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Chapter 6: Problem 69

Consider the dissolution of \(\mathrm{CaCl}_{2} :\) $$ \mathrm{CaCl}_{2}(s) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \quad \Delta H=-81.5 \mathrm{kJ} $$ An 11.0 -g sample of \(\mathrm{CaCl}_{2}\) is dissolved in 125 g water, with both substances at \(25.0^{\circ} \mathrm{C}\) . Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of 4.18 $\mathrm{J} /^{\prime} \mathrm{C} \cdot \mathrm{g} .$

Short Answer

Expert verified
The final temperature of the solution after the dissolution of 11.0 g of CaCl2 in 125 g of water at \(25.0^{\circ} \mathrm{C}\) is 40.4 °C.

Step by step solution

01

Calculate the heat released by the dissolution process

To calculate the heat released by the dissolution process, we need to multiply the moles of CaCl2 with the enthalpy of the process. First, we should find the number of moles of CaCl2 used: Molar mass of CaCl2 = 40.08 (Ca) + 2 * 35.45 (Cl) = 110.98 g/mol Number of moles = (mass of CaCl2) / (molar mass of CaCl2) Number of moles = 11.0 g / 110.98 g/mol = 0.0991 mol Now, we can find the heat released by the dissolution process: Heat released (q) = moles * enthalpy change q = 0.0991 mol * -81.5 kJ/mol q = -8.07 kJ Since q is negative, the heat is released during the process.
02

Calculate the heat absorbed by water

Using the equation \(q = mc \Delta T\), we can find the heat absorbed by water: 8.07 kJ = (125 g) * (4.18 J/g·°C) * ΔT 8070 J = (125 g) * (4.18 J/g·°C) * ΔT Now, we need to solve for ΔT (the temperature change of water).
03

Calculate the final temperature

First, we should isolate ΔT: ΔT = 8070 J / (125 g * 4.18 J/g·°C) ΔT = 15.4 °C Since ΔT is positive, the temperature of the water will increase. So, we can compute the final temperature by adding the initial temperature with ΔT: Final Temperature = Initial Temperature + ΔT Final Temperature = 25°C + 15.4 °C Final Temperature = 40.4 °C The final temperature of the solution after the dissolution of CaCl2 is 40.4 °C.

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Most popular questions from this chapter

Given the following data $$ \begin{array}{ll}{\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-23 \mathrm{kJ}} \\ {3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-39 \mathrm{kJ}} \\ {\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow 3 \mathrm{FeO}(s)+\mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=18 \mathrm{kJ}}\end{array} $$ calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$

The equation for the fermentation of glucose to alcohol and carbon dioxide is: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) $$ The enthalpy change for the reaction is \(-67 \mathrm{kJ} .\) Is this reaction exothermic or endothermic? Is energy, in the form of heat, absorbed or evolved as the reaction occurs?

It has been determined that the body can generate 5500 \(\mathrm{kJ}\) of energy during one hour of strenuous exercise. Perspiration is the body's mechanism for eliminating this heat. What mass of water would have to be evaporated through perspiration to rid the body of the heat generated during 2 hours of exercise? (The heat of vaporization of water is 40.6 \(\mathrm{kJ} / \mathrm{mol.} )\)

A gaseous hydrocarbon reacts completely with oxygen gas to form carbon dioxide and water vapor. Given the following data, determine \(\Delta H_{f}^{\circ}\) for the hydrocarbon: $$ \begin{aligned} \Delta H_{\mathrm{reacion}}^{\circ} &=-2044.5 \mathrm{kJ} / \mathrm{mol} \text { hydrocarbon } \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{CO}_{2}\right) &=-393.5 \mathrm{kJ} / \mathrm{mol} \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right) &=-242 \mathrm{kJ} / \mathrm{mol} \end{aligned} $$ Density of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) product mixture at 1.00 \(\mathrm{atm}\) , \(200 . \mathrm{C}=0.751 \mathrm{g} / \mathrm{L}\) . The density of the hydrocarbon is less than the density of Kr at the same conditions.

Consider the reaction $$ 2 \mathrm{HCl}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{BaCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ $$ \Delta H=-118 \mathrm{kJ} $$ Calculate the heat when 100.0 \(\mathrm{mL}\) of 0.500\(M \mathrm{HCl}\) is mixed with 300.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{Ba}(\mathrm{OH})_{2}\) . Assuming that the temperature of both solutions was initially \(25.0^{\circ} \mathrm{C}\) and that the final mixture has a mass of 400.0 \(\mathrm{g}\) and a specific heat capacity of 4.18 \(\mathrm{J} / \mathrm{C} \cdot \mathrm{g}\) , calculate the final temperature of the mixture.
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