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    can someone plese help me do this problem...

    Calculate the energy change for the formation of 1.19 mol of NaBr(s) given the following information.

    Since Br2is a liquid at room temperature, it requires one more step than is in your text. That step is given here.

    Br2(l) --> Br2(g) : 30.92 KJ/mol

    Na(s) --> Na(g) : 107.32 KJ/mol

    2Br(g) --> Br2(g) : -254.66 KJ/mol

    Na(g) --> Na+(g) + e- : 496 KJ/mol

    Br(g) + e- --> Br-(g) : -324.7 KJ/mol

    Lattice energy for NaBr = -752 KJ/mol

    i would appreciate it greatly!!!
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