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
Answer:
The work done on the system is 17.75_KJ
Explanation:
To solve this question we need to know the required equations from the given variables, thus
Initial volume = 85L
Final volume = 12L
External pressure = 2.4 atm.
work done = - PΔV
2.4×(85-12) = 175.2L×atm
Converting from L•atm to KJ is given by
1 L•atm = 0.1013 kJ
(175.2 L•atm) * (0.1013 kJ / 1 L•atm)
= 17.75 kJ
The work done on the system is 17.75_KJ
The objects may exchange heat between each other if they are in contact with each other.
The two objects will allow flow of heat between them if they have a difference in temperature from each other.
The heat will flow from higher temperature object to lower temperature object. The will reach a final temperature (common to each other) and then there will be no heat transfer between each other.
Elements that belong to the same period which is seen horizontally in the periodic table have different properties. Elements with similar properties are seen in columns or groups. Elements in the same period have similar principal quantum number and belong to the same principal shell. Hence the answer is that t<span>hey have different properties that repeat across the next period</span>.