How many moles of argon are there in 20.0 L, at 25C and 98.8kPa?
Answer:
98.8 kPa = 98800 Pa
101325 Pa = 1atm
98800 / 101325 = 0.975 atm
T = 25 + 273 = 298 K
pV = nRT
n = pV / RT = 0.975 x 20 / 0.0821 x 298 = 0.797 moles
2.36
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