Answer:
Explanation:
Step 1
The question is based on the concept of PH and pOH calculations.
pH is defined as negative logarithmic of hydronium Ion concentration.
while pOH is defined as negative logarithmic of hydroxide ion concentration of the solution.
Step 2
[H+] = 7.7*10-7 M
pH = -log[H+]
= -log ( 7.7*10-7 )
= 6.12
Step 3
pOH = 14 - pH
= 14 - 6.11
= 7.89
Answer:
The fluoride which precipitates first is CaF₂
Explanation:
When F⁻ is added, CaF₂ and BaF₂ are produced following the ksp equation:
For CaF₂:
Ksp = 3.2x10⁻¹¹ = [Ca²⁺] [F⁻]²
<em>Where [Ca²⁺] = 0.075M * {35mL / (25mL + 35mL)} = 0.04375M</em>
3.2x10⁻¹¹ = [0.04375M] [F⁻]²
[F⁻]² = 7.31x10⁻¹⁰
[F⁻] = 2.7x10⁻⁵M
<h3>CaF₂ begins precipitation when [F⁻] = 2.7x10⁻⁵M.</h3>
For BaF₂:
Ksp = 1.5x10⁻⁶ = [Ba²⁺] [F⁻]²
<em>Where [Ba²⁺] = 0.090M * {25mL / (25mL + 35mL)} = 0.0375M</em>
1.5x10⁻⁶ = [0.0375M] [F⁻]²
[F⁻]² = 4x10⁻⁵
[F⁻] = 6.3x10⁻³M
BaF₂ begins precipitation when [F⁻] = 6.3x10⁻³M
Thus, the fluoride which precipitates first is CaF₂
A) They are different states of matter.
1 mole Zn ---------- 6.02 x 10²³ atoms
0.750 moles Zn ----- ?
atoms = 0.750 * ( 6.02 x 10²³ / 1 )
= 4.515 x 10²³ atoms
hope this helps!
I answered that in your last question refer to your answer in section 2
