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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Answer:
A
Explanation:
Argon has 17 protons and 18 neutron
atomic number is the same as the protons number.
Answer:
1
Explanation:
4 HBr + O2 → 2H 20 + 2Br 2
...............
When in water, MgCl2 dissociates into magnesium ions and Cl- ions and NaOH into Na and OH ions. The equation is as follows:
MgCl2 = Mg2+ + 2Cl-
NaOH = Na+ + OH-
The initial concentrations are as follows:
[Mg2+] = .220(<span> 2.47x10^-4) / .220+.180 = 1.36x10^-4 M Mg2+
</span>[OH-] = .180 (3.52x10^-4) / .220+.180 = 1.58x10^-4 M OH-
