Answer:
When [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate
Explanation:
Ksp of BaF₂ is:
BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)
Ksp = 1.7x10⁻⁶ = [Ba²⁺] [F⁻]²
The solution will produce BaF₂(s) -precipitate- just when [Ba²⁺] [F⁻]² > 1.7x10⁻⁶.
As the concentration of [Ba²⁺] is 0.0144M, the product [Ba²⁺] [F⁻]² will be equal to ksp just when:
1.7x10⁻⁶ = [Ba²⁺] [F⁻]²
1.7x10⁻⁶ = [0.0144M] [F⁻]²
1.18x10⁻⁴ = [F⁻]²
0.0109M = [F⁻]
That means, when [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate
A. Chemical energy transforms into electrical and sound energy. Hope this helps I had this question not to long ago
It’s number 3 : decomposition and melting
Here we have to get the temperature and pressure at which helium gas mostly behaves as ideal gas.
Helium (He) behave most like an ideal gas upto 24K temperature and 0 atm pressure.
The deviation of a real gas to ideal gas occurs at high temperature and low pressure.
The deviation of ideal gas to real gas occurs on taking into account the van der waals' force of attraction between the gas molecules. Now, the van der waals' interaction depends upon the polarisibility of the gas molecule.
As helium (He) is non-polarisable and very small (atomic number 2) it mostly behaves as ideal gas upto 24K temperature and 0 atm pressure.
Although the deviation from the ideal gas behavior to real gas is not so prominent at little high temperature also. Upto 50K it mostly behaves like an ideal gas.
