Using the square-bracket notation, express the concentration of ions in a solution that is 0.811 M in aluminum sulfate [Al2(SO4)3]. [Al^3+] = M

[SO4^2−] = M

[Al³⁺] = 1.62 M

[SO₄²⁻] = 2.43 M

Explanation:

Step 1: Given data

Molar concentration of Al₂(SO₄)₃: 0.811 M

Step 2: Write the reaction for the ionization of Al₂(SO₄)₃

Al₂(SO₄)₃(aq) ⇒ 2 Al³⁺(aq) + 3 SO₄²⁻(aq)

Step 3: Calculate the molar concentration of Al³⁺

The molar ratio of Al₂(SO₄)₃ to Al³⁺ is 1:2. The molarity of Al³⁺ is 2/1 × 0.811 M = 1.62 M.

Step 4: Calculate the molar concentration of SO₄²⁻

The molar ratio of Al₂(SO₄)₃ to SO₄²⁻ is 1:3. The molarity of SO₄²⁻ is 3/1 × 0.811 M = 2.43 M

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.