Please I need help with this exercise. Does anyone know the solution to it and please with explanations: Ag(s) + NO3–(aq) + H+(aq) → Ag+(aq) + NO(g) + H2O(l)
When the oxidation-reduction equation above is balanced, what is the coefficient for H+(aq)?
A.1
B.2
C.3
D.4

Option D. 4

Explanation:

Ag(s) + NO3–(aq) + H+(aq) → Ag+(aq) + NO(g) + H2O(l)

To know the coefficient of H+, let us balance the equation. This is can be done as shown below:

Ag(s) + NO3–(aq) + 2H+(aq) → Ag+(aq) + NO(g) + H2O(l)

There are 3 atoms of O on the left side and a total of 2 atoms on the right side. It can be balance by putting 2 in front of H2O as shown below:

Ag(s) + NO3–(aq) + H+(aq) → Ag+(aq) + NO(g) + 2H2O(l)

There are 4 atoms of H on the right side and 1 atom on the left side. It can be balance by putting 4 in front of H+ as shown below:

Ag(s) + NO3–(aq) + 4H+(aq) → Ag+(aq) + NO(g) + 2H2O(l)

Total charge on the left side:

– 1 + 4 = +3

Total charge on the right side:

+1

To balance the charge, we simply put 3 in front of Ag and Ag+ as shown below:

3Ag(s) + NO3–(aq) + 4H+(aq) → 3Ag+(aq) + NO(g) + 2H2O(l)

Now the equation is balanced.

The coefficient of H+ is 4.

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying