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The diagram below shows a common charcoal burner .Assume the burning take place in a room with sufficient supply of air. (a)Explain what happens around:
28 (i)Layer A Sufficient/excess air /oxygen enter through the air holes into the burner .It reacts with/oxidizes Carbon to carbon(IV)oxide Chemical equation C(s) + O2(g) -> CO2 (g) (ii)Layer B Hot carbon(IV)oxide rises up and is reduced by more carbon/charcoal to carbon (II)oxide. Chemical equation C(s) + CO2(g) -> 2CO (g) (ii)Layer C Hot carbon(II)oxide rises up and burns with a blue flame to be oxidized by the excess air to form carbon(IV)oxide. 2CO (g) + O2(g) -> 2CO2(g) (b)State and explain what would happen if the burner is put in an enclosed room. The hot poisonous /toxic carbon(II)oxide rising up will not be oxidized to Carbon(IV)oxide. (c)Using a chemical test , describe how you would differentiate two unlabelled black solids suspected to be charcoal and copper(II)oxide. Method I -Burn/Ignite the two substances separately. -Charcoal burns with a blue flame - Copper(II)oxide does not burn Method II -Add dilute sulphuric(VI)acid/Nitric(V)acid/Hydrochloric acid separately. -Charcoal does not dissolve. - Copper(II)oxide dissolves to form a colourless solution. 5. Excess Carbon(II)oxide was passed over heated copper(II)oxide as in the set up shown below for five minutes. 29 Dry carbon(IV)oxideHEATCopper(II)oxideBlue flame A (a)State and explain the observations made in the combustion tube. Observation Colour change from black to brown Explanation Carbon (II)oxide reduces black copper(II)oxide to brown copper metal itself oxidized to Carbon(IV)oxide. Chemical equation CO(g) + CuO (s) -> Cu(s) + CO2(g) (black) (brown) (b) (i)Name the gas producing flame A Carbon(II)oxide (ii)Why should the gas be burnt?
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29 Dry carbon(IV)oxideHEATCopper(II)oxideBlue flame A (a)State and explain the observations made in the combustion tube. Observation Colour change from black to brown Explanation Carbon (II)oxide reduces black copper(II)oxide to brown copper metal itself oxidized to Carbon(IV)oxide. Chemical equation CO(g) + CuO (s) -> Cu(s) + CO2(g) (black) (brown) (b) (i)Name the gas producing flame A Carbon(II)oxide (ii)Why should the gas be burnt? It is toxic/poisonous (iii)Write the chemical equation for the production of flame A 2CO(g) + O2(g) -> 2CO2(g) (c)State and explain what happens when carbon(IV)oxide is prepared using Barium carbonate and dilute sulphuric(VI)acid. Reaction starts then stops after sometime producing small/little quantity of carbon(IV)oxide gas. Barium carbonate react with dilute sulphuric(VI)acid to form insoluble Barium sulphate(VI) that cover/coat unreacted Barium carbonate stopping further reaction to produce more Carbon(IV)oxide. 30 (d) Using dot () and cross(x) to represent electrons show the bonding in a molecule of : (i) Carbon(II)oxide [email protected]●xxOxxCx ●●●lone pairs of electrons in carbon and oxygen atoms6 bonded pairs of electrons2covalent bonds1 dative bond (ii) Carbon(IV)Oxide. [email protected]●x O●●C●x●x●xlone pairs of electrons in oxygen atom4 bonded pairs of electronsO●●●●●●CO2 (triatomic molecule) (e) Carbon (IV)oxide is an environmental pollutant of global concern. Explain. -It is a green house gas thus causes global warming. -It dissolves in water to form acidic carbonic acid which causes “acid rain” (f)Explain using chemical equation why lime water is used to test for the presence of Carbon (IV) oxide instead of sodium hydroxide.
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Explain. -It is a green house gas thus causes global warming. -It dissolves in water to form acidic carbonic acid which causes “acid rain” (f)Explain using chemical equation why lime water is used to test for the presence of Carbon (IV) oxide instead of sodium hydroxide. 31 Using lime water/calcium hydroxide: - a visible white precipitate of calcium carbonate is formed that dissolves on bubbling excess Carbon (IV) oxide gas Chemical equation Ca(OH)2(aq) + CO2 (g) -> CaCO3 (s) + H2O(l) (white precipitate) CaCO3 (aq) + H2O(l) + CO2 (g) -> Ca(HCO3) 2 (aq) Using sodium hydroxide: - No precipitate of sodium carbonate is formed Both sodium carbonate and sodium hydrogen carbonate are soluble salts/dissolves. Chemical equation 2NaOH (aq) + CO2 (g) -> Na2CO3 (s) + H2O(l) (No white precipitate) Na2CO3 (s) + H2O(l) + CO2 (g) -> 2NaHCO3 (s) (g)Ethan-1,2-dioic acid and methanoic acid may be used to prepare small amount of carbon(II)oxide in a school laboratory. (i) Explain the modification in the set up when using one over the other. Before carbon(II)oxide is collected: -when using methanoic acid, no concentrated sodium/potassium hydroxide is needed to absorb Carbon(IV)oxide. -when using ethan-1,2-dioic acid, concentrated sodium/potassium hydroxide is needed to absorb Carbon(IV)oxide. (ii)Write the equation for the reaction for the formation of carbon(II)oxide from: I.Methanoic acid. Chemical equation HCOOH(aq) -> CO(g) + H2O(l) II. Ethan-1,2-dioic acid Chemical equation HOOCCOOH(aq) -> CO2(g)+CO(g)+H2O(l) (h)Both carbon(II)oxide and carbon(IV)oxide affect the environment. Explain why carbon(II)oxide is more toxic/poisonous. -Both gases are colourless,denser than water and odourless.
