JK-TK/Newdata huggingface dataset:
Collect the copper and dry it. Mg(s) + Cu2+(aq) -> Mg2+(aq) + Cu(s) (ii) Repeat the experiment by attempting to displace copper metal using powdered zinc and iron metal. Note the comparative activity of the metals. (b) REACTION OF SODIUM WITH WATER (i) A very safe way of demonstrating the reaction of sodium and water is to drop a very small piece of sodium into a swimming pool. (ii) Pour a 2 cm layer of kerosene on to the surface of water in a test-tube. Drop a 3 mm diameter piece of sodium into the kerosene. Be careful! Sodium sinks in the kerosene and float in the water. Adjust the layer of kerosene to be shallow enough to allow the top of the sodium to protrude above the surface. This reaction between sodium and the water is much slower than if the sodium had been dropped directly on to the water. You can watch the reaction through a hand lens held at the side, but never at the top. Sodium metal is lighter than water but heavier than kerosene. A small area of the sodium suddenly reacts causing a stream of bubbles to appear. The stream of bubbles at one side causes movement. The irregular shape of the sodium changes to a sphere. The sodium melts because the reaction gives off heat. Note any variations in light refraction and reflection below the sodium that indicates something dissolving in the water. Slight smoke where the hot sodium is above the kerosene level suggests a slight reaction with air. Test the gas bubbles for oxygen or hydrogen. (c) DISPLACEMENT OF HYDROGEN FROM ACIDS BY METALS (i) Pour 5 cm of the acids in the table below into test-tubes. Place a piece of metal foil in each test-tube. Note the formation of hydrogen and compare the different rates at which the bubbles are formed.
(c) DISPLACEMENT OF HYDROGEN FROM ACIDS BY METALS (i) Pour 5 cm of the acids in the table below into test-tubes. Place a piece of metal foil in each test-tube. Note the formation of hydrogen and compare the different rates at which the bubbles are formed. Rate of formation of hydrogen gas - very rapid, rapid, slight, very slight, none Metal (b) 3M hydrochloric acid (c) 3M sulphuric acid Magnesium (b) Very rapid (c) Rapid Aluminium (b) Slight (c) None Zinc (b) Moderate (c) Slight Iron (b) Very slight (c) Very slight Tin (b) None (c) None Lead (b) None (c) None Copper (b) None (c) None (ii) Recover the zinc after the reaction has ceased. Evaporate the solution to leave zinc sulphate crystals. Dissolve the colourless zinc sulphate crystals in water and put two carbon electrodes (central poles of dry cell batteries) in the solution. Connect the electrodes to a 6-volt or 12 volt DC supply. Zinc forms rapidly on the cathode. (d) PREPARE SULPHUR DIOXIDE Do the following preparations in a fume cupboard. (i) Prepare sulphur dioxide by burning sulphur in air. Put powdered sulphur in a porcelain jar; ignite it and collecting the gas formed in a funnel. Aspirate the gas into a bottle containing water. (ii) Prepare sulphur dioxide in a generator which allows dilute sulphuric or hydrochloric acid to drip slowly on to sodium sulphate. The acid is contained in a thistle funnel and a tap controls the flow on to sodium sulphite in a suitable flask. The sulphur dioxide produced can be collected in gas jars covered with cardboard discs, which have central holes for the delivery tube. (e) REDUCTION OF POTASSIUM PERMANGANATE WITH SULFUR DIOXIDE (i) Add 10 mL of 0.1M solution of potassium permanganate and 10 mL 3M solution of dilute sulphuric acid to 200 mL of water containing sulphur dioxide. The solution will gradually become colourless as the sulphur dioxide reacts with the permanganate.
