iPhone 14 ki anokhi SIM #LLAShorts 390

Determine the mass of PbI_(2) that will dissolve in (a) 500mL water (b) 500mL of 0.01M KI solution (c) 500mL of a solution containing 1.33 g Pb(NO_(3))_(2), K_(sp) of PbI = 1.4 xx 10^(-8) .

Mercuric iodate [Hg_5(IO_6)_2] reacts with a mixutre of KI and HCl according to the following equation. Hg_5(IO_6)_2+34KI+24"HCl"to5K_2HgI_4+8I_2+24KCI+12H_2O or Hg_5(IO_6)_2+34I^(ɵ)+24H^(o+)to5HgI_4^(2-)+8I_2+12H_2O The liberated I_2 is titrated with Na_2S_2O_3 solution. 1.0 " mL of " Na_2S_2O_3 is equivalent to 3.992 g of CuSO_4.5H_2O . What volume (in mL) of the Na_2S_2O_3 solution will be required to react with liberated I_2 from 14.485 g Hg_5(IO_6)_2 . [Hg=200.5,Cu=63.5,I=127] ? Mw(Hg_5(IO_6)_2)=1448.5g Mw(CuSO_4.5H_2O)=249.5g

If A=[{:(4i-6,10i),(14i,6+4i):}]and k(1)/(ki), where i- sqrt(-1), then kA is equal to

The equivalent conductivity of N/10 solution of acitic acid at 25^(@)C is 14.3 ohm^(-1) cm^(2) eq^(-1) . Calculate the degree of dissociation of CH_(3)COOH if Lambda_(oo CH_(3)COOH) is 390.71.

Calculate the lattice energy from the following data (given 1 eV = 23.0 kcal mol^(-1) ) i. Delta_(f) H^(ɵ) (KI) = -78.0 kcal mol^(-1) ii. IE_(1) of K = 4.0 eV iii. Delta_("diss")H^(ɵ)(I_(2)) = 28.0 kcal mol^(-1) iv. Delta_("sub")H^(ɵ)(K) = 20.0 kcal mol^(-1) ltbvrgt v. EA of I = -70.0 kcal mol^(-1) vi. Delta_("sub")H^(ɵ) of I_(2) = 14.0 kcal mol^(-1)