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Social ScienceHEAT OF NEUTRALIZATIONNAME:POST LABORATORY QUESTIONS1. Specific to this experiment, list two reasons why your data contains an experimentalerror.a.b.2. A 15.00 gram sample of ammonium chloride (NH4CI) was dissolved in 100.00 gramsof water. How much heat is produced by the dissolution (absorbed from the solution) ifthe solution temperature goes from 22.0 oC to 1.0 oC. The specific heat of the solutionis 4.18 J/g °C3. 300.0 mL of a 2.500 M solution X was mixed with 300.0 mL of a 2.500 M solution Yin a calorimeter. Both of the solutions were at the same temperature initially. Determinethe heat of neutralization (kJ/mole) if the temperature goes from 12.0 °C to 29.0 °C.Assume a molar ratio of 1:1.The specific heat of the solution is 4.281 J/g °C. Thedensity of the mixture is 1.050 g/mL.4. Sketch a crude time-temperature graph for an endothermic reaction.
Question 3
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HEAT OF NEUTRALIZATION NAME: POST LABORATORY QUESTIONS 1. Specific to this experiment, list two reasons why your data contains an experimental error. a. b. 2. A 15.00 gram sample of ammonium chloride (NH4CI) was dissolved in 100.00 grams of water. How much heat is produced by the dissolution (absorbed from the solution) if the solution temperature goes from 22.0 oC to 1.0 oC. The specific heat of the solution is 4.18 J/g °C 3. 300.0 mL of a 2.500 M solution X was mixed with 300.0 mL of a 2.500 M solution Y in a calorimeter. Both of the solutions were at the same temperature initially. Determine the heat of neutralization (kJ/mole) if the temperature goes from 12.0 °C to 29.0 °C. Assume a molar ratio of 1:1.The specific heat of the solution is 4.281 J/g °C. The density of the mixture is 1.050 g/mL. 4. Sketch a crude time-temperature graph for an endothermic reaction.
Expert Answer
The total mass of the mixture (Msol) is calculated as shown in equation (1) where VX and VY are the volume of solution X and Y respectively, and ρsol is the density of the mixture. The values for ρsol, VX, and VY are substituted in equation (1) to calculate the total mass of mixture (Msol). The total mass of the solution is 630 g.
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sol = (300 mL+300 mL)x1.050 g mL =630 g
Total heat produced during mixing is calculated as shown in equation (2) where Qsol is the heat released or absorbed by the solution, CSol is the heat capacity for the total solution, T1, and T2 is the initial temperature and final temperature of the solution. The values for CSol, T1, and T2 are substituted to calculate the heat of the solution. The heat of the solution is 45,849.51 J.
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QdMCT-T)...(2) -630 gx4.281 J g 1C1(29 -12)°C sol 45,849.51 J
The heat of neutralization is the heat released when 300 mL of solution X is mixed with 300 mL of solution Y and the ...
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-x1000 mL/Lx -(3) 300 mL 45,849.51 J 1 x 2.5 M -x 1000 mL/L 300 mL - 61,132.68 Jmol
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