8.00 mL of 5.00 x 103 M Fe(NO3)3 (5.00 times 10 to the minus 3rd power MF e (NO3)3) is added to 4.00 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 2.00 x 10-4 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 exponents). Your Answer: Answer units

8.00 mL of 5.00 x 103 M Fe(NO3)3 (5.00 times 10 to the minus 3rd power MF e (NO3)3) is added to 4.00 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 2.00 x 10-4 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 exponents). Your Answer: Answer units

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter16: Applications Of Neutralization Titrations

Section: Chapter Questions

Problem 16.20QAP

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calculate the mass of Sn when 50 mL of a sample containing Sn2+ is titrated with 42.00 mL of 0.0150M MnO4- to reach endpoint
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