kilograms
Kilograms.
You find a measure of the amount of matter in the rock.
You put it on a scale or a triple-beam balance. It'll give you the mass. However, if you have the density of the rock, divide it by the volume to find the mass mathematically. You can also find the volume by the water displacement method, where you placed the rock in a graduated cylinder/beaker filled partially with water and record the difference between the original water level and the raised water level after placing the rock inside the container.
density = mass divided by volume = 100/25 = 4 grammes per cubic centimetre.
It depends on the size and mass of the rock. For a "normal" rock (as opposed to merely a stone) I suggest a crane with some means for measuring the rock's mass. A number of pulleys, each with a weighing machine should do the trick. Alternatively, you could use a weighbridge, if there is one in the vicinity. To obtain its volume, you probably need a 3-d laser-profiling device. That should give the most precise measurement of its volume.
Kilograms.
Kilograms.
Weigh it! The mass is usually found indirectly, via the weight.
Place a specific amount of water in your bottle. Record the volume. Place the rock in the bottle. Read and record the new volume of water. Subtract the first volume measurement from the measurement after you added the rock. The difference is the volume of the rock.
Yes. Kilogram is the unit of mass.
Balanced
You find a measure of the amount of matter in the rock.
- to determine the chemical nature of a rock - to determine the age of a rock by comparison
The answer will depend on how big the rock is.
The answer depnds on the density of each rock. If they are different denities thenthere is a chance that the 1mL rock may have more mass. If they have the same density then the 4mL rock will have more mass
Determine its volume by how much water it displaces, then divide mass by volume
Find the mass using a scale. Find the volume by the water displacement method. Divide mass by volume and that equals density.