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(a) A cube of osmium metal \(1.500 \mathrm{~cm}\) on a side has a mass of \(76.31 \mathrm{~g}\) at \(25^{\circ} \mathrm{C}\). What is its density in \(\mathrm{g} / \mathrm{cm}^{3}\) at this temperature? (b) The density of titanium metal is \(4.51 \mathrm{~g} / \mathrm{cm}^{3}\) at \(25^{\circ} \mathrm{C}\). What mass of titanium displaces \(125.0 \mathrm{~mL}\) of water at \(25^{\circ} \mathrm{C}\) ? (c) The density of benzene at \(15^{\circ} \mathrm{C}\) is \(0.8787 \mathrm{~g} / \mathrm{mL}\). Calculate the mass of \(0.1500 \mathrm{~L}\) of benzene at this temperature.

Short Answer

Expert verified
(a) The density of osmium metal at 25°C is \(22.6 \frac{\mathrm{g}}{\mathrm{cm}^3}\). (b) The mass of titanium that displaces 125.0 mL of water at 25°C is 564 g. (c) The mass of 0.1500 L of benzene at 15°C is 132 g.

Step by step solution

01

Problem (a)

We're given the size of the cube (1.500 cm on a side) and its mass (76.31 g) and asked to find its density. First, we need to find the volume of the cube. Recall that the volume of a cube (V) is calculated using the formula: V = side^3. \(V = (1.500 \mathrm{~cm})^3 = 3.375 \mathrm{~cm}^3 \) Next, use the formula for density: Density = Mass / Volume. \(\rho = \frac{Mass}{Volume} = \frac{76.31 \mathrm{~g}}{3.375 \mathrm{~cm}^3} = 22.6 \frac{\mathrm{g}}{\mathrm{cm}^3} \) So, the density of osmium metal at 25°C is 22.6 g/cm³.
02

Problem (b)

The density of titanium metal is given as 4.51 g/cm³, and we need to find the mass of titanium that displaces 125.0 mL of water. First, convert the volume of water to cm³ (1 mL = 1 cm³). Volume of water = 125.0 mL = 125.0 cm³ Now, use the formula for density: Density = Mass / Volume. We need to find the mass, so we can rearrange the formula as follows: Mass = Density * Volume. \(Mass = \rho * Volume = 4.51 \frac{\mathrm{g}}{\mathrm{cm}^3} * 125.0 \mathrm{~cm}^3 = 564 \mathrm{~g} \) So, the mass of titanium that displaces 125.0 mL of water at 25°C is 564 g.
03

Problem (c)

The density of benzene at 15°C is 0.8787 g/mL. We need to find the mass of 0.1500 L of benzene at this temperature. First, convert the volume of benzene to mL (1 L = 1000 mL). Volume of benzene = 0.1500 L = 150.0 mL Now, use the formula for density: Density = Mass / Volume. Rearrange the formula to find the mass: Mass = Density * Volume. \(Mass = \rho * Volume = 0.8787 \frac{\mathrm{g}}{\mathrm{mL}} * 150.0 \mathrm{~mL} = 132 \mathrm{~g} \) So, the mass of 0.1500 L of benzene at 15°C is 132 g.

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Key Concepts

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

Understanding the Mass-Volume Relationship
The mass-volume relationship is a fundamental concept in understanding the physical properties of materials. It simply refers to how much space (volume) a certain amount of substance (mass) occupies. This is crucial when determining the density of a substance, which is a measure of how tightly matter is packed together. In the context of the textbook exercise, for the osmium metal cube, we know both the mass (76.31 g) and the volume, which is calculated from the cube's dimensions (1.500 cm on each side).

