The partial stress of H2 is 737.47 mmHg Let's observe the Ideal Gas Law to find out the whole mols.
We count on that the closed vessel has 1L of volume
- P.V=n.R.T
- We must convert mmHg to atm. 760 mmHg.
- 1 atm
- 755 mmHg (755/760) = 0.993 atm
- 0.993 m.1L=n.0.082 L.atm/mol.K .
- 293 K(0.993 atm 1.1L)/(0.082mol.K /L.atm).
- 293K = n
- 0.0413mols = n
These are the whole moles. Now we are able to know the moles of water vapor, to discover the molar fraction of it.
- P.V=n.R.T
- 760 mmHg. 1 atm
- 17.5 mmHg (17.5 mmHg / 760 mmHg)=0.0230 atm
- 0.0230 m.1L=n.0.082 L.atm/mol.K.293 K(0.0230atm.1L)/(0.082mol.K/L.atm .293K)=n 9.58 × 10 ^ 4 mols = n.
- Molar fraction = mols )f gas/general mols.
- Molar fraction water vapor =9.58×10^ -four mols / 0.0413 mols
- Sum of molar fraction =1
- 1 - 9.58 × 10 ^ 4 × mols / 0.0413 ×mols = molar fraction H2
- 0.9767 = molar fraction H2
- H2 pressure / Total pressure =molar fraction H2
- H2 pressure / 55mmHg = =0.9767 0.9767 = h2 pressure =755 mmHg.
- 737,47 mmHg.
<h3>What is a mole fraction?</h3>
Mole fraction is a unit of concentration, described to be identical to the variety of moles of an issue divided through the whole variety of moles of a solution. Because it's miles a ratio, mole fraction is a unitless expression.
Thus it is clear that the partial pressure of H2 is 737,47 mmHg.
To learn more about partial pressure refer to the link :
brainly.com/question/19813237
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Answer:
a. C₄H₁₀/O₂ = 2:13 (example)
b. O₂/CO₂ = 13:8
c. O₂/H₂O = 13:10
d. C₄H₁₀/CO₂ = 2:8
e. C₄H₁₀/H₂O = 2:10
Explanation:
2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O
The molar ratios are the same as the coefficients in front of the formulas.
Answer:
43.96
Explanation:
Graham's law was applied and the rates of effusion of nitrogen and the unknown gas were compared as shown in the image. The unknown gas is heavier than hydrigen hence it effuses slower than hydrogen as anticipated by Graham's law.
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