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CH 301 CH 9 calorim trickier example_answers

# CH 301 CH 9 calorim trickier example_answers - CH301 Notes...

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CH301 Notes: A trickier example of getting a thermochemical equation from data! We will use a coffee-cup calorimeter of heat capacity 27.8 J/ o C to find the thermochemical equation for the acid-base neutralization reaction of NaOH and CH 3 COOH. Data: Add 25.00 mL of NaOH at 23.00 o C to 25.00 mL of CH 3 COOH which is already in the calorimeter and at the same temperature. After mixing, the resulting temperature is observed to be 25.947 o C. Assume that the specific heat of the mixture is the same as that of water, 4.18 J/g 0 C and that the density of the mixture is 1.02 g/mL. temperature change: 25.947 o C – 23 o C = 2.947 o C heat absorbed by calorimeter body = q calorimeter = 2.947 o C * 27.8 J/ o C = 81.9J Now calculate the amount of heat given off in the reaction. We use the volume of the combined solutions and the density of the resulting solution to find its mass, then use its specific heat to find the heat given off in the reaction. Volume of solution in calorimeter = 25.00mL + 25.00mL = 50.00mL Mass of this solution = 50.00mL x 1.02g/mL = 51.0g Heat absorbed by solution = q

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CH 301 CH 9 calorim trickier example_answers - CH301 Notes...

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