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3 (a) Carbon dioxide is one of the gases in the Earth's atmosphere - Edexcel - GCSE Chemistry Combined Science - Question 3 - 2019 - Paper 1

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3 (a) Carbon dioxide is one of the gases in the Earth's atmosphere. The percentage of carbon dioxide in the Earth's atmosphere has changed over time. (i) Which row ... show full transcript

Worked Solution & Example Answer:3 (a) Carbon dioxide is one of the gases in the Earth's atmosphere - Edexcel - GCSE Chemistry Combined Science - Question 3 - 2019 - Paper 1

Step 1

Which row of the table shows the approximate percentage of carbon dioxide thought to be in the Earth's early atmosphere and how this percentage changed to form the Earth's atmosphere today?

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Answer

The row showing 95% carbon dioxide for the Earth's early atmosphere, which decreased to form the Earth's atmosphere today, is the correct answer. This reflects the significant reduction in atmospheric carbon dioxide due to various geological and biological processes over time.

Step 2

Explain two factors that cause the percentage of carbon dioxide in today’s atmosphere to vary.

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Answer

Factor 1: Natural processes such as photosynthesis and respiration impact carbon dioxide levels. Plants absorb CO₂ during photosynthesis, reducing its concentration, while respiration from animals returns CO₂ to the atmosphere.

Factor 2: Human activities, including the burning of fossil fuels and deforestation, increase carbon dioxide concentrations. These activities release significant amounts of CO₂, contributing to variations in atmospheric levels.

Step 3

Explain why carbon dioxide has a low boiling point.

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Answer

Carbon dioxide has a low boiling point due to the weak intermolecular forces between its molecules. As a simple molecular compound, CO₂ experiences only van der Waals forces, which are easily overcome with relatively low energy, resulting in its low boiling point of -78.5°C.

Step 4

Calculate the number of molecules in 0.11 g of carbon dioxide.

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Answer

First, we find the number of moles of CO₂ in 0.11 g:

extNumberofmoles=0.11extg44extg/mol=0.0025extmoles ext{Number of moles} = \frac{0.11 ext{ g}}{44 ext{ g/mol}} = 0.0025 ext{ moles}

Next, using Avogadro's number to find the number of molecules:

extNumberofmolecules=0.0025extmoles×6.02×1023extmolecules/mole=1.51×1021extmolecules ext{Number of molecules} = 0.0025 ext{ moles} \times 6.02 \times 10^{23} ext{ molecules/mole} = 1.51 \times 10^{21} ext{ molecules}

Rounding to two significant figures, the answer is 1.5 × 10²¹ molecules.

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