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Bess [88]
3 years ago
9

Somebody answer this please picture is attached

Chemistry
2 answers:
Nonamiya [84]3 years ago
7 0

I swear i think it is 3.22

Pavlova-9 [17]3 years ago
6 0
The answer is 3.22 because 3.22 is still a sig fig so it would have to be 3.22
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Nitrogen dioxide and water react to form nitric acid and nitrogen monoxide, like this:
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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  

For the given chemical reaction:

2HNO_3(aq)+NO(g)\rightarrow 3NO_2(g)+H_2O(l)

The expression for K_c is written as:

K_c=\frac{[NO_2]^3\times [H_2O]^1}{[HNO_3]^2\times [NO]^1}

K_c=\frac{(0.051)^3\times (1.10)^1}{(0.026)^2\times (0.058)^1}

K_c=3.72

Thus  the value of the equilibrium constant Kc for this reaction is 3.72

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