| List-I (Test) | List-II (Identification) |
|---|---|
| (A) Bayer's test | (I) Phenol |
| (B) Ceric ammonium nitrate test | (II) Aldehyde |
| (C) Phthalein dye test | (III) Alcoholic-OH group |
| (D) Schiff's test | (IV) Unsaturation |
To solve the given matching problem between different chemical tests and their identifications, we need to understand each test's purpose:
Based on this analysis, the correct matching is: (A)-(IV), (B)-(III), (C)-(I), (D)-(II).
To solve the given problem involving matching tests with their corresponding chemical identification, we need to understand the specific tests and what they identify in organic chemistry.
Now, let's match List-I with List-II based on our understanding:
Therefore, the correct answer is:
(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Kjeldahl's method cannot be used for the estimation of nitrogen in which compound? 
In the group analysis of cations, Ba$^{2+}$ & Ca$^{2+}$ are precipitated respectively as
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :
