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uranmaximum [27]
3 years ago
11

Arsine, ash3 is a highly toxic compound used in the electronics industry for the production of semiconductors. its vapor pressur

e is 35 torr at – 111.95°c and 253 torr at – 83.6°c. using these data calculate
Physics
1 answer:
Natalka [10]3 years ago
5 0

Answer:  a. 17.7 KJ/Mol

b. T=210K

Explanation:

Arsine, ash3 is a highly toxic compound used in the electronics industry for the production of semiconductors. its vapor pressure is 35 torr at – 111.95°c and 253 torr at – 83.6°c. using these data calculate.

the question isnt completely originally, but we could look at the likely derivation from the questions

(a) the standard enthalpy of vaporization

using the clausius clapeyron equation

In (PT1vap / PT2vap) = delta H (vap) / R ( (1/T1) - (1/T2) )

In (35Torr/253Torr) = delta H (vap) / 8.3145 ( (1/189.55) - (1/161.2) )

Therefore, Delta H (vap) = 17.7 KJ/Mol

b. Also the boiling point

What is the normal boiling point of arsine?

At the boiling point Pvap = atmospheric pressure = 1 atm=760 torr

substitution into the equation as stated in question 1

ln(760/253)=17700/8.314(1/189.55-1/T)

T=210K

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A boy throws a ball vertically up it returns the ground after 10 seconds find the maximum height reached by the ball
Akimi4 [234]

Answer:

Approximately 122.625\; {\rm m} (assuming that g = 9.81\; {\rm m\cdot s^{-2}}, the ball was launched from ground level, and that the drag on the ball is negligible.)

Explanation:

Let v_{0} denote the velocity at which the ball was thrown upward.

If the drag (air friction) on the ball is negligible, the ball would land with a velocity of exactly (-v_{0}). The velocity of the ball would be changed from v to (-v_{0})\! (such that \Delta v = (-v_{0}) - v_{0} = (-2\, v_{0})) within t = 10\; {\rm s}.

Also because the drag on the ball is negligible, the acceleration of the ball would be a = -g = -9.81\; {\rm m\cdot s^{-2}}. Thus:

\Delta v = a\, t = 10\; {\rm s} \times (-9.81\; {\rm m\cdot s^{-2}}) = -98.1\; {\rm m\cdot s^{-1}}.

Since \Delta v = (-2\, v_{0}):

-2\, v_{0} = \Delta v = -98.1\; {\rm m\cdot s^{-1}.

\begin{aligned}v_{0} &= \frac{-98.1\; {\rm m\cdot s^{-1}}}{-2}= 49.05\; {\rm m \cdot s^{-1}}\end{aligned}.

The ball reaches maximum height when its velocity is v_{1} = 0\; {\rm m\cdot s^{-1}}. Apply the SUVAT equation x = ({v_{1}}^{2} - {v_{0}}^{2}) / (2\, a) to find the displacement x between the original position (ground level, where v_{0} = 49.05\; {\rm m\cdot s^{-1}}) and the max-height position of the ball (where v_{1} = 0\; {\rm m\cdot s^{-1}}.)

\begin{aligned}x &= \frac{(0\; {\rm m\cdot s^{-1}})^{2} - (49.05\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-9.81\; {\rm m\cdot s^{-2}})} \\ &\approx 122.625\; {\rm m\cdot s^{-1}}\end{aligned}.

7 0
2 years ago
This is due by 11:59 PM tonight.
svetoff [14.1K]

Answer:

1. increases

2. increases

3. increases

Explanation:

Part 1:

First of all, since the box remains at rest, the horizontal net force acting on the box must equal zero:

F1 - fs = 0.

And this friction force fs is:

fs = Nμs,

where μs is the static coefficient of friction, and N is the normal force.

Originally, the normal force N is equal to mg, where m is the mass of the box, and g is the constant of gravity. Now, there is an additional force F2 acting downward on the box, which means it increases the normal force, since the normal force by Newton's third law, is the force due to the surface acting on the box upward:

N = mg + F2.

So, F2 is increasing, that means fs is increasing too.

Part 2:

As explained in the part 1, N = mg + F2. F2 is increasing, so the normal force is thus increasing.

Part 3:

In part 1 and part 2, we know that fs = Nμs, and since the normal force N is increasing, the maximum possible static friction force fs, max is also increasing.

6 0
4 years ago
Give an example of where you see the law of conservation of energy in your life
lesya692 [45]

Answer:

The law of conservation of energy can be seen in these everyday examples of energy transference: Water can produce electricity. Water falls from the sky, converting potential energy to kinetic energy.

The cue ball loses energy because the energy it had has been transferred to the 8 ball, so the cue ball slows down

Explanation:

8 0
3 years ago
An example of kinetic energy is a _____.
zhenek [66]

Answer:

An example of kinetic energy is a <u><em>car coming to a stop</em></u>

Explanation:

Kinetic energy is the energy that a body or system possesses due to its movement. In physics this energy is defined as the amount of work necessary to accelerate a body of a certain mass and in rest position, until reaching a certain speed. This energy obtained will remain unchanged as long as this body does not vary its speed. That is, kinetic energy measures how many changes an object that is moving can cause.

<u><em>An example of kinetic energy is a car coming to a stop</em></u>. If the car is moving and comes to a stop, there is a change in speed, therefore in movement, eventually producing a change in kinetic energy. This energy depends on the mass of the body, in this case the car, and the speed.  As the speed decreases, the kinetic energy will decrease.

4 0
4 years ago
To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel
alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0
3 years ago
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