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Chemistry, 27.11.2019 20:31 izzy123abc

Given the thermochemical equations x 2 + 3 y 2 ⟶ 2 xy 3 δ h 1 = − 380 kj x 2 + 2 z 2 ⟶ 2 xz 2 δ h 2 = − 130 kj 2 y 2 + z 2 ⟶ 2 y 2 z δ h 3 = − 260 kj calculate the change in enthalpy for the reaction. 4 xy 3 + 7 z 2 ⟶ 6 y 2 z + 4 xz 2

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Given the thermochemical equations x 2 + 3 y 2 ⟶ 2 xy 3 δ h 1 = − 380 kj x 2 + 2 z 2 ⟶ 2 xz 2 δ h 2...
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