Class 11th

The 1^st such digit is 11 * 19 = 209

Sum = [209 + 220 + 231 + .+ 495] - [231 + 319 + 341 + 418 + 451] = 7744

a + b = 1   $\alpha +\gamma =\frac{10}{3}$

$\alpha \beta =2\lambda$ $\alpha \gamma =9\lambda$

$\frac{\beta }{\gamma }=\frac{2}{9},$ $\beta -\gamma =1-\frac{10}{3}=-\frac{7}{3}$

$\Rightarrow \beta =\frac{2}{3}$ $\gamma =3$

$\alpha =\frac{1}{3},\lambda =\frac{1}{9}$

$\therefore \frac{\beta \gamma }{\lambda }=18$                                  

 = coefficient of x⁴ in (1 + x)²¹ + coefficient of x⁴ in (1 + x)²¹

${=}^{21}{C}_4{+}^{21}{C}_4=2{.}^{21}{C}_4$              

$A_3$ = coefficient of x³ in (1 + x)²¹ + coefficient of x³ in (1 + x)²¹

23 * 3 gm of Na contained by 164 gm of Na₃PO₄

So, 3.45 gm of Na contained by $\left(\frac{164*3.45}{69}\right)$ gm of Na₃PO₄

Therefore mole of Na₃PO₄= $\frac{164*3.45}{69*164}$ = 0.05 mol

Molarity = $\frac{0.05*1000}{100}$ M = 0.5 M

= 50.0 * 10^-2 M

For equation A + B $?C+D$  

  $K_C=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}=100at298K$             

Here;    A +  B  $?$ C    +    D

t = 0       1M        1M

eq.         (1-x)      (1-x)      (1-x)      (1-x)

$Kc=\frac{\left(1+x\right)*\left(1+x\right)}{\left(1-x\right)*\left(1-x\right)}$

$\therefore x=\frac{9}{11}$  

At equilibrium, concentration of D is = 1 + $\frac{9}{11}=\frac{11+9}{11}=\frac{20}{11}$  

= 1.818 = 181.8 * 10^-2 M

Ans. = 182 (the nearest integer)

$\Delta H=3{\left(\Delta H_{comb}\right)}_{C_2{H}_2\left(g\right)}-{\left(\Delta H_{comb}\right)}_{C_6{H}_6\left(l\right)}$

$=-1300*3-\left(-3268\right)kJ/mole$

$=\left(-3900+3268\right)kJ/mole$

$=-632kJ/mole$

In photo-electric effect;

  $\frac{hc}{\lambda }=h{v}_0+\frac{1}{2}m{v}^2$             

  $\frac{6.63*10^{-34}*3*10^8}{500*10^{-9}}=\left(6.63*10^{-34}*4.3*10^{14}\right)+\frac{1}{2}*9.0*10^{-31}*v^2$            

  $v=\sqrt[]{\frac{\left(3.978*10^{-19}-2.85*10^{-19}\right)}{4.5*10^{-31}}}=\sqrt[]{25*10^{10}}=5*10^{5}m/s$             

Ans. = 5

Hydroxyapatite is the calcium minerals whose molecular formula is $3C{a}_3{\left(P{O}_4\right)}_2.Ca{\left(OH\right)}_2$ ${}_{}{{}_{}}_{}{}_{}$ or $C{a}_{10}{\left(P{O}_4\right)}_6{\left(OH\right)}_2.$ ${}_{}{{}_{}}_{}{}_{}$ During contact with water dissolved F ion, hydroxyapatite converted into ${}_{}{{}_{}}_{}{}_{}$ $3C{a}_3{\left(P{O}_4\right)}_2.Ca{F}_2$ .

According to Bohr's model, velocity of electron increases according to atomic number (z).

$v_e\propto \frac{z}{n}$ (z = Atomic number and n = number of shell/principal quantum number)

Similarly, velocity of electron decreases according to number of shell (i.e. principal quantum number).

$\Delta H_{vap}=41KJmo{l}^{-1}$

$\Delta H=\Delta E+\Delta nRT$      

$H_2{O}_{\left(\mathcal{l}\right)}\to H_2{O}_{vap}$        

$\Delta n=1$             

R = 8.3 JK^-1 mol^-1

T = 373 K

$\Delta$ E (Change of internal energy) = $\Delta H-\Delta nRT$  

$?\Delta E=41-\frac{1*8.3*373}{1000}kJmo{l}^{-1}=41-3.09kJmo{l}^{-1}=37.91kJmo{l}^{-1}$  

Nearest integer = 38.