Date: Name: BaCl2 BaCl2*H₂O BaCl2*2H₂O Sum atomic 208.23 208.23 208.23 mass of barium chloride (g) Sum of atomic mass of water 0 18.02 x 1 = 18.02 (g) Sum of atomic mass of hydrate (g) 208.23 (208.23 +18.02) = 226.25 C = C - 2006. 09.06 - 2005 Percent water 0% (18.02/226.25) x in the hydrate 100 = (%) 7.96 % Per. BaCl2*3H₂O 208.23

Chemistry for Engineering Students
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ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter8: Molecules And Materials
Section: Chapter Questions
Problem 8.22PAE
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100%
a) Calculate the theoretical percent water for each value of n--divide the sum of the atomic masses due to the water molecules by the sum of all the atomic masses in the hydrate, and multiply the result by 100. Complete the table. b) Compare the percent water in the hydrate with the theoretical values calculated for different values of n. Complete the table
Date:
Name:
BaCl2
BaCl2*H₂O
BaCl2*2H₂O
Sum atomic
208.23
208.23
208.23
mass of barium
chloride (g)
Sum of atomic
mass of water
0
18.02 x 1 = 18.02
(g)
Sum of atomic
mass of hydrate
(g)
208.23
(208.23 +18.02) =
226.25
C = C - 2006.
09.06 - 2005
Percent water
0%
(18.02/226.25) x
in the hydrate
100 =
(%)
7.96 %
Per.
BaCl2*3H₂O
208.23
Transcribed Image Text:Date: Name: BaCl2 BaCl2*H₂O BaCl2*2H₂O Sum atomic 208.23 208.23 208.23 mass of barium chloride (g) Sum of atomic mass of water 0 18.02 x 1 = 18.02 (g) Sum of atomic mass of hydrate (g) 208.23 (208.23 +18.02) = 226.25 C = C - 2006. 09.06 - 2005 Percent water 0% (18.02/226.25) x in the hydrate 100 = (%) 7.96 % Per. BaCl2*3H₂O 208.23
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