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
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter20: Chemistry Of The Metals
Section: Chapter Questions
Problem 42QAP: Rust, which you can take to be Fe(OH)3, can be dissolved by treating it with oxalic acid. An...
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First of all we will calculate the number of moles of ferric ion added into the solution and then we will calculate the number of moles of ferric ion present as ions. Their difference will give us the value of number of moles of ferric ions present in the solution at equilibrium.
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