AP Chemistry Chapter 14 Answers

AP Chemistry Chapter 14 Answers - AP Chemistry: Chapter 14...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
AP Chemistry: Chapter 14 - Solutions and their Behavior Problems: 1, 5, 9, 11, 13, 17, 19, 23, 25, 31, 35, 39, 48, 51, 55 14.1 2 4 2 2 2 4 2 2 1 mol C H (CO H) 2.56 g · = 0.0217 mol C H (CO H) 118.1 g 3 2 2 3 1 cm 1.00 g 1 mol H O 500. mL · · · = 27.7 mol H O 1 mL 18.02 g 1 cm (a) 3 2 4 2 2 amount of solute 0.0217 mol C H (CO H) 10 g = = · = 0.0434 kg of solvent 500. g 1 kg m m (b) acid 0.0217 mol = = 0.000781 0.0217 mol + 27.7 mol X (c) 2.56 g Weight % = · 100% = 0.509% 2.56 g + 500. g 14.5 2 3 2 2 3 3 2 2 3 0.200 mol Na CO 106.0 g 125 g H O = 2.65 g Na CO 1 kg H O 1 mol Na CO 10 g/1 kg 2 3 2 3 Na CO 3 2 2 3 0.200 mol Na CO = = 0.00359 1 mol H O 0.200 mol Na CO + 10 g 18.02 g X 14.9 (a) 1000 mL 1.18 g Mass of solution = 1 L = 1180 g solution 1 L mL 2 2 2 12.0 mol HCl 36.46 g Mass of HCl = 1 L = 438 g HCl 1 L 1 mol HCl Mass of H O = 1180 g – 438 g = 742 g H O 12.0 mol HCl = = 16.2 0.742 kg H O m m (b) 438 g HCl Weight % = 100% = 37.1%
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

AP Chemistry Chapter 14 Answers - AP Chemistry: Chapter 14...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online