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Luba_88 [7]
3 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]3 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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3 years ago
The volume of a cylindrical tin can with a top and a bottom is to be 16π cubic inches. If a minimum amount of tin is to be used
sergiy2304 [10]

Answer:

h = 4 in

Explanation:

GIVEN DATA:

volume of tin= 16 \pi

we know that

volume of cylinder is v = \pi r^2  h

so,

16 \pi = \pi r^2 h

16 = r^2 h

r = \sqrt{\frac{16}{h}}

construct formula for surface area

S = 2\pi r^2 +  2\pi rh

S = \frac{2v}{h} + 2 \sqrt{v \pi h}

minimize the function wrt  h

S' = \frac{2v}{h^2} + \sqrt{\frac{v \pi}{h} =  0

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8 0
3 years ago
In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor
leonid [27]

Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}.....(I)

Where, F(x) = net force

We know, the net force is the sum of forces.

So, \sum{F}=ma

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

F(x)+(-F(f))=ma

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}

We need to find the value of \int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}

Using newton's second law

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}...(II)

We know that,

Acceleration is rate of change of velocity.

a=\dfrac{dv}{dt}

Put the value of a in equation (II)

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}

Now, the work done by the net force on the block is,

W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

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Answer:

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Explanation:

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