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OpenStudy (anonymous):
How many grams of chlorine gas (Cl2) are in a 17.8 liter sample at 1.1 atmospheres and 29°C? Show all work used to solve this problem.
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OpenStudy (anonymous):
T = 29 + 273 = 302 K moles Cl2 = pV/RT = 1.1 x 17.8 / 0.08206 x 302 =0.790 mass Cl2 = 0.790 mol x 70.906 g/mol=56.0 g
OpenStudy (anonymous):
PV=nRT n= 1.1*17.8/[0.08206*(29+273.15)] n= 0.78969 (do not round yet) mass= 0.78969*(35.453*2)= 55.99g = 56.0 g (3 s.f)
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