A gold ingot weighs 5.50 lbs.If the density of gold is 19.31 g/cm³,and the length and width of the ingot are 12.0 cm and 3.00 cm respectively,what is the height of the ingot?

Question

Quizplus · Accepted Answer

First, we need to find the volume of the gold ingot:

Density of gold = mass/volume
19.31 g/cm^3 = (5.50 lbs) * (453.59 g/lb) / volume

Solving for the volume:
volume = (5.50 lbs) * (453.59 g/lb) / (19.31 g/cm^3) = 0.129 L = 129 cm^3

The volume of a rectangular solid is given by V = l * w * h, where l is the length, w is the width, and h is the height. We know that l = 12.0 cm and w = 3.00 cm. Solving for h:

h = V / (l * w) = 129 cm^3 / (12.0 cm * 3.00 cm) = 3.59 cm

Therefore, the height of the gold ingot is 3.59 cm, which is answer choice B.

A Gold Ingot Weighs 5