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Al,(SO.); Al: S: O: NaOH Na: 0: H: (NH,),S N: H: S: DO IT LIKE THE GIVEN EXAMPLE EXAMPLE 1) CuBr2 Molar Mass = 1(Cu)+2(Br) Cu: _28.45% 1(63.546) + 2(79.904) =223.354 g/mol Br. 71.55% %Cu = (63.546-223.354) x 100 = 28.45% %Br = (159.808-223.354) x 100 = 71.55% %3D

Al,(SO.); Al: S: O: NaOH Na: 0: H: (NH,),S N: H: S: DO IT LIKE THE GIVEN EXAMPLE EXAMPLE 1) CuBr2 Molar Mass = 1(Cu)+2(Br) Cu: _28.45% 1(63.546) + 2(79.904) =223.354 g/mol Br. 71.55% %Cu = (63.546-223.354) x 100 = 28.45% %Br = (159.808-223.354) x 100 = 71.55% %3D

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
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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Al,(SO.); Al: S: O: NaOH Na: O: H: (NH,),S N: H: S: DO IT LIKE THE GIVEN EXAMPLE EXAMPLE 1) CuBr2 Molar Mass = 1(Cu)+2(Br) Cu: 28.45% 1(63.546) + 2(79.904) =223.354 g/mol Br. 71.55% %Cu = (63.546-223.354) x 100 = 28.45% %Br = (159.808-223.354) x 100 = 71.55%
Transcribed Image Text:Al,(SO.); Al: S: O: NaOH Na: O: H: (NH,),S N: H: S: DO IT LIKE THE GIVEN EXAMPLE EXAMPLE 1) CuBr2 Molar Mass = 1(Cu)+2(Br) Cu: 28.45% 1(63.546) + 2(79.904) =223.354 g/mol Br. 71.55% %Cu = (63.546-223.354) x 100 = 28.45% %Br = (159.808-223.354) x 100 = 71.55%
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