The monomers of the following polymers

1.81\(\mathring { A } \)

Thermodynamics help us to understand how coke reduces the oxide and why blast furnace is chosen for this process. The main reduction reaction is:
FeO(s) + C(s) ⟶ Fe(s or l) + CO(g)
This reaction is a combination of two simple oxidation-reduction reactions. In one reaction, reduction of FeO is taking place and in the other carbon is being oxidised to CO as:
\(FeO(s) ⟶ Fe(s)+{1\over 2}O_2{(g})\ \ ΔG(FeO,\ Fe)\)
\((s)+{1\over 2}O_2 \longrightarrow CO(g)\ \ ΔG(C, CO)\)
When both the reactions (ii) and (iii) take place
to give equation (i) the net Gibbs energy change is :
ΔG(C, CO) + ΔG(FeO, Fe) = Δ_rG
The net reaction [Eq. (i)] will occur' when the resultant Δ_rG is negative. In Ellingham diagram ΔG vs T plot representing Eq. (ii) goes upward while that representing the change C⟶ CO (i.e., C, CO line) goes downward. At temperature about 1073 K (approx), the C, CO line falls below Fe, FeO line [G_(C.CO)<∆G_(FeO.Fe)].This means that in this temperature range, coke can reduce FeO to Fe and it will itself be oxidised to CO. Similarly, the reduction of Fe₂O₃ and Fe₃O₄ at relatively lower temperatures by CO can be explained on the basis of lower lying points of intersection of their curves with the COand CO₂ curves
In blast furnace, the reduction of oxides takes place at different temperature ranges. For this, hot air is blown from the bottom of the furnace and coke is burnt to give temperature up to 2200 Kin the lower portion. The burning of coke, therefore, supplies most of the heat required for the working temperature of the furnace. The CO and heat moves to upper part of the furnace. In upper part, the temperature is lower and the iron oxides (Fe₂O₃and Fe₃O₄) coming from the top are reduced to FeO in steps. Thus, the reduction reactions occurring in the lower temperature range and in the higher temperature range depend on the points of corresponding intersections in the ArG vs T
plots. It is observed that at the bottom of the furnace (higher temperature range) the reducing agent is carbon itself but at the top of the furnace (lower temperature range) the reducing agent is carbon monoxide. These reactions are summarised below:
At lower temperature range in the blast furnace (500-800 K)
3Fe₂O₃ + CO ⟶ 2Fe₃O₄+ CO₂
Fe₃O₄ + 4CO ⟶ 3Fe + 4CO₂
Fe₂O₃ + CO ⟶ FeO + CO₂
At higher temperature range (900-1500 K) in the blast furnace
C + CO₂ ⟶ 2CO
FeO + C ⟶ Fe + CO₂

Depression in freezing point is related to the molality, therefore, the molality of the solution with respect to ethylene glycol \(=\frac{\text { moles of ethylene glycol }}{\text { mass of water in kilogram }}\)
\(\text {Moles of ethylene glycol }=\frac{45 \mathrm{~g}}{62 \mathrm{~g} \mathrm{~mol}^{-1}}=0.73 \mathrm{~mol}\)
\(\text {Mass of water in } \mathrm{kg}=\frac{600 \mathrm{~g}}{1000 \mathrm{~g} \mathrm{~kg}^{-1}}=0.6 \mathrm{~kg}\)
\(\text {Hence molality of ethylene glycol }=\frac{0.73 \mathrm{~mol}}{0.60 \mathrm{~kg}}=1.2 \mathrm{~mol} \mathrm{~kg}^{-1}\)
Therefore freezing point depression,
ÄT_f = 1.86 K kg mol^–1 x 1.2 mol kg^–1 = 2.2 K
Freezing point of the aqueous solution = 273.15 K – 2.2 K = 270.95 K

Molar mass of KCI = 39 + 35.5 = 74.35 g mol^-1
A KCI dissociates completely, number of ions produced are 2.There, Van't Hoff factor, i = 2
Mass of KCI solution = 1000 \(\times\)1.04 = 1040 g
Mass of solvent = 1040 - 74.5 = 965.5 g = 0.9655 kg^1/2
Molality of the solution :
\(\\ \frac { No.\quad of\quad moles\quad of\quad solute }{ Mass\quad of\quad solvent\quad in\quad kg } =\frac { 1\quad mol }{ 0.9655\quad kg } =1.0357\quad m\)
T_b = i \(\times\)K_b \(\times\)m
= 2 \(\times\) 0.52 \(\times\)1.0357 = 1.078 \(°\) C
Therefore, boiling point of solution
= 100 = 1.078 = 101.078\(°\)C

It is because reaction does not stop in one step.  It can for disubstituted product also.

