/*! 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 13 Balance the following equations ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Balance the following equations by filling in the blanks. (a) \({ }_{92}^{235} \mathrm{U}+{ }_{0} n \longrightarrow{ }_{54}^{137}=2{ }_{0}^{1} n+\) (b) \({ }_{90}^{232} \mathrm{Th}+{ }_{6}^{12}\) \(\longrightarrow\) \(1 . n+{ }_{96}^{240} \mathrm{Cm}\) (c) \({ }_{2}^{4} \mathrm{He}+{ }_{42}^{96} \mathrm{Mo} \longrightarrow{ }_{43}^{100}\) (d) \(+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{84}^{210}+{ }_{0}^{1} n\)

Short Answer

Expert verified
b) _{90}^{232}Th + _{6}^{12}C → 1_{0}^{1}n + ? c) _{2}^{4}He + _{42}^{96}Mo → ? d) _{?}^{?}At + _{1}^{2}H → _{84}^{210}Po + 1_{0}^{1}n Answer: a) _{92}^{235}U + _{0}^{1}n → _{54}^{137}Xe + 2_{0}^{1}n + _{38}^{94}Sr b) _{90}^{232}Th + _{6}^{12}C → 1_{0}^{1}n + _{96}^{243}Cm c) _{2}^{4}He + _{42}^{96}Mo → _{43}^{100}Tc d) _{85}^{211}At + _{1}^{2}H → _{84}^{210}Po + 1_{0}^{1}n

Step by step solution

01

Balancing Protons

Since the total proton number (also known as the atomic number) must be conserved on both sides of the equation, we can calculate the missing atomic number on the right-hand side by subtracting the proton number of the product from that of the reactants: 92 - 54 = 38.
02

Balancing Neutrons

We know that the total neutron number must also be conserved. Since the right-hand side has two neutrons, the desired neutron number on the right-hand side will be the sum of the reactants' neutron numbers minus two. We can calculate this as follows: 235 + 1 - 1*2 = 233 - 2 = 231.
03

Writing the Balanced Equation

Now that we have balanced the protons and neutrons, we can write the complete balanced equation for the nuclear reaction: \begin{equation}\mathrm{_{92}^{235}U}+\mathrm{_{0}^{1}n}\longrightarrow\mathrm{_{54}^{137}Xe}+2\mathrm{_{0}^{1}n} + \mathrm{_{38}^{94}Sr} \end{equation} #b) Balancing Thorium Reaction#
04

Balancing Protons

The total proton number on the left-hand side is 90 + 6 = 96. So, the proton number on the right-hand side should be 96.
05

Balancing Neutrons

The total neutron number on the left-hand side is 232 + 12 = 244. So, the neutron number on the right-hand side should be 244 - 1 = 243.
06

Writing the Balanced Equation

Now that we have balanced the protons and neutrons, we can write the complete balanced equation for the nuclear reaction: \begin{equation}\mathrm{_{90}^{232}Th}+\mathrm{_{6}^{12}C}\longrightarrow1\mathrm{_{0}^{1}n} + \mathrm{_{96}^{243}Cm} \end{equation} #c) Balancing Helium Reaction#
07

Balancing Protons

The total proton number on the left-hand side is 2 + 42 = 44. So, the proton number on the right-hand side should be 43.
08

Balancing Neutrons

The total neutron number on the left-hand side is 4 + 96 = 100. So, the neutron number on the right-hand side should be 100.
09

Writing the Balanced Equation

Now that we have balanced the protons and neutrons, we can write the complete balanced equation for the nuclear reaction: \begin{equation}\mathrm{_{2}^{4}He}+\mathrm{_{42}^{96}Mo}\longrightarrow\mathrm{_{43}^{100}Tc} \end{equation} #d) Balancing Unnamed Reaction#
10

Balancing Protons

The total proton number on the right-hand side is 84. So, the proton number on the left-hand side should be equal to 84 + 1 = 85.
11

Balancing Neutrons

The total neutron number on the right-hand side is 210 + 1 = 211. So, the neutron number on the left-hand side should be 211.
12

Writing the Balanced Equation

Now that we have balanced the protons and neutrons, we can write the complete balanced equation for the nuclear reaction: \begin{equation}\mathrm{_{85}^{211}At}+\mathrm{_{1}^{2}H}\longrightarrow\mathrm{_{84}^{210}Po}+1\mathrm{_{0}^{1}n} \end{equation}

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Conservation of Atomic Number
Understanding the conservation of atomic number is crucial in the context of balancing nuclear reactions. The atomic number, denoted as 'Z,' represents the number of protons in the nucleus of an atom and thus defines the element itself. For instance, every atom with 1 proton is hydrogen, while every atom with 92 protons is uranium, regardless of the number of neutrons.

In nuclear reactions, the law of conservation of atomic number states that the sum of atomic numbers of the reactants must equal the sum of the atomic numbers of the products. This law is the basis for balancing nuclear equations and ensures that the identity of elements is preserved through the reaction. If an equation shows a different atomic number total on each side, you know that the reaction isn’t balanced correctly.

