This problem is providing us with the isotopic mass of Ga-71 and its percent abundance, so the isotopic mass of the second isotope is required. At the end, the result turns out to be 68.92547 amu.
In chemistry, isotopes are known as atoms of an element with different atomic masses (isotopic mass) but equal number of protons. In addition, they have a natural occurring abundance as a percentage.
Thus, this problem can be solved by writing the following weighted average for the atomic mass of Gallium:
[tex]m_{Ga}=m_{Ga-71}*\%ab_{Ga-71}+m_{Ga-?}*\%ab_{Ga-?}[/tex]
Hence, since both percent abundances must sum 100 %, that of the second isotope will be 60.108 %. However, since its mass is unknown, one can use the average atomic mass of Gallium consigned in the periodic table in order to write the following, after plugging in what we have so far:
[tex]69.723=70. 9247050 amu *0.39892+m_{Ga-?}*0.60108[/tex]
Hence, one can solve for the unknown as follows:
[tex]m_{Ga-?}=\frac{69.723amu-70. 9247050 amu *0.39892}{0.60108} \\\\m_{Ga-?}=68.92547amu[/tex]
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