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The number of protons corresponds to the atomic number. Thus this is element 51, antimony. There are 3 fewer electrons than protons. Therefore, the charge on the cation is 3+. The anion has one-third the number of protons of the cation, which corresponds to 17 protons; this is element 17, chlorine. The number of electrons in this anion of chlorine is 17 + 1 = 18 electrons. The anion must have a charge of 1−. The formula of the compound formed between Sb3+ and Cl– is SbCl3. The name of the compound is antimony(III) chloride.

00 mol Ca3(PO4)2 × 57. STOICHIOMETRY a. 01 g N × = 140. 97 g P × = 619 g P mol P4 O 6 mol P b. 97 g P × = 310. g P mol Ca 3 (PO 4 ) 2 mol P c. 97 g P × = 155 g P mol Na 2 HPO 4 mol P a. 00 g NH3 × b. 88 × 1022 molecules N2H4 CHAPTER 3 STOICHIOMETRY c. 00 g (NH4)2Cr2O7 × × 60. 39 × 1021 formula units (NH4)2Cr2O7 mol ( NH 4 ) 2 Cr2 O 7 a. 88 g mol P4 O 6 b. 94 × 1021 formula units Ca3(PO4)2 c. 24 × 1021 formula units Na2HPO4 61. Using answers from Exercise 59: a. 54 × 1022 molecules NH3 × b. 76 × 1022 atoms N molecule N 2 H 4 c.

A. This is element 52, tellurium. Te forms stable 2! charged ions in ionic compounds (like other oxygen family members). b. Rubidium. Rb, element 37, forms stable 1+ charged ions. c. Argon. Ar is element 18. d. Astatine. At is element 85. 97. From the XBr2 formula, the charge on element X is 2+. Therefore, the element has 88 protons, which identifies it as radium, Ra. 230 − 88 = 142 neutrons. 98. Because this is a relatively small number of neutrons, the number of protons will be very close to the number of neutrons present.

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