The half-life of Cr-51 is 30.0 days. What fraction of a sample of this isotope will remain after 91.8 days?

A) (1/2)^(16.2 / 30.0) = 0.688 b) (1/2)^( z / 20.4) = (0.810 / 2.20) Solve for z algebraically: Take the log of both sides: ( z / 20.4) log (1/2) = log (0.810 / 2.20) Divide by log (1/2): ( z / 20.4) = log (0.810 / 2.20) / log (1/2) Multiply by 20.4: z = 20.4 ( log (0.810 / 2.20) / log (1/2) ) = 29.4 min d) (17.8 g Ti) / (47.8671 g Ti/mol) x (1 mol TiCl4 / 1 mol Ti) x (189.6799 g TiCl4/mol) 70.5 g TiCl4.

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