Question

Balance the following nuclear equations by filling in the blanks: (a) Es- \(240+\) neutron \(\longrightarrow 2\) neutrons+______+ Gd-161 (b)______ \(\longrightarrow \beta\) -particle \(+\) Co-59 (c) \(4_{1}^{1} \mathrm{H} \longrightarrow \underline{ }+2\) positrons (d)_____ \(\mathrm{Mg}-24+\) neutron \(\longrightarrow\) proton \(+\)______

Short answer

Question: Balance the following nuclear equations and identify the missing elements or particles: (a) Es-240 + neutron → 2 neutrons + _____ + Gd-161 (b) _____ → β-particle + Co-59 (c) 4 \({}_1^1\mathrm{H}\) → _____ + 2 positrons (d) _____ Mg-24 + neutron → proton +______ Answer: (a) Es-240 + neutron → 2 neutrons + Br-164 + Gd-161 (b) Fe-59 → β-particle + Co-59 (c) 4 \({}_1^1\mathrm{H}\) → He-4 + 2 positrons (d) Na-24 + Mg-24 + neutron → proton + Na-24

Step-by-step solution

Analyze the initial reactants

The initial reactants are Es-240 and a neutron. We can write these as \({}_{99}^{240}\mathrm{Es}\) and \({}_{0}^{1}\mathrm{n}\).

Analyze the final products

The final products are 2 neutrons, which can be written as \({}_{0}^{1}\mathrm{n}\), and Gd-161, which can be written as \({}_{64}^{161}\mathrm{Gd}\).

Balance the equation

The unknown reactant can be represented as \({}_{Z}^{A}\mathrm{X}\). According to the conservation of mass-energy and the conservation of charge, we can write: A - 1 = 1 + 1 + 161 A = 164 Z = 0 + 99 - 64 Z = 35 So, we have \({}_{35}^{164}\mathrm{X}\), which is an isotope of Bromine (Br). Thus the balanced equation is: Es-240 + neutron → 2 neutrons + Br-164 + Gd-161 (b) _____ → β-particle + Co-59

Analyze the final products

The final products are a β-particle and Co-59. A β-particle is an electron with charge -1 and mass number 0, which can be represented as \({}_{-1}^{0}\beta\), and Co-59 can be written as \({}_{27}^{59}\mathrm{Co}\).

Balance the equation

The unknown reactant can be represented as \({}_{Z}^{A}\mathrm{X}\). According to the conservation of mass-energy and the conservation of charge, we can write: A = 0 + 59 A = 59 Z - (-1) = 27 Z = 26 So, we have \({}_{26}^{59}\mathrm{X}\), which is an isotope of Iron (Fe). Thus the balanced equation is: Fe-59 → β-particle + Co-59 (c) 4 \({}_1^1\mathrm{H}\) → _____ + 2 positrons

Analyze the initial reactants

The initial reactants are 4 hydrogen atoms, each of which can be written as \({}_1^1\mathrm{H}\).

Balance the equation

The unknown product can be represented as \({}_{Z}^{A}\mathrm{X}\). According to the conservation of mass-energy and the conservation of charge, we can write: A = 4 * 1 A = 4 Z = 4 * 1 - 2 Z = 2 So, we have \({}_{2}^{4}\mathrm{X}\), which is an isotope of Helium (He). Thus the balanced equation is: 4 \({}_1^1\mathrm{H}\) → He-4 + 2 positrons (d) _____ Mg-24 + neutron → proton +______

Analyze the initial reactants

The initial reactants are Mg-24 and a neutron. We can write these as \({}_{12}^{24}\mathrm{Mg}\) and \({}_{0}^{1}\mathrm{n}\).

Analyze the final products

The final products are a proton and another unknown product. A proton can be written as \({}_1^1\mathrm{p}\).

Balance the equation

The unknown reactant and final products can be represented as \({}_{Z}^{A}\mathrm{X}\) and \({}_{Z'}^{A'}\mathrm{Y}\). According to the conservation of mass-energy and the conservation of charge, we can write: Z = 12, A = 24 Z' + 1 = Z, A' + 1 = A So, Z' = 11 and A' = 24 Therefore, the missing reactant is \({}_{11}^{24}\mathrm{Na}\) (sodium), and the balanced equation is: Mg-24 + neutron → proton + Na-24