Given :           (1) 2NF3 + 2NO  → N2F4  + 2ONF     ∆H= -82.9 kJ           (2) NO + ½ F2  → ONF                       ∆H = -156.9 kJ           (3) Cu + F2  → CuF2                           ∆ H= -531.0 kJ       Calculate the value of ∆H for the reaction:            2NF3 + Cu → N2F4    +  CuF2             ∆H = ? Select one: a. -400.1 KJ b. -200.1 kJ c. -100.1 kJ d. -300.1 Kj

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8  Given :

          (1) 2NF3 + 2NO  → N2F4  + 2ONF     ∆H= -82.9 kJ

          (2) NO + ½ F2  → ONF                       ∆H = -156.9 kJ

          (3) Cu + F2  → CuF2                           ∆ H= -531.0 kJ

      Calculate the value of ∆H for the reaction:

           2NF3 + Cu → N2F4    +  CuF2             ∆H = ?

Select one:
a.
-400.1 KJ
b.
-200.1 kJ
c.
-100.1 kJ
d.
-300.1 Kj
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