Answer:
CaCO3(aq) → Ca2+(aq ) + CO3 2-(aq)
Explanation:
- Dissolution reactions are reactions that occur when a solute either in gaseous, liquid, or solid form dissolves in a solvent such as water to form a solution.
- In this case we are given Calcium carbonate (CaCO3) which undergoes dissolution according to the equation;
- CaCO3(aq) → Ca2+(aq ) + CO3 2-(aq)
- Then<em><u> the bicarbonate ion combines with two protons from water to form a weak acid H2CO3. The weak acid is then broken down to form CO2 and H2O since its unstable.</u></em>
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
Answer:
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
Explanation:
Step 1: Data given
The combustion reaction of octane produces 5104.1 kJ per mol octane
Step 2: The balanced equation
C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g) ∆H°rxn = -5104.1 kJ/mol
Step 3:
∆H°rxn = ∆H°f of products minus the ∆H° of reactants
∆H°rxn = ∆H°f products - [∆H°f reactants]
-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)
∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol
∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ
/mol
∆H°f C8H18 = -220.1 kJ/mol
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol