Physics, 04.09.2019 17:20 julialombardo53
Given the thermochemical equations x2+3y2⟶2xy3δh1=−370 kj x2+2z2⟶2xz2δh2=−120 kj 2y2+z2⟶2y2zδh3=−270 kj calculate the change in enthalpy for the reaction. 4xy3+7z2⟶6y2z+4xz2
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Given the thermochemical equations x2+3y2⟶2xy3δh1=−370 kj x2+2z2⟶2xz2δh2=−120 kj 2y2+z2⟶2y2zδh3=−270...
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