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lina2011 [118]
3 years ago
8

Naturally occuring boron is 80.20% boron-11 (atomic mass 11.01 amu) and 19.80% of some other isotope. What must the atomic mass

of this second isotope be in order to account for the 10.81 amu average atomic mass of boron?.
Chemistry
1 answer:
Yanka [14]3 years ago
7 0
You will have to do some math. 

<span>80.2 × 11.01 + 19.8 × x = 100 * 10.81. </span>

<span>x is the mass in amu of the second isotope. </span>

<span>Solve for x. </span>

<span>19.8 × x = 100 * 10.81 - 80.2 × 11.01 </span>
<span>19.8 × x = 1081 - 883.00 = 198.00 </span>
<span>x = 1964.00 / 19.8 = 10.00 </span>

<span>The mass of the other isotope is 10.00 amu.</span>
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