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Chapter 1: Problem 32

The volume of a certain bacterial cell is \(2.56 \mu \mathrm{m}^{3} .\) (a) What is its volume in cubic millimeters (mm \(^{3}\) )? (b) What is the volume of \(10^{5}\) cells in liters (L)?

Short Answer

Expert verified
The volume in cubic millimeters is \(2.56 \times 10^{-9} \text{mm}^3\). The volume of \(10^5\) cells in liters is \(2.56 \times 10^{-10} \text{L}\).

Step by step solution

01

- Convert Volume to Cubic Millimeters

First, convert the volume from cubic micrometers (\textmu\text{m}^3) to cubic millimeters (mm^3). We use the conversion factor: \(1 \textmu\text{m} = 1 \times 10^{-3} \text{mm}\). Therefore, \((1 \textmu\text{m})^3 = (1 \times 10^{-3} \text{mm})^3 = 1 \times 10^{-9} \text{mm}^3\). Thus, the volume in cubic millimeters is \(2.56 \text \textmu\text{m}^3 \times 1 \times 10^{-9} \text{mm}^3/\text{\textmu\text{m}^3} = 2.56 \times 10^{-9} \text{mm}^3\).
02

- Calculate Volume of 10^5 Bacterial Cells in Cubic Millimeters

Next, find the total volume for \(10^5\) bacterial cells. Multiply the volume of one cell by \(10^5\): \(2.56 \times 10^{-9} \text{mm}^3 \times 10^5 = 2.56 \times 10^{-4} \text{mm}^3 \).
03

- Convert Volume to Liters

Finally, convert the volume from cubic millimeters (mm^3) to liters (L). Use the conversion factor: \(1 \text{mm}^3 = 1 \times 10^{-6} \text{L}\). Thus, the volume in liters is: \(2.56 \times 10^{-4} \text{mm}^3 \times 1 \times 10^{-6} \text{L}/\text{mm}^3 = 2.56 \times 10^{-10} \text{L}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Volume Conversion
Understanding volume conversion is crucial in many scientific fields, including chemistry and biology. It allows scientists to compare and analyze quantities that are presented in different units. When converting volume between units, it's essential to understand the relationship between these units.
The basic principle of volume conversion involves multiplying or dividing by specific conversion factors. For instance, converting from smaller units to larger units typically involves dividing by a conversion factor, while converting from larger units to smaller units involves multiplying.
Cubic Micrometers to Cubic Millimeters
To convert volume from cubic micrometers (\textmu\text{m}^3) to cubic millimeters (mm^3), you need to know the relationship between these two units. 1 micrometer (\textmu\text{m}) is equal to 1 x 10^-3 millimeters (mm).
When dealing with volume, you cube that relationship: \((1 \textmu\text{m})^3 = (1 \times 10^{-3} \text{mm})^3 = 1 \times 10^{-9} \text{mm}^3\)
Thus, to convert 2.56 \textmu\text{m}^3 to mm^3, we multiply the volume in micrometers by \(1 \times 10^{-9}\).
So, \(2.56 \textmu\text{m}^3 = 2.56 \times 10^{-9} \text{mm}^3\).This little volume is typical for bacterial cells, showing how tiny these units are.
Cubic Millimeters to Liters
Converting from cubic millimeters (mm^3) to liters (L) is another important volume conversion. The conversion factor between these units is: 1 mm^3 equals 1 x 10^-6 liters.
When we have the total volume of bacterial cells in cubic millimeters, we can convert it to liters for a broader perspective. For example, after finding the total volume of \(10^5\) bacterial cells is \(2.56 \times 10^{-4}\text{mm}^3\), next, we use the conversion factor:\(2.56 \times 10^{-4} \text{mm}^3 \times 1 \times 10^{-6} \text{L/mm}^3 = 2.56 \times 10^{-10} \text{L}\).Such conversions are essential, especially in fields where we deal with vastly different scales, enhancing our understanding of these measurements.

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Most popular questions from this chapter

Underline the significant zeros in the following numbers: (d) \(4.0100 \times 10^{4}\) (a) \(0.41 ;\) (b) \(0.041 ;\) (c) 0.0410

Earth's surface area is \(5.10 \times 10^{8} \mathrm{~km}^{2} ;\) its crust has a mean thickness of \(35 \mathrm{~km}\) and a mean density of \(2.8 \mathrm{~g} / \mathrm{cm}^{3}\). The two most abundant elements in the crust are oxygen \(\left(4.55 \times 10^{5} \mathrm{~g} / \mathrm{t},\right.\) where \(\mathrm{t}\) stands for "metric ton"; \(1 \mathrm{t}=1000 \mathrm{~kg}\) ) and silicon ( \(2.72 \times 10^{5} \mathrm{~g} / \mathrm{t}\) ), and the two rarest nonradioactive elements are ruthenium and rhodium, each with an abundance of \(1 \times 10^{-4} \mathrm{~g} / \mathrm{L}\). What is the total mass of each of these elements in Earth's crust?

Asbestos is a fibrous silicate mineral with remarkably high tensile strength. But it is no longer used because airborne asbestos particles can cause lung cancer. Grunerite, a type of asbestos, has a tensile strength of \(3.5 \times 10^{2} \mathrm{~kg} / \mathrm{mm}^{2}\) (thus, a strand of grunerite with a \(1-\mathrm{mm}^{2}\) cross-sectional area can hold up to \(3.5 \times 10^{2} \mathrm{~kg}\) ). The tensile strengths of aluminum and Steel No. 5137 are \(2.5 \times 10^{4} \mathrm{lb} / \mathrm{in}^{2}\) and \(5.0 \times 10^{4} \mathrm{lb} / \mathrm{in}^{2},\) respectively. Calculate the cross-sectional areas (in \(\mathrm{mm}^{2}\) ) of wires of aluminum and of Steel No. 5137 that have the same tensile strength as a fiber of grunerite with a crosssectional area of \(1.0 \mu \mathrm{m}^{2}\).

Bromine is used to prepare the pesticide methyl bromide and flame retardants for plastic electronic housings. It is recovered from seawater, underground brines, and the Dead Sea. The average concentrations of bromine in seawater \((d=1.024 \mathrm{~g} / \mathrm{mL})\) and the Dead Sea \((d=1.22 \mathrm{~g} / \mathrm{mL})\) are \(0.065 \mathrm{~g} / \mathrm{L}\) and \(0.50 \mathrm{~g} / \mathrm{L},\) respectively. What is the mass ratio of bromine in the Dead Sea to that in seawater?

For each pair, which has higher potential energy? (a) The fuel in your car or the gaseous products in its exhaust (b) Wood in a fire or the ashes after the wood burns
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