Using the Born- Haber cycle, the OH_f° of KBr is equal to _ _. A) OH_f°[K (g)] + OH_f°[Br (g)] - I₁- E(Br) + OH_lattice B) OH_f°[K (g)] + OH_f°[Br (g)] + I₁(K) + E(Br) + OH_lattice C) OH_f°[K (g)] - OH_f°[Br (g)] + I₁(K) - E(Br) + OH_lattice D) OH_f°[K (g)] + OH_f°[Br (g)] + I₁(K) + E(Br) - OH_lattice E) OH_f°[K (g)] - OH_f°[Br (g)] - I₁(K) - E(Br) - OH_lattice

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The Answer of Using the Born- Haber cycle, the OH_f° of KBr is equal to _ _. A) OH_f°[K (g)] + OH_f°[Br (g)] - I₁- E(Br) + OH_lattice B) OH_f°[K (g)] + OH_f°[Br (g)] + I₁(K) + E(Br) + OH_lattice C) OH_f°[K (g)] - OH_f°[Br (g)] + I₁(K) - E(Br) + OH_lattice D) OH_f°[K (g)] + OH_f°[Br (g)] + I₁(K) + E(Br) - OH_lattice E) OH_f°[K (g)] - OH_f°[Br (g)] - I₁(K) - E(Br) - OH_lattice

Using the Born- Haber Cycle, the OHf° of KBr Is

Using the Born- Haber Cycle, the OH_f° of KBr Is