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For a reaction, A(x),B(y)hArrxA^(y+)+yB...

For a reaction, `A_(x),B_(y)hArrxA^(y+)+yB^(x-),K_(sp)` xan be represented as

A

`[A^(y+)]^(x)[B^(x-)]^(y)`

B

`[A]^(y)[B]^(x)`

C

`[A]^(x)[B]^(y)`

D

`[A]^(x+y)[B]^(x-y)`

Text Solution

Verified by Experts

The correct Answer is:
A

`A_(x)B_(y)hArrxA^(y+)+yB^(x-)`
`K_(sp)=[A^(y+)]^(x)[B^(x-)]^(y)`
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Knowledge Check

  • In a compound A_(x)B_(y)

    A
    moles of A = moles of B = moles of `A_(x)B_(y)`
    B
    eq.of A = eq of B = eq.of `A_(x)B_(y)`
    C
    y `xx ` moles of A = y `xx` moles of B = `(x+y) xx` moles of `A_(x)B_(y)`
    D
    y `xx` moles of A = y `xx` moles of B
  • In a compound A_(x)B_(y) :

    A
    Mole of `A= "Mole of" B = "Mole of" A_(x)B_(y)`
    B
    Eq. of `A=Eq. "of" B=Eq. "of" A_(x)B_(y)`
    C
    `yxx"mole of" A=yxx"mole of" B=(x+y)xx"mole of" A_(x)B_(y)`
    D
    `yxx"mole of" A= yxx"mole of"B`
  • In a compound A_(X)B_(Y) :-

    A
    `"Mole of "A = "mole of " B" eq of " B="mole of "A_(x)B_(Y)`
    B
    eq of A = eq of B=eq of `A_(X)B_(Y)`
    C
    yx mole of A = yx mole of B = (x+y) mole of `A_(x)B_(y)`
    D
    yx mole of A = yx mole of B
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