Question

agno₃ + hcl → agcl + hno₃
50.0 ml of 0.20 m agno₃ is added to
50.0 ml of 0.20 m hcl. the resulting
temperature change is an increase of
1.60 °c.

cₛₒₗ = 4.20 j/g °c dₛₒₗ = 1.02 g/ml c_cal = 5.60 j/°c
massₛₒₗₙ = 102 g
what is the heat of reaction?
remember: qᵣₓₙ = −(qₛₒₗ + q_cal)
qᵣₓₙ = ? j
enter either a + or - sign and the magnitude.

Explanation:

Step1: Calculate \( q_{sol} \)

The formula for heat absorbed by the solution is \( q_{sol} = c_{sol} \times m_{sol} \times \Delta T \). Given \( c_{sol} = 4.20 \, \text{J/g}^\circ\text{C} \), \( m_{sol} = 102 \, \text{g} \), and \( \Delta T = 1.60^\circ\text{C} \).
\( q_{sol} = 4.20 \, \text{J/g}^\circ\text{C} \times 102 \, \text{g} \times 1.60^\circ\text{C} \)
\( q_{sol} = 4.20 \times 102 \times 1.60 \, \text{J} \)
\( q_{sol} = 685.44 \, \text{J} \)

Step2: Calculate \( q_{cal} \)

The formula for heat absorbed by the calorimeter is \( q_{cal} = c_{cal} \times \Delta T \). Given \( c_{cal} = 5.60 \, \text{J/}^\circ\text{C} \) and \( \Delta T = 1.60^\circ\text{C} \).
\( q_{cal} = 5.60 \, \text{J/}^\circ\text{C} \times 1.60^\circ\text{C} \)
\( q_{cal} = 8.96 \, \text{J} \)

Step3: Calculate \( q_{rxn} \)

Using the formula \( q_{rxn} = -(q_{sol} + q_{cal}) \).
First, find \( q_{sol} + q_{cal} = 685.44 \, \text{J} + 8.96 \, \text{J} = 694.4 \, \text{J} \)
Then, \( q_{rxn} = - 694.4 \, \text{J} \) (The negative sign indicates the reaction is exothermic as the temperature increased, meaning the reaction released heat.)

Answer:

\(-694\) (or more precisely \(-694.4\))