Asked by Peter

Naturally occurring boron is 80.20% boron-11 (of atomic mass 11.01 amu) and 19.80% of some other isotope form of boron. What must be the atomic mass of this second isotope in order to account for the 10.81 amu average atomic mass of boron?

A. 10.99 amu
B. 11.00 amu
C. None of these
D. 11.01 amu
E. 11.99 amu
F. 12.00 amu
G. 10.00 amu
H. 9.00 amu

13 years ago

Answered by DrBob222

0.8020(11.01) + 0.1980(x) = 10.81
Solve for x.

13 years ago

Answered by Tay

X= 5.9089

11 years ago

Answered by Anonymous

Ans is 10.97amu

7 years ago

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