Diana's answer to Diana's Hong Kong question.
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Date Posted:
6 years ago
Mary's method is more accurate. Firstly, she is using an electronic balance accurate to 0.1g. Benson is using a beaker which has scale interval of 1 cm³. If the volume reading is between intervals, it would have to be estimated. and therefore is more inaccurate.
Secondly, Benson is measuring the volume of liquid calcium. It is known that liquid calcium occupies a greater volume than solid calcium of the same mass (read up on chemistry of solids and liquids if you're unsure). So he would be overestimating the volume.
Lastly, it is unlikely that Benson can transfer all the liquid calcium to the beaker from the heating pan. So there is some loss during transfer and accuracy is lowered. Furthermore, by heating the calcium, side reactions may occur with other compounds that may be present in the pan, leading to formation of other compounds that would further reduce the accuracy of the measurement.
Secondly, Benson is measuring the volume of liquid calcium. It is known that liquid calcium occupies a greater volume than solid calcium of the same mass (read up on chemistry of solids and liquids if you're unsure). So he would be overestimating the volume.
Lastly, it is unlikely that Benson can transfer all the liquid calcium to the beaker from the heating pan. So there is some loss during transfer and accuracy is lowered. Furthermore, by heating the calcium, side reactions may occur with other compounds that may be present in the pan, leading to formation of other compounds that would further reduce the accuracy of the measurement.
Using Benson's method, a total volume of __ cm³ is obtained for 20 solid blocks of calcium.
So average volume of a calcium block
= __ cm³ ÷ 20 = __ cm³ ÷ 10 ÷ 2
Dividing by 10 results in one decimal place (0.1 cm³) Dividing by another 2 would result in two decimal places, and the 2nd decimal place is always 0 or 5.
And since the volume measured by the beaker is to the nearest 1 cm (whole number), your average volume of the block can only be to 0.05 cm³ .(let's say it's 63 cm³, 63 cm³ ÷ 20 = 3.15 cm³ per block)
(Unless you're reading the halfway mark of intervals (nearest 0.5 cm). Then estimation to 0.01cm³ is possible. Eg. If the volume is 65.5 cm³, then 65.5 cm³ ÷ 20 = 3.275 cm³
which can be rounded to the nearest 0.01 cm³)
For Mary's method, it is x g ÷ 1.6 g/cm³
The weight of calcium blocks measured using the weighing balance is to 0.1 g (1 decimal place). The density of calcium is also to 1 decimal place.
It is possible to get a density of up to 4 decimal places (eg if weight is 5.1g, 5.1g ÷ 1.6 g/cm³ = 3.1875 cm³, which can be rounded to the nearest 0.01cm³)
So only Mary's method is possible
So average volume of a calcium block
= __ cm³ ÷ 20 = __ cm³ ÷ 10 ÷ 2
Dividing by 10 results in one decimal place (0.1 cm³) Dividing by another 2 would result in two decimal places, and the 2nd decimal place is always 0 or 5.
And since the volume measured by the beaker is to the nearest 1 cm (whole number), your average volume of the block can only be to 0.05 cm³ .(let's say it's 63 cm³, 63 cm³ ÷ 20 = 3.15 cm³ per block)
(Unless you're reading the halfway mark of intervals (nearest 0.5 cm). Then estimation to 0.01cm³ is possible. Eg. If the volume is 65.5 cm³, then 65.5 cm³ ÷ 20 = 3.275 cm³
which can be rounded to the nearest 0.01 cm³)
For Mary's method, it is x g ÷ 1.6 g/cm³
The weight of calcium blocks measured using the weighing balance is to 0.1 g (1 decimal place). The density of calcium is also to 1 decimal place.
It is possible to get a density of up to 4 decimal places (eg if weight is 5.1g, 5.1g ÷ 1.6 g/cm³ = 3.1875 cm³, which can be rounded to the nearest 0.01cm³)
So only Mary's method is possible
Thx I understand
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