1.) Pure air” best described as A.) A mixture because it contains CO 2 which is
ID: 531075 • Letter: 1
Question
1.) Pure air” best described as
A.) A mixture because it contains CO2 which is composed of two different elements.
B.) A pure substance
C.) A mixture because it contains N2, CO2, H2O, and a number of other chemicals in addition to O2.
D.) A pure substance because pure air contains only O2.
2.) The air exhaled from the lungs of a smoker has a CO concentration of 22 ppm. Express this concentration as a percent.
3.) In dry polar regions, water vapor may be present at a level of a mere 10 ppm. Express this as a percent.
4.) Which value, expressed as scientific notation, represents 0.12 nm (nanometers), the approximate diameter of an atom?
2.) The air exhaled from the lungs of a smoker has a CO concentration of 22 ppm. Express this concentration as a percent.
3.) In dry polar regions, water vapor may be present at a level of a mere 10 ppm. Express this as a percent.
4.) Which value, expressed as scientific notation, represents 0.12 nm (nanometers), the approximate diameter of an atom?
Explanation / Answer
Ans. 1. The term “pure air” is NOT a scientific term. It may refer to “the air which is free of dust, aerosols and reactive gaseous contaminants of anthropogenic origin”.
It is a mixture of N2, O2, CO2, and other gases.
Correct option. C.
Ref: https://goldbook.iupac.org/html/C/C01214.html
#2. [CO] = 22 ppm
= 22 mg CO / L of air ; [1 ppm = 1 mg/ L]
= 22 mg CO/ 1000 mL air
= 0.022 g CO/ 1000 mL air ; [1 g = 1000 mg]
= 0.0022 g/ 100 mL air
= 0.0022 % ; [ % (Wt/V) = mass in g per 100 mL]
Hence, [CO] = 0.0022 % (Wt/ V)
#3. [Vapor] = 10 ppm
= 10 mg / 1000 mL
= 0.010 g/ 1000 mL
= 0.0010 g/ 100 mL
= 0.0010 % (Wt/ V)
#4. 0.12 nm
= 0.12 x 10-9 m ; [1 nm = 10-9 m]
= 1.20 x 10-10 m
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