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balu736 [363]
2 years ago
5

I need help with this

Chemistry
2 answers:
cricket20 [7]2 years ago
7 0

Answer:

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Copy and paste to help Bob.

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Copy and paste to help Bob.

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Copy and paste to help Bob.

We can do it together!

Explanation:

Vedmedyk [2.9K]2 years ago
5 0

Answer:

i think its 63 pls do not get mad if not 63

Explanation:

from 60 you add three?(:

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\\ \sf\longmapsto 1mole\:is\:a\:specific\:amount\:of\:particles\:of\:a\:substance .

\\ \sf\longmapsto It\:is\:Represented\:as\:mol

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In the coal-gasification process, carbon monoxide is converted to carbon dioxide via the following reaction: CO (g) + H2O (g) ⇌
Oksana_A [137]

Answer: Equilibrium constant is 0.70.

Explanation:

Initial moles of  CO = 0.35 mole

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Initial moles of  H_2O = 0.40 mole

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equilibrium concentration of CO=\frac{moles}{volume}=\frac{0.18moles}{1L}=0.18M [/tex]

The given balanced equilibrium reaction is,

                            CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

Initial conc.            0.35 M       0.40M       0     0

At eqm. conc.    (0.35-x) M   (0.40-x) M   (x) M    (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CO_2]\times [H_2O]}{[CO]\times [H_2O]}

K_c=\frac{x\times x}{(0.40-x)(0.35-x)}

we are given : (0.35-x)= 0.18

x = 0.17

Now put all the given values in this expression, we get :

K_c=\frac{0.17\times 0.17}{(0.40-0.17)(0.35-0.17)}

K_c=0.70

Thus the value of the equilibrium constant is 0.70.

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