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question 12 (2 points)
a sample of naturally occurring leungstine consists lg - 28 (amu = 27.9769), lg - 29 (amu = 28.9765) and lg - 30 (amu = 29.9738). if the weighted atomic mass of leungstine is 28.0855 and the natural abundance of lg - 29 is 4.67%, what is the natural abundance of lg - 28?
what is the natural abundance of lg - 30?
question 13 (4 points)
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Step1: Set up the equation for weighted - atomic - mass
The formula for the weighted atomic mass \(M\) of an element with isotopes \(X_1,X_2,X_3\) (in this case, isotopes \(Lg - 28\), \(Lg - 29\), \(Lg - 30\)) is \(M=\sum_{i = 1}^{n}m_i\times a_i\), where \(m_i\) is the mass of the \(i -\)th isotope and \(a_i\) is the abundance of the \(i -\)th isotope (in decimal form). Let the abundance of \(Lg - 28\) be \(x\) (in decimal), the abundance of \(Lg - 29\) is \(0.0467\) (since \(4.67\%=0.0467\)), and the abundance of \(Lg - 30\) is \(y\). We know that \(x + 0.0467+y=1\), so \(y = 1-(x + 0.0467)\). And \(M=28.0855\), \(m_1 = 27.9769\), \(m_2=28.9765\), \(m_3 = 29.9738\).
Substitute \(y = 1-(x + 0.0467)\) into the above equation:
Step2: Expand and simplify the equation
First, calculate \(28.9765\times0.0467\approx1.3532\) and \(29.9738\times0.0467\approx1.4008\).
Step3: Solve for \(x\)
Rearrange the equation \(-1.9969x=28.0855 - 29.9262\)
\(-1.9969x=-1.8407\)
Since \(x\) is the abundance of \(Lg - 28\) in decimal, the percentage abundance of \(Lg - 28\) is \(92.18\%\)
Step4: Solve for \(y\) (abundance of \(Lg - 30\))
Since \(y = 1-(x + 0.0467)\) and \(x = 0.9218\)
The percentage abundance of \(Lg - 30\) is \(3.15\%\)
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The natural abundance of \(Lg - 28\) is \(92.18\%\) and the natural abundance of \(Lg - 30\) is \(3.15\%\)