Answer : The correct option is, (b) +115 J/mol.K
Explanation :
Formula used :
where,
= change in entropy
= change in enthalpy of vaporization = 40.5 kJ/mol
= boiling point temperature = 352 K
Now put all the given values in the above formula, we get:
Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K
Answer:
Explanation:
You don't give the reaction, but we can get by just by balancing atoms of Na.
We know we will need the partially balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.
M_r: 142.04
2NaOH + … ⟶ Na₂SO₄ + …
n/mol: 0.75
1. Use the molar ratio of Na₂SO₄ to NaOH to calculate the moles of NaF.
Moles of Na₂SO₄ = 0.75 mol NaOH × (1 mol Na₂SO₄/2 mol NaOH
= 0.375 mol Na₂SO₄
2. Use the molar mass of Na₂SO₄ to calculate the mass of Na₂SO₄.
Mass of Na₂SO₄ = 0.375 mol Na₂SO₄ × (142.04 g Na₂SO₄/1 mol Na₂SO₄) = 53 g Na₂SO₄
The reaction produces of Na₂SO₄.
Answer:
0.03atm
Explanation:
Given parameters:
Total pressure = 780torr
Partial pressure of water vapor = 1.0atm
Unknown:
Partial pressure of radon = ?
Solution:
A sound knowledge of Dalton's law of partial pressure will help solve this problem.
The law states that "the total pressure of a mixture of gases is equal to the sum of the partial pressures of the constituent gases".
Mathematically;
P = P
+ P
+ P
Since the total pressure is 780torr, convert this to atm;
760torr = 1 atm
780torr = atm = 1.03atm
For this problem;
Total pressure = Partial pressure of radon + Partial pressure of water vapor
1.03 = Partial pressure of radon + 1.0
Partial pressure of radon = 1.03 - 1.00 = 0.03atm