Below is the graph that the student attached to her report. Let\'s think how we

Question

Below is the graph that the student attached to her report. Let's think how we can use it to calculate the heat of neutralization reaction. in this experiment, the student measured the change in temperature not for the system itself but for the surroundings (solution). the solution is diluted and we can safely consider its specific heat and its density to be the same as for pure water: c = 4.18 J/g-degree C and d = 1.00 g/cm^3. What is the change in temperature recorded in this experiment? Delta T = degree C Did the system loose or gain energy in this experiment? Explain.

Explanation / Answer

Q4.

a)

find the chang ein T in experiment

Tinitial = 25.25 °C approx

Tfinal = 24.25 °C approx

dT = Tfinal - Tinitial = 24.25 - 25.25 = -1 °C

therefore, the system is losing heat, i.e. it is endothermic, which is INCORRECT, since NaOH + HCl ,neutralization is almost always exothermic

But experimental data states otherwise

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