Answer:
ΔH = - 272 kJ
Explanation:
We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:
N2(g) + 3H2(g) → 2NH3(g) ΔH=−92.kJ Multiplying by 2:
2N2(g) + 6H2(g) → 4NH3(g) ΔH=− 184 kK
plus
4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ΔH=−905.kJ
__________________________________________________
2N2(g) + 6H2(g) + 5O2(g)→ 4NO(g) + 6H2O(g) ΔH = (-184 +(-905 )) kJ
ΔH = -1089 kJ
Notice how the intermediate NH3 cancels out.
As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH for the formation of one mol NO:
-1089 kJ/4 mol NO x 1 mol NO = -272 kJ (rounded to nearest kJ)
3 lol i was thinking about that from the ad
Well, all of this we owe it to Bohr who analyzed the atomic emission spectrum of hydrogen and he could probe matematically that it was a result of movement of e- from an especific energy level to a lower one. The understanding of levels of energy took to the development of the atomic theory
Answer:
3
Explanation:
calcium (Ca)
carbon(C)
oxygen(O)
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For any question comment me
1. Diatomic
2. Products
3. Reactants
4. Coefficient
5. Subscript
6. True
7. False
8. First you must write the reactants. Ensure the valencies of the reactants are correct. Draw an arrow. Write the products. Ensure the valencies of the products are correct and apply subscripts as necessary. Balance the equation by ensuring both sides have equal amounts of each elements - use only coefficients to complete this NOT subscripts.
