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An experiment was designed to determine whether an aquatic plant absorbed iodide ion from water. Iodine-131 \(\left(t_{1 / 2}=8.02\right.\) days) was added as a tracer, in the form of iodide ion, to a tank containing the plants. The initial activity of a \(1.00-\mu \mathrm{L}\) sample of the water was 214 counts per minutc. After 30 days the level of activity in a \(1.00-\mu \mathrm{I}\). sample was \(15.7\) counts per minute. Did the plants absorb iodide from the water?

Short Answer

Expert verified
The number of half-lives that elapsed in 30 days can be calculated as: \( \text{Number of half-lives} = \frac{30}{8.02} \approx 3.74 \) Assuming no plant absorption, the remaining activity after 30 days is: \( \text{Remaining activity} = 214 \times (\frac{1}{2})^{3.74} \approx 22.8 \) The expected activity of 22.8 counts per minute without plant absorption is higher than the observed activity of 15.7 counts per minute. Therefore, we can conclude that the plants absorbed iodide from the water.

Step by step solution

01

Determine the number of half-lives that elapsed in 30 days

To find the number of half-lives that transpired within the given time frame, we can use the following formula: Number of half-lives = Time elapsed (days) ÷ Half-life (days/half-life) Let's calculate this value: \( \text{Number of half-lives} = \frac{30}{8.02} \)
02

Calculate the expected activity after 30 days with no plant absorption

Assuming no plant absorption, the iodium-131 would just decay at its normal rate. In this situation, the remaining activity can be calculated by the following formula: Remaining activity (counts per minute) = Initial activity (counts per minute) × \((\frac{1}{2})\)^{number\:of\:half-lives} Solving for the remaining activity after 30 days: \( \text{Remaining activity} = 214 \times (\frac{1}{2})^{\frac{30}{8.02}} \)
03

Compare the expected and observed activity after 30 days

Evaluate the expected activity obtained in Step 2 and compare it with the observed activity of 15.7 counts per minute. If the expected activity is higher than the observed activity, this suggests that the plants absorbed iodide from the water.
04

Determine whether the plants absorbed iodide

Based on the comparison between the expected and observed activity of iodine-131 in Step 3, make a conclusion about the absorption of iodide by the aquatic plants.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Half-Life Calculation
Half-life is a concept used to describe the time it takes for half of a given amount of a radioactive substance to decay. To determine how much of a substance remains after a certain period, you can use the number of half-lives passed. Typically, you calculate this with the formula: \[\text{Number of half-lives} = \frac{\text{Time elapsed (days)}}{\text{Half-life (days/half-life)}}\] For Iodine-131, with a half-life of 8.02 days, you can calculate the number of half-lives over 30 days with the formula: \[\frac{30}{8.02} \] This gives a clear picture of the decay progression. It helps you predict the remaining activity, crucial to understanding whether additional processes, like plant absorption, affect the system.
Exploring Radioactive Decay
Radioactive decay is a random process in which unstable nuclei release energy by emitting radiation. This natural process follows an exponential decay pattern. When you add a radioactive isotope, like iodine-131, into an environment, it decays over time in a predictable manner. The decay can be calculated using: \[\text{Remaining activity (counts per minute)} = \text{Initial activity (counts per minute)} \times \left(\frac{1}{2}\right)^{\text{number of half-lives}}\] This formula reflects how the initial quantity decreases exponentially. Inferring from changes between expected and observed decay levels, like in our experiment, can help indicate other influences, such as plant absorption, altering expected outcomes.
Designing a Chemistry Experiment
The design of an experiment in chemistry is crucial for obtaining accurate and reliable results. When investigating the absorption of iodide by aquatic plants, careful planning is essential. Here are the key elements for a successful experimental design: - **Objective:** Clearly define what you're testing—here, whether plants absorb iodide. - **Controlled Variables:** Keep the conditions consistent to ensure that any observed changes are due to the independent variable. - **Use of Tracers:** Radioactive tracers like iodine-131 provide a quantifiable way to track substance movement and transformations. - **Data Collection:** Use precise measurements, like counts per minute, to quantify results over time. - **Comparison:** Compare expected decay with actual observations to determine if absorption occurred. This methodical approach allows for determining the effects of aquatic plants on iodide levels, ensuring conclusions are valid and based on solid data.

