(a) Blood is a colloidal sol. When we rub the injured part with moist alum, coagulation of blood takes place.
(b) Coagulation stops bleeding. Moist alum is an electrolyte.

K_sp = 1.8 \(\times\)10^-10, [AgNO₃]= 1 M
\(\mathrm{AgCl}_{(\mathrm{s})} \rightleftharpoons \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-} \)
                    s            s
\(\mathrm{AgNO}_{3(\mathrm{aq})} \rightleftharpoons \mathrm{Ag}_{(\text {aq })}^{+}+\mathrm{NO}_{3_{(\text {aq })}}^{-}\\ 1 \mathrm{M} \quad \quad \quad \quad \quad 1 \mathrm{M} \quad \quad 1 \mathrm{M} \)
\(\left[\mathrm{Ag}^{+}\right]=(\mathrm{s}+1) \approx 1 \quad(\therefore \mathrm{s}<<1) \)
\(\left[\mathrm{Cl}^{-}\right]=\mathrm{s} \)
\(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right] \)
\(1.8 \times 10^{-10}=(1)(s) \)
\(\therefore \mathrm{s}=1.8 \times 10^{-10} \mathrm{M}\)

∴ Two dipeptides structures are possible from glycine and alanine. They are glycyl alanine and Alanyl glycine.

(i) Let A = 100M, [A₀] - [A] = 20M,
For the zero order reaction
\(k=\left( \frac { \left[ { A }_{ 0 } \right] -\left[ A \right] }{ t } \right) \)
(i) 20% completion  \(k=\left( \frac { 20M }{ 20min } \right) \) = 1 mol L^-1 min^-1 
(ii) 80% completion
\(\mathrm{K}=1 \mathrm{~mol} \mathrm{~L}{ }^{-1} \mathrm{~s}^{-1} ;\left[\mathrm{A}_{0}\right]=100 \mathrm{M} ;\left[\mathrm{A}_{0}\right]-[\mathrm{A}]=80 \mathrm{M} ; \mathrm{t}=?\)
\(\therefore t=\left(\frac{\left[A_{O}\right]-[A]}{K}\right)=\frac{80}{1}=80 \mathrm{mins}\)

HNO₃ reacts with basic oxides to form salts and water
ZnO + 2HNO₃\(\longrightarrow \) Zn(NO₃)₂ + H₂O
3FeO + 10HNO₃ \(\longrightarrow \) 3Fe(NO₃)₃ + NO + 5H₂O

\(\text { Density }(\rho)=\frac{\mathrm{nM}}{\mathrm{a}^{3} \mathrm{~N}_{\mathrm{A}}}  \)
\(\mathrm{n}=4, \mathrm{M}=\text { Molar mass of } \mathrm{KF}=58.1 \mathrm{~g} / \mathrm{mol}  \)
\(\rho=2.48 \mathrm{~g} \mathrm{~cm}^{-3}  \)
\(\mathrm{~N}_{\mathrm{A}}=6.023 \times 10^{23} \)
\(a^{3} =\frac{n M}{\rho N_{A}}=\frac{4 \times 58.1}{2.48 \times 6.023 \times 10^{23}}  \)
\(a^{3} =15.55 \times 10^{-23}  \)
\(a^{3} =0.1555 \times 10^{-21}  \)
\(a =\sqrt[3]{0.1555 \times 10^{-21}}  \)
\(a =0.5375 \times 10^{-7} \mathrm{~cm}=5.375 \times 10^{-8} \mathrm{~cm}=537.5 \mathrm{pm}  \)
\(d =\frac{a}{\sqrt{2}}(\text { for fcc })  [\therefore r = \frac{a\sqrt{2}}{4}]\)
\(=\frac{537.5}{1.414}=380.13 \mathrm{pm}\)
\(\therefore\) The distance between K⁺ and F^- in KF = 380.13 pm

