P = 2.30 atm
Volume in liter = 2.70 mL / 1000 => 0.0027 L
Temperature in K = 30.0 + 273 => 303 K
R = 0.082 atm
molar mass O2 = 31.9988 g/mol
number of moles O2 :
P * V = n * R* T
2.30 * 0.0027 = n * 0.082 * 303
0.00621 = n * 24.846
n = 0.00621 / 24.846
n = 0.0002499 moles of O2
Mass of O2:
n = m / mm
0.0002499 = m / 31.9988
m = 0.0002499 * 31.9988
m = 0.008 g
Answer: False, I believe.
Explanation: If a Hypothesis is proven correct, then another experiment to strengthen that Hypothesis is should be done.
Good ventilation as a product of it is pure Cl2 gas
Answer:
0.189 g.
Explanation:
- This problem is an application on <em>Henry's law.</em>
- Henry's law states that the solubility of a gas in a liquid is directly proportional to its partial pressure of the gas above the liquid.
- Solubility of the gas ∝ partial pressure
- If we have different solubility at different pressures, we can express Henry's law as:
<em>S₁/P₁ = S₂/P₂,</em>
S₁ = 0.0106/0.792 = 0.0134 g/L and P₁ = 0.321 atm
S₂ = ??? g/L and P₂ = 5.73 atm
- So, The solubility of the gas at 5.73 atm (S₂) = S₁.P₂/P₁ = (0.0134 g/L x 5.73 atm) / (0.321 atm) = 0.239 g/L.
<em>The quantity in (g) = S₂ x V = (0.239 g/L)(0.792 L) = 0.189 g.</em>
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Explanation:
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