The volume calculation involves cubing the side of the cube, leading to a volume of 3.375 cm³. The relationship between mass and volume is expressed through the density formula, which is essentially the ratio of mass to volume. The cube's mass and volume are directly used in the formula to compute its density.
The Density Formula Explained
Density, symbolically represented as \( \rho \), is a characteristic property of a material. The density formula, \( \rho = \frac{Mass}{Volume} \), relates to how much mass is contained in a given volume. This formula is essential for solving the problems in our textbook exercise. For instance, with the osmium metal cube, after finding the volume, we apply the density formula to find the density at a specific temperature, 25°C.

In the context of problem (b) and (c), we know the density of the materials and are asked to find the mass corresponding to a certain volume. By rearranging the density formula, we can solve for mass as \( Mass = \rho * Volume \). Knowing how to manipulate the density formula allows us to solve a variety of practical problems, such as determining the mass of a material that would displace a certain volume of liquid, as seen with the titanium example.
Conversion of Units for Accurate Calculations
Converting units is a vital step in many scientific calculations. This step ensures that all aspects of a calculation are expressed in compatible units, leading to accurate and meaningful results. For density calculations, it's common to convert volumes between mL (milliliters) and cm³ (cubic centimeters) since 1 mL is equivalent to 1 cm³.

In problem (b) of the exercise, the volume of water (125.0 mL) is converted to cm³ to align with the given density units. Similarly, in problem (c), the volume of benzene (0.1500 L) is converted to mL before applying the density formula. Remember to convert liters to milliliters by multiplying by 1000, as 1 L equals 1000 mL. Mastering unit conversions is crucial when working with various measurements, avoiding errors, and ensuring that your results correctly reflect the physical quantities being measured.

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Most popular questions from this chapter

In 2009, a team from Northwestern University and Western Washington University reported the preparation of a new "spongy" material composed of nickel, molybdenum, and sulfur that excels at removing mercury from water. The density of this new material is \(0.20 \mathrm{~g} / \mathrm{cm}^{3}\), and its surface area is \(1242 \mathrm{~m}^{2}\) per gram of material. (a) Calculate the volume of a 10.0-mg sample of this material. (b) Calculate the surface area for a \(10.0\)-mg sample of this material. (c) A \(10.0\)-mL sample of contaminated water had \(7.748 \mathrm{mg}\) of mercury in it. After treatment with \(10.0 \mathrm{mg}\) of the new spongy material, \(0.001 \mathrm{mg}\) of mercury remained in the contaminated water. What percentage of the mercury was removed from the water? (d) What is the final mass of the spongy material after the exposure to mercury?

Give the chemical symbol or name for the following elements, as appropriate: (a) sulfur, (b) gold, (c) potassium, (d) chlorine, (e) copper, (f) U, (g) Ni, (h) Na, (i) Al, (j) Si.

The inside dimension of a box that is cubic is \(24.8 \mathrm{~cm}\) on each edge with an uncertainty of \(0.2 \mathrm{~cm}\). What is the volume of the box? What do you estimate to be the uncertainty in the calculated volume?

By using estimation techniques, determine which of the following is the heaviest and which is the lightest: a 5-lb bag of potatoes, a 5-kg bag of sugar, or \(1 \mathrm{gal}\) of water (density \(=1.0 \mathrm{~g} / \mathrm{mL}\) ).

Gold is alloyed (mixed) with other metals to increase its hardness in making jewelry. (a) Consider a piece of gold jewelry that weighs \(9.85 \mathrm{~g}\) and has a volume of \(0.675 \mathrm{~cm}^{3}\). The jewelry contains only gold and silver, which have densities of \(19.3\) and \(10.5 \mathrm{~g} / \mathrm{cm}^{3}\), respectively. If the total volume of the jewelry is the sum of the volumes of the gold and silver that it contains, calculate the percentage of gold (by mass) in the jewelry. (b) The relative amount of gold in an alloy is commonly expressed in units of carats. Pure gold is 24 carat, and the percentage of gold in an alloy is given as a percentage of this value. For example, an alloy that is \(50 \%\) gold is 12 carat. State the purity of the gold jewelry in carats.

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