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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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kolezko [41]

Answer:

Doing it right now i’ll pm once done give me 5 minutes:)

Explanation:

5 0
3 years ago
P1: Explain in detail - How can the motion of an object that is already moving change? (What are the different ways a moving obj
Tomtit [17]

Inertia: tendency of an object to resist changes in its velocity. An object at rest has zero velocity - and (in the absence of an unbalanced force) will remain with a zero velocity. Such an object will not change its state of motion (i.e., velocity) unless acted upon by an unbalanced force.

5 0
4 years ago
An icicle falls off of a skyscraper from rest and falls for 34 seconds. How fast will that icicle be moving after that time? Sho
goblinko [34]

Answer:

<em>The icicle will be moving at 333.54 m/s</em>

Explanation:

<u>Free Fall Motion </u>

A free-falling object falls under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. Free-falling objects do not encounter air resistance.

If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is g = 9.81 m/s^2.

The final velocity of a free-falling object after a time t is given by:

vf=g.t

The icicle falls from rest for 34 seconds. We need to find the speed after that time:

vf = 9.81*34

vf = 333.54 m/s

The icicle will be moving at 333.54 m/s

7 0
3 years ago
Help This is Urgent
MaRussiya [10]
I don’t use the metric system, so I used feet and then went back.

25 meters is about 82 feet.
So around 82 feet per minute (kinda slow LOL)

82 (feet per minute) x 90 (minutes) = 7380 feet (in 90 minutes)

7380 feet is equal to 2249.424 meters.

(I hope that helped)

6 0
3 years ago
Read 2 more answers
5. George walks to a friend's house. He walks 750 meters North, then realizes he walked too far.
dedylja [7]

Answer:

Average speed: approximately 76.9\; {\rm m\cdot s^{-1}}.

Average velocity: approximately 38.5\; {\rm m \cdot s^{-1}} (to the north.)

Explanation:

Consider an object that travelled along a certain path. Distance travelled would be equal to the length of the entire path.

In contrast, the magnitude of displacement is equal to distance between where the object started and where it stopped.

In this question, the path George took required him to travel 750\; {\rm m} + 250\; {\rm m} = 1000\; {\rm m} in total. Hence, the distance George travelled would be 1000\; {\rm m}. However, since George stopped at a point (750\; {\rm m} - 250\; {\rm m}) = 500\; {\rm m} to the north of where he started, his displacement would be only 500\; {\rm m} to the north.

Divide total distance by total time to find the average speed.

Divide total displacement by total time to find average velocity.

The total time of travel in this question is 13\; {\rm s}.. Therefore:

\begin{aligned}\text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &= \frac{1000\; {\rm m}}{13\; {\rm s}} \\ &\approx 76.9\; {\rm m\cdot s^{-1}}\end{aligned}.

\begin{aligned}\text{average velocity} &= \frac{\text{total displacement}}{\text{total time}} \\ &= \frac{500\; {\rm m}}{13\; {\rm s}} && \genfrac{}{}{0px}{}{(\text{to the north})}{}\\ &\approx 38.5\; {\rm m\cdot s^{-1}} && (\text{to the north})\end{aligned}.

3 0
2 years ago
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