(i) The extraction of metals from their oxides can be carried out by using different reducing agents.
(ii) Consider the following reaction
\(\frac{2}{\mathrm{y}} \mathrm{M}_{\mathrm{x}} \mathrm{O}_{\mathrm{y}(\mathrm{s})} \rightarrow \frac{2 \mathrm{x}}{\mathrm{y}} \mathrm{M}_{(s)}+\mathrm{O}_{ 2(\mathrm{~g})}\)            (1)
(iii) The above reduction may be carried out with carbon. In this case the reducing agent carbon may be oxidized to either CO or CO₂
\(\mathrm{C}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}  \)                  (2)
\(2 \mathrm{C}+\mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{(\mathrm{g})} \)                (3)
(iv) If CO is used as a reducing agent
\(2 \mathrm{CO}+\mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2(\mathrm{~g})}\)           (4)
(v) A suitable reducing agent is selected based on the thermodynamics considerations.
(vi) We know that for a spontaneous reaction, the change in free energy (\(\triangle\)G) should be negative.
(vii) Therefore, thermodynamically, the reduction of metal oxide with a given reducing agent can occur if the free energy change for the coupled reaction is negative.
(viii) Hence, the reducing agent is selected in such a way that it provides a large negative \(\triangle\)G value for the coupled reaction.

Propanal into butanone:

(b) Hex-3-yne into hexan-3-one.
Hydration of alkynes in 42% sulphuric acid containg H₂SO₄ as catalyst
\(\underset {Hex-3-yne}{\mathrm{CH}_{3}-\mathrm{CH}_{2}}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{3} \frac{\mathrm{HgSO}_{4}}{\mathrm{H}_{2} \mathrm{SO}_{4} / \mathrm{H}_{2} \mathrm{O}} \)

(c) phenylmethanal into benzoic acid

(d) phenylmethanal into benzoin

(i) Oxidation at anode:
\(2H_{2(g)}+4OH^{-}_{(aq)}\rightarrow 4H_{2}O_{(I)}+4e^{-}\)
(ii) 1 mole of hydrogen gas produces 2 moles of electrons at 25⁰C and 1 atm pressure, 1 mole of hydrogen gas occupies = 22.4 litres
\(\therefore \) no. of moles of hydrogen gas produced
= \(\frac{1 mole}{22.4 litres} \times 44.8 litres\) 
= 2 moles of hydrogen
(iii) \(\therefore \)  2 of moles of hydrogen produces 4 moles of electro i.e., 4F charge.
t = 10 min
t = 10 x 60 sec
t = 600s
We know that Q= It
\(I=\frac{Q}{t}\)
\(=\frac{4F}{10 mins}\)
\(=\frac{4\times96500\quad C}{10\times60\quad s}\)
I = 643.33 A
Electro deposition of copper
\(Cu^{2+}_{(aq)}+2e^{-}\rightarrow Cu_{(s)}\)
(iv) 2F charge is required to deposit
(v) 1 mole of copper i.e., 63.5 g
(vi) If the entire current produced in the fuel cell ie., 4 F is utilised for electrolysis, then \(2\times63.5\) i.e., 127.0 g copper will be deposited at cathode.

Elemental copper is more stable than Cu²⁺. The electronic configuration of copper is 3d¹⁰ 4s¹ completely filled 3d orbital with stable configuration.
But Cu²⁺ has configuration as 3d⁹. Hence \(\mathrm{E}_{\mathrm{M}^{2+} / \mathrm{M}}^{0}\) value is positive for Cu²⁺.

Fusion of urea with B(OH)_3' in an atmosphere of ammonia at 800 - 1200 K gives boron nitride.
B(OH)₃ + NH₃ \(\overset { \Delta }{ \longrightarrow } \) BN+ 3H₂O

1. The materials that are readily decomposed by microorganisms in the environment are called biodegradable.
2. Natural polymers degrade on their own after certain period of time but the synthetic polymers do not.
3. It leads to serious environmental pollution. One of the solution to this problem is to produce biodegradable polymers which can be broken down by soil micro organism.
Examples:
(i) Polyhydroxy butyrate (PHB)
(ii) Polyhydroxy butyrate-co-A- hydroxyl valerate (PHBV)
(iii) Polyglycolic acid (PGA), Polylactic acid (PLA)
(iv) Poly ( E caprolactone) (PCL)
(v) Biodegradable polymers are used in medical field such as surgical sutures, plasma substitute etc...
4. these polymers are decomposed by enzyme action and are either metabolized or excreted from the body.

[Ni (H₂O)₆]²⁺

It has two unpaired electrons. So there is d-d transition. Hence it is green coloured.
[Ni(CN)₄]^2-

There is no unpaired electrons. So it is colourless, as there is no d-d transition.