How many grams of iron (density = 7.87 g/mL ) would occupy the same volume as 96.4 g of aluminum (density = 2.70 g/mL)?
2 answers:
Volume of Al:
D = m / V
2.70 = 96.4 / V
V = 96.4 / 2.70 => 35.70 mL
Mass of iron:
D = m / V
7.87 = m / 35.70
m = 7.87 x 35.70 => 280.959 g
hope this helps!
Answer:
280.9881 grams of iron.
Explanation:
In order to solve this we first have to calculate the volume of the aluminium because it is the one that we have all of the information to calculate:
Remember that the formula for density is:

We clear it for volume and the result is:

Now we just insert the values:

Now that we have the volume we calculate how many grams of iron we need to occupy that volume, solving the formula for mas:

So the mass of iron needed to cover the same volume as 96.4 g of aluminium would be 280.9881 grams
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