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Luda [366]
3 years ago
7

fills a 500.mL flask with 3.6atm of carbon monoxide gas and 1.2atm of water vapor. When the mixture has come to equilibrium she

determines that it contains 3.36atm of carbon monoxide gas, 0.96atm of water vapor and 0.24atm of hydrogen gas. The engineer then adds another 1.8atm of carbon monoxide, and allows the mixture to come to equilibrium again. Calculate the pressure of carbon dioxide after equilibrium is reached the second time. Round your answer to 2 significant digits.
Chemistry
1 answer:
enot [183]3 years ago
7 0

Answer:

The answer to the question is

The pressure of carbon dioxide after equilibrium is reached the second time is 0.27 atm rounded to 2 significant digits

Explanation:

To solve the question, we note that the mole ratio of the constituent is proportional to their partial pressure

At the first trial the mixture contains

3.6 atm CO

1.2 atm H₂O (g)

Total pressure = 3.6+1.2= 4.8 atm

which gives

3.36 atm CO

0.96 atm H₂O (g)

0.24 atm H₂ (g)

That is

CO+H₂O→CO(g)+H₂ (g)

therefore the mixture contained

0.24 atm CO₂ and the total pressure =

3.36+0.96+0.24+0.24 = 4.8 atm

when an extra 1.8 atm of CO is added we get Increase in the mole fraction of CO we have one mole of CO produces one mole of H₂

At equilibrium we have 0.24*0.24/(3.36*0.96) = 0.017857

adding 1.8 atm CO gives 4.46 atm hence we have

 (0.24+x)(0.24+x)/(4.46-x)(0.96-x) = 0.017857

which gives x = 0.031 atm or x = -0.6183 atm

Dealing with only the positive values we have the pressure of carbon dioxide = 0.24+0.03 = 0.27 atm

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2 H2O2(aq) ----> 2 H2O(l) + O2(g) in the presence of I-(aq) is proposed to be: Step 1 (slow): H2O2 + I- -----> H2O + OI- S
andrezito [222]

Answer:

Molecularity of the rate determining step = 2

Explanation:

Step 1 (slow): H₂O₂ + I⁻ -----> H₂O + OI⁻

Step 2 (fast): H₂O₂ + OI⁻ -----> H₂O + O₂ + I⁻

The rate determining step in a reaction mechanism is also considered as slowest step.

Slowest step is also considered its highest activation energy in energy profile diagram.

In this case intermediate  (IO⁻) is formed.

Step 1 considered as a slowest step.

So,  Rate = K [H₂O₂][I⁻]

  Molecularity = 2

6 0
2 years ago
For the reaction 2NH3(g) + 2O2(g)N2O(g) + 3H2O(l) H° = -683.1 kJ and S° = -365.6 J/K The standard free energy change for the rea
BlackZzzverrR [31]

Answer:

\Delta G^{0} = -457.9 kJ and reaction is product favored.

Explanation:

The given reaction is associated with 2 moles of NH_{3}

Standard free energy change of the reaction (\Delta G^{0}) is given as:

           \Delta G^{0}=\Delta H^{0}-T\Delta S^{0}   , where T represents temperature in kelvin scale

So, \Delta G^{0}=(-683.1\times 10^{3})J-(273K\times -365.6J/K)=-583291.2J

So, for the reaction of 1.57 moles of NH_{3}, \Delta G^{0}=(\frac{1.57}{2})\times -583291.2J=-457883.592J=-457.9kJ

As, \Delta G^{0} is negative therefore reaction is product favored under standard condition.

6 0
3 years ago
How do you balance this Chemical Equation? <br>C5H12O + O2 = CO2 + H2O​
ICE Princess25 [194]

Answer:

The chemical equation by putting, a 2 on C₅H₁₂O, 15 on O₂, 10 on CO₂ , and 12 on H₂O in the equation;

2C₅H₁₂O + 15O₂ → 10CO₂ + 12H₂O​

Explanation:

  • Chemical equations are balanced by putting coefficients on the reactants and products to ensure the total number of atoms on the left side equal to those on the right side.
  • Balancing chemical equations is done to make chemical equations obey the law of conservation of mass.
  • According to the law of conservation of mass, the mass of the reactants should always be equal to the mass of products.
  • This is done by balancing chemical equations to ensure the total number of atoms on the left side is equal to that on the right side.
  • Therefore, the balanced equation is;

          2C₅H₁₂O + 15O₂ → 10CO₂ + 12H₂O​

3 0
3 years ago
Aqueous solutions of isopropyl alcohol are commonly sold as rubbing alcohol. The boiling point of isopropyl alcohol is 82.4 °C.
Monica [59]

Answer:

This is due to more hydrogen bonding in ethylene glycol than it is in isopropyl alcohol

Explanation:

The boiling point of isopropyl alcohol is 82.4 °C it contains only a single OH group, hence intermolecular hydrogen bonding is solely responsible for it's boiling point, whereas Ethylene glycol (CH2OHCH2OH) contains 2-OH group and both intermolecular and intramolecular hydrogen bonding are responsible for the higher boiling point of ethylene glycol at 198 °C.

8 0
3 years ago
0.00121 mol to grams
Gwar [14]

Answer: I believe the answer you are looking for is 0.02180

7 0
3 years ago
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