A sample of 238U has an activity of 4.300×10-3 Ci. What is the mass of the sample? The half life is 1.41×1017 s, and the atomic mass is 238.0507884 u

To solve this problem, we need to use several steps involving different physics and chemistry concepts. Here's a step-by-step solution:

Step 1: Convert the activity from Curie (Ci) to decays per second (Bq). 1 Ci = 3.7 x 10^10 Bq. So, the activity is 4.300 x 10^-3 Ci x 3.7 x 10^10 Bq/Ci = 1.591 x 10^8 Bq.

Step 2: Use the definition of activity (A) which is the decay rate, or the number of atoms that decay per unit time. A = λN, where λ is the decay constant and N is the number of atoms. We can find λ using the half-life (t1/2) with the formula λ = ln(2) / t1/2. Substituting the given half-life, λ = 0.693 / 1.41 x 10^17 s = 4.91 x 10^-19 s^-1.

Step 3: Now we can find the number of atoms (N) using the formula A = λN. Rearranging for N gives N = A / λ = 1.591 x 10^8 Bq / 4.91 x 10^-19 s^-1 = 3.24 x 10^26 atoms.

Step 4: Convert the number of atoms to moles using Avogadro's number (6.022 x 10^23 atoms/mole). So, the number of moles is 3.24 x 10^26 atoms / 6.022 x 10^23 atoms/mole = 538 moles.

Step 5: Finally, convert moles to mass using the atomic mass. The mass is 538 moles x 238.0507884 g/mole = 1.28 x 10^5 g or 128 kg.