FInal answer:
The amount of heat released when burning 0.148 mol of the solid in the bomb calorimeter is approximately 18.3 kJ.
Step-by-step explanation:
In a bomb calorimeter, the heat released during a reaction is determined by the temperature change of the surrounding water. The formula to calculate the heat released (q) is given by the equation q = mcΔT, where m is the mass of water, c is the specific heat of water, and ΔT is the temperature change.
First, we need to find the mass of water (m) in grams. Given that the density of water is 1.00 g/mL and the volume is 1000.0 mL, the mass of water is 1000.0 g.
Next, we use the temperature change (ΔT), which is 2.28 °C. The specific heat of water (c) is 4.184 J/K/g. Plugging in these values into the formula, we get q_water = (1000.0 g) * (4.184 J/K/g) * (2.28 °C) = 9568.32 J.
Now, since we know the molar mass of the solid (533.4 g/mol), we can find the mass of the solid burned (2.046 g) corresponds to 2.046 g / 533.4 g/mol = 0.00384 mol.
To find the heat released per mole, we divide the heat released by the moles of the substance: q_per_mol = 9568.32 J / 0.00384 mol = 2493787.5 J/mol.
Finally, to find the heat released when burning 0.148 mol, we multiply the heat released per mole by the number of moles: q_final = 2493787.5 J/mol * 0.148 mol = 369172.05 J. Converting this to kilojoules, we get 369172.05 J / 1000 = 369.17 kJ, which can be rounded to approximately 18.3 kJ.