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Chapter 2: Problem 66

What number of protons and neutrons are contained in the nucleus of each of the following atoms? Assuming each atom is uncharged, what number of electrons are present? a. \(^{235}_{92} \mathrm{U}\) b. \(_{13}^{27} \mathrm{Al}\) c. \(_{26}^{57} \mathrm{Fe}\) d. \(_{82}^{208} \mathrm{Pb}\) e. \(_{37}^{86} \mathrm{Rb}\) f. \(_{20}^{41} \mathrm{Ca}\)

Short Answer

Expert verified
a. 92 protons, 143 neutrons, 92 electrons b. 13 protons, 14 neutrons, 13 electrons c. 26 protons, 31 neutrons, 26 electrons d. 82 protons, 126 neutrons, 82 electrons e. 37 protons, 49 neutrons, 37 electrons f. 20 protons, 21 neutrons, 20 electrons

Step by step solution

01

a. \(^{235}_{92} \mathrm{U}\)#

For Uranium-235, the atomic number Z is 92, which means there are 92 protons. The mass number A is 235. To find the number of neutrons, subtract the atomic number Z from the mass number A: 235 - 92 = 143 neutrons. As the atom is uncharged, the number of electrons will be equal to the number of protons, so there are 92 electrons.
02

b. \(_{13}^{27} \mathrm{Al}\)#

For Aluminum-27, the atomic number Z is 13, which means there are 13 protons. The mass number A is 27. To find the number of neutrons, subtract the atomic number Z from the mass number A: 27 - 13 = 14 neutrons. As the atom is uncharged, the number of electrons will be equal to the number of protons, so there are 13 electrons.
03

c. \(_{26}^{57} \mathrm{Fe}\)#

For Iron-57, the atomic number Z is 26, which means there are 26 protons. The mass number A is 57. To find the number of neutrons, subtract the atomic number Z from the mass number A: 57 - 26 = 31 neutrons. As the atom is uncharged, the number of electrons will be equal to the number of protons, so there are 26 electrons.
04

d. \(_{82}^{208} \mathrm{Pb}\)#

For Lead-208, the atomic number Z is 82, which means there are 82 protons. The mass number A is 208. To find the number of neutrons, subtract the atomic number Z from the mass number A: 208 - 82 = 126 neutrons. As the atom is uncharged, the number of electrons will be equal to the number of protons, so there are 82 electrons.
05

e. \(_{37}^{86} \mathrm{Rb}\)#

For Rubidium-86, the atomic number Z is 37, which means there are 37 protons. The mass number A is 86. To find the number of neutrons, subtract the atomic number Z from the mass number A: 86 - 37 = 49 neutrons. As the atom is uncharged, the number of electrons will be equal to the number of protons, so there are 37 electrons.
06

f. \(_{20}^{41} \mathrm{Ca}\)#

For Calcium-41, the atomic number Z is 20, which means there are 20 protons. The mass number A is 41. To find the number of neutrons, subtract the atomic number Z from the mass number A: 41 - 20 = 21 neutrons. As the atom is uncharged, the number of electrons will be equal to the number of protons, so there are 20 electrons.

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Most popular questions from this chapter

How many protons and neutrons are in the nucleus of each of the following atoms? In a neutral atom of each element, how many electrons are present? a. \(^{79} \mathrm{Br}\) b. \(^{81} \mathrm{Br}\) c. \(^{239} \mathrm{Pu}\) d. \(^{133} \mathrm{Cs}\) e. \(^{3} \mathrm{H}\) f. \(^{56} \mathrm{Fe}\)

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Give the names of the metals that correspond to the following symbols: Sn, Pt, Hg, Mg, K, Ag.

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Name the compounds in parts a–d and write the formulas for the compounds in parts e–h. a. \(\mathrm{Hg}_{2} \mathrm{O}\) b. \(\mathrm{FeBr}_{3}\) c. \(\mathrm{CoS}\) d. \(\mathrm{TiCl}_{4}\) e. tin(II) nitride f. cobalt(III) iodide g. mercury(II) oxide h. chromium(VI) sulfide
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