Section 1.7 showed that in 1997 los angeles county air had carbon monoxide (co) levels of 15.0 ppm. an average human inhales abo

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Section 1.7 showed that in 1997 los angeles county air had carbon monoxide (co) levels of 15.0 ppm. an average human inhales abo

ut 0.50 l of air per breath and takes about 20 breaths per minute. part a how many milligrams of carbon monoxide does the average person inhale in an 6 hour period for this level of carbon monoxide pollution? assume that the carbon monoxide has a density of 1.2 g/l. (hint: 15.0 ppm co means 15.0 lco per 106l air.)

Chemistry

2 answers:

castortr0y [3K] · Answer 1 · 2026-03-19T00:11:11+03:00

Given data:

CO concentration in air = 15 ppm

Volume of air inhaled per breath = 0.50 L

Breaths taken per minute = 20

CO density = 1.2 g/L

Objective:

milligrams of CO inhaled over 6 hours

Clarification:

A concentration of 15 ppm of CO means that there are 15 liters of CO per 10⁶ liters of air

. Consequently, the volume of CO inhaled through 0.50 L of air is

= 15 L CO * 0.50 L air/10⁶ L air = 7.5 *10⁻⁶ L CO/breath

Next, considering there are 20 breaths in a minute,

the total number of breaths in 360 minutes (or 6 hours) will be

= 360 min * 20 breaths/1 min = 7200 breaths

Thus, the total volume of CO inhaled in that time frame is

= 7200 breaths * 7.50*10⁻⁶L/1 breath = 0.054 L

Given that the density of CO is 1.2 g/L

the mass of CO inhaled equals Density*Volume

= 0.054 * 1.2 = 0.0648 g = 64.8 mg

Thus, the mass of CO inhaled over 6 hours is 64.8 mg

castortr0y [3K] · Answer 2 · 2026-03-19T00:09:47+03:00

To start, let’s calculate the amount of air a person breathes in each minute:

0.5 L/breath * (20 breaths/min) = 10 L/min

For a duration of 6 hours, the total volume would be:

6 hrs = 360 min
10 L/min (360 min amounts to a total of 3,600 L of air inhaled

Next, we convert the concentration to grams per liter. The calculation is as follows:
15 L CO/10⁶ L air * 1.2 g CO/L = 18 g CO/L of air
. Thus, we multiply this by the inhaled air volume.

Total CO inhaled = (18 g CO/L air)(3,600 L of air)(1,000 mg/1 g)
Total CO inhaled = 64.8×10⁶ mg of CO inhaled