Answer:
pH 8.89
Explanation:
English Translation
If the MgCl₂ solution of 0.2 M has its pH raised by adding NH₄OH, the precipitate will begin to form at a pH of approximately.
Given the solubility product (Ksp) of Mg(OH)₂ = 1.2 x 10⁻¹¹
Assuming all of the salts involved all ionize completely
MgCl₂ ionizes to give Mg²⁺ and Cl⁻
MgCl₂ ⇌ Mg²⁺ + 2Cl⁻
1 mole of MgCl₂ gives 1 moles of Mg²⁺
Since the concentration of Mg²⁺ is the same as that of MgCl₂ = 0.2 M
Mg(OH)₂ is formed from 1 stoichiometric mole of Mg²⁺ and 2 stoichiometric moles of OH⁻
Ksp Mg(OH)₂ = [Mg²⁺][OH⁻]²
(1.2 x 10⁻¹¹) = 0.2 × [OH⁻]²
[OH⁻]² = (6×10⁻¹¹)
[OH⁻] = √(6×10⁻¹¹)
[OH⁻] = 0.000007746 M
p(OH) = - log [OH⁻] = - log (0.000007746)
pOH = 5.11
pH + pOH = 14
pH = 14 - pOH = 14 - 5.11 = 8.89
Hope this Helps!!!
Use the following equations to fill the chart.
E = hf
where
h = 6.63 x 10⁻³⁴ J/s, Planck's constant
f = frequency Hz
E = quanta of energy, J
c = fλ
where
c = 3 x 10⁸ m/s, the velocity of light
λ = wavelength, m
If energy is given in J/mmol, divide by Avogadro's number, N = 6.02 x 10²³, to convert it to J.
The completed table is shown below.
Answer:
5.37 × 10⁻⁴ mol/L
Explanation:
<em>A chemist makes 660. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a 0.00154 mol/L stock solution of magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.</em>
Step 1: Given data
- Initial concentration (C₁): 0.00154 mol/L
- Initial volume (V₁): 230. mL
- Final concentration (C₂): ?
- Final volume (V₂): 660. mL
Step 2: Calculate the concentration of the final solution
We want to prepare a dilute solution from a concentrated one. We can calculate the concentration of the final solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
C₂ = C₁ × V₁ / V₂
C₂ = 0.00154 mol/L × 230. mL / 660. mL = 5.37 × 10⁻⁴ mol/L
Answer:
Unevenly
Explanation:
Fresh water is distributed unevenly in both time and space.
