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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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4 0
3 years ago
The graph above shows the changes in temperature recorded for the 2.00 l of h2o surrounding a constant-volume container in which
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Answer:

25.2 kJ

Explanation:

The complete question is presented in the attached image to this answer.

Note that, the heat gained by the 2.00 L of water to raise its temperature from the initial value to its final value comes entirely from the combustion of the benzoic acid since there are no heat losses to the containing vessel or to the environment.

So, to obtained the heat released from the combustion of benzoic acid, we just calculate the heat required to raise the temperature of the water.

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Hope this Helps!!!

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