/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 111 A radioactive metal atom decays ... [FREE SOLUTION] | 91Ó°ÊÓ

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A radioactive metal atom decays (goes to another kind of atom) by emitting an alpha particle (He \({ }^{2+}\) ion). The alpha particles are collected as helium gas. A sample of helium with a volume of \(9.050 \mathrm{~mL}\) was obtained at 765 \(\mathrm{mmHg}\) and \(23^{\circ} \mathrm{C}\). How many atoms decayed during the period of the experiment?

Short Answer

Expert verified
Approximately \(2.27 \times 10^{20}\) atoms decayed during the experiment.

Step by step solution

01

Understand the Ideal Gas Law

To solve this problem, we need to first understand the Ideal Gas Law, which is given by \( PV = nRT \). In this equation: \( P \) represents pressure, \( V \) is volume, \( n \) is the number of moles of the gas, \( R \) is the ideal gas constant \( (0.0821 \text{ } \mathrm{L} \cdot \mathrm{atm}/\mathrm{mol} \cdot \mathrm{K}) \), and \( T \) is the temperature in Kelvin.
02

Convert Units

First, we need to convert the pressure from \( \mathrm{mmHg} \) to \( \mathrm{atm} \).\[ P = \frac{765 \mathrm{\ mmHg}}{760 \mathrm{\ mmHg/atm}} = 1.0066 \mathrm{\ atm} \]Next, convert the volume from \( \mathrm{mL} \) to \( \mathrm{L} \).\[ V = \frac{9.050 \mathrm{\ mL}}{1000 \mathrm{\ mL/L}} = 0.00905 \mathrm{\ L} \]Finally, convert the temperature from Celsius to Kelvin.\[ T = 23^{\circ} \mathrm{C} + 273.15 = 296.15 \mathrm{\ K} \]
03

Calculate Moles of Helium

Now apply the ideal gas law to solve for \( n \), the number of moles of helium gas.\[ PV = nRT \rightarrow n = \frac{PV}{RT} \]Substitute the known values into the equation:\[ n = \frac{(1.0066 \mathrm{\ atm})(0.00905 \mathrm{\ L})}{(0.0821 \mathrm{\ L} \cdot \mathrm{atm}/\mathrm{mol} \cdot \mathrm{K})(296.15 \mathrm{\ K})} \approx 3.77 \times 10^{-4} \mathrm{\ mol} \]
04

Calculate Number of Atoms

To find the number of atoms, multiply the moles of helium by Avogadro's number \( (6.022 \times 10^{23} \mathrm{\ atoms/mol}) \).\[ \text{Number of atoms} = (3.77 \times 10^{-4} \mathrm{\ mol})(6.022 \times 10^{23} \mathrm{\ atoms/mol}) \approx 2.27 \times 10^{20} \mathrm{\ atoms} \]

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Key Concepts

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

Radioactive Decay
Radioactive decay is a natural process in which an unstable atomic nucleus loses energy by emitting radiation. This can happen in several ways, one of which is the emission of alpha particles. When a radioactive element undergoes decay, it transforms into a different element altogether. This transformation releases energy and particles.
The type of decay that occurs depends on the element's nucleus and its configuration. Since it leads to the formation of a more stable nucleus, the decay process is random and can vary between elements. Despite this randomness, statistical predictions about a large number of atoms enable scientists to understand decay rates and processes.
Alpha Particles
Alpha particles are a type of ionizing radiation ejected by the nuclei of some radioactive atoms. These particles are made up of two protons and two neutrons bound together, similar to a helium nucleus. Due to their charge and mass, they are relatively heavy and do not penetrate materials very deeply.
In our exercise, alpha particles are emitted from the radioactive metal, eventually accumulating as helium gas. It's important to note that while alpha particles can be hazardous to health if ingested or inhaled, they are generally stopped by simple barriers like a sheet of paper or even the skin's surface.
However, understanding the role and behavior of alpha particles in radioactive decay helps us predict the outcomes of such processes.
Avogadro's Number
Avogadro's Number is a fundamental constant in chemistry, given by approximately \( 6.022 \times 10^{23} \) entities per mole. This number serves as a bridge, connecting the macroscopic measurements we observe with the microscopic world of atoms and molecules.
In this exercise, we use Avogadro's Number to convert the moles of helium collected into the actual number of atoms. This highlights the essential role Avogadro's Number plays in quantifying the massive number of atoms present in a macroscopic sample.
With Avogadro's Number, chemists can easily convert quantities like the amount of substance in moles to the number of molecules or atoms, vital for precise calculations in chemical reactions and equations.
Unit Conversion
Unit conversion is an essential skill in chemistry and physics, enabling us to translate various measurements into common units for consistent calculations. In many scientific exercises, such as ours, data may be presented in different units, necessitating conversion for compatibility.
For example, the pressure measurement given in \( \mathrm{mmHg} \) was converted to \( \mathrm{atm} \), the standard unit used in the Ideal Gas Law. Likewise, milliliters (mL) were converted to liters (L), and temperature from Celsius to Kelvin.
These conversions ensure that our equations, such as those involving the Ideal Gas Law, are dimensionally consistent and can be solved accurately. Mastering unit conversions is key to solving scientific problems and avoiding potential errors in calculations.

