6.00 mL of 5.00 x 10-3 M Fe(NO3)3 (5.00 times 10 to the minus 3rd power M F e (NO3)3) is added to 9.00 mL of 7.00 x 10-3 M KSCN (7.00 times 10 to the minus 3rd power M KSCN) along with 5.00 mL of water. The concentration of [F e(SCN)2+] was found to be 2.00 x 10-3 M (2.00 times 10 to the minus 3rd power M) at equilibrium. How many initial moles of SCN^- are present in the solution? Express your answer as a decimal number (no exponents).
6.00 mL of 5.00 x 10-3 M Fe(NO3)3 (5.00 times 10 to the minus 3rd power M F e (NO3)3) is added to 9.00 mL of 7.00 x 10-3 M KSCN (7.00 times 10 to the minus 3rd power M KSCN) along with 5.00 mL of water. The concentration of [F e(SCN)2+] was found to be 2.00 x 10-3 M (2.00 times 10 to the minus 3rd power M) at equilibrium. How many initial moles of SCN^- are present in the solution? Express your answer as a decimal number (no exponents).
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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6.00 mL of 5.00 x 10-3 M Fe(NO3)3 (5.00 times 10 to the minus 3rd power M F e (NO3)3) is added to 9.00 mL of 7.00 x 10-3 M KSCN (7.00 times 10 to the minus 3rd power M KSCN) along with 5.00 mL of water. The concentration of [F e(SCN)2+] was found to be 2.00 x 10-3 M (2.00 times 10 to the minus 3rd power M) at equilibrium. How many initial moles of SCN^- are present in the solution? Express your answer as a decimal number (no exponents).
Expert Solution
Step 1
Volume of Fe(NO3)3 = 6ml
Concentratio of Fe(NO3)3 = 5×10-3M
Volume of KSCN = 9ml
Concentratio of KSCN = 7×10-3M
Volume of water = 5ml
Equilibrium concentration of Fe(SCN)2+ = 2×10-3
We are asked to calculate the initial moles of SCN- present.
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