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Chapter 1: Problem 70

Pheromones are compounds secreted by females of many insect species to attract mates. Typically, \(1.0 \times 10^{-8} \mathrm{~g}\) of a pheromone is sufficient to reach all targeted males within a radius of 0.50 mi. Calculate the density of the pheromone (in grams per liter) in a cylindrical air space having a radius of \(0.50 \mathrm{mi}\) and a height of \(40 \mathrm{ft}\).

Short Answer

Expert verified
The density of the pheromone in the given cylindrical airspace is \( \frac{1.0 \times 10^{-8} \, \text{g}}{\pi \times (0.5)^{2} \times \frac{40}{5280} \times 4.16818183 \times 10^{9}}\) g/L.

Step by step solution

01

Convert Units of Cylinder Dimensions to Common Units

The dimensions of the cylinder are given in different units, so the first step is to convert them to a common unit. Convert height from feet to miles considering that one mile is approximately 5280 feet. So, \( 40 \, \text{ft} = \frac{40}{5280} \, \text{miles} \).
02

Calculate the Volume of the Cylinder in Miles

The volume of a cylinder is given by the formula \( \pi \times r^{2} \times h \), where \( r \) is the radius and \( h \) is the height. Substituting the given radius and the converted height into the formula, the volume in miles would be \( \pi \times (0.5)^{2} \times \frac{40}{5280} \) miles\(^{3}\)
03

Convert Volume in Miles to Liters

To get the density in grams per liter, convert the volume from miles \(\,^{3}\) to liters. The exact conversion is complex and not a round number, but for practical purposes we can say that 1 mile\(^{3}\) is approximately equal to \(4.16818183 \times 10^{9}\) liters. So, the volume in liters would be \( \pi \times (0.5)^{2} \times \frac{40}{5280} \times 4.16818183 \times 10^{9}\) liters.
04

Calculate the Density

Finally, with the mass and volume in the appropriate units, the density can be calculated using the formula density = mass/volume. So, the density of the pheromones in this cylinder is \( \frac{1.0 \times 10^{-8} \, \text{g}}{\pi \times (0.5)^{2} \times \frac{40}{5280} \times 4.16818183 \times 10^{9}}\) g/L.

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Most popular questions from this chapter

Suppose that a new temperature scale has been devised on which the melting point of ethanol \(\left(-117.3^{\circ} \mathrm{C}\right)\) and the boiling point of ethanol \(\left(78.3^{\circ} \mathrm{C}\right)\) are taken as \(0^{\circ} \mathrm{S}\) and \(100^{\circ} \mathrm{S}\), respectively, where \(\mathrm{S}\) is the symbol for the new temperature scale. Derive an equation relating a reading on this scale to a reading on the Celsius scale. What would this thermometer read at \(25^{\circ} \mathrm{C} ?\)

A slow jogger runs a mile in \(13 \mathrm{~min} .\) Calculate the speed in (a) in/s, (b) \(\mathrm{m} / \mathrm{min}\), (c) \(\mathrm{km} / \mathrm{h} .(1 \mathrm{mi}=1609 \mathrm{~m}\) \(1 \mathrm{in}=2.54 \mathrm{~cm} .)\)

In the determination of the density of a rectangular metal bar, a student made the following measurements: length, \(8.53 \mathrm{~cm}\); width, \(2.4 \mathrm{~cm}\); height, \(1.0 \mathrm{~cm}\); mass, 52.7064 g. Calculate the density of the metal to the correct number of significant figures.

A 250 -mL glass bottle was filled with 242 mL of water at \(20^{\circ} \mathrm{C}\) and tightly capped. It was then left outdoors overnight, where the average temperature was \(-5^{\circ} \mathrm{C}\). Predict what would happen. The density of water at \(20^{\circ} \mathrm{C}\) is \(0.998 \mathrm{~g} / \mathrm{cm}^{3}\) and that of ice at \(-5^{\circ} \mathrm{C}\) is \(0.916 \mathrm{~g} / \mathrm{cm}^{3}\)

Calculate the mass of each of these: (a) a sphere of gold of radius \(10.0 \mathrm{~cm}\) [the volume of a sphere of radius \(r\) is \(V=\left(\frac{4}{3}\right) \pi r^{3} ;\) the density of gold \(\left.=19.3 \mathrm{~g} / \mathrm{cm}^{3}\right]\) (b) a cube of platinum of edge length \(0.040 \mathrm{~mm}\) (the density of platinum \(\left.=21.4 \mathrm{~g} / \mathrm{cm}^{3}\right),\) (c) \(50.0 \mathrm{~mL}\) of ethanol (the density of ethanol \(=0.798 \mathrm{~g} / \mathrm{mL})\)
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