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Gelneren [198K]
3 years ago
8

2. A sample of helium gas has a volume of 200.0 mL at 0.960 atm. What pressure, in atmospheres, is needed to reduce the volume a

t constant tem-perature to 50.0 mL?
Chemistry
1 answer:
brilliants [131]3 years ago
4 0

Answer:

3.840 atm

Explanation:

We use Boyle's law, which relates the pressure (P) and the volume (V) of a gas at a constant temperature. The change from initial P and V (P₁ and V₁) to final P and V (P₂ and V₂) is expressed as:

P₁V₁ = P₂V₂

We have the following data:

P₁= 0.960 atm

V₁= 200.0 mL

V₂ = 50.0 mL

Thus, we introduce the data in the equation and calculate the final pressure P₂:

P₂ = P₁V₁/V₂= (0.960 atm x 200.0 mL)/50.0 mL = 3.840 atm

Therefore, a pressure of 3.840 atm is needed to reduce the volume of the gas from 200.0 mL to 50.0 mL.

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m(KCl) = 15 g.
n(KCl) = m(KCl) ÷ M(KCl).
n(KCl) = 15 g ÷ 74,55 g/mol.
n(KCl) = 0,2 mol
m(H₂O) = 1650 g ÷ 1000 g/kg = 1,65 kg.
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