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Chapter 1: Problem 16

After 44 minutes, a sample of \(_{19}^{44} \mathrm{K}\) is found to have decayed to 25 percent of the original amount present. What is the half-life of \(_{19}^{44} \mathrm{K}?\) (A) 11 minutes (B) 22 minutes (C) 44 minutes (D) 66 minutes

Short Answer

Expert verified
The half-life of \(_{19}^{44} \mathrm{K}\) is 22 minutes, hence, the correct answer is (B) 22 minutes.

Step by step solution

01

Identify the Knowns

We know that after 44 minutes, the sample has decayed to 25 percent of the original amount present. Since a half-life is the amount of time it takes for half the quantity of a substance to decay, when a sample has decayed to 25 percent, it means two half-lives have passed.
02

Calculate the Half-life

Since 44 minutes represent two half-lives, we can find the duration of one half-life by dividing 44 by 2. So, the half-life of \(_{19}^{44} \mathrm{K}\) is \( \frac{44}{2} \) minutes.
03

Simplification

After simplifying the above calculation, we get 22 minutes. This is the half-life of \(_{19}^{44} \mathrm{K}\)

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Most popular questions from this chapter

For a reaction involving nitrogen monoxide inside a sealed flask, the value for the reaction quotient \((Q)\) was found to be \(1.1 \times 10^{2}\) at a given point. If, after this point, the amount of NO gas in the flask increased, which reaction is most likely taking place in the flask? (A) \(\operatorname{NOBr}(g) \rightarrow \operatorname{NO}(g)+1 / \operatorname{Br}_{2}(g) \quad K_{\mathrm{C}}=3.4 \times 10^{-2}\) (B) \(2 \mathrm{NOCl}(g) \mapsto 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \quad K_{\mathrm{c}}=1.6 \times 10^{-5}\) (C) \(2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \quad K_{\mathrm{c}}=4.0 \times 10^{6}\) (D) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{NO}(g) \quad K_{\mathrm{c}}=4.2 \times 10^{2}\)

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