Answer:
Water ______________________________________________
Answer:
ΔH
°fCaO= -655.09 KJ/mol = ΔH
°fCaO
Explanation:
ΔH
°(reaction)a=ΔH
°f(product)-ΔH
°f(reactant)
where,
ΔH
°f(product)=is the standard heat of formation of products
-ΔH
°f(reactant)=is the standard heat of formation of reactants
The standard heat of formation of H
2
O
(
l
) is −
285.8
k
J
/
m
o
l
.
The standard heat of formation of Ca(OH
)2 is −
986.09
k
J
/
m
o
l
.
For the given reaction, the standard enthalpy change is calculated by the expression shown below.
ΔH
°(reaction)a=ΔH
°f(CaO+H2O)-ΔH
°f(Ca(OH)₂
65.2 = ΔH
°fCaO+( -265.8) -(-986.09)
= ΔH
°fCaO-265.8+986.09
65.2 = ΔH
°fCaO+ 720.29
65.2-720.29 = ΔH
°fCaO
-655.09 KJ/mol == ΔH
°fCaO
When KOH and Cu(NO3)2 are mixed it yields copper(II) hydroxide and potassium nitrate. The balanced reaction is:
Cu(No3)2(aq) + 2 KOH(aq)<span> = Cu(OH)2(s) + 2 KNo3(aq)
</span>
As we can see from the equation a solid is formed therefore a
precipitate is formed which is the copper (II) hydroxide which has a blue to
purple appearance. This can be observed since copper
(II) hydroxide has a low
solubility in aqueous solution.
