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

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

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We start by noting that potassium‑40 (K‑40) decays to argon‑40 (Ar‑40) with a half‐life of 1.3×10⁹ years. For a rock that is 1.3×10⁹ years old (i.e., one half‑life), half of the original K‑40 will have decayed to form Ar‑40.

Let X be the original amount (in moles) of K‑40 present. After one half‑life, half of that amount remains as K‑40 and the other half has decayed to Ar‑40. That means the amount of Ar‑40 produced is X/2.

We’re told that the rock contains 3.14×10⁻⁷ mol of Ar‑40. Therefore, we have:

X/2 = 3.14×10⁻⁷ mol

Solve for X:

X = 2 × (3.14×10⁻⁷ mol) = 6.28×10⁻⁷ mol

Thus, for the rock to be 1.3×10⁹ years old, it would have had to originally contain 6.28×10⁻⁷ moles of K‑40.
To determine the original amount of potassium-40 (K-40) in the rock, we use the fact that after one half-life (1.3×10⁹ years), half of the original K-40 decays to form argon-40 (Ar-40). Given that the rock contains 3.14×10⁻⁷ moles of Ar-40, the original amount of K-40 would be twice that, which is 6.28×10⁻⁷ moles.

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

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