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alexira [117]
3 years ago
13

In a first–order reaction (A → B and rate = k [A]), tripling the concentration of A will have what effect on the reaction rate?

Chemistry
1 answer:
Natali [406]3 years ago
5 0

Answer:

Tripling the concentration of A will triple the reaction rate.

Explanation:

  • For a first–order reaction hat has a rate law:

<em>Rate = k[A].</em>

  • It is clear that the reaction rate is directly proportional to the concentration of A.

<em>Rate ∝ [A].</em>

<u><em>So, Tripling the concentration of A will triple the reaction rate.</em></u>

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The compound that would have the highest osmotic pressure when dissolved in water is FeCl_{3} .

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The dissociation of one molecule of  FeCl_{3} gives the maximum number of ions when dissolved in water ( 4 ions ) . Osmotic pressure is a colligative property and depends upon number of solute particles present in the solution . The solution having maximum number of solute particles will have maximum number of the osmotic pressure .

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Sodium sulfate is slowly added to a solution containing 0.0500 M Ca 2 + ( aq ) and 0.0390 M Ag + ( aq ) . What will be the conce
levacccp [35]

The given question is incomplete. The complete question is as follows.

Sodium sulfate is slowly added to a solution containing 0.0500 M Ca^{2+}(aq) and 0.0390 M Ag^{+}(aq). What will be the concentration of Ca^{2+}(aq) when Ag_{2}SO_{4}(s) begins to precipitate? What percentage of the Ca^{2+}(aq) can be separated from the Ag(aq) by selective precipitation?

Explanation:

The given reaction is as follows.

      Ag_{2}SO_{4} \rightleftharpoons 2Ag^{+} + SO^{2-}_{4}

[Ag^{+}] = 0.0390 M

When Ag_{2}SO_{4} precipitates then expression for K_{sp} will be as follows.

         K_{sp} = [Ag^{+}]^{2}[SO^{2-}_{4}]

        1.20 \times 10^{-5} = (0.0390)^{2} \times [SO^{2-}_{4}]

       [SO^{2-}_{4}] = 0.00788 M

Now, equation for dissociation of calcium sulfate is as follows.

         CaSO_{4} \rightleftharpoons Ca^{2+} + SO^{2-}_{4}

      K_{sp} = [Ca^{2+}][SO^{2-}_{4}]

     4.93 \times 10^{-5} = [Ca^{2+}] \times 0.00788

           [Ca^{2+}] = 0.00625 M

Now, we will calculate the percentage of Ca^{2+} remaining in the solution as follows.

               \frac{0.00625}{0.05} \times 100

                 = 12.5%

And, the percentage of Ca^{2+} that can be separated is as follows.

                     100 - 12.5

                     = 87.5%

Thus, we can conclude that 87.5% will be the concentration of Ca^{2+}(aq) when Ag_{2}SO_{4}(s) begins to precipitate.

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