Answer:
a) 1.34 x 10 ³ j
b) 5.35 x 10⁴ j
c) See explanation below.
Explanation:
a) q rxn = m mix x C x ΔT = ( 100 mL x 1.00 g/mL) 4.18 J/g/K x (3.2 K) =
1.34 x 10³ j
b) mol HCl = (50mL x 1L/1000 mL) x 0.500 mol/L = 0.0025 mol
1.33 x 10³ j/ 0.0025 mol x 1mol = 5.35 x 10⁴ joule
c) The heat of reaction is heat released in the neutralization, if we change the amount of HCl to neutralize but do not increase the equivalent number of moles of NaOH, the heat released will be the same.