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1.198 g of an unknown hydrocarbon (111.8 g/mol) burns in a bomb calorimeter in excess oxygen. The heat capacity of the calorimeter, [tex]C_v[/tex], is 5.135 kJ/ºC, and [tex]\Delta T = 7.869[/tex] ºC.

1. Find [tex]\Delta E[/tex] per mole for this hydrocarbon in kJ.

2. Calculate the change in entropy (in J/K) when 98.2 g of water is heated from 28.7 ºC to 76.7 ºC at 1 atm. (The specific heat is 4.184 J/(g-K).)

Answer :

Final answer:

To find ΔE per mole for the hydrocarbon, we can calculate the energy change by using the equation ΔE = Cv × ΔT. The change in entropy can be calculated using the equation ΔS = m × C × ΔT.

Explanation:

First, we need to calculate the energy change (ΔE) for the combustion of the hydrocarbon. We can use the equation × Δ = v × Δ to find ΔE, where Δ is the change in temperature and v is the heat capacity of the calorimeter. Given that the mass of the hydrocarbon is 1.198 g and its molar mass is 111.8 g/mol, we can calculate the number of moles of the hydrocarbon.

Then, we can use the molar ratio between the hydrocarbon and oxygen to determine the amount of oxygen consumed in the reaction. With this information, we can calculate the amount of energy released in the combustion of 1 mole of the hydrocarbon. Finally, we can convert this energy change to kJ by dividing it by 1000.

The change in entropy when heating the water can be calculated using the equation ΔS = m × C × Δ, where m is the mass of water, C is the specific heat capacity, and Δ is the change in temperature. Substituting the given values, we can calculate ΔS and convert it to J/K.

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