Question

question 15
potassium has two main isotopes, k-39 and k-41. the average atomic mass of potassium is 39.13 amu. the relative abundance for k-41 is blank 1 %.
blank 1
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Explanation:

Step1: Define variables

Let \( x \) be the relative abundance of K - 41 (in decimal form), so the relative abundance of K - 39 is \( 1 - x \) (since the sum of relative abundances of all isotopes is 1). The mass of K - 39 is 39 amu and the mass of K - 41 is 41 amu. The average atomic mass \( A \) is given by the formula:

$$ A=( \text{mass of K - 39} \times \text{abundance of K - 39})+( \text{mass of K - 41} \times \text{abundance of K - 41}) $$

We know that \( A = 39.13 \) amu, mass of K - 39 \( = 39 \) amu, mass of K - 41 \( = 41 \) amu. Substituting the values into the formula:

$$ 39.13=39\times(1 - x)+41\times x $$

Step2: Solve the equation

First, expand the right - hand side:

$$ 39.13 = 39-39x + 41x $$

Combine like terms:

$$ 39.13=39 + 2x $$

Subtract 39 from both sides:

$$ 39.13-39=2x $$

$$ 0.13 = 2x $$

Then, solve for \( x \) by dividing both sides by 2:

$$ x=\frac{0.13}{2}=0.065 $$

Step3: Convert to percentage

To convert the decimal abundance to a percentage, multiply by 100:

$$ 0.065\times100 = 6.5\% $$

Answer:

6.5