Answer:
The standard enthalpy of formation of methanol is, -238.7 kJ/mole
Explanation:
The formation reaction of CH_3OH will be,
The intermediate balanced chemical reaction will be,
..[1]
..[2]
..[3]
Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:
We get :
..[1]
..[2]
[3]
The expression for enthalpy of formation of will be,
The standard enthalpy of formation of methanol is, -238.7 kJ/mole
part 1 : the final volume : 1.404 L
part 2 : the initial concentration : 4.06 M
<h3>Further explanation </h3>Dilution is the process of adding a solvent to get a more dilute solution.
The moles(n) before and after dilution are the same.
Can be formulated :
M₁V₁=M₂V₂
M₁ = Molarity of the solution before dilution
V₁ = volume of the solution before dilution
M₂ = Molarity of the solution after dilution
V₂ = Molarity volume of the solution after dilution
part 1 :
M₁=44.8%
V₁=0.73 L
M₂=23.3%
part 2 :
V₁=739 ml=0.739 L
V₂=1.5 L
M₂=2