Que volumen ocuparan 3,5 moles de amoniaco a -13 °c y 4,9 atm
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P V = n R T
V = ?
P = 4.9 atm
n = 3.5 moles
T = -13 + 273 = 260 K
R = 0.0821 (L atm / mol K)
V = n R T
````````
p
V = 3.5 moles x 0.0821 /L atm / mol K) x 260 K
``````````````````````````````````````````````````````````````
4.9 atm
V = 15.25 Litros
V = ?
P = 4.9 atm
n = 3.5 moles
T = -13 + 273 = 260 K
R = 0.0821 (L atm / mol K)
V = n R T
````````
p
V = 3.5 moles x 0.0821 /L atm / mol K) x 260 K
``````````````````````````````````````````````````````````````
4.9 atm
V = 15.25 Litros
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