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17. When 13.8 ml of 0.870 M lead(II) nitrate reacts with 90.0 ml of 0.777 Msodium chloride, 0.279 kJ of heat is released at c
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11) moles of Pb(NO3), = 13.8 X0-870/1000 0.012 S moles of Nad , 900X0.777/1000 = 0.06993 Pb(NO₃)₂ + 2 Nad obel, + 2NaNO3 0.01

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17. When 13.8 ml of 0.870 M lead(II) nitrate reacts with 90.0 ml of 0.777 Msodium chloride, 0.279 kJ of heat is rel...
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