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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Calculate pCu in a 0.10 M CuSO4(aq) solution that is also 6.0 M in NH3. Cu²+ (aq) + 4 NH3(aq) [Cu(NH3)4]²+ (aq) Kf = 1.1 x 1013
A 1.00-L solution contains 2.50×10-4 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+2(aq)Kf=[Cu(en)2+2]/[Cu2+][en]2 What is the concentration of Cu2+(aq) in the solution?
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
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Can you explain the answer below?
A 1.00 L solution contains 4.00×10-4 M Cu(NO3)2 and 3,000×10-3 Methylenediamine (en). The Kf for Cu(en)22+ is 1.00x1020. What is the concentration of Cu2+(aq) in the solution?
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