Balance the following skeleton reaction and identify the oxidizing and reducing agents: Fill in all blanks with numbers so if the term is not in the equation make it 0. CrO42- (aq) + N2O(g)+ H+(aq) + OH-(aq) + H2O(l) LaTeX: \longrightarrow⟶ Cr3+(aq) + NO(g) + H+(aq) + OH-(aq) + H2O(l) oxidizing agent is (enter just the formula of the species, e. g. CrO42-,N2O, Cr3+,NO, H2O, H+, OH-) reducing agent is (enter just the formula of the species, e. g. CrO42-,N2O, Cr3+,NO, H2O, H+, OH-)

Reducing agent: nitrogen.

Oxidizing agent: chromium.

Explanation:

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In this case, the first step is to rewrite the undergoing chemical reaction indicating the oxidation state for each atom as shown below:

In such a way, we observe the following half-reactions:

* Reduction as chromium undergoes a decrease in its oxidation state:

* Oxidation as nitrogen undergoes an increase in its oxidation state:

In such a way, we state that the nitrogen is the reducing agent as it is oxidized and chromium the oxidizing agent as it is reduced, therefore, the balanced chemical reaction turns out:

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the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.

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