The equation representing the enthalpy of formation of methane is:
\(C(s)_{ 2 }(g)\longrightarrow CH_{ 4 }(g); \ \triangle HH=-78.84 \ KJ\)
\(\triangle H=78.84\quad KJ; \ \triangle ^{ ng }=1-2=-1 \ mol\)
\(R=8.314\times { 10 }^{ -3 }KJ \ { K }^{ -1 }mol^{ -1 },T=300K\)
According to the relation, \(\triangle H=\triangle U+\triangle ^{ ng }RT\)
\(\triangle U=\triangle H+\triangle ^{ ng }RT\)
= (-78.84 KJ) - (1 mol) x 8.314 x 10^-3 kJ K^-1 mol^-1) x 300K
= - 78.84 - 2.49 = - 81.35 kJ

Chromatography: Chromatography is based on the principle of selective distribution of the components of a mixture between two phases, a stationary phase and a moving phase. The stationary phase can be a solid or liquid, while the moving phase is a liquid or a gas. When the stationary phase is solid the basis is adsorption and when it is a liquid the basis is partition. Chromatography is generally used for the separation of coloured substances such as plant pigments or dyestuffs.

According to uncertainty principle,
\(\Delta x.m\Delta v=\frac { h }{ 4\pi } \) or \(m=\frac { h }{ 4\pi \Delta x\Delta v } \);h = 6.626 x 10^-34 kg m²s^-1 
\(\Delta \)x = 10^-10,m; \(\Delta \)v = 5.27 x 10^-24 ms-1
m = \(\frac { (6.626\times10^{ -34 }kgm^{ 2 }s^{ -1 }) }{ 4x3.143\times (10^{ -10 }m)\times (5.27\times10^{ -24 }ms^{ -1 }) } \)= 0.1kg

(a) 2, 2, 4-trimethylpentane 
Condensed formula: (CH₃)₂CHCH₂C (CH₃)₃
Bond line formula:

(b) 2-hydroxy-1, 2, 3-propanetricarboxylic acid
Condensed Formula:  (COOH)CH₂C(OH) (COOH)CH₂(COOH)
Bond line formula:

The functional groups present in the given compound are carboxylic acid (-COOH) and alcoholic (-OH) groups.
(c) Hexanedial Condensed Formula: (CHO) (CH₂)₄ (CHO)
Bond line Formula:

The functional group present in the given compound is aldehyde (-CHO).

The electrons may be assigned to the following orbitals :
(i) 4s
(ii) 3d
(iii) 4p
(iv) 3d
(v) 3p
(vi) 4p.
The increasing order of energy is :
(v) < (ii) = (iv) < (vi) = (iii) < (i)

As coke burns in a closed room, oxygen is used up and amount of oxygen in the room decreases. In the presence of carbon monoxide which is very poisonous because it combines with haemoglobin to form carboxy haemoglobin. This reaction takes place much more easily than reaction with oxygen to form oxyhaemoglobin. Thus supply of oxygen to body parts is hindered and this could be fatal.

When SiO₂ reacts with HF, silicon tetrafluoride is formed which dissolves in HF to from hydrofluorosilicic acid.
SiO₂ + 4HF\(\longrightarrow \)SiF₄ + 2H₂O
SiF₄ + 2HF \(\longrightarrow \)H₂SiF₆

sodium ash is obtained
2 NaHCO₃ \(\rightarrow\) Na₂Co₃ + H₂O + CO₂

(a) \(2NOCl(g)\rightleftharpoons 2NO(g)+Cl_{ 2 }(g);\)
\( { K }_{ p }=1.8\times { 10 }^{ -2 }at500k\)
\(2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) ; K_{p}=1.8 \times 10^{-2} \text { at } 500 \mathrm{~K}\)
(b) \(CaCO_{ 3 }(s)\leftrightharpoons CaO(s)+CO_{ 2 }(g);\)
\({ K }_{ p }=167 \ at \ 1073 \ K\)
\( { \Delta n }_{ g }={ n }_{ p }-{ n }_{ R }=1\)
\( { K }_{ c }=\frac { { k }_{ p } }{ (RT)^{ \Delta n_{ g } } } =\frac { 167 }{ 0.0821\times 1073 } =1.89\)

