(b) boron occurs in nature in the form of two isotopes whose atomic masses are 10.01 u and 11.01 u. the atomic mass of natural boron is 10.81 u. calculate the percentage of each isotope in natural boron.
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x-percentage of 10.01u boron
y-percentage of 11.01u boron
x, y=not known
to know that solve the below equation
x/100*10.01+y/100*11.01=10.81
y-percentage of 11.01u boron
x, y=not known
to know that solve the below equation
x/100*10.01+y/100*11.01=10.81
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