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The set up below was used to investigate some properties of two gases M and N
When beaker A was filed with gas M, the level of water in the glass tube rose to
point II. When the experiment was repeated using gas N, the level of water dropped to point III. Explain these observations.
The pressure of nitrogen gas contained in a 1dm3 cylinder at -196°C was 107 Pascals.
Calculate the: a) Volume of the gas at 25°C and 105 Pascals. b) Mass of nitrogen gas(Molar volume of gas is 24dm3, N = 14.0)
ANSWERS
The graph below shows the relationship between pressure and the temperature of a gas in a fixed volume container.
a) State the relationship between pressure and temperature that can be deduced from the graph.
b) Using kinetic theory, explain the relationship shown in the graph.
ANSWERS
(a) Temperature and pressure are directly proportional (l) IR words towards that of real
(b) With increase in temperature, the gas particles gain more Kinetic energy They move faster and collide with the walls of the container more frequently hence increasing pressure.
A sealed glass tube containing air at s.t.p was immersed in water at 1000c. Assuming that there was no increase in the volume of the glass tube due to the expansion of the glass, calculate the pressure of the inside tube.
(standard pressure = 760mmHg,)
Expected Response
a) State the Graham's law diffusion.
b) The molar masses of gases W and X are 16.0 and 44.0 respectively. If the rate of diffusion of W through a porous material is 12cm3s-1 calculate the rate of diffusion of X through the same material.
ANSWERS
​Related Chemistry Questions and Answers on Gas Laws Form 3 Level
A small crystal of potassium manganate (VII) was placed in a beaker water. The beaker was left standing for two days without shaking. State and explain the observations that were made.
ANSWERS
a) State the Charles law
b) The volume of a sample of nitrogen gas at a temperature of 291 K and 1.0x105 Pascal‟s was 3.5 x 10-2m3. Calculate the temperature at which the volume of the gas would be 2.8 x 10-2m3 at 1.0 x 105 Pascal.
ANSWERS
A gas occupies a volume of 400cm3 at 500k and 1 atmosphere pressure. What will be the temperature of the gas when the volume and pressure of the gas is 100cm3 and 0.5 atmospheres respectively.
Expected Response
60cm3 of oxygen gas diffused through a porous partition in 50 seconds. How long would it take 60cm3 of sulphur (IV) oxide gas to diffuse through the same partition under the same conditions? (S= 32.0, 0 = 16.0)
ANSWERS
Study the set – up below and answer the questions that follow
a) What observation would be made in the tube?
b) Indicate with across (x) on the diagram the likely position where the observation stated in (a) above would be made. NB. Not to touch the cotton wool.
Expected Response
The graph below shows the behaviour of a fixed mass of a gas at constant temperature.
a) What is the relationship between the volume and the pressure of the gas?
b) 3 litres of oxygen gas at one atmosphere pressure were compressed to two atmospheres at constant temperature. Calculate the volume occupied by the oxygen gas
answers
A few crystals of potassium permanganate were carefully placed into water in a beaker at one spot. The beaker was left undisturbed for two hours. State and explain the observation that was made.
answer
In an experiment to study the diffusion of gases, a student set up the apparatus shown in diagram 1. After sometime the student noticed a change in the water level as shown in the diagram 
Give an explanation for the change in water level (2mks)
expected response
Air is less dense than carbon dioxide and so it enters the porous pot faster than carbon dioxide out of it. This sets up a higher pressure; in the pot and the level rises as shown:
A fixed mass of a gas has a volume of 250cm3 at a temperature of 270C and 750mm Hg pressure. Calculate the volume the gas would occupy at 420C (2mks)
expected response
A mixture containing equal volumes of hydrogen and carbon dioxide was introduced one end of a tube as shown below. 
Which gas would be detected at point 'C' first? Explain (2mks)
Expected Response
Hydrogen, because it is lighter/ less denser / diffuses faster (2mks)
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December 2024
AuthorMaurice Atika is a teacher and a netprenuer. |
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