Answer:
4. E) Both A) and C) are true.
5. D) 1.39 J·°C⁻¹g⁻¹
6. C) 5.11 × 10² kJ
Step-by-step explanation:
4. Heat of reaction
H₂(g) + ½O₂(g) ⟶ H₂O(ℓ); ΔH° = -286 kJ
A) The negative sign tells you that energy has gone out of the system. Therefore, the reaction is exothermic.
B) is wrong. The reaction is exothermic.
C) If energy has left the system (and the products are part of the system), the enthalpy of the products is less than that of the reactants,
D) is wrong. Energy is released from the system.
E) Both A) and C) are correct, so E) is the correct answer.
5. Specific heat capacity
q = mCΔT
C = q/(mΔT)
Data:
q = 166.7 J
m = 15.0 g
T₁ = 25.00 °C
T₂ = 33.00 °C
Calculation:
ΔT = T₂ - T₁ = (33.00 - 25.00) °C = 8.00 °C
C = 166.7/(15.0× 8.00) = 1.39 J·°C⁻¹g⁻¹
The specific heat capacity of mercury is 1.39 J·°C·g⁻¹.
6. Heat of combustion
M_r: 46.1
C₂H₅OH + 3O₂ ⟶ 2CO₂ + 3H₂O; ΔH = -1.37 × 10³ kJ
m/g: 17.2
n =17.2/46.1 = 0.3731 mol
q = nΔH = 0.3731 × (-1.37 × 10³) = -511 kJ = -5.11 × 10² kJ
The reaction releases 5.11 × 10² kJ of heat.