The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each. (a) BaSeO4, 0.0118 g/100 mL
(b) Ba(BrO3)2 H20, 0.30 g/100 mL
(c) NH4MgAsO4-6H20, 0.038 g/100 mL
(d) La2(MoOs)3, 0.00179 g/100 mL

(a)

(b)

(c)

(d)

Explanation:

Hello,

In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:

(a)

In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:

(B)

In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:

(C)

In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:

(D)

In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:

Best regards.

2b2t

explanation:

2b2t

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.