The sulphur dioxide produced can be collected in gas jars covered with cardboard discs, which have central holes for the delivery tube. (e) REDUCTION OF POTASSIUM PERMANGANATE WITH SULFUR DIOXIDE (i) Add 10 mL of 0.1M solution of potassium permanganate and 10 mL 3M solution of dilute sulphuric acid to 200 mL of water containing sulphur dioxide. The solution will gradually become colourless as the sulphur dioxide reacts with the permanganate. The experiment can be continued further by stirring in a 0.25M solution of barium chloride when the solution will become "milky" due to the formation of barium sulphate. (ii) The generator used in experiment 2.75B is a convenient piece of apparatus for giving a continuous supply of sulphur dioxide for bleaching Bowers and other plants. The gas from the generator is passed through a jar containing the plant, and excess gas is absorbed in water. The colour of the bleached plant can easily be regenerated by placing the plant in a solution of hydrogen peroxide. This experiment could be used as an introduction to the processes of reduction and oxidation. (f) REACTION OF MAGNESIUM WITH CARBON DIOXIDE Fill a gas jar with carbon dioxide as described in experiment 2.38. Hold a piece of clean magnesium ribbon in a pair of tongs; ignite the magnesium with a Bunsen burner flame and plunge it into the carbon dioxide gas. The magnesium continues to burn. If the magnesium is taking oxygen from the carbon dioxide for burning then you would find carbon in the gas jar. Look for carbon specks in the gas jar. To make the carbon more visible, you can add drops of sulphuric acid to remove the magnesium oxide and any unburnt magnesium. (g) TITRATION OF ACIDS AND BASES Measure exactly 20 drops of a dilute acid such as vinegar and put these into a test-tube. Add one drop of indicator; either methyl orange or phenolphthalein is satisfactory. To this mixture of acid and indicator, add a dilute base drop by drop, and count the drops. Within experimental error, it will always take the same number of drops to neutralize the 20 drops of acid provided that the same dropper is used. A teat pipette makes a satisfactory dropper.
To this mixture of acid and indicator, add a dilute base drop by drop, and count the drops. Within experimental error, it will always take the same number of drops to neutralize the 20 drops of acid provided that the same dropper is used. A teat pipette makes a satisfactory dropper. If the concentration of the acid is known, the concentration of the base can be estimated by comparing the numbers of drops of acid and drops of base that just react. (h) MAKE SOAP FROM FATS Soap can be made from many oils and fats. The reaction is a double displacement involving a strong base such as sodium hydroxide and fats. (a) Obtain animal fat from a butcher. Boil this fat in water and the oil will separate on the surface. When cold, the fat will solidify and it can be separated from the water. Melt the fat again and strain through several layers of cloth. (b) Weigh this fat and then weigh out about one third as much sodium hydroxide pellets. Take care not to touch either the solid sodium hydroxide or the solution, because it is very caustic. Heat the fat in an iron saucepan or dish and, when it is molten, slowly add the sodium hydroxide solution with continuous stirring. Heat with a small flame to avoid boiling over. Allow the fat and the sodium hydroxide to boil for 30 minutes. Stir the mixture frequently. (c) The next stage is to weigh out common salt, sodium chloride; about twice the weight of sodium hydroxide used in (b) is needed. After the 30 minutes boiling, stir this salt well into the mixture. Then allow to cool. The soap separates as a layer at the top. Separate this soap from the liquid below, melt and pour into matchboxes where it will solidify again as small bars of soap. 17. ENERGY FROM CHEMICAL REACTIONS The following group of reactions involve ions in aqueous solution. When the water containing the reacting ions becomes hotter, then we have gained this heat and we can make it do work for us. During the reaction the ions have lost this heat, which we have gained. On the other hand, when the water containing the ions becomes colder, it is the ions which have gained the energy and the water has lost an equivalent amount. (a) REACTIONS THAT GIVE OUT HEAT ENERGY Be careful! The reaction is vigorous so do not do the experiment in a stoppered bottle!
