Gold, which has a density of 19.32g/cm3, is the most ductile metal and can be pressed into a thin
leaf or drawn out into a long fiber.
(a) If a sample of gold, with a mass of 27.63g, is pressed into a leaf of 1.00 μm thickness,
what is the area of the leaf?
(b) If, instead, the gold is drawn out into a cylindrical fiber of radius 2.5 μm, what is
the length of the fiber?

To solve these problems, we need to use the relationship between mass, density, and volume, as well as the geometric formulas for the volume of a thin sheet (a rectangular prism) and a cylindrical fiber.

### (a) Area of the Leaf

**Given:**
- Density of gold: \(\rho = 19.32 \, \text{g/cm}^3\)
- Mass of gold: \(m = 27.63 \, \text{g}\)
- Thickness of the leaf: \(t = 1.00 \, \mu\text{m} = 1.00 \times 10^{-4} \, \text{cm}\)

**Step 1: Calculate the volume of gold.**

We know the relationship:
\[
\text{Volume} = \frac{\text{Mass}}{\text{Density}}
\]
\[
V = \frac{27.63 \, \text{g}}{19.32 \, \text{g/cm}^3} \approx 1.430 \, \text{cm}^3
\]

**Step 2: Calculate the area of the leaf.**

Since the leaf is a rectangular prism, the volume can also be expressed as:
\[
V = \text{Area} \times \text{Thickness}
\]
\[
\text{Area} = \frac{V}{t}
\]
Substituting the values:
\[
\text{Area} = \frac{1.430 \, \text{cm}^3}{1.00 \times 10^{-4} \, \text{cm}} = 1.430 \times 10^4 \, \text{cm}^2
\]

### (b) Length of the Cylindrical Fiber

**Given:**
- Radius of the fiber: \(r = 2.5 \, \mu\text{m} = 2.5 \times 10^{-4} \, \text{cm}\)

**Step 1: Calculate the cross-sectional area of the cylinder.**

The cross-sectional area \(A\) of a cylinder is given by:
\[
A = \pi r^2
\]
Substituting the radius:
\[
A = \pi \times (2.5 \times 10^{-4} \, \text{cm})^2 \approx 1.9635 \times 10^{-7} \, \text{cm}^2
\]

**Step 2: Calculate the length of the fiber.**

The volume of the cylindrical fiber can also be expressed as:
\[
V = \text{Cross-sectional Area} \times \text{Length}
\]
\[
\text{Length} = \frac{V}{\text{Cross-sectional Area}}
\]
Substituting the values:
\[
\text{Length} = \frac{1.430 \, \text{cm}^3}{1.9635 \times 10^{-7} \, \text{cm}^2} \approx 7.284 \times 10^6 \, \text{cm} = 72.84 \, \text{km}
\]

### Final Answers:
- **(a)** The area of the leaf is **1.43 \(\times\) 10\(^4\) cm\(^2\)**.
- **(b)** The length of the fiber is **72.84 km**.

Would you like more details or have any questions?

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Here are 5 related questions:
1. How does the density of a material affect its volume for a given mass?
2. What are some practical applications of gold’s ductility in industry?
3. How would changing the thickness of the leaf affect its area?
4. What would happen to the length of the fiber if the radius were halved?
5. Can you calculate the volume of gold required to create a leaf with a specific area and thickness?

**Tip:** Always double-check unit conversions, especially when dealing with very small or very large values.

Gold, which has a density of 19.32g/cm3, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber. (a) If a sample of gold, with a mass of 27.63g, is pressed into a leaf of 1.00 μm thickness, what is the area of the leaf? (b) If, instead, the gold is drawn out into a cylindrical fiber of radius 2.5 μm, what is the length of the fiber?

This problem involves calculating the area of a gold leaf and the length of a gold fiber using the density of gold (19.32g/cm3) and its mass (27.63g). It explores key concepts in volume, density, and geometric formulas, making it suitable for students in Grades 10-12.

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