Question

when 19.55 g of solid k dissociates in 500.0 g of water in a coffee cup calorimeter, the temperature increases from 20.0 °c to 61.5 °c.
$c_{sol} = 4.18 \\, j/g \\, ^\circ c$
$mass_{sol} = 519.6 \\, g$ $q_{sol} = +90,100 \\, j$
what is the the heat of reaction, $q_{rxn}$?
$q_{rxn} = ? \\, j$
because the heat capacity of the calorimeter cannot be determined, $q_{rxn} = -q_{sol}$.
enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall the relationship between \( q_{rxn} \) and \( q_{sol} \)

The problem states that \( q_{rxn} = -q_{sol} \). We know \( q_{sol} = +90100 \, \text{J} \).

Step2: Calculate \( q_{rxn} \)

Substitute \( q_{sol} \) into the formula: \( q_{rxn} = -90100 \, \text{J} \) (since the reaction releases heat when the solution absorbs heat, the sign of \( q_{rxn} \) is negative relative to \( q_{sol} \)).

Answer:

\(-90100\)