\[ \text{CH}_4\,(g) + 2\text{O}_2\,(g) \rightarrow \text{CO}_2\,(g) + 2\text{H}_2\text{O}\,(l) \]
\[ n_{\text{CO}_2} = \frac{22}{44} = 0.5 \, \text{moles} \]
\[ \text{So moles of CH}_4 \text{ required} = 0.5 \, \text{moles} \]
\[ \text{i.e., } 50 \times 10^{-2} \, \text{mole} \]
\[ x = 50 \]
List-I | List-II | ||
(A) | [Co(NH3)5(NO2)]Cl2 | (I) | Solvate isomerism |
(B) | [Co(NH3)5(SO4)]Br | (II) | Linkage isomerism |
(C) | [Co(NH3)6] [Cr(CN)6] | (III) | Ionization isomerism |
(D) | [Co(H2O)6]Cl3 | (IV) | Coordination isomerism |
List-I | List-II | ||
(A) | 1 mol of H2O to O2 | (I) | 3F |
(B) | 1 mol of MnO-4 to Mn2+ | (II) | 2F |
(C) | 1.5 mol of Ca from molten CaCl2 | (III) | 1F |
(D) | 1 mol of FeO to Fe2O3 | (IV) | 5F |
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32