I'm curious on what it is but I wish I could help
Answer:
1367.7 g of ethylene glycol was added to the solution
Explanation:
In order to find out the mass of glycol we added, we apply the colligative property of lowering vapor pressure: ΔP = P° . Xm
ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution(P')
525.8 mmHg - 451 mmHg = 451 mmHg . Xm
74.8 mmHg / 451 mmHg = Xm → 0.166 (mole fraction of solute)
Xm = Mole fraction of solute / Moles of solute + Moles of solvent
We can determine the moles of solvent → 2000 g . 1 mol/18 g = 111.1 mol
(Notice we converted the 2kg of water to g)
0.166 = Moles of solute / Moles of solute + 111.1 moles of solvent
0.166 (Moles of solute + 111.1 moles of solvent) = Moles of solute
18.4 moles = Moles of solute - 0.166 moles of solute
18.4 = 0.834 moles of solute → Moles of solute = 18.4/0.834 = 22.06 moles
Let's convert the moles to mass → 62 g/mol . 22.06 mol = 1367.7 g
4 moles of water are produced
Explanation:
- 4 moles of water are produced when 5 moles of hydrogen is reacted with 2 moles of oxygen gas
- The balanced equation given is when 2 moles of hydrogen reacts with 1 mole of oxygen and it forms 2 moles of water.
- The equation we have to solve is the 5 moles of hydrogen is reacting with 2 moles of oxygen gas, we can write the equation as
- This is the balanced equation when 5 moles of hydrogen reacts with 2 moles of oxygen. The balanced equation means the number of hydrogen atoms and oxygen atoms on both sides would be equal in number.
The unit of momentum is the product of the units of mass and velocity.
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