Question

1) how many feet are contained in 0.344 km, given that 1 mi=1.609 km and 5280 ft= 1 mile

2) if the density of carbon dioxide in air is 0.65g/L, what volume is occupied by 25 kg of carbon dioxide?

Answer 1

1)

1 mile = 1.609 km

1 mile = 5280 feet

so

1.609 km = 5280 feet

1 km = 5280 / 1.609 feet

1 km    = 3281.54 feet

now

0.344 km = 0.344 x 3281.54 feet

                 = 1128.85 feet

0.344 km = 1128.85 feet

2)

density (d) = 0.65 g /L

mass of CO2 = 25 kg = 25000 g

density = mass / volume

volume = mass / density

             = 25000 / 0.65

            = 38461.54 L

volume occupied by 25 kg of carbon dioxide = 38461.54 L