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  • Question #f84c2 - Socratic
    K_"sp" is given by the ion product we gets K_"sp"=3 33xx10^-25 We set up the solubility equilibrium Cr (OH)_3 (s)rightleftharpoonsCr^ (3+) + 3HO^- Where K_"sp"= [Cr^ (3+)] [HO^-]^ (3) But we have been given [HO^-]=4xx10^-6*mol*L^-1 Now for a saturated solution, WE KNOW from the stoichiometry that [Cr^ (3+)]=1 3 [HO^-] And so we substitute this value back into the K_"sp
  • Question #05998 - Socratic
    {a in RR|a!=1} Given: 1 = 1 (1-a) + a (a-1) Restrict the value of "a" to avoid division by 0: 1 = 1 (1-a) + a (a-1); a!=1 Multiply the first fraction by 1 in the form of (-1) -1: 1 = (-1) -1 1 (1-a) + a (a-1); a!=1 Follow the rules of multiplication of fractions: 1 = (-1) (a-1) + a (a-1); a!=1 The two denominators are the same so combine the numerator over the common denominator: 1 = (a-1) (a
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