0.0045 x 0.000789 sig fig
2 answers:
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You must use 134 g O₂ to produce 118 g H₂O.
M_r: 32.00 18.02
2H₂ + O₂ ⟶ 2H₂O
Moles of H₂O = 150.9 g H₂O × (1 mol H₂O/18.02 g H₂O) = 8.374 mol H₂O
Moles of O₂ = 8.374 mol H₂O × (1 mol O₂/2 mol H₂O) = 4.187 mol O₂
Mass of O₂ = 4.1877 mol O₂ × (32.00 g O₂/1 mol O₂) = 134 g O₂
Answer:
<h2>The answer is 1.48 L</h2>Explanation:
In order to find the original volume we use the same for Boyle's law which is
where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume
Since we are finding the original volume
From the question
P1 = 172 kPa = 172000 Pa
P2 = 85 kPa = 85000 Pa
V2 = 3 L
We have
We have the final answer as
<h3>1.48 L</h3>Hope this helps you