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(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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(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 of... show full transcript

Worked Solution & Example Answer:(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 correct row is 5% for the early atmosphere and it has increased to form the Earth’s atmosphere today. Therefore, the answer is option A: 5% increased.

Step 2

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

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Answer

One factor that causes variation in carbon dioxide levels is human activity, particularly the burning of fossil fuels, which releases additional CO₂ into the atmosphere.

Another factor is natural processes, such as volcanic activity or ocean absorption, which can both release and absorb carbon dioxide, affecting 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 because it exists as a simple molecular compound with weak intermolecular forces (Van der Waals forces) between its molecules. These weak forces require little energy to break, hence allowing CO₂ to vaporize at low temperatures.

Step 4

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

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Answer

First, calculate the number of moles of CO₂:

Number of moles = mass / molar mass = 0.11 g / 44 g/mol = 0.0025 moles.

Using Avogadro's constant, we can then find the number of molecules:

Number of molecules = number of moles × Avogadro's constant = 0.0025 mol × 6.02 × 10²³ molecules/mol = 1.51 × 10²¹ molecules.

Therefore, the answer is approximately 1.5 × 10²¹ molecules, rounded to two significant figures.

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