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Chemistry

Given:

Cu2O(s) + 1/2 O2(g) → 2CuO(s) deltaH°= -144 kJ
2Cu2O(s) → 2Cu(s) + 2CuO(s) deltaH°= 22 kJ

Calculate the standard enthalpy of formation of CuO(s).

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  1. multiply 1 by 2 and add to the reverse of 2. That will give you 2Cu + O2 ==> 2CuO which is just twice what you want so divide by 2 for the equation and divide the dH by 2. Remember to multiply dH by 2 when multiplying the equation and remember to reverse the sign of dH when reversing the equation.

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