It uses elimination againLet A be 15% juice and B is 5% juice
A+B = 100.15A + 0.05B = 0.11*10 = 1.1Multiply 2nd equation by 100 to get rid of decimals
A+B = 1015A + 5B = 110
Answer:
Average atomic mass = 15.86 amu.
Explanation:
Given data:
Number of atoms of Z-16.000 amu = 205
Number of atoms of Z-14.000 amu = 15
Average atomic mass = ?
Solution:
Total number of atoms = 205 + 15 = 220
Percentage of Z-16.000 = 205/220 ×100 = 93.18%
Percentage of Z-14.000 = 15/220 ×100 = 6.82 %
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (93.18×16.000)+(6.82×14.000) /100
Average atomic mass = 1490.88 + 95.48 / 100
Average atomic mass = 1586.36 / 100
Average atomic mass = 15.86 amu.
