Answer:
fH = - 3,255.7 kJ/mol
Explanation:
Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.
Qsystem + Qcombustion = 0
Qsystem = heat capacity*ΔT
10000*(25.000 - 20.826) + Qc = 0
Qcombustion = - 41,740 J = - 41.74 kJ
So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:
n = mass/molar mass = 1/ 78
n = 0.01282 mol
fH = -41.74/0.01282
fH = - 3,255.7 kJ/mol
Answer:
Molality of the solution = 0.7294 M
Explanation:
Given:
Number of magnesium arsenate = 1.24 moles
Mass of solution = 1.74 kg
Find:
Molality of the solution
Computation:
Molality of the solution = Mole of solute / Mass of solution = 1.74 kg
Molality of the solution = 1.24 / 1.7
Molality of the solution = 0.7294 M
Color change, temperature change, bubbling, state change
green to blue, hot to cold, bubbles (lol), and liquid to gas
