Electrochemical cell is a device in which the redox reaction is carried indirectly and the decrease in free energy appears as electrical energy

Poise.

The amount of oxygen required by bacteria to break down the organic matter present in a certain volume of a sample of water is called biochemical oxygen demand (BOD). The amount of BOD in the water is a measure of the amount of organic matter in water, in terms of how much oxygen will be required to break it down biologically. Clean water would have BOD value of less than 5 ppm.

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.

Rotation around C-C single bond is not completely free and it is restricted due to repulsions between the electron clouds of C-H bonds in the adjacent carbon atoms. Therefore, ethane exixts in infinate number of conformations. Out of these, two extreme conformations are staggered and eclipsed.

NH₄Cl + NaOH \(\rightarrow\) NaCl + NH₃ + H₂O
Thus, A is ammonium chloride.

High dipole moment and high dielectric constant, these are the two properties of water which make it useful as a solvent It can dissolve both ionic compounds as well as those covalent compounds which can form hydrogen bonds with water such as ethyl alchol, sugar, glucose etc.

(a) MnO^-₄  loses oxygen, therfore it is an oxidant while C₂H₅OH gains oxygen, therefore it is a reductant.
(b) Zn is reductant whilr hydrogen (H) is oxidant
(c) pbO is oxidant while C is reductant

\({ NO }_{ 2 }^{ + }\)and CO₂ are sp hybridized, therefore, they have linear shape. 

 \({ { NO }_{ 2 }^{ - } }\) and O₃ are sp² hybridised.Both have lone pair of electrons but in former case it is not donated. Thus, their shape is angular.
   

Mass of proton
\(\frac { 1.008\times 10^{ -3 } }{ 6.02\times 10^{ 23 } } kg=1.67\times 10^{ -27 }kg\)
\(KE=\frac { 1 }{ 2 } mv^{ 2 }\)
\(v=\frac { 2KE }{ m } =\frac { 2\times 5\times 10^{ -27 } }{ 1.67\times 10^{ -27 } } =5.98\)

Step 1: 
From the formula, we can say that,
1 mol of Na₂CO₃.10H₂O contains 10+3 = 13 moles O atoms
we have 0.1 moles of Na₂CO₃.10H₂O
So, we will have 13 x 0.1 = 1.3mol O atoms
Step 2:
Since, 1 mole O atoms = 6.022 x 10²³atoms
1.3 mol O atoms = 1.3 x 6.022 x 10²³ = 0.78 x 10²⁴ O atoms.
Hence, the number of oxygen atoms in 0.10 mole of Na₂CO₃.10H₂O are 0.78×10²⁴.

Electronic configuration of carbon atom: C: σ1s² 2s² 2p².
In the excited state, the orbital picture of carbon can be represented as:

Hence, carbon atom undergoes sp³ hybridization in CH₄ molecule and takes a tetrahedral shape.

For a square planar shape, the hybridization of the central atom has to be dsp³. However, an atom of carbon does not have d-orbitals to undergo dsp3 hybridization. Hence, the structure of CH₄ is tetrahedral.

Step I: Calculation of K_p for the reaction
Let the total mass of the gaseous mixture = 100 g
Mass of CO in the mixture = 90.55 g
Mass of CO₂ in the mixture = (100 - 90.55) = 9.45 g
No. of moles of CO = \(\frac { 90.55g }{ (28g\ mol^{ -1 }) } \) = 3.234 mol
No. of moles of CO₂ = \(\frac { 9.45 }{ (44g\ mol^{ -1 }) } \) = 0.215 mol
P_co in the mixture = \(\frac { (3.234\ mol) }{ (3.234 + 0.215) } \times 1atm\) = \(\frac { (3.234\ mol) }{ (3.449\ mol) } \times 1atm\) = 0.938 atm
P_^co2 in the mixture = \(\frac { (0.215\ mol) }{ (3.449\ mol) } \times 1atm\) = 0.062 atm
C(s) + CO₂(g) ⇌ 2CO(g)
Eqm. pressure 0.062atm 0.938atm
K_p = \(\frac { { p }^{ 2 }CO }{ pCO_{ 2 } } =\frac { (0.938\ atm{ ) }^{ 2 } }{ (0.062\ atm) } \) = 14.19 atm
Step II. Calculation of K₂ for the reaction.
\(K_{ c }=\frac { K_{ p } }{ (RT)^{ \Delta ng } } \)
K_p = 14.19 atm, R = 0.0821 L atm K^-1 mol^-1, T = 1127 K ; Δ^ng = 2 - 1 = 1
K_c = \(K_{ c }=\frac { (14.19\ atm) }{ (0.0821\ L\ atm\ K^{ -1 }\ mol^{ -1 })\times (1127K)^{ 1 } } \) = 6.46

