N2<CO<Ch₄

mass of Zn rod = 23.725 g, molarity of SO₄^2- remains same.

a. Tertiary butyl ketone does not give precipitate with sodium bisulphate whereas acetone does.
b. Dialkyl cadmium is considered superior to grignard reagent for the preparation of a ketone from an acid chloride.

It decomposes at baking or cooking temperatures and hence can be used only in cold foods and drinks. 

Open structure of D-glucose could not explain the following reactions:
(i) Despite having the aldehyde group, glucose does not give Schiffs test and 2, 4-DNP test.
(ii) Glucose does not react with sodium hydrogen sulphite to form addition product.
(iii) The pentaacetate of glucose does not react with hydroxyl amine showing the absence of free -CHO group.
(iv) When glucose is heated with methanol in the presence of dry HCI gas, it forms two isomeric monomethyl derivatives known as \(\alpha \)-D-glucoside(m.p. = 165°C) and \(\beta \)-D-glucoside (m.p = 107°C). Since only one molecule of methanol is used for the formation of methyl glucoside, these must be hemiacetals. 
These results show that glucose does not have open chain form structure.

(a) In presence of moisture or water, Cl₂ gives nascent oxygen which is responsible for its oxidising and bleaching properties as shown below :
Cl₂ + H₂O ⟶ [HCI + HCIO] ⟶ 2 HCI + \(\underset{Nascent\ \ oxygen}{O}\)
(i) It oxidises acidified ferrous sulphate to ferric sulphate
                CI₂+ H₂O ⟶ 2 HCl + O
\(\underline{2 FeSO_4 + H_2SO_4 +O ⟶ Fe_2(SO_4)_3 + H_2O}\)
2 FeSO₄ + H₂SO₄ + Cl₂ ⟶ Fe₂(SO₄)₃ + 2 HCl
_{(Green)                                           (Brown)}
(ii) It bleaches vegetable and organic colouring matter to colourless substances by oxidation.
Cl₂ + H₂O ⟶ 2 HCI + [0]
Vegetable colouring matter +O ⟶ Oxidised colourless substances.
(b) The structures of H₃PO₂,·H₃PO₃ and H₃PO₄ are:
Due to the presence of P - H bonds, both H₃PO₂ and H₃PO₃ act as reducing agents. In contrast, H₃PO₄ does not have any P - H bonds and hence it does not act as a reducing agent.
(c) 03 is a powerful oxidising agent. Therefore, it oxidises aq. Kl to violet vapours of I₂
                            O₃ (g) ⟶ 0₂ (g) + O(g)
\(\underline{2 Kl (aq) + H_2O (I) + O(g) ⟶ 2 KOH (aq) + I_2(s)}\)_{__________}
2 KI (aq) + H₂O (I) + O₃ (g) ⟶ 2 KOH (aq) + O₂ (g) +I₂(s)
 potassium iodide                                                                Idoine       
colourless                                                                                violet​​

There are three t_2g orbitals with lower energy and two e_g orbitals with higher energy. Three electrons with parallel spins can first enter into the lower energy three t_2g orbitals, i.e., for complexes with d¹. d² and d³ ion. For d ⁴, d ⁵, d⁶ etc. ions, the electrons can either enter into t_2g orbitals and pair up or they may enter into e_g orbitals. The former gives rise to low spin complex while the latter gives rise to high spin complex. If crystal field splitting energy (difference of energy between t_2g and eg orbitals) is large, low spin state is more stable and if it is small, high spin state is more stable.

For a 1st order reaction, \(k=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
In terms of pressures, \(a\propto { P }_{ 0 }\)  (initial pressure)
 \(a-x\propto \)(pressure of \({ N }_{ 2 }{ O }_{ 5 }\) at time t)
\(\therefore \ k=\frac { 2.303 }{ t } \log { \frac { { P }_{ 0 } }{ P } } \) or \(\frac { kt }{ 2.303 } =\log { { P }_{ 0 } } -\log { P } \) or \(\log { P } =\frac { kt }{ 2.303 } +\log { { P }_{ 0 } } \)
Thus, plot of \(\log { P } \) vs. t will be linear if the reaction is of  1st order. Further, from the straight line plot, slope\(=-\frac { kt }{ 2.303 } \) and intercept on \(\log { P } axiz=\log { { P }_{ 0 } } \) . Thus, rate constant k and initial pressure \({ P }_{ 0 }\) can be found.

219.17pm

Diamond is a covalent crystal in which each C-atom is sp³ hybridized.Thus, each C-atom is covalently bonded to four other C-atom tetrahedrally.These tetrahedra are linked together into a three dimensional giant molecule.
The structure of diamond is similar to that o Zns.
In diamond, we have C-atoms in place of Zn²⁺ and S^2- ions.Thus, diamond has face-centred cubic structure in which C-atoms are present at the corners as well as face-centres and alternate tetrahedral voids.
Contribution of C-atoms at the corners = \(8\times{1\over 8}=1\)
Contribution of C-atoms at face-centres = \(6\times{1\over 2}=3\)
As four C-atoms are present on the body diagonals i.e., in the alternate voids, their contribution = 4
Total no. of C-atoms in the unit cell = 1 + 3 + 4 = 8

Liquid A with lower boiling point is more volatile and hence will have higher vapour pressure.

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.