Question

1. michelangelo’s statue of david occupies a volume of 2.098 m³. the average density of marble is 2.65 g/cm³. what is the mass of the statue of david in kg? (it’s slightly more than 6 tons!!)
2. the density of tin is 7.31 g/cm³. what is the mass, in kg, of the sample pictured below? image of rectangular prism with dimensions 2.5 cm, 1.2 cm, 5.50 cm
3. the mass of the pebble shown below is 2.25 g. what is the density of the pebble? images of graduated cylinders
4. angie received a gold pendant of her initial “a” from her boyfriend. she was very excited about her new gold pendant! angie placed the pendant in a graduated cylinder and the volume of the water rose from 5.50 ml to 9.85 ml. she then found that the mass of the ring was 39.0 g. is angie’s new pendant pure gold? if not, what is her ring made from? (density of gold = 19.3 g/cm³, density of platinum = 21.45 g/cm³, copper = 8.96 g/cm³, silver = 10.5 g/cm³) image of pendant
5. challenge: a sample of an unknown substance has been found to have a density of 3.85 g/cm³ and a volume of 2.53 × 10⁻⁴ m³. what is the mass of the sample in kilograms?

Explanation:

Problem 6

Step1: Convert density units

Density of marble is \(2.65\ \text{g/cm}^3\). Since \(1\ \text{kg} = 1000\ \text{g}\) and \(1\ \text{m}^3 = 10^6\ \text{cm}^3\), convert density to \(\text{kg/m}^3\):
\(2.65\ \text{g/cm}^3=\frac{2.65\ \text{g}}{1\ \text{cm}^3}\times\frac{1\ \text{kg}}{1000\ \text{g}}\times\frac{10^6\ \text{cm}^3}{1\ \text{m}^3}=2650\ \text{kg/m}^3\)

Step2: Calculate mass using \(m =

ho V\)
Volume \(V = 2.098\ \text{m}^3\), density \(
ho = 2650\ \text{kg/m}^3\).
\(m=
ho V = 2650\ \text{kg/m}^3\times2.098\ \text{m}^3\)
\(m = 2650\times2.098 = 5560.7\ \text{kg}\) (approx)

Step1: Calculate volume of tin sample (rectangular prism)

Volume of a rectangular prism: \(V = l\times w\times h\).
Given \(l = 5.50\ \text{cm}\), \(w = 2.5\ \text{cm}\), \(h = 1.2\ \text{cm}\).
\(V = 5.50\times2.5\times1.2 = 16.5\ \text{cm}^3\)

Step2: Calculate mass in grams

Density \(
ho = 7.31\ \text{g/cm}^3\), so \(m=
ho V = 7.31\ \text{g/cm}^3\times16.5\ \text{cm}^3 = 120.615\ \text{g}\)

Step3: Convert mass to kilograms

\(1\ \text{kg} = 1000\ \text{g}\), so \(m=\frac{120.615\ \text{g}}{1000} = 0.120615\ \text{kg}\)

Step1: Determine volume of the pebble (water displacement)

Initial volume: \(4.2\ \text{mL}\) (from first graduated cylinder), final volume: \(5.0\ \text{mL}\) (after adding pebble).
Volume of pebble \(V = 5.0 - 4.2 = 0.8\ \text{mL}\). Since \(1\ \text{mL} = 1\ \text{cm}^3\), \(V = 0.8\ \text{cm}^3\).

Step2: Calculate density using \(

ho=\frac{m}{V}\)
Mass \(m = 2.25\ \text{g}\), volume \(V = 0.8\ \text{cm}^3\).
\(
ho=\frac{2.25\ \text{g}}{0.8\ \text{cm}^3}=2.8125\ \text{g/cm}^3\)

Answer:

\(5560.7\ \text{kg}\) (or more precise calculation: \(2650\times2.098 = 5560.7\))

Problem 7