What is the mass, in grams of a pure iron cube that has a volume of 4.20cm^3

asked Aug 20, 2017 88.1k views

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$d_(Fe)=7874(kg)/(m^(3))=7,874(g)/(cm^(3))\\ V=4,2cm^(3)\\\\ m=dV=7,784(g)/(cm^(3))*4,2cm^(3)\approx 32,7g$

Enoktate answered Aug 25, 2017

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