Rain water normally has a pH of 5.6 due to dissolution of atmospheric CO₂ into it. Oxides of sulphur and nitrogen in the atmosphere may be absorbed by droplets of frater that make up clouds and get chemically converted into sulphuric acid and nitric acid respectively as a results of pH of rain water drops below the level 5.6, hence it is called acid rain. Acid rain is a by-product of a variety of sulphur and nitrogen oxides in the atmosphere. Burning of fossil fuels (coal and oil) in power stations, furnaces and petrol, diesel in motor engines produce sulphur dioxide and nitrogen oxides The main contributors of acid rain are SO₂ and NO₂.They are converted into sulphuric acid and nitric acid respectively by the reaction with oxygen and water.
2SO₂ + O₂ + 2H₂O ⟶ 2H₂SO₄
4NO₂ + O₂ + 2H₂O ⟶ 4HNO₃
Harmful effects of acid rain
Some harmful effects are discussed below :
(i) Acid rain causes extensive damage to buildings and structural materials of marbles. This attack on marble is termed as Stone leprosy.
CaCO₃ + H₂SO₄ ⟶ CaSO₄ + H₂O +CO₂ ↑
(ii) Acid rain affects plants and animal life in aquatic ecosystem.
(iii) It is harmful for agriculture, trees and plants as it dissolves and removes the nutrients needed for their growth.
(iv) It corrodes water pipes resulting in the leaching of heavy metals such as iron, lead and copper into the drinking water which have toxic effects.
(v) It causes respiratory ailment in humans and animals.

compound	Structural formula	Name
A	CH2 = CH2	Ethene
B		1,2 - dichloroethane
C	HCHO	Methanal
D	\(\mathrm{CH} \equiv \mathrm{CH}\)	Ethyne

(k_H)_bonzene = 4.2 x 10^–5 mm Hg
Solubility of methane = ?
P = 750mm Hg           
P = 840 mm Hg
According to Henrys Law,
P = K_H . x_{in solution}.
750 mm Hg = 4.2 x 10^–5 mm Hg . x_{in solution}
\(\Rightarrow X_{insolution}={750\over 4.2\times 10^{-5}}\)
i.e, solubility = 178.5 x 10⁵
similarly at P = 840 mm Hg
solubility = \({840\over 4.2\times 10^{-5}}\)
= 200 x 10^-5.

A₂(g) + B₂(g) ⇌ 2AB(g)

	A2	B2	AB
Initial Concentration	1	1	-
No.of moles reacted	x	x	-
No.of moles at equilibrium	1 - x	1 - x	2x

Total no. of moles = 1 – x + 1 – x + 2x = 2
\(K_p={(P_{AB})^2\over (P_{A_2})(P_{B_2})}={({2x\over 2}\times p)^2\over ({(1-x)\over 2}\times p)({1-x\over 2}\times p)}\)
\(K_p={4x^2\over (1-x)^2}\)
Given that K_p = 1; \({4x^2\over (1-x)^2}=1\)
\(\Rightarrow \) 4x² = (1 - x)²
\(\Rightarrow \) 4x² = 1 + x² - 2x
3x² + 2x - 1 = 0
\(X={-2\pm\sqrt{4-4\times 3\times -1}\over 2(3)}\)
\(X={-2\pm\sqrt{4+12}\over 6}\)
\(={-2\pm\sqrt{16}\over 6}\)
\(={-2+4\over 6};{-2-4\over 6}\)
\(={2\over 6};{-6\over 6}\)
X = 0.33 ; -1 (not possible)
\(\therefore\) [A₂]_eq = 1 - x = 1 - 0.33 = 0.67
[B₂]_eq = 1 - x = 1 - 0.33 = 0.67
[AB]_eq = 2X = 2 x 0.33 = 0.66.

The balanced chemical equation is
\(\mathrm{MgCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{MgO}+\mathrm{CO}_{2}\)
Molar mass of MgCO₃ is 84 g mol^–1.
84 g MgCO₃ contain 24 g of Magnesium.
∴ 100 g of MgCO₃ contain

= 28.57 g Mg.
i.e. percentage of magnesium
= 28.57 %.
84 g MgCO₃ contain 12 g of carbon

∴ 100 g MgCO₃ contain

= 14.29 g of carbon.
∴ Percentage of carbon
= 14.29 %.
84 g MgCO₃ contain 48 g of oxygen
∴ 100 g MgCO₃ contains

