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2 years ago

15

When air is heated, its density A. Increases B. Decreases C. stays the same

2 answers:

soldier1979 [14.2K]2 years ago

6 0

Answer:

a

Explanation:

Stels [109]2 years ago

4 0

Answer:

A. Increases

Hope this helps!

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2 years ago

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3 years ago

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

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

s = $7.14 \times 10^{-7}$

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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3 years ago

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3 years ago

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Wewaii [24]

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3 years ago