# SOLUTION: If 23.00 grams of a 45.00 gram sample of unknown isotope remain after 17 days 12 hours, what is the half-life of the unknown isotope to the nearest minute?

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 Question 1130: If 23.00 grams of a 45.00 gram sample of unknown isotope remain after 17 days 12 hours, what is the half-life of the unknown isotope to the nearest minute?Answer by AnlytcPhil(1278)   (Show Source): You can put this solution on YOUR website!```If 23.00 grams of a 45.00 gram sample of unknown isotope remain after 17 days 12 hours, what is the half-life of the unknown isotope to the nearest minute? The formula is A = Pert When time t=0 days, the amount A = 45g. Substitute these two values 45 = Per(0) 45 = Pe0 45 = P(1) P = 45 grams so substitute 45 for P in A = Pert A = 45ert When time t=17.5 days, the amount A = 23g. Substitute these two values: 23 = 45er(17.5) 23 = 45e17.5r Divide both sides by 45 23/45 = e17.5r .5111111111 = e17.5r Rewrite this equation using the fact that " N = eM can be rewritten as M = ln(N) " 17.5r = ln(.5111111111) Use a calculator to find ln(.5111111111) = -.6711682738 17.5r = -.6711682738 Solve for r r = -.6711682738/17.5 r = -.0383524728 So substitute this value of r in A = 45ert A = 45e-.0383524728t We wish to know how many minutes is required for it to become half its original amount, or 1/2 of 45g, or 22.5 g. So substitute this for A in A = 45e-.0394804867t 22.5 = 45e-.0383524728t Divide both sides by 45 22.5/45 = e-.0383524728t .5 = e-.0383524728t Rewrite this as ln(.5) = -.0383524728t Solve for t -.6931471806 = -.0383524728t -.0383524728t = -.6931471806 t = -.6931471806/(-.0383524728) t = 18.07307665 days To find hours, multiply .07307665 by 24 18 days 1.753839658 hours To find minutes, multiply .753839658 by 60 18 days 1 hour 45.23037947 minutes or to nearest minute: 18 days 1 hour, 45 minutes. Edwin ```