= 57.14 g of oxygen.
∴ Percentage of oxygen
= 57.14 %.
As per the stoichiometric equation,
84 g of 100 % pure MgCO₃ on heating gives 44 g of CO_2.
∴ 1000 g of 90 % pure MgCO₃ gives
\(\frac{\text { wt. of } \mathrm{MgCO}_{3}}{84 \mathrm{~g}} \frac{\text { % Purities }}{100 \%} \frac{\text { wt.of } \mathrm{CO}_{2}}{44 \mathrm{~g}}\)
\(x=44 \times \frac{100}{84} \times \frac{90}{100}\)
= 471.43 g CO₂
= 0.471 kg CO₂

T=175⁰ C=175+273=448K
Concentration of l-pentyne [A] = 1.3%
Concentration of2-pentyne [B] = 95.2%
Concentration of 1, 2-pentadiene [C] = 3.5%
At equilibrium
B\(\rightleftharpoons \)A
95.2%1.3%\(\Rightarrow \)
\({ K }_{ 1 }=\frac { 1.3 }{ 95.2 } =0.0136\)
B\(\rightleftharpoons \)​​C
95.2%3.5%\(\Rightarrow \)
\({ K }_{ 2}=\frac { 3.5 }{ 95.2 } =0.0367\)
\(\Rightarrow \)\(\Delta { G }_{ 1 }^{ 0 }\) =-2.303RT log K₁
\(\Delta { G }_{ 1 }^{ 0 }\)=-2.303X8.314X448Xlog0.0136
\(\Delta { G }_{ 1 }^{ 0 }\) =+16010J
\(\Delta { G }_{ 1 }^{ 0 }\)=+16 kJ
\(\Rightarrow \)\(\Delta { G }_{ 2 }^{ 0 }\)=-2.303X8.314X448Xlog0.0367
\(\Delta { G }_{ 2 }^{ 0 }\) =+12312J
\(\Delta { G }_{ 2 }^{ 0 }\)=+12.312 kJ.

Potential difference = 100V
= 100 x 1.6 x 10-¹⁹J
\(\lambda=\frac{h}{\sqrt{2mev}}\)
\(=\frac { 6.626\times { 10 }^{ -34 }Kg{ m }^{ 2 }{ s }^{ -1 } }{ \sqrt { 2\times 9.1\times { 10 }^{ -31 }Kg\times 100\times 1.6\times { 10 }^{ -19 }J } } \)
\(\lambda=1.22\times10^{-10}m\)

(i) Magnesium and calcium also play a vital role in biological functions. A typical adult human body contains about 25 g of magnesium and 1200 g of calcium. Magnesium plays an important· role in many biochemical reactions catalysed by enzymes. It is the co-factor of all enzymes that utilize ATP in phosphate transfer and energy release. It is also essential for DNA synthesis and is responsible for the stability and proper functioning of DNA.It is also used for balancing electrolytes in our body. Deficiency of magnesium results into convulsion and neuromuscular irritation.
(ii) Calcium is a major component of bones and teeth. It is also present in blood and its concentration is maintained by hormones (calcitonin and parathyroid hormone). Deficiency of calcium in blood causes it to take longer time to clot. It is also important for muscle contraction.
(iii) The main pigment that is responsible for photosynthesis, chlorophyll, contains magnesium which plays an important role in photosynthesis.

