(3) Calculate the heat absorbed by the reaction mixture. You may assume the dens
ID: 590700 • Letter: #
Question
(3) Calculate the heat absorbed by the reaction mixture. You may assume the density of the reaction mixture was 1.02 g mL-1 and the specific heat of the reaction mixture was 3.822 Jg"K answer (4) Calculate the amount of heat absorbed by the calorimeter, thermometer, and stirrer, using the calorimeter constant calculated in Question 2 (4) (5) Calculate the total heat absorbed. answer (6) Find the total heat releasecd by the reaction of 100.0 mL of 0.8500M NaOH and 100 mL of 0,8404M acetic acid. answerExplanation / Answer
Q3
Q = m*C*(Tf-Ti)
m = D*V = 1.02 * V
V = 100 mL + 100 mL = 200 mL
Tinitial average = Tbase + Tacid = (17.58+17.80)/2 = 17.70
d T= Tf- Ti = 23.28-17.70 = 5.58
Q = 1.02*200*3.822*(5.58)
Q = 4350.659 J
Q4
heat absorbed by calorimeter + thermometer + stirrer
Qcal = Ccal *dT = CCal*(5.58)
no data on calorimeter...
Q5
Ttoal Heat will be:
Qtotal = 4350.659 + CCal*(5.58)
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