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Chapter 1: Problem 71

The density of osmium (the densest metal) is \(22.57 \mathrm{~g} / \mathrm{cm}^{3} .\) If a \(1.00-\mathrm{kg}\) rectangular block of osmium has two dimensions of \(4.00 \mathrm{~cm} \times 4.00 \mathrm{~cm}\), calculate the third dimension of the block.

Short Answer

Expert verified
The third dimension (height) of the 1.00-kg rectangular block of osmium with dimensions of \(4.00 \mathrm{~cm} \times 4.00 \mathrm{~cm}\) is 2.79 cm.

Step by step solution

01

Write down the given information.

We are given the following information: Density of osmium, \(蟻 = 22.57\) g/cm鲁. Mass of osmium block, m = 1.00 kg (We need to convert this to grams). Dimensions of osmium block (Length 脳 Width 脳 Height): \(4.00 \mathrm{~cm} \times 4.00 \mathrm{~cm} \times h\). (Note: The third dimension (height) is what we need to calculate.)
02

Convert the mass into grams.

Given that 1 kg = 1000 g, we can convert the mass of the osmium block into grams: Mass, m = 1.00 kg 脳 (1000 g/kg) = 1000 g
03

Write down the formula relating mass, density, and volume.

The formula we will use is: mass = density 脳 volume. In this case, mass (m) = 1000 g, density (蟻) = 22.57 g/cm鲁, and volume (V) = Length 脳 Width 脳 Height.
04

Calculate the volume of the osmium block.

Using the mass and density values, we can calculate the volume of the osmium block. Rearrange the density formula to solve for volume: Volume, V = mass / density = m / 蟻 Now, plug in the values for mass and density: V = 1000 g / 22.57 g/cm鲁 = 44.31 cm鲁
05

Calculate the third dimension (height).

Now, we know the volume (44.31 cm鲁) and two dimensions (4.00 cm and 4.00 cm) of the block. We can find the third dimension (height) by dividing the volume by the product of the known dimensions: Height, h = V / (Length 脳 Width) Plugging in the values, we get: h = 44.31 cm鲁 / (4.00 cm 脳 4.00 cm) = 2.79 cm
06

State the final answer.

The third dimension (height) of the 1.00-kg rectangular block of osmium with dimensions of \(4.00 \mathrm{~cm} \times 4.00 \mathrm{~cm}\) is 2.79 cm.

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Key Concepts

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

Mass to Volume Ratio
Understanding the 'mass to volume ratio' is crucial in executing various scientific calculations, especially when dealing with substances鈥� physical properties. This ratio, most commonly referred to as density, is a measure of how much mass is contained within a unit volume of a material. It can be intuitively understood as how tightly packed the matter within a substance is.

For example, in the given exercise, we deal with osmium, which is known as the densest metal. Its mass to volume ratio is given as 22.57 g/cm鲁. This number tells us that for every cubic centimeter of osmium, there is 22.57 grams of mass. Knowing this ratio enables us to determine one of the most enigmatic properties: the actual volume that a certain mass of a substance will occupy, which is foundational in solving problems related to material properties.

Let's put this into practice. With the osmium example, if we have a 1.00 kg block, we can find out how much space it occupies through the density value. Using the formula for volume (V = m/蟻), where m is mass and 蟻 (rho) is density, we can calculate the volume in cubic centimeters for the given mass. This part of the calculation is central since the mass to volume ratio is directly linked to the concept of density.
Converting Mass Units
In many scientific problems, 'converting mass units' from one system to another is an essential skill, as it allows for the consistent use of formulas that require specific units. Conversions are necessary because measurements can be recorded in different units, and using them incorrectly could lead to errors in calculations and results.

For example, in the exercise, the mass of the osmium block is provided in kilograms (kg), but the density is given in terms of grams per cubic centimeter (g/cm鲁). It鈥檚 important to convert the mass into grams to use the density formula effectively because the units must match for proper computation. This translates to multiplying the mass by 1000 because one kilogram equals 1000 grams.

In this context, after converting 1.00 kg into 1000 g, we align the units with those of density, thus enabling us to apply the formula for density without hitches. Mastery of unit conversion is a fundamental skill in not just chemistry or physics, but across all sciences.
Density Formula Application
Applying the 'density formula' is at the core of many problems in physics and chemistry. The formula is simply density (蟻) = mass (m) / volume (V). It is used to find any of the three variables when the other two are known. Applying this formula correctly is vital for both theoretical understanding and practical experiments.

