Class 11th

Power of bulb = 662 watt = 662 Js^-1

Energy of one photon $=\frac{6.62*{10}^{-34}*3*{10}^8}{300*{10}^{-9}}$ $\frac{{}^{}{}^{}}{{}^{}}$

$=6.62*{10}^{-19}J$

Number of photons emitted $=\frac{662}{6.62*{10}^{-19}}$ $\frac{}{{}^{}}$

$={10}^{21}{s}^{-1}$

Inside a spherical shell gravitational field is zero, so net force acting on a particle placed inside it is also zero.

Total energy of a bound system is always negative.

$N{H}_3$ is recovered when solution containing $N{H}_{4}Cl$ is treated with $Ca{\left(OH\right)}_2$   (slaked lime).

$2N{H}_{4}Cl+Ca{\left(OH\right)}_2\to 2N{H}_3+CaC{l}_2+H_{2}O$ .

$u_{rm{s}_1}=\sqrt[]{\frac{3RT}{M}}$

When the temperature is tripled,

$u_{rm{s}_2}=\sqrt[]{\frac{3R\left(3T\right)}{M}}=\sqrt[]{3}*u_{rm{s}_1}=1.73*u_{rm{s}_1}$

Elastic potential energy stored in the spring

$U=\frac{1}{2}k{x}^2$

$=\frac{1}{2}Fx=\frac{F^2}{2k}$

$Hence\frac{U_1}{U_2}=\frac{k_2}{k_1}\left. [Fissameforbothsprings\right]$

$=\frac{500}{1200}=\frac{5}{12}$

Kindly consider the following figure

$pH=12,pOH=14-12=2$

$\left. [O{H}^{-}\right]={10}^{-2}$

$K_{sp}=\left. [C{a}^{2+}\right]{\left. [O{H}^{-}\right]}^2=\frac{\left({10}^{-2}\right)}{2}{\left({10}^{-2}\right)}^2=0.5*{10}^{-6}$

$K_{sp}=5*{10}^{-7}$

  $2Cl\left(g\right)\to C{l}_2\left(g\right)$ is exothermic process so $?H<0$ $$

$?n_g<0so,?S<0$

$Ca{H}_2+2H_{2}O\to Ca{\left(OH\right)}_2+2H_2\left(g\right)$

(Hydrolith)