Calcium fluoride, CaF2, has a molar solubility of 2.1 x 10^-4 mol/L at pH = 7.00. By what factor does its molar solubility increase in a solution with pH = 3.00? The pKa of HF is 3.17
(A) 1.48
(B) 1.83
(C) 2.48
(D) 4.96

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

personal, she chose to do it.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2