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12345 [234]
3 years ago
10

Lithium, is used in dry cells and storage batteries and in high temperature lubricants, it has two naturally occurring isotopes,

6Li and 7Li. Calculate the atomic mass of lithium.
Chemistry
1 answer:
erastovalidia [21]3 years ago
7 0

<u>Answer:</u> The average atomic mass of lithium is 6.9241 u.

<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 _3^6\textrm{Li} isotope:</u>

Mass of _3^6\textrm{Li} isotope = 6 u

Percentage abundance of _3^6\textrm{Li} isotope = 7.59 %

Fractional abundance of _3^6\textrm{Li} isotope = 0.0759

  • <u>For _3^7\textrm{Li} isotope:</u>

Mass of _3^7\textrm{Li} isotope = 7 u

Percentage abundance of _3^7\textrm{Li} isotope = 92.41%

Fractional abundance of _3^7\textrm{Li} isotope = 0.9241

Putting values in equation 1, we get:

\text{Average atomic mass of Lithium}=[(6\times 0.0759)+(7\times 0.9241)]

\text{Average atomic mass of Lithium}=6.9241u

Hence, the average atomic mass of lithium is 6.9241 u.

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<h3>What are iron ores?</h3>

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2 years ago
Determine the energy of the electron in the 1s orbital of a helium ion (He+ )
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He Rydberg formula can be extended for use with any hydrogen-like chemical elements. 
<span>1/ λ = R*Z^2 [ 1/n1^2 - 1/n2^2] </span>
<span>where </span>
<span>λ is the wavelength of the light emitted in vacuum; </span>
<span>R is the Rydberg constant for this element; R 1.09737x 10^7 m-1 </span>
<span>Z is the atomic number, for He, Z =2; </span>
<span>n1 and n2 are integers such that n1 < n2 </span>
<span>The energy of a He+ 1s orbital is the opposite to the energy needed to ionize the electron that is </span>
<span>taking it from n = 1 (1/n1^2 =1) to n2 = ∞ (1/n2^2 = 0) </span>
<span>.: 1/ λ = R*Z^2 = 1.09737x 10^7*(2)^2 </span>
<span>λ = 2.278*10^-8 m </span>
<span>E = h*c/λ </span>
<span>Planck constant h = 6.626x10^-34 J s </span>
<span>c = speed of light = 2.998 x 10^8 m s-1 </span>
<span>E = (6.626x10^-34*2.998 x 10^8)/(2.278*10^-8) = 8.72*10^-18 J ion-1 </span>
<span>Can convert this value to kJ mol-1: </span>
<span>(8.72*10^-18*6.022 x 10^23)/1*10^3 = 5251 kJ mol-1 </span>
<span>Lit value: RP’s secret book: 5240.4 kJ mol-1 (difference is due to a small change in R going from H to He+) </span>
<span>So energy of the 1s e- in He+ = -5251 kJ mol-1</span>
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