We’re being asked to determine the energy of the combustion per mole of C_{14}H_{10}.

We will use the **heat released** by the sample of steam to calculate the final temperature of the mixture.

Recall that **heat** (**q**) can be calculated using the following equation:

$\overline{){\mathbf{q}}{\mathbf{=}}{\mathbf{mc}}{\mathbf{\u2206}}{\mathbf{T}}}$

q = heat, J

• **+q** → **absorbs **heat

• **–q** → **l****oses **heat

m = mass (g)

c = specific heat capacity = J/(g·°C)

ΔT = T_{f} – T_{i} = (°C)

When 0.550 g of C_{14}H_{10} (molar mass = 178.22 g mol ^{–1} ) is combusted in a bomb calorimeter that has a water jacket containing 400. g of water, the temperature of the water increases by 10.54 °C (specific heat of water = 4.184 J g^{–1} °C^{–1} ). Assuming the heat absorbed by the walls of the calorimeter is negligible, calculate the energy change (ΔE) for the combustion reaction per mole of C_{14}H_{10}.

A. –5.43 × 10^{–2} kJ mol^{–1}

B. +17.6 kJ mol^{–1}

C. –17.6 kJ mol^{–1}

D. –5.70 × 10^{3} kJ mol^{–1}

E. +5.70 × 10^{3} kJ mol^{–1}

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