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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Question
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please solve for question 2 using the data above the reslt table.

Transcribed Image Text:2. A student is working to find the formula for an iron fluoride hydrate. After following the same
procedure as the in this lab, they collected the following data. Use their data to complete
the results table below.
Data
Mass (g)
Cooled crucible
29.986
Crucible + hydrate before heating
30.896
Crucible + anhydrous after heating
30.456
Filter paper
0.890
Watch glass
23.671
Watch glass, filter paper, and dried iron
24.789
Results
Mass (g)
Moles
iron fluoride hydrate
H2O(g) removed
anhydrous iron fluoride
iron
fluoride
Formula of the unknown iron fluoride hydrate
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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Follow-up Questions
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Follow-up Question
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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