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mario62 [17]
3 years ago
9

What is the pressure of 0.5 mol of nitrogen gas in a 5 L container at 203 K

Chemistry
1 answer:
vlada-n [284]3 years ago
5 0

Answer:

=1.666 liters

Explanation:

1 mole of a has at standard temperature and pressure occupies a volume of 22.4 liters.

0.5 moles of nitrogen occupy a volume of (0.5 moles×22.4 dm³/mol)/ 1

=11.2 liters.

Standard pressure= 1 atmosphere (Atm)

Standard temperature = 273.15 Kelvin

According to Combined gas equation, P₁V₁/T₁=P₂V₂/T₂

Let us take the conditions under standard conditions as the reference, with the subscript 1 and the conditions under the 5L container to be scenario 2 with subscript 2.

Therefore P₂ =P₁V₁T₂/T₁V₂

Substituting for the values we get:

P₂= (1 atm× 11.2L ×203K)/ (273K×5L)

=1.666 atm

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3 years ago
Identify the information that can be included in a chemical equation.
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Which of the following atoms or ions does not have a filled outer subshell?CaSZn2+S2Ca2+
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S and S²⁻ do not have the outer subshell fully filled with electrons.

Explanation:

We look at electronic configurations:

Ca   1s² 2s² 2p⁶ 3s² 3p⁶ 4s² - the outer subshell 4s² is fully-filled with electrons

S      1s² 2s² 2p⁶ 3s² 3p⁴ - the outer subshell 3p⁴ is not fully-filled with electrons

Zn²⁺  1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s⁰ - here the 4s subshell is higher in energy than 3d subshell so will consider 3d¹⁰ the out subshell which is fully-filled with electrons

S²⁻  1s² 2s² 2p⁶ 3s² 3p² - the outer subshell 3p² is not fully-filled with electrons

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4 0
3 years ago
A container is at a pressure of 3 atm and a temperature of 280K. What is the new temperature when the pressure is reduced to 1.5
GenaCL600 [577]

Answer:

140 K

Explanation:

Step 1: Given data

  • Initial pressure of the gas (P₁): 3 atm
  • Initial temperature of the gas (T₁): 280 K
  • Final pressure of the gas (P₂): 1.5 atm
  • Final temperature of the gas (T₂): ?

Step 2: Calculate the final temperature of the gas

We have a gas whose pressure is reduced. If we assume an ideal behavior, we can calculate the final temperature of the gas using Gay-Lussac's law.

T₁/P₁ = T₂/P₂

T₂ = T₁ × P₂/P₁

T₂ = 280 K × 1.5 atm/3 atm = 140 K

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3 years ago
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