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A sample of sodium-24 chloride contains \(0.050 \mathrm{mg}\) of \(\mathrm{Na}-24\) to study the sodium balance of an animal. After \(24.9 \mathrm{~h}, 0.016 \mathrm{mg}\) of \(\mathrm{Na}-24\) is left. What is the half- life of \(\mathrm{Na}-24 ?\)

Short Answer

Expert verified
Answer: The half-life of sodium-24 is approximately 14.97 hours.

Step by step solution

01

Write down the radioactive decay formula.

The radioactive decay formula is given by: \(N(t) = N_0 \times (\frac{1}{2})^{\frac{t}{T}}\) Where: \(N(t)\) is the remaining amount of the substance after a given time \(t\), \(N_0\) is the initial amount of the substance, \(T\) is the half-life, and \(t\) is the time elapsed.
02

Plug in values into the radioactive decay formula.

We are given \(N_0 = 0.050mg\), \(N(t) = 0.016mg\), and \(t = 24.9h\). We will plug these values into the radioactive decay formula: \(0.016 = 0.050 \times (\frac{1}{2})^{\frac{24.9}{T}}\)
03

Solve for the half-life (T).

First, divide both sides by \(0.050\): \((\frac{1}{2})^{\frac{24.9}{T}} = \frac{0.016}{0.050}\) Then, take the logarithm of both sides (base 0.5) to isolate the exponent: \(\log_{\frac{1}{2}} (\frac{1}{2})^{\frac{24.9}{T}} = \log_{\frac{1}{2}} (\frac{0.016}{0.050})\) Now, we can simplify by applying the logarithm rules: \(\frac{24.9}{T} = \log_{\frac{1}{2}} (\frac{0.016}{0.050})\) Next, solve for \(T\): \(T=\frac{24.9}{\log_{\frac{1}{2}} (\frac{0.016}{0.050})}\)
04

Calculate the half-life (T).

Now we will calculate the half-life of sodium-24 using the expression found in step 3: \(T=\frac{24.9}{\log_{\frac{1}{2}} (\frac{0.016}{0.050})} \approx 14.97h\) The half-life of sodium-24 is approximately \(14.97\) hours.

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Key Concepts

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

half-life
Half-life is a fundamental concept in radioactive decay describing how long it takes for half of a radioactive substance to decay into another form. This period is constant for any given substance and doesn't depend on the initial amount. It allows scientists and researchers to predict the decay pattern of a substance over time.

In practical terms, knowing the half-life helps in understanding how long a radioactive isotope will remain active or potentially hazardous. For instance, when treating or diagnosing a disease using a radioactive tracer like sodium-24 (Na-24), understanding the half-life ensures effective and safe application in medical practices.
  • Half-life is symbolized by "T" in equations.
  • It provides a clear timeline to predict when a material will reduce to a certain amount.
  • This concept is vital for fields like archaeology, medicine, and nuclear physics.
The half-life equation is simple but powerful in application, making it an essential part of radioactive decay analysis.
sodium-24
Sodium-24 is a radioactive isotope of sodium used in a variety of scientific and medical applications. Its primary application lies in the study of sodium management in biological systems, like in our given exercise dealing with an animal's sodium balance.

Sodium-24 emits beta particles and gamma rays, which are detectable using specific equipment. This property makes it an effective tracer in diagnostic tests and in understanding physiological processes. In labs and medical studies, small, controlled amounts of Na-24 are used, given its radioactive nature.
  • Sodium-24 has a relatively short half-life, approximately 15 hours, meaning it decays swiftly compared to other isotopes.
  • It safely exits biological systems before its radioactivity could pose any health hazard.
  • Its decay helps researchers track the distribution of sodium in the body.
Understanding the role and behavior of sodium-24 in tests is crucial, particularly in medicine, due to its efficiency in providing results before decaying entirely.
radioactive decay formula
The radioactive decay formula is an essential tool for calculating how much of a radioactive substance remains after a given period. It is represented mathematically as:\[ N(t) = N_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T}} \]This formula helps in predicting the remaining quantity of a substance, such as sodium-24, after a specific time has elapsed.

The components of the formula include:
  • \( N(t) \): the amount of the substance left after time \( t \)
  • \( N_0 \): the initial quantity of the substance
  • \( T \): the half-life of the substance
  • \( t \): the time over which decay is observed
To use this formula effectively, substitute the known values for the variables, as demonstrated in solving for the half-life of sodium-24. By rearranging and solving the equation, scientists can determine unknowns like the half-life \( T \) through logarithmic calculations.

