MCQ
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- A

- ✓

- C

- D







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$Zn | ZnSO_4 \,(0.01\, M) | | CuSO_4\,(1.0\, M) | Cu$
When the concentration of $ZnSO_4$ is $1.0\,M$ and that of $CuSO_4$ is $0.01\,M$, the $emf$ changed to $E_2$ What is the relation between $E_1$ and $E_2$?
$CH \equiv CH\xrightarrow[{(excess)}]{{NaN{H_2}/liq.N{H_3}}}A\xrightarrow{{2C{H_3} - I}}B\xrightarrow[{quinolene}]{{{H_2}/Pd - BaS{O_4}}}C$
$C\xrightarrow[{quinolene}]{{{H_2}/Pd - BaS{O_4}}}\xrightarrow[{liq.\,N{H_3}}]{{NaN{H_2}}}D$

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The frequency of the wave is $\mathrm{x} \times 10^{19} \mathrm{~Hz}$. $x=$ . . . . (nearest integer)