Aussie Maths & Science Teachers: Save your time with SmarterEd
A geochemist analyzes a rock sample and finds it contains 15 600 kg of ore. Chemical analysis shows that the ore contains 3.8% w/w of chromium metal.
Calculate the mass of chromium in the rock sample. Express your answer in scientific notation. (2 marks)
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\(5.93 \times 10^2\ \text{kg}\)
| Mass of chromium | \(= 3.8\% \times 15\,600\ \text{kg}\) | |
| \(=0.038 \times 15\,600\ \text{kg}\) | ||
| \(=5.93 \times 10^2\ \text{kg}\) |
250,000 tonnes of iron ore contains 3.2% w/w of magnetite, \(\ce{Fe3O4}\).
What is the mass of magnetite?
\(B\)
\(\Rightarrow B\)
A mixture of sand and salt was provided to a group of students for them to determine its percentage composition by mass.
They added water to the sample before using filtration and evaporation to separate the components.
During the evaporation step, the students noticed white powder ‘spitting’ out of the basin onto the bench, so they turned off the Bunsen burner and allowed the remaining water to evaporate overnight.
After filtering, they allowed the filter paper to dry overnight before weighing. An electronic balance was used to measure the mass of each component to two decimal places.
The results were recorded as shown:
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a. % Sand = 70.88%, % Salt = 19.71%
b. Experiment validity:
| a. | \(\text{Sand mass}\ \) | \(\text{ = Mass of dried filter paper – Mass of filter paper}\) |
| \(= 11.95-0.80 = 11.15\ \text{g}\) |
| \(\text{Salt mass}\) | \(\text{ = Mass of dried filter paper – Mass of filter paper}\) | |
| \(= 36.60-33.50 = 3.10\ \text{g}\) |
\(\text{% sand} = \left(\dfrac{\text{Mass of sand}}{\text{Mass of original mixture}}\right) \times 100= \left(\dfrac{11.15 \ \text{g}}{15.73 \ \text{g}}\right) \times 100 = 70.88\%\)
\(\text{% salt} = \left(\dfrac{\text{Mass of salt}}{\text{Mass of original mixture}}\right) \times 100 = \left(\dfrac{3.10 \ \text{g}}{15.73 \ \text{g}}\right) \times 100 = 19.71\%\)
b. Experiment validity:
Silver oxide, when heated, decomposes according to the following chemical equation:
\(\ce{2Ag2O -> 4Ag + O2}\)
231.74 grams of silver oxide (\(\ce{Ag2O}\)) is heated, and the metallic silver (\(\ce{Ag}\)) residue is weighed and recorded in the table below.
\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{Ag2O} \rule[-1ex]{0pt}{0pt} & \text{231.74 grams} \\
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{Ag} \rule[-1ex]{0pt}{0pt} & \text{216.0 grams} \\
\hline
\end{array}
Calculate the percentage of oxygen (\(\ce{O2}\)), by weight, released from the silver oxide during this process. (2 marks)
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\(\%\ce{O2} = 6.8\%\)
\(\text{Initial mass of}\ \ce{Ag2O} = 231.74\ \text{grams}\)
\(\text{Mass of}\ \ce{Ag} = 216.0\ \text{grams}\)
\(\text{Mass of}\ \ce{O2} = 231.74-216.0 = 15.74\ \text{grams}\)
\(\%\ce{O2}\ \text{released}\ = \left(\dfrac{15.74}{231.74} \times 100\%\right) = 6.8\%\)
Calcium carbonate, when heated, decomposes according to the following chemical equation:
\(\ce{CaCO3 -> CaO(s) + CO2(g) }\)
60.0 grams of calcium carbonate is heated and the calcium oxide residue produced is weighed and recorded in the table below.
\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{CaCO3} \rule[-1ex]{0pt}{0pt} & \text{60.0 grams} \\
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{CaO} \rule[-1ex]{0pt}{0pt} & \text{33.6 grams} \\
\hline
\end{array}
Determine the percentage of carbon dioxide, by weight, released from the calcium carbonate during this process. (2 marks)
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\(\%\ce{CO2} = 44.0\%\)
\(\text{Mass of}\ \ce{CaCO3} = 60.0\ \text{grams}\)
\(\text{Mass of}\ \ce{CO2} = 60.0-33.6 = 26.4\ \text{grams}\)
\(\%\ce{CO2}\ \text{released}\ = \dfrac{26.4}{60.0} \times 100\% = 44.0\%\)
15.00 grams of Silver(\(\text{I}\)) oxide is heated until it decomposes and the silver produced is weighed and recorded in the table below.
\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{Ag2O} \rule[-1ex]{0pt}{0pt} & \text{15.00 grams} \\
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{Ag} \rule[-1ex]{0pt}{0pt} & \text{13.96 grams} \\
\hline
\end{array}
Determine the percentage of oxygen, by weight, in the sample of Silver(\(\text{I}\)) oxide. (2 marks)
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\(\%\ce{O} = 6.93\%\)
\(\text{Mass of}\ \ce{Ag} = 13.96\ \text{grams}\)
\(\text{Mass of}\ \ce{O} = 15.00-13.96 = 1.04\ \text{grams}\)
\(\%\ce{O} = \dfrac{1.04}{15.00} \times 100\% = 6.93\%\)
Calculate the mass of \(\ce{H}\) in 15.2 grams of \(\ce{H2O}\). Give your answer in grams correct to two decimal places. (2 marks)
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\(1.70\ \text{grams}\)
\(\ce{MM(H2O) = 2 \times 1.008 + 16.00 = 18.016\ \text{g mol}^{-1}}\)
\(\ce{\% of H = \dfrac{2.016}{18.016} \times 100 = 11.19\%}\)
\(\ce{m(H)\ \text{within 15.2 grams}\ H2O = 11.19\% \times 15.2 = 1.70\ \text{grams (to 2 d.p.)}} \)
Calculate the mass of \(\ce{H}\) within 100 grams of the compound \(\ce{H2CO3}\). Give your answer in grams correct to two decimal places. (2 marks)
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\(3.25\ \text{grams}\)
\(\ce{MM(H2CO3) = 2 \times 1.008 + 12.01 + 3 \times 16.00 = 62.026\ \text{g mol}^{-1}}\)
\(\ce{\% of H = \dfrac{2.016}{62.026} \times 100 = 3.25\%}\)
\(\ce{m(H)\ \text{within 100g}\ H2CO3 = 3.25\% \times 100 = 3.25\ \text{grams}} \)
Calculate the mass of \(\ce{O}\) in 8.4 grams of \(\ce{CO2}\). Give your answer in grams correct to two decimal places. (2 marks)
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\(6.11\ \text{g}\)
\(\ce{MM(CO2) = 12.01 + 2 \times 16.00 = 44.01\ \text{g mol}^{-1}}\)
\(\ce{\% of O = \dfrac{32.00}{44.01} \times 100 = 72.71\%}\)
| \(\ce{m(O\ \text{in 8.4 g})}\) | \(= 8.4 \times 72.71\% \) | |
| \(=6.11\ \text{g (2 d.p.)}\) |