Kala chana , soya bean and sprouted lentils are the sources of iron potassium and proteins

0.2 L(or 200 mL) of 5M solution will contain 
\(\frac { 340\times0.2 }{ 2 } =34g{ H }_{ 2 }{ O }_{ 2 }\)
\(\underset { 2\times 34=68g }{ { 2H }_{ 2 }O_{ 2 } } \rightarrow { 2H }_{ 2 }O+\underset { 2\times 16=32g }{ { O }_{ 2 } } \)
\(\because \ 68g \ { H }_{ 2 }O_{ 2 }\) on decomposition will give 32g\(O_{ 2 }\)
\(\therefore \ 34g \ { H }_{ 2 }O_{ 2 }\) on decomposition will give
\(\frac { 32\times 34 }{ 68 } =16g{ O }_{ 2 }\)

The reaction of H₂O₂ with acidified KI Solution.
\(2KI+{ H }_{ 2 }{ SO }_{ 4 }+{ H }_{ 2 }{ O }_{ 2 }\longrightarrow { K }_{ 2 }{ SO }_{ 4 }+{ I }_{ 2 }+2{ H }_{ 2 }O\)
\( 34 \ g \ 254 \ g\)
\( 5 \ { cm }^{ 3 } \ 0.508 \ g\)
\(254 \ g \ af \ { I }_{ 2 } \ is \ liberated \ by \ { H }_{ 2 }{ O }_{ 2 }=34g\)
\(0.508 \ g \ { I }_{ 2 } \ is \ liberated \ by \ { H }_{ 2 }{ O }_{ 2 }=\frac { 34 }{ 254 } \times 0.508=0.068g\)
\(Now,5{ cm }^{ 3 } \ of \ { H }_{ 2 }{ O }_{ 2 } \ contains=0.068g \ of\quad { I }_{ 2 }\)
\(then,1{ cm }^{ 3 } \ of \ { H }_{ 2 }{ O }_{ 2 } \ contains=\frac { 0.068 }{ 5 } =0.0136g \ of \ { I }_{ 2 }\)
\( now,\ 2{ H }_{ 2 }{ O }_{ 2 }\longrightarrow 2{ H }_{ 2 }O+{ O }_{ 2 }\)
\( 68g \ 22400mL \ at \ STP\)
\( 68g \ of \ { H }_{ 2 }{ O }_{ 2 } \ gives \ 22400mL \ { O }_{ 2 } \ at \ STP\)
\(1{ cm }^{ 3 } \ or \ 0.0136g \ of \ { H }_{ 2 }{ O }_{ 2 } \ give \ { O }_{ 2 } \ at \ STP\)
\( =\frac { 22400 }{ 68 } \times 0.0136\)
\( =4.48mL \ of \ STP\)
\( \therefore \ Strength \ of \ { H }_{ 2 }{ O }_{ 2 }=4.48 \ volumes\)

(i) A\(\rightarrow \)B (Temperature and pressure are constant).
\(\therefore \) It is isothermal and isobaric process.
(ii) B\(\rightarrow \) C It is adiabatic expansion in which temperature falls from T₁ and T₂.
(iii) C\(\rightarrow \)D (Temperature and volume are constant)
\(\therefore \) This process is isothermal and isochoric.
(iv) D \(\rightarrow \)E (Temperature and pressure are constant)
\(\therefore \) It is isothermal and isobaric contraction.
(v) E\(\rightarrow \)F (It is adiabatic compression in which temperature increases from T₂ to T₁)
(vi) F\(\rightarrow \)A (Temperature and volume are constant)
\(\therefore \) It is isothermal and isochoric process.

(i) Total work done in an expansion when the state of an ideal gas is changed reversibly and isothermally from (p _i ,V _i) to ( p _f ,V _f). Reversible work is represented by the combined areas ABC and BCV_iV_f.

