(a) If you had 1 mole of M&M’s, how big would the box need to be to hold all of them? Assume the
volume of a single M&M candy is about 2.2 cm3
, and calculate the volume of the box in km3
.
(b) Assuming the box from part (a) is a cube with all sides of equal length, is the side of the box
closest in length to the diameter of the Sun (1,400,000 km), the diameter of the Earth (13,000 km),
the length of the state of Texas (1200 km), the length of Utah Lake (39 km) or the width of the BYU
Marriott Center (120 m)?

Let's break this problem down step by step.

### Part (a) - Finding the volume of the box in km³

We are given the following:

- Volume of a single M&M candy: \( 2.2 \, \text{cm}^3 \)
- 1 mole of M&Ms: \( 6.022 \times 10^{23} \) (Avogadro's number)

To find the total volume of 1 mole of M&Ms, we multiply the number of M&Ms by the volume of a single M&M:

\[
V_{\text{total}} = 6.022 \times 10^{23} \times 2.2 \, \text{cm}^3
\]

Next, we convert this volume from cubic centimeters (\(\text{cm}^3\)) to cubic kilometers (\(\text{km}^3\)). The conversion factor is:

\[
1 \, \text{km}^3 = 10^{15} \, \text{cm}^3
\]

Thus, the total volume of the box in cubic kilometers will be:

\[
V_{\text{box}} = \frac{V_{\text{total}}}{10^{15}}
\]

Let's calculate this now.

### Part (b) - Side length of the box and comparison with various lengths

For part (b), we are assuming the box is a cube, meaning all sides have equal length. The volume \(V_{\text{box}}\) of a cube is related to the side length \(s\) by the formula:

\[
V_{\text{box}} = s^3
\]

To find the side length \(s\), we take the cube root of the volume:

\[
s = \sqrt[3]{V_{\text{box}}}
\]

After finding the side length, we will compare it to:

- Diameter of the Sun: \(1,400,000 \, \text{km}\)
- Diameter of the Earth: \(13,000 \, \text{km}\)
- Length of Texas: \(1200 \, \text{km}\)
- Length of Utah Lake: \(39 \, \text{km}\)
- Width of BYU Marriott Center: \(120 \, \text{m} = 0.12 \, \text{km}\)

Let's calculate both parts.### Part (a) - Volume of the box in cubic kilometers:
The total volume needed to hold 1 mole of M&Ms is approximately \( 1.32 \times 10^9 \, \text{km}^3 \).

### Part (b) - Side length of the cube:
The side length of the cube-shaped box is approximately \( 1098.3 \, \text{km} \).

### Comparison with given lengths:
- The side length of the box (1098.3 km) is closest to the **length of the state of Texas (1200 km)**.
  
### Would you like to explore more details or have further questions?

Here are five related questions:
1. How does the size of a mole of M&Ms compare to the volume of Earth?
2. What if M&Ms had a different volume, how would that affect the side length of the box?
3. How could this method be adapted to different objects besides M&Ms?
4. What are the largest man-made objects in comparison to this box?
5. How would you calculate the mass of a mole of M&Ms, assuming a certain density?

**Tip**: Always ensure you use proper unit conversions when working with extremely large or small values to maintain accuracy.

(a) If you had 1 mole of M&M’s, how big would the box need to be to hold all of them? Assume the volume of a single M&M candy is about 2.2 cm³, and calculate the volume of the box in km³. (b) Assuming the box from part (a) is a cube with all sides of equal length, is the side of the box closest in length to the diameter of the Sun (1,400,000 km), the diameter of the Earth (13,000 km), the length of the state of Texas (1200 km), the length of Utah Lake (39 km) or the width of the BYU Marriott Center (120 m)?

Discover how to calculate the volume of a box needed to hold 1 mole of M&M's using scientific notation, volume formulas, and unit conversions. Compare the side length of the cube to the diameter of the Sun, Earth, and other well-known lengths. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation