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Alpha particles produced by radioactive decay eventually pick up electrons from their surroundings to form helium atoms. Calculate the volume (in mL) of He collected at STP when \(1.00 \mathrm{~g}\) of pure \({ }^{226} \mathrm{Ra}\) is stored in a closed container for 125 years. (Assume that there are five \(\alpha\) particles generated per \({ }^{226} \mathrm{Ra}\) as it decays to \({ }^{206} \mathrm{~Pb}\). \()\)

Short Answer

Expert verified
495 mL of He is produced.

Step by step solution

01

Determine the number of moles of Ra

Using the atomic mass of \({ }^{226} \text{Ra}\), convert 1.00 gram of \({ }^{226} \text{Ra}\) to moles. The atomic mass of \({ }^{226} \text{Ra}\) is approximately 226 g/mol. Therefore, the number of moles is calculated as follows:\[\text{moles of } { }^{226} \text{Ra} = \frac{1.00 \text{ g}}{226 \text{ g/mol}} = 0.00442 \text{ mol}.\]
02

Calculate the number of moles of He produced

Since every atom of \({ }^{226} \text{Ra}\) produces five alpha particles, and each alpha particle corresponds to one helium nucleus, the number of moles of helium is given by:\[\text{moles of He} = 5 \times 0.00442 = 0.0221 \text{ mol}.\]
03

Use ideal gas law to find the volume at STP

At Standard Temperature and Pressure (STP), the volume occupied by one mole of gas is 22.4 liters. Therefore, the volume of helium gas produced is:\[\text{Volume of He} = 0.0221 \text{ mol} \times 22.4 \text{ L/mol} = 0.495 \text{ L}.\]
04

Convert volume from liters to milliliters

Since the volume in milliliters is requested, convert from liters to milliliters:\[\text{Volume in mL} = 0.495 \text{ L} \times 1000 \text{ mL/L} = 495 \text{ mL}.\]

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Key Concepts

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

Helium formation
On a cosmic scale, helium is one of the simplest and most abundant elements in the universe. However, its formation on Earth through radioactive decay is a fascinating process. When certain heavy elements undergo radioactive decay, they emit alpha particles. An alpha particle is essentially a helium nucleus, consisting of two protons and two neutrons. Over time, these alpha particles pick up electrons and transform into stable helium atoms.

This transformation is significant in nuclear reactions, such as the decay of \(^{226}\mathrm{Ra}\), which emits five alpha particles as it decays into \(^{206}\mathrm{Pb}\). Each alpha particle eventually becomes a helium atom, leading to the formation of helium gas.
Radioactive decay
Radioactive decay is a natural process where unstable atomic nuclei lose energy by emitting radiation. This is a spontaneous event as atoms transition from a higher energy state to a lower one.

There are various types of radioactive decay, including gamma decay, beta decay, and alpha decay. In alpha decay, an atom emits an alpha particle – or essentially a helium nucleus.
  • The decay process alters the parent atom, reducing its atomic number by two and its mass number by four, leading to the eventual formation of a different element.
  • In the case of \(^{226}\mathrm{Ra}\), it decays into \(^{206}\mathrm{Pb}\), a stable lead isotope.
Understanding this process is key in computing the amount of helium produced during decay.
Standard Temperature and Pressure (STP)
Standard Temperature and Pressure (STP) is an essential concept in chemistry used to express the conditions under which gas volumes are usually compared. By convention, STP is set at a pressure of 1 atmosphere (atm) and a temperature of 273.15 Kelvin (0°C).

At these conditions, 1 mole of an ideal gas occupies 22.4 liters of volume. This standardization simplifies calculations and allows chemists to predict how gases will behave under different conditions. For example, when calculating the volume of helium produced from the decay of \(^{226}\mathrm{Ra}\), STP conditions are used to estimate how much space the resulting helium gas will occupy.
Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry that describes the behavior of ideal gases. The formula is expressed as \(PV = nRT\), where \(P\) stands for pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) stands for temperature.

Using the Ideal Gas Law allows scientists to calculate how changes in pressure, temperature, and the amount of gas can impact the volume. At STP, \(R\) is often taken as 0.0821 L*atm/(mol*K). In computations involving the decay of \(^{226}\mathrm{Ra}\), the Ideal Gas Law explains how we can predict the helium volume generated and collected at STP.

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Most popular questions from this chapter

In the thorium decay series, thorium- 232 loses a total of six \(\alpha\) particles and four \(\beta\) particles in a 10 -stage process. What is the final isotope produced?

Nuclei with an even number of protons and an even number of neutrons are more stable than those with an odd number of protons and/or an odd number of neutrons. What is the significance of the even numbers of protons and neutrons in this case?

The constituents of wine contain, among others, carbon, hydrogen, and oxygen atoms. A bottle of wine was sealed about 6 years ago. To confirm its age, which of the isotopes would you choose in a radioactive dating study? The half-lives of the isotopes are: \({ }^{14} \mathrm{C}: 5715\) years; \({ }^{15} \mathrm{O}: 124 \mathrm{~s} ;{ }^{3} \mathrm{H}: 12.5\) years. Assume that the activities of the isotopes were known at the time the bottle was sealed.

An electron and a positron are accelerated to nearly the speed of light before colliding in a particle accelerator. The resulting collision produces an exotic particle having a mass many times that of a proton. Does this result violate the law of conservation of mass? Explain.

Complete the following nuclear equations, and identify \(\mathrm{X}\) in each case: (a) \({ }_{12}^{26} \mathrm{Mg}+{ }_{1}^{1} \mathrm{p} \longrightarrow \alpha+\mathrm{X}\) (b) \({ }_{27}^{59} \mathrm{Co}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{27}^{60} \mathrm{Co}+\mathrm{X}\) (c) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{36}^{94} \mathrm{Kr}+{ }_{56}^{139} \mathrm{Ba}+3 \mathrm{X}\) (d) \(\frac{53}{24} \mathrm{Cr}+{ }_{2}^{4} \alpha \longrightarrow{ }_{0}^{1} \mathrm{n}+\mathrm{X}\) \((\mathrm{e}){ }_{8}^{20} \mathrm{O} \longrightarrow{ }_{9}^{20} \mathrm{~F}+\mathrm{X}\)

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