In an experiment to calculate heat of neutralization reaction using a calorimete

Question

In an experiment to calculate heat of neutralization reaction using a calorimeter uVmg a coffee cup, a group of students added 1.969 g of solid NaOH to 100.0 ml (100.0 g) of 0.50 M HCl solution. The initial temperature of the solution was 22.5 degree C and final temperature measured was 33.9 degree C. Calculate the heat of neutralization (q) in this reaction. (use SH of water as 4.18 J/g- degree C). Show all steps of calculations. Give final value of q in J Finish with KJ. Show Table values Mass of solution Delta t degree C = H = Show formula of q and substitute Catenation step with units

Explanation / Answer

mass of solution = 100 g.

Specific heat = 4.18 J/0C.g

Change in temperature = 33.9 - 22.5 = 11.4 0C

Therefore, heat change = m * s * (t2-t1) = 100 * 4.18 * 11.4 = 4765.2 J = 4.76 kJ

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