Incorrect. If a rock sample was found to contain \( 3.14 \times 10^{-7} \mathrm{~mol} \) of argon- 40 , how much potassium \( -40\left(\mathrm{t}_{1 / 2}=1.3 \times 10^{9}\right. \) yr) would also have to be present for the rock to be \( 1.3 \times 10^{9} \) years old? eTextbook and Media Hint

The relationship between potassium-40 and argon-40 is essential for dating rocks. For every 1 mole of potassium-40 that decays, it produces 1 mole of argon-40. Since the half-life of potassium-40 is 1.3 billion years, the amount of potassium-40 present when the rock was formed can be calculated using the formula for half-life. Considering that your rock is 1.3 billion years old and currently has \( 3.14 \times 10^{-7} \) mol of argon-40, you would actually need 2 times that amount of potassium-40 at the beginning, which equals \( 6.28 \times 10^{-7} \) mol. To solve this problem effectively, it can be beneficial to understand that the amount of the original potassium-40 can be determined by noting that the ratio of the remaining substance to the substance produced is linked through the equation of exponential decay. A common mistake here is failing to double the amount of the current argon-40 because the decay means some potassium-40 has transformed into argon-40! So, always remember - for each mole of argon produced, an equal amount of potassium has decayed. Happy calculating!