Answer:
The answer to your question is: letter c (96%)
Explanation:
Indium -113 (112-9040580 amu) ₁₁₃In
Indium-115 (114.9038780 amu) ₁₁₅In
Atomic mass of Indium is 114.82 amu ₁₁₄.₈₂In
Formula
Atomic mass = m₁(%₁) +m₂(%₂) / 100
%₁ = x I established this is an equation
%₂ = 100 - x
Substituting values
114.82 = 112.8040x + 114.9039(100-x) /100 and know we expand and simplify
114.82 = 112.8040x + 11490.39 - 114.9039x /100
11482 = 112.8040x -114.9039x +11490.39
11482 - 11490.39 = 112.8040x -114.9039x
-8.39 = -2.099x
x = 3.99
Then % of Indium-115 = 100 - 3.99 = 96
c)95.7%
Indium -113 (112-9040580 amu) ₁₁₃In
Indium-115 (114.9038780 amu) ₁₁₅In
Atomic mass of Indium is 114.82 amu ₁₁₄.₈₂In
Formula to be used to calculate the percentage for isotopes will be:
Atomic mass = m₁(%₁) +m₂(%₂) / 100
%₁ = x
%₂ = 100 - x
On substituting the values:
[tex]114.82 = \frac{112.8040x + 114.9039(100-x)}{100} \\\\ 114.82 = \frac{ 112.8040x + 11490.39 - 114.9039x }{100} \\\\ 114.82 = 112.8040x -114.9039x +11490.39\\\\ 114.82 - 11490.39 = 112.8040x -114.9039x\\\\ -8.39 = -2.099x\\\\ x = 3.99 [/tex]
Thus, % of Indium-115 = 100 - 3.99 = 96%.
Therefore, the correct option is c.
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