Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.
Explanation :
As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.
Let us assume that the radius of Cl⁻ be, (x) pm
So, the radius of Na⁺ = 
In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.
Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.
Given:
Distance between Na⁺ nuclei = 566 pm
Thus, the relation will be:
The radius of Cl⁻ ion = (x) pm = 181 pm
The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm
Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.
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Answer:
Mass of oxygen = 2.2 g
Explanation:
Given data:
Mass of benzoic acid= 8.20 g
Mass of oxygen= ?
Solution:
Molar mass of oxygen = 16×2 g/mol
Molar mass of C₆H₅COOH = 7×12 + 1×6 + 2×16
Molar mass of C₆H₅COOH = 84 + 6 + 32
Molar mass of C₆H₅COOH = 122g/mol
Mass of oxygen in 8.20 g of C₆H₅COOH :
Mass of oxygen = 32 g.mol⁻¹/122 g.mol⁻¹ × 8.20 g
Mass of oxygen = 2.2g
Explanation:
Transcribed image text: H26.25 - Level 2 Homework. Unanswered Match reagents as starting materials for the synthesis of ethyl 3-phenyl-3-oxopropanate. You can draw out the structures on your own to help you answer this question. Premise Response Drag and drop to match 1 Methyl benzoate = A Acetophenone 2 Diethyl carbonate = B Ethyl acetate
