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What is the approximate concentration of free Cu* ion at equilibrium when 1.57x102 mol copperII) nitrate is added to 1.00 L oWhat is the approximate concentration of free Fe2* ion at equilibrium when 1.42x102mol iron(II) nitrate is added to 1.00 L of

please helpIn the presence of excess OH, the Al*(aq) ion forms a hydroxide complex ion, Al(OH)4. Calculate the concentration of free Al3

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Answer #1

Gliven Copper () minele solation 57x 10m added to ooL NH3380 M 2.X1o3 - 1360) (1) 138 Males Veaction The 2t Cy 27 (ag) 1.57x1-31 =>2.X3 S7X102 7(3-0102) (2-1X 1053.01017) = 5410 =7 13 G.32 X 107 = 157X10 1.53X 10 2 G32x1o3 2.48X10-14 M cu 2-J = 248X[FeCeN Fe 6CA J380 2- - 6X -42Y10 2 C -42 (Fe (en)V kp 2 0xb35 [2948) 35 }.y2 x 10 2 2948x 1o3 29y8X 1035 -37 -37 096x 10 M3 Given -56 X10mal o al (CH00)3 i ad ded to hol of Salution PH- 12.10 kp= x103 heveaction i 3+ Al -ןיx D -3 CalCou Alow, pHFrom Eg kp [al3Cou 33 0-0156 [al30n5) 2 44XID =7 2.68 X/o5 [A130015 0.o156 2 b8x25 [al37 5.82X10-28

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