1. Review for Unit 4 Assessment

2. Write the formula for the following compounds? L.T. #5 Carbon tetrabromide Dioxygen tetranitride Heptasulfur pentaiodide Triphosporus monoselenide CBr 4 O2N4O2N4 S7I5S7I5 P 3 Se

3. Name the following compoundsL.T. #4 Br 2 O 6 C 3 N 5 S 8 P 4 NCl Dibromine hexaoxide Tricarbon pentanitride Octasulfur tetraphosphide Nitrogen monochloride

4. How many moles are in 100 g of C 3 N 5 ? L.T. = #10, 11 100 g (___________) = 0.94 moles C 3 N 5 1 mole From periodic table: (C 3 N 5 ) C = 12.01 g (3) = 36.03 g N= 14.01 g (5) = + 70.05 g molar mass = 106.08 g 106.08 g

5. How many molecules are in 4.1 moles of Br 2 O 6 ? L.T. = #10, 11 = 2.468 x 10 24 mc 4.1 mol (____________)6.02 x 10 23 mc 1 mol

6. What is the percent composition of each element in triphosphorus pentachloride? / 270.16 = 34.4 % P From periodic table: P = 30.97 g (3) = 92.91 g Cl= 35.45 g (5) = + 177.25 g molar mass = 270.16 g P 3 Cl 5 / 270.16 = 65.6 % Cl L.T. = #8

7. What is the bond type for each compound? C = 2.5 Br = 2.8 Difference = 0.3 Bond type = nonpolar covalent CBr 4 Al 2 S 3 LiF Al = 1.5 S = 2.5 Bond type = Polar covalent Li = 1.0 F = 4.0 Bond type = Ionic L.T. = #3

8. What might be the properties for each compound? Nonpolar = liquid or gas, low m.p., low b.p., soluble in nonpolar substances CBr 4 LiF Ionic = solid, high m.p., high b.p., soluble in ionic substances, excellent conductor L.T. = #2

9. What is the molecular shape for each compound? tetrahedral CBr 4 SO 2 PCl 3 AlBr 3 angular triangular pyramidal triangular planar L.T. = #6,7

10. What is the percent composition the salt in the following hydrate compound; CuSO 4 * 5H 2 O? Cu = 63.55 g (1) = 63.55 g S= 32.07 g (1) = 32.07 g O = 16.00 g (4) = + 64.00 g molar mass salt= 159.62 g/mol / 249.72 = 63.69 % salt L.T. = #9 H 2 O = 18.02 (5) = +90.10 g/ mol molar mass hydrate = 249.72 g/mol