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how many moles of 18L of NH3 at 30 celcius and 912 mmHg? how many grams?
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Dejanras: Answer is: mass of ammonia is 14.76 grams and amount is 0.87 moles.
V(NH₃) = 18 L.
T = 30°C = 303.15 K.
p = 912 mmHg ÷ 760 mmHg/atm= 1.2 atm.
R = 0.08206 L·atm/mol·K.
Ideal gas law: p·V = n·R·T.
n = p·V / R·T.
n(NH₃) = 1.2 atm ·18 L / 0.08206 L·atm/mol·K · 303.15 K.
n(NH₃) = 0.87 mol.
m(NH₃) = n(NH₃) · M(NH₃).
m(NH₃) = 0.87 mol · 17 g/mol.
m(NH₃) = 14.76 g.
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