Carbon monoxide (CO) detectors sound an alarm when peak levels of carbon monoxide reach 100 parts per million (ppm). This level roughly corresponds to a composition of air that contains 400,000 /(/mu/)g carbon monoxide per cubic meter of air (400,000 /(/mu/)g/m3). Assuming the dimensions of a room are 18 ft x 12 ft x 8 ft, estimate the mass (in grams) of carbon monoxide in the room that would register 100 ppm on a carbon monoxide detector. (1 meter = 3.28 feet)

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javillamizarp javillamizarp · Answer 1 · 2019-09-13T17:57:28+02:00

Answer:

19.61kg

Explanation:

The first thing to do is divide the amount of feet into 3.28:

[tex]18ft*12ft*8ft=5.49m*3.66m*2.44m[/tex]

With this information, we calculate the volume:

[tex]V=5.49m*3.66m*2.44m=49.03m^{3}[/tex]

As we have the concentration and the volume, we can calculate the mass of CO:

[tex]m=400,000 \mu g/m^{3}*49.03m^{3}=19611158.4 \mu g[/tex]

As this is a very big number, we can express it in kg by dividing in 10^6:

[tex]19611158.4 \mu g/10^{6}=19.61 kg[/tex]