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Nutka1998 [239]
3 years ago
11

For a particular redox reaction, NO-2 is oxidized to NO-3 and Ag+ is reduced to Ag . Complete and balance the equation for this

reaction in basic solution. The phases are optional. balanced reaction: NO-2 + Ag^{+} -> NO-3+ Ag
Chemistry
1 answer:
Salsk061 [2.6K]3 years ago
3 0

The following are the steps  to complete and balnce the equation for the given reaction

<u>Explanation:</u>

We are given, NO2– is oxidized to NO3– and Ag is reduced to Ag

NO2– + Ag+ -----> NO3– + Ag(s)

Step 1) Assign the oxidation state to each element reaction

NO2– + Ag+ -----> NO3– + Ag(s)

N= +3                           N = +5                        

O = -2                            O = -2

Ag = +1                         Ag = 0

NO2– -----> NO3– ………oxidation half reaction

Ag+ -----> Ag(s) ……….reduction half reaction

Step 2) Balance the element other than O and H

     NO2– -----> NO3–

     Ag+ -----> Ag(s)

Step 3) Balance the O by adding 1 H2O for 1 O

     NO2– + H2O -----> NO3–

     Ag+ -----> Ag(s)

Step 4) Balance the H by adding H+

    NO2– + H2O -----> NO3– + 2H+

     Ag+ -----> Ag(s)

Step 5) Balance the charge by adding electron

    NO2– + H2O -----> NO3– + 2H+ + 2e-

     Ag+ + 1e------> Ag(s)

Step 6) Balance the electron in both half reaction

    NO2– + H2O -----> NO3– + 2H+ + 2e-

     2 Ag+ + 2e------> 2 Ag(s)

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(a) Here, we are going to compare the the strength, length, and order of nitrogen-nitrogen bonds in hydrazine, diazene, and N₂.

Total valence electrons in N₂H₄ = 2 × 5 + 4×1 = 4e⁻

Thus, single N-N bond is present in hydrazine (N₂H₄) molecule. Therefore, the bond order is =1

Total valence electrons in N₂H₄ = 2× 5 + 2 ×1 = 12e⁻

Thus,  one  N=N bond is present in diazene  (N₂H₂) molecule. Therefore, the bond order is =2

Total valence electrons in N₂  = 2 × 5 = 10e⁻

Thus,  one  N-N triple bond is present in nitrogen  (N₂) molecule. Therefore, the bond order is =3.

Therefore, the order of bond order is given below:

N₂ (Bond order =3) > diazene  (N₂H₂, bond order =2) > hydrazine (N₂H₄, bond order=1)

We know that, with the increasing bond order, the strength of the bond  increases.

Therefore, the order of the strength of the bond is given below:

N₂ (Bond order =3) > diazene  (N₂H₂, bond order =2) > hydrazine (N₂H₄, bond order=1)

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Phosphine, an extremely poisonous and highly reactive gas, will react with oxygen to form tetraphosphorus decaoxide and water. P
pantera1 [17]

Answer:

The mass of P₄O₁₀ (s) formed is 471.03 g

Explanation:

4PH₃(g) + 8O₂(g) →  P₄O₁₀ (s)  + 6H₂O (g)

This is the ballanced equation.

Every 4 moles of phosphine, 1 mol of tetraphosphorus decaoxide is generated.

Moles of PH₃ = Mass PH₃ / Molar mass PH₃

Moles PH₃ = 225 g / 33.9 g/m

Moles PH₃ = 6.63 moles

So the rule of three, will be:

If 4 moles of PH₃ generate 1 mol  P₄O₁₀

Then, 6.63 moles of  PH₃ generate (6.63  .1)/4 = 1.66 moles

If we want to know the mass:

Moles . Molar mass = mass

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Which of the following families would have the highest electronegativity
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halogens

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Electronegativity increases as you move to the right and up the periodic table. Thus, halogens would have the greatest electronegativity.

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For the chemical reaction 2HBr(aq)+Ba(OH)2(aq)⟶2H2O(l)+BaBr2(aq) write the net ionic equation, including the phases.
Crazy boy [7]

Answer:

2H⁺(aq) + 2OH⁻(aq)  --> 2H2O(l)

Explanation:

2HBr(aq)+Ba(OH)2(aq)⟶2H2O(l)+BaBr2(aq)

We break the compounds into ions. Only compounds in the aqueous form can be turned into ions.

The ionic equation is given as;

2H⁺(aq)  +  2Br⁻(aq)  + Ba²⁺(aq) + 2OH⁻(aq)   --> 2H2O(l)  +  Ba²⁺(aq)  + 2Br⁻(aq)

Upon eliminating the spectator ions; The net equation is given as;

2H⁺(aq) + 2OH⁻(aq)  --> 2H2O(l)

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