Answer:
NiCO3 (s) + 2H+ (aq) → H2O (l) + CO2 (g) + Ni2+ (aq)
Explanation:
To write the complete ionic equation:
1. Start with a balanced molecular equation.
2. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions
3. indicate the correct formula and charge of each ion
4. indicate the correct number of each ion
5. write (aq) after each ion
6. Bring down all compounds with (s), (l), or (g) unchanged.
Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bn%5Ctimes%20H_f_%7Bproducts%7D%5D-%5Bn%5Ctimes%20H_f_%7Breactants%7D%5D)
Putting the values we get :
![\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B8%5Ctimes%20H_f_%7BCO_2%7D%2B10%5Ctimes%20H_f_%7BH_2O%7D%5D-%5B2%5Ctimes%20H_f_%7BC_4H_%7B10%7D%2B13%5Ctimes%20H_f_%7BO_2%7D%7D%5D)
![\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%288%5Ctimes%20-393.5%29%2B%2810%5Ctimes%20-241.82%29%5D-%5B%282%5Ctimes%20-125.7%29%2B%2813%5Ctimes%200%29%5D)
2 moles of butane releases heat = 5314.8 kJ
1 mole of butane release heat = 
Thus enthalpy of combustion per mole of butane is -2657.4 kJ
Answer:
Whether you get the metal or hydrogen during electrolysis depends on the position of the metal in the reactivity series: the metal will be produced if it is less reactive than hydrogen. hydrogen will be produced if the metal is more reactive than hydrogen.
