To find the percentage composition of the isotope with mass number 37, given that the mean relative atomic mass is 35, we can set up a system of equations. Let:
- \( x \) be the fraction (or percentage) of isotope 35.
- \( y \) be the fraction (or percentage) of isotope 37.
Since there are only two isotopes, we have:
\[ x + y = 1 \]
The mean relative atomic mass equation is:
\[ 35x + 37y = 35 \]
Now, we can substitute \( y \) from the first equation into the second equation.
From \( x + y = 1 \), we find:
\[ y = 1 - x \]
Now substituting \( y \) into the second equation:
\[ 35x + 37(1 - x) = 35 \]
Expanding the equation:
\[ 35x + 37 - 37x = 35 \]
Combining like terms gives:
\[ -2x + 37 = 35 \]
Subtracting 37 from both sides:
\[ -2x = 35 - 37 \]
\[ -2x = -2 \]
Dividing by -2:
\[ x = 1 \]
Now, substitute back to find \( y \):
\[ y = 1 - x = 1 - 1 = 0 \]
It seems there was a misunderstanding; \( x \) and \( y \) represent fractions of the isotopes. If we denote:
Let \( A \) be the fraction of isotope 37 and \( B \) be the fraction of isotope 35.
Then:
- \( A + B = 1 \)
- \( 35B + 37A = 35 \)
Substitute \( B = 1 - A \) into the second equation:
\[ 35(1 - A) + 37A = 35 \]
Expanding gives:
\[ 35 - 35A + 37A = 35 \]
Combining the terms gives:
\[ 2A = 0 \]
Hence \( A = 0 \) and \( B = 1 \), meaning that only the isotope with mass number 35 contributes to the average atomic mass of 35, so the proportion of the isotope with mass number 37 is 0%.
Conclusion
The percentage composition of the isotope with a mass number of 37 is 0%.
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