How does the negative control of the lac operon by glucose function through inducer exclusion? See Section 18.3 (Page 373) . Vie
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The solubility in g/l of lead (ii) fluoride in water at 25°C if ksp of PbF2 is 4. 1 x 10-8. is 5.3 ×10^(-1).
When PbF₂ dissolves, it dissociates as follows;
PbF₂ --> Pb²⁺ + 2F⁻
If molar solubility of PbF₂ is x, then molar solubility of Pb²⁺ is x and F⁻ is 2x.
ksp is solubility product constant and it can be calculated as follows;
ksp = [Pb²⁺][F⁻]²
ksp = (x)(2x)²
ksp = 4x³
To calculate solubility in terms of g/L, we have to know the molar mass of PbF₂.
Solubility of PbF₂ is mol/L x 248 g/mol =
g/L
Thus, we concluded that the solubility of given solution is 5.14 × 10^(-1). which is approx 5.3 × 10^(-1).Hence option D is the correct option.
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