Suppose 0.100 mol of a metal (M) is combusted in excess O2lg) in a bomb calorimeter. 4M(s) + 302 – 2M2O3(s) The heat capacity of the empty calorimeter is 458 J/°C and the calorimeter contains 450 mL of water. As a result of the reaction, the temperature of the water rises from 25.0°C to 40.8°C. Calculate AE for the reaction in KJ/molpxn: Assume, DH,0 = 1.00 mL SH,0 = 4.18 °C -1.48 X 103 KJ/mol O +1.48 X 10³ KJ/mol +3.35 X 103 KJ/mol O -3.35 X 103 KJ/mol

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Suppose 0.100 mol of a metal (M) is combusted in excess O2(g) in a bomb calorimeter.
4M(s) + 302
2M203(s)
>
The heat capacity of the empty calorimeter is 458 J/°C and the calorimeter contains 450 mL of water. As a result of the
reaction, the temperature of the water rises from 25.0°C to 40.8°C.
Calculate AE for the reaction in KJ/molxn: Assume,
DH20
= 1.00 8_
mL
J
SH20 = 4.18
g.°C
O -1.48 X 103 KJ/mol
+1.48 X 103 KJ/mol
+3.35 X 103 KJ/mol
-3.35 X 103 KJ/mol
Transcribed Image Text:Suppose 0.100 mol of a metal (M) is combusted in excess O2(g) in a bomb calorimeter. 4M(s) + 302 2M203(s) > The heat capacity of the empty calorimeter is 458 J/°C and the calorimeter contains 450 mL of water. As a result of the reaction, the temperature of the water rises from 25.0°C to 40.8°C. Calculate AE for the reaction in KJ/molxn: Assume, DH20 = 1.00 8_ mL J SH20 = 4.18 g.°C O -1.48 X 103 KJ/mol +1.48 X 103 KJ/mol +3.35 X 103 KJ/mol -3.35 X 103 KJ/mol
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