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Luba_88 [7]
2 years ago
11

A flask contains equal masses of f2 and cl2 with a total pressure of 2.00 atm at 298k. What is the partial pressure of cl2 in th

e flask?
Physics
1 answer:
Sindrei [870]2 years ago
7 0

Answer:

Partial Pressure of F₂ = 1.30 atm

Partial pressure of Cl₂ = 0.70 atm

Explanation:

Partial pressure for gases are given by Daltons law.

Total pressure of a gas mixture = sum of the partial pressures of individual gases

Pt = P(f₂) + P(cl₂)

Partial pressure = mole fraction × total pressure

Let the mass of each gas present be m

Number of moles of F₂ = m/38 (molar mass of fluorine = 38 g/Lol

Number of moles of Cl₂ = m/71 (molar mass of Cl₂)

Mole fraction of F₂ = (m/38)/((m/38) + (m/71)) = 0.65

Mole fraction of Cl₂ = (m/71)/((m/38) + (m/71)) = 0.35 or just 1 - 0.65 = 0.35

Partial Pressure of F₂ = 0.65 × 2 = 1.30 atm

Partial pressure of Cl₂ = 0.35 × 2 = 0.70 atm

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6 0
3 years ago
A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient
gavmur [86]

Answer:

a) The minimum force required to start moving the box is 352.86 N

b) i) The friction force for the box in motion is 147.025 N

ii) The acceleration of box is 4.21625 m/s²

Explanation:

The parameters of the box at rest and the floor are;

The mass of the box = 60 kg

The static coefficient friction of the floor = 0.6

The kinetic coefficient friction of the floor = 0.25

Frictional force = Normal force × Friction coefficient

For an horizontal floor and the box laying on the floor, we have;

The normal force = The weight of the box = Mass of the box × Acceleration due to gravity, g

The acceleration due to gravity, g = 9.81 m/s²

The weight of the box  = 60 × 9.81 = 588.6 N

a) The static coefficient gives the frictional force observed by the box and which must be surpassed to bring about motion

Therefore;

The minimum force required to start moving the box = The static frictional force = Weight of the box × The static coefficient of friction

The minimum force required to start moving the box = 588.1 × 0.6 = 352.86 N

The minimum force required to start moving the box = 352.86 N

b) i) When an horizontal force of 400 N is applied, the applied force is larger than the static friction force, and the box will be in motion with the kinetic coefficient of friction being the source of friction

The friction force for the box in motion = 588.1 × 0.25 = 147.025 N

ii) The force, F with the box is in motion, is given as follows;

F = Mass of box × Acceleration of box, a = Applied force - Kinematic friction force

F = 60 × a = 400 - 147.025 = 252.975 N

60 × a = 252.975 N

a = 252.975 N/(60 kg) = 4.21625 m/s²

Acceleration of box, a = 4.21625 m/s².

6 0
2 years ago
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