80.25 mol^-1

\({ P }_{ c }=\frac { a }{ 27{ b }^{ 2 } } =\frac { 0.751{ L }^{ 2 }atm \ mol^{ -2 } }{ 27\times (0.0226L \ mol^{ -1 })^{ 2 } } =54.5 \ atm\)
\( T_{ c }=\frac { 8a }{ 27Rb } =\frac { 8\times 0.75L^{ 2 }atm \ mol^{ -2 } }{ 27\times 0.082L \ atm \ K^{ -1 }mol^{ -1 } } \times 0.0226Lmol^{ -1 }=120K\)
\({ V }_{ c }=3b=3\times 0.0226L \ mol^{ -1 }=0.0678Lmol^{ -1 }\)

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

Pesticides are the substances used to kill unwanted pests.
For example, DDT.

NaCl, CaCO₃ and NH₃

Boron in BCl³ is sp² hybridised, due to this the shape of BCl³ molecule is trigonal planar. It is symmetrical in shape. The net dipole moment for symmetrical molecule is zero (because individual dipole moments cancel out due to the symmetry of the molecule).

μ = 0
Thus, dipole moment of BCl³ is zero.

The metals with low ionisation enthalpies are used in photoelectric cells. Caesium (Cs), an alkali metal belonging to group 1 is the most commonly used metal.

Let the oxidation number of CaO₂ be x
2 + 2x = 0
x = -1
Thus, oxidation number of in CaO₂ = -1

(i) 0.0048 = 4.8× 10^–3
(ii) 234, 000 = 2.34 ×10⁵
(iii) 8008 = 8.008 ×10³
(iv) 500.0 = 5.000 × 10²
(v) 6.0012 = 6.0012

For exhibiting cis-trans (or geometrical isomerism, a molecule must fulfil the following condition.
The groups attached to each double bonded carbon atom must be different.

Percentage of carbon =\(\frac { 12\times 0.198\times 100 }{ 44\times 0.246 } =21.95%\) %
Percentage of hydrogen =\(\frac { 2\times 0.1014\times 100 }{ 18\times 0.246 } =4.58%\) %

Interaction of H⁺ and OH^- ions of H₂O with the anion and the cation of a salt respectively to yield the original acid and the original base is called hydrolysis.
eg.
\(Na2​​​​​​​CO3+2H2O\rightarrow 2NaOH+H_{ 2 }CO_{ 3 }\\ Salt\quad \quad \quad \quad \quad \quad Base\quad \quad \quad Acid\)
Hydration, on the other hand, means addition of H₂O to ions or moleclues to form hydrated ions or hydrated.
\(KCl(s)+H_{ 2 }O(l)\rightarrow K^{ + }(aq)+cl^{ - }(aq)\\ Salt\)
\(\\ CuSo_{ 4 }(s)+5H_{ 2 }O(l)\rightarrow \ CuSO_{ 4 }.5H_{ 2 }O(s)\\ Colourless\quad \quad \quad \quad \quad \quad \quad \quad Blue\)

NaCl is neutral, it simply dilutes the HCl solution from 1 ml to 1000mL so that
[H+] = \(\frac { 0.01 }{ 1000 } ={ 10 }^{ 5 }M \)
pH = log (10^-5) = 5

We know,
\(\log \ K=\frac { -\Delta _{ r }G^{ \circ } }{ 2.303RT } =\frac { (-13.6 \times 10^{3}J \ mol^{-1}) }{ 2.303( 8.314 JK^{-1} mol^{-1}) (298 K) } =2.38\)
Hence, K = antilog 2.38 = 2.4 x 10²

Within a group, electron gain enthalpy becomes less negative down a group. but electron gain enthalpy of chlorine is more negative  (-349 kJ\(mol^{ -1 }\)than that of the fluorine (-328 kJ\(mol^{ -1 }\)).
This is due to small size of fluorine as the electron-electron repulsions in relatively compact 2p-orbital is greater than that in the larger 3p-orbital and hence, the incoming electron feels greater repulsion in fluorine than in chlorine. That's why chlorine have more negative electron gain enthalpy that that of fluorine.

