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
Answer:
In general, the energy needed to remove an electron (the ionization energy) increases as the atomic number increases across a period.
Explanation:
Each additional proton increases the attraction between the nucleus and the electron.
Thus, it takes more energy to remove an electron as you add extra protons.
Data:
solute: ethylene glicol => not ionization
molar mass of ethylene glicol (from internet) = 62.07 g/mol
solute = 400 g
solvent = water = 4.00 kg
m =?
ΔTf = ?
Kf = 1.86 °C/mol
Formulas:
m: number of moles of solute / kg of solvent
ΔTf = Kf*m
number of moles of solute = mass in grams / molar mass
Solution
number of moles of solute = 400 g / 62.07 g/mol = 6.44 moles
m = 6.44 mol / 4 kg = 1.61 m <-------- molality (answer)
ΔTf = 1.86 °C / m * 1.61 m = 2.99 °C <---- lowering if freezing point (answer)
Answer:
The value of partial pressure of oxygen 
= 83.66 K pa
Explanation:
Volume of oxygen = 80 liters
Temperature = 50°c
The total pressure = 96 K pa
The partial pressure of water at 50°c is calculated from the tables.
= 12.344 K pa
The total pressure is given by
Put all the values in the above equation we get
96 = 12.344 +
96 - 12.344
= 83.66 K pa
This is the value of partial pressure of oxygen.
