According to the US Mint, the standard gold bar is 17.78 cm by 20.75 cm by 4.445 cm. Calculate the volume of the gold bar in units of cubic centimeters. Write Equation Rearrange Equation Substitute Numbers with Units Calculate Final Answer with Correct Number, Units; SF The density of pure gold is 19.3 g/cm^3. Use the volume of a gold bar from question 3 to calculate the mass in units of grams for a single gold bar. Write Equation Rearrange Equation Substitute Numbers with Units Calculate Final Answer with Correct Number, Units.
The Correct Answer and Explanation is:
Volume and Mass of a Standard Gold Bar
Part 1: Calculate the Volume of the Gold Bar
Given:
- Length = 17.78 cm
- Width = 20.75 cm
- Height = 4.445 cm
Step 1: Write the Equation
Volume=Length×Width×Height\text{Volume} = \text{Length} \times \text{Width} \times \text{Height}Volume=Length×Width×Height
Step 2: Substitute the Numbers (with Units)
Volume=17.78 cm×20.75 cm×4.445 cm\text{Volume} = 17.78\ \text{cm} \times 20.75\ \text{cm} \times 4.445\ \text{cm}Volume=17.78 cm×20.75 cm×4.445 cm
Step 3: Calculate
Volume=17.78×20.75×4.445=1643.997075 cm3\text{Volume} = 17.78 \times 20.75 \times 4.445 = 1643.997075\ \text{cm}^3Volume=17.78×20.75×4.445=1643.997075 cm3
Step 4: Round to Correct Significant Figures
All measurements have 4 significant figures, so:Volume=1644 cm3\text{Volume} = \boxed{1644\ \text{cm}^3}Volume=1644 cm3
Part 2: Calculate the Mass of the Gold Bar
Given:
- Volume = 1644 cm³
- Density of gold = 19.3 g/cm³
Step 1: Write the Equation
Mass=Density×Volume\text{Mass} = \text{Density} \times \text{Volume}Mass=Density×Volume
Step 2: Substitute the Numbers (with Units)
Mass=19.3 g/cm3×1644 cm3\text{Mass} = 19.3\ \text{g/cm}^3 \times 1644\ \text{cm}^3Mass=19.3 g/cm3×1644 cm3
Step 3: Calculate
Mass=19.3×1644=31729.2 g\text{Mass} = 19.3 \times 1644 = 31729.2\ \text{g}Mass=19.3×1644=31729.2 g
Step 4: Round to Correct Significant Figures
Both values have 3 or 4 significant figures; answer should have 3 significant figures (based on the least precise value, 19.3):Mass=3.17×104 g\text{Mass} = \boxed{3.17 \times 10^4\ \text{g}}Mass=3.17×104 g
Explanation (Textbook-Style, ~300 Words)
To determine the volume of a standard gold bar, we multiply its three dimensions (length, width, and height), each given in centimeters, because volume in cubic centimeters is calculated from multiplying three linear measurements in centimeters. The formula for volume of a rectangular solid is:Volume=Length×Width×Height\text{Volume} = \text{Length} \times \text{Width} \times \text{Height}Volume=Length×Width×Height
Using the values provided—17.78 cm (length), 20.75 cm (width), and 4.445 cm (height)—the product of these three gives the volume:17.78×20.75×4.445=1643.997075 cm317.78 \times 20.75 \times 4.445 = 1643.997075\ \text{cm}^317.78×20.75×4.445=1643.997075 cm3
When rounding for significant figures, we consider the measurement with the fewest significant figures, which in this case is four, so the final volume is rounded to 1644 cm³.
Next, using the density formula, which relates mass to volume through:Mass=Density×Volume\text{Mass} = \text{Density} \times \text{Volume}Mass=Density×Volume
and the known density of gold (19.3 g/cm³), we multiply this by the volume to get:19.3 g/cm3×1644 cm3=31729.2 g19.3\ \text{g/cm}^3 \times 1644\ \text{cm}^3 = 31729.2\ \text{g}19.3 g/cm3×1644 cm3=31729.2 g
The mass is then rounded to 3 significant figures because 19.3 has 3 significant figures. Therefore, the mass of one standard gold bar is 3.17 × 10⁴ grams, or 31.7 kilograms, since 1,000 grams = 1 kilogram. This calculation is important in finance and science, where precision matters in valuing precious metals.
