Que volumen ocuparan 10 gramos de oxigeno gaseoso a una temperatura de 22 °C y una presion de 740 torr?
Recuerde que= 1 atm = 760 torr y K=°C+273
Respuestas
Respuesta dada por:
17
Ley de los gases Ideales
V x P = n x R x T
Mm O2 = 32 g/mol
n = masa/Mm
n = 10 g / 32 g/mol
n = 0.313 moles de O2
P = 760 Torr
1 atm --- 760 torr
x --- 740 torr
x = 0.97 atm
T = 22 ºC + 273 = 295 K
R = 0.082 (L atm / mol K)
Calcular volumen
V = 0.313 mol x 0.082 (L atm/mol K) x 295 K
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0.97 atm
V = 7.805 L
V x P = n x R x T
Mm O2 = 32 g/mol
n = masa/Mm
n = 10 g / 32 g/mol
n = 0.313 moles de O2
P = 760 Torr
1 atm --- 760 torr
x --- 740 torr
x = 0.97 atm
T = 22 ºC + 273 = 295 K
R = 0.082 (L atm / mol K)
Calcular volumen
V = 0.313 mol x 0.082 (L atm/mol K) x 295 K
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0.97 atm
V = 7.805 L
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