This preview shows page 1. Sign up to view the full content.
Unformatted text preview: . 149‐151; 9th ed. p. 146‐148. p. 160: 4.74. Mixing solutions is diluting solution. New concentration = mole of solute/ total volume of solutions. Thus, M = (0.568x46.2+1.396 x 80.5)/(46.2+80.5) = 1.09. 9. What volume (in mL) of a 0.500 M HCl solution is needed to neutralize 10.0 mL of a 0.2000 M Ba(OH)2 solution? (a) 8.00 (b) 4.00 (c) 2.00 (d) 1.00 (e) 0.50 Hint: 11th ed.: p.p. 151‐154 and Example 4.12; 10th ed. p.p. 153‐156 Example 4.11 ; 9th ed. p.p. 150‐
153. Solution stoichioimetry. Example 4.11. Short‐cut formula of acid‐base neutralization: ia x Ma x Va = ib x Mb x Vb where a refers to acid and b refers to base. ia refers to number of H in the chemical formula of acid and ib refers to the number of OH in the chemical formula of base. 10. What volume (in L) does a sample of air occupy at 6.6 atm when 1.2 atm, 3.8 L of air is compressed? (a) 0.34 (b) 0.57 (c) 0.69 (d) 0.77 (e) 0.86 Hint: Boyle’s law: 11th ed.: p.p. 178‐181 and p. 216: 5.19; 10th ed. p.p. 179‐182 Equation (5.2); 9th ed. p.p. 175‐178. Equation (5.2). 11. What is the final temperature (in K), under constant‐pressure condition, when a sample of hydrogen gas initially at 88oC and 9.6 L is cooled until its finial v...
View Full Document
This note was uploaded on 01/25/2014 for the course CHEM 1411 taught by Professor None during the Fall '11 term at HCCS.
- Fall '11