The table shows the amount of radioactive element remaining in a sample over a period of time. radioactive decay rateamount of radioactive sample(grams) 56.0 47.1 39.6 33.3 28 time(years)04008001,2001,600each correspond w eachotherpart 1: what is the half-life of the element? explain how you determined this. part 2: how long would it take 312 g of the sample to decay to 9.75 grams? show your work or explain your answer

8,000 years.

Explanation:

It is known that the decay of a radioactive isotope isotope obeys first order kinetics.Half-life time is the time needed for the reactants to be in its half concentration.If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

Part 1: What is the half-life of the element? Explain how you determined this.

The half-life of the element is 1,600 years.

Half-life time is the time needed for the reactants to be in its half concentration.

The sample stats with 56.0 g and reaches its half concentration (28.0 g) after 1,600 years.

So, the half-life of the sample is 1,600 years.

Part 2: How long would it take 312 g of the sample to decay to 9.75 grams? Show your work or explain your answer.

For, first order reactions:

k = ln(2)/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(1,600 years) = 4.33 x 10⁻⁴ year⁻¹.

Also, we have the integral law of first order reaction:

kt = ln([A₀]/[A]),

where, k is the rate constant of the reaction (k = 4.33 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of the sample ([A₀] = 312.0 g).

[A] is the remaining concentration of the sample ([A] = 9.75 g).

∴ t = (1/k) ln([A₀]/[A]) = (1/4.33 x 10⁻⁴ year⁻¹) ln(312.0 g/9.75 g) = 8,000 years.

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