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GarryVolchara [31]
3 years ago
12

A particular reaction, A- products, has a rate that slows down as the reaction proceeds. The half-life of the reaction is found

to depend on the initial concentration of A. Determine whether each statement is likely to be true or false for this reaction.
a. A doubling of the concentration of A doubles the rate of the reaction.
b. A plot of 1/[A] versus time is linear.
c. The half-life of the reaction gets longer as the initial concen- tration of A increases.
d. A plot of the concentration of A versus time has a constant slope.
Chemistry
1 answer:
Thepotemich [5.8K]3 years ago
8 0

Explanation:

Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.

a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement  a is wrong.

b. Expression for second order reaction is as follows:

\frac{1}{[A]} =\frac{1}{[A]_0} +kt

the above equation can be written in the form of Y = mx + C

so, the plot between 1/[A] and t is linear. So the statement b is true.

c.

Expression for half life is as follows:

t_{1/2}=\frac{1}{k[A]_0}

As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.

d.

Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong

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Sodium phosphate is added to a solution that contains 0.0030 M aluminum nitrate and 0.016 M calcium chloride. The concentration
gavmur [86]

Explanation:

It is given that aluminium nitrate and calcium chloride are mixed together with sodium phosphate.

And, K_{sp} of AlPO_{4} = 9.84 \times 10^{-21}

        K_{sp} of Ca_{3}(PO_{4})_{2} = 2.0 \times 10^{-29}

Let us assume that the solubility be "s". And, the reaction equation is as follows.

        AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}

     9.84 \times 10^{-21} = s \times s

             s = 9.92 \times 10^{-11}

Also,     Ca_{3}(PO_{4})_{2} \rightleftharpoons 3Ca^{2+} + 2PO^{3-}_{4}

                2 \times 10^{-29} = (3s)^{3} \times (2s)^{2}

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This means that first, aluminium phosphate will precipitate.

Now, we will calculate the concentration of phosphate when calcium phosphate starts to precipitate out using the K_{sp} expression as follows.

         K_{sp} = [Ca^{2+}]^{3}[PO^{3-}_{4}]^{2}

          2.0 \times 10^{-29} = (0.016)^{3}[PO^{3-}_{4}]^{2}

       2.0 \times 10^{-29} = 4.096 \times 10^{-6} \times [PO^{3-}_{4}]^{2}

       [PO^{3-}_{4}]^{2} = 4.88 \times 10^{-24}

                             = 2.21 \times 10^{-12} M

Similarly, calculate the concentration of aluminium at this concentration of phosphate as follows.

             AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}

           K_{sp} = [Al^{3+}][PO^{3-}_{4}]

       9.84 \times 10^{-21} = [Al^{3+}] \times 2.21 \times 10^{-12}

                [Al^{3+}] = 4.45 \times 10^{-9} M

Thus, we can conclude that concentration of aluminium will be 4.45 \times 10^{-9} M when calcium begins to precipitate.

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2 years ago
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Answer:

Explanation:

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How many grams of calcium phosphate are theoretically produced if we start with 552.2 grams of Ca(NO3)2 and 285.4 grams of Li3PO
Sedbober [7]
<span>molecular weight of Ca(NO3)2 --> 164 Ca=40 N=14 O=16 molecular weight of Li3PO4 --> 115.8 molecular weight of LiNO3 --> 68.9 Ca3(PO4)2 produced ---> 310.2 g/mol molecular weight Hence the answer would be 310.2 g/mol of calcium phosphate will be produced</span>
7 0
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