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At the end of the experiment, it was discovered that the thermometer had not been calibrated. When it was calibrated, it was found that the thermometer read 0.50 C low. What effect would thus thermometer reading have on the reported change in H neoutzn calculated above?

At the end of the experiment, it was discovered that the thermometer had not been calibrated. When it was calibrated, it was found that the thermometer read 0.50 C low. What effect would thus thermometer reading have on the reported change in H neoutzn calculated above?

Chemistry
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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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At the end of the experiment, it was discovered that the thermometer had not been calibrated. When it was calibrated, it was found that the thermometer read 0.50 C low. What effect would thus thermometer reading have on the reported change in H neoutzn calculated above?
© 1988 Cengage Learning oriw olemholso odt ot quo otonorli rol 1esod eenlg 6 gntutedue to gilidig oieno.E alubom eirli ni becioeob inomiopo srit anob 2. When a neutralization reaction was carried out using 100.0 mL of 0.7890M NH3 water and 100.0 mL of 0.7940M acetic acid, AT was found to be 4.76 °C. The specific heat of the reaction mixture was 4.104 J gK and its density was 1.03 g mL-1. The calorimeter constant was 3.36 J K-1. (1) Calculate AHneutzn for the reaction of NH3 and acetic acid. heat absorbed. by reaction= (volume)( density) (AT) (specific hcat) = (200 mL)(1.03)(4.76) (4.104) = 4,0 24. 22 heat absor bed by calorimeter =kalorimet er constant)(AT) = 3.36 (4.76) =15. 99 total heat = (heat absorbed by mixture) (heat absor bed bu cealorimeter) W (S) = 4. 04 E 3J 4,04 WJ 1o bisoter limiting Moks of NHs=0.789..130.0789 mol agent AH ntu ten 6L/2 KJ /mol .0774 Moles of CHg COOH=.794-. =,0794 m answer (2) At the end of the experiment, it was discovered that the thermometer had not been calibrated. When it was calibrated, it was found that the thermometer read 0.50 °C low. What effect would this thermometer reading have on the reported AHneutzn calculated in (2) above?
Transcribed Image Text:© 1988 Cengage Learning oriw olemholso odt ot quo otonorli rol 1esod eenlg 6 gntutedue to gilidig oieno.E alubom eirli ni becioeob inomiopo srit anob 2. When a neutralization reaction was carried out using 100.0 mL of 0.7890M NH3 water and 100.0 mL of 0.7940M acetic acid, AT was found to be 4.76 °C. The specific heat of the reaction mixture was 4.104 J gK and its density was 1.03 g mL-1. The calorimeter constant was 3.36 J K-1. (1) Calculate AHneutzn for the reaction of NH3 and acetic acid. heat absorbed. by reaction= (volume)( density) (AT) (specific hcat) = (200 mL)(1.03)(4.76) (4.104) = 4,0 24. 22 heat absor bed by calorimeter =kalorimet er constant)(AT) = 3.36 (4.76) =15. 99 total heat = (heat absorbed by mixture) (heat absor bed bu cealorimeter) W (S) = 4. 04 E 3J 4,04 WJ 1o bisoter limiting Moks of NHs=0.789..130.0789 mol agent AH ntu ten 6L/2 KJ /mol .0774 Moles of CHg COOH=.794-. =,0794 m answer (2) At the end of the experiment, it was discovered that the thermometer had not been calibrated. When it was calibrated, it was found that the thermometer read 0.50 °C low. What effect would this thermometer reading have on the reported AHneutzn calculated in (2) above?
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