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Reika [66]
3 years ago
6

Suppose 0.981 g of iron (II) iodide is dissolved in 150. mL of a 35.0 m M aqueous solution of silver nitrate. Calculate the fina

l molarity of iodide anion in the solution, You can assume the volume of the solution doesn't change shen th s sove m Be sure your answer has the correct number of significant digits
Chemistry
1 answer:
yaroslaw [1]3 years ago
8 0

Answer:

Final molarity of iodide ion C(I-) = 0.0143M

Explanation:

n = (m(FeI(2)))/(M(FeI(2))

Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol

So n = 0.981/309.85 = 0.0031 mol

V(solution) = 150mL = 0.15L

C(AgNO3) = 35mM = 0.035M = 0.035m/L

n(AgNO3) = C(AgNO3) x V(solution)

= 0.035 x 0.15 = 0.00525 mol

(AgNO3) + FeI(3) = AgI(3) + FeNO3

So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol

C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M

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ICE Princess25 [194]

Answer:

54.7°C is the new temperature

Explanation:

We combine the Ideal Gases Law equation to solve this.

P . V = n. R. T

As moles the balloon does not change and R is a constant, we can think this relation between the two situations:

P₁ . V₁ / T₁ = P₂ . V₂ / T₂

T° is absolute temperature (T°C + 273)

68.7°C + 273 = 341.7K

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1.63 atm.L/K = 532.5 atm.L / T₂

T₂ = 532.5 atm.L / 1.63 K/atm.L → 326.7K

T° in C = T°K - 273 → 326.7K + 273 = 54.7°C

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