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Given the thermochemical equations X2+3Y2⟶2XY3ΔH1=−390 kJX2+3Y2⟶2XY3ΔH1=−390 kJ X2+2Z2⟶2XZ2ΔH2=−180 kJX2+2Z2⟶2XZ2ΔH2=−180 kJ 2Y2+Z2⟶2Y2ZΔH3=−230 kJ2Y2+Z2⟶2Y2ZΔH3=−230 kJ Calculate the change...

Given the thermochemical equations

X2+3Y2⟶2XY3ΔH1=−390 kJX2+3Y2⟶2XY3ΔH1=−390 kJ

X2+2Z2⟶2XZ2ΔH2=−180 kJX2+2Z2⟶2XZ2ΔH2=−180 kJ

2Y2+Z2⟶2Y2ZΔH3=−230 kJ2Y2+Z2⟶2Y2ZΔH3=−230 kJ

Calculate the change in enthalpy for the reaction.

4XY3+7Z2⟶6Y2Z+4XZ24XY3+7Z2⟶6Y2Z+4XZ2

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Given the thermochemical equations X2+3Y2⟶2XY3ΔH1=−390 kJX2+3Y2⟶2XY3ΔH1=−390 kJ X2+2Z2⟶2XZ2ΔH2=−180 kJX2+2Z2⟶2XZ2ΔH2=−180 kJ 2Y2+Z2⟶2Y2ZΔH3=−230 kJ2Y2+Z2⟶2Y2ZΔH3=−230 kJ Calculate the change...
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