On the other hand, when the water containing the ions becomes colder, it is the ions which have gained the energy and the water has lost an equivalent amount. (a) REACTIONS THAT GIVE OUT HEAT ENERGY Be careful! The reaction is vigorous so do not do the experiment in a stoppered bottle! (i) Put white anhydrous copper sulphate powder to a depth of about 1 cm in a test-tube. Hold a thermometer with the bulb in the powder. Add water drop by drop. Record the changes of the thermometer reading. (ii) Put about 10 mL of strong aqueous copper sulphate solution into a wide test-tube or small beaker. Support a thermometer with the bulb in the solution. Add magnesium powder, or ribbon, a little at a time until the blue colour disappears. Note any changes in the thermometer reading. (iii) To a little water in a wide test-tube, add concentrated sulphuric acid, drop by drop, down the side of the tube. Stir gently with a thermometer after the addition of each drop. Note any changes in the thermometer reading. (b) REACTIONS THAT TAKE IN HEAT ENERGY Put 10 mL of water in a test-tube. Read the temperature of the water. Dissolve about 2 g of potassium nitrate in the water. The temperature should fall through 90oC. This means that, in the process of dissolving in the water, the particles have absorbed energy. This energy has been taken from the surrounding water in the form of heat. A similar result can be obtained by using potassium chloride instead. (c) HEAT OF A NEUTRALIZATION REACTION Dissolve 40 g of sodium hydroxide pellets in water and make up to 500 mL. This is a 2M solution Also prepare 500 mL of a 2M hydrochloric acid solution. Leave the solutions to cool to room temperature. Note the actual temperature of the solutions when cool. Then add the acid to the base quite rapidly and stir with a thermometer. Note the maximum temperature reached. The increase of temperature should be about 13oC. (iii) Since the volume of water has been doubled by adding one solution to the other, the final solution contains 1 mole of OH- (aq) ions which reacted with 1 mole of H+(aq) ions to form 1 mole of water molecules. We must assume that the specific heat of this moderately weak solution is the same as that of water.
The increase of temperature should be about 13oC. (iii) Since the volume of water has been doubled by adding one solution to the other, the final solution contains 1 mole of OH- (aq) ions which reacted with 1 mole of H+(aq) ions to form 1 mole of water molecules. We must assume that the specific heat of this moderately weak solution is the same as that of water. (d) HEAT OF A COPPER DISPLACEMENT REACTION (i) Put 25 mL 0.2 M copper sulphate solution in a 100 mL polythene fitted with a 1-hole stopper and thermometer. Replace the stopper, invert the bottle and shake it gently. Record the temperature of this solution. Turn the bottle the right way up, remove the stopper and add 0.5 g of zinc dust. The quantity of zinc powder is in excess to ensure that all the copper sulphate is used up in the reaction, so some zinc will remain when the reaction stops. Replace the stopper, invert the bottle, and shake gently. Record the highest temperature reached. Calculate the rise of temperature. This rise of temperature in not affected by the volume of 0.2 M copper sulphate used for the experiment. For a 1 M solution, multiply the rise in temperature by 5 (5 X 0.2M = 1.0 M). The reactants lost energy to the solution. The temperature change is usually between 9oC and 10oC. Zn(s) + Cu2+(aq) --> Zn2+(aq) + Cu(s) (ii) Repeat the experiment with 0.5 g of iron powder or iron filings. This amount is again an excess so that all the copper sulphate will be used up in the reaction. The temperature change is usually between 6oC and 7oC. (e) ELECTRICAL ENERGY FROM CHEMICAL REACTIONS - ELECTROCHEMICAL CELLS In the preceding experiment zinc metal became zinc ions and copper ions became copper metal due to transfer of electrons from zinc metal to the copper ion. To get electrical energy these electrons must flow in an external conductor from the zinc to copper. The potential or voltage will reflect the greater activity of zinc over copper. The current flowing will depend on the extent and rate of the reaction. 2.84 Electrical energy from the displacement of copper by zinc Put concentrated copper sulphate solution in a beaker.
The potential or voltage will reflect the greater activity of zinc over copper. The current flowing will depend on the extent and rate of the reaction. 2.84 Electrical energy from the displacement of copper by zinc Put concentrated copper sulphate solution in a beaker. Connect copper foil to the positive terminal of a 5 V voltmeter. Connect zinc foil to the negative terminal. Dip the two metals briefly into the copper sulphate solution. Note any changes in the voltmeter reading. Note the maximum reading. Note any changes at the copper foil and the zinc foil. The voltage falls to zero after a short time because copper deposited on the zinc and caused the reaction to stop. 18. RATE OF REACTION(a) SIZE OF PARTICLES AND RATE OF REACTION Marble chips can be broken up with a hammer and graded into 3 or 4 sizes: (a) coarse powder; (b) pieces about half the size of a rice grain; (c) pieces as large as rice grains; and (d) the original lumps of marble chips. Place four 100 X 16 mm test tubes in a stand. Weigh approximately 2 g of each grade, size, of marble chips and put the four grades separately into each of the four tubes. Obtain four balloons and blow them up several times to stretch them. Put 5 mL of bench hydrochloric acid into each of the four balloons and slip the mouth of the balloon over the top of the tube without letting any acid into the tube. When each balloon is in place, tip the acid into each test-tube at the same time and observe which balloon is the fastest and the slowest to be blown up. The smallest should give the carbon dioxide in the shortest time. Instead of using marble chips you can use granulated zinc. Do not use metals in powder form because the reaction may be too vigorous and even cause an explosion. Be careful! This reaction produces hydrogen gas! Instead of collecting the gas in a balloon or plastic bag, a more accurate method would be to collect the gas in a burette inverted over water and compare the volume of gas given off in unit time for each grade of marble chips. Another accurate method is to stand a conical flask containing the marble chips and acid on a balance and record the loss in mass every half minute. Carbon dioxide is a heavy gas and most balances will enable the loss in mass to be found as the gas escapes.
