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).