Question

when 10.0 g of barium chloride (bacl2) is dissolved in 102 g of water in a styrofoam calorimeter of negligible heat capacity, the temperature increases from 25.00 to 27.01 °c. based on this observation, calculate q for the water and δh° for the process, assuming that the heat absorbed by the salt is negligible. bacl2(s)→ba2+(aq)+2cl−(aq) the specific heat of water is 4.184 j °c−1 g−1. give the answers in kj. qh2o = kj δh° = kj 3 item attempts remaining try another version submit answer use the references to access important values if needed for this question.

Explanation:

Step1: Identify the formula for heat

The formula for heat absorbed by a substance is $q = mc\Delta T$, where $m$ is the mass, $c$ is the specific - heat capacity, and $\Delta T$ is the change in temperature.

Step2: Calculate the change in temperature

$\Delta T=T_{final}-T_{initial}=27.01^{\circ}C - 25.00^{\circ}C=2.01^{\circ}C$. The mass of water $m = 102g$ and the specific heat of water $c = 4.184J^{\circ}C^{-1}g^{-1}$.

Step3: Calculate $q_{H_2O}$

Substitute the values into the formula: $q_{H_2O}=mc\Delta T=102g\times4.184J^{\circ}C^{-1}g^{-1}\times2.01^{\circ}C$.
$q_{H_2O}=102\times4.184\times2.01J\approx852.9J = 0.853kJ$.

Step4: Determine $\Delta H^{\circ}$

Since the heat absorbed by water is equal to the heat released by the dissolution process (assuming no heat loss to the surroundings and negligible heat absorption by the salt), $\Delta H^{\circ}=-q_{H_2O}=- 0.853kJ$.

Answer:

$q_{H_2O}=0.853kJ$
$\Delta H^{\circ}=-0.853kJ$