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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
  • How do you find the vertical, horizontal and slant . . . - Socratic
    horizontal asymptote at y = 0 The denominator of y cannot be zero as this would make y undefined Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes solve : x^2+4=0rArrx^2=-4 This has no real solutions hence there are no vertical asymptotes Horizontal asymptotes occur as lim_(xto+
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  • How do you find the vertical, horizontal and slant . . . - Socratic
    "horizontal asymptote at " y=2 The denominator of y cannot be zero as this would make y undefined Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes "solve " x^2+9=0rArrx^2=-9 "there are no real solutions hence no vertical asymptotes" "horizontal asymptotes occur as" lim_(xto+-oo
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