Port Said International Schools
Better Education for Future Generations
Science Department
Grade: 9
Some substances decomposes by heat:
Tools: red mercuric oxide – copper hydroxide - copper carbonate – copper sulphate –
sodium nitrate – test tubes – flame.
Procedures:
a. Put a little amount of red mercuric oxide in a test tube.
b. Heat the mercuric oxide.
c. Observe the change that happens to the red mercuric.
d. Repeat the previous steps with the other compounds.
Observations:
a. Red mercuric oxide turns into silvery mercury and oxygen gas evolves.
b. Blue copper hydroxide turns into black copper oxide and water vapor evolves.
c. Green copper carbonate turns into black copper oxide and carbon dioxide gas
evolves.
d. Blue copper sulphate turns into black copper oxide and sulphur trioxide gas
evolves.
e. White sodium nitrate turns into yellowish white and oxygen gas evolves.
Conclusions:
a. Metal oxides decompose by heat into metal and oxygen.
b. Metal hydroxides decompose by heat into metal oxide and water vapor.
c. Metal carbonates decompose by heat into metal oxide and carbon dioxide.
d. Metal sulphates decompose by heat into metal oxide and sulphur trioxide gas.
The reaction of water with sodium:
Tools: beaker – water – a piece of sodium.
Procedure:
a. Add the piece of sodium into a beaker filled with water.
Observation:
a. Sodium reacts with water rapidly.
Conclusion:
a. Sodium reacts with water producing heat and hydrogen gas evolves.
Port Said International Schools
Better Education for Future Generations
Science Department
Grade: 9
The reaction of diluted hydrochloric with copper, aluminum and
zinc:
Tools: 3 beakers – diluted hydrochloric acid – zinc – copper – aluminum.
Procedures:
a. Put diluted hydrochloric acid in each beaker.
b. Add copper to the 1st beaker, aluminum to the 2nd beaker and zinc to the 3rd
beaker.
c. Observe the reaction in each beaker.
Observations:
a. No reaction between copper and the acid.
b. Strong late reaction between aluminum and the acid.
c. Medium reaction between zinc and the acid.
Conclusions:
a. Elements follow hydrogen in the chemical activity series can’t replace hydrogen
such as copper “no reaction”.
b. Elements precede hydrogen in the chemical activity series replaces hydrogen such
as zinc and aluminum.
c. Higher elements in the series react stronger.
d. Although the reaction of aluminum is stronger than the reaction of zinc it is
delayed as aluminum forms thin layer of aluminum oxide on its surface.
Substitution of a material instead of another one of its salt solution:
Tools: beaker – blue copper sulphate solution – pieces of magnesium
Procedures:
a. Add the piece of magnesium to the beaker filled with blue copper sulphate
solution.
Observation:
a. The blue color of copper sulphate solution disappears.
Conclusion:
a. Magnesium replaces copper in its salt forming magnesium sulphate and copper
metal precipitate.
Port Said International Schools
Better Education for Future Generations
Science Department
Grade: 9
The reaction of hydrochloric acid with sodium carbonate:
Tools: hydrochloric acid – sodium carbonate powder – a bottle – a balloon.
Procedures:
a. Put the hydrochloric acid in the bottle.
b. Add the sodium carbonate to the bottle and cover the opening with a balloon.
Observation:
a. The size of the balloon starts to increase.
b. When this gas is tested it turbid the clear lime water.
Conclusions:
a. Hydrochloric acid reacts with sodium carbonate producing sodium chloride,
carbon dioxide and water vapor.
Effect of surface area on the speed of a chemical reaction:
Tools: equal amounts of diluted hydrochloric acid – equal masses of iron one of the
samples is in the form of iron fillings and the other is a piece – two conical flasks – two
syringes.
Procedures:
1. Put iron fillings in one conical flasks and the piece of iron in the other flask.
2. Add equal amount of diluted hydrochloric acid the two flasks.
3. Observe the gas collected in the two syringes.
Observations:
1. The amount of hydrogen gas collected in the 1st flask (the iron fillings) is more than
the hydrogen gas collected from the 2nd flask (the piece of iron).
Conclusion:
1. The reaction rate (speed of reaction) is faster when the surface area exposed to
the reaction is bigger (such as in the case of iron fillings).
The effect of reactants concentration on the speed of reaction:
Tools: 2 pieces of magnesium of the same size – 2 test tubes - diluted hydrochloric acid concentrated hydrochloric acid.
Procedures:
1. Put diluted hydrochloric acid in tube “A”. Put the same amount of concentrated
hydrochloric acid in tube “B”.
2. Put a piece of magnesium in each tube.
Observations:
Port Said International Schools
Better Education for Future Generations
Science Department
Grade: 9
1. The reaction in tube B is faster than tube A.
Conclusions:
1. The speed of reaction increases when the concentration of reactants increases.
The effect the temperature on the speed of reaction:
Tools: 2 glasses – 2 effervescent tablets – hot water – cold water
Procedures:
1. Fill half of the first glass with hot water and the second one with cold water.
2. Add an effervescent tablet to each of the glasses.
Observations:
1. The reaction in the hot water is faster than the reaction in the cold water.
Conclusions:
1. The speed of reaction increases as the temperature of the reaction increases.
The decomposition of hydrogen peroxide (the effect of a catalyst )
on the chemical reaction:
Tools: Hydrogen peroxide – manganese dioxide – 2 test tubes.
Procedures:
1. Put an equal amount of hydrogen peroxide in the two test tubes.
2. Put a small amount of manganese dioxide in one test tube.
Observation:
1. The chemical reaction in the test tube with the catalyst is faster than the reaction
in the other test tube.
