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michaelbromley9759
18.03.2020 •
Chemistry
A calorimeter contains 32.0 mLmL of water at 15.0 ∘C∘C . When 1.20 gg of XX (a substance with a molar mass of 54.0 g/molg/mol ) is added, it dissolves via the reaction X(s)+H2O(l)→X(aq)X(s)+H2O(l)→X(aq) and the temperature of the solution increases to 29.0 ∘C∘C . Calculate the enthalpy change, ΔHΔHDelta H, for this reaction per mole of XX. Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(g⋅∘C)J/(g⋅∘C)], that density of water is 1.00 g/mLg/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings.
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Ответ:
The enthalpy change ΔH for this reaction per mole of X is -181.10 mol/kJ.
Explanation:
Volume of water in calorimeter = 32.0 mL
Mass of water in calorimeter = M
Density of the water = d = 1.00 g/mL
Mass of substance = 1.20 g
Mass of solution ,m= 1.20 g+ 32.0 g = 33.20 g
Specific heat of solution = c = 4.18 J/g°C
Change in temperature of the calorimeter = ΔT = 29.0°C
Heat absorbed by the solution = Q
Moles of substance X added to water ,n=![\frac{1.20 g}{54.0 g/mol}=0.02222 mol](/tpl/images/0551/5281/fb64a.png)
Heat released when 1.20 grams of X was dissloved = Q'= -Q = -4,024.504 J
Q' = -4,024.504 J = -4.024504 kJ ≈ 4.0245 kJ
1 J = 0.001 kJ
Enthalpy change for this reaction per mole of X:
Ответ:
Yes 91 lol but 9+1 is 10.