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Chapter 7: Problem 133

Complete and balance the equations for the following reactions. a. \(\mathrm{Li}(s)+\mathrm{N}_{2}(g) \rightarrow\) b. \(\mathrm{Rb}(s)+\mathrm{S}(s) \rightarrow\)

Short Answer

Expert verified
The balanced equations for the given reactions are as follows: a. \(3\mathrm{Li}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Li}_{3}\mathrm{N}(s)\) b. \(2\mathrm{Rb}(s)+\mathrm{S}(s) \rightarrow \mathrm{Rb}_{2}\mathrm{S}(s)\)

Step by step solution

01

a. Write the complete chemical equation.

Lithium (Li) is a metal from Group 1 (Alkali metals), and Nitrogen (N) is a nonmetal from Group 15. The metal and nonmetal will form an ionic compound. Lithium has a +1 charge and Nitrogen gas exists as N2, which means it will form Lithium Nitride, Li3N. This results in the following chemical equation: \[\mathrm{Li}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Li}_{3}\mathrm{N}(s)\] Now, we need to balance this equation.
02

a. Balance the chemical equation.

First, count the number of atoms for each element on both sides of the equation: On the left side, there are 1 Li and 2 N atoms. On the right side, there are 3 Li and 1 N atom. To balance the Li atoms, we place a coefficient of 3 in front of Li on the left side: \[3\mathrm{Li}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Li}_{3}\mathrm{N}(s)\] Now, we have 3 Li and 2 N atoms on both sides. The equation is balanced.
03

b. Write the complete chemical equation.

Rubidium (Rb) is a metal from Group 1 (Alkali metals), while sulfur (S) is a non-metal. Rb has a +1 charge and sulfur (S) has a -2 charge because it is from Group 16 (Chalcogens). They will form the ionic compound Rb2S, which is Rubidium Sulfide. The full equation is: \[\mathrm{Rb}(s)+\mathrm{S}(s) \rightarrow \mathrm{Rb}_{2}\mathrm{S}(s)\] Now, we need to balance this equation.
04

b. Balance the chemical equation.

First, we count the number of atoms for each element on both sides: On the left side, there is 1 Rb and 1 S atom. On the right side, there are 2 Rb atoms and 1 S atom. To balance the Rb atoms, we place a coefficient of 2 in front of Rb on the left side: \[2\mathrm{Rb}(s)+\mathrm{S}(s) \rightarrow \mathrm{Rb}_{2}\mathrm{S}(s)\] Now, we have 2 Rb and 1 S atom on both sides. The equation is balanced.

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Most popular questions from this chapter

The wave function for the \(2 p_{z}\) orbital in the hydrogen atom is $$ \psi_{2 p_{i}}=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a_{0}}\right)^{3 / 2} \sigma \mathrm{e}^{-\sigma / 2} \cos \theta $$ where \(a_{0}\) is the value for the radius of the first Bohr orbit in meters \(\left(5.29 \times 10^{-11}\right), \sigma\) is \(Z\left(r / a_{0}\right), r\) is the value for the distance from the nucleus in meters, and \(\theta\) is an angle. Calculate the value of \(\psi_{2 p_{z}}^{2}\) at \(r=a_{0}\) for \(\theta=0^{\circ}(z\) axis \()\) and for \(\theta=90^{\circ}\) \((x y\) plane).

Which of the following sets of quantum numbers are not allowed? For each incorrect set, state why it is incorrect. a. \(n=3, \ell=3, m_{\ell}=0, m_{s}=-\frac{1}{2}\) b. \(n=4, \ell=3, m_{\ell}=2, m_{s}=-\frac{1}{2}\) c. \(n=4, \ell=1, m_{\ell}=1, m_{s}=+\frac{1}{2}\) d. \(n=2, \ell=1, m_{\ell}=-1, m_{s}=-1\) e. \(n=5, \ell=-4, m_{\ell}=2, m_{s}=+\frac{1}{2}\) f. \(n=3, \ell=1, m_{\ell}=2, m_{s}=-\frac{1}{2}\)

In the ground state of mercury, \(\mathrm{Hg}\), a. how many electrons occupy atomic orbitals with \(n=3\) ? b. how many electrons occupy \(d\) atomic orbitals? c. how many electrons occupy \(p_{z}\) atomic orbitals? d. how many electrons have spin "up" \(\left(m_{s}=+\frac{1}{2}\right)\) ?

Calculate the de Broglie wavelength for each of the following. a. an electron with a velocity \(10 . \%\) of the speed of light b. a tennis ball \((55 \mathrm{~g})\) served at \(35 \mathrm{~m} / \mathrm{s}(\sim 80 \mathrm{mi} / \mathrm{h})\)

In each of the following sets, which atom or ion has the smallest radius? a. \(\mathrm{H}, \mathrm{He}\) b. \(\mathrm{Cl}, \mathrm{In}, \mathrm{Se}\) c. element 120, element 119, element 116 d. \(\mathrm{Nb}, \mathrm{Zn}, \mathrm{Si}\) e. \(\mathrm{Na}^{-}, \mathrm{Na}, \mathrm{Na}^{+}\)
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