4.00 mL of 4.00 x 103 M. Fe(NO3)3 (4.00 times 10 to the minus 3rd power M Fe (NO3)3) is added to 9.00 mL of 6.00 x 103 M KSCN (6.00 times 10 to the minus 3rd power M KSCN) along with 3.00 mL of water. The concentration of [F e(SCN)2*] was found to be 1.00 x 104 M (1.00 times 10 to the minus 4th power M) at equilibrium.How many moles of [F e(SCN)2*] are present in the solution? Express your answer as a decimal number (no exponents). Your Answer:

4.00 mL of 4.00 x 103 M. Fe(NO3)3 (4.00 times 10 to the minus 3rd power M Fe (NO3)3) is added to 9.00 mL of 6.00 x 103 M KSCN (6.00 times 10 to the minus 3rd power M KSCN) along with 3.00 mL of water. The concentration of [F e(SCN)2] was found to be 1.00 x 104 M (1.00 times 10 to the minus 4th power M) at equilibrium.How many moles of [F e(SCN)2] are present in the solution? Express your answer as a decimal number (no exponents). Your Answer:

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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