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Ethan-1,2-dioic acid Chemical equation HOOCCOOH(aq) -> CO2(g)+CO(g)+H2O(l) (h)Both carbon(II)oxide and carbon(IV)oxide affect the environment.
Explain why carbon(II)oxide is more toxic/poisonous.
-Both gases are colourless,denser than water and odourless.
-Carbon(II)oxide is preferentially absorbed by human/mammalian haemoglobin when inhaled forming stable carboxyhaemoglobin instead of oxyhaemoglobin.This reduces the free haemoglobin in the blood leading to suffocation and quick death.
-Carbon(IV)oxide is a green house gas that increases global warming.
-Carbon(II)oxide is readily oxidized to carbon(IV)oxide 6.Study the flow chart below and use it to answer the questions that follow.
32 (a)Name: (i)the white precipitate A Calcium carbonate (ii) solution B Calcium hydrogen carbonate (iii) gas C Carbon(IV)oxide (iv) white residue B Calcium oxide (v) solution D Calcium hydroxide/lime water (b)Write a balanced chemical equation for the reaction for the formation of: (i) the white precipitate A from solution D Chemical equation Ca(OH)2(aq) + CO2 (g) -> CaCO3 (s) + H2O(l) (ii) the white precipitate A from solution B Chemical equation Ca(HCO3)2(aq) -> CO2 (g) + CaCO3 (s) + H2O(l) (iii) solution B from the white precipitate A Chemical equation CO2 (g) + CaCO3 (s) + H2O(l) -> Ca(HCO3)2(aq)
33 (iv) white residue B from the white precipitate A Chemical equation CaCO3(s) -> CO2 (g) + CaO (s) (iv) reaction of white residue B with water Chemical equation CaO (s) + H2O(l) -> Ca(OH)2(aq)
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PRACTICAL CHEMISTRY
Introduction
Scientific subjects are, by their nature, experimental. It is accordingly important that an assessment of a student’s knowledge and understanding of Chemistry should contain a component relating to practical work and experimental skills. This booklet has been produced to help students preparing for and taking practical Examinations. The material contained in this booklet does not extend the curriculum specification content. Rather it seeks to help the candidate succeed in practical examination by explaining in more depth what is required of him or her in carrying out the exercises, making observations and measurements with appropriate precision and recording these methodically. This booklet advices candidate on how he or she should interpret, explain, evaluate and communicate the results of the exercises clearly and logically using relevant chemical knowledge and understanding and using appropriate specialist vocabulary. 1.1APPARATUS IN A CHEMISTRY LABORATORY
1.2 SPECIAL LABORATORY APPARATUS AND TECHNIQUES
(a) USING THE BURETTE
A burette is used to deliver solution in precisely measured, variable volumes. Burettes are used primarily for titration, to deliver one reactant until the precise end point of the reaction is reached. To fill a burette, close the stopcock at the bottom and use a funnel. You may need to lift up on the funnel slightly, to allow the solution to flow in freely. You can also fill a burette using a disposable transfer pipette. This works better than a funnel for the small, 10 mL burettes. Be sure the transfer pipette is dry or conditioned with the titrant, so the concentration of solution will not be changed. Before titrating, condition the burette with titrant solution and check that the burette is flowing freely. To condition a piece of glassware, rinse it so that all surfaces are coated with solution, then drain. Conditioning two or three times will insure that a stray drop of water does not change the concentration of titrant.Rinse the tip of the burette with water from a wash bottle and dry it carefully. After a minute, check for solution on the tip to see if your burette is leaking. The tip should be clean and dry before you take an initial volume reading.When your burette is conditioned and filled, with no air bubbles or leaks, take an initial volume reading. A burette-reading card with a black rectangle can help you to take a more accurate reading. Read the bottom of the meniscus.
