Challenge Problems (required): 1. An object in the shape of a rectangular prism with dimensions of \( 1.50 \mathrm{dm} \times 2.00 \mathrm{dm} \times 0.250 \mathrm{dm} \) is placed into a tank of water ( \( 1.00 \mathrm{~kg} / \mathrm{L} \) ) and sinks to the bottom. A scale placed under the object in the tank identifies that the object has a weight \( 20.0 \% \) less than it does out of the water. Determine the density of the object. Recall that \( 1 \mathrm{~L}=1 \mathrm{dm}^{3} \).

To determine the density of the object, we can follow these steps: 1. **Calculate the volume of the object**: The volume \( V \) of a rectangular prism is given by the formula: \[ V = \text{length} \times \text{width} \times \text{height} \] Substituting the given dimensions: \[ V = 1.50 \, \text{dm} \times 2.00 \, \text{dm} \times 0.250 \, \text{dm} = 0.750 \, \text{dm}^3 \] 2. **Calculate the weight of the object in air**: The weight \( W \) of the object in air can be expressed as: \[ W = \text{mass} \times g \] where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)). However, we will focus on the mass for density calculation. 3. **Determine the apparent weight in water**: The problem states that the weight of the object in water is \( 20.0\% \) less than its weight in air. If we denote the weight in air as \( W_a \) and the weight in water as \( W_w \), we have: \[ W_w = W_a - 0.20 W_a = 0.80 W_a \] 4. **Calculate the buoyant force**: The buoyant force \( F_b \) acting on the object when submerged in water is equal to the weight of the water displaced by the object. The density of water is given as \( 1.00 \, \text{kg/L} = 1.00 \, \text{kg/dm}^3 \). The volume of water displaced is equal to the volume of the object: \[ F_b = \text{density of water} \times V \times g = 1.00 \, \text{kg/dm}^3 \times 0.750 \, \text{dm}^3 \times g \] Thus, the buoyant force is: \[ F_b = 0.750 \, \text{kg} \times g \] 5. **Relate the weights**: The apparent weight in water can also be expressed as: \[ W_w = W_a - F_b \] Substituting the expressions for \( W_w \) and \( F_b \): \[ 0.80 W_a = W_a - 0.750 \, \text{kg} \times g \] Rearranging gives: \[ 0.20 W_a = 0.750 \, \text{kg} \times g \] Therefore: \[ W_a = \frac{0.750 \, \text{kg} \times g}{0.20} = 3.75 \, \text{kg} \times g \] 6. **Calculate the mass of the object**: Since \( W_a = \text{mass} \times g \), we can equate: \[ \text{mass} \times g = 3.75 \, \text{kg} \times g \] Thus, the mass of the object is: \[ \text{mass} = 3.75 \, \text{kg} \] 7. **Calculate the density of the object**: The density \( \rho \) of the object is given by: \[ \rho = \frac{\text{mass}}{V} \] Substituting the values we have: \[ \rho = \frac{3.75 \, \text{kg}}{0.750 \, \text{dm}^3} = 5.00 \, \text{kg/dm}^3 \] Thus, the density of the object is \( \boxed{5.00 \, \text{kg/dm}^3} \).