Azeotrope, in chemistry, a mixture of liquids that has a constant boiling point because the vapour has the same composition as the liquid mixture. The boiling point of an azeotropic mixture may be higher or lower than that of any of its components.

Conductivity of a solution at any given concentration is the conductance of one unit volume of the solution kept between two platinum electrodes with the unit cross-sectional area at a distance of unit length. Its SI unit is Sm⁻¹

Molar conductivity of a solution at a given concentration, on the other hand is the conductance of volume V of a solution containing 1 mole of the electrolyte, kept between two electrodes with the cross-sectional area of A and distance of unit length. It is denoted as Λ_m. The SI unit of molar conductivity is Sm²mol^-1.

Electrolysis of aqueous sodium chloride yields hydrogen at cathode  and chlorine at anode, with aqueous sodium hydroxide remaining in solution.

Thanks u both

The acid produced in the reaction acts as a catalyst and makes the reaction faster. Substances that act as catalysts in the same reaction in which they are obtained as products are known as autocatalysts

Ester hydrolysis can be represented as: Ester + Water → Acid + Alcohol

The acid produced in the reaction acts as a catalyst and makes the reaction faster. Substances that act as catalysts in the same reaction in which they are obtained as products are known as autocatalysts.

In case of  isotonic electrolytes ,

* the concentrations of both the electrolytes are equimolecular

*  their  osmotic pressures are equal

*  both the electrolytes have same temperature 

Therefore , considering the given electrolytes ,  

{tex}\pi{/tex}_{(K2SO4)   =}  {tex}\pi{/tex}_(NaOH ) 

i_(K2SO4) C_(K2SO4) RT  =  i_(NaOH)  C (_NaOH)RT

but  ,  C_{(K2SO4)  =}  C_(NaOH)

 {tex}\therefore{/tex}   i _{(K2SO4)  =   i ( NaOH )}

Now,  we know that  i (ie .  vant Hoff' 's  factor )

 =   No. of particles in dissociated molecule ( ie. after dissociation ) /   No. of particles in undissociated molecule  ( ie before dissociation )

ie.  =   degree of dissociation  ({tex}\alpha{/tex})

So, let us calculate degree of dissociation for both the given electrolytes-

K₂ SO₄----------------->  2K⁺    +    SO₄^2-                    ;        NaOH ----------->   Na⁺     +    OH^- 

( 1 - {tex}\alpha{/tex} ).........................( 2 {tex}\alpha{/tex} )...........({tex}\alpha{/tex}  ) -----------------------(1- {tex}\alpha{/tex} )..................({tex}\alpha{/tex}).............({tex}\alpha{/tex})

Therefore total number of  particles ( ie ions / molecules )  after dissociation in both cases  are -

( 1 - {tex}\alpha{/tex}  +  2{tex}\alpha{/tex} +  {tex}\alpha{/tex} ).............................................................................( 1 - {tex}\alpha{/tex}  +  {tex}\alpha{/tex}   +    {tex}\alpha{/tex}  )

=  (1  +  2{tex}\alpha{/tex} )................................. ...............................................................=  (  1  +    {tex}\alpha{/tex}  )

..........................................substituting the given value for degree of dissociation ({tex}\alpha{/tex}) in case of NaOH as = 1

.......................................................................................................................= 2

Further  ,   i( K₂ S O₄ )   =  [ (1 + 2{tex}\alpha{/tex} ) /  1  ] ........................&    i (NaOH )   =  ( 2 / 1 )    

.......................................= [ ( 1  +  2{tex}\alpha{/tex} (K₂ SO₄ )  ] .......................................= 2....................................

           Equating i_(K2SO4)   and  i _{( NaOH)} ,   as inferred above we get -

1  +  2{tex}\alpha{/tex} _(K2SO4)    =   2

{tex}\therefore{/tex}   2 {tex}\alpha{/tex} _{(K2SO4)   =  1}

&     {tex}\alpha{/tex}  _(K2SO4)   =  1/ 2

...........................=  0.50

Hence , the degree of dissociation of  given 17.4% K₂ SO₄  is  0.50  and is ,therefore ,50%  dissociated  when it is isotonic with  given 4% NaOH solution .

1

The reaction of Propene with HBr  can  adopt any one of the following paths yielding  different products.

The anti- Markonikov addition occurs under following  conditions,

Thus , the mechanism of anti - Markonikov reaction  follows a free radical mechanism as depicted below,

Mechanism:-- As in the presence of any peroxide ,the negative part of the alkyl halide( here-Br) always gets attached to that carbon atom of alkene which has more no.of H atom therefore,

CH₃CH=CH₂ + HBr (In peroxide) 

----------> CH₃CH₂CH₂Br 

Adsorption is the adhesion of atoms, ions, or molecules from a gas, liquid, or dissolved solid to a surface. This process creates a film of the adsorbate on the surface of the adsorbent.

