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2AgBr(s) ⇒ 2Ag(s) + Br₂(l) ΔH = -220kJ ⇒ for 2 mol Ag
mol Ag = 7.08 : 108 g/mol = 0.065
the enthalpy change = 0.065/2 x -220 kJ = -7.15 kJ
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