0.1 nm.

The balanced chemical equation is:
\(\mathrm{BaCl}_2(a q)+\mathrm{Na}_2 \mathrm{SO}_4(a q) \rightarrow \mathrm{BaSO}_4(\mathrm{~s})+2 \mathrm{NaCl}(a q)\)
Let us first calculate moles of \(\mathrm{Na}_2 \mathrm{SO}_4\) and \(\mathrm{BaCl}_2\)
0.5 M solution of \(\mathrm{Na}_2 \mathrm{SO}_4\) means that \(0.5 \mathrm{~mol}\) of \(\mathrm{Na}_2 \mathrm{SO}_4\) are present in 1000 mL of solution.
1000 mL of solution contain \(\mathrm{Na}_2 \mathrm{SO}_4=\frac{0.5}{1000} \times 250=0.125 \mathrm{~mol}\)
Moles of \(\mathrm{BaCl}_2\) in solution \(=\frac{10}{208}\)
(Mol. mass of \(\left.\mathrm{BaCl}_2=137+2 \times 35.5=208\right)=0.048\)
According to the balanced equation, \(1 \mathrm{~mol}\) of \(\mathrm{BaCl}_2\) reacts with \(1 \mathrm{~mol}\) of \(\mathrm{Na}_2 \mathrm{SO}_4\) Therefore, \(\mathrm{BaCl}_2\) is the limiting reactant, so only \(0.048 \mathrm{~mol}\) of \(\mathrm{Na}_2 \mathrm{SO}_4\) reacts with \(0.048 \mathrm{~mol}\) of \(\mathrm{Na}_2 \mathrm{SO}_4\).
Now, according to the equation, \(1 \mathrm{~mol}\) of \(\mathrm{BaCl}_2\) produces \(\mathrm{BaSO}_4=1 \mathrm{~mol}\)
\(0.048 \mathrm{~mol}\) of \(\mathrm{BaCl}_2\) produces \(\mathrm{BaSO}_4=1 \times 0.048=0.048 \mathrm{~mol}\)
Amount of \(\mathrm{BaSO}_4\) obtained \(=0.048 \times 233 =11.18 \mathrm{~g}\)

Halogens have a strong tendency to accept electrons. Therefore, they are strong oxidising agents. Their relative oxidising power is, however, measured in terms of their electrode potentials. Since the electrode potentials of halogens decrease in the order: F2(+2.87V) > Cl2 (+1.36V) > Br2(+1.09V) > 12(+0.54V), therefore, their oxidising power decreases in the same order.
This is evident from the observation that F2 oxidises Cl^- to Cl₂' Br^- to Br₂, I^- - to I₂; Cl₂ oxidises Br^- to Br₂ and r to I₂ but not F- to F₂. Br₂, however, oxidises 1- to I₂but not F^- to F2^' and Cl^- to Cl₂
F2(g) + 2Cl^-(aq) \(\longrightarrow \) 2F^-(aq) + CI₂(g); F2(g) + 2Br^-(aq) \(\longrightarrow \) 2F^-(aq) + Br₂(l)
F2(g) + 2I^-(aq) \(\longrightarrow \) 2F^-(aq) + I₂(s);Cl₂(g) + 2Br^-(aq) \(\longrightarrow \) 2Cl^-(aq) + Br₂(l)
Cl₂(g) + 2l^-(aq) \(\longrightarrow \) 2Cl^-(aq) +I₂(s) and Br₂(l) + 2I^- \(\longrightarrow \) 2Br^-(aq) + I₂(s)
Thus, F2 is the best oxidant.
Conversely, halide ions have a tendency to lose electrons and hence can act as reducing agents. Since the electrode potentials of halide ions decreases in the order: F (-0.54 V) > Br" (-1.09 V) > Cl" (-1.36 V) > F (-2.87 V), therefore, the reducing power of the halide ions or their corresponding hydrohalic acids decreases in the same order: HI > HBr > HCl > HF. Thus, hydroiodic acid is the best reductant. This is supported by the following reactions. For example, HI and HBr reduce H₂SO₄ to SO₂ while HCI and HF do not.
2HBr + H₂SO₄ \(\longrightarrow \) Br₂ + SO₂ + 2H₂O; 2HI + H₂SO₄ \(\longrightarrow \) I₂+ SO₂ + 2H₂O
Further T reduces Cu²⁺ to Cu⁺ but Br" does not.
2Cu²⁺(aq) + 4I^-(aq) \(\longrightarrow \) Cu₂I₂(s)+ 12(aq);Cu2+(aq) + 2Br^- \(\longrightarrow \) No reaction. Thus, HI is a stronger reductant than HBr
Further among HCl and HF, HCl is a stronger reducing agent than HF because HCI reduces MnO₂ to Mn²⁺ but HF does not.
MnO₂(s) + 4HCl(aq) \(\longrightarrow \) MnCl₂(aq) + Cl₂(g) + 2H₂O
MnO₂(s) + 4HF(I) \(\longrightarrow \) No reaction
Thus, the reducing character of hydrohalic acids decreases in the order: HI > HBr > HCl > HF

