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Science · 2024 · 5 marks
CBSE 2024 · Region 4 · Set 2 · Q34
When lead nitrate is heated strongly in a boiling tube, two gases are liberated and a solid residue is left behind in the test tube.(i)Name the type of chemical reaction and define it.(ii)Write the name and formula of the coloured gas liberated.(iii)Write the balanced chemical equation for the reaction.(iv)Name the residue left in the test tube and state the method of testing its nature (acidic/basic).(b)Write balanced chemical equation for the following word equation. Lead nitrate + Potassium iodide → Lead iodide + Potassium nitrate Is this a double displacement reaction? Justify your answer. Name the compound precipitated and write the ions present in it.(ii)Write the method of preparation of $\displaystyle \mathrm{Ca}(\mathrm{OH})_{2}$. What happens when $\displaystyle \mathrm{CO}_{2}$ is passed through it ? Write balanced chemical equation for the reaction involved.
When lead nitrate is heated strongly in a boiling tube, two gases are liberated and a solid residue is left behind in the test tube.
(i)
Name the type of chemical reaction and define it.
(ii)
Write the name and formula of the coloured gas liberated.
(iii)
Write the balanced chemical equation for the reaction.
(iv)
Name the residue left in the test tube and state the method of testing its nature (acidic/basic).
(b)
Write balanced chemical equation for the following word equation. Lead nitrate + Potassium iodide → Lead iodide + Potassium nitrate Is this a double displacement reaction? Justify your answer. Name the compound precipitated and write the ions present in it.
(ii)
Write the method of preparation of $\displaystyle \mathrm{Ca}(\mathrm{OH})_{2}$. What happens when $\displaystyle \mathrm{CO}_{2}$ is passed through it ? Write balanced chemical equation for the reaction involved.
Marking-scheme solution
(i)
• Decomposition reaction
(ii) Nitrogen dioxide, \(\displaystyle \mathrm{NO}_{2}\)
(iii)
Residue left - Lead oxide.
Dissolve the residue in water and test the solution using litmus paper/Universal indicator. The colour of the litmus paper changes to blue indicating that lead oxide is basic in nature.
(b) (i) \(\displaystyle \mathrm{Pb}\left(\mathrm{NO}_{3}\right)(\mathrm{aq})+2 \mathrm{KI}(\mathrm{aq}) \longrightarrow \mathrm{PbI}_{2(\mathrm{ppt})}+2 \mathrm{KNO}_{3(\mathrm{aq})}\)
Yes, it is a double displacement reaction.
In this reaction, exchange of ions between the reactants (Lead nitrate and potassium iodide) is taking place.
Lead iodide; \(\displaystyle \left[P b^{2+}\right]\left[\mathrm{I}^{-}\right]\)
(ii) Calcium hydroxide is prepared on adding water to quicklime (calcium oxide) /
\[\underset{\text { (Quick lime) }}{\mathrm{CaO}(\mathrm{~s})}+\mathrm{H}_{2} \mathrm{O}(1) \rightarrow \underset{\text { (Slaked lime) }}{\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{aq})}+\text { Heat }
\]
When \(\displaystyle \mathrm{CO}_{2}\) is passed through \(\displaystyle \mathrm{Ca}(\mathrm{OH})_{2}\) It turns milky white/ calcium carbonate is formed.
(Calcium
hydroxide)
(ii) \(\displaystyle \mathrm{f}=-20 \mathrm{~cm} ; \mathrm{h}=5 \mathrm{~cm} ; \mathrm{v}=-15 \mathrm{~cm}\)
\[\begin{aligned}
\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \quad \text { or } & \\
\frac{1}{u}=\frac{1}{v}-\frac{1}{f} & =\frac{1}{(-15)}-\frac{1}{(-20)} \\
& =\frac{-1}{60 c m}
\end{aligned}
\]
or \(\displaystyle u=-60 \mathrm{~cm}\) object is at a distance of $\displaystyle 60$ cm from the lens
Size of the image(magnification): \(\displaystyle \mathrm{m}=\frac{h^{\prime}}{h}=\frac{v}{u}\)
\[h^{\prime}=\frac{v}{u} \times h=\frac{(-15)}{(-60)} \times 5=1.25 \mathrm{~cm}
\]
(c)
\(\displaystyle 16 \Omega\)
Justification: According to Ohm's law when same current flows, the potential difference across a higher resistance is always higher./
Potential difference across \(\displaystyle 16 \Omega=\mathrm{V}=\mathrm{IR} \quad=0.2 \times 16=3.2 \mathrm{~V}\)
Potential difference across \(\displaystyle 8 \Omega=\mathrm{V}=\mathrm{IR}_{(\text {total })}=0.2 \times 4=0.8 \mathrm{~V}\)
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