To determine the enthalpy of combustion of octane, we first need to calculate the heat absorbed by the water in the calorimeter. We can do this using the formula:q = mcTwhere q is the heat absorbed, m is the mass of the water, c is the specific heat capacity of water, and T is the change in temperature.Given:Mass of water m = 200 mL assuming the density of water is 1 g/mL, this is equivalent to 200 g Specific heat capacity of water c = 4.18 J/gCInitial temperature T_initial = 25.0 CFinal temperature T_final = 45.0 CFirst, we calculate the change in temperature T :T = T_final - T_initial = 45.0 C - 25.0 C = 20.0 CNow, we can calculate the heat absorbed by the water q :q = mcT = 200 g 4.18 J/gC 20.0 C = 16720 JNext, we need to determine the moles of octane burned. The molecular weight of octane C8H18 is: 12.01 g/mol x 8 + 1.01 g/mol x 18 = 114.23 g/molGiven that 1.50 grams of octane was burned, we can calculate the moles of octane:moles of octane = 1.50 g / 114.23 g/mol = 0.0131 molNow, we can calculate the enthalpy of combustion of octane H_combustion per mole:H_combustion = -q / moles of octane = -16720 J / 0.0131 mol = -1274800 J/molThe enthalpy of combustion of octane is approximately -1,274,800 J/mol.