[Answer] ΔH for the reaction IF5 (g) → IF3 (g) + F2 (g) is ________ kJ given the data below. IF (g) + F2 (g) → IF3 (g) ΔH = -390 kJ IF (g) + 2F2 (g) → IF5 (g) ΔH = -745 kJ+355+35-35-1135+1135

Answer: +355

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ΔH for the reaction IF5 (g) → IF3 (g) + F2 (g) is ________ kJ given the data below. IF (g) + F2 (g) → IF3 (g) ΔH = -390 kJ IF (g) + 2F2 (g) → IF5 (g) ΔH = -745 kJ+355+35-35-1135+1135 Wed Apr 07 2010 · You have to flip the second reation because you have to get the IF5 on the reactant side so when you do that you flip the sign on that reaction s ΔH . So then you have IF(g ) + F2 ( g ) -> IF3 ( g ) ΔH for the reaction IF5 ( g ) → IF3 ( g ) + F2 ( g ) is _____ kJ give the data below . IF (g ) + F2 ( g ) → IF3 ( g ) ΔH = - 390 kJ IF (g ) + 2F2 ( g ) → IF5 ( g ) ΔH = - 745 kJ ΔH for the reaction IF5 (g) → IF3 ( g ) + F2 ( g ) is _____ kJ give the data below . IF (g ) + F2 ( g ) → IF3 ( g ) ΔH = - 390 kJ IF (g ) + 2F2 ( g ) → IF5 ( g ) ΔH = - 745 kJ A) + 355 B) - 1135 C) + 1135 D) +35 E) -35 Scxb commented over 1 year ago ΔH for the reactionIF5( g ) → IF3 ( g ) + F2 ( g ) is _____ kJ give the data below . IF(g ) + F2 ( g ) → IF3 ( g ) ΔH = - 390 kJIF( g ) + 2F2 ( g ) → IF5 ( g ) ΔH = - 745 kJ Wed Apr 07 2010 · You have to flip the second reation because you have to get the IF5 on the reactant side so when you ...

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