Homework Help: Pre-AP Chemistry

Posted by Anonymous on Thursday, October 7, 2010 at 3:09pm.

Volume (cm3 Area(cm2) Thickness (cm)
.20 cm3 100 cm2 0.002 cm
.28 cm3 122 cm2 0.002 cm
.24 cm3 144 cm2 0.017 cm
.41 cm3 225 cm2 0.0018cm

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?

Pre-AP Chemistry - Samuel Li, Sunday, October 7, 2012 at 12:37am
1. Basically, it would influence your results, making them either too high, or too low. The effect of your crude balance would be compounded. Your results likely would be too high, compared to the true value.

2. Yes, assuming the spill is uniformly spread on top of the water surface, and the surface is perfectly still. As long as these conditions persist, the thickness of the spill can be calculated if one knows the mass of the total value of the oil spilt, the density of the oil, and the surface area that the oil covers.

3. You would have to tell us what you're graphing for an adequate answer to be given.

4. d_oil = m_oil/v_oil
m_oil = d_oil * v_oil

T_oil = m_oil/(d_oil * A_oil)
T_oil = (d_oil * v_oil)/(d_oil * A_oil)
T_oil = = v_oil / A_oil
T_oil = (1000cm^3)/(2 500cm * 3 000cm)
T_oil = 1.33 x 10^-4cm

5. This method might be used in finding the size of molecules, by layering it on a flat surface with a known mass, such that it is very thin (approximately 1 molecule thin). Then, knowing the difference in mass, and the area, the size of molecules can be easily determined.

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