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The bromine- 82 nucleus has a half-life of \(1.0 \times 10^{3}\) min. If you wanted \(1.0 \mathrm{g}^{82} \mathrm{Br}\) and the delivery time was 3.0 days, what mass of NaBr should you order (assuming all of the Br in the NaBr was \(\left.^{82} B r\right) ?\)

Short Answer

Expert verified
You should order 3.13g of NaBr to have 1.0g of 82Br after a delivery time of 3 days.

Step by step solution

01

Calculate the number of half-lives during the delivery time

First, we need to find how many half-lives have passed during the delivery time. The delivery time is given as 3 days, and we need to convert it to minutes. There are 24 hours in a day and 60 minutes in an hour, so 3 days is equal to \(3 \times 24 \times 60 = 4320\) minutes. The half-life of 82Br is given as \(1.0 \times 10^3\) minutes. To find the number of half-lives during the delivery time, we can divide the time by the half-life: Number of half-lives = \(\frac{4320}{1.0 \times 10^3} = 4.32\).
02

Calculate the remaining mass of 82Br after delivery time

Next, we need to find out how much 82Br will remain after these 4.32 half-lives have passed. Since the mass of 82Br decreases by half with each half-life, the remaining mass of 82Br can be calculated using the formula: Remaining mass = Initial mass × \((0.5)^{\text{number of half-lives}}\) We want 1.0g of 82Br at the end of the delivery time, so we can set the remaining mass as 1.0g and solve for the initial mass. 1.0g = Initial mass × \((0.5)^{4.32}\) Now, solving for the initial mass of 82Br: Initial mass = \(\frac{1.0}{(0.5)^{4.32}} = 2.44\, \text{g}\)
03

Calculate the mass of NaBr needed

The final step is to find the mass of NaBr required to have 2.44g of 82Br. The balanced chemical equation for the synthesis of NaBr from 82Br is given below: \({}^{82}Br + Na \rightarrow NaBr\) From the equation, it is clear that 1 mole of NaBr is produced from 1 mole of 82Br. To determine the mass of NaBr required, we need to find the moles of 82Br and then use the molar mass of NaBr to find the mass. First, find the moles of 82Br: Moles of 82Br = \(\frac{\text{mass of 82Br}}{\text{molar mass of 82Br}} = \frac{2.44\, \text {g}}{82\, \text{g/mol}} = 0.0298\, \text{mol}\) Now, using the moles of 82Br and the molar mass of NaBr (22.99 g/mol for Na and 82 g/mol for Br), we can find the mass of NaBr: Mass of NaBr = Moles of 82Br × (molar mass of Na + molar mass of 82Br) = \(0.0298\, \text{mol} \times (22.99 + 82)\, \text{g/mol} = 3.13\, \text{g}\) So, you should order 3.13g of NaBr to have 1.0g of 82Br after a delivery time of 3 days.

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Key Concepts

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

Half-Life Calculation
Understanding the half-life of a radioactive isotope is crucial for applications in medicine, archaeology, and nuclear physics. The half-life of a substance is the time it takes for half of the radioactive atoms to undergo decay. We calculate the remaining mass of a radioactive isotope after a given period using the formula:
Remaining mass = Initial mass \times (0.5)^{\text{number of half-lives}}
In the given problem, the half-life of bromine-82 (\(^{82}Br\)) is 1,000 minutes. If we desire 1.0 g of \(^{82}Br\) after 3 days, we must first convert 3 days to minutes (3 days = 4320 minutes) and then determine how many half-lives have passed in that period by dividing the total time by the half-life duration. With the number of half-lives, we can use the formula above to find out the initial mass of \(^{82}Br\) needed.
Radioactive Decay
Radioactive decay is a natural process by which an unstable atomic nucleus loses energy by emitting radiation. This decay process reduces the number of radioactive isotopes over time and is a random but statistically predictable process. In the classroom or lab, we use the concept of half-life to make predictions about the behavior of a radioisotope over time. During each half-life, exactly half of the remaining radioactive atoms will decay. In our exercise, after 4.32 half-lives, the mass of \(^{82}Br\) will have decreased significantly, which is why we must start with a larger amount to ensure we have the desired mass at the end of the delivery period.
Molar Mass
The molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is a physical property defined as the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. In our problem, we use the molar masses of \(^{82}Br\) and sodium (Na) to determine the mass of sodium bromide (NaBr) that we need to order.
To find the required mass of NaBr, we first calculate the moles of \(^{82}Br\) that would correspond to the initial mass of 2.44 g. After finding the number of moles, we use the combined molar mass of sodium and bromine in NaBr to convert moles back to grams. This final mass represents how much NaBr should originally be ordered to ensure that 1.0 g of \(^{82}Br\) is present after three days, accounting for the decay that will occur during that time.

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Most popular questions from this chapter

In each of the following radioactive decay processes, supply the missing particle. a. \(^{73} \mathrm{Ga} \rightarrow^{73} \mathrm{Ge}+?\) b. \(^{192} \mathrm{Pt} \rightarrow^{188} \mathrm{Os}+?\) c. \(^{205} \mathrm{Bi} \rightarrow^{205} \mathrm{Pb}+?\) d. \(^{241} \mathrm{Cm}+? \rightarrow^{241} \mathrm{Am}\)

In 1994 it was proposed (and eventually accepted) that element 106 be named seaborgium, \(\mathrm{Sg}\), in honor of Glenn T. Seaborg, discoverer of the transuranium elements.a. \(^{263} \mathrm{Sg}\) was produced by the bombardment of \(^{249} \mathrm{Cf}\) with a beam of \(^{18} \mathrm{O}\) nuclei. Complete and balance an equation for this reaction. b. \(^{263}\) Sg decays by \(\alpha\) emission. What is the other product resulting from the \(\alpha\) decay of \(^{263} \mathrm{Sg} ?\)

Write an equation describing the radioactive decay of each of the following nuclides. (The particle produced is shown in parentheses, except for electron capture, where an electron is a reactant.) a. \(^{68}\) Ga (electron capture) b. \(^{62} \mathrm{Cu}\) (positron) c. \(^{212} \operatorname{Fr}(\alpha)\) d. \(^{129} \operatorname{Sb}(\beta)\)

Why are the observed energy changes for nuclear processes so much larger than the energy changes for chemical and physical processes?

Write balanced equations for each of the processes described below. a. Chromium- \(51,\) which targets the spleen and is used as a tracer in studies of red blood cells, decays by electron capture. b. Iodine-131, used to treat hyperactive thyroid glands, decays by producing a \(\beta\) particle. c. Phosphorus- \(32,\) which accumulates in the liver, decays by \(\beta\) -particle production.

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