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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)?

Respuesta :

Willchemistry
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!
david8644

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:

[tex]Density=\frac{mass}{volume}[/tex]

We clear it for volume and the result is:

[tex]Volume=\frac{Mass}{Density}[/tex]

Now we just insert the values:

[tex]Volume=\frac{Mass}{Density}\\Volume=\frac{96,4g}{2,7}\\Volume= 35,70ml[/tex]

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

[tex]Volume=\frac{Mass}{Density}\\Mass=Volume*Density\\Mass= 35,70ml*7.87g/ml\\Mass=280.9881 ml[/tex]

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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