09 Question: A 1.00-L solution contains 2.25×104 M Cu(NO3)2 and 1.00×103 M ethylenediamine (en). The Kf for Cu(en) 22+ is 1.00 × 1020. Cu2+(aq) + 2en(aq) — Cu(en)+(aq) [Cu(en)+] K₁= [Cu²+] [en]² See page 795

09 Question: A 1.00-L solution contains 2.25×104 M Cu(NO3)2 and 1.00×103 M ethylenediamine (en). The Kf for Cu(en) 22+ is 1.00 × 1020. Cu2+(aq) + 2en(aq) — Cu(en)+(aq) [Cu(en)+] K₁= [Cu²+] [en]² See page 795

Chemistry

10th Edition

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

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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A 1.00-L solution contains 3.50×10-4 M Cu(NO3)2 and 2.00×10-³ M ethylenediamine (en). The Kf for Cu(en) 22+ is 1.00 × 1020. Cu²+(aq) +2en(aq) — Cu(en)2+ (aq) [Cu(en)2+] Ke= [Cu²+][en]²
Please choose the correct option. Thank you.
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Can you explain the answer below?
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Consider the reaction: Cu2+(aq) + 4NH3(aq) 2 Cu(NH3)4²*(aq) Kf = 1.7x 1013 %3D What is [Cu2+] remaining in solution after 0.020 mol of CuSO4 is dissolved in 1.00 L of 0.70 M NH3? Assume no change of volume. O 3.8 x 10-14 M. 1.6 x 10-14 M 1.1 x 10-13 M 8.0 x 10-15 M 5.0 x 10-13 M
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