The van der Waals equation for n moles is
\(\left( P+\frac { { an }^{ 2 } }{ { v }^{ 2 } } \right) \left( v-nb \right) =nRT\) ....(1)
For 1 mole
\(\left( P+\frac { { a }^{ 2 } }{ { v }^{ 2 } } \right) \left( v-b \right) =RT\) ...(2)
From the equation we can derive the values of critical constants P_c, V_c and T_c, in terms of a and b, the van der Waals constants, On expanding the above equation
\(pv+\frac { a }{ v } -pb-\frac { ab }{ { v }^{ 2 } } -RT=0\)    ...(3)
Multiply equation (3) by V² / P
\(\frac { { v }^{ 2 } }{ p } \left( Pv+\frac { a }{ v } -pb-\frac { ab }{ { v }^{ 2 } } -RT \right) =0\)
\({ v }^{ 3 }+\frac { av }{ p } +-{ bv }^{ 2 }-\frac { ab }{ { v }^{ 2 } } -\frac { { RTV }^{ 2 } }{ p } \)    ...(4)
When the above equation is rearranged in powers of Y.
\({ v }^{ 3 }-\left[ \frac { RT }{ P } +b \right] { v }^{ 2 }+\left[ \frac { a }{ p } \right] v-\left[ \frac { ab }{ p } \right] =0\)        ...(5)
The equation (5) is a cubic equation in V. On solving this equation,
we will get three solutions. At the critical point all these three solutions of V are equal to the critical volume V_C. The pressure and temperature becomes Pc and T_c respectively
i.e., V = V_c
V - V_c = 0
(V - V_C)³ = 0
V3 - 3V_CV² + 3V_c²V - V_c³ = 0 .....(6)
As equation (5) is identical with equation (6), we can equate the coefficients of V², V and constant terms in (5) and (6).
\(-3{ v }_{ c }{ v }^{ 2 }=-\left[ \frac { { RT }_{ c } }{ { p }_{ c } } +b \right] { v }^{ 2 }\)
\(3{ v }_{ c }=\frac { { RT }_{ c } }{ { p }_{ c } } +b\)           .....(7)
\(3{ v }_{ c }^{ 2 }=\frac { a }{ { p }_{ c } } \)           ......(8)
\(3{ v }_{ c }^{ 2 }=\frac { ab }{ { p }_{ c } } \)           ...(9)
Divide equation (9) by equation (8)
\(\frac { { v }_{ c }^{ 3 } }{ 3{ v }_{ c }^{ 2 } } =\frac { ab/{ p }_{ c } }{ a/{ p }_{ c } } \)
\(\frac { { V }_{ c } }{ 3 } =b\)
i.e. V_c = 3b          .....(10)
when equation (10) is substituted in (8)
\(3{ v }_{ c }^{ 2 }=\frac { a }{ { p }_{ c } } \)
\({ p }_{ c }=\frac { a }{ 3{ v }_{ c }^{ 2 } } =\frac { a }{ 3\left( { 3b }^{ 2 } \right) } =\frac { a }{ { 3\times 9b }^{ 2 } } =\frac { a }{ { 27 }b^{ 2 } } \)
\({ p }_{ c }=\frac { a }{ { 27 }b^{ 2 } } \)            ...(11)
substituting the values of V_c and P_c in equation (7),
\(3vc=b+\frac { { RT }_{ c } }{ p } \)
\(3\left( 3b \right) =b+\frac { { RT }_{ c } }{ \left( \frac { a }{ { 27b }^{ 2 } } \right) } \)
\(9b-b=\left( \frac { { RT }_{ c } }{ a } \right) 27{ b }^{ 2 }\)
\(8b=\frac { { t }_{ c }R{ 27b }^{ 2 } }{ a } \)
\(\therefore { T }_{ c }=\frac { 8ab }{ 27R{ b }^{ 2 } } =\frac { 8a }{ 27Rb } \)
\({ T }_{ c }=\frac { { 8 }_{ a } }{ 27Rb } \)           ......(12)
The critical constants can be calculated using the values of van der Waals constant of a gas and vice versa.
\(a=3{ V }_{ C }^{ 2 }{ P }_{ C }\quad and\quad b=\frac { { V }_{ C } }{ 3 } \)

The electronic configuration of Sodium (Z = 11) Is²2s²2p⁶3s¹.
Magnesium (Z = 12) 1s²2s²2p⁶3s²
Magnesium atom has a smaller radius and higher nuclear charge than a sodium atom, thus more energy will be required to remove the electron from the same orbital (3s), making the first ionisation energy of magnesium higher than that of sodium.
However, the second ionization enthalpy of sodium is higher than that of magnesium. This is because after losing 1 electron, sodium attains the stable noble gas configuration of neon (1s²2s²2p⁶). On the other hand, magnesium, after losing 1 electron still has one electron in the 3s-orbital(1s²2s²2p⁶3s¹). In order to attain the stable noble gas configuration, Thus, the energy required to remove the second electron in case of sodium is much higher than that required in case of magnesium. Hence, the second ionization enthalpy of sodium is higher than that of magnesium.

High purity hydrogen (> 99.9%) is obtained by the electrolysis of water containing traces of acid or alkali or the electrolysis of aqueous solution of sodium hydroxide or potassium hydroxide using a nickel anode and iron cathode. However, this process is not economical for large-scale production.
 At anode: 2OH^- ➝ H₂O + 1/2O₂ + 2e^-
At cathode: 2H₂O + 2e^- ➝ 2OH^- + H₂
Overall reaction: H₂O ➝ H₂ + 1/2O₂