Question

28. A sample of oxygen is collected over water at 22°C and 752 mmHg in a 125 ml flask. The vapor pressure of water at 22°C is 19.8 mmHg. a. What is the partial pressure of oxygen? b. How many moles of dry gas are collected? C. How many moles of wet gas are in the flask? d. If 0.0250 g of N, are added to the flask at the same temperature, what is the partial pressure of nitrogen in the flask? e. What is the total pressure in the flask after nitrogen is added? 29. When potassium chlorate is burned, potassium chloride and oxygen are formed. a. Write a balanced equation for the reaction. b. How much potassium chlorate must be burned to produce 198.5 g of oxygen? The yield is found to be 83.2%. 30. If 0.0129 mol of N, O, effuses through a pinhole in a certain amount of time, how much NO would effuse in that same amount of time under the same conditions? 31. It takes 12.6 s for 1.73 x 10 mol of Co to effuse through a pinhole. Under the same conditions, how long will it take for the same amount of Co, to effuse through the same pinhole? 32. What is the average speed of a. a bromine molecule at 28*C? b. a krypton atom at -28°C? 33. Which of the following represents the net ionic equation for NH, (aq) + HBr (aq) ? a. NH, (aq) + HBr (aq) + NH(aq) + Br(aq) b. NH, (aq) + HBr (aq) NH,Br (aq) C. NH, (aq) +H* (aq) + Br (aq) - NH Br (aq) d. NH, (aq) +H* (aq) + NH,* (aq) 34. Which is a spectator ion in the following reaction in water? ZnCl, + NaOH - a. Zn²+ (aq) b. Cl(aq) C. OH(aq) d. There is no spectator ion

Answer 1

28. A sample of oxygen is collected over water at 22°C and 752 mmHg in a 125 ml flask. The vapor pressure o water at 22°C is 19.8 mmHg. a. What is the partial pressure of oxygen? b. How many moles of dry gas are collected? C. How many moles of wet gas are in the flask? d. If 0.0250 g of N, are added to the flask at the same temperature, what is the partial pressure of nitrogen in the flask? e. What is the total pressure in the flask after nitrogen is added?

a) partial pressure of O2 = total pressure - vapor pressure of H2O

= 752 - 19.8 =732.2 mmHg

b) moles of dry gas are calculated using ideal gas equation PV = nRT

n = PV/RT

P = 732.2 mm = 732.2/760 atm

V = 125mL = 0.125L

T= 22C = 22+273 = 295 K

Thus n = [732.2/760]atmx 0.125L /0.0821l.atm/mol.K x295K

=0.00497 mol

c) moles of wet gas calculated using ideal gas equation similarly using P = 752mm

thus

n = [752/760]atmx 0.125L /0.0821l.atm/mol.K x295K

=0.00510mol

d) 0.0250g of N2 added

moles of N2 = 0.0250g/28g/mol =8.92x10^-4 mol

partial pressure of N2 = nRT/V

= 8.92x10^-4 molx0.0821 L.atm/mol.K x295K/ 0.125 L

=0.173 atm =131.48 mmHg

e)Total pressure after N2 is added =[ Partial pressure of O2 + vapor pressure of H2O] + partial presure of N2

= pressure before additon + partial pressure of N2

= 752 + 131.48 =883.48 mmHg