Class 11th

$\Delta U=-742.24kJ/mole\Delta H_{298}=?$

NH₂CN (s) + $\frac{3}{2}$ O₂ (g) ® N₂ (g) + O₂ (g) + H₂O (l)

= $-742.24+\frac{1}{2}*\frac{8.314}{1000}*298$

$-742.24+1.238=-741.002\approx -741kJ/mol$

So, magnitude of $\Delta H$ = 741 kJ/mol.

(1) $C{r}^{+1}=4s^{0}3d^{5}M{n}^{+2}=4s^{0}3d^5$ ${}^{}{}^{}{}^{}{}^{}{}^{}{}^{}$

(2) $N{i}^{+2}=4s^{0}3d^{8}C{u}^{+}=4s^{0}3d^{10}$ ${}^{}{}^{}{}^{}{}^{}{}^{}{}^{}$

(3) $F{e}^{+2}=4s^{0}3d^{6}C{o}^{+}=4s^{1}3d^7$ ${}^{}{}^{}{}^{}{}^{}{}^{}{}^{}$

(4) $V^{+2}=4s^{0}3d^{3}C{r}^{+}=4s^{o}3d^5$

(1) $H{e}_2^{-}{\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^{1}B.O=\frac{3-2}{2}=\frac{1}{2}$ ${}_{}^{}{}_{}^{}{}_{}^{}{}_{}^{}\frac{}{}\frac{}{}$

(2) $H{e}_2^{+}{\sigma }_{1s}^2{\sigma }_{1s}^{*1}B.O=\frac{2-1}{2}=\frac{1}{2}$ ${}_{}^{}{}_{}^{}{}_{}^{}\frac{}{}\frac{}{}$

(3) $B{e}_2{\sigma }_{1s}^2{\sigma }_{1s}^{*2}{\sigma }_{2s}^2{\sigma }_{2s}^{*2}B.O=\frac{4-4}{2}=0$ ${}_{}{}_{}^{}{}_{}^{}{}_{}^{}{}_{}^{}\frac{}{}$

So Be2 does not exits

Kindly go throuigh the solution

Given   ${\displaystyle \sum _{i=1}^{18}\left(x_i-\alpha \right)=36i.e}{\displaystyle \sum _{i=1}^{18}{x}_i-18\alpha =36..........\left(i\right)}$

&       ${\displaystyle \sum _{i=1}^{18}{\left(x_i-\beta \right)}^2=90i.e{\displaystyle \sum _{i=1}^{18}{x_i}^2}}-2\beta {\displaystyle \sum _{}^{}{x}_i+18{\beta }^2=90.............\left(ii\right)}$      

(i) & (ii)   ${\displaystyle \sum _{i=1}^{18}{x_i}^2}=90-18\beta +36\beta \left(\alpha +2\right).............\left(iii\right)$

Now variance ${\sigma }^2=\frac{\sum {x_i}^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2$ = 1 given

->(a - b) (a - b + 4) = 0

Since $\alpha \ne \beta so\left|\alpha -\beta \right|=4$  

Kindly consider the following figure

$Given\left|z+5\right|\le 4..........\left(i\right)$

->Represent a circle ${\left(x+5\right)}^2+y^2\le 16$  

$=z\left(1+i\right)+\overline{z}\left(1-i\right)\ge -10..........\left(ii\right)$           

->Represent a line X – y $\ge -5$  

So max |z + 1|² = AQ²

$={\left(-4-2\sqrt[]{2}\right)}^2+8$         

$\begin{array}{l}=32+16\sqrt[]{2}\\ =\alpha +\beta \sqrt[]{2}\end{array}$        

Hence a + b = 48

Kindly consider the following figure

Kindly consider the following figure

18 = 3² * 2

For G.C.D to be 3. no. of four digits should be only multiple of 3, but not multiple of 9 & also should not be even.

As we know no. of the form                         

9 k -> 1000

9 k + 1 -> 1000

9 k + 2 -> 1000

9 k + 3 -> 1000  -> Total no. = 2000

9 k + 4 -> 1000

9 k + 5 -> 1000

9 k + 6 ->1000

9 k + 7 -> 1000

9 k + 8 -> 1000

In which half will be even & half be odd so Required no. = 1000