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hram777 [196]
3 years ago
15

A sample of nitrogen occupies 10.0 liters at 25°c and 98.7 kpa. What would be the volume at 20°c and 102.7 kpa?

Chemistry
1 answer:
Fynjy0 [20]3 years ago
5 0

Answer:

  •    9.4 liter

Explanation:

<u>1) Data:</u>

  •  V₁ = 10.0 L
  •  T₁ = 25°C = 25 + 273.15 K = 298.15 K
  •   P₁ = 98.7 Kpa
  •   T₂ = 20°C = 20 + 273.15 K = 293.15 K
  •    P₂ = 102.7 KPa
  •    V₂ = ?

<u>2) Formula:</u>

Used combined law of gases:

  •    PV / T = constant

  •    P₁V₁ / T₁ = P₂V₂ / T₂

<u>3) Solution</u>:

Solve the equation for V₂:

  •    V₂ = P₁V₁ T₂ / (P₂ T₁)

Substitute and compuite:

  • V₂ = P₁V₁ T₂ / (P₂ T₁)

  • V₂ = 98.7 KPa × 10.0 L × 293.15 K / (102.7 KPa × 298.15 K)

  • V₂ =  9.4 liter ← answer

You can learn more about gas law problems reading this other answer on Brainly: brainly.com/question/12732788.

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Answer: Equilibrium constant is 0.70.

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Initial concentration of H_2O=\frac{moles}{volume}=\frac{0.40moles}{1L}=0.40M

equilibrium concentration of CO=\frac{moles}{volume}=\frac{0.18moles}{1L}=0.18M [/tex]

The given balanced equilibrium reaction is,

                            CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

Initial conc.            0.35 M       0.40M       0     0

At eqm. conc.    (0.35-x) M   (0.40-x) M   (x) M    (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CO_2]\times [H_2O]}{[CO]\times [H_2O]}

K_c=\frac{x\times x}{(0.40-x)(0.35-x)}

we are given : (0.35-x)= 0.18

x = 0.17

Now put all the given values in this expression, we get :

K_c=\frac{0.17\times 0.17}{(0.40-0.17)(0.35-0.17)}

K_c=0.70

Thus the value of the equilibrium constant is 0.70.

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2 years ago
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