In our problem, we use the formula to find the unknown dimension of a block of osmium. With the mass and density known, we rearrange the formula to find the volume (V = m/蟻). After determining the volume, if the material is of a regular shape, like the rectangular block in our problem, we can then calculate any missing dimension by manipulating the formula further. We divide the volume by the product of the known dimensions to tease out the unknown third dimension.

The iterative application of the density formula is a powerful tool, allowing students to navigate complex problems by breaking them down into more manageable steps. This particular application brings in a sense of accomplishment as students use it not only to comprehend theoretical concepts but also to tackle everyday practical questions.

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Most popular questions from this chapter

An experiment was performed in which an empty \(100-\mathrm{mL}\) graduated cylinder was weighed. It was weighed once again after it had been filled to the \(10.0-\mathrm{mL}\) mark with dry sand. A \(10-\mathrm{mL}\) pipet was used to transfer \(10.00 \mathrm{~mL}\) of methanol to the cylinder. The sand- methanol mixture was stirred until bubbles no longer emerged from the mixture and the sand looked uniformly wet. The cylinder was then weighed again. Use the data obtained from this experiment (and displayed at the end of this problem) to find the density of the dry sand, the density of methanol, and the density of sand particles. Does the bubbling that occurs when the methanol is added to the dry sand indicate that the sand and methanol are reacting? Mass of cylinder plus wet sand \(\quad 45.2613 \mathrm{~g}\) Mass of cylinder plus dry sand \(\quad 37.3488 \mathrm{~g}\) Mass of empty cylinder \(22.8317 \mathrm{~g}\) Volume of dry sand \(10.0 \mathrm{~mL}\) Volume of sand plus methanol \(\quad 17.6 \mathrm{~mL}\) Volume of methanol \(\quad 10.00 \mathrm{~mL}\)

The density of an irregularly shaped object was determined as follows. The mass of the object was found to be \(28.90 \mathrm{~g} \pm 0.03 \mathrm{~g}\). A graduated cylinder was partially filled with water. The reading of the level of the water was \(6.4 \mathrm{~cm}^{3} \pm 0.1 \mathrm{~cm}^{3}\). The object was dropped in the cylinder, and the level of the water rose to \(9.8 \mathrm{~cm}^{3} \pm 0.1 \mathrm{~cm}^{3}\). What is the density of the object with appropriate error limits? (See Appendix 1.5.)

In each of the following pairs, which has the greater volume? a. \(1.0 \mathrm{~kg}\) of feathers or \(1.0 \mathrm{~kg}\) of lead b. \(100 \mathrm{~g}\) of gold or \(100 \mathrm{~g}\) of water c. \(1.0 \mathrm{~L}\) of copper or \(1.0 \mathrm{~L}\) of mercury

How many significant figures are there in each of the following values? a. \(6.07 \times 10^{-15}\) e. \(463.8052\) b. \(0.003840\) f. 300 c. \(17.00\) g. 301 d. \(8 \times 10^{8}\) h. 300 .

You have two beakers, one filled to the \(100-\mathrm{mL}\) mark with sugar (the sugar has a mass of \(180.0 \mathrm{~g}\) ) and the other filled to the \(100-\mathrm{mL}\) mark with water (the water has a mass of \(100.0 \mathrm{~g}\) ). You pour all the sugar and all the water together in a bigger beaker and stir until the sugar is completely dissolved. a. Which of the following is true about the mass of the solution? Explain. i. It is much greater than \(280.0 \mathrm{~g}\). ii. It is somewhat greater than \(280.0 \mathrm{~g}\). iii. It is exactly \(280.0 \mathrm{~g}\). iv. It is somewhat less than \(280.0 \mathrm{~g}\). v. It is much less than \(280.0 \mathrm{~g}\). b. Which of the following is true about the volume of the solution? Explain. i. It is much greater than \(200.0 \mathrm{~mL}\). ii. It is somewhat greater than \(200.0 \mathrm{~mL}\). iii. It is exactly \(200.0 \mathrm{~mL}\). iv. It is somewhat less than \(200.0 \mathrm{~mL}\). v. It is much less than \(200.0 \mathrm{~mL}\).
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