Respuesta :

this 273 K and 101.3 kPa is Standart conditions, so 22.4 L is 1 mol, 1 mol anything contains 6.023 x 10^23, 
AnswerA
ow many oxygen molecules are in 22.4 liters of oxygen gas at 273k and 101.3kpa
First solve the number of moles of the oxygen gas by using the ideal gas equation:
PV = nRT
Where n is the number of moles
n = PV/RT
n = (101 300 Pa) (22.4 L) (1 m3/1000 L ) / ( 8.314 Pa m3 / mol K) ( 273 K)
n = 1 mol O2
the number of molecules can be solve using avogrados number 6.022x10^23 molecule / mole
molecules of one mole O2 = 6.022x 10^23 molecules