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When one mole of C_6 H_6 is burned, 3.27 MJ of hea

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Answer #1

Heat energy produced by combustion of 6.05 g of Benzene ( C6H6 ):

(6.05 g C6H6) / (78.11236 g C6H6/mol) x (3.27 x 10^6 J/mol) = 253269.7 J

This heat enegy is absorbed by the water . Thus the tempature raised is

delta T = Q / mc

(253269.7 J ) / (4.183 J/g·°C) / (5690 g) = 10.6 °C

Thus, the temperature of water will be

21.0°C + 10.6 °C = 31.6°C

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