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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
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
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can you help me and line it exactly like this because i keep doing it and its says its wrong

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 +
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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