The curie (Ci) is a commonly used unit for measuring nuclear radioactivity: 1 curie of radiation is equal to \(3.7 \times 10^{10}\) decay events per second (the number of decay events from 1 g radium in 1 s). a. What mass of \(\mathrm{Na}_{2}^{38} \mathrm{SO}_{4}\) has an activity of 10.0 \(\mathrm{mCi}^{2}\) Sulfur- 38 has an atomic mass of 38.0 \(\mathrm{u}\) and a half-life of 2.87 \(\mathrm{h} .\) b. How long does it take for 99.99\(\%\) of a sample of sulfur- 38 to decay?

First, let's convert 10.0 mCi to Ci by dividing by 1000: \(10.0\,\mathrm{mCi} = 10.0 \times 10^{-3}\,\mathrm{Ci}\) Now, we will calculate the number of decay events per second for this activity: \(N = (10.0 \times 10^{-3}\,\mathrm{Ci}) \times (3.7 \times 10^{10}\,\mathrm{decay\, events/s}\) Total decay events = \(10.0 \times 10^{-3}\) * \(3.7 \times 10^{10}\) = \(3.7 \times 10^7\) Now, we will use the fact that 1 mole of Sulfur-38 will have an atomic mass of 38.0 g, and with Avogadro's number (\(6.02 \times 10^{23}\, \mathrm{atoms}\)), we can find the number of decay events per second for 1 mole of Sulfur-38, \(N_{1\,\mathrm{mole}} = \frac{3.7 \times 10^7\,\mathrm{decay\,events/s}}{(6.02 \times 10^{23}\,\mathrm{atoms})}\) Now, we need to find the number of moles of \(\mathrm{S}^{38}\) in the compound with 10.0 mCi activity. \(n = \frac{3.7 \times 10^7}{N_{1\,\mathrm{mole}}}\) Consider the molar mass of \(\mathrm{Na}_{2}^{38}\mathrm{SO}_{4}\) as, Molar mass = 2 x (mass of Na) + mass of Sulfur-38 + 4 x (mass of Oxygen) g/mol Finally, the mass of \(\mathrm{Na}_{2}^{38}\mathrm{SO}_{4}\) with 10.0 mCi activity is given by: Mass = n * Molar mass

We are given the half-life of Sulfur-38, so we will use that information to find the time it takes for 99.99% of a sample to decay. The decay formula is given by: \(N = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{t_{\frac{1}{2}}}}\) \(N\) = final amount \(N_0\) = initial amount \(t_{\frac{1}{2}}\) = half-life \(t\) = time Here, we want 99.99% decay, meaning only 0.01% of the initial amount remains after time t. So, \(\frac{N}{N_0} = 0.0001\), as we are given \(t_{\frac{1}{2}} = 2.87\,\mathrm{h}\) We can find the time t. \(0.0001 = \left(\frac{1}{2}\right)^{\frac{t}{2.87}}\) Now solve for t.