Conclusion:
1. The speed of the chemical reaction increases when we use a catalyst.
The effect of enzymes on the speed of chemical reaction:
Tools: Pro hydrogen oxide – a piece of sweet potato – a glass.
Procedures:
1. Fill a half of the glass with pro hydrogen oxide. What do you observe?
2. Put the piece of the sweet potato in the glass. What do you observe?
Observations:
1. In step 1 no a chemical reaction takes place.
2. In step 2 a chemical reaction takes place.
Conclusions:
1. The speed of reaction increases as we use a catalyst.
Port Said International Schools
Better Education for Future Generations
Science Department
Grade: 9
Discover types of mixtures in terms of homogeneity:
Tools: 3 glasses – salt – sand – some oil – a stirring rod – spoon.
Procedure:
1. Put some pure water in each glass.
2. Put a spoon of salt in the first glass, a spoon of sand in the second and a spoon of
oil in the third.
3. Predict which of the previous substance mixes or does not mix with water?
4. Stir each glass and record your observations.
Observations:
1. In some beakers (salt) the molecules of solute disappear, while in other beakers
(oil and sand) the molecules of the solute can be distinguished.
Conclusions:
Mixtures can be classified into:
1. Homogenous mixtures in which the solute molecules are regularly distributed in
the solvent – the molecules can’t be distinguished.
2. Heterogeneous (non-homogenous) mixtures in which the molecules are irregularly
distributed in the solvent – the molecules can be distinguished.
Discover types of solutions in terms of solute concentration:
Tools: Table salt - a glass – flame – stirring rod.
Procedures:
1. Pour 100m distilled water in the glass.
2. Put a little amount of table salt in the glass and stir well.
3. What do you observe? What is the solution called in this case?
4. Continue adding more table salt with stirring so that no more additional amounts
of table salt can be dissolved in the water.
5. What do you observe? What is the name of the solution in this case?
6. Heat the glass and observe what happens to the precipitated table salt?
7. After checking the dissolution of the precipitated table salt, add additional
amounts of table salt and continue heating.
8. What do you observe? What is the name of the solution in this case?
Observations:
1. In step No. 2 we still can dissolve more salt to the water at the same temperature.
2. In step No. 5 no more salt can be dissolved in water at the same temperature.
3. In step No. 8 more table salt can be dissolved in water with increasing
temperature.
Port Said International Schools
Better Education for Future Generations
Science Department
Grade: 9
Conclusions:
1. The unsaturated solution is a mixture in which more solute molecules can be
dissolved in the solvent in the same temperature.
2. The saturated solution is a mixture in which no more solute molecules can be
dissolved in the solvent in the same temperature.
3. Super saturated solution is a mixture in which more solute molecules can be
dissolved in the solvent by rising the temperature.
How the Ammeter is used? And why?
Tools: Ammeter – wires – key – battery – lamp.
Procedures:
1. Create an electric circuit as shown in
2. Close the electric current key.
Observations:
1. The pointer of the Ammeter moves
the circuit is closed.
Conclusions:
1. The reading of the Ammeter is an
of the current intensity.
figure.
when
indicator
Discover the relationship between the current intensity and potential difference:
Activity to determine the value of an unknown resistance:
Tools: Ammeter – voltmeter – battery – wires – unknown resistance – Rheostat.
Procedures:
1. Connect an electric circuit as
shown.
2. Switch On electric current to the
circuit
and observe the current intensity
in the
circuit, and the potential
difference.
3. Change the resistance by using
the
Rheostat and record the readings
of the
Ammeter and voltmeter.
4. Repeat the previous step several times and record your observations.
Observations:
Port Said International Schools
Better Education for Future Generations
Science Department
Grade: 9
1. Every time we change the value of the Rheostat the values of the Ammeter and
voltmeter change.
Conclusions:
𝑉
1. Dividing = constant value = the value of the constant resistance.
𝐼
2. The electric current intensity passing through a conductor is directly proportional
with the difference between its ends when the temperature is constant.
Measuring the electromotive force (e.m.f) of cells connected in
series:
Tools: electric cells – wires – voltmeter.
Procedures:
1. Make an electric circuit consisting of one cell and a voltmeter. Let e.m.f. reading in
the voltmeter in this case be (E1).
2. Connect another cell similar to the first electrode in series. Let e.m.f. reading in
this case be (E2).
3. Connect another similar cell in series. Let us assign the e.m.f. reading in this case
be (E3).
Observations:
1. The e.m.f. in the second case is twice the e.m.f. in the first cast i.e (E2) is twice the
value of (E1).
2. The e.m.f. in the third case is three times the e.m.f. in the first case i.e (E3) equals
three times the value of (E1).
Conclusions:
1. e.m.f. of a battery made up of cells connected in series = the sum of e.m.f. of these
cells i.e E= E1+ E2+ E3
2. e.m.f. of a battery made up of similar cells connected in parallel = e.m.f. of one cell
.i.e E = E1 = E2 = E3
Measuring e.m.f. of cells connected in parallel:
Tools: electric cells – wires – voltmeter.
Procedures:
1. Make an electric circuit consisting of one cell and a voltmeter. Let e.m.f. reading in
the voltmeter in this case be (E1).
Port Said International Schools
Better Education for Future Generations
Science Department
Grade: 9
2. Connect another cell similar to the first electrode in parallel. Let e.m.f. reading in
this case will be (E2).
3. Connect another similar cell in parallel. Let us assign the e.m.f. reading in this case
be (E3).
Observations:
1. The reading in the third case is the same as in the second case and the same in the
first case, i.e E1 = E2 = E3.
Conclusions:
1. The e.m.f. of several cells connected in parallel equals the e.m.f. of one. i.e F (of
the battery) = E1 (of one cell).