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4. The solubility product of PbCl2m is 1.7 x 10-5, The AGⓇ for PbCl2() and Pb2e are -314 kJ moland -24.3 kJ mol! Calculate t
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PbCl₂ls) 2 Pbeta + 2018) Ksp = 1:7x10-5 and for reaction. Alb= Abo & RT ln Q at equilibrium - Q = key -06-0 so, AGO = -RT lnkard condition R=8.313/mol-k_$T=273\/stand- 50 (06°xw = 24, 919 I 24.92 K1 ] ..our one so = E (svif products -(come ractant (| 24.923-24.3 + 2x outlet - (314) exon (C1) = -264.78 k] *1 047(01-)= - 132-39 %) min

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