When Ag₂S dissolves, it dissociates as follows;
Ag₂S ---> 2Ag⁺ + S²⁻
First we need to calculate molar solubility which gives the number of moles dissolved in 1 L of solution.
If molar solubility of Ag₂S is y, then molar solubility of Ag²⁺ and S²⁻ is 2y and y respectively.
ksp gives the solubility constant
ksp = [Ag⁺]²[S²⁻]
ksp = [2y]²[y]
4y³ = 8.00 x 10⁻⁵¹
y³ = 2 x 10⁻⁵¹
y = 1.26 x 10⁻¹⁷ mol/L
molar mass = 247.8 g/mol
solubility of Ag₂S = 1.26 x 10⁻¹⁷ mol/L x 247.8 g/mol = 3.12 x 10⁻¹⁵ g/L
Solubility of Ag₂S = 3.12 x 10⁻¹⁵ g/L
Answer:
temporary changes are the changes which are there only for a short period if time.
Explanation:
generally temporary changes are reversible. permanent changes are the changes which remain for a longer time and are not reversible
5 degrees because five - 5=0 which is where you need to get to
Ionic bonding is the answer