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What is the approximate concentration of free Cu2+ ion at equilibrium when 1.22x102 mol copper(II) nitrate is added to 1.00 L
Determine ion concentration when separating ions in a mixture. A solution contains 1.58x102 M lead acetate and 1.58x10 ? Mman
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Answer #1

Cu 2+ (aq) + 4 NH3(ar) - [Cu(NH3)4] 2+ [NA](M) [[Cu(NH3 Ju I t] (M) [Cu²+ ] (M) 1.210 Initial 1. 22x102 change 1.210-40 Equil

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