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A lung holding a residual air volume of 1.50 L at 36 ºC has a pressure of 1.00 atm. How many moles of residual air is the lung holding? R = 0.0821 L.atm/mol.K.

Respuesta :

Answer:

0.05909882564 mol

Explanation:

PV=nRT

n = 1.00atm*1.50L /(0.0821L.atm/mol.k * 309.15 k)

= 0.05909882564 mol

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