Chemistry, asked by nehasonawane8404, 3 months ago

D/d=[1-x(y-1)]
plzzz....tell solution
& this eq^n answer is x=D-d/d(y-1)
plzz tell na how to explain this eq^n
plzzz.....this is from chemistry... how to explain this

Answers

Answered by anchitsingh40
1

Answer:

Given,

(x

2

D

2

+3xD+1)y=

(1−x)

2

1

......(1)

Let, x=e

z

,z=lnx,x>0,

dz

d

≡θ

⇒xD≡θ,x

2

D

2

≡θ(θ−1)

substituting these in equation (1)

(θ(θ−1)+3θ+1)y=

(1−e

z

)

2

1

⇒(θ

2

+2θ+1)y=

(1−e

z

)

2

1

....(2)

The general solution is given by, y=y

c

+y

p

y

c

=c

1

e

−z

+c

2

ze

−z

y

p

=

θ

2

+2θ+1

1

(1−e

z

)

2

1

=

θ+1

1

{

θ+1

1

(1−e

z

)

2

1

}

=

θ+1

1

{e

−z

(1−e

z

)

2

1

e

z

dz}

=

θ+1

1

{

1−e

z

e

−z

}

=e

−z

1−e

z

e

−z

e

z

dz

=−e

−z

ln∣e

−z

−1∣

y

p

=−e

−z

ln∣e

−z

−1∣

∴y=c

1

e

−z

+c

2

ze

−z

−e

−z

ln∣e

−z

−1∣

⇒y=c

1

x

−1

+c

2

(lnx)x

−1

−x

−1

ln∣x

−1

−1∣

Explanation:

please mark me as brainliest

Answered by sohanveers245
1

Answer:

α=(n−1)d(D−d)→(1)

α=Degree of dissociation

d=Vapor density

n= no of moles of gaseous product for 1 mole of reactant

  Any equation in which the total number of moles on product side =n and on reactant side =1.

A) A⇌nB/2+nC/3

  here at reactant side n=1 

  But product side n=2n+3n=65n=n

B)A⇌nB/3+(2n/3)C

     here at reactant side n=1

     product side n=2n+32n=66n=n

C)A→(n/2)B+(n/4)C

   here at reactant side n=1 

    But product side n=2n+n=42n=n

D)A⇌(n/2)B+C

   here at reactant side n=1 

   But product side n=

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