Answer:
[tex]303,882.84649 kg\times 3=9.12\times 10^5 kg[/tex] of carbon dioxide gas.
Explanation:
Average distance covered by Americans in a day= [tex]4.0\times 10^9 km[/tex]
1 day = 24 × 60 min = 1,440 min
Average distance covered by Americans in a minute= [tex]\frac{4.0\times 10^9 km}{1,440}=2,777,777.78 km[/tex]
Average mileage of the car = 20 miles/gal = 32.18 km/gal
1 mile = 1.609 km
20 miles = 20 × 1.609 km = 32.18 km
Volume of gasoline used in minute = [tex]\frac{2,777,777.78 km}{32.18 km/gal}[/tex]
[tex]V=86,320.00 gal[/tex]
[tex]V=86,320.00\times 3.7854 L[/tex]
(1 L = 1000 mL)
[tex]V=86,320.00\times 3.7854 \times 1000 mL=326,755,748.91 mL[/tex]
Mass of 86,320.00 gallons of gasoline = m
Density of the gasoline = d = [tex]0.93 g/cm^3=0.93 g/mL[/tex]
[tex]1 mL= 1 cm^3[/tex]
[tex]m=d\times V=0.93 g/mL\times 326,755,748.91 mL[/tex]
[tex]m=303,882,846.49 g=303,882.84649 kg[/tex]
1 kilogram of gasoline gives 3 kg of carbon dioxde gas .
Then 303,882.84649 kg of gasoline will give :
[tex]303,882.84649 kg\times 3=9.12\times 10^5 kg[/tex] of carbon dioxide gas.