HNO3 + S --> H2SO4 + NO Now identify the element oxidized and the element reduced. Which element is oxidized? Which element is reduced? Which elements undergo no oxidation or reduction?

S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

Explanation:

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when oxidation number of a species decreases.

For the given chemical reaction:

On the reactant side:

Oxidation number of H = +1

Oxidation number of N = +5

Oxidation number of O = -2

Oxidation number of S = 0

On the product side:

Oxidation number of H = +1

Oxidation number of N = +2

Oxidation number of O = -2

Oxidation number of S = +6

As the oxidation number of S is increasing from 0 to +6. Thus, it is getting oxidized. Similarly, the oxidation number of N is decreasing from +5 to +2. Thus, it is getting reduced.

The oxidation numbers of O and H remain the same on both sides of the reaction. Thus, they are neither getting oxidized or reduced.

Hence, S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

a:

by the combination of an unlimited number of elements

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24