How do you balance the equation NO2 + H2 →-> NH3 +H2O using the oxidation number method?

Chemistry

2 answers:

Ksivusya [100]3 years ago

6 0

$NO_2 + H_2 \to NH_3 + H_2 O$

<u /> $2NO_2 + 7H_2 \to 2NH_3 + 4H_2 O$

taurus [48]3 years ago

4 0

NO2 + H2 ----> NH3 + H2O

N(+IV) + 7e -----> N(-III) |2
H2(0) - 2e ----> 2H(+I) |7

after: 2NO2 + 7H2 ----> 2NH3 + 4H2O

You might be interested in

9. What are the advantages of using an indicator to inform pH measurements? What are the advantages of using a pH meter?

Llana [10]

Answer:

The advantages of using an indicator to inform pH measurements:

It gives a mathematically result of the pH, in addition, it gives the precise pH of solvent, and it also gives an idea of the straight of the solution also.

Now, the advantage of using a pH meter:

It is a rapid method to characterize between acids, bases. However, this method does not show how strong acid or base actually are, plus it tends to gives a range of acidity or basicity not quite accurate as a result.

3 0

3 years ago

In this picture below one object is attractive to another by-

Ne4ueva [31]

I think it'd be gravity.

5 0

3 years ago

Read 2 more answers

EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?

AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

$\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq)$ .

The equilibrium constant for this reaction is called the self-ionization constant of water:

$K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]$ .

Note that water isn't part of this constant.

The value of $K_w$ at 25 °C is $10^{-14}$ . How to memorize this value?

The pH of pure water at 25 °C is 7.
$[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$
However, $[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$ for pure water.
As a result, $K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}$ at 25 °C.