(i) Glucose does not give Schiff‘s Test although it contains aldehyde group.
(ii) Glucose does not form crystaline product with NaHSO₃.

CH₃CH₂CH₂CH₃  <  C₂H₅OC₂H₅  <  CH₃CH₂CH₂CHO  <  CH₃ (CH₂ )₂OH
\((hydrocarbon)\overset { (ether) }{ \underset { increase\quad in\quad bond\quad polarity }{ \longrightarrow } } (aldehyde)(alcohol)\).

(a)
  \(4N{ H }_{ 3 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \xrightarrow [ 1100K ]{ Pt } 4NO\left( g \right) +6{ H }_{ 2 }O\left( l \right) ;\ 2NO\left( g \right) +{ O }_{ 2 }\left( g \right) \longrightarrow 2{ NO }_{ 2 }\left( g \right) \)
\(\\ 3{ NO }_{ 2 }\left( g \right) +{ H }_{ 2 }O\left( l \right) \longrightarrow 2HN{ O }_{ 3 }\left( aq \right) +NO\left( g \right) ;\ { NH }_{ 3 }\left( g \right) +HN{ O }_{ 3 }\left( aq \right) \longrightarrow { NH }_{ 4 }{ NO }_{ 3 }\left( aq \right) \)
\(\\ { NH }_{ 4 }{ NO }_{ 3 }\left( aq \right) \xrightarrow [ at\ room\quad temperature ]{ Vacuum\ evaporation } { NH }_{ 4 }{ NO }_{ 3 }\left( s \right) ;\ { NH }_{ 4 }{ NO }_{ 3 }\left( s \right) \xrightarrow { 523K } { N }_{ 2 }O\left( g \right) +2{ H }_{ 2 }O\left( l \right) \)
(b) Animals need large amount of energy to move around and maintain the body temperature. Therefore, to obtain the required energy, it is much easier for them to break weaker double bond (493-4 kJ \({mol }^{ -1 }\) ) of  \({ O }_{ 2 }\) than breaking the much stronger triple bond (941·4 kJ \({ mol}^{ -1 }\) ) of \({ N }_{ 2 }.\) 

\(Step \ 1.{ O }_{ 3 }\rightleftharpoons { O }_{ 2 }+O\quad (fast)\quad \quad \quad ..(i)\)
\(Step \ 2.O+{ O }_{ 3 }\longrightarrow 2{ O }_{ 2 }\quad (slow)\quad \quad \quad...(ii)\)
\(From \ slow \ step, \ Rate \ = \ k\left[ { O }_{ 3 } \right] \left[ O \right] \ \ \ \ \ \ \ \ \ \ \ \ \ ...(iii)\)
\(From \ eqn. \ (i), \ { K }_{ eqm. }=\frac { \left[ { O }_{ 2 } \right] \left[ O \right] }{ \left[ { O }_{ 3 } \right] } or \ \left[ O \right] ={ K }_{ eqm. }\frac { \left[ { O }_{ 3 } \right] }{ \left[ { O }_{ 2 } \right] } \)
Substituting this value in eqn. (iii), we get
\(Rate=k\left[ { O }_{ 3 } \right] \times { K }_{ eqm. }{ \left[ { O }_{ 3 } \right] }/{ \left[ { O }_{ 2 } \right] }=k\prime { \left[ { O }_{ 3 } \right] }^{ 2 }{ \left[ { O }_{ 2 } \right] }^{ -1 }\) 
Thus, order w.r.t  \({ O }_{ 2 }\) = -1

\(CH=CH+{ H }_{ 2 }\underrightarrow { Pd } { CH }_{ 2 }={ CH }_{ 2 }\overset { { +H }_{ 2 } }{ \underset { Pd }{ \rightarrow } } { CH }_{ 3 }\_ { CH }_{ 3 }\)
\(\\ \quad \quad Acetylene\quad \quad \quad \quad \quad \quad Ethane\quad Ethane\)
\(\\ CH=CH+{ H }_{ 2 }\overset { Pd+Ba{ SO }_{ 4 } }{ \underset { +quinoline/S }{ \longrightarrow } } { CH }_{ 2 }={ CH }_{ 2 }\)
                                                Ethane
BaSO₄ + quinoline/S poison the catalyst. Hence, the efficiency of the catalyst decreases and the reaction stops at the first stage of reduction.

0.676g cm^-3

Copper matte chiefly consists of Cu₂S along with some unchanged FeS. When a blast of hot air is passed through molten matte taken in a silica lined converter, FeS present in matte is oxidised to FeO which combines with silica (SiO₂) to form FeSiO₃ slag.
\(2FeS+3{ O }_{ 2 }\longrightarrow 2FeO+2S{ O }_{ 2 }\uparrow \quad ;\quad FeO+\underset { Silica }{ Si{ O }_{ 2 } } \longrightarrow \underset { Slag }{ FeSi{ O }_{ 3 } } \)
When whole of iron has been removed as slag, some of the CU₂S undergoes oxidation to form Cu ₂O which then reacts with more Cu₂S to form copper metal.
\(2{ Cu }_{ 2 }S+3{ O }_{ 2 }\longrightarrow 2{ Cu }_{ 2 }S+2S{ O }_{ 2 }\uparrow \quad ;\quad 2{ Cu }_{ 2 }O+{ Cu }_{ 2 }S\longrightarrow 6Cu+2S{ O }_{ 2 }\uparrow \) 
Thus, copper matte is heated in silica lined converter to remove FeS present in matte as FeSiO₃ slap.