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3 years ago

URGENT is... Mg(ClO3)2

Chemistry

2 answers:

stiv31 [10]3 years ago

6 0

That's a ionic compound because it has a metal and polyatomic parts within the chemical formula.

iogann1982 [59]3 years ago

6 0

Answer:

Ionic

Explanation:

The corresponding Compound Mg(ClO3)2 is an ionic compound. By the structure of it be say that there is donation of electron by Magnesium to the Chloarate ion.

Moreover, Magnesium chlorate is soluble in water. It also has low density and low melting point. The structure is crystalline. Hence we can say that Mg(ClO3)2 is ionic in nature.

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3 years ago

A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would

Olenka [21]

Answer:

C. $Ca_3(PO_4)_2$ will precipitate out first

the percentage of $Ca^{2+}$ remaining = 12.86%

Explanation:

Given that:

A solution contains:

$[Ca^{2+}] = 0.0440 \ M$

$[Ag^+] = 0.0940 \ M$

From the list of options , Let find the dissociation of $Ag_3PO_4$

$Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}$

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Solubility product constant Ksp of $Ag_3PO_4$ is $8.89 \times 10^{-17}$

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$Ksp = [Ag^+]^3[PO_4^{3-}]$

replacing the known values in order to determine the unknown ; we have :

$8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]$

$\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]$

$[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}$

$[PO_4^{3-}] =1.07 \times 10^{-13}$

The dissociation of $Ca_3(PO_4)_2$

The solubility product constant of $Ca_3(PO_4)_2$ is $2.07 \times 10^{-32}$

The dissociation of $Ca_3(PO_4)_2$ is :

$Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}$

Thus;

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2$

$\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2$

$[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}$

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Therefore:

$Ca_3(PO_4)_2$ will precipitate out first

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$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

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$[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}$

$[Ca^{2+}]^3 =1.808 \times 10^{-7}$

$[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}$

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This implies that when the second cation starts to precipitate ; the concentration of $[Ca^{2+}]$ in the solution is 0.00566

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the percentage of $Ca^{2+}$ remaining = concentration remaining/initial concentration × 100%

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