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Consider the following equations. 3 A + 6 B → 3 D, ΔH = -440. kJ/mol...

Consider the following equations.

3 A + 6 B → 3 D, ΔH = -440. kJ/mol

E + 2 F → A, ΔH = -102.5 kJ/mol

C → E + 3 D, ΔH = +64.5 kJ/mol

Suppose the first equation is reversed and multiplied by 1/6, the second and third equations are divided by 2, and the three adjusted equations are added. What is the net reaction?

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Ans: suppose the first equation is reversed and multiply by 16, we get 3D 3A +6Bar - +44 Okimo a (when the reaction is revers

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