The following rate data were obtained for different concentration of the reactan

Question

The following rate data were obtained for different concentration of the reactants, where P is phenolthalein: Determine the order in phenolphalein and in hydroxide, and determine the rate constant, k. Bananas are radioactive. A banana has typically 0.42 g of potassium. Of this potassium. 0.0117% of it is radioactive^40 K. Its half-life Is 1.248 times 10^9 years. (a) How many grams of^46 K are in a banana? (b) Assuming the atomic mass of this isotope to be 40.0 g/mol, how many moles of this isotope are there? (c) How many atoms of the isotope are there? (d) What is the decay constant of potassium, in s^-1? (e) Finally, knowing that radioactive decays are first order, how many atoms per second decay in the banana?

Explanation / Answer

Q2.

m = 0.42 g of potassium

mass of Kradioactiv = 0.0117/100 * 0.42 = 0.00004914 g of radioactive material

b)

MW = 40 g/mol

mol = mass/MW = 0.00004914/40 = 0.0000012285 mol of K40

c)

atoms -->

1 mol = 6.022*10^23

0.0000012285 mol = 0.0000012285*6.022*10^23 = 7.3980*10^17 atoms of K40

d)

decay constant -->

t 1/2 = ln(2) / k

1.248*10^9 = ln(2)/k

k = ln(2) / (1.248*10^9) = 5.554*10^-10 1/year

5.554*10^-10 1/year * 1 year / 365 days * 1 days / 24 h * 1 h / 3600 s = 1.7611*10^-17 1/s

e)

find aotms per second decay:

7.3980*10^17 --> 1.7611*10^-17 1/s

Rate = (7.3980*10^17) * (1.7611*10^-17 ) = 13.20 atoms per second

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