Resources and water: choose treatment, measure residue and recover copper
Diperkenalkan| English | Makna |
|---|---|
| potable water/ˈpəʊtəbl ˈwɔːtə/ | air minum |
| finite resource | sumber daya terbatas |
| renewable resource/rɪˈnjuːəbl rɪˈzɔːs/ | sumber daya terbarukan |
| sustainable development/səˈsteɪnəbl dɪˈveləpmənt/ | pembangunan berkelanjutan |
| sterilise/ˈsterəlaɪz/ | sterilisasi |
| desalination/dɪˌsælɪˈneɪʃn/ | desalinasi |
| reverse osmosis | osmosis terbalik |
| sedimentation/ˌsedɪmənˈteɪʃn/ | pengendapan |
| sludge/slʌdʒ/ | lumpur |
| effluent | air limbah |
| anaerobic digestion | pencernaan anaerobik |
| aerobic biological treatment | perlakuan biologis aerobik |
| phytomining | penambangan tumbuhan |
| bioleaching | biopengurasan |
| leachate/ˈliːtʃeɪt/ | cairan percolate |
Can clear water be unsafe?
- Clear-looking water may contain dissolved salts, harmful microbes or dissolved chemicals. Choose treatment for what must be removed, rather than assuming one filter solves every problem.
- This lesson covers resources, potable water 饮用水, waste-water branches and HT biological copper recovery. RP8 written knowledge does not replace a supervised practical record.
Match use to replenishment
- Earth's resources provide warmth, shelter, food and transport. Agriculture supplements food, timber and natural fibres; synthetic fibres can supplement or replace cotton or wool. Synthetic does not automatically mean more sustainable.
- A finite resource 有限资源 is not replenished on the time scale of use. A renewable resource 可再生资源 can be replenished, but harvest can still exceed growth. Sustainable development 可持续发展 meets present needs without compromising future generations' ability to meet theirs.
Check units and scale
- Stocks of 20 tonnes and 20 000 tonnes differ by $20\,000/20=1000=10^3$, three orders of magnitude. A stock figure alone does not show sustainability.
- A forest growing 1 200 tonnes yearly but harvested at 2 000 tonnes loses $2000-1200=800$ tonnes per year. Long-term management must address that imbalance and wider environmental effects.
Separate solids, microbes and salts
- Potable water has sufficiently low dissolved salts and harmful microbes to be safe to drink; it is not chemically pure. UK fresh supplies collect from rainfall in ground, rivers and lakes. Choose a suitable source, filter suspended solids, then sterilise 消毒 with chlorine, ozone or ultraviolet.
- Desalination 海水淡化 removes salts by distillation or reverse osmosis 反渗透. Distillation boils and condenses water vapour; reverse osmosis uses pressure and a suitable membrane. Both require substantial energy. Ordinary filtration and sterilising do not remove all dissolved salt.
Which route can remove dissolved salt from seawater?
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Distillation separates water by boiling and condensation, leaving most dissolved salts. Ordinary filters remove suspended solids and UV controls microbes; neither alone desalinate. Reverse osmosis is another appropriate salt-removal route.
RP8: measure residue and collect distilled water
- Compare water pH and dissolved-solid residue. Measure a known sample volume, weigh a dry container, evaporate safely, cool and reweigh; repeat to constant mass where appropriate. Residue mass equals final mass minus empty mass.
- A 50.0 cm³ sample leaving 0.120 g gives $c=\dfrac{0.120\ \mathrm{g}}{0.0500\ \mathrm{dm^3}}=2.40\ \mathrm{g/dm^3}$. Distillation must collect cooled vapour; boiling a sample dry alone is not distillation. Never drink laboratory samples or treat residue as a microbial test.
Which two statements make a water investigation defensible?
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Mass subtraction and known volume support the residue result. Safety needs additional evidence; never drink lab samples. Distillation requires collection and condensation of vapour.
Keep sewage branches distinct
- Screen large objects, remove grit, then use sedimentation 沉降 to separate sludge 污泥 from liquid effluent 出水. Sludge undergoes anaerobic digestion 厌氧消化 without oxygen; the effluent undergoes aerobic biological treatment 好氧生物处理 with oxygen.
- Industrial waste can require extra chemical treatment. Release-quality water is not automatically potable. Suitable ground water often needs less treatment than waste water under comparable local conditions, but source contamination can change the comparison.
Match the term with its precise meaning.
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Use the chemical structure, process or comparison boundary to distinguish the terms.
HT: concentrate compounds, then recover metal
- Phytomining 植物采矿 uses plants to absorb metal compounds; harvest and burn them to get ash containing compounds. Bioleaching 生物浸出 uses bacteria to produce a leachate 浸出液 containing metal compounds. Neither directly yields a pure copper object.
- Recover copper from a suitable solution by iron displacement, $Fe+CuSO_4\rightarrow FeSO_4+Cu$, or electrolysis. Compare reduced rock excavation with processing time, yield, energy and waste. “Biological” does not mean free or impact-free.
Burning phytomining plants directly produces pure copper metal in the ash.
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The ash contains metal compounds, not automatically pure copper. Further processing to a suitable solution and recovery by iron displacement or electrolysis are needed.
Name the bacterial process producing a leachate containing metal compounds.
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Bioleaching uses bacteria to produce a solution containing metal compounds. It is distinct from phytomining, which concentrates compounds in plants before harvest and burning.
Transfer to an unfamiliar water sample
- Independent sample: 25.0 cm³ leaves 0.050 g residue, giving $0.050/0.0250=2.00\ \mathrm{g/dm^3}$. This measurement alone cannot establish drinking safety; investigate microbes and any relevant dissolved chemicals.
- Explain a treatment using the impurity and separation mechanism. Then state its limitation: filters remove suspended solids, sterilising controls microbes, and salt removal needs a suitable desalination process.
Choose by source and required quality. Preserve separate waste-water branches, units in residue calculations and the distinction between metal compounds and recovered metal.
A 25.0 cm³ water sample leaves 0.050 g dry residue. Calculate residue concentration in g/dm³.
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25.0 cm³ = 0.0250 dm³. Concentration = residue mass/sample volume = 0.050 g / 0.0250 dm³ = 2.00 g/dm³. This does not measure microbial safety.