Question

answer the question​

Answer 1

Dronacharya ko Dhristdhum ne chal se mara ..pandwo ne aswasthama mara gya bol kr unka shashtra niche rkhwa diye thae

Answer 2

Answer:

Given:

Concentration of OH- is decreased up to 1/4 time

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Need to find:

Change in concentration of Fe3+

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Answer:

》We know, Fe(OH)3 is a solid.

thus [Fe(OH)3] =1

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Stoichiometric coeff of:

Fe =1

OH= 3

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let concentration be Fe initially

be : Fe1

and finally be Fe2

now,

Kc = \dfrac{{concentration \: of \: product}^{coeff} }{{concentration \: of \: product}^{coeff} }Kc=

concentrationofproduct

coeff

concentrationofproduct

coeff

\implies \: Kc1 = {[Fe1]\: ({OH})^{3} }⟹Kc1=[Fe1](OH)

3

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and

Kc2 = {[Fe2] \times ({\dfrac{1}{4}[ OH]})^{3} }Kc2=[Fe2]×(

4

1

[OH])

3

now

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Kc of a reversible reaction can't be different for same reaction at same temperature,

thus

\begin{gathered}Kc1 = Kc2 \\ \implies \: [Fe1] {[OH]}^{3} =[ Fe2] \dfrac{ {[OH]}^{3} }{64}\end{gathered}

Kc1=Kc2

⟹[Fe1][OH]

3

=[Fe2]

64

[OH]

3

\red{\bold{\boxed{\large{ \implies \: [Fe2] = 64[Fe1]}}}}

⟹[Fe2]=64[Fe1]

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Thus The new concentration of Fe will be 64times of initial concentration.