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Chemistry - Air Quality and Climate

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The composition of clean, dry air is approximately 78%78\% Nitrogen (N2N_2), 21%21\% Oxygen (O2O_2), 0.9%0.9\% Argon (ArAr), and 0.04%0.04\% Carbon Dioxide (CO2CO_2).

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Air pollutants include Carbon Monoxide (COCO) from incomplete combustion, Sulfur Dioxide (SO2SO_2) from sulfur-containing fossil fuels, and Nitrogen Oxides (NOxNO_x) from high-temperature reactions in car engines.

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Acid rain is primarily caused by SO2SO_2 reacting with water and oxygen to form sulfuric acid (H2SO4H_2SO_4), and NOxNO_x forming nitric acid (HNO3HNO_3), which damages buildings and aquatic life.

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The greenhouse effect is caused by gases like Carbon Dioxide (CO2CO_2) and Methane (CH4CH_4) absorbing thermal energy (infrared radiation) and re-radiating it towards Earth, leading to global warming.

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Catalytic converters in vehicles reduce pollution by converting harmful gases into less harmful ones via redox reactions, such as turning COCO into CO2CO_2 and NONO into N2N_2.

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Fractional distillation of liquid air separates components based on their boiling points: Nitrogen boils at −196∘C-196^\circ C and Oxygen boils at −183∘C-183^\circ C.

📐Formulae

6CO2+6H2O→C6H12O6+6O26CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2

2CO+2NO→2CO2+N22CO + 2NO \rightarrow 2CO_2 + N_2

S+O2→SO2S + O_2 \rightarrow SO_2

SO2+H2O→H2SO3SO_2 + H_2O \rightarrow H_2SO_3

CH4+2O2→CO2+2H2OCH_4 + 2O_2 \rightarrow CO_2 + 2H_2O

N2+O2→2NON_2 + O_2 \rightarrow 2NO

💡Examples

Problem 1:

Calculate the volume of oxygen present in a 400 cm3400 \text{ cm}^3 sample of dry air.

Solution:

Volume of oxygen = 21%21\% of total volume 400×0.2184\begin{array}{r} 400 \\ \times 0.21 \\ \hline 84 \end{array} Oxygen volume = 84 cm384 \text{ cm}^3

Explanation:

Since dry air is 21%21\% oxygen by volume, we multiply the total sample volume by 0.210.21 to find the specific volume of oxygen.

Problem 2:

Explain how a catalytic converter removes Nitrogen Monoxide (NONO) and Carbon Monoxide (COCO) simultaneously.

Solution:

The reaction is: 2CO+2NO→2CO2+N22CO + 2NO \rightarrow 2CO_2 + N_2 In this reaction, COCO is oxidized to CO2CO_2 and NONO is reduced to N2N_2.

Explanation:

The catalytic converter provides a surface (usually platinum or palladium) where the nitrogen oxides and carbon monoxide can react. The oxygen from NONO is transferred to COCO, resulting in non-toxic nitrogen gas and carbon dioxide.

Problem 3:

If a sample of 250 cm3250 \text{ cm}^3 of air is passed over excess hot copper turnings, what is the decrease in volume?

Solution:

Decrease in volume = 250×0.21=52.5 cm3250 \times 0.21 = 52.5 \text{ cm}^3 250.0−197.552.5\begin{array}{r} 250.0 \\ - 197.5 \\ \hline 52.5 \end{array}

Explanation:

Copper reacts with oxygen to form copper(II) oxide (2Cu+O2→2CuO2Cu + O_2 \rightarrow 2CuO). Since oxygen makes up 21%21\% of air, that volume is removed from the gas phase, resulting in a decrease of 52.5 cm352.5 \text{ cm}^3.