What volume of oxygen gas at 25 degrees celsius and 1.04 atm is needed for the complete combustion of 5.53g of propane

14.7 L of oxygen gas at 25 degrees celsius and 1.04 atm is needed for the complete combustion of 5.53g of propane

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number of particles.

According to stoichiometry :

1 mole of propane requires 5 moles of oxygen  

Thus 0.125 moles of propane require= moles of oxygen  

According to the ideal gas equation:'

P = Pressure of the gas = 1.04 atm

V= Volume of the gas = ?

T= Temperature of the gas = 25°C = 298 K  

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas= 0.625

8 m/s because for him to get the same amount of energy as ray is putting out, bobby, who weighs half as much as ray, needs to run twice as fast.