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1. What would be the AE (in kJ) for the following reaction if 6.300 moles of H20 were decomposed? 2 H2O + 2H2 + O AE = 483.6 please answer the full thing
7. Two solutions, initially at 24.60 °C, are mixed in a coffee cup calorimeter (Ccal = 15.5 J/°C). When a 100.0 mL volume of
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Answer #1

AE = 483.6 KJ js 2th0 2t2 +02 Whan 2 md H,0deiompoar 483.6 KJ AE Rer 63 mol KJ 936x6-2ph) 2mp )/ 11 Be 1523.3 KT fi 6.3mol eftoten, loa.5 c anad RCI fosmad DE is -336KJ 120 -336 KJX13 /1 109.59 739.8901 KT AE=-39.990y KJ okan HCt is emed M 1 Cal 4.18

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