Chemistry

Correct order : Na? > Li? = Mg? > Be²?

$C{H}_4+2O_2\to C{O}_2+2H_{2}O$

2 moles of water produced by 1 mole of methane

Or 36 gm of water produced by 1 mole of CH₄

$\therefore$  81 gm of water produced = $\frac{1*81}{36}$  = 2.25 mole of CH₄

Mole of CH₄ required = 225 * 10^-2

the nearest integer = 225

C? H? + 5O? → 3CO? + 4H? O
ΔH = − {6 (C = O) + 8 (OH)} + 2 (C – C) + 8 (C – H) + 5 (O = O)
= 8a? + 5a? + 2a? - 6a? - 8a?

${\left[Co{\left(H_{2}O\right)}_6\right]}^{2+}+N{H}_{3\left(excess\right)}\to \underset{Diamagneticnature}{{\left[Co{\left(N{H}_3\right)}_6\right]}^{3+}+6H_{2}O}$

$C{o}^{3+}\to 3d^{6}4s^0$

$\Rightarrow {t_{2g}}^{6}e{g}^0$

$\Rightarrow {t_{2g}}^{6}e{g}^0$ (form of paired electron)

              No. of electron in t_2g = 6

Cl? > F? > O? > N?
3p-3p 2p-2p B.O=2 B.O=3

$K_{cat}=A{e}^{-}\frac{{\left(Ea\right)}_{cat}}{RT}and{K}_{uncat}=A{e}^{-}\frac{{\left(Ea\right)}_{uncat}}{RT}$

 $\frac{K_{cat}}{K_{uncat}}=e^{\frac{{\left(Ea\right)}_{uncat}-{\left(Ea\right)}_{cat}}{RT}}=e^{\frac{10*1000}{8.314*300}}=e^{4.009}=e^x$

$\therefore x=4$

FeO (Impurities) + SiO? (Flux) → FeSiO? (Iron silicate) Slag

Ease of Carbocation formation

cell reaction is ;

              $Cu\left(s\right)+S{n}_{\left(aq\right)}^{2+}\to C{u}_{\left(aq\right)}^{++}+S{n}_{\left(s\right)}$  

              $Q=\frac{\left[C{u}^{++}\right]}{\left[S{n}^{++}\right]}=\frac{0.01}{0.001}andn=2$  

              $E_{cell}=E_{cell}^o-\frac{0.059}{2}log\frac{0.01}{0.001}$  

              $=-0.48-\frac{0.059}{2}log10=-0.509$  

              $=983.33*10^{-1}KJ/mole$  

              Nearest integer = 983