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Darya [45]
3 years ago
12

A balloon is filled to a volume of 1.50 L with 3.00 moles of gas at 25 °C. With pressure and temperature held constant, what wil

l be the volume of the balloon if 0.20 moles of gas are added?
Chemistry
2 answers:
Vaselesa [24]3 years ago
5 0

Answer: 1.6L

Explanation:

V1 = 1.50 L,

V2 =?

n1 = 3mol

n2 = 3 + 0.2 = 3.2mol

From PV = nRT

V1 /n1 = V2/n2

1.5/3 = V2 /3.2

V2 = (1.5/3/) x 3.2 = 1.6L

n200080 [17]3 years ago
3 0

Answer:

V = 1.6 L

Explanation:

assuming ideal gas:

  • PV = RTn

∴ R = 0.082 atm.L/K.mol

∴ V1 = 1.50 L

∴ n1 = 3.00 mol

∴ T1 = 25°C ≅ 298 K

⇒ P1 = (RT1n1)/(V1) = ((0.082 atm.L/K.mol)(298 K)(3.00 mol))/(1.50 L)

⇒ P1 = 48.872 atm

with pressure and temperature held constant:

∴ T2 = T1 = 298 K

∴ P2 = P1 = 48.872 atm

∴ n2 = 0.20 mol + 3.00 mol = 3.20 mol

⇒ V2 = (RT2n2)/P2

⇒ V2 = ((0.082 atm.L/K.mol)(298 K)(3.20 mol))/(48.872 atm)

⇒ V2 = 1.6 L

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The chemical reactions that occur in a closed container can reach a state of <u>chemical equilibrium</u> that is characterized because the concentrations of the reactants and products remain constant over time. The <u>equilibrium constant</u> of a chemical reaction is the value of its reaction quotient in chemical equilibrium.

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[H_{2}O] = \frac{400 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 8.12 \frac{mol}{L}

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