Chemistry, asked by nishantbhadauria, 1 year ago

Oxidation no of Ni in Ni(CO)4

Answers

Answered by Anonymous
0
the oxidation state of Ni is 0
Answered by MrVampire01
1

Answer:

GIVEN EQUATION IS

\bold{\boxed{\boxed{ ❲\frac{4x - 3}{2x + 1}❳ - 10 (\frac{2x + 1}{4x - 3} ) = 3}}}

❲ 2x+14x−3 ❳−10( 4x−3 2x+1)=3

\bold{( \frac{4x - 3}{2x + 1} ) - 10 (\frac{2x + 1}{4x - 3} ) = 3}</p><p>⟹( 2x+14x−3)−10( 4x−32x+1 )=3

\bold{\frac{ {(4x - 3)}^{2} - 10 {(2x + 1)}^{2} }{(2x + 1)(4x - 3)} = 3}⟹ (2x+1)(4x−3)(4x−3) 2−10(2x+1) 2 =3

\bold⟹(16 {x}^{2} - 24x + 9) - 10(4 {x^{2} + 4x + 1)}

⟹(16x 2 −24x+9)−10(4x2 +4x+1)

\bold{= 3(8 {x}^{2} - 6x + 4x - 3)}=3(8x </p><p>2−6x+4x−3)

\bold{16 {x}^{2} - 24x + 9 - 40 {x}^{2} - 40x - 10}16x 2 −24x+9−40x 2−40x−10

\bold{ = 24 {x}^{2} - 18x + 12x - 9}=24x </p><p>2−18x+12x−9

\bold{⟹- 24 {x}^{2} - 64x - 1 = 24 {x}^{2} - 6x - 9}⟹−24x 2 −64x−1=24x 2 −6x−9

⟹\bold{- 24 {x}^{2} - 24 {x}^{2} - 64x + 6x - 1 + 9 = 0}⟹−24x 2−24x 2−64x+6x−1+9=0

⟹\bold{- 48 {x}^{2} - 58x + 8 = 0}

⟹−48x 2−58x+8=0

⟹\bold{24 {x}^{2} + 29x - 4 = 0}⟹24x </p><p>2+29x−4=0

⟹\bold{24 {x}^{2} + 32x - 3x - 4 = 0}

⟹24x 2 +32x−3x−4=0

⟹\bold{8x(3x + 4) - 1(3x + 4) = 0}

⟹8x(3x+4)−1(3x+4)=0

⟹\bold{(3x + 4)(8x - 1) = 0}

⟹(3x+4)(8x−1)=0

⟹\bold{3x + 4 = 0}⟹3x+4=0

⟹\bold{8x - 1 = 0}⟹8x−1=0

\bold{\boxed{\red{x = - \frac{4}{3} }}}

x=−3/4

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