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Most popular questions from this chapter

Gas Laws and Kinetic Theory of Gases I Shown here are two identical containers labeled \(\mathrm{A}\) and \(\mathrm{B}\). Container A contains a molecule of an ideal gas, and container B contains two molecules of an ideal gas. Both containers are at the same temperature. (Note that small numbers of molecules and atoms are being represented in these examples in order that you can easily compare the amounts. Real containers with so few molecules and atoms would be unlikely.) How do the pressures in the two containers compare? Be sure to explain your answer. Shown below are four different containers \((\mathrm{C}, \mathrm{D}, \mathrm{E}\) and \(\mathrm{F}\) ), each with the same volume and at the same temperature. How do the pressures of the gases in the containers compare? Container \(\mathrm{H}\) below has twice the volume of container G. How will the pressure in the containers compare? Explain your reasoning. How will the pressure of containers \(\mathrm{G}\) and \(\mathrm{H}\) compare if you add two more gas molecules to container \(\mathrm{H}\) ? Consider containers I and J below. Container J has twice the volume of container \(\mathrm{I}\). Container \(\mathrm{I}\) is at a temperature of \(100 \mathrm{~K},\) and container \(\mathrm{J}\) is at \(200 \mathrm{~K}\). How does the pressure in container I compare with that in container \(\mathrm{J} ?\) Include an explanation as part of your answer.

5.161 Power plants driven by fossil-fuel combustion generate substantial greenhouse gases (e.g., \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ) as well as gases that contribute to poor air quality (e.g., \(\mathrm{SO}_{2}\) ). To evaluate exhaust emissions for regulation purposes, the generally inert nitrogen supplied with air must be included in the balanced reactions; it is further assumed that air is composed of only \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) with a volume/volume ratio of \(3.76: 1.00,\) respectively. In addition to the water produced, the fuel's \(\mathrm{C}\) and \(\mathrm{S}\) are converted to \(\mathrm{CO}_{2}\) and \(\mathrm{SO}_{2}\) at the stack. Given these stipulations, answer the following for a power plant running on a fossil fuel having the formula \(\mathrm{C}_{18} \mathrm{H}_{10} \mathrm{~S}\). Including the \(\mathrm{N}_{2}\) supplied in air, write a balanced combustion equation for the complex fuel assuming \(120 \%\) stoichiometric combustion (i.e., when excess oxygen is present in the product gases). Except for \(\mathrm{N}_{2},\) use only integer coefficients. (See problem \(3.141 .\) ) What gas law is used to effectively convert the given \(\mathrm{N}_{2} / \mathrm{O}_{2}\) volume ratio to a molar ratio when deriving the balanced combustion equation in (a)? (i) Boyle's law (ii) Charles's law (iii) Avogadro's law (iv) ideal gas law c) Assuming the product water condenses, use the result from (b) to determine the stack gas composition on a "dry" basis by calculating the volume/volume percentages for \(\mathrm{CO}_{2}, \mathrm{~N}_{2},\) and \(\mathrm{SO}_{2}\) Assuming the product water remains as vapor, repeat (c) on a "wet" basis.Assuming the product water remains as vapor, repeat (c) on a "wet" basis. Some fuels are cheaper than others, particularly those with higher sulfur contents that lead to poorer air quality. In port, when faced with ever-increasing air quality regulations, large ships operate power plants on more costly, higher-grade fuels; with regulations nonexistent at sea, ships' officers make the cost-saving switch to lower-grade fuels. Suppose two fuels are available, one having the general formula \(\mathrm{C}_{18} \mathrm{H}_{10} \mathrm{~S}\) and the other \(\mathrm{C}_{32} \mathrm{H}_{18} \mathrm{~S} .\) Based on air quality concerns alone, which fuel is more likely to be used when idling in port? Support your answer to (d) by calculating the mass ratio of \(\mathrm{SO}_{2}\) produced to fuel burned for \(\mathrm{C}_{18} \mathrm{H}_{10} \mathrm{~S}\) and \(\mathrm{C}_{32} \mathrm{H}_{18} \mathrm{~S},\) assuming \(120 \%\) stoichiometric combustion.

The combustion method used to analyze for carbon and hydrogen can be adapted to give percentage \(\mathrm{N}\) by collecting the nitrogen from combustion of the eompound as \(\mathrm{N}_{2}\). A sample of a compound weighing \(8.75 \mathrm{mg}\) gave \(1.59 \mathrm{~mL} \mathrm{~N}_{2}\) at \(25^{\circ} \mathrm{C}\) and \(749 \mathrm{mmHg} .\) What is the percent- age \(\mathrm{N}\) in the compound?

The density of air at \(20^{\circ} \mathrm{C}\) and 1.00 atm is \(1.205 \mathrm{~g} / \mathrm{L}\). If this air were compressed at the same temperature to equal the pressure at \(50.0 \mathrm{~m}\) below sea level, what would be its density? Assume the barometric pressure is constant at \(1.00 \mathrm{~atm} .\) The density of seawater is \(1.025 \mathrm{~g} / \mathrm{cm}^{3} .\)

Sulfur hexafluoride, \(\mathrm{SF}_{6},\) is an extremely dense gas. How does its density compare with the density of air? Use a molar mass for air of \(29.0 \mathrm{~g} / \mathrm{mol}\).

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