 mol
At Standard Temperature and Pressure (STP), which is defined as 0 degrees Celsius (273.15 Kelvin) and 1 atmosphere pressure, the molar volume of an ideal gas is approximately 22.4 liters/mol. The molar mass of nitrogen gas (Nā) is approximately 28.02 grams/mol. To calculate the density (D) of nitrogen gas at STP, you can use the ideal gas law: ļæ½ = Molar mass Molar volume at STP D= Molar volume at STP Molar mass ā ļæ½ = 28.02 ā g/mol 22.4 ā L/mol D= 22.4L/mol 28.02g/mol ā ļæ½ ā 1.25 ā g/L Dā1.25g/L Therefore, the density of nitrogen gas at STP is approximately 1.25 grams per liter.
Because oxygen gas (O2) has a molar mass of 32g/mol, 11.3 g * 1/32 mol/g gives about .35 moles. An ideal gas has a volume of 22.4 L/mol at STP, so 11.3 g O2 would have a volume of 7.91 L at STP.
2.24 L O2 (= 0.100 mol O2) is needed to react with 0.200 moles of SO2 to form SO3
The volume of 10.9 mol of helium at STP is 50 litres.
816 g C12H22O11 x (1 mol C12H22O11 / 342.0 g C12H22O11) x (12 mol O2 / 1 mol C12H22011) x (22.4 L O2 / 1 mol O2) = 641 L O2
1mol of a gas occupies 24 dm3 at STP, so 2.2mol X 24 mol/dm3 =52.8dm3 or 5280cm3
what is the volume of a balloon containing 50.0 moles of O2 gas at a pressure of 15.0 atm at 28 degrees
Density of CO2 at STP = 44.01 g/mol divided by the 22.4 liters. 1.96 grams/Liter
12.54 (g O2) / 2*15.99 (g/mol O2) = 0.3921 mol O2 -->0.3921 (mol O2) * 6.022*1023 (molecules O2)/(molO2) == 2.361*1023 molecules O2= 4.723*1023 atoms O
108kJ/mol
The estimated value of the density of francium at STP is 1,87 g/cm3.
13 L of FeO2 can be produced from 50.0 L of O2 at STP.