Instead of collecting the gas in a balloon or plastic bag, a more accurate method would be to collect the gas in a burette inverted over water and compare the volume of gas given off in unit time for each grade of marble chips. Another accurate method is to stand a conical flask containing the marble chips and acid on a balance and record the loss in mass every half minute. Carbon dioxide is a heavy gas and most balances will enable the loss in mass to be found as the gas escapes. (b) CONCENTRATION AND RATE OF REACTION The reaction between sodium thiosulphate and hydrochloric acid can take a noticeable time. Sulphur is produced during the reaction making the solution cloudy. The rate of reaction can be found by finding the time taken to reach a certain degree of cloudiness in the solution. The degree of cloudiness in this case may be defined as the point at which a black cross marked below the reaction vessel can no longer be seen by looking through the solution from above. In this experiment the concentration of sodium thiosulphate is made variable, whilst the concentration of acid is kept constant. Sodium thiosulphate may be bought as "hypo" which is used in photography. Make up 500 mL of aqueous solution containing 20 g sodium thiosulphate. 2 M hydrochloric acid is also needed. Bench dilute acid is usually of this strength. Using a measuring cylinder, put 50 mL of thiosulphate solution into a 100 mL beaker. Place the beaker on a black cross marked on a sheet of paper. Add 5 mL of the acid and note the time given by the second hand of a clock. Stir the acid into the solution. Note the time when the cross is no longer visible through the sulphur in the solution. Repeat the experiment with a smaller concentration of thiosulphate. Take 40 mL of thiosulphate solution and add 10 mL of distilled water. Stir and then add 5 mL of acid as before. The time for the cross to become invisible should be greater than for the last experiment. Repeat the experiment using 30 mL, 20 mL and 10 mL of this sulphate mixed with 20 mL, 30 mL and 40 mL of distilled water. Plot concentration of the thiosulphate solution against time taken for the reaction.
The time for the cross to become invisible should be greater than for the last experiment. Repeat the experiment using 30 mL, 20 mL and 10 mL of this sulphate mixed with 20 mL, 30 mL and 40 mL of distilled water. Plot concentration of the thiosulphate solution against time taken for the reaction. Concentration values may be taken as the volume of the original thiosulphate solution used. Since I/time, reciprocal of time, is the measure of the rate of the reaction, plot thiosulphate concentrations against I/time. The equation for the reaction can be written as: Na2S2O3 (aq) + 2HCl(aq) --> H2O (l) + SO2 (g) + S(s)
(c) TEMPERATURE AND RATE OF A REACTION Use the reaction in experiment 2.92 to investigate the effect of temperature. Put 10 mL of sodium thiosulphate solution into the 100 mL beaker and stir in 40 mL of water. Use this concentration for the series of experiments with the temperature of the solution as the variable. Add 5 mL of acid as before and record the initial time and the temperature of the solution. Record the final time when the black cross below the beaker is no longer visible. Repeat the experiment, each time warming the thiosulphate solution to just over 30oC
(c) CATAL YSTS AND RATE OF REACTION The variable in this reaction is the substance used as a catalyst in the decomposition of an aqueous solution of hydrogen peroxide. Set up the burette filled with water as in a standard water displacement experiment. 2 mL of 20-volume hydrogen peroxide will give enough oxygen almost to fill the burette. Weigh out 1 g each of copper (II) oxide, nickel oxide, manganese (lV) oxide and zinc oxide. Put 50 mL of water in the flask and add 2 mL of hydrogen peroxide solution. Add the 1 g of copper oxide. Immediately insert the bung with the delivery tube into the flask. Time the volume of oxygen given off at intervals of 15 seconds. Plot the volume of oxygen produced every 15 seconds against the time of the reaction. Repeat the experiment using the other oxides as catalysts. Plot a graph for each experiment. Manganese (IV) oxide is usually used as a catalyst in this reaction. The catalyst is not used up during the reaction.
