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Activity 1.1 (Experiment):
Aim: To show the reaction between magnesium and oxygen present in the air.
Apparatus: Mg ribbon, watch – glass, pair of tongs, burner, sand paper
Procedure: Clean a magnesium ribbon about 2 cm long by rubbing it with sandpaper. Hold it with a pair of tongs Burn it using a burner and collect the ashes so formed in a watch – glass as shown in the fig.
while burning the magnesium ribbon keep it as far as possible from your eyes.
Observation: magnesium ribbon burns with a dazzling white flame and changes into a white powder.
Activity 1.2 (Experiment):
Aim: To show change in state and colour when a chemical reaction takes place.
Apparatus: lead nitrate solution, potassium iodide solution, test tube, and a conical flask.
Procedure: Take lead nitrate solution in a test tube Add potassium iodide solution to this.
Observation: lead nitrate reacts with potassium iodide to give lead iodide and potassium nitrate.
The colour of the products is different from that of the reactants.
Conclusion: Pb(NO3)2 + 2KI → PbI2 + 2KNO3.
Aim: To show the reaction between magnesium and oxygen present in the air.
Apparatus: Mg ribbon, watch – glass, pair of tongs, burner, sand paper
Procedure: Clean a magnesium ribbon about 2 cm long by rubbing it with sandpaper. Hold it with a pair of tongs Burn it using a burner and collect the ashes so formed in a watch – glass as shown in the fig.
while burning the magnesium ribbon keep it as far as possible from your eyes.
Observation: magnesium ribbon burns with a dazzling white flame and changes into a white powder.
This powder is magnesium oxide Conclusion: MgO is formed due to the reaction between magnesium and oxygen present in the air.
Equation, 2Mg + O2 → 2MgO
Aim: To show change in state and colour when a chemical reaction takes place.
Apparatus: lead nitrate solution, potassium iodide solution, test tube, and a conical flask.
Procedure: Take lead nitrate solution in a test tube Add potassium iodide solution to this.
Observation: lead nitrate reacts with potassium iodide to give lead iodide and potassium nitrate.
The colour of the products is different from that of the reactants.
Conclusion: Pb(NO3)2 + 2KI → PbI2 + 2KNO3.
The change in colour is because lead iodide and potassium nitrate have been formed after the reaction. PbI2 is a yellow ppt. KNO3 is a colourless solution
Aim: To show the interaction between zinc granules and hydrochloric acid or sulphuric acidformation of H gas)
Apparatus: Zinc granules, conical flask, test tube, hydrochloric acid / sulphuric acid
Procedure: Take a few zinc granules in a conical flask or a test tube. Add dilute hydrochloric acid or sulphuric acid to this.
Touch the conical flask or test tube.
Observation: A gas is evolved and the conical flask is warm
Conclusion: From this we conclude that when a chemical reaction takes place there is a change in the state, colour, and temperature.
Zn + 2HCl → ZnCl2 + H2↑ + heat
Zn + H2SO4 → ZnSO4 + H2↑ + heat
Activity 1.4 (Experiment):
Aim: Formation of slaked lime by the reaction of calcium oxide with water
Apparatus: Calcium oxide, beaker, and water
Procedure: Take a small amount of calcium oxide or quick lime in a beaker. Slowly add water to this. Touch the beaker
Observation: The beaker becomes hot (exothermic reaction). It reacts vigorously with water.
Conclusion: Slaked lime is formed by the reaction of the combination of calcium oxide with water
CaO(s) + H2O(l) → Ca(OH)2(aq) + heat
(quick lime) (Slaked lime) (exothermic reaction)
This is a combination reaction, and also an exothermic reaction.
Aim: To show a decomposition reaction
Apparatus: ferrous sulphate crystals, dry boiling tube and burner
Procedure: Take about 2g ferrous sulphate crystals in a dry boiling tube. Note the colour of the ferrous sulphate crystals. Heat the boiling tube over the flame of a burner or spirit lamp observe the colour of the crystals after heating
Observation: The green colour of the ferrous sulphate crystals has changed there is also the characteristic odour of burning sulphur.
Conclusion: This is a Decomposition reaction, shown by the reaction
2FeSO4(s) + Heat → Fe2O3 + SO2(g) + SO3(g)
Aim: Heating of lead nitrate and emission of nitrogen dioxide
Apparatus: lead nitrate powder, boiling tube, pair of tongs and burner Procedure: Take about 2 g lead nitrate powder in a boiling tube. Hold the boiling tube with a pair of tongs and heat it over a flame. Observe the change if any.
Observation: We observe the emission of brown fumes.
