Chapter 5: Problem 27
Naturally occurring zirconium exists as five stable isotopes: \({ }^{90} \mathrm{Zr}\) with a mass of \(89.905 \mathrm{u}(51.45 \%) ;{ }^{91} \mathrm{Zr}\) with a mass of \(90.906 \mathrm{u}\) \((11.22 \%) ;{ }^{92} \mathrm{Zr}\) with a mass of \(91.905 \mathrm{u}(17.15 \%) ;{ }^{94} \mathrm{Zr}\) with a mass of \(93.906 \mathrm{u}(17.38 \%)\); and \({ }^{96} \mathrm{Zr}\) with a mass of \(95.908 \mathrm{u}\) \((2.80 \%)\). Calculate the average mass of zirconium.
Short Answer
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