The question is incomplete , complete question is:
Hydrogen, a potential future fuel, can be produced from carbon (from coal) and steam by the following reaction:
![C(s)+ 2 H_2O(g)\rightarrow 2H_2(g)+CO_2(g).\Delta H=?](https://tex.z-dn.net/?f=C%28s%29%2B%202%20H_2O%28g%29%5Crightarrow%202H_2%28g%29%2BCO_2%28g%29.%5CDelta%20H%3D%3F)
Note that the average bond energy for the breaking of a bond in CO2 is 799 kJ/mol. Use average bond energies to calculate ΔH of reaction for this reaction.
Answer:
The ΔH of the reaction is -626 kJ/mol.
Explanation:
![C(s)+ 2 H_2O(g)\rightarrow 2H_2(g)+CO_2(g).\Delta H=?](https://tex.z-dn.net/?f=C%28s%29%2B%202%20H_2O%28g%29%5Crightarrow%202H_2%28g%29%2BCO_2%28g%29.%5CDelta%20H%3D%3F)
We are given with:
![\Delta H_{H-O}=459 kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H_%7BH-O%7D%3D459%20kJ%2Fmol)
![\Delta H_{H-H}=432 kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H_%7BH-H%7D%3D432%20kJ%2Fmol)
![\Delta H_{C=O}=799 kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H_%7BC%3DO%7D%3D799%20kJ%2Fmol)
ΔH = (Energies required to break bonds on reactant side) - (Energies released on formation of bonds on product side)
![\Delta H=(4\times \Delta H_{O-H})-(2\times \Delta H_{H-H}+2\times\Delta H_{C=O})](https://tex.z-dn.net/?f=%5CDelta%20H%3D%284%5Ctimes%20%5CDelta%20H_%7BO-H%7D%29-%282%5Ctimes%20%5CDelta%20H_%7BH-H%7D%2B2%5Ctimes%5CDelta%20H_%7BC%3DO%7D%29)
![=(4\times 459 kJ/mol)-(2\times 432 kJ/mol+2\times 799 kJ/mol](https://tex.z-dn.net/?f=%3D%284%5Ctimes%20459%20kJ%2Fmol%29-%282%5Ctimes%20432%20kJ%2Fmol%2B2%5Ctimes%20799%20kJ%2Fmol)
![\Delta H=-626 kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H%3D-626%20kJ%2Fmol)
The ΔH of the reaction is -626 kJ/mol.
Metallic bonds are found in metals like zinc.
Answer:
u already know the answer
Explanation:answer
Covalent
bonds = sharing of electrons between two atoms of the same elements or elements
close to each other on the periodic table. Usually they are metals sometimes
non-metals. In polar bonds electrons are
shared unequally. Non polar bonds share electrons equally.