Units of rate = mol L^-1s^-1
\(\mathrm{k}=\frac{\text { Rate }}{[A][B]^2}=\frac{m a l L^{-1} s^{-1}}{\left(m o l L^{-1}\right)\left(m o l L^{-1}\right)^2}\)

\(\underset{Pyrolusite\\ (A)\ Blackish\ brown}{MnO_2 }+ 4KOH + O_2\xrightarrow{Fuse}\underset{Potassium\ magnate\\ (B)-Green\ coloured}{2K_2MnO_4}+2H_2O\)
\(\underset B{2 KMnO_4} + H20 + (0)_2 \xrightarrow[medium]{Alkaline}\underset{Potasium\ permanganate\\ (C)-Purple\ coloured}{2KMnO_4}+2KOH\)
or \(MnO^{-2}_4 \xrightarrow{} MnO_4^- + e^-\)
(b) When acidic solution of green compound (B), i·.e., potassium manganate is allowed to stand for some time, it disproportionates to give permanganate as follows:
\(3 MnO^{2-}_4+ 4H^+\longrightarrow2 MnO^-_4 + MnO_2 + 2 H_2O\)
This reaction is called disproportionation reaction.

It is due to osmosis.

Conductivity is the conductance of a solution of 1 cm length and having 1 sq.cm as the area of cross-section i.e., of one-centimeter cube.
Direct current results in the electrolysis of the electrolytic solution. As a result, the concentration of the electrolyte near the electrodes changes and these result change in the resistance of the solution.

1.123

In the presence of catalyst , activation energy barrier is lowerd.

(i) HOCI in presence of an acid generates the reactive electrophile, chloronium ion «r, which attacks benzene to give chlorobenzene (A).
(ii) Chlorobenzene (A) on treatment with NaNH2/liq. NH₃ undergaes-dehydrohalogenation via benzyne to afford aniline (B).
(iii) Aniline (B) on treatment with HBF4 forms the corresponding salt anilinium tetrafluoroborate (C) which on heating with NaNO₂ undergoes Balz-Schiemann reaction through the intermedium formation of benzenediazonium tetrafluoroborate to afford fluorobenzene (D).

\(\Delta { T }_{ f }=\frac { { K }_{ f }{ W }_{ b }\times 1000 }{ { M }_{ b }\times { W }_{ a } } \)
\(0.383=\left( \frac { 3.83\times 2.56 }{ M\times 100 } \right) \times 1000\)
M = 256
S \(\times \) x = 256
32 \(\times \) x = 256
x = 8
(b) (i) If we place blood cells in solution containing 1.2% sodium chloride solution, they shrink.
(ii) It we place blood cells in solution containing 0.4% sodium chloride solution, they swell.

(i) Friendly, caring nature, supportive and has good awareness towards health. (Any two)
(ii) Antacids are the chemical substances which neutralize excess acid in the gastric juices and give relief from acid indigestion, heart bums, acidity and gastric ulcers. e.g., Histamine, Ranitidine, Cimetidine.
(iii) It is not advisable to take antacids for a long time as they control only symptoms and not the cause. Therefore, the patients cannot be treated easily and completely through them. The production of more acid They may also lead to.

(ii) Alcohols have higher boiling point as compared to that of amines, because oxygen being more electronegative atom, forms strong hydrogen bond as compared to that of nitrogen. In tertiary amine, there is no hydrogen bond formation due to the absence of H-atoms and hence, has the lowest boiling point. Therefore, increasing order of boiling point (CH₃)₃N < C₂H₅NH₂ < C₂H₅OH .
(iii) (CH₃)₃NH and (CH₃)₃N are secondary and tertiary arnines respectively. These are distinguished by Hinsberg's reagent which gives sulphonamide with secondary amines and no reaction carried out with tertiary arnines. (CH₃)₃NH reacts with benzene sulphonyl chloride as follows:

(i) The plot [N₂O₅] against time is shown below :

(ii) Initial concentration [at t = 0] of N₂O₅
= 1.63 x 10^-2 M
Half of its concentration = 1.63X 10^-2/ 2
= 0.815 x 10^-2 M
From, the above graph, the time which corresponds to 0.815 X 10^-2 M concentration is 1450 s.
Hence, t_1/2 = 1450 s
(iii)

The plot of log [N₂O₅] and time can be represented as :

(iv) As the plot of log [N₂O₅] versus time obtained is a straight line, hence it is a reaction of first order.
ஃ Rate = k [N₂O₅] 
(v) For first order reaction, the slope of the line
\({c} =-\frac{k}{2.303} \)
\(\\ \therefore\text { Slope }=\frac{-2.46-(-1.79)}{3200-0}=\frac{-0.67}{3200} \)
Equating Eqs. (i) and (ii), we get
\({c} -\frac{k}{2.303}=\frac{-0.67}{3200} \)
\(\text {or } k=\frac{0.67 \times 2.903}{3200} \text { or } k=4.82 \times 10^{-1} \mathrm{~s}^{-1} \)
Half-life period (t_1/2) are calculated from the formula and slope are approximately the same.