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Ymorist [56]
4 years ago
14

For the equilibrium PCl5(g) PCl3(g) + Cl2(g), Kc = 2.0 × 101 at 240°C. If pure PCl5 is placed in a 1.00-L container and allowed

to come to equilibrium, and the equilibrium concentration of PCl3(g) is 0.27 M, what is the equilibrium concentration of PCl5(g)?
Chemistry
1 answer:
Cloud [144]4 years ago
6 0

Answer:

the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M

Explanation:

for the reaction

PCl₅(g) → PCl₃(g) + Cl₂(g)

where

Kc= [PCl₃]*[Cl₂]/[PCl₅] = 2.0*10¹ M = 20 M

and [A] denote concentrations of A

if initially the mixture is pure PCl₅ , then it will dissociate according to the reaction and since always one mole of PCl₃(g) is generated with one mole of Cl₂(g) , the total number of moles of both at the end is the same → they have the same concentration → [PCl₃(g)] = [Cl₂]=0.27 M

therefore

Kc= [PCl₃]*[Cl₂]/[PCl₅] = 0.27 M* 0.27 M /[PCl₅] = 20 M

[PCl₅]  =  0.27 M* 0.27 M / 20 M = 3.64*10⁻³ M

[PCl₅]  = 3.64*10⁻³ M

the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M

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Explain why the sand on a beach cools down at night more quickly than the ocean water.
Lesechka [4]

The sand on the beach cools down at night then the ocean is because the sand is on dry land.

7 0
3 years ago
How many grams of AgNO3 are needed to prepare a 75.00 mL solution that is 6.5% (m/v)?
FrozenT [24]

Answer:

Volume of AgNO₃ = 4.9 ml (Approx)

Explanation:

Given:

Total solution = 75 ml

Volume of AgNO₃ = 6.5%

Find:

Volume of AgNO₃

Computation:

Volume of AgNO₃ = Total solution x Volume of AgNO₃

Volume of AgNO₃ = 75 x 6.5%

Volume of AgNO₃ = 4.875

Volume of AgNO₃ = 4.9 ml (Approx)

5 0
3 years ago
Determine how many moles are present in 0.23kg of SO(2)
Rama09 [41]

Answer:

3.59 moles

Explanation:

Hopefully this helps! :)

Mark as brainliest if right!

6 0
3 years ago
What effect does observing a substance's physical properties have on the substance?
Phantasy [73]
There is no effect. When you observe a substance's physical properties, you're not doing anything that would catalyze a chemical change. For example, you most often observe with your eyes, and/or, if safe, you waft the substance's smell to your nose. Neither of these observations will effect the substance's physical properties.
8 0
3 years ago
the isotope 146c has a half life of 5730 years. what fraction of 146c in a sample with mass ,m, after 28650 years
defon

Answer:

3.1% is the fraction of the sample after 28650 years

Explanation:

The isotope decay follows the equation:

Ln[A] = -kt + Ln[A]₀

<em>Where [A] could be taken as fraction of isotope after time t, k is decay constant and [A]₀ is initial fraction of the isotope = 1</em>

<em />

k could be obtained from Half-Life as follows:

K = Ln 2 / Half-life

K = ln 2 / 5730 years

K = 1.2097x10⁻⁴ years⁻¹

Replacing in isotope decay equation:

Ln[A] = -1.2097x10⁻⁴ years⁻¹*28650 years + Ln[1]

Ln[A] = -3.4657

[A] = 0.0313 =

<h3>3.1% is the fraction of the sample after 28650 years</h3>

<em />

3 0
3 years ago
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