Answer:
3.18 L
Explanation:
Step 1: Given data
- Initial pressure (P₁): 0.985 atm
- Initial volume (V₁): 3.65 L
- Final pressure (P₂): 861.0 mmHg
Step 2: Convert P₁ to mmHg
We will use the conversion factor 1 atm = 760 mmHg.
0.985 atm × 760 mmHg/1 atm = 749 mmHg
Step 3: Calculate the final volume of the gas
Assuming ideal behavior and constant temperature, we can calculate the final volume using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁/P₂
V₂ = 749 mmHg × 3.65 L/861.0 mmHg = 3.18 L
Answer:
its 0.163 g
Explanation:
From the total pressure and the vapour pressure of water we can calculate the partial pressure of O2
PO 2 =P t −P H 2 O
= 760 − 22.4
= 737.6 mmHg
From the ideal gas equation we write.
W= RT/PVM = (0.0821Latm/Kmol)(273+24)K(0.974atm)(0.128L)(32.0g/mol/) =0.163g
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