You find a ring that has a mass of 96.6 mathrm(~g) . You fill a jar up with 15 mathrm(~mL) of water and after you drop in the ring, the water level rises to 20 mathrm(~mL) . What is your ring made of? Justify your answer. multicolumn(2)(|c|)( Table of Densilies ) Solids & Density (mathrm(g) / mathrm(cm)^3) Marble & 2.56 Quartz & 2.64 Diamond & 3.52 Copper & 8.92 Gold & 19.32 Platinum & 21.4 The ring is made of Choose. The ratio Choose. and Choose. gives the density of the ring.

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You find a ring that has a mass of 96.6 mathrm(~g) . You fill a jar up with 15 mathrm(~mL) of water and after you drop in the ring, the water level rises to 20 mathrm(~mL) . What is your ring made of? Justify your answer. multicolumn(2)(|c|)( Table of Densilies ) Solids & Density (mathrm(g) / mathrm(cm)^3) Marble & 2.56 Quartz & 2.64 Diamond & 3.52 Copper & 8.92 Gold & 19.32 Platinum & 21.4 The ring is made of Choose. The ratio Choose. and Choose. gives the density of the ring.

Answer 1

# Explanation To determine what the ring is made of, we need to calculate its density and compare it with the densities provided in the table. Density is defined as mass per unit volume. The formula for density (\(\rho \)) is: \[ \rho = \frac{m}{V} \] where \( m \) is the mass and \( V \) is the volume. From the given information, we have: - Mass of the ring, \( m = 96.6 \) g - Initial volume of water, \( V_{\text{initial}} = 15 \) mL - Final volume of water after the ring is dropped in, \( V_{\text{final}} = 20 \) mL The volume of the ring can be determined by the displacement of the water, which is the difference between the final and initial volumes: \[ V_{\text{ring}} = V_{\text{final}} - V_{\text{initial}} \] Now, we can calculate the density of the ring using the mass and the volume we just found. # Answer First, calculate the volume of the ring: \[ V_{\text{ring}} = V_{\text{final}} - V_{\text{initial}} = 20 \text{ mL} - 15 \text{ mL} = 5 \text{ mL} \] Since \( 1 \text{ mL} = 1 \text{ cm}^3 \), the volume of the ring is \( 5 \text{ cm}^3 \). Next, calculate the density of the ring: \[ \rho_{\text{ring}} = \frac{m}{V_{\text{ring}}} = \frac{96.6 \text{ g}}{5 \text{ cm}^3} = 19.32 \text{ g/cm}^3 \] Comparing the calculated density with the table of densities, we find that the density of the ring matches the density of gold. The ring is made of \(\boxed{\text{Gold}}\). The ratio of mass to volume (\( \frac{96.6 \text{ g}}{5 \text{ cm}^3} \)) gives the density of the ring, which is \( 19.32 \text{ g/cm}^3 \).

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