Question

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

Short answer

(a) Mass of gold sphere: 80893.65 g (b) Mass of platinum cube: 1.37 μg (c) Mass of ethanol: 39.9 g

Step-by-step solution

Calculate the Volume of the Gold Sphere

Use the formula for the volume of a sphere: \[ V = \frac{4}{3} \pi r^3 \]Given that the radius \(r\) is \(10.0\, \text{cm}\), substitute this into the formula:\[ V = \frac{4}{3} \pi (10.0)^3 = \frac{4}{3} \pi \times 1000 = \frac{4000}{3} \pi \ approximate 4188.79\, \text{cm}^3 \]

Calculate the Mass of the Gold Sphere

Using the volume calculated in Step 1 and the density of gold (\(19.3\, \text{g/cm}^3\)), apply the mass formula:\[ \text{Mass} = \text{Density} \times \text{Volume} \]\[ \text{Mass} = 19.3\, \text{g/cm}^3 \times 4188.79\, \text{cm}^3 \approx 80893.65\, \text{g} \]

Convert Edge Length of Platinum Cube to Centimeters

The edge length of the platinum cube is given as \(0.040\, \text{mm}\). Convert this to centimeters by noting that \(1\, \text{mm} = 0.1\, \text{cm}\):\[ 0.040\, \text{mm} = 0.0040\, \text{cm} \]

Calculate the Volume of the Platinum Cube

Use the formula for the volume of a cube \[ V = a^3 \]Where \(a\) is the edge length. For the platinum cube with edge \(0.0040\, \text{cm}\):\[ V = (0.0040)^3 = 6.4 \times 10^{-8}\, \text{cm}^3 \]

Calculate the Mass of the Platinum Cube

Using the volume calculated in Step 4 and the density of platinum \(21.4\, \text{g/cm}^3\):\[ \text{Mass} = 21.4\, \text{g/cm}^3 \times 6.4 \times 10^{-8}\, \text{cm}^3 \approx 1.37 \times 10^{-6}\, \text{g} \]

Calculate the Mass of Ethanol

Given the volume of ethanol as \(50.0\, \text{mL}\) and the density as \(0.798\, \text{g/mL}\), use the formula:\[ \text{Mass} = \text{Density} \times \text{Volume} \]\[ \text{Mass} = 0.798\, \text{g/mL} \times 50.0\, \text{mL} = 39.9\, \text{g} \]

Key concepts

Volume of a Sphere

The volume of a sphere is an essential concept when dealing with spherical objects. To calculate the volume of a sphere, use the formula: \[ V = \frac{4}{3} \pi r^3 \]where \( V \) is the volume and \( r \) is the radius of the sphere. This formula arises from integrating the surface area of a sphere.

• Importance: Knowing the volume helps in understanding how much space a sphere occupies.
• Application: Used in physics, engineering, and various fields where spherical objects are involved.

For example, if you have a sphere with a radius of 10 cm, you simply substitute this into the formula to find the sphere's volume, which would be approximately 4188.79 cubic centimeters.
Calculating volume is an essential skill in science and engineering tasks, helping to determine quantities or design parameters.

Density of Substances

Density is a fundamental property of materials that indicates how much mass an object has in a given volume. It is calculated using the formula: \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]

• Units: Typically expressed in grams per cubic centimeter (g/cm³) or kilograms per cubic meter (kg/m³).
• Significance: Determines whether an object will float or sink in a fluid.

For instance, gold has a high density of 19.3 g/cm³, meaning it is quite heavy for its size.
Understanding density is vital for applications in material science, chemistry, and mechanical engineering.
This property helps in comparing different substances and understanding their suitability for various uses.

Conversion of Units

Converting units is a crucial skill in solving problems across different measurement systems.
To convert accurately, you must multiply by the conversion factor that accompanies your current unit:

• Examples: Convert mm to cm by multiplying by 0.1.
• Why: Consistent units are necessary for accurate calculations.

For example, converting 0.040 mm to cm involves multiplying by 0.1, resulting in 0.004 cm.
These conversions are useful when dealing with international standards or when measuring instruments are calibrated in different units.
Mastering this skill ensures accuracy and clarity in scientific and engineering communication.

Volume of a Cube

The volume of a cube is easy to compute with its simple geometric shape.
The formula for the volume of a cube is: \[ V = a^3 \]

• Definition: \( a \) represents the length of one edge of the cube.
• Application: Useful in packaging, material use, and space planning tasks.

For example, if a cube has an edge length of 0.004 cm, substituting this into the formula yields a volume of \( 6.4 \times 10^{-8} \) cubic centimeters.
Calculating the volume of a cube is fundamental in various disciplines involving regular geometric shapes.
It is a building block for more complex mathematical and engineering solutions.

Mass Formula

The mass formula is a crucial tool in determining the mass of an object from its volume and density. The formula is expressed as: \[ \text{Mass} = \text{Density} \times \text{Volume} \]

• Purpose: It allows calculation of mass when given density and volume.
• Application: Used in chemistry, physics, and engineering.

For instance, with a known volume of 50 mL and density of 0.798 g/mL for ethanol, the mass can be calculated as approximately 39.9 g.
This formula is indispensable in practical applications like solution preparations where specific amounts of substances are required.
Understanding this helps predict how substances will behave in different environments based on their mass properties.