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12345 [234]
3 years ago
10

How many liters of 0.37 M solution can be made with 29.53 grams of lithium fluoride. (LiF)?

Chemistry
1 answer:
dsp733 years ago
4 0

Answer:

V = 3.1 L      

Explanation:

Given data:

Molarity of solution = 0.37 M

Mass of LiF = 29.53 g

Volume of solution = ?

Solution:

Number of moles of LiF:

Number of moles = mass/molar mass

Number of moles = 29.53 g/ 25.94g/mol

Number of moles = 1.14 mol

Volume:

Molarity = number of moles of solute / Volume in L

0.37 M = 1.14 mol / V

V = 1.14 mol / 0.37 M

V = 3.1 L         (M = mol/L)

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C) by using ICE table:

              Br2(g) + Cl2(g) → 2BrCl (g)
initial       1               1               3
change  -X              -X            +X
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when Kc = [Brcl]^2/[Cl2][Br2]
by substitution:
7 = (3+X)^2 / (1+X) (1+X)  by solving this equation for X
∴X = 0.215
so at equilibrium:
∴ [Br2] = [Cl2] = 1-0.215 = 0.785 m 
    [BrCl] = 3+0.215 = 3.215 m
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