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Basile [38]
2 years ago
9

The naturally occurring isotopes of magnesium are magnesium-24. magnesium-25, and magnesium-26. Magnesium-24 has an abundance of

78.994% and a mass of 23.985 amu. Magnesium-25 has an abundance of 10.001% and a mass of 24.986 amu. Magnesium-26 has an abundance of 11.013% and a mass of 25.983 amu. Calculate the average atomic mass of magnesium.
Chemistry
2 answers:
iren2701 [21]2 years ago
8 0

<u>Answer:</u> The average atomic mass of magnesium is 24.307 amu.  

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.  

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

  • <u>For _{12}^{24}\textrm{Mg} isotope:</u>

Mass of _{12}^{24}\textrm{Mg} isotope = 23.985 amu

Percentage abundance of _{12}^{24}\textrm{Mg} isotope = 78.994 %

Fractional abundance of _{12}^{24}\textrm{Mg} isotope = 0.78994

  • <u>For _{12}^{25}\textrm{Mg} isotope:</u>

Mass of _{12}^{25}\textrm{Mg} isotope = 24.986 amu

Percentage abundance of _{12}^{25}\textrm{Mg} isotope = 10.001 %

Fractional abundance of _{12}^{25}\textrm{Mg} isotope = 0.10001

  • <u>For _{12}^{26}\textrm{Mg} isotope:</u>

Mass of _{12}^{26}\textrm{Mg} isotope = 25.983 amu

Percentage abundance of _{12}^{26}\textrm{Mg} isotope = 11.013 %

Fractional abundance of _{12}^{26}\textrm{Mg} isotope = 0.11013

Putting values in equation 1, we get:  

\text{Average atomic mass of magnesium}=[(23.985\times 0.78994)+(24.986\times 0.10001)+(25.983\times 0.11013)]

\text{Average atomic mass of magnesium}=24.307amu

Hence, the average atomic mass of magnesium is 24.307 amu.

RoseWind [281]2 years ago
5 0
69.420 would be your answer

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2 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
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C. Ca_3(PO_4)_2  will precipitate out first

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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}

[PO_4^{3-}]^2 = 2.43 \times 10^{-29}

[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}

[PO_4^{3-}] =4.93 \times 10^{-15}

Thus; the phosphate anion needed for precipitation is smaller i.e 4.93 \times 10^{-15} in Ca_3(PO_4)_2 than  in  Ag_3PO_4  1.07 \times 10^{-13}

Therefore:

Ca_3(PO_4)_2  will precipitate out first

To determine the concentration of [Ca^+] when  the second cation starts to precipitate ; we have :

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

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

[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}}

[Ca^{2+}] =0.00566

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 = 0.00566/0.0440  × 100%

the percentage of Ca^{2+} remaining = 0.1286 × 100%

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

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Answer:

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Explanation:

Step 1: Data given

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Step 2: Calculate molar mass of KHCO3

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