Chemistry Unit 7 Review Answers Assignment

University:
Felician University
Course:
CHEM-105 | Principles of Chemistry
Academic year:

2023
Views:

475

Pages:

4
Author:

StellarStrategist

58) a) 15.5 g SiO₂ × (1 mol SiO₂ / 60.079 g SiO₂) = 0.252 mol SiO₂ b) 0.0688 g AgCl × (1 mol AgCl / 143.32 g AgCl) = 4.800 × 10⁻⁴ mol AgCl c) 79.3 g Cl₂ × (1 mol Cl₂ / 70.9 g Cl₂) = 1.12 mol Cl₂ d) 5.96 g KOH × (1 mol KOH / 56.11 g KOH) = 0.106 mol KOH e) 937 g Ca(C₂H₃O₂)₂ × (1 mol Ca(C₂H₃O₂)₂ / 158.14 g Ca(C₂H₃O₂)₂) = 5.93 mol Ca(C₂H₃O₂)₂ f) 0.800 g Ca × (1 mol Ca / 40.08 g Ca) = 0.0200 mol Ca 59) a) 3.21 × 10⁻² mol Ni × (58.69 g Ni / 1 mol Ni) = 1.88 g Ni 60) b) 0.44 mol C₂H₆ × (22.4 L / 1 mol) = 9.9 L C₂H₆ 61) a) C₃H₈: 44.11 g C₃H₈ × (1 mol C₃H₈ / 1 mol C₃H₈) × (22.4 L C₃H₈ / 1 mol C₃H₈) = 1.97 g/L b) Ne: 20.18 g Ne × (1 mol Ne / 20.18 g Ne) × (22.4 L Ne / 1 mol Ne) = 22.4 L/mol = 0.900 g/L c) NO₂: 45.99 g NO₂ × (1 mol NO₂ / 45.99 g NO₂) × (22.4 L NO₂ / 1 mol NO₂) = 2.05 g/L 63) a) H₂S H = 1.01 x 2 / (2(1.01) + 32.07) x 100 = 5.9% H S = 32.07 / (2(1.01) + 32.07) x 100 = 94.1% S b) (NH₄)₂C₂O₄ N = 14.01 x 2 / 124.08 x 100 = 22.6% N H = 1.01 x 8 / 124.08 x 100 = 6.5% H C = 12.01 x 2 / 124.08 x 100 = 19.4% C O = 15.99 x 4 / 124.08 x 100 = 51.5% O c) Mg(OH)₂ Mg = 24.31 / 58.31 x 100 = 41.7% Mg O = 15.99 x 2 / 58.31 x 100 = 54.8% O H = 1.01 x 2 / 58.31 x 100 = 3.5% H 64) b) nitrogen in 25 g (NH₄)₂C₂O₄ 22.6% N 25 g x 0.226 = 5.65 g N c) magnesium in 97.4 g Mg(OH)₂ 41.7% Mg 97.4 g x 0.417 = 40.6 g Mg