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From your answer to (vi) and the mass of FB 1 used in (a), calculate the percentage by mass of calcium carbonate in the limestone - CIE - A-Level Chemistry - Question 2 - 2014 - Paper 1

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From your answer to (vi) and the mass of FB 1 used in (a), calculate the percentage by mass of calcium carbonate in the limestone. [A; C, 12.0; O, 16.0; Ca, 40.1] ... show full transcript

Worked Solution & Example Answer:From your answer to (vi) and the mass of FB 1 used in (a), calculate the percentage by mass of calcium carbonate in the limestone - CIE - A-Level Chemistry - Question 2 - 2014 - Paper 1

Step 1

Calculate the percentage by mass of calcium carbonate in the limestone

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Answer

To calculate the percentage by mass of calcium carbonate in the limestone, use the formula:

extPercentagebymassofCaCO3=(mass of CaCO3mass of limestone)×100 ext{Percentage by mass of } CaCO_3 = \left( \frac{\text{mass of } CaCO_3}{\text{mass of limestone}} \right) \times 100

Where you gather the mass of calcium carbonate from the analysis in (vi) and the total mass of FB1 from (a). Plug in the values to find the percentage.

Step 2

Smallest volume of FB 2 added

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Answer

The smallest volume that could have been added by Student X would be the recorded volume minus the maximum error:

Smallest volume=48.50cm30.05cm3=48.45cm3\text{Smallest volume} = 48.50 \, cm^3 - 0.05 \, cm^3 = 48.45 \, cm^3

Step 3

Largest volume of FB 2 added

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The largest volume that could have been added by Student X would be the recorded volume plus the maximum error:

Largest volume=48.50cm3+0.05cm3=48.55cm3\text{Largest volume} = 48.50 \, cm^3 + 0.05 \, cm^3 = 48.55 \, cm^3

Step 4

How would the mass used for Student Y compare?

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Answer

In Student Y's experiment, even though the mass of FB 1 remains constant as in Student X's experiment, a different volume of FB 2 is used. When comparing the two students:

  • Lesser volume used by Student Y (47.70 cm³) may suggest a different calculation for the mass by analyzing the setup and results of their experiments.
  • It is crucial to ensure that all components react adequately to determine the correctness of the procedure.

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