General Chemistry 10th Edition Solution Manual by Ralph H. Petrucci, F. Geoffrey Herring, Jeffry D. Madura,

By Ralph H. Petrucci, F. Geoffrey Herring, Jeffry D. Madura, Carey Bissonnette

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M) Let Z = # of protons, N = # of neutrons, E = # of electrons, and A = # of nucleons = Z + N. (a) Z + N = 234 The mass number is 234 and the species is an atom. 0% more neutrons than protons. Next we will substitute the second expression into the first and solve for Z. 600 Thus this is an atom of the isotope 234 Th . Z (b) Z = E + 2 The ion has a charge of +2. 0% more protons than electrons. By equating these two expressions and solving for E, we can find the number of electrons. 100 E 2 E Z 20 2 22, (titanium).

55 g of Rb. 350 g of rock 106 = 159 ppm Rb SELF-ASSESSMENT EXERCISES 94. , total of protons and neutrons equalsA). (b) particle: An electron produced as a result of the decay of a neutron (c) Isotope: Nuclei that share the same atomic number but have different atomic masses (d) 16O: An oxygen nucleus containing 8 neutrons (e) Molar Mass: Mass of one mole of a substance 57 Chapter 2: Atoms and the Atomic Theory 95. (E) (a) The total mass of substances present after the chemical reaction is the same as the total mass of substances before the chemical reaction.

Thus, we can determine the percent natural abundance of the second isotope by difference. 904 u. We use this value in the expression for determining the weighted-average atomic mass, along with the isotopic mass of 79 Br and the fractional abundances of the two isotopes (the percent abundances divided by 100). 4931 42 81 Br, the other isotope Chapter 2: Atoms and the Atomic Theory 47. (M) Since the three percent abundances total 100%, the percent abundance of difference. 0117% 40 K is found by Then the expression for the weighted-average atomic mass is used, with the percent abundances converted to fractional abundances by dividing by 100.

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