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Gekata [30.6K]
3 years ago
15

Suppose that a fictitious element, X , has two isotopes: 59 X ( 59.015 amu ) and 62 X ( 62.011 amu ) . The lighter isotope has a

n abundance of 91.7 % . Calculate the average atomic mass of the element X .
Chemistry
1 answer:
Gemiola [76]3 years ago
8 0

<u>Answer:</u> The average atomic mass of element X is 59.3 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 isotope 1 (X-59) :</u>

Mass of isotope 1 = 59.015 amu

Percentage abundance of isotope 1 = 91.7 %

Fractional abundance of isotope 1 = 0.917

  • <u>For isotope 2 (X-62) :</u>

Mass of isotope 2 = 62.011 amu

Percentage abundance of isotope 2 = (100 - 91.7) % = 8.3 %

Fractional abundance of isotope 2 = 0.083

Putting values in equation 1, we get:

\text{Average atomic mass of X}=[(59.015\times 0.917)+(62.011\times 0.083)]

\text{Average atomic mass of X}=59.3amu

Hence, the average atomic mass of element X is 59.3 amu.

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Substitute all these values in the above formula:

ln K_e_q=\frac{2mol* 96485 C/mol}{8.314 J.K^-^1.mol^-^1x298K} \\\\lnK_e_q=77.8\\K_e_q=e^7^7^.^8\\=>K_e_q=6.13x10^3^3

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Answer:

2445 L

Explanation:

Given:  

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Using ideal gas equation as:

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R is Gas constant having value = 08206 L.atm/K.mol

Applying the equation as:

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<u>⇒V = 2445.39 L</u>

Answer to four significant digits, Volume = 2445 L

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