Two types of stoichiometry are; molar mass and coefficients from balanced equation.
1 mole=6.02 x 10^23 atoms so how many moles are there in 3.0 x 10^23 we will cross multiply, 1 x 3.0 x 10^23 / 6.02 x 10 ^23. Which will give us 0.498 moles.
Hope this helped
Answer:
1.12 × 10⁻⁴ M
Explanation:
Step 1: Write the reaction for the solution of Mg(OH)₂
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Step 2: Make an ICE chart
We can relate the solubility product constant (Ksp) with the solubility (S) through an ICE chart.
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
I 0 0
C +S +2S
E S 2S
The solubility product constant is:
Ksp = 5.61 × 10⁻¹² = [Mg²⁺] × [OH⁻]² = S × (2S)² = 4S³
S = 1.12 × 10⁻⁴ M
Answer:
B. 8
Explanation:
because I learned that in 6th