(ii) Work against constant pressure, p_f is represented by the area BCV_iV_f .Work(i) > Work (ii) 

(i) Potassium, K (Z = 19)
(ii) Strontium, Sr (Z = 38)
(iii) Ruthenium, Ru (Z = 44)
(iv) Praseodymium, Pr (Z = 59 ) 
(v) Xenon, Xe(Z = 54 )

According to Einstein's equation \(KE=\frac { 1 }{ 2 } { m_{e}v }^{ 2 }=h(v-{ v }_{ o })\)
\(=\left( 6.626\times { 10 }^{ -34 }Js \right) (1.0\times { 10 }^{ 15 }{ s }^{ -1 }-7.0\times { 10 }^{ 14 }{ s }^{ -1 })\)
\( =\left( 6.626\times { 10 }^{ -34 }Js \right) (100\times { 10 }^{ 14 }{ s }^{ -1 }-7.0\times { 10 }^{ 14 }{ s }^{ -1 })\)
\(=\left( 6.626\times { 10 }^{ -34 }Js \right) \times (3.0\times { 10 }^{ 14 }{ s }^{ -1 })=1.988\times { 10 }^{ -19 }J\)

Step I As the question is already in skeletal form of the equation, so, a step I is completed.
Step II changing the elementary substance hydrogen to atomic form
\(Fe+{ H }_{ 2 }O\rightarrow { Fe }_{ 3 }{ O }_{ 4 }+{ 2H }\)
Step III \({ Fe }_{ 3 }{ O }_{ 4 }\) has the largest number of atoms, To balance this, multiply \({ H }_{ 2 }O\) by 4 balance oxygen atoms. In 4 molecules of \({ H }_{ 2 }O\)there are 8 atoms of H which are balanced by multiplying H on RHS by 8.
\(3Fe+{ 4H }_{ 2 }O\rightarrow { Fe }_{ 3 }{ O }_{ 4 }+{ 8H }\)
Step IV Converting H-atom to molecular form.
\(3Fe+{ 4H }_{ 2 }O\rightarrow { Fe }_{ 3 }{ O }_{ 4 }+{ 8H }\)
Step V The number of Fe, H and O atoms on both the sides of the equation respectively 3,8 and 4. The equation obtained is a balanced chemical equation.

(a) Distillation under reduced pressure or vacuum distillation
(b) Sublimation.

For preparation of alkanes containing odd number of carbon atoms, a mixture of two alkyl halides has to be used. Since two alkyl halides can react in three different ways, therefore, a mixture of three alkanes instead of the desired alkane would be formed. For example, the Wurtz reaction between 1-bromopropane and 1-bromobutane gives a mixture of three alkanes i.e., hexane, heptane and octane as shown below:

(a) Cr is getting oxidised and Mn\({ O }_{ 4 }^{ - }\) is getting reduced.
(b) Cr is negative electrode, Pt in Mn\({ O }_{ 4 }^{ - }\)-acts as positive electrode

32.39

(a) \(B_{ 2 }H_{ 6 }+2NH_{ 3 } \rightarrow \ 2BH_{ 3 }.NH_{ 3 }\)
                     Borane-ammonia complex (adduct)
or \([BH_{ 2 }(NH_{ 3 })_{ 2 }]^{ + }[BH_{ 4 }]^{ - }\)
(b) \(3{ B }_{ 2 }{ H }_{ 6 }+6NH_{ 3 } \ \rightarrow \) \(3[BH_{ 2 }(NH_{ 3 })_{ 2 }]^{ + }[BH_{ 4 }]^{ - } \ \underrightarrow { \Delta } \ 2B_{ 3 }{ N }_{ 3 }{ H }_{ 6 }+12{ H }_{ 2 }\)
                    Borazine (inorganic enzene)

The name of the compound is hydrogen peroxide, \({ H }_{ 2 }O_{ 2 }\) It acts as an oxidising agent as well as reducing agent in both acidic and basic nedium.

(a) Pressure will be doubled (2p. atm).
(b) Temperature will remain same
(c) Density will remain same  (d g cm^-3).
(d) Number of moles will be doubled i.e.2n.

Bond order is defined as one half the difference between the number of electron present in the bonding and anti-bonding orbitals, i.e
Bond order (BO) \(=\frac { 1 }{ 2 } ({ N }_{ b }-{ N }_{ a })\)
A positive bond order means a stable molecule while a negative or zero bond order means an unstable molecule.
Stability of a molecule \(\propto \) bond order
Bond length \(\propto \) \(\frac { 1 }{ bond \ order } \)
Bond order values 1,2 or 3 corresponding to single, double or triple bonds respectively.

\(\lambda _{ O_{ 2 } }/\lambda _{ SO_{ 2 } }=2 \ (because \ \lambda =\frac { h }{ mv } ,i.e.\lambda \propto \frac { 1 }{ m } \) and mass of SO₂ molecule viz.64 u is double than that of O₂ molecule viz.32 u).