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Tems11 [23]
3 years ago
15

Generally, A loses electrons when reacting with nonmetals and gains electrons when reacting with metals. A is a(n):

Chemistry
1 answer:
lukranit [14]3 years ago
4 0
Sorry I don’t exactly know I have exams rn I need points
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Carbon monoxide replaces oxygen in oxygenated hemoglobin according to the reaction: HbO2(aq) + CO(aq) ∆ HbCO(aq) + O2(aq) a. Use
gayaneshka [121]

The question is incomplete, complete question is;

Carbon monoxide replaces oxygen in oxygenated hemoglobin according to the reaction:

HbO_2(aq) + CO(aq)\rightleftharpoons HbCO(aq) + O_2(aq)

Use the reactions and associated equilibrium constants at body temperature to find the equilibrium constant for the above reaction.

Hb(aq) + O_2(aq)\rightleftharpoons HbO_2(aq) ,K_1=1.8

Hb(aq) + CO(aq)\rightleftharpoons HbCO(aq) ,K_2=306

Answer:

The equilibrium constant for the given reaction is 170.

Explanation:

Hb(aq) + O_2(aq)\rightleftharpoons HbO_2(aq) ,K_1=1.8

K_1=\frac{[HbO_2]}{[Hb][O_2]}

[Hb]=\frac{[HbO_2]}{[K_1][O_2]}..[1]

Hb(aq) + CO(aq)\rightleftharpoons HbCO(aq) ,K_2=306

K_2=\frac{[HbCO]}{[Hb][CO]}..[2]

HbO_2(aq) + CO(aq)\rightleftharpoons HbCO(aq) + O_2(aq)

K_c=\frac{[HbCO][O_2]}{[HbO_2][CO]}..[3]

Using [1] in [2]:

K_2=\frac{[HbCO]}{\frac{[HbO_2]}{[K_1][O_2]}\times [CO]}

K_2=K_1\times \frac{[HbCO][O_2]}{[HbO_2][CO]}

K_2=K_1\times K_c ( using [3])

306=1.8\times K_c

K_c=\frac{306}{1.8}=170

The equilibrium constant for the given reaction is 170.

7 0
3 years ago
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