Um try B? i think that would be correct. either B or C
Answer:
Explanation:
Data:
H₃O⁺ + OH⁻ ⟶ 2H₂O
V/mL: 50.0 50.0
c/mol·L⁻¹: 1.0 1.0
ΔT = 6.5 °C
ρ = 1.210 g/mL
C = 4.18 J·°C⁻¹g⁻¹
C_cal = 12.0 J·°C⁻¹
Calculations:
(a) Moles of acid

(b) Volume of solution
V = 50.0 mL + 50.0 mL = 100.0 mL
(c) Mass of solution

(d) Calorimetry
There are three energy flows in this reaction.
q₁ = heat from reaction
q₂ = heat to warm the water
q₃ = heat to warm the calorimeter
q₁ + q₂ + q₃ = 0
nΔH + mCΔT + C_calΔT = 0
0.0500ΔH + 1.10×4.18×6.5 + 12.0×6.5 = 0
0.0500ΔH + 2989 + 78.0 = 0
0.0500ΔH + 3067 = 0
0.0500ΔH = -3067
ΔH = -3067/0.0500
= -61 000 J/mol
= -61 kJ/mol

Note: The answer can have only two significant figures because that is all you gave for the change in temperature.
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