Chemistry

$?G^0=-2.303RT\log ?K$

$=-2.303*R*300*\log ?{10}^{10}$

$=-2.303*3000R$

=-6909 R

$N{a}_{2}C{O}_3+HCl\stackrel{Phenolphthalein}{\to }NaHC{O}_3+NaCl$

$E_{N{a}_{2}C{O}_3}=\frac{M_{N{a}_{2}C{O}_3}}{n-factor}=\frac{106}{1}=106$

Kindly consider the following figure

For esters, prefix used is alkoxycarbony.

Kindly consider the following figure'

$\left. [Co{\left(N{H}_3\right)}_{5}Cl\right]C{l}_2+AgN{O}_3\left(excess\right)\to \left. [Co{\left(N{H}_3\right)}_{5}Cl\right]{\left(N{O}_3\right)}_2+2AgCl\downarrow$

Secondary valence of $Co$ $$ in $\left. [Co{\left(N{H}_3\right)}_{5}Cl\right]C{l}_2$ $\left. [{\left({}_{}\right)}_{}\right]{}_{}$ is six.

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 $=\frac{662}{6.62*{10}^{-19}}$ $\frac{}{{}^{}}$

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

$N{H}_3$ 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,

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

Kindly consider the following figure