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what is the weight in si units and us customary units of one of the federal reserve gold bars (7""x 3.525""x 1.25"")? provide reference for the density of gold

User Efthymis
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Final answer:

The weight of the federal reserve gold bar is 21.612 ounces (US customary units) or 612.45 grams (SI units).

Step-by-step explanation:

To calculate the weight of the federal reserve gold bar, we need to know its volume and density. The given dimensions of the gold bar are 7"" x 3.525"" x 1.25"". To calculate the volume, we multiply the length, width, and height: 7"" x 3.525"" x 1.25"" = 30.984375 cubic inches. The density of gold is 19.3 g/cm³, which is equivalent to 0.6972 ounces/cubic inch (since 1 ounce = 28.3495 grams). Therefore, the weight of the gold bar is 30.984375 cubic inches x 0.6972 ounces/cubic inch = 21.612 ounces (in US customary units) or 612.45 grams (in SI units).

User Lulijeta
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