What is the mass of a piece of iron that has a volume of 1.44 × 10^2 mL and a density of 7.90 g/mL. Be sure answer has correct number of significant figures.

To calculate the mass of the piece of iron, we use the formula:

\[
\text{Mass} = \text{Density} \times \text{Volume}
\]

Given:
- Density of iron \( \rho = 7.90 \, \text{g/mL} \)
- Volume of iron \( V = 1.44 \times 10^2 \, \text{mL} \)

Now, calculate the mass:

\[
\text{Mass} = 7.90 \, \text{g/mL} \times 1.44 \times 10^2 \, \text{mL}
\]

Multiplying:

\[
\text{Mass} = 7.90 \times 144 = 1137.6 \, \text{g}
\]

Since the volume has 3 significant figures and the density has 3 significant figures, the result should also be rounded to 3 significant figures:

\[
\text{Mass} = 1.14 \times 10^3 \, \text{g}
\]

Thus, the mass of the iron is \( 1.14 \times 10^3 \, \text{g} \) or 1140 g.

Would you like more details or have any questions?

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Here are 5 related questions for further exploration:
1. What is the mass of a piece of lead with a similar volume and a density of 11.34 g/mL?
2. How does temperature affect the density of metals like iron?
3. If the

Learn how to calculate the mass of a piece of iron with a volume of 1.44 × 10^2 mL and a density of 7.90 g/mL. This detailed solution explains the use of the mass formula, significant figures, and key concepts of density. Suitable for high school students in Grades 9-12.

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