155 mL solution X STARTING AMOUNT 155 mL solution x 0.762 mol FeCl 0.762 L solution x 0.762 mol-FECT 0.762 L solution 162.2 g FeCl₂ 162.2 mol FeCl, 0.1 = 6.60 L solution 6.60 If you have 155 mL solution of a 0.762 M FeCl, solution, how many grams of FeCl, are contained in this sample? X 0.762 124 g FeCl, L solution 162.2 g FeCl₂ 162.2 100 0.01 1 Question 6 of 17 molFECT f g Cl M FeCl 25.1 1.92 155 0.118 12.6 7.28 x 10¹ 6.60 35.45 4.92 162.2 mol FeCI/L solution mol FeCl mL solution L solution g Fe³ 2 55.85 0.203 1000 19.2 g FeCl/mol FeCl₂ Submit +
155 mL solution X STARTING AMOUNT 155 mL solution x 0.762 mol FeCl 0.762 L solution x 0.762 mol-FECT 0.762 L solution 162.2 g FeCl₂ 162.2 mol FeCl, 0.1 = 6.60 L solution 6.60 If you have 155 mL solution of a 0.762 M FeCl, solution, how many grams of FeCl, are contained in this sample? X 0.762 124 g FeCl, L solution 162.2 g FeCl₂ 162.2 100 0.01 1 Question 6 of 17 molFECT f g Cl M FeCl 25.1 1.92 155 0.118 12.6 7.28 x 10¹ 6.60 35.45 4.92 162.2 mol FeCI/L solution mol FeCl mL solution L solution g Fe³ 2 55.85 0.203 1000 19.2 g FeCl/mol FeCl₂ Submit +
Introductory Chemistry: A Foundation
9th Edition
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter15: Solutions
Section: Chapter Questions
Problem 45QAP
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can you help me and line it exactly like this because i keep doing it and its says its wrong
Transcribed Image Text:155
mL solution X
STARTING AMOUNT
155 mL solution x
0.762 mol FeCl₂
0.762 L solution
X
0.762
0.762
molFeCT
L solution
162.2 g FeCl₂
162.2 mol FeCl₂
3
0.1
= 6.60 L solution
6.60
X
g FeCl
If you have 155 mL solution of a 0.762 M FeCl solution, how many
grams of FeCl are contained in this sample?
X
:( )
0.762
124
L solution
162.2 g FeCl₂
162.2
100
0.01
Question 6 of 17
1
molFeci
X
t
g Cl
M FeCl₂
25.1
1.92
155
= 6.60
0.118
12.6
7.28 × 10-4
mol FeCl /L solution
mol FeCl₂
35.45
4.92
162.2
L solution
g Fe³+
mL solution
S
55.85
0.203
1000
19.2
g FeCl/mol FeCl₂
Submit
+
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