Question

the calculation of the density of a metal using water displacement. density is equal to the mass of a substance divided by its volume: density = \frac{mass of substance}{volume of substance}. water displacement can be used to find the volume of a solid by placing the object in a known volume of water. when the object is added to water, the measured total volume increases. this final volume is equal to the volume of the initial water plus the volume of the solid object added. to determine the volume of the water that is displaced by the object, subtract the initial volume from the final volume. the volume of water displaced is equal to the volume of the solid object. using a scale to measure the mass of the object, you can now calculate the density of the object as its mass divided by its volume. use the simulation to explore this method.

part a
follow the steps provided in the simulation to add water to the graduated cylinder, select one of the three samples (copper, silver, or gold), set its mass to the values given in the statements below, find its volume, and calculate its density.
to save time, you can approximate the initial mass of the solid to the nearest ±1 g. for example, if you are asked to add 14.3 g of copper, add between 13 g and 15 g.
which of the following sets include two samples with an equal density?
check all that apply.
view available hint(s)
15.4 g gold and 18.7 g silver
11.2 g gold and 14.9 g gold
20.2 g silver and 20.2 g copper
15.2 g copper and 50.0 g copper
submit

part b
you add 5.3 g of iron to 27.80 ml of water and observe that the volume of iron and water together is 28.47 ml. calculate the density of iron. express your answer to two significant figures with the appropriate units.
view available hint(s)
value
units

Explanation:

Step1: Recall density property

Density is a characteristic property of a substance. Substances of the same type have the same density regardless of mass.

Step2: Analyze Part A options

• Gold has a characteristic density, silver has a different one, and copper has its own. So, samples of the same metal will have equal density.
• For "15.4 g gold and 18.7 g silver": Different metals, different densities.
• For "11.2 g gold and 14.9 g gold": Same metal (gold), so equal densities.
• For "20.2 g silver and 20.2 g copper": Different metals, different densities.
• For "15.2 g copper and 50.0 g copper": Same metal (copper), so equal densities.

Step3: Solve Part B

The volume of iron is the final - volume of water and iron minus the initial volume of water.
The initial volume of water $V_1 = 27.60\ mL$, the final volume of water and iron $V_2=28.47\ mL$. So the volume of iron $V = V_2 - V_1=28.47 - 27.60=0.87\ mL$.
The mass of iron $m = 5.3\ g$.
Using the density formula $
ho=\frac{m}{V}$, we substitute $m = 5.3\ g$ and $V = 0.87\ mL$ to get $
ho=\frac{5.3\ g}{0.87\ mL}\approx6.1\ g/mL$.

Answer:

Part A

• B. 11.2 g gold and 14.9 g gold
• D. 15.2 g copper and 50.0 g copper

Part B

• Value: 6.1
• Units: g/mL