9.00 mL of 4.00 x 10-3 M Fe(NO3)3 (4.00 times 10 to the minus 3rd power M F e (NO3)3) is added to 5.00 mL of 4.00 x 10-3 M KSCN (4.00 times 10 to the minus 3rd power M KSCN) along with 6.00 mL of water. The concentration of [F e(SCN)2+] was found to be 1.00 x 10-3 M (2.00 times 10 to the minus 4th power M) at equilibrium.How many moles of F e3+ are present in the solution at equilibrium? Express your answer as a decimal number (no
9.00 mL of 4.00 x 10-3 M Fe(NO3)3 (4.00 times 10 to the minus 3rd power M F e (NO3)3) is added to 5.00 mL of 4.00 x 10-3 M KSCN (4.00 times 10 to the minus 3rd power M KSCN) along with 6.00 mL of water. The concentration of [F e(SCN)2+] was found to be 1.00 x 10-3 M (2.00 times 10 to the minus 4th power M) at equilibrium.How many moles of F e3+ are present in the solution at equilibrium? Express your answer as a decimal number (no
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter15: Complex Ion And Precipitation Equilibria
Section: Chapter Questions
Problem 26QAP: At 25C, 10.24 mg of Cr(OH)2 are dissolved in enough water to make 125 mL of solution. When...
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9.00 mL of 4.00 x 10-3 M Fe(NO3)3 (4.00 times 10 to the minus 3rd power M F e (NO3)3) is added to 5.00 mL of 4.00 x 10-3 M KSCN (4.00 times 10 to the minus 3rd power M KSCN) along with 6.00 mL of water. The concentration of [F e(SCN)2+] was found to be 1.00 x 10-3 M (2.00 times 10 to the minus 4th power M) at equilibrium.How many moles of F e3+ are present in the solution at equilibrium? Express your answer as a decimal number (no
Expert Solution
Step 1
Volume of Fe(NO3)3 = 9ml
Concentratio of Fe(NO3)3 = 4×10-3M
Volume of KSCN = 5ml
Concentratio of KSCN = 4×10-3M
Volume of water = 6ml
We are asked to calculate moles of Fe3+ at equilibrium.
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