Calculate the lattice energy for NaF (s) given the following
1 answer:
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Answer:
How many grams Acetic Acid in 1 mol? The answer is 60.05196.
We assume you are converting between grams Acetic Acid and mole?
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Explanation:
You can view more details on each measurement unit:
molecular weight of Acetic Acid or mol
The molecular formula for Acetic Acid is CH3COOH.
The SI base unit for amount of substance is the mole.
1 grams Acetic Acid is equal to 0.016652245821785 mole.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between grams Acetic Acid and mole.
Examples
1 grams Acetic Acid to mol = 0.01665 mol
10 grams Acetic Acid to mol = 0.16652 mol
20 grams Acetic Acid to mol = 0.33304 mol
30 grams Acetic Acid to mol = 0.49957 mol
40 grams Acetic Acid to mol = 0.66609 mol
50 grams Acetic Acid to mol = 0.83261 mol
100 grams Acetic Acid to mol = 1.66522 mol
200 grams Acetic Acid to mol = 3.33045 mol
Answer:
Explanation:
One way to calculate the lattice energy is to use Hess's Law.
The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:
Li⁺(g) + Cl⁻(g) ⟶ LiCl(s); U = ?
We must generate this reaction rom the equations given.
(1) Li(s) + ½Cl₂ (g) ⟶ LiCl(s); ΔHf° = -409 kJ·mol⁻¹
(2) Li(s) ⟶ Li(g); ΔHsub = 161 kJ·mol⁻¹
(3) Cl₂(g) ⟶ 2Cl(g) BE = 243 kJ·mol⁻¹
(4) Li(g) ⟶Li⁺(g) +e⁻ IE₁ = 520 kJ·mol⁻¹
(5) Cl(g) + e⁻ ⟶ Cl⁻(g) EA₁ = -349 kJ·mol⁻¹
Now, we put these equations together to get the lattice energy.
<u>E/kJ </u>
(5) Li⁺(g) +e⁻ ⟶ Li(g) 520
(6) Li(g) ⟶ Li(s) -161
(7) Li(s) + ½Cl₂(g) ⟶ LiCl(s) -409
(8) Cl(g) ⟶ ½Cl₂(g) -121.5
(9) Cl⁻(g) ⟶ Cl(g) + e⁻ <u>+349</u>
Li⁺(g) + Cl⁻(g) ⟶ LiCl(s) -862
The lattice energy of LiCl is .