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This question is about metals and the reactivity series - AQA - GCSE Chemistry - Question 2 - 2020 - Paper 1

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This question is about metals and the reactivity series. 02.1 Which two statements are properties of most transition metals? Tick (✓) two boxes. - They are soft m... show full transcript

Worked Solution & Example Answer:This question is about metals and the reactivity series - AQA - GCSE Chemistry - Question 2 - 2020 - Paper 1

Step 1

Which two statements are properties of most transition metals?

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Answer

The properties of most transition metals include:

  • They form ions with different charges.
  • They have high melting points.

Step 2

Explain how these observations show that silver is less reactive than copper.

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Answer

The pale grey crystals observed in the reaction indicate that silver has been displaced from the silver nitrate solution. This suggests that copper is more reactive than silver, as it can displace silver ions from the solution. Additionally, the blue color of the solution suggests the formation of copper ions, which further confirms that copper is replacing silver.

Step 3

Plan an investigation to identify the three metals by comparing their reactions with dilute hydrochloric acid.

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Answer

  1. Add the metals X, Y, and Z separately to a test tube containing dilute hydrochloric acid.
  2. Observe and record the reaction of each metal.
    • For copper, expect no reaction; there should be no bubbles or change.
    • For magnesium, look for vigorous bubbling and a temperature increase, and note that it forms a colourless solution.
    • For iron, note bubbles, but expect a less vigorous reaction than magnesium, and observe a coloured solution.
  3. Control variables to ensure accurate results:
    • Use the same concentration and volume of hydrochloric acid for each test.
    • Ensure each metal sample has the same mass.
    • Maintain the same temperature during the experiments.

Step 4

Calculate the relative atomic mass (A_r) of metal M.

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Answer

To calculate the relative atomic mass (A_r) of metal M, use the formula:

Ar=(203×30)+(205×70)100A_r = \frac{(203 \times 30) + (205 \times 70)}{100}

Calculating:

  • Numerator: (203 × 30) + (205 × 70) = 6090 + 14350 = 20440
  • Denominator: 100

Thus, the relative atomic mass A_r = \frac{20440}{100} = 204.4.

Therefore, the answer to 1 decimal place is 204.4.

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