Chipa Maimela GRADE 12 PHYSICAL SCIENCES, PHYSICAL SCIENCES GRADE 12

Alkanes and alkenes react differently with bromine and potassium permanganate.


Reaction of Alkanes with Bromine:

Alkanes typically do not react with bromine in the absence of light or heat. However, when exposed to light or heat, alkanes can undergo a free radical substitution reaction with bromine, which results in the formation of a bromoalkane. The equation for this reaction is:

CnH2n+2 + Br2 → CnH2n+2Br

Reaction of Alkenes with Bromine:

Alkenes can react with bromine in the absence of light or heat. This reaction is called an electrophilic addition reaction, and it results in the formation of a dibromoalkane. The equation for this reaction is:

CnH2n + Br2 → CnH2nBr2

Reaction of Alkanes with Potassium Permanganate:

Alkanes do not react with potassium permanganate in the absence of heat or an acid catalyst. However, when heated with an acid catalyst, alkanes can undergo an oxidation reaction that results in the formation of carboxylic acids or alcohols. The equation for this reaction is:

CnH2n+2 + KMnO4 + H+ → CnH2n+1COOH or CnH2n+1OH

Reaction of Alkenes with Potassium Permanganate:

Alkenes do not react with potassium permanganate in the absence of heat or an acid catalyst. However, when heated with an acid catalyst, alkenes can undergo an oxidation reaction that results in the formation of carboxylic acids or alcohols. The equation for this reaction is:

CnH2n + KMnO4 + H+ → CnH2n-1COOH or CnH2n-1OH

It’s important to note that the reagents used and the conditions for the reaction can affect the products formed in these reactions.

Advertisement
Chipa Maimela EXPERIMENT, GRADE 12 PHYSICAL SCIENCES, PHYSICAL SCIENCES GRADE 12, PRACTICAL INVESTIGATION

Position vs. Time and velocity vs. time for a free-falling object:


The position of a free-falling object is measured downward from a certain starting point.

The position of the object increases with time as it falls.

The graph would be a straight line with a negative slope, starting at the initial position and ending at the final position.

Velocity vs. Time graph for a free-falling object:

The velocity of a free-falling object increases with time as it falls.

The graph would be a straight line with a positive slope, starting at zero and ending at the final velocity.

To determine the acceleration due to gravity, we can use the equation:

a = v^2/s or a = 2s/t^2

Where a is the acceleration, v is the final velocity, s is the final position, and t is the time of fall.

Since the acceleration due to gravity is constant, the slope of the velocity vs time graph is the acceleration.

Position vs. Time:

The graph for position vs. time for a free-falling object would be a straight line that starts at a non-zero position (the initial height of the object) and decreases as time goes on. The slope of this line would represent the velocity of the object. The equation for this line would be:

y = -g*t + h

where g is the acceleration due to gravity, t is the time, and h is the initial height of the object.

Velocity vs. Time:

The graph for velocity vs. time for a free-falling object would be a straight line that starts at a non-zero velocity (the initial velocity of the object) and decreases as time goes on. The slope of this line would represent the acceleration of the object. The equation for this line would be:

y = -g*t + v

where g is the acceleration due to gravity, t is the time, and v is the initial velocity of the object.

Determining the Acceleration due to Gravity:

To determine the acceleration due to gravity, we can use the slope of either the position vs. time or velocity vs. time graph. For the position vs. time graph, the slope is -g, and for the velocity vs. time graph, the slope is -g. Therefore, we can conclude that the acceleration due to gravity is -g.

Chipa Maimela GRADE 12 PHYSICAL SCIENCES, PHYSICAL SCIENCES GRADE 12

Momentum


Momentum is a measure of an object’s inertia or its resistance to changes in its motion. It is calculated as the product of an object’s mass (m) and its velocity (v). So, the formula for momentum (p) is:

p = m x v

Where: p = momentum m = mass of the object (in kg) v = velocity of the object (in m/s)

Example: If an object has a mass of 5 kg and is moving at a velocity of 10 m/s, then its momentum would be: p = 5 kg x 10 m/s = 50 kg m/s

Momentum is a vector quantity, which means it has both magnitude and direction. The magnitude of momentum is simply the product of an object’s mass and velocity, as described by the formula p = mv. The direction of momentum is the same as the direction of an object’s velocity.