Following  are the points  of  differences  between  vitamins   and  halogen containing  drugs ,

No because the threshold energy remains the same irrespective of it's speed. Hence speed do not depends upon the activation energy 

No because the threshold energy remains the same irrespective of it's speed.. Hence speed do not depends upon the activation energy 

Rate constant k=AeEa/RT

_{AS temperature is infinite, K=A.}

At 338K the value i <font size="+1">2.40 x 10^-3</font>

Activation energy is always a positive quantity.It never have a negative energy.

Yes sometime it is negative in case of backward reaction 

Reaction never have negative activation energy.Its always a positive quantity.

There is no effect of altering the concentrations of reactant on zero order reaction,since they are concentration independent.

Never

This is because  the speed of the molecules in the gaseous mixture  at room temperature does not provide enough kinetic energy  to activate the reaction during  collision between the reactants . A mixture of  gases is formed with potential to  react or explode violently  , if sufficient energy is given. to such a gaseous mixture .Even introduction of an spark to this mixture results in raised temperature among some of the hydrogen and oxygen molecules causing  their faster collision .  This , then results into an exothermic reaction proceeding violently .

2H_2(g)   +   O_{2(g)   --------->}    2 H₂ O _(g)         

The activation energy barrier to the reaction is indeed very large. The Arrhenius equation allows the rate constant, k, for a reaction to be determined: 

k = Ae^(-Ea/RT) 

where A = Arrhenius pre-exponential factor (depends on the system in question) 
Ea = activation energy barrier in J/mol 
R = 8.314 J/(K·mol) 
T = absolute temperature in K 

If Ea is sufficient large, k can be so close to zero that the reaction essentially will not occur at the temperature

It is because  practically we fail to determine electrode  potential of a half cell consisting of an electrode dipped in its own salt solution. We, can measure only the difference between two half - cell potentials. that gives emf of the cell . If we arbitrarily choose the potential of one of the electrode ( half cell ) then  that of the other can be determined  with respect to this.  Therefore  ,   we need to set up a galvanic cell  consisting of two half cells , and measuring  out its  EMF.  This is possible only when we construct a cell with two half cells  & calculate its  EMF  (  in volts ) using the expression

E^o _cell  =  E^o_R  -  E^o_L 

Hence ,   electrode potential  of  one of the half cells  must be  known .  As per IUPAC norms  Hydrogen electrode  has been chosen  under standard conditions (ie. H₂   with  1 bar  press . ,[ H⁺ ]  =  1 M  &   at 298 K  )  with its arbitrarily fixed electrode potential as   equal to zero.

So , the reduction electrode potentials of all the different electrodes have been determined  and reported as electrode potential  relative to that of  hydrogen electrode .  

An electrode is an electrical conductor that makes contact with the nonmetallic circuit parts of a circuit, such as an electrolyte, semiconductor or vacuum.

A half cell is one of the two electrodes in a galvanic cell or simple battery. For example, in the Zn-Cu battery, the two half cells make an oxidizing-reducing couple. Placing a piece of reactant in an electrolyte solution makes a half cell.

Face centred  cubic  (fcc)   or   Cubic closed pack (ccp) -  these are the two names for the same lattice.

We can thhink of this cell as being made by another atom into each face of a simple cubic lattice -  hence the " face centred cubic "  name .

In such a unit cell there are atoms at all the corners , and at the centre of all the faces of the cube.

Each atom located at the face - centre is shared between two adjacent cells and only half of the atom belongs to a unit cell. Thus ,the number of atoms  in this cell  are ,

(i)    8 corner atoms  x  {tex}\frac{1}{8}{/tex} atom per unit cell   =   8  x  {tex}\frac{1}{8}{/tex} =  1 atom

(ii)   6 atoms  at the faces x{tex}\frac{1}{2}{/tex} atom per unit cell =  6 x {tex}\frac{1}{2}{/tex} =  3 atoms

{tex}\therefore{/tex} Total number of atoms per unit cell                               =   4  atoms

N_{​​​​​​2} is exist as diatomic molecule in which two nitrogen atoms are attached to each other by triple bond this triple bond is due to p(π)-p(π) multiple bonding (sidewise overlapping) and such bonding is not possible in phosphorus due to it's large size but possible in nitrogen due to it's small size.

In case of phosphorus four atoms are associated so it have high molecular mass, large magnitude of venderwall forces so phosphorus is solid and nitrogen is gas.