Class 11th

$T_{r+1}{=}^{15}\text{?}{C}_r\cdot {\left(2x^{1/5}\right)}^{15-r}{\left(-x^{-1/5}\right)}^r$  

= ¹⁵C_{r $\cdot 2^{15-r}\cdot x^{\frac{15-2r}{5}}\cdot {\left(-1\right)}^r$}

Coefficient of x^-1 -> r = 10 ->m = ¹⁵C_{10 $\cdot 2^5$}

$x^{-3}\Rightarrow r=15\Rightarrow n{=}^{15}\text{?}{C}_{15}\cdot 2^0=1$  

now mn² = ¹⁵C_{r $\cdot 2^r$}  

r = 5

For expansion in vacuum, workdone, w = 0

For isothermal process,   $\Delta U=0$  

According to first law of thermodynamics,

$\begin{array}{l}\Delta U=q+w\\ q=0\end{array}$  

sin x = 1 – sin² x

sin x =  $\frac{-1+\sqrt[]{5}}{2},\frac{-1-\sqrt[]{5}}{2}\left(rejected\right)$  

draw y = sin x

y =  $\frac{\sqrt[]{5}-1}{2},$ find their pt. of intersection.

Hybridization of P in PF₅ is sp³d, so value of y = 1

Work function, ${?}_0=6.63*10^{-19}J$  

Threshold wavelength  ${\lambda }_0=?$  

Using ;                 ${?}_0=\frac{hc}{{\lambda }_0}$  

${\lambda }_0=\frac{hc}{{?}_0}$  

=300 * 10^-9 m = 300 nm

$Mn{O}_2+4HCl\to MnC{l}_2+C{l}_2+2H_{2}O$

$\begin{array}{l}C{l}_2+2Kl\to 2KCl+I_2\\ I_2+2N{a}_2{S}_2{O}_3\to 2Nal+N{a}_2{S}_4{O}_6\end{array}$

Here, meq of MnO₂ = meq of Na₂S₄O₆

$\frac{w}{E}*1000=M*n-factor*V$

$\frac{w}{87/2}*1000=0.1*1*60$

Mass of MnO₂ in sample = 0.261 g

Percentage of MnO₂ in sample = $\frac{0.261}{2}*100$  

 = 13.05%

First common term to both AP's is 9

t₇₈ of $\left(3,6,9,......\right)=78*3=234$  

t₅₉ of $\left(5,9,13,........\right)=5+\left(5-1\right)4=237$  

n^th common term $\le 234$  

9 + (n – 1) 12   234

n <  $\frac{237}{12}\Rightarrow n=19$  

Now sum of 19 terms with a = 9, d = 12

$=\frac{19}{2}\left(2.9+\left(19-1\right)\cdot 12\right)=2223$  

Kindly consider the following Image

Moles of N₂O = $\frac{2.2}{44}=\frac{1}{20}$  

$\Delta H=n{C}_p\Delta T=\frac{1}{20}*100\left(-40\right)=-200J$  

$\Delta U=q_p+w$

$w=-P_{ext}.\Delta V$  

$w=-1\frac{\left(167.75-217.1\right)}{1000}*101.3J=+5J$  

$\Delta U=-200+5=195J$