Question

1. given that when, in the reaction above, 100.0 g of pbcl2(s) is used with excess s^2-(aq) in a calorimeter with heat capacity of 4095 j/°c. the temperature rises from 21.66 to 31.96°c.

Explanation:

Step1: Identify the heat - capacity formula

The heat absorbed by the calorimeter is given by $q = C\Delta T$, where $C$ is the heat - capacity of the calorimeter and $\Delta T$ is the change in temperature.

Step2: Calculate the change in temperature

$\Delta T=T_2 - T_1$. Given $T_1 = 21.66^{\circ}C$ and $T_2=31.96^{\circ}C$, so $\Delta T=31.96 - 21.66=10.3^{\circ}C$.

Step3: Calculate the heat absorbed by the calorimeter

$C = 4095\ J/^{\circ}C$, and using $q = C\Delta T$, we substitute the values: $q=4095\ J/^{\circ}C\times10.3^{\circ}C$.
$q = 4095\times10.3=42178.5\ J = 42.1785\ kJ$.

Answer:

$42.1785\ kJ$