Final answer:
The calculated volume of the object is larger than the water volume in which it is supposed to be submerged, suggesting a discrepancy since the object cannot be fully submerged in the given water volume. Therefore, the scenario provided does not allow for a simple additive calculation of the resulting volume.
Step-by-step explanation:
To determine the resulting volume after placing an object in water, we use the principle of displacement. The object's volume is determined by how much water it displaces. Since the object has a density of 0.75 g/dm³ and a mass of 2 g, we can calculate its volume using the formula density = mass/volume, which rearranges to volume = mass/density. So, the object's volume is 2 g / 0.75 g/dm³ = 2.6667 dm³ or 2666.7 mL (as 1 dm³ = 1000 mL). When this object is placed into 25.0 mL of water, we add its volume to the original water volume to get the resulting volume.
However, since the object's volume is much larger than the container's capacity, it suggests that the object will not be fully submerged, and thus the scenario is physically inconsistent. We could only calculate the resulting volume if the object's volume were less than or equal to the container's capacity.