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marissa [1.9K]
3 years ago
12

Nitrogen monoxide, NO, is formed in automobile exhaust by the reaction of nitrogen and oxygen in the air: N2 (g) + O2 (g) ↔ NO (

g) BALANCE THE EQUATION The equilibrium constant Kc is 0.0025 at 2127 oC. If an equilibrium mixture at this temperature contains 0.023 M N2 and 0.031 M O2, what is the equilibrium concentration of NO? __________ M NO
Chemistry
1 answer:
Solnce55 [7]3 years ago
6 0

Answer:

The equilibrium concentration of NO is 0.001335 M

Explanation:

Step 1: Data given

The equilibrium constant Kc is 0.0025 at 2127 °C

An equilibrium mixture contains 0.023M N2 and 0.031 M O2,

Step 2: The balanced equation

N2(g) + O2(g) ↔ 2NO(g)

Step 3:  Concentration at the equilibrium

[N2] = 0.023 M

[O2] = 0.031 M

Kc = 0.0025 = [NO]² / [N2][O2]

Kc = 0.0025 = [NO]² / (0.023)(0.031)

[NO] = 0.001335 M

The equilibrium concentration of NO is 0.001335 M

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allochka39001 [22]

Answer:

30, 50

Explanation:

Hello,

In this since an element's atomic number is equal to the number of protons in its atom, we can infer that selenium's atomic number is 30. Moreover, due to the fact the the neutrons are equal to the atomic mass minus the atomic number or the number of protons, by knowing the number of neutrons we compute the atomic as follows:

neutrons=mass-protons\\\\mass=neutrons+protons\\\\mass=30+20\\\\mass=50a.m.u

Thus, answer is 30, 50.

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8 0
3 years ago
The conversion of cyclopropane to propene occurs with a first-order rate constant of . How long will it take for the concentrati
Gennadij [26K]

This is an incomplete question, here is a complete question.

The conversion of cyclopropane to propene occurs with a first-order rate constant of 2.42 × 10⁻² hr⁻¹. How long will it take for the concentration of cyclopropane to decrease from an initial concentration 0.080 mol/L to 0.053 mol/L?

Answer : The time taken will be, 17.0 hr

Explanation :

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 2.42\times 10^{-2}\text{ hr}^{-1}

t = time passed by the sample  = ?

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a - x = concentration left = 0.053 M

Now put all the given values in above equation, we get

t=\frac{2.303}{2.42\times 10^{-2}}\log\frac{0.080}{0.053}

t=17.0\text{ hr}

Therefore, the time taken will be, 17.0 hr

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