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1 year ago

Nitrogen and hydrogen combine to form ammonia.

Chemistry

1 answer:

zhenek [66]1 year ago

6 0

The balanced chemical equation for the formation ammonia is

N2(g) + 3H2(g) ----> 2NH3(g) .

The balanced chemical equations explains that the same number of each element exist as reactants and products. The coefficients in a balanced equation must be the simplest whole number ratio. Mass is always conserved in chemical reactions.

For the formation of ammonia, the chemical equation is

N2(g) + H2(g) ----> NH3(g)

Balancing the chemical reaction, we can write,

N2(g) + 3H2(g) ----> 2NH3(g) .

This equation shows two nitrogen entering the reaction together and two hydrogens entering the reaction together. Since NH3 is multiplied by a coefficient of 2 there are now 2 nitrogen and 6 hydrogens. The 6 hydrogens come from the 2 multiplied by the subscript of 3. This is the balanced chemical reaction.

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Answer : The balanced chemical equation in a basic solution are,

(A) $2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr(OH)_3+4OH^-+3Cu(OH)_2$

(B) $2NO_2+2OH^-\rightarrow NO_3^-+H_2O+NO_2^-$

(C) $4Zn+7OH^-+NO_3^-+6H_2O\rightarrow 4[Zn(OH)_4]^{2-}+NH_3$

(D) $Br_2+12OH^-+5Br_2\rightarrow 2BrO_3^-+6H_2O+10Br^-$

Explanation :

(A) The given chemical reaction is :

$CrO_4^{2-}(aq)+Cu(s)\rightarrow Cr(OH)_3(s)+Cu(OH)_2(s)$

The oxidation-reduction half reaction will be :

Oxidation : $Cu\rightarrow Cu^{2+}+2e^-$

Reduction : $CrO_4^{2-}+4H_2O+3e^-\rightarrow Cr^{3+}+8OH^-$

In order to balance the electrons, we multiply the oxidation reaction by 3 and reduction reaction by 2 then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr^{3+}+16OH^-+3Cu^{2+}$

or,

$2CrO_4^{2-}+8H_2O+3Cu\rightarrow 2Cr(OH)_3+4OH^-+3Cu(OH)_2$

(B) The given chemical reaction is :

$NO_2(g)\rightarrow NO_3^-(aq)+NO_2^-(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $NO_2+2OH^-\rightarrow NO_3^-+H_2O+1e^-$

Reduction : $NO_2+1e^-\rightarrow NO_2^-$

The electrons in both the reactions are balanced. Now added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$2NO_2+2OH^-\rightarrow NO_3^-+H_2O+NO_2^-$

(C) The given chemical reaction is :

$Zn(s)+NO_3^-(aq)\rightarrow Zn(OH)_4^{2-}(aq)+NH_3(g)$

The oxidation-reduction half reaction will be :

Oxidation : $Zn+4OH^-\rightarrow [Zn(OH)_4]^{2-}+2e^-$

Reduction : $NO_3^-+6H_2O+8e^-\rightarrow NH_3+9OH^-$

In order to balance the electrons, we multiply the oxidation reaction by 4 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$4Zn+7OH^-+NO_3^-+6H_2O\rightarrow 4[Zn(OH)_4]^{2-}+NH_3$

(D) The given chemical reaction is :

$Br_2(l)\rightarrow BrO_3^-(aq)+Br^-(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $Br_2+12OH^-\rightarrow 2BrO_3^-+6H_2O+10e^-$

Reduction : $Br_2+2e^-\rightarrow 2Br^-$

In order to balance the electrons, we multiply the reduction reaction by 5 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$Br_2+12OH^-+5Br_2\rightarrow 2BrO_3^-+6H_2O+10Br^-$

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gogolik [260]

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Explanation:

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A Reductant thus exactly the opposite.

Note that the equation provided shows that Iodine (I2) received an electron to become NEGATIVELY CHARGED:

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