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3 years ago

9

A gas cylinder is filled with 4.00 moles of oxygen gas at 300.0 k. the piston is compressed to yield a pressure of 400.0 kpa. wh

at is the volume inside the cylinder?

1 answer:

Radda [10]3 years ago

8 0

To answer this item, it is assumed that the gas in the cylinder is ideal such that it follows the equation,
PV = nRT
when V is to be calculated,
V = nRT/P
V = (4)(0.0821 L.atm/molK)(300 K) / (400 kPa/101.325 kPa/atm)
V = 24.95 L
Thus, the volume of gas in the cylinder is 24.95 L.

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a proton of mass 1 u travelling with a speed of 3.6 x 10 ^4 m/s has an elastic head on collision with a helium nucleus initially

CaHeK987 [17]

Answer:

Velocity of the helium nuleus = 1.44x10⁴m/s

Velocity of the proton = 2.16x10⁴m/s

Explanation:

From the conservation of linear momentum of the proton collision with the He nucleus:

$P_{1i} + P_{2i} = P_{1f} + P_{2f]$ (1)

<em>where $P_{1i}$ : is the proton linear momentum initial, $P_{2i}$ : is the helium nucleus linear momentum initial, $P_{1f}$ : is the proton linear momentum final, $P_{2f}$ : is the helium nucleus linear momentum final </em>

<u>From (1):</u>

$m_{1}v_{1i} + 0 = m_{1}v_{1f} + m_{2}v_{2f}$ (2)

<em>where m₁ and m₂: are the proton and helium mass, respectively, $v_{1i}$ and $v_{2i}$ : are the proton and helium nucleus velocities, respectively, before the collision, and $v_{1f}$ and $v_{2f}$ : are the proton and helium nucleus velocities, respectively, after the collision </em>

By conservation of energy, we have:

$K_{1i} + K_{2i} = K_{1f} + K_{2f}$ (3)

<em>where $K_{1i}$ and $K_{2i}$ : are the kinetic energy for the proton and helium, respectively, before the colission, and $K_{1f}$ and $K_{2f}$ : are the kinetic energy for the proton and helium, respectively, after the colission </em>

<u>From (3):</u>

$\frac{1}{2}m_{1}v_{1i}^{2} + 0 = \frac{1}{2}m_{1}v_{1f}^{2} + \frac{1}{2}m_{2}v_{2f}^{2}$ (4)

<u>Now we have two equations: (2) ad (4), and two incognits: $v_{1f}$ and $v_{2f}$ . </u>

Solving equation (2) for $v_{1f}$ , we have:

$v_{1f} = v_{1i} -\frac{m_{2}}{m_{1}} v_{2f}$ (5)

<u>From getting (5) into (4) we can obtain the $v_{2f}$ :</u>

$v_{2f}^{2} \cdot (\frac{m_{2}^{2}}{m_{1}} + m_{2}) - 2v_{2f}v_{1i}m_{2} = 0$

$v_{2f}^{2} \cdot (\frac{(4u)^{2}}{1u} + 4u) - v_{2f}\cdot 2 \cdot 3.6 \cdot 10^{4} \cdot 4u = 0$

From solving the quadratic equation, we can calculate the velocity of the helium nucleus after the collision:

$v_{2f} = 1.44 \cdot 10^{4} \frac{m}{s}$ (6)

Now, by introducing (6) into (5) we get the proton velocity after the collision:

$v_{1f} = 3.6 \cdot 10^{4} -\frac{4u}{1u} \cdot 1.44 \cdot 10^{4}$

$v_{1f} = -2.16 \cdot 10^{4} \frac{m}{s}$

The negative sign means that the proton is moving in the opposite direction after the collision.

I hope it helps you!

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Which of the following are likely to form a covalent bond? 1.two gold atom 2. two neon atoms 3. hydrgen and oxygen atoms 4. magn

Gwar [14]

Number 3.hydrogen and oxygen atoms because covalent bonds have both hydrogen and oxygen in them

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3 years ago

How are electromagnetic waves produced? A) An accelerating electric point charge (electron) emits electromagnetic waves. B) A ba

shusha [124]

Hello,

Here is your answer:

The proper answer to this question will be option A "An accelerating electric point charge (electron) emits electromagnetic waves." That's because electromagnetic waves are formed by motion of electrically charged particles (electrons)!

Your answer is A!

If you need anymore help feel free to ask me!

Hope this helps!

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3 years ago

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What is the motion of the particles in this kind of wave? A hand holds the left end of a set of waves. The waves themselves make

almond37 [142]

Answer:

D

Explanation:

The particles will move side to side over large areas

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Answer:

Kevin wants to make the firm private

Explanation:

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If you have any more questions feel free to ask away at Brainly.

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