5. Calculate AHrxn (28) using bess's law2+Ca(s)+2H2Om→ Ca (ag)+ 20H (ag) + H2(g)_AHrxn(28) = ?H (ag) +OH (ag)H20(1)C = -56.4 kJ/molCa(s)+2H*(ag) → Ca²*(aq}+H2(g)AHrsn(27) = -643.83 kJ/molReverse equation:Multiply the reverse of equation 26 by 2 to obtain equation 31The reverse of equation 26 is H2O(1) → H* (ag)+2H* (ag)AHr(31)= 2 × -1 × AHrxn(26)Multiply equation 26 by two which is 2H2O →2H* +2OH to obtain AHenfB1)b). compare AHrxn(28) with AHHess(28) using a percentage differencePercentage difference = absolute e1-e2x 1000.5 (erte2)

We would use Hess's law to calculate ∆Hrxn of above reaction , using the given values of ∆Hrxn for…

Answered: 5. Calculate AHrxn (28) using hess's…

5. Calculate AHrxn (28) using hess's law 2+ Ca(s)+2H2Om - Са (aд) + 20 (ад) + H2(g) AHrxu(28) = ? H (ag) +OH- (ag)H20(1) C = -56.4 kJ/mol Ca(s)+2H*(ag) Ca2 (ag)+H2(g) AHrxn(27) = -643.83 kJ/mol

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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Look at the image.
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