the density of gold is 19.3 g/cm3. suppose a certain gold wedding ring deplaced 0.55mL of liquid when dropped in a glass of sparkling cider. how much did this wedding ring cost

19.3

Explanation:

Assuming we have to find Specific gravity of gold.

As we know that specific gravity is defined as the ratio of weight of the object and weight of the water displaced by the object

so it is given by

specific gravity = weight of the object/weight of the water displaced

now we have

weight of the object = (density) (volume) g

weight of object = (19.3) (0.55) g

now weight of the liquid displaced is given by

weight of water displaced = (1 g/cm^3) (0.55ml) g

now we have

specific gravity = (19.3*0.55) / (1*0.55)

specific gravity = 19.3