From the available data: V₁ = 100 L, V₂ = 175L
P₁ = 2 bar, P₂ = 2 x 1/3 = 2/3 bar
T₁ = 20 + 273 = 293 K, T₂=?
According to Gas equation,\(\frac { { P }_{ 1 }V_{ 1 } }{ { T }_{ 1 } } =\frac { { P }_{ 2 }V_{ 2 } }{ { T }_{ 2 } } \)  or T₂ = \(\frac { { P }_{ 2 }V_{ 2 }{ T }_{ 2 } }{ { T }_{ 2 } } \)
By substituting the values, T₂ = \(\frac { (2/3bar)\times (175L)\times (293K) }{ (2bar)\times (100L) } \)
= 170.9 K = 170.9 - 273.0 = -102.1^oC

I. Alkali metals are  strong reducing agents, hence cannot be extracted by reduction of their oxides and other compounds.
II. Being highly positive in nature it is not possible to displace them from their salt solutions by any other elements.
III. Alkali metals cannot be obtained by the electrolysis of the aqueous solution of their salts because H₂ is liberated at cathode instead of alkali metal.
That's why alkali metals are prepared by electrolysis of their fused chloride.
NaCl  \(\underrightarrow { Fusion } \) Na⁺ + Cl^-
During electrolysis
At anode, 2 Cl^- \(\longrightarrow \) Cl₂ + 2 e^-
At cathode, 2 Na⁺ + 2 e^-  \(\longrightarrow \)  2 Na

\({ N }_{ 2 }{ O }_{ 4 }(g)\leftrightharpoons { 2NO }_{ 2 }(g)\)
\( Pressure \ at \ equilibrium \ 0.28\ \ \ 1.1\)
\( { K }_{ P }=\frac { p{ { (NO }_{ 2 }) }^{ 2 } }{ p({ N }_{ 2 }{ O }_{ 4 }) } =\frac { { (1.1) }^{ 2 } }{ (0.28) } =4.32atm\)
If volume of the container is doubled, the pressure will be reduced to half
\( { N }_{ 2 }{ O }_{ 4 }\leftrightharpoons { 2NO }_{ 2 }\)
\( New \ pressure \ \left( \frac { 0.28 }{ 2 } -p \right) \ \left( \frac { 1.1 }{ 2 } +2p \right)\)
\( { K }_{ p }=\frac { { \left( \frac { 1.1 }{ 2 } +2p \right) }^{ 2 } }{ \left( \frac { 0.28 }{ 2 } -p \right) } =4.32\)
\( On \ solving,\)
\( p=0.045\)
\( \therefore \ p({ N }_{ 2 }{ O }_{ 4 })=0.14-0.0045=0.095atm\)
\( p({ NO }_{ 2 })=0.55+0.045=0.64 \ atm\)

Enthalpy change for the reaction, 
\(\triangle { H }^{ o }=\sum { { \triangle }_{ f } } { H }^{ o }_{ (products) }-\sum { { \triangle }_{ f } } { H }^{ o }_{ (reactants) }\\ [{ \triangle }_{ f }{ H }^{ 0 }N{ O }_{ 2 }(g)]-[{ \triangle }_{ f }{ H }^{ 0 }NO(g)+\frac { 1 }{ 2 } { \triangle }_{ f }{ H }^{ 0 }{ O }_{ 2 }(g)]\)
\( =[33.8]-\left[ 904+\frac { 1 }{ 2 } \times0 \right] =-56.6KJ \ { mol }^{ -1 }\)
\(\triangle { S }^{ 0 }=-70J{ K }^{ -1 }{ mol }^{ -1 } ...(i)\)
\(Now, \ \triangle { G }^{ o }=\triangle { H }^{ o }-T\triangle { S }^{ o },T=298K\)
\( \triangle { G }^{ o }= \ 56600-298\times(-70)=-35740J{ mol }^{ -1 }\)
\(logk=-\frac { \triangle G }{ 2.303RT } \)
\(or \ logK=-\frac { -35740 }{ 2.303\times8.31\times298 } =6.267\)
∴ K = Antilog (6.267) = 1.85 x 10⁶

All the hybrid orbitals have same shape. However, their sizes are in the order : sp < sp² < sp³

Cu is largest due to less effective nuclear charge. It has 29 elements,29 protons.Cu⁺ has 28 elements and 29 protons, Cu²⁺ has 27 electrons and 29 protons.

(i) \({ 1s }^{ 2 }2s^{ 2 },{ 2p }_{ x }^{ 2 },{ 2p }_{ y }^{ 2 },{ 2p }_{ z }^{ 2 },3s^{ 2 }\)
(ii) \({ 1s }^{ 2 }2s^{ 2 },{ 2p }_{ x }^{ 1 },{ 2p }_{ y }^{ 1 },{ 2p }_{ z }^{ 1 }\)
(iii) \({ 1s }^{ 2 }2s^{ 2 },{ 2p }^{ 6 },3s^{ 2 },{ 3p }^{ 6 },4s^{ 2 },3d^{ 3 }\)
(iv) \({ 1s }^{ 2 }2s^{ 2 },{ 2p }^{ 6 },3s^{ 2 },{ 3p }^{ 6 },3d^{ 5 },4s^{ 1 }\)

Given, (5.7 x 10^-5) \(\div \) (4.2 x 10^-3)
(5.7 \(\div \) 4.2) x (10^-5-(-3)) = 23.94 x 10^-2

(a) 

(b) 

(c)

(d)

(e)

(f) 

22.4litres of CO₂ at N.T.P. = 1 gram mol
11.2 litres of CO₂ at N.T.P. =\(\frac { (1 \ gram \ mol) }{ (22.4 \ liters) } \times \)(11.2 litres) = 0.5 gram mol
Now 1 gram mole of CO₂ contain molecules = 6.022 x 10²³
\(\therefore \) 0.5 gram mole of CO₂ contain molecules = 6.022 x 10²³ x 0.5 = 3.011 x 10²³
Step II. Number of carbon and oxygen atoms in 3.011 x 1&3 molecules of CO₂
1 molecule of CO₂ contains carbon atoms = 1
\(\therefore \)3.011 x 10²³ molecules of CO₂ will contain carbon atoms = 3.011 x 10²³
Similarly, 1 molecule of CO₂ contains oxygen atoms = 2
\(\therefore \) 3.011 x 10²³molecules of CO₂ will contain oxygen atoms = 2 x 3.011 x 10²³
= 6.022 x 10²³ atoms.