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The tip should be clean and dry before you take an initial volume reading.When your burette is conditioned and filled, with no air bubbles or leaks, take an initial volume reading. A burette-reading card with a black rectangle can help you to take a more accurate reading. Read the bottom of the meniscus. Be sure your eye is at the level of meniscus, not above or below. Reading from an angle, rather than straight on, results in a parallax error. Deliver solution to the titration flask by turning the stopcock. The solution should be delivered quickly until a couple of mL from the endpoint. The endpoint should be approached slowly, a drop at a time. Use a wash bottle to rinse the tip of the burette and the sides of the flask. Your TA can show you how to deliver a partial drop of solution, when near the endpoint. (b) Volumetric (measuring) Flasks
Erlenmeyer flasks and beakers are used for mixing, transporting, and reacting, but not for accurate measurements. The volumes stamped on the sides are approximate and accurate to within about 5%. (c) Graduated Cylinders
Graduated cylinders are useful for measuring liquid volumes to within about 1%. They are for general-purpose use, but not for quantitative analysis. If greater accuracy is needed, use a pipette or volumetric flask. (d) Pipette
A pipette is used to measure small amounts of solution very accurately. A pipette bulb is used to draw solution into the pipette. Start by squeezing the bulb in your preferred hand. Then place the bulb on the flat end of the pipette.Place the tip of the pipette in the solution and release your grip on the bulb to pull solution into the pipette. Draw solution in above the mark on the neck of the pipette. If the volume of the pipette is larger than the volume of the pipette bulb, you may need to remove the bulb from the pipette and squeeze it and replace it on the pipette a second time, to fill the pipette volume completely. Quickly, remove the pipette bulb and put your index finger on the end of the pipette. Gently release the seal made by your finger until the level of the solution meniscus exactly lines up with the mark on the pipette. Practice this with water until you are able to use the pipette and bulb consistently and accurately.
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Quickly, remove the pipette bulb and put your index finger on the end of the pipette. Gently release the seal made by your finger until the level of the solution meniscus exactly lines up with the mark on the pipette. Practice this with water until you are able to use the pipette and bulb consistently and accurately. (e) Volumetric Flask
A volumetric flask is used to make up a solution of fixed volume very accurately. This volumetric flask measures 500 mL ± 0.2 mL. This is a relative uncertainty of 4 x 10-4 or 400 parts per million. To make up a solution, first dissolve the solid material completely, in less water than required to fill the flask to the mark. After the solid is completely dissolved, very carefully fill the flask to the 500 mL mark. Move your eye to the level of the mark on the neck of the flask and line it up so that the circle around the neck looks like a line, not an ellipse. Then add distilled water a drop at a time until the bottom of the meniscus lines up exactly with the mark on the neck of the flask. Take care that no drops of liquid are in the neck of the flask above the mark. After the final dilution, remember to mix your solution thoroughly, by inverting the flask and shaking. (f) Analytical Balance
An analytical balance measures masses to within 0.0001 g. Use these balances when you need this high degree of precision.Turn the balance on by pressing the control bar. The display lights up for several seconds, then resets to 0.0000.Place creased, small weighing paper on the balance pan.Close the sliding glass doors. Wait for the green dot on the left to go out. This is the stability indicator light, indicating that the weight is stable.Press the control bar to cancel out the weight of the container or paper. The display will again read 0.0000. Carefully add the substance to be weighed up to the desired mass. Do not attempt to reach a particular mass exactly. Before recording the mass, close the glass doors and wait until the stability detector lamp goes out. Record mass of solid. Don't pick up tare containers with bare hands since your fingerprints add mass. Use tongs to prevent this.Don't lean on the bench while weighing.Do record the mass of your container, if you will need it later.Do check the level indicator bubble before weighing.
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Record mass of solid. Don't pick up tare containers with bare hands since your fingerprints add mass. Use tongs to prevent this.Don't lean on the bench while weighing.Do record the mass of your container, if you will need it later.Do check the level indicator bubble before weighing. The two rear balance feet serve as levelling screws. Use the brush provided to clean spills in the weighing chamber. Discard any disposable tare containers or weighing paper, in the nearest wastebasket.(g) Calorimetry
Calorimetry is used to determine the heat released or absorbed in a chemical reaction. The calorimeters shown here can determine the heat of a solution reaction at constant (atmospheric) pressure. The calorimeter is a double Styrofoam cup fitted with a plastic top in which there is a hole for a thermometer. (It's crude, but very effective!) Key techniques for obtaining accurate results are starting with a dry calorimeter, measuring solution volumes precisely, and determining change in temperature accurately. (h) Using a Calorimeter
Solutions volumes should be carefully measured with a graduated cylinder. Add solution completely, to a dry calorimeter. Don't forget to add the spin bar each time! Set up the calorimeter with the thermometer (0° to 50°C, graduated every 0.1°C) supported from a stand so that the bulb does not touch the bottom of the cup. Note that the thermometer used for calorimetry differs from the less accurate one in your glassware drawer. Clamp the calorimeter so that it rests on the stirrer. Be careful not to turn on the heat or you will melt the Styrofoam. The change in temperature is determined by measuring the initial temperature, T1, of the reactants, and the maximum temperature, T2, of the contents of the calorimeter during the exothermic reaction. Use a magnifying glass to measure temperature values precisely.Interpolate between the divisions of the thermometer and record temperatures to +/- 0.01 °C. See your lab manual for a discussion of how to determine accurately the change in temperature from your graph of temperature vs. time. (i) Top-loading Balance
Use a top loading balance to weigh solid material when a precision of 0.1 g is adequate. For more accurate mass measurements or small amounts, use an analytical balance. Using a Top-loading Balance
Check if the balance is turned on.
