Answer:
region 2 and region 3
Explanation:
you can tell by the color of the land my friends^^
Answer:
The concentration of this sodiumhydroxide solutions is 0.50 M
Explanation:
Step 1: Data given
Mass of sodium hydroxide (NaOh) = 8.0 grams
Molar mass of sodium hydroxide = 40.0 g/mol
Volume water = 400 mL = 0.400 L
Step 2: Calculate moles NaOH
Moles NaOH = mass NaOH / molar mass NaOH
Moles NaOH = 8.0 grams / 40.0 g/mol
Moles NaOh = 0.20 moles
Step 3: Calculate concentration of the solution
Concentration solution = moles NaOH / volume water
Concentration solution = 0.20 moles / 0.400 L
Concentration solution = 0.50 M
The concentration of this sodiumhydroxide solutions is 0.50 M
Answer:
a. Ksp = 4s³
b. 5.53 × 10⁴ mol³/dm⁹
Explanation:
a. Obtain an expression for the solubility product of AB2(S),in terms of s.
AB₂ dissociates to give
AB₂ ⇄ A²⁺ + 2B⁻
Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as
AB₂ ⇄ A²⁺ + 2B⁻
1 : 1 : 2
Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s
So, we have
AB₂ ⇄ A²⁺ + 2B⁻
[s] [s] [2s]
So, the solubility product Ksp = [A²⁺][B⁻]²
= (s)(2s)²
= s(4s²)
= 4s³
b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³
Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³
Substituting the value of s into the equation, we have
Ksp = 4s³
= 4(2.4 × 10³ mol/dm³)³
= 4(13.824 × 10³ mol³/dm⁹)
= 55.296 × 10³ mol³/dm⁹
= 5.5296 × 10⁴ mol³/dm⁹
≅ 5.53 × 10⁴ mol³/dm⁹
Ksp = 5.53 × 10⁴ mol³/dm⁹
