APC Chemistry
posted by Anonymous on .
Please correct me if any wrong info is given
Volume(cm3)Area (cm2)Thickness (cm)
.20 cm3 100 cm2 0.02 cm
.28 cm3 122 cm2 0.002 cm
.24 cm3 144 cm2 0.016 cm
.41 cm3 225 cm2 0.0182 cm
If you had used a very crude balance that allowed only one significant figure, how would this have affected your results for: Area? Volume? Thickness?
Could this method be used to determine the thickness of an oil spill? What information would be needed?
slope y = 49.4x  30
What does the slope tell about the thickness
By mistake, a quart of oil (1.06 quarts = 1 liter = 1000 cm3) was dumped into a
swimming pool that measures 25.0 m by 30.0 m. The density of the oil was 0.750 g/cm3. How thick was the resulting oil slick? Be careful with significant figures and exponential notation. Density is not needed to calculate the answer for this problem.
How might this method of finding thickness be used in finding the size of molecules?

First, if I multiply area by thickness I should arrive at the volume.
100 cm^2 x .02 cm = 2 cc (you have 0.2 cc)
122 cm^2 x 0.002 cm = 0.244 cc(you have 0.28 cc)
etc.
Since V = area x thichness, then
V/A = thickness. for the first one,
0.20/100 = 0.0020 cm
Second one is
0.28/122 = 0.00230. You are allowed 2 s.f.; therefore, this should be rounded to 0.0023
Third one.
0.24/144 = 0.01666. You are allowed 2 s.f.; therefore, round to 0.017
etc.