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Alex73 [517]
3 years ago
8

The equilibrium constant for the decomposition of PCl5 at 250 celcius is 1.05. PCl5(g)-->PCl3(g)+Cl2(g) If the equilibrium pr

essures of PCl5 and PCl3 are .875 atm and .463 atm, respectively, then what is the equilibrium partial pressure of Cl2 at 250 celsius.
Chemistry
1 answer:
mrs_skeptik [129]3 years ago
8 0

Answer : The equilibrium partial pressure of Cl₂ at 25°C is, 1.98 atm

Explanation :

For the given chemical reaction:

PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)

The expression of K_p for above reaction follows:

K_p=\frac{P_{PCl_3}\times P_{Cl_2}}{P_{PCl_5}}

We are given:

P_{PCl_5}=0.875atm

P_{PCl_3}=0.463atm

K_p=1.05

Putting values in above equation, we get:

1.05=\frac{0.463\times P_{Cl_2}}{0.875}\\\\P_{Cl_2}=1.98atm

Thus, the equilibrium partial pressure of Cl₂ at 25°C is, 1.98 atm

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climate

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6 0
1 year ago
A flexible container at an initial volume of 6.13 L contains 2.51 mol of gas. More gas is then added to the container until it r
aleksley [76]

Answer:

2.12 moles of gas were added.

Explanation:

We can solve this problem by using<em> Avogadro's law</em>, which states that at constant temperature and pressure:

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Where in this case:

  • V₁ = 6.13 L
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We <u>input the data</u>:

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5 0
3 years ago
0.415 g of an unknown triprotic acid are used to make a 100.00 mL solution. Then 25.00 mL of this solution is transferred to an
Arturiano [62]

Answer:

Explanation:

Initial burette reading = 1.81 mL

final burette reading = 39.7 mL

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No of moles contained by 37.89 mL of .1029 M NaOH

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6 0
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