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determine the enthalpy change when 7.08g of Ag is obtained. use the abbreviated form of the unit in your answer.
2AgBr(s) -> 2Ag(s) + Br2(l) deltaH = -220kJ

Respuesta :

indrawboy

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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