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tia_tia [17]
3 years ago
5

A mixture containing initially 3.90 mole of NO(g) and 0.88 mole of CO2(g) was allowed to react in a flask of volume 1.00 L at a

certain temperature according to the reaction CO2(g) + NO(g) ↔ CO(g) + NO2(g) At equilibrium 0.11 mole of CO2(g) was found present in the reaction mixture. Calculate the equilibrium constant Kc for the reaction at the temperature of the experiment.
Chemistry
1 answer:
Ymorist [56]3 years ago
6 0

Answer:

Equilibrium constant Kc for the reaction will be 1.722

Explanation:

O2(g)+NO(g)→CO(g)+ NO2(g)

0.88    3.9        ---         ---

0.88x   3.9-x     x           x

GIVEN:

0.88X-X= 0.11

⇒ X=0.77

CO2(g)+NO(g) → CO(g) + NO2(g)

0.88       3.9         ---          ---

0.88-x    3.9-x      x            x

=              3.13       0.77      0.77

=0.11

Kc = \frac{[CO] *[NO2]} {[CO2]*[NO]}

     ={{0.77}×0.77÷{{0.11×3.13}}

     = 1.722

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Answer:

  • 602 mg of CO₂ and 94.8 mg of H₂O

Explanation:

The<em> yield</em> is measured by the amount of each product produced by the reaction.

The chemical formula of <em>fluorene</em> is C₁₃H₁₀, and its molar mass is 166.223 g/mol.

The <em>oxidation</em>, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

        2C_{13}H_{10}+31O_2\rightarrow 26CO_2+10H_2O

To calculate the yield follow these steps:

<u>1. Mole ratio</u>

          2molC_{13}H_{10}:31molO_2:26molCO_2:10molH_2O

<u />

<u>2. Convert 175mg of fluorene to number of moles</u>

  • 175mg/times 1g/1,000mg=0.175g

  • Number of moles = mass in grams / molar mass

  • \text{number of moles}=0.175g/166.223g/mol=0.0010528mol

<u>3. Set a proportion for each product of the reaction</u>

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i) number of moles

         2molC_{10}H_{13}/26molCO_2=0.0010528molC_{10}H{13}/x

x=0.0010528molC_{10}H_{13}\times 26molCO_2/2molC_{10}H_{13}=0.013686molCO_2

ii) mass in grams

The molar mass of CO₂ is 44.01g/mol

  • mass = number of moles × molar mass
  • mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg

b) <u>For H₂O</u>

i) number of moles

0.0010528molC_{10}H_{13}\times10molH_2O/2molC_{10}H_{13}=0.00526molH_2O

ii) mass in grams

The molar mass of H₂O is 18.015g/mol

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  • mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg
4 0
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s344n2d4d5 [400]
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3 0
3 years ago
If the half-life of a radioisotope is 10,000 years, the amount remaining after 20,000 years is
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Answer:

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t=10,000 years____________x/2

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During the first 10,000 years the radioisotope lost half of its mass. After 10,000 years more (which means 2 half-lives), the remaining amount also lost half of its mass. Therefore, after 20,000 years, the we will have a fourth of the initial amount.

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Answer:

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Explanation:

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3 0
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