Utiliza la ecuación de los gases ideales PV= n.r.t
para determinar :
a) El volumen de 1,20 moles de O2 a 27 grados c y 1 atmósfera de presión
b) El numero de moles en 10 litros de CO2 a 20 grados c y 800 torr
c) el peso molecular de un gas cuya densidad es de 1,62 g/L a 200 grados kelvin y 1,89 atmósferas.
Respuestas a la pregunta
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1. v = ?
P = 1 atm
n = 1.20 moles
T = 27 ºC + 273 = 300 K
R = 0.0821 (L x atm / mol x K)
V = 1.20 moles x 300 K x 0.0821 (L x atm / mol K)
``````````````````````````````````````````````````````````````````
1 atm
V = 29.56 L
2. n = ?
V = 10 L
T = 20 ºC + 273 = 293 K
P = 800 torr
1 torr ---- 0.001316 atm
800 torr ---- x
x = 1.0528 atm
R = 0.0821 (L x atm / mol x K)
n = 10 L X 1.0528 atm
``````````````````````````````````````````
293 K x 0.0821 (L x atm / mol K)
n = 0.437 moles
C) Mm = ?
d = P x Mm
````````````````
R x T
Mm = d x R x T
````````````````
P
Mm = 1.62 g/L x 0.0821 (L x atm / mol x K) x 200 K
````````````````````````````````````````````````````````````````
1.89 atm
Mm = 14.07 g/mol
P = 1 atm
n = 1.20 moles
T = 27 ºC + 273 = 300 K
R = 0.0821 (L x atm / mol x K)
V = 1.20 moles x 300 K x 0.0821 (L x atm / mol K)
``````````````````````````````````````````````````````````````````
1 atm
V = 29.56 L
2. n = ?
V = 10 L
T = 20 ºC + 273 = 293 K
P = 800 torr
1 torr ---- 0.001316 atm
800 torr ---- x
x = 1.0528 atm
R = 0.0821 (L x atm / mol x K)
n = 10 L X 1.0528 atm
``````````````````````````````````````````
293 K x 0.0821 (L x atm / mol K)
n = 0.437 moles
C) Mm = ?
d = P x Mm
````````````````
R x T
Mm = d x R x T
````````````````
P
Mm = 1.62 g/L x 0.0821 (L x atm / mol x K) x 200 K
````````````````````````````````````````````````````````````````
1.89 atm
Mm = 14.07 g/mol
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