Rip bro but I need the point
False cause there particles arent large enough to be filtered
Answer:
= 0.014 g of BaCO3
Explanation:
Let x = mol/L of BaCO3 that dissolve.
This will give;
x mol/L Ba2+ and x mol/L CO32-
But;
Ksp = 5.1x10^-9.
Therefore;
Ksp = 5.1 x 10^-9 = (x)(x)
Thus;
x = molar solubility
= √ (5.1 x 10^-9)
= 7.1 x 10^-5 M
Therefore;
Mass BaCO3 = 7.1 x 10^-5 M x 1 L x 197.34 g/mol
= 0.014 g