This is because of high solubility of potassium hydrogen carbonate due to which it does not precipitate out when CO₂ is pased through an ammoniated solution of KCl.

Given, q = +701 J (heat is absorbed, hence q is positive).
W = -394 J (work is done by the system, hence W is negative). By first law of thermodynamics;
Internal energy change,
\(\Delta U=q+W\)
\( =+701J+(-394J)=+307J\)
Hence, internal energy of the system increases by 307J.

From the given chemical equation of the reaction is
MnO₂ + 4HCl→ MnCl₂+Cl₂+2H₂O
So, in this reaction we can understand that the 1 mole of  MnO₂ reacts with 4 moles of  HCl
Now, Mass of MnO₂ = atomic mass of Manganese + 2(atomic mass of oxygen)
Therefore, we can write the mass of  MnO₂ is = 55 +2(16)55 +2(16) = 87 gms/mole
This means, 1 mole of MnO₂ comprises 87 gm of MnO₂.
Similarly, Mass of HCl = atomic mass of Hydrogen + atomic mass of Chlorine
Therefore, we can write as mass of  HCl = 1 + 35.451 + 35.45 =36.46 gms/mole
This means, 1 mole of  HCl comprises a mass of 36.46 gm of MnO₂.
So, 4 moles of HCl = 4 ×36.46 gm. = 145.84 gm of HCl
Thus, according to given chemical equation,145.84 gm. of HCl reacts with 87 gm. of MnO₂ to accomplish the reaction.
So now we can find the mass of HCl reacts with 5gm of MnO₂ to accomplish the reaction.
∴ Mass of  HCl (gm.) required = \(\frac { 4\times 36.5\times 5 }{ 87 } \)= 8.39 g HCI.
Hence, the answer is, 8.38 gm. of HCl will require to react with 5 gm. of MnO₂

It is defined as the ratio of product of molar concentration of products to the product of molar concentration of reactants at any stage of reaction.
Q_c = \(\frac { \left[ C \right] ^{ c }\left[ D \right] ^{ d } }{ \left[ A \right] ^{ a }\left[ B \right] ^{ b } } \)
for the reaction 
aA + bB ⇌ cC + dD

Bond order of N₂ (= 3) is greater than that of 0₂ (= 2).

\(\overset { +1 }{ K } \overset { +3 }{ Al } \left( \overset { x }{ S } \overset { -2 }{ { O }_{ 4 } } \right) 12\left( \overset { +1\quad +2 }{ { H }_{ 2 }O } \right) \)or
+ 1 + 3 + 2x + 8 (-2) + 12 (2 x 1 - 2) or x = + 6
Alternatively, since H20 is a neutral molecule, therefore, sum of oxidation numbers of all the atoms in H20 may be taken as zero. As such water molecules may be ignored white computing the oxidation number of 5. 
\(\therefore\) + 1 + 3 + 2x - 16 = 0 or x = +6
Thus, the oxidation number of S in KAI(SO₄)₂.12H₂0 = +6

Methane is the limiting reactant because the other reactant is oxygen of the air which is always present in excess. Thus, the amounts of CO₂ and H₂O formed depend upon the amount of methane burnt.

Butanoic acid,
\(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COO}^{-} \mathrm{Na}^{+}+\mathrm{NaOH} \stackrel{\mathrm{CaO}}{\longrightarrow}  \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3+\mathrm{Na}_2 \mathrm{CO}_3\)

Raw materials used for the manufacture of washing soda by Solvay process are NaCl, CaCO₃ and NH₃.

(i) According to Le Chatelier’s principle, the equilibrium will shift in the backward direction.
(ii) According to Le Chatelier’s principle, as the reaction is endothermic, the equilibrium will shift in the forward direction.
(iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.

NaH₂P O₄
Let the oxidation number of P be x. Writing the oxidation number of each atom above its symbol, we get \(\stackrel{+1}{\mathrm{Na}} \stackrel{+1}{\mathrm{H}}{ }_2 \stackrel{\mathrm{P}-2}{\mathrm{P}} \mathrm{O}_4\)
In neutral compounds the sum of the oxidation numbers of all the atoms is zero.
1(+1)+2(+1)+x+4(-2)=0
3+x+(-8) =0
x=8-3=+5
Hence, the oxidation number of P in NaH₂PO₄ is +5.
Calculate oxidation number of other elements in the same way as shon in(i). You will get oxidation number of S,P,Mn,O,B,S,S,Cr,S in the given compound as
+6,+5,+6,-1,+3,+6,+6,+6,+2 respectively.
Note 
H₂ O is a neutral molecule, therefore sun of oxidation numbers of all atoms in H₂O is Zero.
Hence, the oxidation number of S in
KAI(SO₄)₂.12H₂O is +6.

△_subH^∘=△_fusH^∘+△_vapH^∘ = 2.6+98.2
= 100.8KH mol^-1

V = 9.82 cm³