Exam 5 Practice Problem Answers

Exam 5 Practice Problem Answers

1. 812 mm Hg * (1 atm/760 mm Hg) = 1.07 atm

812 mm Hg * (760 torr/760 mm Hg) = 812 torr (value in torr = value in mm Hg)

2. rearrange the ideal gas law to solve for volume: V = nRT/P

V = (3.52 mol)(0.0821 L*atm/mol*K)(30+273 K) / (777 torr * 1 atm/760 torr)

= 85.6 L

3. Pressure of gas 1 = 730 mm Hg

Pressure of gas 2 = 784 torr * (760 mm Hg/760 torr) = 784 mm Hg

Pressure of gas 3 = 1.2 atm * (760 mm Hg/1 atm) = 912 mm Hg

Total pressure = 730 mm Hg + 784 mm Hg + 912 mm Hg = 2426 mm Hg

(in terms of atmospheres, total pressure = 3.19 atm)

(in terms of torr, total pressure = 2426 torr)

4. Pressure of gas 1 + Pressure of gas 2 = Total pressure = 2.8 atm

900 mm Hg * (1 atm/760 mm Hg) = 1.18 atm = Pressure of gas 1

Therefore, 1.18 atm + Pressure of gas 2 = 2.8 atm

Subtract 1.18 from both sides, so Pressure of gas 2 = 1.62 atm (or 1231 mm Hg)

5. 211.2 g HCl * (1 mol/36.5 g) = 5.79 mol HCl 500 mL * (1 L/1000 mL) = 0.500 L

Molarity = moles solute/L solution

= 5.79 mol/0.500 L = 11.6 M

6. 2.5 M = x mol/0.750 L x = 1.875 mol HBr

1.875 mol HBr * (80.9 g/1 mol) = 151.7 g

7. a) acid b) base c) acid

8. low pH value = more acidic, so increasing acidity value (lowest to highest): 9, 7, and 4.

9. Orange juice contains citric acid, so it is acidic. Ammonia is a base. Pure water is neutral.

So, pH = 3.0 should be on the bottle containing orange juice. The pH = 7.0 should go on the

bottle containing water. The pH = 11.0 should go on the bottle containing base, the ammonia.

10.a) pH = -log[H+] pH = -log(3*10-6)= 5.52

b) A pH = 5.52 is acidic since the pH of acids range from pH = 0-7.0.