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A student used the internet to find information about the percentages of different gases in the Earth’s early atmosphere - Edexcel - GCSE Chemistry - Question 1 - 2013 - Paper 1

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A student used the internet to find information about the percentages of different gases in the Earth’s early atmosphere. She was surprised to find the information g... show full transcript

Worked Solution & Example Answer:A student used the internet to find information about the percentages of different gases in the Earth’s early atmosphere - Edexcel - GCSE Chemistry - Question 1 - 2013 - Paper 1

Step 1

a) One of the gases in the table is present in a much larger amount in today’s atmosphere. State the name of this gas.

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Answer

The gas present in a much larger amount today is nitrogen.

Step 2

b) A gas not named in the table makes up about 21% of today’s atmosphere. State the name of this gas.

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Answer

The gas that constitutes about 21% of today's atmosphere is oxygen.

Step 3

c) Complete the sentence by putting a cross (X) in the box next to your answer. The amount of carbon dioxide in the early atmosphere was reduced by

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Answer

The correct answer is D) the gas dissolving in oceans.

Step 4

d) The information given on two websites is very different. Explain why it is difficult to be certain about the composition of the Earth’s early atmosphere.

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Answer

It is challenging to determine the composition of the Earth's early atmosphere because:

  • There were no humans present to directly observe or record the atmosphere.
  • No reliable measurements were taken at that time.
  • Different sources may conflict due to varying evidence or interpretations.
  • Scientific data may be limited or based on insufficient records, leading to inconsistencies.

Step 5

e) i) Calculate the percentage of oxygen in this sample of air.

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Answer

To calculate the percentage of oxygen:

The initial volume of air = 50.0 cm³ Volume of gas remaining after the reaction = 41 cm³

Therefore, the volume of oxygen reacted = 50.0 cm³ - 41 cm³ = 9.0 cm³.

The percentage of oxygen is calculated as: 9.050.0×100=18%\frac{9.0}{50.0} \times 100 = 18\%

Step 6

e) ii) The word equation for the reaction is. Balance the equation for this reaction by putting numbers in the spaces provided.

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Answer

The balanced equation for the reaction is:

2Cu+O22CuO2 \text{Cu} + \text{O}_2 \rightarrow 2 \text{CuO}

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