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Review for Unit 4 Assessment

Review for Unit 4 Assessment. Write the formula for the following compounds? L.T. #5. Carbon tetrabromide Dioxygen tetranitride Heptasulfur pentaiodide Triphosporus monoselenide. CBr 4. O 2 N 4. S 7 I 5. P 3 Se. Name the following compounds L.T. #4. Br 2 O 6 C 3 N 5 S 8 P 4 NCl.

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Review for Unit 4 Assessment

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  1. Review for Unit 4 Assessment

  2. Write the formula for the following compounds? L.T. #5 Carbon tetrabromide Dioxygentetranitride Heptasulfurpentaiodide Triphosporusmonoselenide CBr4 O2N4 S7I5 P3Se

  3. Name the following compounds L.T. #4 Br2O6 C3N5 S8P4 NCl Dibrominehexaoxide Tricarbonpentanitride Octasulfurtetraphosphide Nitrogen monochloride

  4. How many moles are in 100 g of C3N5? L.T. = #10, 11 1 mole 100 g (___________) 106.08 g = 0.94 moles C3N5 From periodic table: (C3N5) C = 12.01 g (3) = 36.03 g N= 14.01 g (5) = + 70.05 g molar mass = 106.08 g

  5. How many molecules are in 4.1 moles of Br2O6 ? L.T. = #10, 11 4.1 mol (____________) 6.02 x 1023 mc 1 mol = 2.468 x 1024mc

  6. L.T. = #8 What is the percent composition of each element in triphosphoruspentachloride? P3Cl 5 From periodic table: P= 30.97 g (3) = 92.91 g Cl= 35.45 g (5) = + 177.25 g molar mass = 270.16 g / 270.16 = 34.4 % P / 270.16 = 65.6 % Cl

  7. L.T. = #3 What is the bond type for each compound? Difference = 0.3 Bond type = nonpolar covalent CBr4 Al2S3 LiF C = 2.5 Br = 2.8 Al = 1.5 S = 2.5 Bond type = Polar covalent Li = 1.0 F = 4.0 Bond type = Ionic

  8. L.T. = #2 What might be the properties for each compound? CBr4 LiF Nonpolar = liquid or gas, low m.p., low b.p., soluble in nonpolar substances Ionic = solid, high m.p., high b.p., soluble in ionic substances, excellent conductor

  9. L.T. = #6,7 What is the molecular shape for each compound? CBr4 SO2 PCl3 AlBr3 tetrahedral angular triangular pyramidal triangular planar

  10. L.T. = #9 What is the percent composition the salt in the following hydrate compound; CuSO4 * 5H2O? 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 H2O = 18.02 (5) = + 90.10 g/ mol molar mass hydrate= 249.72 g/mol

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