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Chapter 1: Problem 45

Calculate the mass of each of these: (a) a sphere of gold of radius \(10.0 \mathrm{~cm}\) [the volume of a sphere of radius \(r\) is \(V=\left(\frac{4}{3}\right) \pi r^{3} ;\) the density of gold \(\left.=19.3 \mathrm{~g} / \mathrm{cm}^{3}\right]\) (b) a cube of platinum of edge length \(0.040 \mathrm{~mm}\) (the density of platinum \(\left.=21.4 \mathrm{~g} / \mathrm{cm}^{3}\right),\) (c) \(50.0 \mathrm{~mL}\) of ethanol (the density of ethanol \(=0.798 \mathrm{~g} / \mathrm{mL})\)

Short Answer

Expert verified
The mass of (a) the gold sphere, (b) the platinum cube, and (c) the ethanol are calculated respectively based on their volumes and densities. Apply the provided formulas and appropriate unit conversions where necessary.

Step by step solution

01

Calculate the mass of the sphere of gold

Firstly, we have to find the volume of the sphere. The formula for the volume of a sphere is \(V= \left( \frac{4}{3}\right) \pi r^{3}\). Replace \(r\) with 10.0 cm and calculate the volume. Then, to get the mass, multiply the volume by the density of gold, which is 19.3 g/cm³.
02

Calculate the mass of the cube of platinum

The volume of a cube is given by \(V= a³\), where 'a' is the length of an edge. Replace \(a\) by the given value of the edge length, which is 0.040 mm. Convert mm to cm by dividing by 10, as 1 cm = 10 mm. Then, multiply this volume by the density of platinum, which is 21.4 g/cm³, to get the mass.
03

Calculate the mass of 50.0 mL ethanol

This one is simpler, as the volume is already given. The mass is given by the product of density and volume. Simply multiply the volume of ethanol, which is 50.0 mL, by its density, which is 0.798 g/mL.

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Most popular questions from this chapter

Give an example of a homogeneous mixture and an example of a heterogeneous mixture.

Give the SI units for expressing these: (a) length, (b) area, (c) volume, (d) mass, (e) time, (f) force, (g) energy, (h) temperature.

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