Question

Consider a beaker of water sitting on the pan of an electronic scale that has been tared. A metal weight hanging from a string is suspended in the water and the mass reading on the scale reads 0.13 g. The metal weight falls from the string, rests on the bottom of the beaker, and the reading becomes 0.70 g. (Density water = 0.9982 g/ml)
a. What is the mass of water displaced by the metal weight?
b. What is the volume of water displaced by the metal weight?
c. What is the volume of the metal weight?
d. What is the mass of the metal weight?
e. What is the density of the metal weight?

Answer 1

Answer:

See explanation below for answers

Explanation:

We know that the balance is tared, so the innitial weight would be zero. Now, let's answer this by parts.

a) mass of displaced water.

In this case all we need to do is to substract the 0.70 with the 0.13 g. so:

mW = 0.70 - 0.13

mW = 0.57 g of water

b) Volume of water.

In this case, we have the density of water, so we use the formula for density and solve for volume:

d = m/V

V = m/d

Replacing:

Vw = 0.57/0.9982

Vw = 0.5710 mL of water

c) volume of the metal weight

In this case the volume would be the volume displaced of water, which would be 0.5710 mL

d) the mass of the metal weight.

In this case, it would be the mass when the metal weight hits the bottom which is 0.70 g

e) density.

using the above formula of density we calculate the density of the metal

d = 0.70 / 0.5710

d = 1.2259 g/mL