ater and in nitric acid. It therefore dissolves
in the undecomposed acid, and imparts a yellowish or reddish color to
it. Concentrated nitric acid highly charged with this substance is
called _fuming nitric acid_.

3. _Oxidizing action._ According to its formula, nitric acid contains a
large percentage of oxygen, and the reaction just mentioned shows that
the compound is not a very stable one, easily undergoing decomposition.
These properties should make it a good oxidizing agent, and we find that
this is the case. Under ordinary circumstances, when acting as an
oxidizing agent, it is decomposed according to the equation

2HNO_{3} = H_{2}O + 2NO + 3O.

The oxygen is taken up by the substance oxidized, and not set free, as
is indicated in the equation. Thus, if carbon is oxidized by nitric
acid, the oxygen combines with carbon, forming carbon dioxide (CO_{2}):

C + 2O = CO_{2}.

4. _Action on metals._ We have seen that when an acid acts upon a metal
hydrogen is set free. Accordingly, when nitric acid acts upon a metal,
such as copper, we should expect the reaction to take place which is
expressed in the equation

Cu + 2HNO_{3} = Cu(NO_{3})_{2} + 2H.

This reaction does take place, but the hydrogen set free is immediately
oxidized to water by another portion of the nitric acid according to the
equation

HNO_{3} + 3H = 2H_{2}O + NO.

As these two equations are written, two atoms of hydrogen are given off
in the first equation, while three are used up in the second. In order
that the hydrogen may be equal in the two equations, we must multiply
the first by 3 and the second by 2. We shall then have

3Cu + 6HNO_{3} = 3Cu(NO_{3})_{2} + 6H,

2HNO_{3} + 6H = 4H_{2}O + 2NO.

The two equations may now be combined into one by adding the quantities
on each side of the equality sign, canceling the hydrogen which is given
off in the one reaction and used up in the other. We shall then have the
equation

3Cu + 8HNO_{3} = 3Cu(NO_{3})_{2} + 2NO + 4H_{2}O.

A number of other reactions may take place when nitric acid acts upon
metals, resulting in the formation of other oxides of nitrogen, free
nitrogen, or even ammonia. The reaction just given is, however, the
usual one.

~Importance of steps in a reaction.~ This complete equation has
the advantage of making it possible to calculate very easily
the proportions in which the various substances enter into the
reaction or are