Calculate the thickness of the monolayer assuming that the volume of the monolayer is 7.41×10−6 mL and diameter of the watch glass is 5 cm.

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• Think of it as a tiny cylinder.

A mL is equal to 1 cubic cm

V = pi * r^2 * h

7.41 * 10^(-6) cubic cm = pi * (5/2 cm)^2 * h

Solve for h

7.41 * 10^(-6) / (pi * (5/2)^2) = h

7.41 * 10^(-6) / (pi * (25/4)) = h

h = 4 * 7.41 * 10^(-6) / (pi * 25)

h = 4 * 4 * 7.41 * 10^(-6) / (pi * 25 * 4)

h = 16 * 7.41 * 10^(-6) / (pi * 100)

h = (16 * 7.41 / pi) * 10^(-6 - 2)

h = (16 * 7.41 / (22/7)) * 10^(-8)

An approximation of pi = 22/7 at this point isn't going to change much

h = (16 * 7 * 7.41 / 22) * 10^(-8) cm

h = (8 * (49 + 2.87) / 11) * 10^(-8) cm

h = (8 * 51.87 / 11) * 10^(-8) cm

h = ((408 + 6.96)/11) * 10^(-8) cm

h = (414.96 / 11) * 10^(-8) cm

h = ((407 + 7.96)/11) * 10^(-8) cm

h = (37 + (7.7 + 0.22 + 0.04)/11) * 10^(-8) cm

h = (37 + 0.7 + 0.02 + 0.0036363636....) * 10^(-8) cm

h = 37.7236363636... * 10^(-8) cm

h = 3.772363636... * 10^(1) * 10^(-8) cm

h = 3.772363636... * 10^(-7) cm

h = 3.77 * 10^(-7) cm

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• The volume of a thin layer is (area)*(thickness), and the area inside a circle of diameter d is πd²/4, so:

7.41 * 10^-6 mL = [π (5 cm)² / 4] * x . . . . where x is the unknown thickness

4 * (7.41 * 10^-6 cm³) = πx * (25 cm²)

x = (2.964 * 10^-5 cm³) / (25π cm²) ~~ 3.77 * 10-7 cm

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