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Scorpion4ik [409]
2 years ago
10

A gaseous mixture of 10.0 volumes of CO₂, 1.00 volume of unreacted O₂, and 50.0 volumes of unreacted N₂ leaves an engineat 4.0 a

tm and 800. K. Assuming that the mixture reaches equilibrium, what are (b) the concentration (in picograms per liter, pg/L) of CO in this exhaust gas?2CO₂(g) ⇄ 2CO(g) + O₂(g) Kp = 1.4×10⁻²⁸ at 800K(The actual concentration of CO in exhaust gas is much higher because the gases do not reach equilibrium in the short transit time through the engine and exhaust system.)
Chemistry
1 answer:
ycow [4]2 years ago
3 0

The masses of CO and CO2​ are 90.55g and 100−90.55=9.45 g respectively.

<h3>Total mass.</h3>

Let the mixture has 100g as total mass.

The number of moles of CO is 2890.55​=3.234.

The number of moles of CO2​ is 449.45​=0.215.

The mole fraction of CO is 3.234+0.2153.234​=0.938.

The mole fraction of CO2​ is 1−0.938=0.062.

The partial pressure of CO is the product of the mole fraction of CO and the total pressure.

It is 0.938×1=0.938 atm.

The partial pressure of carbon dioxide is 0.062×1=0.042 atm.

The expression for the equilibrium constant is:

Kp​=PCO2​​PCO2​​=0.062(0.938)2​=14.19

Δng​=2−1=1

Kc​=Kp​(RT)−Δn=14.19×(0.0821×1127)−1=0.153.

To learn more about equilibrium constant visit the link

brainly.com/question/15118952

#SPJ4

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A

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The answer A is the best answer because it contains the most general characteristic of a chemical change.

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This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): An analytical chemist has det
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Answer:

0.054 mol O

Explanation:

<em>This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): CH₃CO₂H. An analytical chemist has determined by measurements that there are 0.054 moles of carbon in a sample of acetic acid. How many moles of oxygen are in the sample?</em>

<em />

Step 1: Given data

  • Chemical formula of acetic acid: CH₃CO₂H
  • Moles of carbon in the sample: 0.054 moles

Step 2: Establish the appropriate molar ratio

According to the chemical formula, the molar ratio of C to O is 2:2.

Step 3: Calculate the moles of oxygen in the sample

We will use the molar ratio to determine the moles of oxygen accompanying 0.054 moles of carbon.

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6 0
3 years ago
Calculate the pH and fraction of dissociation ( α ) for each of the acetic acid ( CH 3 COOH , p K a = 4.756 ) solutions. A 0.002
marysya [2.9K]

Answer:

The degree of dissociation of acetic acid is 0.08448.

The pH of the solution is 3.72.

Explanation:

The pK_a=4.756

The value of the dissociation constant = K_a

pK_a=-\log[K_a]

K_a=10^{-4.756}=1.754\times 10^{-5}

Initial concentration of the acetic acid = [HAc] =c = 0.00225

Degree of dissociation = α

HAc\rightleftharpoons H^++Ac^-

Initially

c

At equilibrium ;

(c-cα)                                cα        cα

The expression of dissociation constant is given as:

K_a=\frac{[H^+][Ac^-]}{[HAc]}

1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha}{(c-c\alpha)}

1.754\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha)}

1.754\times 10^{-5}=\frac{0.00225 \alpha ^2}{(1-\alpha)}

Solving for α:

α = 0.08448

The degree of dissociation of acetic acid is 0.08448.

[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M

The pH of the solution ;

pH=-\log[H^+]

=-\log[0.0001901 M]=3.72

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