Question

when 17.6 g hcl dissociates in 55.0 g of water in a coffee cup calorimeter, the temperature rises from 28.1 °c to 66.5 °c.
$c_{sol} = 4.04 \\, \text{j/g °c}$ $c_{cal} = 4.90 \\, \text{j/°c}$
$q_{cal} = +188 \\, \text{j}$ $q_{sol} = +11,300 \\, \text{j}$
what is the heat of reaction, $q_{rxn}$?
$q_{rxn} = ? \\, \text{j}$
remember : $q_{rxn} = -(q_{sol} + q_{cal})$
enter either a + or - sign and the magnitude.

Explanation:

Step1: Identify given values

We know \( q_{sol} = 11300 \, \text{J} \) and \( q_{cal} = 188 \, \text{J} \). The formula for \( q_{rxn} \) is \( q_{rxn} = -(q_{sol} + q_{cal}) \).

Step2: Calculate the sum of \( q_{sol} \) and \( q_{cal} \)

First, find the sum: \( q_{sol} + q_{cal} = 11300 + 188 = 11488 \, \text{J} \).

Step3: Apply the formula for \( q_{rxn} \)

Using the formula \( q_{rxn} = -(q_{sol} + q_{cal}) \), substitute the sum we found: \( q_{rxn} = -11488 \, \text{J} \).

Answer:

\(-11488\)