A piston containing 0.120moles of methane gas, CH4, has a volume of 2.12liters. If methane is added until the volume is increase
1 answer:
Answer:
2.83 g
Explanation:
At constant temperature and pressure, Using Avogadro's law
Given ,
V₁ = 2.12 L
V₂ = 3.12 L
n₁ = 0.120 moles
n₂ = ?
Using above equation as:
n₂ = 0.17660 moles
Molar mass of methane gas = 16.05 g/mol
So, Mass = Moles*Molar mass = 0.17660 * 16.05 g = 2.83 g
<u>2.83 g are in the piston.</u>
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Answer:
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Explanation:
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Answer:
Molar percent of sodium in original mixture is 88,50%
Explanation:
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BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl
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0,132L × 3,80M = 0,502 moles of BaCl₂
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I hope it helps