To measure the volume of a small non-uniform piece of an unknown metal, you fill

Question

To measure the volume of a small non-uniform piece of an unknown metal, you fill a graduated cylinder with a small amount of water, and carefully place the metal into the cylinder. Before the metal was added, the cylinder had 6.87 mL of water. After you added the metal, the water line was at 11.46 mL mark. After you had finished, you discovered that the liquid was not water (density of 0.9986 g/mL) but was an oil (density of 0.8985 g/mL) Will the density you calculated using the change in volume be too big or too small Explain how you can tell.

Explanation / Answer

Answer:

So, first thing we need to calculate the volume change:

Vchange= 11.46mL-6.87mL=4.59mL, this would be the volume of the metal piece.

Now, if weight the liquid with metal toguether in order to calculate the metal density, You will have the total weight=liquid+metal due the fact the the density of the water is higher than the density of the oil, you by mistake will believe that more weight correspond to the liquid that is really from the metal, so at the moment of calculate the metal density you will use a lower weight than real, therefore, you will obtain a smaller density than the real one.

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