Conclusion: These fumes are of nitrogen dioxide (NO2). The reaction that takes place is
2Pb(NO3)2(s) ---Heat→ 2PbO(s) + 4NO2(g) + O2(g)
(Lead nitrate) (Lead oxide) (Nitrogen dioxide) (Oxygen)
Activity 1.7 (Experiment):
Aim: To show that water is a compound containing two atoms of hydrogen and 1 atom
of oxygen
Apparatus: Plastic mug, rubber stoppers, carbon electrodes, 6 volt battery, water, dilute sulphuric acid, burning candle
Procedure: Take a plastic mug. Drill two holes at its base and fit rubber stoppers in these
holes. Insert carbon electrodes in these rubber stoppers as shown in fig then connect these
electrodes to a 6 volt battery. Fill the mug with water such that the electrodes are immersed.
Add a few drops of dilute sulphuric acid to the water. Take two test tubes filled with water and
invert them over the two carbon electrodes. Switch on the current and leave the apparatus undisturbed for some time. Observe the formation of bubbles at both the electrodes. These bubbles displace water in the test tubes. The volume of the gas collected is not the same in both the test tubes. Once the test tubes are filled with the respective gases, remove them carefully.
Test these gases one by one by bringing a burning candle close to the mouth of the test tubes.
Observation: We observe the formation of bubbles at both the electrodes. These bubbles displace water in the test tubes. The volume in one test tube is twice that in the other test tube. `The test tube containing hydrogen gas has double the volume of the test tube containing oxygen gas. Hydrogen gas burns with a light blue flame with a pop sound.
Conclusion: Water is a compound containing two atoms of hydrogen and 1 atom of oxygen.
Aim: To show how silver chloride is affected by sunlight
Apparatus: silver chloride and a china dish.
Procedure: Place this china dish containing silver chloride in sunlight for some time. Observe the colour of the silver chloride after some time.
Observation: White silver chloride turns grey in sunlight.
Conclusion: This is due to the decomposition of silver chloride into silver and chlorine by light
2AgCl(s) ----Sunlight→ 2Ag(s) + Cl2(g)
Activity 1.9 (Experiment):
Aim: To show displacement reaction
Apparatus: two iron nails, sand paper, copper sulphate solution, thread and two test tubes.
Procedure: Take two iron nails and clean them by rubbing them with sand paper. Take two test
tubes marked as (A) and (B). In each test tube, take about 10 ml copper sulphate solution.
tubes marked as (A) and (B). In each test tube, take about 10 ml copper sulphate solution.
Tie one iron nail with a thread and immerse it carefully in the copper sulphate solution in test tube A
for about 20 minutes. Keep one iron nail aside for comparison.
for about 20 minutes. Keep one iron nail aside for comparison.
After 20 minutes, take out the iron nail from the copper sulphate solution. Compare the intensity of the blue colour of copper sulphate solutions in the test tubes (A) and (B). Also, compare the colour of the iron nail dipped in the copper sulphate solution with the one kept aside.
Observation: the iron nail dipped in the copper sulphate solution in test tube A becomes brownish in colour and the blue colour of copper sulphate solution fades in test tube A. While the blue colour of copper sulphate solution in test tube B remains the same.
Conclusion: iron has displaced or removed another element, copper, from copper sulphate solution in test tube A. This reaction is known as displacement reaction.
Fe(s) + CuSO4(aq) → FeSO4(aq) + Cu(s)
[copper sulphate] [Iron sulphate]
[copper sulphate] [Iron sulphate]
Aim: To show double displacement reaction
Apparatus: 3 ml of sodium sulphate, 2 test tubes, 3 ml of barium chloride solution.
Procedure: Take about 3 ml of sodium sulphate solution in a test tube. In another test tube, take about 3 ml of barium chloride solution, Mix the two solutions.
Observation: A white substance, which is insoluble in water, is formed. This insoluble substance formed is known as a precipitate. Any reaction that produces a precipitate can be called a precipitation reaction.
Conclusion: This is a double displacement reaction shown by the equation below where Na+ being more reactive than Ba+2 displaces Ba+2 from its compound BaCl2 and takes its place to form NaCl.
Na2SO4(aq) + BaCl2(aq) → BaSO4(s) + 2NaCl(aq)
(sodium sulphate) (Barium chloride) (Barium sulphate) (Sodium chloride)
(sodium sulphate) (Barium chloride) (Barium sulphate) (Sodium chloride)
Aim: To show oxidation and reduction reaction
Apparatus: China dish 1g copper powder, burner, wire gauze, tripod stand and Hydrogen gas.
Procedure:
Heat a china dish containing about 1 g copper powder.
Now pass hydrogen gas over this heated material.
Heat a china dish containing about 1 g copper powder.
Now pass hydrogen gas over this heated material.
Observation:
The surface of copper powder becomes coated with black copper oxide.
Hydrogen gas is passed over this heated material [CuO] the black coating on the surface turns brown as the reverse reaction takes place and copper is obtained.
Conclusion: In the first case copper, was oxidized to copper oxide by the following reaction
2Cu + O2 -----Heat→ 2CuO
this is an oxidation reaction (black coating)
In the second case when hydrogen gas was passed over heated [CuO] copper metal was obtained because copper oxide got reduced to copper metal by the following reaction
