One of the main components of hard water is caco3. when hard water evaporates, some of the caco3 is left behind as a white mineral deposit. if a hard water solution is saturated with caco3, what volume of the solution has to evaporate ot deposit 1*10^{2} mg of caco3?

My source says the Ksp of CaCO3 at 25 C is 8.7 X 10^-9 

Ksp = [Ca++][CO3--] = 8.7 X 10^-9 

Let "X" = the solubility; then: 

X^2 = 8.7 X 10^-9 

X = 9.33 X 10^-5 M 

Mol. mass CaCO3 = 40 + 12 + 3(16) = 100 g / mole 

Solubility CaCO3 = (100 g / mole)(9.33 X10^-5 mole / liter) = 9.33 X 10^-3 g liter or 9.33 mg / liter 

Volume of CaCO3 solution = 1.00 X 10^2 mg / 9.33 mg / liter = 10.7 liters 

10.7 liters

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There is one s orbital, and there are three p orbitals, five d orbitals, and seven f orbitals. thus, the answer is b. do not confuse the number of orbitals in a subshell with the number of electrons the subshell can hold. each orbital can hold two electrons, so the capacity of an nf subshell is 14 electrons.