Calculate the molar solubility of AgI in the following. Ksp for AgI is 8.30 ×10−17.(a) pure water×10MEnter your answer in scientific notation.(b) 0.0400 M NaI×10MEnter your answer in scientific notation.

(a) solubility in pure water =9.11 x 10⁻⁹  (M)

(b) solubility in 0.08 (M) NaI = 1.04 x 10⁻¹⁵ (M)

Explanation:

(1)                              

                             AgI (s) + H₂O ⇄ Ag⁺(aq) + I⁻ (aq)

                                                          s             s

Solubility product of AgI

Ksp = [Ag⁺] [I⁻]

⇒Ksp = s²

⇒s

solubility in pure water

(2)      

                             Ksp = [Ag⁺] [I⁻]

                          ⇒ 8.3 X 10⁻¹⁷ = S x (0.08+S)

                          and assuming S is small relative to 0.08, due to common          ion effect (I⁻) solubility of salt decreases

                                                            S + 0.08 = 0.08

                                                 ⇒S X 0.08 = 8.3 X 10⁻¹⁷

                                        ⇒S = 1.04 X 10⁻¹⁵ (M) solubility in 0.08 (M) NaI

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2