Question

6. STATEANSWERIN SIG FIGSOR g/cm3. What mass of silver in grams (g) would be required to cover a dimensions of 120.0 ft x 60.0 ft to a depth of 0.33 ft? SCIENTIFIC NOTATION. Silver has a densityof 10.5

Answer 1

Answer:

$7.0643\times 10^8 grams$ of silver would be required.

Explanation:

Dimensions of silver mas :

Length = l =  120.0 ft

breadth = b = 60.0 ft

Height = h = 0.33 ft

Volume of silver mass ,V= l × b × h

$V=120.0 ft\times 60.0ft\times 0.33 ft=2,376 ft^3=67,278,816 cm^3$

$1ft^3=28316.8 cm^3$

Density of silver,d = $10.5 g/cm^3$

$Mass = Density\times Volume$

$m=10.5 g/cm^3\times 67,278,816 cm^3=7.0643\times 10^8 g$

$7.0643\times 10^8 grams$ of silver would be required.