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What is the mass of a sample of fool's gold that, when placed into a graduated cylinder raises the water level from 25.5 m. to 37.9 m.? The density of fool's gold is 5.90 g/cm3 O 151 g O 218 g O 1.95

Answer :

Final answer:

The mass of the sample of fool's gold is 7306 g.

Explanation:

To calculate the mass of the sample of fool's gold, we can use the formula:

mass = density x volume

First, we need to determine the volume of the sample. The change in water level in the graduated cylinder represents the volume of the sample. The initial water level is 25.5 m and the final water level is 37.9 m. Therefore, the change in water level is 37.9 m - 25.5 m = 12.4 m.

Since the density of fool's gold is given as 5.90 g/cm3, we need to convert the change in water level from meters to centimeters. There are 100 centimeters in 1 meter, so the change in water level is 12.4 m x 100 cm/m = 1240 cm.

Now, we can calculate the volume of the sample:

volume = change in water level = 1240 cm3

Finally, we can calculate the mass of the sample:

mass = density x volume = 5.90 g/cm3 x 1240 cm3 = 7306 g

Learn more about calculating the mass of a sample using density and volume here:

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