When 50.0 mL of 0.400 mol L^-1 Ca(NO₃)₂ is added to 50.0 mL of 0.800 mol L^-1 NaF,CaF₂ precipitates,as shown in the net ionic equation below.The initial temperature of both solutions is 24.0 °C.Assuming that the reaction goes to completion,and that the resulting solution has a mass of 100.00 g and a specific heat of 4.18 J $g ^ { - 1 }$ ${ } ^ { \circ } \mathrm { C } ^ { - 1 }$ ,calculate the final temperature of the solution. Ca²⁺(aq)+ 2F^-(aq)? CaF₂(s) $\Delta _ { \mathrm { r } }$ H = -11.5 kJ

Question

Quizplus · Accepted Answer

First, let's calculate the heat released by the reaction using the given enthalpy change and the moles of limiting reactant (Ca(NO3)2) used:

n(Ca(NO3)2) = 0.0500 L * 0.400 mol/L = 0.0200 mol
ΔH = -11.5 kJ/mol
q = ΔH * n = -11.5 kJ/mol * 0.0200 mol = -0.230 kJ

Since q = m * c * ΔT, we can solve for ΔT:

m = 100.00 g
c = 4.18 J/(g°C)
ΔT = q / (m * c) = -0.230 kJ / (100.00 g * 4.18 J/(g°C)) = -0.0549°C

Finally, we can calculate the final temperature by adding the change in temperature to the initial temperature:

Tf = Ti + ΔT = 24.0°C - 0.0549°C = 23.95°C ≈ 24.55°C (rounded to two decimal places)
Therefore, the answer is B.

When 500 ML of 0 g−1g ^ { - 1 }g−1

When 500 ML of 0 $g ^ { - 1 }$