The preparation of lead (ii) sulphate from lead (ii) carbonate occurs in two steps:
- insoluble lead carbonate is converted to soluble lead (ii) nitrate
- soluble lead (ii) nitrate is reacted with sulphuric acid to produce lead (ii) sulphate.
<h3>How can a solid sample of lead (ii) sulphate be prepared from lead (ii) carbonate?</h3>
Lead (ii) carbonate and lead (ii) sulphate are both insoluble salts of lead.
In order to prepare lead (ii) sulphate, a two step process is performed.
In the first step, Lead (ii) carbonate is reacted with dilute trioxonitrate (v) acid to produce lead (ii) nitrate.
- PbCO₃ + 2HNO₃ → Pb(NO₃)₂ + CO₂ + H₂O
In the second step, dilute sulfuric acid is reacted with the lead (ii) nitrate to produce insoluble lead (ii) sulphate which is filtered and dried.
- Pb(NO₃)₂ + H₂SO₄ → PbSO₄ + 2HNO₃
In conclusion, lead (ii) sulphate is prepared in two steps.
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In general terms, exothermic reactions release energy, so the energy goes from the system to the surroundings.
Let's eliminate these one by one.
The first pair would not be the same, as X would most likely be in group IA, and Y would be in group VIIA, because of their tendency to gain and lose electrons.
The second pair would also violate the same rule, but X would most likely be in group IIA, and Y would most likely be in group VIA.
The third pair would not be the same, as X is most likely in group VIIA, and since Y has eight valence electrons, it is most likely a noble gas.
The final pair has X with atomic number 15, making it phosphorous. Phosphorous wants to gain 3 electrons to have a full octet of 8 outer "valence" electrons, and Y would also like to gain 3 electrons. This means it is possible that the final pair would be in the same group.
Answer:
The pH of the solution is 8.0.
Explanation:
taking the test rn
