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Alex Ar [27]
3 years ago
13

Nitrogen and oxygen do not react appreciably at room temperature, as illustrated by our atmosphere. But at high temperatures, th

e reaction below can proceed to a measurable extent. N2(g) + O2(g) ⇔ 2 NO(g) At 3000 K, the reaction above has Keq = 0.0153. If 0.3152 mol of pure NO is injected into an evacuated 2.0-L container and heated to 3000K, what will be the equilibrium concentration of NO?
Chemistry
1 answer:
Gre4nikov [31]3 years ago
5 0

Answer : The equilibrium concentration of NO is, 0.0092 M.

Solution :

First we have to calculate the concentration of NO.

\text{Concentration of NO}=\frac{\text{Moles of }NO}{\text{Volume of solution}}=\frac{0.3152mol}{2.0L}=0.1576M

The given equilibrium reaction is,

                           N_2(g)+O_2(g)\rightleftharpoons 2NO(g)

Initially conc.      0        0           0.1576

At eqm.               (x)       (x)        (0.1576-2x)

The expression of K_c will be,

K_c=\frac{[NO]^2}{[N_2][O_2]}

0.0153=\frac{(0.1576-2x)^2}{(x)\times (x)}

By solving the term, we get:

x=0.0742,0.0839

Neglecting the 0.0839 value of x because it can not be more than initial value.

Thus, the value of 'x' will be, 0.0742 M

Now we have to calculate the equilibrium concentration of NO.

Equilibrium concentration of NO = (0.1576-2x) = [0.1576-2(0.0742)] = 0.0092 M

Therefore, the equilibrium concentration of NO is, 0.0092 M.

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What is the % of each element in Al2(CrO4)3
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Please give brainliest.

7 0
3 years ago
Select the correct answer.
Alex

no:A

is a correct answer

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A 275 g sample of a metal requires 10.75 kJ to change its temperature from 21.2 oC to its melting temperature, 327.5 oC. What is
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Answer:

c=0.127\ J/g^{\circ} C

Explanation:

Given that,

Mass of the sample, m = 275 g

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c is specific heat of the metal

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So, the specific heat of metal is 0.127\ J/g^{\circ} C.

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