$$\begin{gathered} \mathit{H}\mathit{C}\mathit{O}\mathit{O}\mathit{H}\mathbf{ }\underset{\mathbf{D}\mathbf{e}\mathbf{h}\mathbf{y}\mathbf{d}\mathbf{r}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{n}\mathbf{g}\mathbf{ }\mathbf{A}\mathbf{g}\mathbf{e}\mathbf{n}\mathbf{t}}{\overset{{\mathbf{H}}_{\mathbf{2}}\mathbf{S}{\mathbf{O}}_{\mathbf{4}}}{\mathbf{\to }}}\mathbf{ }\mathit{C}\mathit{O}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{ }\mathbf{ }{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{ } \\ Water is abosrbed by sulphuric acid \\ (Moles{)}_i = \frac{2.3}{46} = \frac{1}{20} 0 0 \\ (moles{)}_f = 0 \frac{1}{20} \frac{1}{20} \\ {\mathit{H}}_{\mathbf{2}}{\mathit{C}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{4}}\mathbf{ }\stackrel{\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{S}{\mathbf{O}}_{\mathbf{4}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }}}{\mathbf{\to }}\mathbf{ }\mathit{C}\mathit{O}\mathbf{ }\mathbf{+}\mathbf{ }\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{ }\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O} \\ Water is abosrbed by sulphuric acid \\ (Moles{)}_i = \frac{4.5}{90} = \frac{1}{20} 0 0 0 \\ (moles{)}_f = 0 \frac{1}{20} \frac{1}{20} \frac{1}{20} \end{gathered}$$

CO₂ is absorbed by KOH.
So the remaining product is only CO.
moles of CO formed from both reaction
= $=\frac{1}{20}+\frac{1}{20} = \frac{1}{10}$

Let mass of CO = moles x molar mass

 = $= \frac{1}{10} x 28$

Smaller the size cation, more will be its polarising power. Hence, BeH₂ will be least ionic.

Or

On moving down the group metallic character of metals increases so ionic character of metal hydride increases.

Due to high electronegativity and small size, F forms only one oxoacid, HOF known as Fluoric (I) acid the oxidation number of F is +1 in HOF.

$$\begin{gathered} \stackrel{+7}{Mn}{O}_4^{-} \stackrel{ }{\to } \stackrel{+2}{M}{n}^{2+} ; 5e^{-} gain ... \left(i\right) \\ {\stackrel{+3}{C}}_2{O}_4^{2-} \stackrel{ }{\to } \stackrel{+4}{C}{O}_2 ;2e^{-} loss ... \left(ii\right) \\ Multiplying \left(i\right) and 2 and \left(ii\right)\hspace{0.17em}by 5 to balance e^{-} \\ 2Mn{O}_4^{-} + 5C_2{O}_4^{2-}\stackrel{ }{\to } 2M{n}^{2+} + 10C{O}_2 \\ On balancing charge; \\ 2Mn{O}_4^{-} + 5C_2{O}_4^{2-} + 16H^{+}\stackrel{ }{\to } 2M{n}^{2+} + 10 C{O}_2 + 8 H_{2}O \end{gathered}$$

The metal which is more reactive than 'Al' can reduce alumina. Also, according to the Ellingham diagram, Mg has more –ΔG value then alumina. Metals which has more –ΔG value can reduce those metals oxide which has less –ΔG value.

The element M in the complex-ion $M{F}_6^{-3}$ has a coordination number of six.

B has only s  and p-orbitals and no d – orbitals available, therefore, at the maximum it can show a coordination number of 4. Thus, B cannot form the complex of the type $M{F}_6^{-3}$

Co³⁺ = [Ar] 3d⁶, Unpaired e^–(n) = 4
Spin magnetic moment $\sqrt{4(4+2)} = \sqrt{24} BM$

Cr³⁺ = [Ar] 3d³, Unpaired e^–(n) = 3
Spin magnetic moment = $\sqrt{3(3+2)} = \sqrt{15} BM$

Fe³⁺ = [Ar] 3d⁵, Unpaired e–(n) = 5
Spin magnetic moment =$\sqrt{5(5+2)} = \sqrt{35} BM$

Ni²⁺ = [Ar] 3d⁸, Unpaired e^–(n) = 2
Spin magnetic moment = $\sqrt{2(2+2)} = \sqrt{8} BM$ 

-I effect increases on increasing electronegativity of the atom. So, the correct order of -I effect is

-NH₂ <-OR < -F

Also, 

-NR₂<-OR<-F

The structure of ClF₃ is

The number of lone pair of electrons on central Cl is 2.

a) Mass of water = 18 x 1 = 18 g

Molecules of water = mole x N_A

$=\frac{18}{18}{N}_A = N_A$

b) Molecules of water = mole x N_A

$=\frac{0.18}{18} N_A ={10}^{-2} N_A$

c) Molecules of water = mole x N_A = 10^-3 N_A
d) Moles of water = 0.00224/22.4 = 10^-4

Molecules of water = mole x N^A = 10^-4 N_A