Problem 118
Question
\(\mathrm{A} 1.0 \mathrm{cm}^{3}\) block of gold can be flattened to a thin sheet that averages \(3.0 \times 10^{-8} \mathrm{cm}\) thick. What is the area (in \(\mathrm{cm} 2 )\) of the flattened gold sheet? (Chapter 2 )
Step-by-Step Solution
Verified Answer
The area of the flattened gold sheet is approximately \(3.33 \times 10^7 \, cm^2\).
1Step 1: Identify the given information
The volume of the gold block is given as \(1.0 \, cm^3\), and the thickness of the flattened gold sheet is given as \(3.0 \times 10^{-8} \, cm\).
2Step 2: Write the equation relating volume and area
Volume is the product of the area and thickness. For the gold block, we have the following equation:
Original volume = Flattened sheet's volume
\(1.0 \, cm^3 = A \times (3.0 \times 10^{-8} \, cm)\)
Here, A is the area of the flattened gold sheet, which we need to find.
3Step 3: Solve for the area of the flattened gold sheet
To solve for the area A, we need to divide both sides of the equation by the thickness of the flattened sheet:
\(A = \frac{1.0 \, cm^3}{3.0 \times 10^{-8} \, cm}\)
4Step 4: Calculate the area A
Now, we can calculate the area A by dividing the given numbers:
\(A = \frac{1.0}{3.0 \times 10^{-8}}\)
\(A = 3.33 \times 10^7 \, cm^2\)
The area of the flattened gold sheet is approximately \(3.33 \times 10^7 \, cm^2\).
Key Concepts
Density and VolumeThin Film ThicknessUnit ConversionMathematical Problem-Solving
Density and Volume
Understanding the relationship between density and volume is fundamental in many scientific calculations. Density is defined as mass per unit volume, denoted as \( \rho = \frac{m}{V} \), where \( m \) is mass and \( V \) is volume. When analyzing objects like the gold block, knowing its density can help us calculate how much it will weigh given its volume.
For solids like gold, you may often work in
For solids like gold, you may often work in
- Cubic centimeters (\( cm^3 \)) for volume
- Grams per cubic centimeter (\( g/cm^3 \)) for density
Thin Film Thickness
Thin film thickness is a term frequently used in technology and physics. It refers to the thickness of materials that are much thinner compared to their width and length, like our gold sheet.
This concept is critical in the task because knowing the thickness allows you to use the volume equation to find other properties:
This concept is critical in the task because knowing the thickness allows you to use the volume equation to find other properties:
- The mathematical relationship: \( V = A \times d \) where \( V \) is volume, \( A \) is area, and \( d \) is thickness.
Unit Conversion
Unit conversion is indispensable in scientific problem-solving. It ensures all measurements align in compatible units for accurate calculations. For instance, ensure all dimensions are in the same units, typically in centimeters when dealing with small objects like our gold block.
In this example:
In this example:
- The volume is in \( cm^3 \)
- The thickness in \( cm \)
- The desired area also in \( cm^2 \)
Mathematical Problem-Solving
Effective mathematical problem-solving involves clearly understanding the problem, formulating relevant equations, and systematically solving for unknowns. In this instance:- Identify what is given (volume and thickness) and what to find (area).- Construct the equation from known relationships \( V = A \times d \).- Solve for the missing variable (area):\[ A = \frac{V}{d} \]Once values are plugged in, perform the calculations accurately.The step-by-step breakdown provides a logical flow, turning complex-seeming questions into manageable tasks. Mastery in translating physical problems into mathematical terms elevates problem-solving from confusing to intuitive. Embrace clear organization, practice regularly, and reflect on mistakes to enhance your skills.
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