Diamond has a three-dimensional network with strong C---C bonds, which are very difficult to break and thus, diamond has high melting point.

Mg₃N₂ is a salt of strong base, Mg(OH)₂and weak acid (NH₃). Thats's why on hydrolysis it gives ammonia (NH₃)
But MgCl₂ (magnisium chloride) is a salt of strong base Mg (OH)₂ and a strong acid (HCL). That's why it dose not hydrolysed to give HCl.

It produce more energy per unit mass of fuel and does not create pollution.

\(2K_{ 4 }[Fe(CN)_{ 6 }]+{ H }_{ 2 }SO_{ 4 }+{ H }_{ 2 }O_{ 2 }\rightarrow 2K_{ 3 }[Fe(CN)_{ 6 }]+{ K }_{ 2 }SO_{ 4 }+2{ H }_{ 2 }O\)

\({ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }+{ H }_{ 2 }O\rightleftharpoons { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 3 }^{ + }+{ OH }^{ - }\)
\({ K }_{ b }=\frac { \left[ { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 3 }^{ + } \right] \left[ { OH }^{ - } \right] }{ \left[ { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 } \right] } =\frac { \left[ { OH }^{ - } \right] ^{ 2 } }{ \left[ { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 } \right] } \)
\(\left[ { OH }^{ - } \right] =\sqrt { { K }_{ a }.C } =\sqrt { 4.27\times { 10 }^{ -10 }\times 0.001 }\)
\( \left[ { OH }^{ - } \right] =6.534\times { 10 }^{ -7 }\)
\(pOH=-log\left[ { 0H }^{ - } \right] =-log\left[ 6.534\times { 10 }^{ -7 } \right]\)
\( pOH=-0.8152+7=6.18\)
\(From,\ pH+pOH=14\)
\( pH=14-6.18=7.82\)
\({ C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }+{ H }_{ 2 }O\rightleftharpoons { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 3 }^{ + }+{ OH }^{ - }\)
\(Initial \ conc.\ C \ \ \quad 0\quad 0\)
\( Equili \ conc.C-C\alpha \quad C\alpha \quad C\alpha\)
\( { K }_{ b }=\frac { C\alpha .C\alpha }{ C(1+\alpha ) } \ [(1-\alpha )\approx 1 \ for \ weak \ base]\)
\( { K }_{ b }=C{ \alpha }^{ 2 }\)
\( or \ \alpha =\sqrt { \frac { { K }_{ b } }{ C } } \)
Degree of ionisation,
\(\alpha =\sqrt { \frac { 4.27\times { 10 }^{ -10 } }{ 0.001 } } =6.53\times { 10 }^{ -4 }\)
\( { K }_{ a } \ of \ conjugate \ acid \ of \ aniline,\)
\( { K }_{ a }=\frac { { K }_{ w } }{ { K }_{ b } } =\frac { { 10 }^{ -14 } }{ 4.27\times { 10 }^{ -10 } } =2.34\times { 10 }^{ -5 }\)

The given 3 processes comprises a cyclic system ABC which is shown in the graph below.

