<u>Answer:</u> For the given equation, only iron has the value of 
equal to 0 kJ.
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%5Eo_f%28reactant%29%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(Fe(s))})+(3\times \Delta H^o_f_{(CO_2(g))})]-[(3\times \Delta H^o_f_{(CO(g))})+(2\times \Delta H^o_f_{(Fe_2O_3(s))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28Fe%28s%29%29%7D%29%2B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO_2%28g%29%29%7D%29%5D-%5B%283%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28CO%28g%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28Fe_2O_3%28s%29%29%7D%29%5D)
The enthalpy of formation for the substances present in their elemental state is taken as 0.
Here, iron is present in its elemental state which is solid.
Hence, for the given equation, only iron has the value of equal to 0 kJ.
Answer:
Barium and sulpher are the elements that forms barium sulphide .
And its molecular fomula is BaS.
Explanation:
The answer is 12g of Carbon
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For N₂,
E.N of Nitrogen = 3.04
E.N of Nitrogen = 3.04
________
E.N Difference
0.00 (Non Polar Covalent)
For Na₂O,
E.N of Oxygen = 3.44
E.N of Sodium = 0.93
________
E.N Difference 2.51 (Ionic)
For CO₂,
E.N of Oxygen = 3.44
E.N of Carbon = 2.55
________
E.N Difference 0.89 (Polar Covalent)
