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Mothballs are composed primarily of the hydrocarbon naphthalene (C10H8 ). When 1.312 g of naphthalene burns in a bomb calorimeter, the temperature rises from 24.710 ∘C to 30.188 ∘C .Find ΔrH for the combustion of naphthalene at 298 K . When considering phase, assume all reactants and products are at 298 K.

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Step-by-step explanation:

To find ΔrH for the combustion of naphthalene, we can use the equation: ΔrH = qp / n

where qp is the heat absorbed by the bomb calorimeter and n is the number of moles of naphthalene.

First, we need to find qp. We can use the formula: qp = mcΔT

where m is the mass of naphthalene, c is the specific heat capacity of the calorimeter, and ΔT is the change in temperature.

Given that m = 1.312 g, c = 4.184 J/g°C, and ΔT = 30.188 - 24.710 = 5.478 °C

qp = (1.312 g)(4.184 J/g°C)(5.478 °C) = 27.948 J

Next, we need to convert the mass of naphthalene to moles. We can use the molar mass of naphthalene, which is 128.17 g/mol

n = m / M = 1.312 g / 128.17 g/mol = 0.0103 mol

Finally, we can use the equation ΔrH = qp / n to find ΔrH:

ΔrH = qp / n = 27.948 J / 0.0103 mol = 2713.8 J/mol

Therefore, the ΔrH for the combustion of naphthalene at 298 K is 2713.8 J/mol.

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