The addition of 1.713 g of Ba(OH)2 to a solution of 0.88 g of NH4CN in 100 g of water in a calorimeter caused the temperature to fall by 3.1 °C. Assuming the specific heat of the solution and products is 4.18 J/g K, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation: Ba(OH)2(s) + 2 NH4CN(aq) → Ba(CN)2(aq) + 2 NH4OH(aq) O a. +129.58 kJ O b. +1472.5 J O-129.58 kJ O d.-1295.8 J O e. +1295.8 J

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Question 21
The addition of 1.713 g of Ba(OH)2 to a solution of 0.88 g of NH4CN in 100 g of water in a calorimeter caused the temperature to fall by 3.1 °C. Assuming the
specific heat of the solution and products is 4.18 J/g K, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the
following equation:
Ba(OH)2(s) + 2 NH4CN(aq) Ba(CN)2(aq) + 2 NH4OH(aq)
Save Answer
O a. +129.58 kJ
O b. +1472.5 J
Oc.-129.58 kJ
O d.-1295.8 J
O e. +1295.8 J
Transcribed Image Text:Question 21 The addition of 1.713 g of Ba(OH)2 to a solution of 0.88 g of NH4CN in 100 g of water in a calorimeter caused the temperature to fall by 3.1 °C. Assuming the specific heat of the solution and products is 4.18 J/g K, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation: Ba(OH)2(s) + 2 NH4CN(aq) Ba(CN)2(aq) + 2 NH4OH(aq) Save Answer O a. +129.58 kJ O b. +1472.5 J Oc.-129.58 kJ O d.-1295.8 J O e. +1295.8 J
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