Calculate
∣xyx+yyx+yxx+yxy∣ \begin{vmatrix} x & y & x + y \\ y & x + y & x \\ x + y & x & y \end{vmatrix} xyx+yyx+yxx+yxy
2HC=CH2Cl2→ANH4Cl2HC = CH_2\text{Cl}_2 \xrightarrow{\text{A}} \text{NH}_4\text{Cl}2HC=CH2Cl2ANH4Cl →B\xrightarrow{\text{B}}B Identify the compounds A and B\text{Identify the compounds A and B}Identify the compounds A and B
Consider the following reactions, C(s) + O2(g) → CO2(g), Δ\DeltaΔ H = -94 kcal\text{ kcal} kcal 2CO(g) + O2 → 2CO2(g), Δ\DeltaΔ H = -135.2 kcal\text{ kcal} kcal Then, the heat of formation of CO (g) is:
NH3NH_3NH3 molecule attract H+H^+H+ ion towards itself to form ammonium ion NH4+{NH}_4^+NH4+ through:
In the redox reaction MnO4−+C2O42−→Mn2++CO2+H2O \text{MnO}_4^- + \text{C}_2\text{O}_4^{2-} \to \text{Mn}^{2+} + \text{CO}_2 + \text{H}_2\text{O} MnO4−+C2O42−→Mn2++CO2+H2O The correct coefficients of C2O42− \text{C}_2\text{O}_4^{2-} C2O42− and H+ \text{H}^+ H+ are
For the reactions C+O2→CO2C + O2 \rightarrow CO2C+O2→CO2; ΔH=−393 kJ2Zn+O2→2ZnO\, \Delta H = -393 \, \text{kJ} 2Zn + O2 \rightarrow 2ZnOΔH=−393kJ2Zn+O2→2ZnO, ΔH=−412 kJ\, \Delta H = -412 \, \text{kJ}ΔH=−412kJ the correct statement is:\text{the correct statement is:}the correct statement is:
If f(x)={1−sinx(n−2x)2ifx≠π2log(sinx)⋅log(1+π4x+x2)ifx=π2 f(x) = \begin{cases} \frac{1 - \sin x}{(n - 2x)^2} & \text{if} \quad x \neq \frac{\pi}{2} \log (\sin x) \cdot \log \left( 1 + \frac{\pi}{4x + x^2} \right) & \text{if} \quad x = \frac{\pi}{2} \end{cases} f(x)={(n−2x)21−sinxifx=2πlog(sinx)⋅log(1+4x+x2π)ifx=2π is continuous at x=π2 x = \frac{\pi}{2} x=2π, then k k k is equal to
The inverse of matrix (01−14−343−34) \begin{pmatrix} 0 & 1 & -1 \\ 4 & -3 & 4 \\3 & -3 & 4 \end{pmatrix} 0431−3−3−144 is
Choose the most appropriate option. If A A A is a square matrix such that A2=A A^2 = A A2=A and B=I B = I B=I, then AB+BA+I−(I−A)2 AB + BA + I - (I - A)^2 AB+BA+I−(I−A)2 is equal to:
Choose the most appropriate option. The value of limx→alogx−1x−a\lim_{x \to a} \frac{\log x - 1}{x - a}limx→ax−alogx−1 is equal to