Class 12th

Required answer:

$=(350*16)+\frac{1}{5}(150*11)=5600+330=5930$

Required number of females:

$=\frac{32}{100}*15000=4800$

$\left|2A\right|=2^3\left|A\right|$

replace A by adj 2A

  $\left|2adj2A\right|=2^3\left|adj2A\right|$           

$=2^3{\left|2A\right|}^2=2^3{\left(2^3\left|A\right|\right)}^2$

$=2^9{\left|A\right|}^2$              

Again replace A by (adj A)

|2adj 2 (adj (adj 2A)| = 2⁹ |adj 2A|⁴

= 2⁹ (|2A|²)⁴

->|A²| = 4

Elemente- gain enthalpy

F-333

Cl-349

Te-190

Po-174

$f\left(x\right)=2x^3-3x^2-12x$  

  $f\text{'}\left(x\right)=6x^2-6x-12$             

= 6 (x – 2) (x + 1)

a = -1, b = 2

$A={\displaystyle \underset{-1}{\overset{0}{\int }}\left(2x^3-3x^2-12x\right)dx-{\displaystyle \underset{0}{\overset{2}{\int }}\left(2x^3-3x^2-12x\right)dx=\frac{57}{2}}}$

->4A = 114

Kindly consider the following figure

Equation of required plane

$\left(x+y+4z-16\right)+\lambda \left(-x+y+z-6\right)=0$ it passes (1, 2, 3)

$\Rightarrow -1+\lambda \left(-2\right)=0$

$\Rightarrow \lambda =-\frac{1}{2}$

$\therefore$ Equation of plane

$\left(1-\lambda \right)x+\left(1+\lambda \right)y+\left(4+\lambda \right)z-16-6\lambda =0$              

$\frac{3}{2}x+\frac{1}{2}y+\frac{7}{2}z-13=0$              

$\Rightarrow 3x+y+7z=26$              

  (4, 2, 2) not satisfying the plane

Reactivity order of halogens and inter halogens are F₂ > ClF > Cl₂ and ClF forms HOCl and HF after the reaction with water.

Required answer: $=150*38:\left(35000*\frac{22}{100}-15000*\frac{11}{100}\right)$ \

5700 : 6050 = 114 : 121

Statement-I is true, because metal reduced by decreasing $\Delta G$ in Elligham diagram but statement-II is false, Ellingham diagram is $\Delta G$ vs temperature graph has no relation with $\Delta S.$