Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 89

Naturally found uranium consists of \(99.274 \%{ }^{238} \mathrm{U},\) $0.720 \%^{235} \mathrm{U},\( and \)0.006 \%^{233} \mathrm{U}\(. As we have seen, \){ }^{235} \mathrm{U}$ is the isotope that can undergo a nuclear chain reaction. Most of the ${ }^{235} \mathrm{U}$ used in the first atomic bomb was obtained by gaseous diffusion of uranium hexafluoride, \(\mathrm{UF}_{6}(g) .(\mathbf{a})\) What is the mass of \(\mathrm{UF}_{6}\) in a 30.0 -L vessel of \(\mathrm{UF}_{6}\) at a pressure of 695 torr at \(350 \mathrm{~K} ?\) (b) What is the mass of \({ }^{235} \mathrm{U}\) in the sample described in part (a)? (c) Now suppose that the UF \(_{6}\) is diffused through a porous barrier and that the change in the ratio of ${ }^{238} \mathrm{U}\( and \){ }^{235} \mathrm{U}$ in the diffused gas can be described by Equation \(10.23 .\) What is the mass of \({ }^{235} \mathrm{U}\) in a sample of

Short Answer

Expert verified
(a) The mass of UF6 in the 30.0-L vessel is 35.2 g. (b) The mass of \({ }^{235}\mathrm{U}\) in the sample is 0.173 g. (c) We cannot solve part (c) without knowing Equation 10.23.

Step by step solution

01

Use the Ideal Gas Law to find the moles of UF6 in the gas

To calculate the moles of UF6 in the gas, we will use the Ideal Gas Law equation: \(PV = nRT\) Where P is the pressure in atm, V is the volume in liters, n is the moles of UF6, R is the gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin. First, we need to convert the pressure from torr to atm: 695 torr * (1 atm/760 torr) = 0.914 atm Now, plug in the values and solve for n: \( (0.914 \mathrm{atm})(30.0 \mathrm{L}) = n(0.0821\frac{\mathrm{ L\cdot atm}}{\mathrm{mol\cdot K}})(350 \mathrm{K})\)
02

Calculate the moles of UF6

We can now solve for n and find the number of moles: \(n = \frac{(0.914 \mathrm{atm})(30.0 \mathrm{L})}{(0.0821\frac{\mathrm{ L\cdot atm}}{\mathrm{mol\cdot K}})(350 \mathrm{K})} = 0.100 \mathrm{mol}\)
03

Convert moles of UF6 to mass

To convert the moles of UF6 to mass, we need to find the molar mass of UF6, which is: 1 U: 238 g/mol 6 F: 6 * 19 g/mol = 114 g/mol Total: 238 + 114 = 352 g/mol Now we can multiply the moles of UF6 by its molar mass to find the mass: (0.100 mol)(352 g/mol) = 35.2 g
04

Calculate the mass of \({ }^{235}\mathrm{U}\) in the sample

First, calculate the total molar mass of uranium in the sample by multiplying the mass of UF6 by the mass fraction of uranium: (35.2 g) * (238 g/mol) / (352 g/mol) = 24.0 g Next, find the mass of \({ }^{235}\mathrm{U}\) in the sample, knowing that natural uranium contains 0.720% of \({ }^{235}\mathrm{U}\): (24.0 g) * (0.00720) = 0.173 g
05

Calculate the mass of \({ }^{235}\mathrm{U}\) in the sample of diffused gas

We are not given a direct method for finding the mass of \({ }^{235}\mathrm{U}\) after diffusion has occurred. We are however told the change in the ratio can be described by Equation 10.23. Without knowing this equation, we cannot solve part (c).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(65-\mathrm{kg}\) person is accidentally exposed for \(240 \mathrm{~s}\) to \(\mathrm{a}\) 15-mCi source of beta radiation coming from a sample of ${ }^{90}$ Sr. (a) What is the activity of the radiation source in disintegrations per second? In becquerels? (b) Each beta particle has an energy of \(8.75 \times 10^{-14} \mathrm{~J} .\) and \(7.5 \%\) of the radiation is absorbed by the person. Assuming that the absorbed radiation is spread over the person's entire body, calculate the absorbed dose in rads and in grays. \((\mathbf{c})\) If the RBE of the beta particles is \(1.0,\) what is the effective dose in mrem and in sieverts? (d) Is the radiation dose equal to, greater than, or less than that for a typical mammogram \((3 \mathrm{mSv}) ?\)

A laboratory rat is exposed to an alpha-radiation source whose activity is \(14.3 \mathrm{mCi}\). (a) What is the activity of the radiation in disintegrations per second? In becquerels? (b) The rat has a mass of $385 \mathrm{~g}\( and is exposed to the radiation for \)14.0 \mathrm{~s}$, absorbing \(35 \%\) of the emitted alpha particles, each having an energy of $9.12 \times 10^{-13} \mathrm{~J}$. Calculate the absorbed dose in millirads and grays. (c) If the RBE of the radiation is \(9.5,\) calculate the effective absorbed dose in mrem and Sv.

What do these symbols stand for? (a) \({ }_{0}^{0} \gamma,(\mathbf{b}){ }_{2}^{4} \mathrm{He},\) (c) \({ }_{0}^{1} \mathrm{n} .\)

In 2002 , a team of scientists from Russia and the United States reported the creation of the first atom of element 118 , which is named oganesson, and whose symbol is Og. The synthesis involved the collision of californium- 249 atoms with accelerated ions of an atom which we will denote X. In the synthesis, an oganesson-294 is formed together with three neutrons. $$ { }_{98}^{249} \mathrm{Cf}+\mathrm{X} \longrightarrow{ }_{118}^{294} \mathrm{Og}+3{ }_{0}^{1} \mathrm{n} $$ (a) What are the identities of isotopes X? (b) Isotope \(X\) is unusual in that it is very long-lived (its half-life is on the order of \(10^{19} \mathrm{yr}\) ) in spite of having an unfavorable neutron-to-proton ratio (Figure 21.1). Can you propose a reason for its unusual stability? (c) Oganesson-294 decays into livermorium-290 by alpha decay. Write a balanced equation for this.

Tests on human subjects in Boston in 1965 and \(1966,\) following the era of atomic bomb testing, revealed average quantities of about \(2 \mathrm{pCi}\) of plutonium radioactivity in the average person. How many disintegrations per second does this level of activity imply? If each alpha particle deposits $8 \times 10^{-13} \mathrm{~J}\( of energy and if the average person weighs \)75 \mathrm{~kg},$ calculate the number of grays and sieverts of radiation in 1 yr from such a level of plutonium.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free