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# Gasoline has a density of ##"0.73 g/cm"^3##. How many liters of gasoline would be required to increase the mass of an automobile from 1271 kg to 1303 kg?

##"44 L"##

The first thing to do here is to make sure that we have the proper units needed to calculate the mass of gasoline required to increase the mass of the car by

##m_"gas" = "1303 kg" - "1271 kg" = "32 kg"##

Notice that the of gasoline is given in grams per cubic centimeter, ##"g cm"^(-3)##, but that the problem wants you to find the volume **in liters**.

One approach that we could use would be to convert the mass from kilograms to grams by using the fact that

##color(purple)(bar(ul(|color(white)(a/a)color(black)("1 kg" = 10^3"g")color(white)(a/a)|)))##

We will have

##32 color(red)(cancel(color(black)("kg"))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = "32000 g"##

Use the density of gasoline to find the volume in cubic centimeters first

##32000 color(red)(cancel(color(black)("g"))) * "1 cm"^3/(0.73color(red)(cancel(color(black)("g")))) = "43835.6 cm"^3##

Now, to convert this to liters, we must use the fact that

##color(purple)(bar(ul(|color(white)(a/a)color(black)("1 L" = "1 dm"^3)color(white)(a/a)|)))" "## and ##" "color(purple)(bar(ul(|color(white)(a/a)color(black)("1 dm"^3 = 10^3"cm"^3)color(white)(a/a)|)))##

In this case, you will have

##43853.6 color(red)(cancel(color(black)("cm"^3))) * (1color(red)(cancel(color(black)("dm"^3))))/(10^3color(red)(cancel(color(black)("cm"^3)))) * "1 L"/(1color(red)(cancel(color(black)("dm"^3)))) = color(green)(bar(ul(|color(white)(a/a)color(black)("44 L")color(white)(a/a)|)))##

The answer is rounded to two .