What is the oxidation state of an individual nitrogen atom in nh4oh ? express the oxidation state numerically (e.g., +1)?

1 answer:

6 0

In NH4OH, the compounds that make it up are NH4+ and OH-
Therefore N exists in the ammonium form.
In the ammonium ion 4H atoms are connected to N.
N is more electronegative than H, therefore when H bonds to N, H is the more positive atom therefore each H has a charge of +1, since there are 4 H atoms the charge contributed by the 4H atoms are +1 * 4 = +4
the overall charge of NH4 is +1
Charge of N (+) +4 = +1
Charge of N = +1 - 4
Therefore oxidation state of N in NH4 is = -3

You might be interested in

Nitrogen dioxide and water react to form nitric acid and nitrogen monoxide, like this:

posledela

Answer: The value of the equilibrium constant Kc for this reaction is 3.72

Explanation:

Equilibrium concentration of $HNO_3$ = $\frac{15.5g}{63g/mol\times 9.5L}=0.026M$

Equilibrium concentration of $NO$ = $\frac{16.6g}{30g/mol\times 9.5L}=0.058M$

Equilibrium concentration of $NO_2$ = $\frac{22.5g}{46g/mol\times 9.5L}=0.051M$

Equilibrium concentration of $H_2O$ = $\frac{189.0g}{18g/mol\times 9.5L}=1.10M$

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$