For example, consider a car moving east with a mass of 1000 kg and a velocity of 20 m/s. The magnitude of its momentum would be: p = mv = 1000 kg x 20 m/s = 20,000 kg m/s

The direction of the momentum would be in the east, the same as the direction of the car’s velocity.

Another example is a ball thrown upwards with a mass of 0.2 kg and a velocity of 10 m/s upward. The magnitude of its momentum would be: p = mv = 0.2 kg x 10 m/s = 2 kg m/s

The direction of the momentum would be upward, the same as the direction of the ball’s velocity.

It’s important to note that in collisions and other interactions between objects, the total momentum of a closed system (systems with no external forces) is conserved, which means that the momentum of all objects before and after the collision will be equal. This is the principle of conservation of momentum.

Chipa Maimela GRADE 12 PHYSICAL SCIENCES, PHYSICAL SCIENCES GRADE 12

Oxidation and Reduction


Chipa Maimela EXPERIMENT, GRADE 12 PHYSICAL SCIENCES, GRADE 12 QUESTION PAPERS, PHYSICAL SCIENCES GRADE 12, PRACTICAL INVESTIGATION

GRADE 12 PRESCRIBED EXPERIMENT 2: ACID-BASE REACTIONS 


                                 

WORK SHEET 
TOTAL MARK: 50

ACTIVITY

Titration of oxalic acid against sodium hydroxide to determine the concentration of the sodium hydroxide.

In this investigation you will prepare an acidic solution accurately and thus you will know its exact concentration. You will then react this acid with a base of an unknown concentration to determine the concentration of the base.

What you will need:

Erlenmeyer flasks,
Burettes,
Burette clamp,
Medicine  dropper,
Retort stand,
White tile /paper,
Measuring cylinders,
Mass meter		Oxalic acid,
Sodium Hydroxide,
Phenolphthalein as indicator,
Funnel,
Beaker,
Spatula,
Glass rod,
Pipette with sucker

What to do:

  1. Prepare a standard solution of oxalic acid which has a concentration of approximately 1mol.dm-3.
  2. Now prepare a sodium hydroxide solution by dissolving approximately 2g of dry sodium hydroxide in 500ml of water.
  3. Add two drops of the indicator solution.
  4. Place the burette in the clamp.
  5. Using the funnel, fill the burette to above zero mark with the acid solution.
  6. Then, holding the beaker, with which you used to pour the acid, beneath the burette, gradually open the tap.
  7. Allow the level of the base to come down to exactly zero (reading from the bottom of the meniscus).
  8. Pipette using the sucker exactly 25ml of oxalic acid solution in a volumetric flask.
  9. Add a few drops of phenolphthalein to the acid.
  10. Hold the conical flask beneath the burette with your right hand and gradually open the tap with your left.
  11. Swirl the conical flak continuously and watch it closely for the first sign of a colour change.
  12. As you see that you are approaching the point of neutralization, close the tap slightly so that you are adding drop by drop.
  13. When the colour changes completely the titration is finished.
  14. Close the tap and read from the burette how much acid was used.
  15. Repeat this procedure at least twice so that you have three readings for the volume of NaOH (of unknown concentration) required to neutralize exactly 25ml of oxalic acid (of known concentration).
  16. Take an average of these three and use it to calculate the concentration of the NaOH.
  17. Now calculate the concentration of the sodium hydroxide solution.
  18. Make a neat labeled sketch to represent the apparatus
  19. Now write a report using the format learnt in class.

     Questions

  1. What is the appropriate concentration of NAOH (2g in 500ml of water)
  2. Calculate the theoretical concentration of NaOH from the actual mass of NaOH you measured.
  3. How does your theoretical value for NaOH concentration (from the actual mass you measured) differ from the actual concentration you calculated (from the titration procedure)? Can you think of some reasons why your values may differ?