Free SACE Chemistry Flashcard Review

They absorb some outgoing infrared radiation and emit infrared radiation in different directions. This alters the energy transfers that determine Earth’s temperature; it is not simply reflection of all incoming sunlight.

At an initially unchanged temperature, greater infrared absorption reduces energy escaping to space in relevant bands. The climate system responds until outgoing and incoming energy are brought toward balance at changed conditions.

CO₂(aq) + H₂O(l) ⇌ H₂CO₃(aq). Subsequent acid dissociation contributes H⁺; dissolving CO₂ does not release H⁺ by atom creation.

H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq). Increased dissolved CO₂ can increase hydrogen-ion concentration through the connected equilibria, lowering pH.

CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + CO₂(g) + H₂O(l). Two hydrogen ions are needed per carbonate ion in this overall representation.

pH=−log₁₀[H⁺]=−log₁₀(3.20×10⁻⁴)≈3.495. Use the base-10 logarithm and the concentration expressed in mol L⁻¹.

Using K_w=1.00×10⁻¹⁴, [H⁺]=1.00×10⁻¹¹ mol L⁻¹ and pH=11.00. The numerical K_w and pH+pOH=14 relationship here are temperature-specific.

H⁺ reacts with CO₃²⁻ to form HCO₃⁻. Removing carbonate shifts equilibria and can favour dissolution of calcium carbonate; describe both the species change and its consequence.

At high temperature, N₂(g)+O₂(g)→2NO(g). Nitrogen’s strong bond makes appreciable formation dependent on high-energy conditions; ordinary cool air does not react rapidly this way.

2NO(g)+O₂(g)→2NO₂(g). This oxidation supplies NO₂ involved in photochemical reactions and air pollution.

NO₂ can absorb light and dissociate to NO and an oxygen atom. The oxygen atom can combine with O₂ to form O₃; the overall smog chemistry also depends on other pollutants and reactions.

2CO(g)+2NO(g)→2CO₂(g)+N₂(g). CO is oxidised while nitrogen in NO is reduced; the catalyst facilitates the reaction but is regenerated overall.

There are 0.250 mol of the stated solute per litre of solution, not per litre of solvent before mixing. Amount is calculated with n=cV using the final solution volume.

Mass concentration=cM=0.0200×58.5=1.17 g L⁻¹. Molar mass converts amount of solute to mass without changing the volume basis.

It means 2.5 g solute per 100 mL of final solution. This equals 25 g L⁻¹; it is not 2.5 g per 100 g unless a mass-per-mass convention is specified.

For a dilute aqueous solution with density close to 1.00 kg L⁻¹, 1 mg kg⁻¹ is approximately 1 mg L⁻¹. The equivalence is not universal for concentrated liquids or other densities.

4.0 μg L⁻¹=0.0040 mg L⁻¹. At density near 1 kg L⁻¹ this is approximately 4.0 ppb by mass.

The solute amount stays constant, n=cV. The relation assumes no solute reaction or loss and uses the final total volume, not just the volume of water added.

The 1:1 reaction gives n_acid=0.1000×0.02500=0.002500 mol. Dividing by 0.01000 L gives 0.2500 mol L⁻¹.

Residual water would dilute the measured aliquot. Rinsing with that solution removes dilution from the inner surface while preserving the intended concentration.