Question

a piece of aluminum has a volume of 12 cm³ and a mass of 32 g. what is its density? do not forget the units.
solution:
editor
a gold nugget has a density of 38.6 g/cm³ and a mass of 270.2. what is its volume?
this problem is a little different. it gives you the density and the mass. you need to find the volume. we need to rearrange the formula d = m/v. move the volume and density to the other side: v = m/d.
solution:
editor

Explanation:

First Problem (Aluminum Density)

Step1: Recall density formula

The formula for density \( D \) is \( D=\frac{m}{v} \), where \( m \) is mass and \( v \) is volume.

Step2: Substitute values

Given \( m = 32\space g \) and \( v=12\space cm^3 \), substitute into the formula: \( D=\frac{32}{12} \space g/cm^3 \).

Step3: Simplify the fraction

Simplify \( \frac{32}{12}=\frac{8}{3}\approx2.67 \space g/cm^3 \).

Step1: Recall rearranged formula

We know the density formula \( D = \frac{m}{v} \), rearranged for volume \( v=\frac{m}{D} \).

Step2: Check units and substitute

Given \( D = 38.6\space g/cm^3 \) and \( m = 270.2\space g \) (assuming the mass unit is grams as density is in \( g/cm^3 \)), substitute into the formula: \( v=\frac{270.2}{38.6} \space cm^3 \).

Step3: Calculate the division

\( \frac{270.2}{38.6}=7 \space cm^3 \).

Answer:

The density of the aluminum is approximately \( 2.67\space g/cm^3 \) (or \( \frac{8}{3}\space g/cm^3 \)).

Second Problem (Gold Nugget Volume)