25.-/4 14.5.014 Complete the table for the radioactive isotope.(Round to 2 decim

Question

25.-/4 14.5.014 Complete the table for the radioactive isotope.(Round to 2 decimal places.) Isotope Half-Life (Years) Initial Quantity Amount After 1000 Years9 14C 5715 10 9 Need Help? RTatk to a Tuter 26.-/4 pointstarCAABLS1 4.5.028 The number N of bacteria in a culture is given by the model N- 200e*, where t is the time (in hours), with t-O corresponding to the time when N- 200. When t10, there are 270 bacteria. (a) How long does it take the bacteria population to double in size? (Round your answer to two decimal places.) (b) How long does it take the bacteria population to triple in size? (Round your answer to two decimal places.) Use the acidity model given by pH-log:o(H 1, where acidity (pH) is a measure of the given hydrogen ion concentration [H (measured in moles of hydrogen per liter) of a solution The pH of a solution is decreased by two units. The hydrogen ion concentration is increased by what factor?

Explanation / Answer

25.Thye formula for radioactice decay is N(t) = N0 e-kt whre N0 is the initial quantity, N(t) is the quantity left after t units of time and k is the constant of decay. The half-life of the isotope 14C is 5715 years so that N0/2 = N0 e-5715k or, e-5715k = ½. On taking natural log of both the sides, we get -5715k = ln 0.5 = -0.69314718 so that k = 0.69314718/5715 = 0.0001212855959. Now, when t = 10000, we have N(10000) = 10e-0.0001212855959*10000 = 10 e-1.212855959 = 10*0.297346855 = 2.97 g (on rounding off to 2 decimal places).

26. We have N = 200ekt, where k is the constant of growth. When t = 10, N = 270 so that 270 = 200e10k or, e10k =270/200 = 1.35. On taking natural log of both the sides, we get 10k = ln 1.35 = 0.300104592 so that k = 0.030010459.

(a). When N = 400, we have 400 = 200 e0.030010459t or, = e0.030010459t = 2. On taking natural log of both the sides, we get 0.030010459t = ln 2 = 0.69314718 so that t = 0.69314718/0.030010459 = 23.10 hours (on rounding off to 2 decimal places).

(b). When N = 600, we have 600 = 200 e0.030010459t or, = e0.030010459t = 3. On taking natural log of both the sides, we get 0.030010459t = ln 3 = 1.098612289 so that t =1.098612289/0.030010459 = 36.61 hours (on rounding off to 2 decimal places).

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