Name_____Average 6.1/10_________________________
Chemistry 101 
Quiz #4—Chapters 4 and 5(part)
1. A typical bacterial cell has a diameter of 1 uM and a volume of
5.2 x 10 –19 m3 and contains 10 molecules of DNA polymerase III, the enzyme responsible for copying the cell’s DNA. During the course of purification of the DNA polymerase, 10 9 bacterial cells are broken up or lysed and mixed with water such that the final volume is 15.0 mL.
What is the final molar concentration of DNA polymerase III?
This is a lesson in when
not to use M1V1 = M2V2
10 DNA pol /
cell x 10 9 cells = 10 10 DNA pol
10 10
DNA pol x 1 mol / 6.02 x 10 23 DNA pol = 1.66 x 10 –14
moles DNA pol
[DNA pol] = 1.66
x 10 –14 moles DNA pol / 0.015 L = 1.1 x 10 –12
M
2. Some commercial
drain cleaners contain two components: sodium hydroxide and aluminum
powder. When the mixture is poured
down a clogged drain, the following reaction occurs:
2 NaOH (aq) + 2 Al (s) + 2 H2O à 2 NaAl(OH)4 (aq) + 2 H2 (g)
The heat generated in this
reaction helps melt away obstructions such as grease, and the hydrogen gas
released stirs up the solids clogging the drain.
Calculate the volume of H2
formed at 25 °C and 1.0 atm if 3.12 g of Al is treated with an excess of
NaOH and water.
Strategy 1) find
the number of moles of Al 2)find the number of moles of H2 3)Find the volume of
H2
3.12 g x 1 mol/
26.98 g = 0.116 moles Al
0.116 moles Al x
2 moles H2 / 2 moles Al = 0.116 moles H2
V = nRT /P =
0.116 moles x 0.0802 L atm/ K mol x (273 + 25 K) / 1 atm
V = 2.8 L (2 sf)
Potentially Useful
Information-- Na = 6.022 x 10 23 atoms or molecules / mole.
R = 0.082 L. atm./ K. mol = 8.314 J/ K.mol
1 L = 1 dm 3 T in Kelvin = 273.15 + T (°C)
|
Element |
Atomic Number |
Mass ( g /mol) |
|
H |
1 |
1.008 |
|
Al |
13 |
26.98 |
Good luck and have a great Fall break!