Introductory Chemistry: A Foundation
9th Edition
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter14: Liquids And Solids
Section: Chapter Questions
Problem 18ALQ
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please solve for question 2 using the data above the reslt table.
Expert Solution
Step 1
Mass of hydrate = (30.896 - 29.986) g
= 0.91 g
Mass of anhydrous salt = (30.456-29.986) g
= 0.47 g
Mass of water removed = Mass of hydrate - Mass of anhydrous
= (0.91-0.47) g
= 0.44 g
Now,
No of moles = Mass / Molar Mass
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I don't undertaand how you calculated the final ratio of H(2)O at the end of the explaination. How did you come to 6?
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