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Most popular questions from this chapter

The Sun radiates energy into space at the rate of \(3.9 \times 10^{26} \mathrm{~J} / \mathrm{s}\). (a) Calculate the rate of mass loss from the Sun in \(\mathrm{kg} / \mathrm{s}\). (b) How does this mass loss arise? (c) It is estimated that the Sun contains \(9 \times 10^{56}\) free protons. How many protons per second are consumed in nuclear reactions in the Sun?

How much energy must be supplied to break a single \({ }^{21} \mathrm{Ne}\) nucleus into separated protons and neutrons if the nucleus has a mass of \(20.98846\) amu? What is the nuclear binding energy for \(1 \mathrm{~mol}\) of \({ }^{21} \mathrm{Ne}\) ?

Radon-222 decays to a stable nucleus by a series of three alpha emissions and two beta emissions. What is the stable nucleus that is formed?

A rock contains \(0.257 \mathrm{mg}\) of lead-206 for every milligram of uranium-238. The half-life for the decay of uranium-238 to lead-206 is \(4.5 \times 10^{5} \mathrm{yr}\). How old is the rock? SOLUTION Analyze We are told that a rock sample has a certain amount of lead206 for every unit mass of uranium-238 and asked to estimate the age of the rock. Plan Lead-206 is the product of the radioactive decay of uranium-238. We will assume that the only source of lead-206 in the rock is from the decay of uranium-238, with a known half-life. To apply firstorder kinetics expressions (Equations \(21.19\) and 21.20) to calculate the time elapsed since the rock was formed, we first need to calculate how much initial uranium-238 there was for every \(1 \mathrm{mg}\) that remains today. Solve Let's assume that the rock currently contains \(1.000 \mathrm{mg}\) of uranium-238 and therefore \(0.257 \mathrm{mg}\) of lead-206. The amount of uranium-238 in the rock when it was first formed therefore equals \(1.000 \mathrm{mg}\) plus the quantity that has decayed to lead-206. Because the mass of lead atoms is not the same as the mass of uranium atoms, we cannot just add \(1.000 \mathrm{mg}\) and \(0.257 \mathrm{mg}\). We have to multiply the present mass of lead-206 \((0.257 \mathrm{mg})\) by the ratio of the mass number of uranium to that of lead, into which it has decayed. Therefore, the original mass of \({ }_{92}^{239} \mathrm{U}\) was $$ \text { Original } \begin{aligned} { }_{98}^{238} \mathrm{U} &=1.000 \mathrm{mg}+\frac{238}{206}(0.257 \mathrm{mg}) \\ &=1.297 \mathrm{mg} \end{aligned} $$ Using Equation 21.20, we can calculate the decay constant for the process from its half-life: $$ k=\frac{0.693}{4.5 \times 10^{9} \mathrm{yr}}=1.5 \times 10^{-10} \mathrm{yr}^{-1} $$ Rearranging Equation \(21.19\) to solve for time, \(t\), and substituting known quantities gives $$ t=-\frac{1}{k} \ln \frac{N_{t}}{N_{0}}=-\frac{1}{1.5 \times 10^{-10} \mathrm{yr}^{-1}} \ln \frac{1.000}{1.297}=1.7 \times 10^{9} \mathrm{yr} $$ Comment To check this result, you could use the fact that the decay of uranium-235 to lead-207 has a half-life of \(7 \times 10^{8} \mathrm{yr}\) and measure the relative amounts of uranium-235 and lead-207 in the rock.

Cobalt-60 is a strong gamma emitter that has a half-life of \(5.26 \mathrm{yr}\). The cobalt- 60 in a radiotherapy unit must be replaced when its radioactivity falls to \(75 \%\) of the original sample. If an original sample was purchased in June 2013, when will it be necessary to replace the cobalt-60?

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