To solve this using unit analysis, we'll convert all units consistently and proceed step-by-step:

Given:

• Density of gold: $19.3 \, \text{g/cm}^3$
• Side length of the cube: $2.0 \, \text{inches}$
• Goal: Find the mass in kilograms (kg)

Steps:

1. Convert the side length from inches to centimeters (cm):

   • $1 \, \text{inch} = 2.54 \, \text{cm}$
   • $2.0 \, \text{inches} \times 2.54 \, \text{cm/inch} = 5.08 \, \text{cm}$

2. Calculate the volume of the cube:

   • The volume $V$ of a cube is given by $\text{side}^3$.
   • $V = (5.08 \, \text{cm})^3 = 131.79 \, \text{cm}^3$

3. Determine the mass in grams (g):

   • The density of gold is $19.3 \, \text{g/cm}^3$, so the mass is: $\text{Mass} = \text{density} \times \text{volume}$ $\text{Mass} = 19.3 \, \text{g/cm}^3 \times 131.79 \, \text{cm}^3 = 2543.55 \, \text{g}$

4. Convert grams to kilograms:

   • $1 \, \text{kg} = 1000 \, \text{g}$, so: $\text{Mass in kg} = \frac{2543.55 \, \text{g}}{1000} = 2.54355 \, \text{kg}$

Thus, the mass of the cube of gold is approximately 2.54 kg.

Would you like any further details or explanations?

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Here are 5 related questions to expand on this:

1. How does the mass change if the side length of the cube doubles?
2. What would the mass be if the density were 10 g/cm³ instead?
3. How would you calculate the mass if the shape was a sphere instead of a cube?
4. Can you use this method to find the mass of other metals with known densities?
5. What is the weight of the cube in Newtons if measured on Earth?

Tip: Always ensure units are consistent when applying unit analysis, as incorrect conversions can lead to significant errors.