Question

104 Experiment 14 Heat Efects and Calorimetry Mass of beaker plus solid 126.9 49.7 Mass of water, Mis of solid, m. 21-2 Initial temperature Final temperature joules Ho for the reaction (Eq. 1) (S.H. - 4.18 Je") joules AH for the reaction (Eq. 5) The quantity you have just calculated is approximately equal to the heat of solution of your sample Calculate the heat of solution per gram of solid sample. _joules/g The solution reaction is endothermic exothermic. (Underline the correct answer Give your reasoning Solid unknown #1352 Molar mass g/mol Optional Formula of compound used (if furnished) Heat of solution per mole of compound kJ/mol C. Heat of Neutralization Initial temperature of HCl solution Initial temperature of NaOH solution 22. 0 °C 22.0°C 28.6 °C Final temperature of neutralized mixture Change in temperature. A/ (use the average of the initial temperatures of HCl and NaOH) 6.6 c 46,0 (assume 50.0 mL of solution and use average density of 1.02 g/mL.) Total AH for the neutralization reaction AH per mole of Hand OH ions reacting kJ/mol

Answer 1

The heat released by the water is calculated:

q H2O = m * cp * ΔT = 49.7 g * 4.18 * (15.7 - 21.2) = - 1142.6 J

You have the reaction enthalpy change:

ΔH = - q = - (-1142.6) = 1142.6 J

The enthalpy change per gram is calculated:

ΔH = ΔH / m = 1142.6 J / 5 g = 228.52 J / g

Neutralization:

The heat of neutralization is calculated:

q = V * d * cp * ΔT = 50 mL * 1.02 g / mL * 4.18 J / g * ° C * 6.6 ° C = 1407 J

You have the enthalpy change:

ΔH = - q = - 1407 J

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