28. The reaction of 50 mL of acid and 50 mL of base described in Example 5.5 increased the temperature of thesolution by 6.9 degrees C. How much would the temperature have increased if 100 mL of acid and 100 mL of basehad been used in the same calorimeter starting at the same temperature of 22.0 °C? Explain your answer. For context, here is Example 5.5:Example 5.5Heat Produced by an Exothermic ReactionWhen 50.0 mL of 1.00 M HCI(aq) and 50.0 mL of 1.00 M NAOH(aq), both at 22.0 °C, are added to a coffeecup calorimeter, the temperature of the mixture reaches a maximum of 28.9 °C. What is the approximateamount of heat produced by this reaction?HСC(ag) + NaOН(ад) —NaCl(aq) + H2O(1)SolutionTo visualize what is going on, imagine that you could combine the two solutions so quickly that no reactiontook place while they mixed; then after mixing, the reaction took place. At the instant of mixing, you have100.0 mL of a mixture of HCl and NaOH at 22.0 °C. The HCl and NaOH then react until the solutiontemperature reaches 28.9 °C.The heat given off by the reaction is equal to that taken in by the solution. Therefore:9 reaction =-9 solution(It is important to remember that this relationship only holds if the calorimeter does not absorb any heatfrom the reaction, and there is no heat exchange between the calorimeter and the outside environment.)Next, we know that the heat absorbed by the solution depends on its specific heat, mass, and temperaturechange:9 solution = (c Xmx AT)olutionTo proceed with this calculation, we need to make a few more reasonable assumptions or approximations.Since the solution is aqueous, we can proceed as if it were water in terms of its specific heat and massvalues. The density of water is approximately 1.0 g/mL, so 100.0 mL has a mass of about 1.0 x 102 g (twosignificant figures). The specific heat of water is approximately 4.184 J/g °C, so we use that for the specificheat of the solution. Substituting these values gives:9 solution = (4.184 J/g °C)(1.0x 10 g)(28.9 °C- 22.0 °C) = 2.9 x 10 JFinally, since we are trying to find the heat of the reaction, we have:9reaction - solution =-2.9x 10 JThe negative sign indicates that the reaction is exothermic. It produces 2.9 kJ of heat.Check Your LearningWhen 100 mL of 0.200 M NaCl(ag) and 100 mL of 0.200 M AGNO3(aq), both at 21.9°C, are mixed ina coffee cup calorimeter, the temperature increases to 23.5 °C as solid AgCl forms. How much heat isproduced by this precipitation reaction? What assumptions did you make to determine your value?Answer: 1.34 x 10 J; assume no heat is absorbed by the calorimeter, no heat is exchanged between thecalorimeter and its surroundings, and that the specific heat and mass of the solution are the same as those forwater

Given: Volume of acid used = 50 mL Volume of base used = 50 mL And change in temperature of solution…

28. The reaction of 50 mL of acid and 50 mL of base described in Example 5.5 increased the temperature of the solution by 6.9 degrees C. How much would the temperature have increased if 100 mL of acid and 100 mL of base had been used in the same calorimeter starting at the same temperature of 22.0 °C? Explain your answer.

28. The reaction of 50 mL of acid and 50 mL of base described in Example 5.5 increased the temperature of the solution by 6.9 degrees C. How much would the temperature have increased if 100 mL of acid and 100 mL of base had been used in the same calorimeter starting at the same temperature of 22.0 °C? Explain your answer.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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