In our example, using this formula allowed the determination that sodium-24 has a half-life of approximately 14.97 hours, illustrating the formula's utility in real-world applications.

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Most popular questions from this chapter

Experimental data are listed for the hypothetical reaction $$\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D}$$ $$\begin{array}{lcccccc}\hline \text { Time (s) } & 0 & 10 & 20 & 30 & 40 & 50 \\\\{[\mathrm{~A}]} & 0.32 & 0.24 & 0.20 & 0.16 & 0.14 & 0.12 \\ \hline\end{array}$$ (a) Plot these data as in Figure \(11.2\). (b) Draw a tangent to the curve to find the instantaneous rate at \(30 \mathrm{~s}\). (c) Find the average rate over the 10 to \(40 \mathrm{~s}\) interval. (d) Compare the instantaneous rate at \(30 \mathrm{~s}\) with the average rate over the thirty-second interval.

Diethylhydrazine reacts with iodine according to the following equation: $$\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}(\mathrm{NH})_{2}(l)+\mathrm{I}_{2}(a q) \longrightarrow\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{~N}_{2}(l)+2 \mathrm{HI}(a q)$$ The rate of the reaction is followed by monitoring the disappearance of the purple color due to iodine. The following data are obtained at a certain temperature. $$ \begin{array}{cccc} \hline \text { Expt. } & {\left[\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}(\mathrm{NH})_{2}\right]} & {\left[\mathrm{I}_{2}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{h}) \\ \hline 1 & 0.150 & 0.250 & 1.08 \times 10^{-4} \\ 2 & 0.150 & 0.3620 & 1.56 \times 10^{-4} \\ 3 & 0.200 & 0.400 & 2.30 \times 10^{-4} \\ 4 & 0.300 & 0.400 & 3.44 \times 10^{-4} \\ \hline\end{array}$$ (a) What is the order of the reaction with respect to diethylhydrazine, iodine, and overall? (b) Write the rate expression for the reaction. (c) Calculate \(k\) for the reaction. (d) What must \(\left[\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2}(\mathrm{NH})_{2}\right]\) be so that the rate of the reaction is \(5.00 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{h}\) when \(\left[\mathrm{I}_{2}\right]=0.500 ?\)

Hydrogen bromide is a highly reactive and corrosive gas used mainly as a catalyst for organic reactions. It is produced by reacting hydrogen and bromine gases together. $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{HBr}(g)$$ The rate is followed by measuring the intensity of the orange color of the bromine gas. The following data are obtained: $$\begin{array}{cccc}\hline \text { Expt. } & {\left[\mathrm{H}_{2}\right]} & {\left[\mathrm{Br}_{2}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \\ \hline 1 & 0.100 & 0.100 & 4.74 \times 10^{-3} \\ 2 & 0.100 & 0.200 & 6.71 \times 10^{-3} \\ 3 & 0.250 & 0.200 & 1.68 \times 10^{-2} \\ \hline\end{array}$$(a) What is the order of the reaction with respect to hydrogen, bromine, and overall? (b) Write the rate expression for the reaction. (c) Calculate \(k\) for the reaction. What are the units for \(k ?\) (d) When \(\left[\mathrm{H}_{2}\right]=0.455 \mathrm{M}\) and \(\left[\mathrm{Br}_{2}\right]=0.215 M\), what is the rate of the reaction?

The decomposition of sulfuryl chloride, \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), to sulfur dioxide and chlorine gases is a first-order reaction. $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ At a certain temperature, the half-life of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(7.5 \times 10^{2} \mathrm{~min}\). Consider a sealed flask with \(122.0 \mathrm{~g}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) (a) How long will it take to reduce the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in the sealed flask to \(45.0 \mathrm{~g}\) ? (b) If the decomposition is stopped after \(29.0 \mathrm{~h}\), what volume of \(\mathrm{Cl}_{2}\) at \(27^{\circ} \mathrm{C}\) and \(1.00\) atm is produced?

The following reaction is second-order in A and first-order in B. $$\mathrm{A}+\mathrm{B} \longrightarrow \text { products }$$ (a) Write the rate expression. (b) Consider the following one-liter vessel in which each square represents a mole of \(\mathrm{A}\) and each circle represents a mole of \(\mathrm{B}\). What is the rate of the reaction in terms of \(k ?\) (c) Assuming the same rate and \(k\) as (b), fill the similar one-liter vessel shown in the figure with an appropriate number of circles (representing B).

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