For example, when uranium-235 captures a neutron and subsequently undergoes fission, the total number of protons before and after the reaction must remain the same. This can be clearly seen in the balanced equation, reflecting that the elements xenon (Z=54) and strontium (Z=38) sum up to the original atomic number of uranium (Z=92).
Conservation of Neutron Number
Parallel to the conservation of atomic number, there is also a principle known as the conservation of neutron number in nuclear reactions. Neutrons, represented by the letter 'N', are neutral particles found within the atomic nucleus along with protons. When balancing nuclear reactions, it is as important to account for the total number of neutrons as it is for the protons.

This principle dictates that the total number of neutrons present in the reactants must be equal to the sum of the neutrons found in the products, after accounting for any released or absorbed neutrons during the reaction. Neutron conservation is critical for understanding nuclear stability and the occurrence of nuclear processes such as fission, fusion, and transmutation.

In our example with uranium-235, we add the number of neutrons in uranium (143) with the one neutron that is absorbed to get a total of 144 neutrons. After the reaction, these 144 neutrons are distributed among the resulting isotopes and any free neutrons. Here, xenon-137 has 83 neutrons and strontium-94 has 56 neutrons. Adding the two free neutrons, the total neutron count is maintained.
Nuclear Equation Writing
Writing nuclear equations is analogous to writing chemical equations, but with focus on the atomic nuclei involved instead of the entire atoms. A properly balanced nuclear equation will respect both the conservation of atomic number and neutron number. This involves making sure that the charges (protons) and mass numbers (protons plus neutrons) are balanced on both sides of the reaction.

In the context of the given exercise, let's look at the Helium and Molybdenum reaction. Initially, Helium (with an atomic number of 2) and Molybdenum (atomic number 42) have a combined atomic number of 44. After the reaction, the product has an atomic number of 43 (technetium). To balance the equation, an additional reaction product with an atomic number of 1 is needed, which would be an element like Hydrogen.

Moreover, to correctly write nuclear equations, all reactants and products must be clearly denoted with their mass numbers and atomic numbers. This indicates not only which elements are involved but also which isotopes, providing a complete picture of the nuclear transformation.

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

It is possible to estimate the activation energy for fusion by calculating the energy required to bring two deuterons close enough to one another to form an alpha particle. This energy can be obtained by using Coulomb's law in the form \(E=8.99 \times 10^{9} q_{1} q_{2} / r\), where \(q_{1}\) and \(q_{2}\) are the charges of the deuterons \(\left(1.60 \times 10^{-19} \mathrm{C}\right), r\) is the radius of the He nucleus, about \(2 \times 10^{-15} \mathrm{~m}\), and \(E\) is the energy in joules. (a) Estimate \(E\) in joules per alpha particle. (b) Using the equation \(E=m v^{2} / 2\), estimate the velocity (meters per second) each deuteron must have if a collision between the two of them is to supply the activation energy for fusion \((m\) is the mass of the deuteron in kilograms).

Classify the following statements as true or false. If false, correct the statement to make it true. (a) The mass number increases in beta emission. (b) A radioactive species with a large rate constant, \(k\), decays very slowly. (c) Fusion gives off less energy per gram of fuel than fission.

The cleavage of ATP (adenosine triphosphate) to ADP (adenosine diphosphate) and \(\mathrm{H}_{3} \mathrm{PO}_{4}\) may be written as follows: It is interesting to determine which bond (the \(\mathrm{P}-\mathrm{O}\) bond marked a or the \(\mathrm{O}\) - \(\mathrm{P}\) bond marked \(\mathbf{b}\) ) is cleaved by hydrolysis (reaction with water). (a) Outline an experiment (using radioactivity) that can be used to determine where the cleavage results. (b) Describe the results that would lead you to conclude that cleavage results at a, (c) Describe the results that would lead you to conclude that cleavage results at \(\mathbf{b}\). Results show that the cleavage occurs at \(\mathrm{b}\).

Iodine-131 is used to treat thyroid cancer. It decays by beta emission and has a half-life of \(8.1\) days. (a) Write a balanced nuclear reaction for the decay of iodine-131. (b) What is the activity (in Ci) of a 2.50-mg sample of the isotope?

The amount of oxygen dissolved in a sample of water can be determined by using thallium metal containing a small amount of the isotope Tl- 204\. When excess thallium is added to oxygen-containing water, the following reaction occurs. $$2 \mathrm{Tl}(s)+\frac{1}{2} \mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{Tl}^{+}(a q)+2 \mathrm{OH}^{-}(a q)$$ After reaction, the activity of a 25.0-mL water sample is 745 counts per minute (cpm), caused by the presence of \(\mathrm{Tl}^{+}-204\) ions. The activity of Tl-204 is \(5.53 \times 10^{5} \mathrm{cpm}\) per gram of thallium metal. Assuming that \(\mathrm{O}_{2}\) is the limiting reactant in the above equation, calculate its concentration in moles per liter.

See all solutions

Recommended explanations on Chemistry Textbooks

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.