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7 March, 04:11

When 1.010 of sucrose (C12H22O11) undergoes combustions in a bomb calorimeter, the temperature rises from 24.92 ◦ C to 28.33 ◦ C. Find DErxn for the combustion of sucrose in kJ/mol sucrose. The heat capacity of the bomb calorimeter determined in a separate experiment is 4.90 kJ/◦ C. (Answer: - 5.66 x 103 kJ/mole).

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  1. 7 March, 07:06
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    Heat of combustion = 5.6 * 10³ kj/mol

    Explanation:

    Given dа ta:

    Mass of sucrose = 1.010 g

    Initial temperature = 24.92 °C

    Final temperature = 28.33 °C

    Heat capacity of calorimeter = 4.90 KJ/°C

    Heat of combustion = ?

    Solution:

    ΔT = 28.33 °C - 24.92 °C = 3.41 °C

    Q = - c. ΔT

    Q = 4.90 KJ/°C. 3.41 °C

    Q = - 16.7 kj

    Number of moles of sucrose:

    Number of moles of sucrose = mass / molar mass

    Number of moles of sucrose = 1.010 g / 342.3 g/mol

    Number of moles of sucrose = 0.003 mol

    Heat of combustion:

    Heat of combustion = Q/n

    Heat of combustion = - 16.7 kj/0.003 mol

    Heat of combustion = - 5.6 * 10³ kj/mol
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