Chem 171 F09 Q4 Emp NIE Key

Chem 171 F09 Q4 Emp NIE Key - Chem 171 Quiz 4 Version A 1....

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Unformatted text preview: Chem 171 Quiz 4 Version A 1. Combustion of a 5.50 g sample of a compound produces 18.59 g CO 2 and 3.81 g H 2 O. Determine the empirical formula and the molecular formula of the compound given that its molar mass is approximately 78 g / mol? (C = 12.01, O = 16.00, H=1.008) Mass of C = 18.59 g CO 2 x 1 mol CO 2 / 44.01 g CO 2 x 1 mol C / 1 mol CO 2 x 12.01 g C/ 1 mol C = 5.073 g C Mass of H = 3.81 g H 2 O x 1 mol H 2 O/18.02 g H 2 O x 1.008 g H / 1 mol H = 0.426 g H 5.073 g + 0.426 g = 5.499 g = 5.50 g so there is no O 5.073 g C / 12.01 g / mol = 0.4224 mol C 0.426 g H / 1.008 g H / mol = 0.423 mol H C 0.4224 H 0.423 = CH empirical formula Empirical formula mass = 12.01 + 1.008 = 13.02 78 / 13.02 = 6 C 6 H 6 2. Predict whether a reaction will occur in the following case. If there is a reaction, write the balanced net ionic equation. Show the phases. KOH(aq) + MgCl 2 (aq) AgNO 3 (aq) + NaBr (aq) AgNO 3 (aq) + KI (aq) Yes. 2 OH- (aq) + Mg 2+ (aq) Mg(OH) 2 (s) (Net Ionic Equation) Yes. Ag + (aq) + Br- (aq) AgBr(s) (Net Ionic Equation) Yes. Ag + (aq) + I- (aq) AgI(s) (Net Ionic Equation) Chem 171 Quiz 4 Version B 1. 1....
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Chem 171 F09 Q4 Emp NIE Key - Chem 171 Quiz 4 Version A 1....

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