| 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)
Among the following cations, the number of cations which will give characteristic precipitate in their identification tests with
\(K_4\)[Fe(CN)\(_6\)] is : \[ {Cu}^{2+}, \, {Fe}^{3+}, \, {Ba}^{2+}, \, {Ca}^{2+}, \, {NH}_4^+, \, {Mg}^{2+}, \, {Zn}^{2+} \]
Match List I with List II:
Choose the correct answer from the options given below:
Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to:
The expression given below shows the variation of velocity \( v \) with time \( t \): \[ v = \frac{At^2 + Bt}{C + t} \] The dimension of \( A \), \( B \), and \( C \) is:
The dimensions of a physical quantity \( \epsilon_0 \frac{d\Phi_E}{dt} \) are similar to [Symbols have their usual meanings]
