# 16 8 o and 17 are isotopes of oxygen and 16 has 16 8

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16 8 O and 17 are isotopes of oxygen, and 16 has 16 – 8 = 8 neutrons whereas 17 has 17 – 8 = 9 neutrons. Same Z value 8 O 8 O 8 O b) and 41 19 have the same number of neutrons (Ar: 40 – 18 = 22; K: 41 – 19 = 22) but different numbers of protons and electrons (Ar = 18 protons and 18 electrons; K = 19 protons and 19 electrons). Same N value 40 18 Ar K c) and have different numbers of protons, neutrons, and electrons. Co: 27 protons, 27 electrons, 60 27 Co 60 28 Ni and 60 – 27 = 33 neutrons; Ni: 28 protons, 28 electrons and 60 – 28 = 32 neutrons. However, both have a mass number of 60. Same A value 2.42 Plan: The superscript is the mass number ( A ), the sum of the number of protons and neutrons; the subscript is the atomic number ( Z , number of protons). The mass number – the number of protons = the number of neutrons. For atoms, the number of protons = the number of electrons. Solution: a) ) and have different numbers of protons, neutrons, and electrons. H: 1 proton, 1 electron, and 3 1 H 3 2 He 3 – 1 = 2 neutrons; He: 2 protons, 2 electrons, and 3 – 2 = 1 neutron. However, both have a mass number of 3. Same A value b) 14 and 15 have the same number of neutrons (C: 14 – 6 = 8; N: 15 – 7 = 8) but different numbers of protons and electrons (C = 6 protons and 6 electrons; N = 7 protons and 7 electrons). Same N value 6 C 7 N c) 19 and 18 have the same number of protons and electrons (9), but different numbers of neutrons. 9 F 9 F 19 9 F and 18 are isotopes of oxygen, and 19 has 19 – 9 = 10 neutrons whereas 18 has 18 – 9 = 9 neutrons. Same Z value 9 F 9 F 9 F 2.43 Plan: Combine the particles in the nucleus (protons + neutrons) to give the mass number (superscript, A ). The number of protons gives the atomic number (subscript, Z ) and identifies the element. Solution: a) A = 18 + 20 = 38; Z = 18; 38 18 Ar b) A = 25 + 30 = 55; Z = 25; 55 25 Mn c) A = 47 + 62 = 109; Z = 47; 109 47 Ag 2-8
2.44 Plan: Combine the particles in the nucleus (protons + neutrons) to give the mass number (superscript, A ). The number of protons gives the atomic number (subscript, Z ) and identifies the element. Solution: a) A = 6 + 7 = 13; Z = 6; 13 6 C b) A = 40 + 50 = 90; Z = 40; 90 40 Zr c) A = 28 + 33 = 61; Z = 28; 61 28 Ni 2.45 Plan: Determine the number of each type of particle. The superscript is the mass number ( A ) and the subscript is the atomic number ( Z , number of protons). The mass number – the number of protons = the number of neutrons. For atoms, the number of protons = the number of electrons. The protons and neutrons are in the nucleus of the atom. Solution: a) 48 22 b) c) 11 Ti 79 34 Se 5 B 22 protons 34 protons 5 protons 22 electrons 34 electrons 5 electrons 48 – 22 = 26 neutrons 79 – 34 = 45 neutrons 11 – 5 = 6 neutrons 22e 22p + 26n 0 34e 34p + 45n 0 5e 5p + 6n 0 2.46 Plan: Determine the number of each type of particle. The superscript is the mass number ( A ) and the subscript is the atomic number ( Z , number of protons). The mass number – the number of protons = the number of neutrons. For atoms, the number of protons = the number of electrons. The protons and neutrons are in the nucleus of the atom.
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