Explanation:
Since liquid isopropanol is a polar liquid and water is also a polar solvent. So, when both of them are added together then according to the like dissolves like principle they get dissolved.
At the molecular level, the polar molecules of isopropanol get attracted towards the polar molecules of water at the surface of water.
As a result, water molecules get surrounded by isopropanol. Thus, water molecules enter the solution and evenly spread into the solution.
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
<h3 />
Answer:
1. K<10−3
Explanation:
Equilibrium Constant is an expression which involves the concentration of the product divided by the concentration of the reactant molecules.
However the concentration of the pure liquid and pure solid is regarded as 1.
Equilibrium expression for the equation 2H2(g)+O2(g)⇌2H2O(g)
Equilibrium Constant = [H2O]^2/[H2]^2 x [O2]
Since H2O is a pure liquid, its concentration = 1
There fore;
Equilibrium Constant = 1/[H2]^2 x [O2]
This shows that the Equilibrium Constant of the equation will be less than 1 and greater than 0.
