Class 11th

NF₃ is stable due to high N-F bond energy.

Other halides of nitrogen (NCl₃ NBr₃, Nl₃) are explosive

Heating energy $=\frac{1}{4}th$ energy of the hammer

$=\frac{1}{4}*\frac{1}{2}m{v}^2$

$=\frac{1}{4}*\frac{1}{2}*1.5*60*60$

$=\frac{15*6*60}{4*2}$

= 675 J

$\Delta T=\frac{675}{42}=16.07°C$

Energy of 1 photon =  $\frac{hc}{\lambda }$

Energy of 1 mole of photon = $N_A*\frac{hc}{\lambda }$ ${}_{}\frac{}{}$

$=\frac{6.022*10^{23}*6.63*10^{-34}*3*10^8}{300*10^{-9}}$

= 0.399 * 106 J = 399 KJ

$\theta ={\omega }_{0}t+\frac{1}{2}\alpha t^2$

$5=0*1+\frac{1}{2}*\alpha *1^2$

= 10 rad/sec²

(ln 2 seconds) = ${\omega }_{0}t+\frac{1}{2}\alpha t^2$

${\theta }_2=20rad$

Angle rotated by wheel in 2^nd second

= ${\theta }_2-\theta$

= 20 – 5

= 15 rad

B₂H₆ has 4 2 c -2e bonds and 2 3c-2e bonds.

Bridging (B-H) bonds have more value of bond- length then terminal (B – H) bonds

Bridging bonds are in one plane, but terminal bonds are in perpendicular plane.

Due to presence of (3c-2e) bonds, it behaves as electrons deficient and prone to get attached by lewis base.

$3NaB{H}_4+4B{F}_3\underset{}{\overset{\Delta }{\to }}3NaB{F}_4+2B_2{H}_6$                

Due to higher extent of polarization by Li⁺ and Mg²⁺, LiCl and Mgcl₂ have covalent character. Therefore they are soluble in ethanol.

Due to very high value of lattice energy, LiF is having very less solubility in water.

$a=\frac{mg+10}{g}=\frac{5*10+10}{10}$

$\Rightarrow a=6m/se{c}^2$

v = u + at

0 = u – 6t

$t_A=\frac{u}{6}$  (time of Ascent)

$=u*\frac{u}{6}-\frac{u}{2}*6*\frac{u}{6}*\frac{u}{6}$

= $\frac{u^2}{6}-\frac{u^2}{12}$

$h=\frac{u^2}{12}$

for time of descent

$a=\frac{50-10}{10}$

a = 4 m /sec²

$\Rightarrow \frac{u^2}{12}=0*t+\frac{1}{2}*4t^2$

$t_B=\frac{u}{\sqrt[]{24}}$

$\frac{t_A}{t_B}=\frac{u*\text{exception 25:}}{6*u}=$$\frac{\sqrt[]{2}}{}$

The highest industrial consumption of hydrogen gas is in the synthesis of ammonia gas (Having use in manufacturing of N-based fertizers)

Number of moles of C = Number of moles of CO₂ = $\frac{330}{44}$ moles

              Number of moles of H = 2 * no. of moles of H₂O = $\left(\frac{270}{18}*2\right)$ moles

              Mass of C = $\frac{330}{44}*12$ gm = 90 gm

              Mass of H = $\frac{270}{18}*2*1gm=30gm$

$\begin{array}{l}\mathrm{\%}ofC=\frac{90}{120}*100\mathrm{\%}=75\mathrm{\%}\\ \mathrm{\%}ofH=\frac{30}{120}*100\mathrm{\%}=25\mathrm{\%}\end{array}$  

Acidic oxide => Cl₂O₇

Neutral oxide => N₂O, NO

Basic oxide => Na₂O

Amphoteric oxide => As₂O₃