Aluminum, gold, and iron have densities of 2.7 g/cc, 19.2 g/cc, and 7.874 g/cc, respectively. Suppose 5.0g of each metal were ob
tained.
1 answer:
The question above is missing key elements of the problem to solve it properly. I have attached an image below detailing the true aim of this question.
We must match each metal to the correct cylinder corresponding to it. The idea here is that knowing the density of the metals along with the mass of each metal present, we can determine the volume of each metal. If each cylinder has the same width, then we can assume that the volume of each metal will be directly proportional to the height of the various cylinders. Now we can simply solve for the volumes, in cubic centimeters (cc/cm³).
Aluminum, Al:
5.0 g/ 2.7 g/cc = 1.9 cc
Gold, Au:
5.0 g/ 19.2 g/cc = 0.26 cc
Iron, Fe:
5.0 g/7.874 g/cc = 0.64 cc
From the following calculations, we have found that 5.0 g of aluminum has the greatest volume of 1.9 cc, followed by iron with 0.64 cc and then gold with 0.26 cc.
Therefore, aluminum will correspond to cylinder II with the greatest volume, iron will correspond to cylinder III and gold will correspond to cylinder I with the least volume.
We must match each metal to the correct cylinder corresponding to it. The idea here is that knowing the density of the metals along with the mass of each metal present, we can determine the volume of each metal. If each cylinder has the same width, then we can assume that the volume of each metal will be directly proportional to the height of the various cylinders. Now we can simply solve for the volumes, in cubic centimeters (cc/cm³).
Aluminum, Al:
5.0 g/ 2.7 g/cc = 1.9 cc
Gold, Au:
5.0 g/ 19.2 g/cc = 0.26 cc
Iron, Fe:
5.0 g/7.874 g/cc = 0.64 cc
From the following calculations, we have found that 5.0 g of aluminum has the greatest volume of 1.9 cc, followed by iron with 0.64 cc and then gold with 0.26 cc.
Therefore, aluminum will correspond to cylinder II with the greatest volume, iron will correspond to cylinder III and gold will correspond to cylinder I with the least volume.
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