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andreyandreev [35.5K]
4 years ago
5

"A solution with a total volume of 850.0 mL contains 43.5 g Mg(NO3)2. If you remove 25.0 mL of this solution and then dilute thi

s 25.0 mL sample with water until the new volume equals 600.0 mL, what is the concentration of nitrate ions in the 600.0 mL solution
Chemistry
1 answer:
ycow [4]4 years ago
7 0

Answer: C = 0.014M

Explanation:

From n= m/M= CV

m =43.5 M= 148, V=850ml

43.5/148= C× 0.85

C= 0.35M

Applying dilution formula

C1V1=C2V2

C1= 0.35, V1= 25ml, C2=?, V2= 600ml

0.35× 25 = C2× 600

C2= 0.014M

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E° = 0.00 V

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Explanation:

We can identify both half-reactions occurring in a concentration cell.

Anode (oxidation): Al(s) → Al³⁺(1.0 × 10⁻⁵ M) + 3 e⁻   E°red = -1.66 V

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E = 0.00 V - (0.05916/3) .log (1.0 × 10⁻⁵/0.100)

E = 0.079 V

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