What is the structure of C ?
What is the structure of C ?
The Correct Option is A
Solution and Explanation
Thus the correct answer is Option 1.
Top Questions on Reaction Mechanisms & Synthesis
- The reaction \( \text{A}_2 + \text{B}_2 \to 2\text{AB} \) follows the mechanism: \[ \text{A}_2 \xrightarrow{k_1} \text{A} + \text{A} \, (\text{fast}), \quad \text{A} + \text{B}_2 \xrightarrow{k_2} \text{AB} + \text{B} \, (\text{slow}), \quad \text{A} + \text{B} \to \text{AB} \, (\text{fast}). \] The overall order of the reaction is:
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Consider the gas phase reaction: \[ CO + \frac{1}{2} O_2 \rightleftharpoons CO_2 \] At equilibrium for a particular temperature, the partial pressures of \( CO \), \( O_2 \), and \( CO_2 \) are found to be \( 10^{-6} \, {atm} \), \( 10^{-6} \, {atm} \), and \( 16 \, {atm} \), respectively. The equilibrium constant for the reaction is ......... \( \times 10^{10} \) (rounded off to one decimal place).
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- Consider the following gas-phase reaction: \[ 2 {SO}_2 + {O}_2 \rightleftharpoons 2 {SO}_3 \] If the enthalpy of reaction is negative, which one of the following conditions promotes a higher equilibrium concentration of SO\(_3\)?
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Molten steel at 1900 K having dissolved hydrogen needs to be vacuum degassed. The equilibrium partial pressure of hydrogen to be maintained to achieve 1 ppm (mass basis) of dissolved hydrogen is ......... Torr (rounded off to two decimal places). Given: For the hydrogen dissolution reaction in molten steel \( \left( \frac{1}{2} {H}_2(g) = [{H}] \right) \), the equilibrium constant (expressed in terms of ppm of dissolved H) is: \[ \log_{10} K_{eq} = \frac{1900}{T} + 2.4 \] 1 atm = 760 Torr.
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Questions Asked in JEE Main exam
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\( S_1 = \{A = [a_{ij}] \in M : A = A^T \text{ and } a_{ij} \in S, \forall i, j\} \),
\( S_2 = \{A = [a_{ij}] \in M : A = -A^T \text{ and } a_{ij} \in S, \forall i, j\} \),
\( S_3 = \{A = [a_{ij}] \in M : a_{11} + a_{22} + a_{33} = 0 \text{ and } a_{ij} \in S, \forall i, j\} \).
If \(n(S_1 \cup S_2 \cup S_3) = 125\), then \( \alpha \) equals:- JEE Main - 2025
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Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
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