I believe the answer is compound B may have a lower molecular weight compared to compound A.
At the same temperature, lighter particles of a compound have a higher average speeds than do heavier particles of another compound. Thus, particles of compound B are lighter than those of compound A and thus they have a higher average speed, hence evaporating faster compared to compound A.
<>"Refraction is the bending of the path of a light wave as it passes from one material into another material. The refraction occurs at the boundary and is caused by a change in the speed of the light wave upon crossing the boundary. The tendency of a ray of light to bend one direction or another is dependent upon whether the light wave speeds up or slows down upon crossing the boundary. The speed of a light wave is dependent upon the optical density of the material through which it moves. For this reason, the direction that the path of a light wave bends depends on whether the light wave is traveling from a more dense (slow) medium to a less dense (fast) medium or from a less dense medium to a more dense medium. In this part of Lesson 1, we will investigate this topic of the direction of bending of a light wave.
Predicting the Direction of Bending
Recall the Marching Soldiers analogy discussed earlier in this lesson. The analogy served as a model for understanding the boundary behavior of light waves. As discussed, the analogy is often illustrated in a Physics classroom by a student demonstration. In the demonstration, a line of students (representing a light wave) marches towards a masking tape (representing the boundary) and slows down upon crossing the boundary (representative of entering a new medium). The direction of the line of students changes upon crossing the boundary. The diagram below depicts this change in direction for a line of students who slow down upon crossing the boundary.
On the diagram, the direction of the students is represented by two arrows known as rays. The direction of the students as they approach the boundary is represented by an incident ray (drawn in blue). And the direction of the students after they cross the boundary is represented by a refracted ray (drawn in red). Since the students change direction (i.e., refract), the incident ray and the refracted ray do not point in the same direction. Also, note that a perpendicular line is drawn to the boundary at the point where the incident ray strikes the boundary (i.e., masking tape). A line drawn perpendicular to the boundary at the point of incidence is known as a normal line. Observe that the refracted ray lies closer to the normal line than the incident ray does. In such an instance as this, we would say that the path of the students has bent towards the normal. We can extend this analogy to light and conclude that:
Light Traveling from a Fast to a Slow Medium
If a ray of light passes across the boundary from a material in which it travels fast into a material in which travels slower, then the light ray will bend towards the normal line.
The above principle applies to light passing from a material in which it travels fast across a boundary and into a material in which it travels slowly. But what if light wave does the opposite? What if a light wave passes from a material in which it travels slowly across a boundary and into a material in which it travels fast? The answer to this question can be answered if we reconsider the Marching Soldier analogy. Now suppose that the each individual student in the train of students speeds up once they cross the masking tape. The first student to reach the boundary will speed up and pull ahead of the other students. When the second student reaches the boundary, he/she will also speed up and pull ahead of the other students who have not yet reached the boundary. This continues for each consecutive student, causing the line of students to now be traveling in a direction further from the normal. This is depicted in the diagram below.
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According to ideal gas equation, we know for 1 mole of gas: PV=RT
where P = pressure, T = temperature, R = gas constant, V= volume
If '1' and '2' indicates initial and final experimental conditions, we have
Given that: V1 = 100.0 kPa, T1 = 100.0 K, V1 = 2.0 m3, T2 = 400 K, P2 = 200.0 kPa
∴ on rearranging above eq., we get V2 =
∴ V2 = 4 m3
Answer:
FALSE
Explanation:
Assuming that the gas is ideal
Therefore the gas obeys the ideal gas equation
<h3>Ideal gas equation is </h3><h3>P × V = n × R × T</h3>
where
P is the pressure exerted by the gas
V is the volume occupied by the gas
n is the number of moles of the gas
R is the ideal gas constant
T is the temperature of the gas
Here volume of the gas will be the volume of the container
Given the volume of the container and number of moles of the gas are constant
As R will also be constant, the pressure of the gas will be directly proportional to the temperature of the gas
P ∝ T
∴ Pressure will be directly proportional to the temperature
Answer:
Power plants are not located in cities. Power has to travel a long distance over power lines to reach cities.
Explanation:
