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Chapter 2: Problem 44

The measured density of lead is \(11.4 \mathrm{~g} / \mathrm{mL}\). What volume in milliliters will \(1.50\) pounds of lead occupy? \([1\) pound \(=453.6 \mathrm{~g}]\)

Short Answer

Expert verified
1.50 pounds of lead will occupy 59.7 milliliters of volume.

Step by step solution

01

Convert mass from pounds to grams

To convert the mass of lead from pounds to grams, use the conversion factor given in the question: 1 pound = 453.6 grams. Multiply the mass in pounds by the conversion factor: \(1.50 \textrm{ pounds} \times \frac{453.6 \textrm{ grams}}{1 \textrm{ pound}}\).
02

Calculate the mass of lead in grams

After converting the mass of lead to grams, the equation looks like the following: \(1.50 \textrm{ pounds} \times \frac{453.6 \textrm{ grams}}{1 \textrm{ pound}} = 680.4 \textrm{ grams}\). So, the mass of lead is 680.4 grams.
03

Use the density formula to calculate the volume of lead

The density formula is defined as Density = Mass / Volume, or \(D = \frac{M}{V}\). We can rearrange this formula to solve for the volume: Volume = Mass / Density, or \(V = \frac{M}{D}\).
04

Substitute the values and calculate the volume

Now, substitute the values of mass and density into the volume formula: \(V = \frac{680.4 \textrm{ grams}}{11.4 \textrm{ g/mL}}\).
05

Calculate the volume of lead in milliliters

After substituting the values, the equation looks like the following: \(V = \frac{680.4 \textrm{ grams}}{11.4 \textrm{ g/mL}} = 59.7 \textrm{ mL}\). So, 1.50 pounds of lead will occupy 59.7 milliliters of volume.

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Most popular questions from this chapter

Indicate the correct number of significant figures for each answer, given that all values are measurements: (a) \(\left(6.350 \times 10^{-8}\right) \times(0.0080)\) (b) \(\left(5.30 \times 10^{2}\right)+\left(22.1 \times 10^{2}\right)\) (c) \(\left(5.830 \times 10^{2}\right)+\left(22.100 \times 10^{2}\right)\) (d) \(\frac{100.0 \times 0.1500}{58.443}\) (e) \(\frac{100.0 \times 0.15}{58.4}\)

Convert: (a) \(23.0{ }^{\circ} \mathrm{C}\) to \(\mathrm{K}\) (b) \(98.6^{\circ} \mathrm{F}\) to \({ }^{\circ} \mathrm{C}\) (c) \(296 \mathrm{~K}\) to \({ }^{\circ} \mathrm{F}\) (d) \(32^{\circ} \mathrm{F}\) to \(\mathrm{K}\) (e) \(523 \mathrm{~K}\) to \({ }^{\circ} \mathrm{C}\) (f) \(38^{\circ} \mathrm{C}\) to \({ }^{\circ} \mathrm{F}\)

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Explain how determining the number of significant figures allowed in an answer when measured values are multiplied or divided is different from determining the number of significant figures allowed in an answer when measured values are added or subtracted.
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