Let $ f: \mathbb{R} \to \mathbb{R} $ be a twice differentiable function such that $$ f''(x)\sin\left(\frac{x}{2}\right) + f'(2x - 2y) = (\cos x)\sin(y + 2x) + f(2x - 2y) $$ for all $ x, y \in \mathbb{R} $. If $ f(0) = 1 $, then the value of $ 24f^{(4)}\left(\frac{5\pi}{3}\right) $ is:
Let A = \(\begin{bmatrix} \log_5 128 & \log_4 5 \log_5 8 & \log_4 25 \end{bmatrix}\) \). If \(A_{ij}\) is the cofactor of \( a_{ij} \), \( C_{ij} = \sum_{k=1}^2 a_{ik} A_{jk} \), and \( C = [C_{ij}] \), then \( 8|C| \) is equal to:
A molecule with the formula $ \text{A} \text{X}_2 \text{Y}_2 $ has all it's elements from p-block. Element A is rarest, monotomic, non-radioactive from its group and has the lowest ionization energy value among X and Y. Elements X and Y have first and second highest electronegativity values respectively among all the known elements. The shape of the molecule is:
A transition metal (M) among Mn, Cr, Co, and Fe has the highest standard electrode potential $ M^{n}/M^{n+1} $. It forms a metal complex of the type $[M \text{CN}]^{n+}$. The number of electrons present in the $ e $-orbital of the complex is ... ...
Consider the following electrochemical cell at standard condition. $$ \text{Au(s) | QH}_2\text{ | QH}_X(0.01 M) \, \text{| Ag(1M) | Ag(s) } \, E_{\text{cell}} = +0.4V $$ The couple QH/Q represents quinhydrone electrode, the half cell reaction is given below: $$ \text{QH}_2 \rightarrow \text{Q} + 2e^- + 2H^+ \, E^\circ_{\text{QH}/\text{Q}} = +0.7V $$
0.1 mol of the following given antiviral compound (P) will weigh .........x $ 10^{-1} $ g.