The three stable isotopes of neon: _{10}^{20} | vedclass

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(A) The average atomic mass is calculated using the weighted average formula:$m_{avg} = \frac{m_1 \eta_1 + m_2 \eta_2 + m_3 \eta_3}{\eta_1 + \eta_2 +

Vedclass · Accepted Answer

$20.1771$

Vedclass · Answer

(A) The average atomic mass is calculated using the weighted average formula:$m_{avg} = \frac{m_1 \eta_1 + m_2 \eta_2 + m_3 \eta_3}{\eta_1 + \eta_2 + \eta_3}$Given:$m_1 = 19.99\;u, \eta_1 = 90.51\%$$m_2 = 20.99\;u, \eta_2 = 0.27\%$$m_3 = 21.99\;u, \eta_3 = 9.22\%$Substituting the values:$m_{avg} = \frac{(19.99 	imes 90.51) + (20.99 	imes 0.27) + (21.99 	imes 9.22)}{90.51 + 0.27 + 9.22}$$m_{avg} = \frac{1809.2949 + 5.6673 + 202.7478}{100}$$m_{avg} = \frac{2017.71}{100} = 20.1771\;u$

The three stable isotopes of neon: $_{10}^{20}Ne$,$_{10}^{21}Ne$,and $_{10}^{22}Ne$ have respective abundances of $90.51\%$,$0.27\%$,and $9.22\%$. The atomic masses of the three isotopes are $19.99\;u$,$20.99\;u$,and $21.99\;u$ respectively. Obtain the average atomic mass of neon. (in $;u$)

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