Chemistry 103-Fall 1999

Quiz 10

Nov. 19, 1999 Name_____average 5.9/10____

 

R= 8.3145 J/(K.mol) R= 0.082 L.atm/(K.mol)

0°C = 273.15 K PV = nRT v = (3RT/M)1/2

ATOM

Z

Mass(g/mol)

H

1

1.008

C

6

12.01

O

8

16.00
  1. The average (rms) speed of molecular oxygen, O2, is 480 m/s at 25°C. Calculate the average speed of ozone, O3, and hydrogen, H2 in the troposphere where the temperature is -50°C. (2 sig. fig).

    2. V ozone = (3 x 8.34 J/K.mol x 223 K/.048 kg/mol )^1/2 = 340 m/s

    UNITS–remember that a joule is kgm^2/s^2 so your molar mass needs to be in kg. Avoid unit problems by using only SI or mks (meter kg sec) units.

    V H2 = 340 m/s x (2 /48)^12 = 1600 m/s

    Or calculate the speed of O2 at —50°C, then O3 and H2 at —50°C. This avoids units problems.

    V O2 at —50°C = 480 m/s (223/298)^1/2 = 415 m/s

    VO3 at —50°C = 415 m/s (32/48)^1/2 = 340 m/s

    VH2 at — 50°C = 415 m/s (32/2)^1/2 = 1600 m/s

    Does your answer make sense? Are heavier molecules slower? Hotter ones faster? Do your answers make sense in light of the 340 m/s for oxygen?

    v O3 = _340 m/s_

    v H2 = _1600 m/s____

    See text 6-69
  2. Calculate the volume of oxygen gas released by 5.0 mL of a 2% hydrogen peroxide solution whose density is 1.01 g/mL. You may assume that the reaction below goes to completion during the disinfection process. You may assume 1.00 atm and 25.C °C

2 H2O2 (aq) à 2 H2O + O2 (g)

Read the problem carefully 2 or 3 times before starting!!. This is meant to review several concepts from earlier in the semester. Many of you need much more problem-solving practice. Don’t mix up gases and solutions.

Set up a 3 part strategy

  1. Use solution info to find the number of moles of H2O2 in solution.
  2. Use stoichiometry to find the number of moles of O2
  3. Use the ideal gas law to find the volume of O2
  4. Does the answer make sense?

Mass H2O2 = 5.0 mL x 1.01 g/mL x .02= .101 g

Moles H2O2 = .101 g x (1 mole/34 g) = .00297 moles H2O2

Moles O2 = 1/2x .00297 = .00148

Volume O2 = .00148 moles x .082 L atm/K mol x 298 K / 1 atm = .036 L

See text 6-57

Volume = __36 mL____

  1. Explain the origin of the saying "hot air rises".

Increasing the temperature of air decreases its density. Air which is less dense will rise; this is how hot air balloons work. Using PV = nRT we can see that at constant pressure a temperature increase will cause an increase in volume so one mole of gas will now take up more space. Therefore density, mass/ volume, will be smaller.

Many of you wrote about hot molecules travelling faster. This is true, but not directly relevant. The best answers are concise and to the point and do not contain irrelevant information. To get full credit, you needed to mention "density".

See text 6-49, 50.