Chemistry Practice Problems Thermochemical Equation Practice Problems Solution: The addition of 3.15 g of Ba(OH) 2∙8H2O to a solut...

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Solution: The addition of 3.15 g of Ba(OH) 2∙8H2O to a solution of 1.52 g of NH 4SCN 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 4.20 J/g°C, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation:Ba(OH)2∙8H2O(s) + 2NH4SCN(aq) ⟶ Ba(SCN)2(aq) + 2NH3(aq) + 10H2O(l)

Problem

The addition of 3.15 g of Ba(OH) 2∙8H2O to a solution of 1.52 g of NH 4SCN 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 4.20 J/g°C, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation:

Ba(OH)2∙8H2O(s) + 2NH4SCN(aq) ⟶ Ba(SCN)2(aq) + 2NH3(aq) + 10H2O(l)