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Anestetic [448]
3 years ago
8

A student conducting the iodine clock experiment accidentally makes an S2O32- stock solution that is too concentrated. How will

this affect the rate measurement and the calculated value of k
Chemistry
1 answer:
zysi [14]3 years ago
4 0

The question is incomplete the complete question is :

Reaction 1: 3I⁻ (aq) + S₂O₈²⁻ (aq) → I₃⁻ (aq) + 2SO₄²⁻ (aq) slow

Reaction 2: I₃⁻ (aq) + 2S₂O₃²⁻ (aq) → 3I⁻ (aq) + S₄O₆²⁻ (aq) fast

Reaction 3: I₃⁻ (aq) + starch (aq) → 3I⁻ ---- starch (bluish black) fast

A. Not all of the 2S₂O₃²⁻ will react, which will increase the calculated rate.

B. Not all of the 2S₂O₃²⁻ will react, which will decrease the calculated rate.

C. It will take longer for the color to change, as the reaction takes longer when more reactant is present.

D. It will take less time for the color to change, as rate increases with concentration of reactant.

Answer: Option D

Explanation:

The reaction time for the experiment will be reduced because the concentrated form of S2O32- reacts very fast with the iodine solution.

The color will change very fast as the reactant is concentrated.Also, the reaction rate(K) of this step would occur faster than the original rate.

This is based on Le Chatelier's Prinicple which states that a corresponding change would happen to the equilibrium of a reaction when pressure, concentration of the substances or temperature is changed.

Here, the concentration has changed.

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Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate
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Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: m_{(AcO)_2Cu} = 0.972 g

Volume of the sodium chromate solution: V_{Na_2CrO_4} = 150.0 mL

Molarity of the sodium chromate solution: c_{Na_2CrO_4} = 0.0400 M

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)

Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:

n_{(AcO)_2Cu} = 0.0053515 mol

According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:

n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol

The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:

c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M

8 0
4 years ago
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