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beks73
2 days ago
15

In a coffee-cup calorimeter experiment, 10.00 g of a soluble ionic compound was added to the calorimeter containing 75.0 g h2o i

nitially at 23.2°c. the final temperature of the solution was 31.8°c. what was the change in enthalpy for the dissolution of this compound? give your answer in units of joules per gram of compound. assume that the specific heat of the solution is the same as that of pure water, 4.18 j ⁄ g ⋅ °c.
A) 2.7 x 10^2 J/g
B) -3.1 x 10^3 J/g
C) -3.1 x 10^2 J/g
D) 3.1 x 10^2 J/g
Chemistry
2 answers:
Alekssandra [2.8K]2 days ago
8 0
The enthalpy change for dissolving this compound is 2.70 x 10²J/g. The mass of the soluble ionic compound is 10.00g, the mass of water is 75.0g, the initial water temperature is 23.2⁰C (T1), and the final temperature is 31.8⁰C (T2). The specific heat of water is 4.18 J/g.⁰C. To compute the heat gained by the water, use the formula for heat transfer: Q = mcΔT. Calculating Q gives you 2696 J. The heat gained by water matches the heat lost by the ionic compound, expressed as q(ionic) = 2696 J. To find ΔH: ΔH = q(ionic)/m, leading to ΔH = 2696 J/10.00g, which results in 2.70 x 10²J/g. Hence, the enthalpy change for this compound's dissolution is 2.70 x 10²J/g.
VMariaS [2.8K]2 days ago
7 0
Given: Mass of the ionic compound = 10.00 g Mass of water = 75.0 g Initial temperature of water T1= 23.2 C Final temperature of water T2 = 31.8 C Specific heat of water c = 4.18 J/gC To determine: Enthalpy of dissolution of the ionic compound Heat gained by water equation: Q = mcΔT m = mass of water c = specific heat ΔT = change in temperature (T2-T1) Q = 75.0 g * 4.18 J/gC * (31.8-23.2)C = 2696 J Thus, the heat gained by water equals heat lost by the ionic compound (enthalpy of dissolution) Therefore, q(ionic) = 2696 J ΔH = q(ionic)/mass of ionic compound = 2696 J/10.00 g = 2.7 *10² J/g Answer: A) enthalpy change = 2.7*10² J/g
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