A rock contains 0.37 mg of Pb-206 and 0.95 mg of U-238. Approximately how many U-238 atoms were in the rock when it was formed billions of years ago? (The half life for ²³⁸U → ²⁰⁶Pb is 4.51 × 10⁹ yr.)

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First, we need to find the ratio of Pb-206 to U-238 in the rock:
0.37 mg Pb-206 / 206 g/mol = 1.79 x 10^-6 mol Pb-206
0.95 mg U-238 / 238 g/mol = 3.99 x 10^-6 mol U-238

Now, we can use the ratio of U-238 to the total number of atoms in the rock to find the number of U-238 atoms:
3.99 x 10^-6 mol U-238 / (1 mol/6.022 x 10^23 atoms) = 2.4 x 10^12 atoms

However, we need to take the half-life of U-238 into account. We can use the formula for radioactive decay:
N = N0 (1/2)^(t/T)
where N is the final number of atoms, N0 is the initial number of atoms, t is the time elapsed, and T is the half-life.

We know that t is billions of years, but we don't need to use the exact number. Instead, we can use the fact that after one half-life, the number of U-238 atoms will be halved. We can keep halving the number until we get to the initial number of atoms (N0) that we're looking for:

N = 2.4 x 10^12 atoms / 2 = 1.2 x 10^12 atoms (after 1 half-life)
N = 1.2 x 10^12 atoms / 2 = 6.0 x 10^11 atoms (after 2 half-lives)
N = 6.0 x 10^11 atoms / 2 = 3.0 x 10^11 atoms (after 3 half-lives)
N = 3.0 x 10^11 atoms / 2 = 1.5 x 10^11 atoms (after 4 half-lives)
N = 1.5 x 10^11 atoms / 2 = 7.5 x 10^10 atoms (after 5 half-lives)
N = 7.5 x 10^10 atoms / 2 = 3.75 x 10^10 atoms (after 6 half-lives)

We can see that after 6 half-lives (6 x 4.51 billion years = 27.06 billion years), we get to about 3.75 x 10^10 atoms, which is close to the initial number of atoms we calculated earlier. Therefore, the best choice is D, 3.3 x 10^18 atoms.

A Rock Contains 0