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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 approximate percentage of carbon dioxide thought to be in the Earth's early atmosphere is 95% (D), which decreased to form the Earth's atmosphere today.

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 is human activity, specifically the burning of fossil fuels which increases CO₂ levels in the atmosphere. Another factor is natural processes, such as photosynthesis in plants which decreases CO₂ 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 weak intermolecular forces (Van der Waals forces) between the molecules. These forces require less energy to overcome, allowing CO₂ to transition to gas at lower temperatures.

Step 4

Calculate the number of molecules in 0.11 g of carbon dioxide. Give your answer to two significant figures.

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Answer

First, calculate the number of moles:

extmoles=0.11extg44extg/mol=0.0025extmol ext{moles} = \frac{0.11 ext{ g}}{44 ext{ g/mol}} = 0.0025 ext{ mol}

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

extnumberofmolecules=0.0025extmol×6.02×1023 molecules/mol=1.51×1021 ext{number of molecules} = 0.0025 ext{ mol} \times 6.02 \times 10^{23} \text{ molecules/mol} = 1.51 \times 10^{21}

Thus, the answer is approximately 1.5×10211.5 \times 10^{21} molecules.

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