Plot a graph for each experiment. Manganese (IV) oxide is usually used as a catalyst in this reaction. The catalyst is not used up during the reaction. A catalyst may slow down a reaction as well as speed it up. 19. BREAKDOWN OF STARCH TO SUGAR FEHLING'S TEST FOR REDUCING SUGARS Starch can be recognized by the deep blue colour which develops when it is in contact with iodine solution. This is a very sensitive test. Sugar does not react with iodine, but sugar will reduce copper (II) in Fehling's solution to red copper (I) oxide, and this is also a sensitive test. Starch does not react with Fehling's solution. Saliva contains enzyme catalysts, which convert starch to sugar. This experiment investigates the progress of this reaction. Put about 10 mL of dilute starch solution into a test-tube. Add to this 1 mL of saliva and stir this into the starch solution. Record the time of adding the saliva. At 5-minute intervals, remove 2 or 3 drops by means of a dropper and put them on a clean white tile, taking care to keep them from running into each other. The dropper must be washed well between each test. Put a little iodine solution on each drop. The decreasing intensity of the blue colour indicates that starch is being used up. Test for increasing amounts of sugar at the same time as testing for starch. To do this, put 2 or 3 drops of the reaction mixture into a small test-tube. Add 3 mL of Fehling's solution and warm this mixture almost to boiling point. The test should show that the amount of sugar is increasing. The enzyme in the saliva is therefore slowly breaking starch down into sugar, which is a smaller molecule. In a previous experiment yeast was used to break down sugar into ethanol, which is an even smaller molecule. Living yeast, which is a variety of fungus, produces enzymes which act as catalysts in the conversion: C6H12O6 -> 2C2H5OH + 2CO2
20. BREAKDOWN OF ETHANOL TO ETHENE (ETHYLENE) Absorb ethanol on to cotton wool or asbestos wool and push this to the bottom of a hard glass test-tube. In the middle of the test-tube pack small pieces of unglazed porcelain. Fit a delivery tube to collect ethene gas over water.
BREAKDOWN OF ETHANOL TO ETHENE (ETHYLENE) Absorb ethanol on to cotton wool or asbestos wool and push this to the bottom of a hard glass test-tube. In the middle of the test-tube pack small pieces of unglazed porcelain. Fit a delivery tube to collect ethene gas over water. Have 3 test-tubes ready to collect ethene. First heat the porous pot strongly and then gently warm the cotton wool to produce ethanol vapour. This vapour will break down over the hot porous pot to produce ethene gas and water vapour. The ethene is insoluble in water, unlike ethanol, and will collect in the test-tubes. Test the 3 samples by: (a) burning ethene; (b) shaking with a few drops of dilute potassium permanganate solution made alkaline with sodium carbonate solution, the colour should disappear; (c) shaking with a little bromine water, the colour again disappears. Disconnect the delivery tube when you stop heating to avoid a suck back of water onto the hot porous pot. 20. BREAKDOWN OF POLYMERS Usually the smallest molecules are gaseous or liquid at room temperature and the large molecules are solids. Perspex and polystyrene are solid polymers, which can be broken down to smaller molecules by heat. Put pieces of Perspex or polystyrene in a hard glass test-tube. Connect a delivery tube as 2.97 From large molecules to small molecules A Perspex or polystyrene B receiving tube c cold water D a liquid collects. The collecting test-tube must be cooled thoroughly with cold water, as the fumes are harmful. Gently heat the test-tube containing the Perspex. The polymer will melt and give off vapours, which are collected in the receiving tube. Heating must be carefully controlled to enable all the fumes to be condensed in the receiving tube. A liquid is obtained. This suggests that the polymer has been broken down by heat to smaller molecules. The liquid does not return to the solid state unless a catalyst is used. The specific catalysts are not usually available in school laboratories. GENERAL EXAMINATION REQUIREMENTS
Apparatus List
This list given below has been drawn up in order to give guidance to schools concerning the apparatus that is expected to be generally available per student for examination purposes.