Jshek
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Report Thread starter 7 months ago
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https://pmt.physicsandmathstutor.com...s%201%20QP.pdf

can someone explain 5a,5b and 5c i dont get it and why does a precipitate form... i never learn magnesium sulphate + NaOH or H2S04
and for 3a why do you divide by 74.1 and times 10

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https://pmt.physicsandmathstutor.com...s%201%20MS.pdf
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8013
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Report 7 months ago
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For 5a, you can first eliminate compound 2, as MgSO4 is soluble in water. Then you can find that it is compound 1 as both have sulfate ions, so there is no visible change. Mg(OH)2 solid is white, so it gives a white precipitate (insoluble salt).
For 5b, the compound has a carbonate ion provided that it has CO2 gas released. BaSO4 is less soluble than MgSO4 as solubility decreases down Group 2, so it has barium ion. So it is BaCO3.
For 5c, you can find that only compound 4 has no observable change with OH- ion as it has OH- ions itself, so compound 4 is the answer.
For 3a, g/dm3 divided by g/mol is mol/dm3
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