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Shkiper50 [21]
2 years ago
14

Ar(g) deviates more from ideal behavior at extremely high pressures than Ne(g) does. Which of the following is one reason for th

is difference?
A
The particle volume of Ar is greater than that of Ne.
B
Ar atoms have more valence electrons than Ne atoms have, so Ar atoms have greater interparticle forces.
The intermolecular forces between Ne atoms are greater than those between Ar atoms.
D
Ar atoms are more attracted to the walls of the container than Ne atoms are.
Chemistry
1 answer:
KengaRu [80]2 years ago
7 0

Ar(g) deviates more from ideal behavior at extremely high pressures than Ne(g) does because A. The particle volume of Ar is greater than that of Ne.

When considering the ideal behavior of a gas, we make 2 assumptions:

  • The volume of the gas is negligible in comparison to the container.
  • The attraction forces between the particles are negligible.

Ar(g) deviates more from ideal behavior at extremely high pressures than Ne(g) does. Which of the following is one reason for this difference?

  • A. The particle volume of Ar is greater than that of Ne. YES. Since Ar particles are bigger than Ne ones, their volume is harder to neglect.
  • B. Ar atoms have more valence electrons than Ne atoms have, so Ar atoms have greater interparticle forces. NO. They have the same number of valence electrons (8).
  • C. The intermolecular forces between Ne atoms are greater than those between Ar atoms. NO. The intermolecular forces between Ne atoms are lower than those between Ar atoms.
  • D. Ar atoms are more attracted to the walls of the container than Ne atoms are. NO. There are no appreciable differences between the attractions to the walls.

Ar(g) deviates more from ideal behavior at extremely high pressures than Ne(g) does because A. The particle volume of Ar is greater than that of Ne.

Learn more: brainly.com/question/16110985

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At what time will the pressure of SO₂Cl₂ decline to 0.50 its initial value? Express your answer using two significant figures.
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Answer: The time is 0.69/k seconds

Explanation:

The following integrated first order rate law

ln[SO₂Cl₂] - ln[SO₂Cl₂]₀ = - k×t

where

[SO₂Cl₂] concentration at time t,

[SO₂Cl₂]₀ initial concentration,

k rate constant

Therefore, the time elapsed after a certain concentration variation is given by:

t=\frac{ln[SO_{2}Cl_{2}]_{0} - ln[SO_{2}Cl_{2}]}{k}=\frac{ln\frac{[SO_{2}Cl_{2}]_{0}}{[SO_{2}Cl_{2}]} }{k}

We could assume that SO₂Cl₂ behaves as a ideal gas mixture so partial pressure is proportional to concentration:

p_{(SO_{2}Cl_{2})}V = n_{(SO_{2}Cl_{2})}RT

[SO_{2}Cl_{2}]= \frac{n_{(SO_{2}Cl_{2})}}{V}}=\frac{p_{(SO_{2}Cl_{2})}}{RT}}

In conclusion,

t = ln( p(SO₂Cl₂)₀/p(SO₂Cl₂) )/k

t=\frac{ln\frac{p_{(SO_{2}Cl_{2})}_{0}}{p_{(SO_{2}Cl_{2})}} }{k}

for

p_{(SO_{2}Cl_{2})}=0.5p_{(SO_{2}Cl_{2})}_{0}

t=\frac{ln\frac{p_{(SO_{2}Cl_{2})}_{0}}{0.5p_{(SO_{2}Cl_{2})_{0}}} }{k}

t=\frac{ln\frac{1}{0.5} }{k}

t=\frac{ln(2)}{k}

t=\frac{0.69}{k}}

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Answer:

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