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What amount of moles does 60.5 L of carbon dioxide occupy at 50.5 kPa and 20 degrees C?

Sagot :

n = PV=RT

where

n = number of mole

p = pressure = 2.15 ATM

v = volume = 60.5L

r = 0.0821 atm (L)/ mol(K)

t = temperature = 25°+273.15=298.15 kelvin

Replacing with the values given

n = 2.15(60.5)/0.0821(298.15)=5.3 mole.