3 (a) Carbon dioxide is one of the gases in the Earth's atmosphere - Edexcel - GCSE Chemistry Combined Science - Question 3 - 2019 - Paper 1
Question 3
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 row A: 5%. This percentage has increased to a higher level in 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
Factor 1: Human Activity - The burning of fossil fuels, deforestation, and industrial processes release significant amounts of carbon dioxide into the atmosphere, contributing to higher concentrations.
Factor 2: Natural Processes - Natural activities such as volcanic eruptions and oceanic absorption of CO₂ can influence the levels of carbon dioxide, leading to variations.
Step 3
Explain why carbon dioxide has a low boiling point.
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Answer
Carbon dioxide has a low boiling point primarily due to the weak intermolecular forces (Van der Waals forces) present between its molecules. These forces require less energy to overcome, allowing CO₂ to change from liquid to gas at low temperatures.
Step 4
Calculate the number of molecules in 0.11 g of carbon dioxide.
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Answer
To find the number of molecules in 0.11 g of carbon dioxide, use the following steps:
Calculate the number of moles:
Number of moles=molar massmass=44 g/mol0.11 g=0.0025 mol
Use Avogadro's number to find the number of molecules:
Number of molecules=0.0025 mol×6.02×1023 molecules/mol=1.51×1021 molecules
Thus, the answer to two significant figures is approximately 1.5×1021 molecules.