Answer:
Ionic character
A. PF₃ > PBr₃ > PCl₃
B. BF₃ > CF₄ > NF₃
C. TeF₄ > BrF₃ > SeF₄
Explanation:
The most electronegative element is fluorine, followed chlorine, phosphorous nitrogen etc.
- Atoms with high electronegativity tend to form negative ions.
- Ionic compounds formed between elements with high electronegativity difference.
- % ionic character is directly proportional to electronegativity difference.
- According to Pauling Scale E.n for F(4.0), O(3.5), N(3.0), C(2.5), B(2.0), P(2.19), Se(2.55) , Te (2.1), Cl(3.16) and Br(2.96)
- ΔE.N (Electronegativity difference) between( P and F = 4 - 2.19 = 1.81), (P and Br = 2.96 - 2.19 = 0.77) , (P and Cl = 3.16 - 2.96 = 0.2 )
- ΔE.N (Electronegativity difference) between( N and F = 4 - 3 = 1), (B and F = 4 - 2 = 2) , (C and F = 4 - 2.5 = 1.5 )
- ΔE.N (Electronegativity difference) between( Se and F = 4 - 2.55 = 1.45), (F and Te = 4 - 2.1 = 1.9) , (F and Br = 4 - 2.19 = 1.81 )
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Answer:
111.15 g are required to prepare 500 ml of a 3 M solution
Explanation:
In a 3 M solution of Ca(OH)₂ there are 3 moles of Ca(OH)₂ per liter solution. In 500 ml of this solution, there will be (3 mol/2) 1.5 mol Ca(OH)₂.
Since 1 mol of Ca(OH)₂ has a mass of 74.1 g, 1.5 mol will have a mass of
(1.5 mol Ca(OH)₂ *(74.1 g / 1 mol)) 111.15 g. This mass of Ca(OH)₂ is required to prepare the 500 ml 3 M solution.
Answer:
B. To Identify the half reactions for the equation
