Question

Cuál es el volumen de 0.25 moles de gas amoníaco (nh3) a 1.00 atm y 273 k?

Answer 1

Ley de los gases ideales

V x P = n x R x T

T = 273 K

P =  1 atm

n = 0.25 moles

R = 0.0821 (L atm / mol K)

V = 0.25 moles  x  0.0821 (L atm / mo K) x 273 atm
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                             1 atm

V = 5.603 Litros

Answer 2

¿Cuál será el volumen que ocupa 4.70 moles de gas de amoniaco NH3 a 100 °C de temperatura y 0.98 atm?

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