n = ?
V = 112 L
CONDICIONES NORMALES:
P = 1 atm
R = 0.08206 atm L / mol °K
T = 273.15 °K
FORMULA DE LOS GASES IDEALES
PV = nRT; despejando n
n = PV/RT
n = (1 atm * 112 L)/(0.08206 atm L/(mol °K)*273.15 °K)
n = 112/ 22.4147 moles
n = 5 moles de O2
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¿Cuantos moles de oxigeno contienen 112 litros de este gas en condiciones normales?
n = ?
V = 112 L
CONDICIONES NORMALES:
P = 1 atm
R = 0.08206 atm L / mol °K
T = 273.15 °K
FORMULA DE LOS GASES IDEALES
PV = nRT; despejando n
n = PV/RT
n = (1 atm * 112 L)/(0.08206 atm L/(mol °K)*273.15 °K)
n = 112/ 22.4147 moles
n = 5 moles de O2
TuProfesor17