Total workdone in cyclic process, W=W_A→B+W_B→C+W_C→A
Workdone in A→B process :
It is a isobaric expansion (at constant pressure, P = 1 atm)
Initial volume, VA= 20 L
Final volume, VB = 40 L
Work done, W_A→B=−P(V_B−V_A)
=−1(40−20)=−20 L.atm
Workdone in B→C process :
It is a reversible isochoric process. The volume is constant at 40 L so there is zero workdone during this process.
W_B→C=0
Workdone in C→A process :
It is an isothermal compression (at constant temperature,T) process.
Initial volume, V_C= 20 L
Final volume, V_A = 40 L
Wordone for isothermal compression,
\( W_{C \rightarrow A}=2.303 \mathrm{nRT} \log \frac{V_C}{V_A} \)
\( =2.303 P_A V_A \log \frac{40}{20}  =13.87 \text { L. atm }\)
Total workdone in cyclic process, \(\mathrm{W}=\mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}+\mathrm{W}_{\mathrm{B} \rightarrow \mathrm{C}}+\mathrm{W}_{\mathrm{C} \rightarrow \mathrm{A}}\)
W=-20+0+13.87
\( =-6.13 \mathrm{~L} . \mathrm{atm} \)
\( =-\frac{6.13 \times 8.314}{0.0821} \mathrm{~J} \text { (in joules) }  =-620.77 \mathrm{~J}\)
 

(i) The second law of thermodynamics states that complete conversion of energy of one kind into another is not possible as some energy is always lost in the form of some other energy.
(ii) Because some energy may be used up in overcoming friction of wheels.
(iii) In refrigerator, heat flows from higher temperature to lower temperature in a compressor with mechanical work which is done by the compressor on the system.
The system consists of four components: compressor, condenser, expansion, value and evaporator. the following cycle is repeated over and again.
Refrigerant \(\overset { compressed }{ \underset { in \ compressor }{ \longrightarrow } } \) Gas form \(\overset { condensed }{ \underset { in \ condenser }{ \longrightarrow } } \)
Liquid form \(\overset { Allowed }{ \underset { to \ throttle \ through \ expansion \ valve }{ \longrightarrow } } \)
Gas form \(\overset { Allowed }{ \underset { to \ evaporate \ in \ evaporator }{ \longrightarrow } } \)Refrigerant. 
(iv) Mr.sharma is scientific and practical

According to Gay-Lussac's law, \(\frac { { p }_{ 1 } }{ { T }_{ 1 } } =\frac { { p }_{ 2 } }{ { T }_{ 2 } } \)
\(\frac { 250 }{ 300 } =\frac { { p }_{ 2 } }{ 1800 } or{ \ p }_{ 2 }=1500kPa\)
As the cylinder can withstand a pressure of \({ 10 }^{ 6 }\)
\(Pa=10^{ 3 }kPa\) = 1000kPa, hence, it will blow up.

There are numerous physical properties of elements such as melting points, boiling points, heats of fusion and vaporisation, energy of atomisation, etc., which show periodic variations. The cause of periodicity in properties is the repetition of similar outer electronic configuration after certain regular intervals. e.g. all the elements of 1s group (alkali metals) have similar outer electronic configuration, i.e. ns¹.
₃Li = 1s², 2s¹
₁₁Na = 1s², 2s², 2p⁶ , 3s¹
₁₉K = 1s², 2s², 2p⁶ , 3s², 3p⁶, 4s¹
Therefore, due to similar outermost shell electronic configuration all alkali metals have similar properties. e.g., sodium and potassium both are soft and reactive metals. They all form basic oxides and their basic character increases down the group. They all form unipositive ion by the loss of one electron. Similarly, all the elements of 17th group (halogens) have similar outermost shell electronic configuration, i.e. ns²  np⁵ and thus possess similar properties.
₉F = 1s², 2s¹ , 2p⁵
₁₇Cl = 1s², 2s², 2p⁶ , 3s² , 3p⁵
₃₅Br = 1s², 2s², 2p⁶ , 3s², 3p⁶, 3d¹⁰, 4s² , 4p⁵