Volume of 2.00 M HC1 Volume of 2.00 M NaOH Volume of solution Tinitial T AT (T final final - Tinitial) fic heat capacity of NaCl solution 2x 25m1 25ml 50 ml 12 31.30 10.3°C Trial 2 3.87 J/g °C 25ml 21°C 2C 21°C 31.3°C 10.30 25m) 50 ml 3.87 J/g °C

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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qsoln (Trial 2)
moles of water = moles of HCI=
Average quoin
soln (Trial 1)
3. Calculate q rxn
4. Calculate the number of moles of water formed. write the chemical equation before attempting to answer this question.
Explain your reasoning determined
5. Calculate AHneutralization (AH°) per mole of water (kJ/mole) formed. Make sure the sign is correct.
Transcribed Image Text:qsoln (Trial 2) moles of water = moles of HCI= Average quoin soln (Trial 1) 3. Calculate q rxn 4. Calculate the number of moles of water formed. write the chemical equation before attempting to answer this question. Explain your reasoning determined 5. Calculate AHneutralization (AH°) per mole of water (kJ/mole) formed. Make sure the sign is correct.
Pain and Calculations:
Volume of 2.00 M HC1
Volume of 2.00 M NaOH
Volume of solution
Tinitial
T final
AT (T final - Tinitial)
Specific heat capacity of NaCl solution
Trial 1
2x25ml
25ml
50ml
31.3⁰0
10.3°C
Trial 2
3.87 J/g °C
25ml
2.1°C
122°C 21°C
31.3°C
10.30
3.87 J/g °C
25m)
50 ml
Transcribed Image Text:Pain and Calculations: Volume of 2.00 M HC1 Volume of 2.00 M NaOH Volume of solution Tinitial T final AT (T final - Tinitial) Specific heat capacity of NaCl solution Trial 1 2x25ml 25ml 50ml 31.3⁰0 10.3°C Trial 2 3.87 J/g °C 25ml 2.1°C 122°C 21°C 31.3°C 10.30 3.87 J/g °C 25m) 50 ml
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