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(i) Top-loading Balance
Use a top loading balance to weigh solid material when a precision of 0.1 g is adequate. For more accurate mass measurements or small amounts, use an analytical balance. Using a Top-loading Balance
Check if the balance is turned on. If not, press the on/off button and wait until the display reads 0.0 g. Place a container or large, creased weighing paper on the balance pan. Push tare button to zero the balance. Carefully add substance to the container or paper. Record the mass. Use the brush provided to clean any spills. Discard any disposable tare containers or weighing paper in the nearest wastebasket. GENERAL EXPERIMENTAL PROCESSES
Quantitative Transfer
Quantitative Transfer simply means that all the material to be transferred from one place to another must make the trip. For example, every particle of solid must be transferred from the weighing paper to the (clean) beaker.This can be done be carefully tipping the creased weighing paper to pour the solid into the beaker. Tapping the paper with a spatula will knock particles into the beaker.Finally, the paper should be rinsed into the beaker, to remove all traces of the solid. Transferring a Solution or Mixture
If you are transferring a solution or heterogeneous mixture to another vessel, rinse the container with solvent to be sure the transfer is quantitative. The rinsing should be transferred to the second vessel along with the rest of the mixture or solution. Titration
A titration is a method of analysis that will allow you to determine the precise endpoint of a reaction and therefore the precise quantity of reactant in the titration flask. A burette is used to deliver the second reactant to the flask and an indicator or pH Meter is used to detect the endpoint of the reaction. COMMON LABORATORY EXPERIMENTAL PROCEDURES
1. BUNSEN BURNER (i) Close the air hole and turn the gas tap full on. Light the gas and hold a piece of wire in different parts of the flame, moving it from the bottom to the top. Note the hottest place in the flame. Open the air hole. Again hold the wire in the flame, moving from the bottom to the top. Note the hottest place in the flame. Compare the two flames and note which has the hottest point. (ii) Close the air hole. Hold a test-tube with its bottom end just above the flame. Carbon deposits on the glass.
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(ii) Close the air hole. Hold a test-tube with its bottom end just above the flame. Carbon deposits on the glass. To test whether unburned carbon gives the yellow colour to the flame, sprinkle powdered charcoal into the flame to see if this gives the same effect. (iii) Open the air hole again. Note whether carbon deposits on a test-tube held in this flame. Air mixing with the gas helps it to burn more rapidly and efficiently. To test what is in the cooler inner cone hold a splint of wood in the flame so that it passes through the inner cone. Note which part of the splint burns. Hold a piece of glass tubing with one end in the inner cone then ignite the gas that comes out of the other end. (iv) Investigate a candle flame and the flame of a spirit lamp in a similar way. Find the hottest part of the flame. Test for unburned carbon particles in the flame. Look for an inner cone of unburned gases. IDENTIFICATION OF PURE SUBSTANCES
2. Melting points, mp, of naphthalene (i) Put a very small amount of naphthalene in a capillary tube sealed at one end. You can pull out the capillary tube from heated glass tubing. Use a rubber band to attach the capillary tube, sealed end down, to a thermometer. Heat a thermometer and capillary tube in a beaker of water on a tripod. Use the thermometer to stir the water but do not let water enter the capillary tube. Slowly raise the temperature of the water. Note the temperature of the naphthalene when it melts. Let it cool and note the temperature when the naphthalene solidifies Calculate the average of these two values. (ii) Use a clean test-tube and thermometer to repeat the experiment using stearic acid, mp 69o, or use any other substance, mp < 100oC. 3. Impurities affect the melting point of a substance Mix stearic acid with the naphthalene, thus making the naphthalene impure. Look for changes in the melting point. Impurities lower the melting point. 4. Boiling point of water We can identify a pure substance from its melting point or boiling point. Put water in a test-tube and hold a thermometer with the bulb just under the water. Add boiling chips to prevent bumping. Bring the water to the boil with a very small flame and read the thermometer.
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Put water in a test-tube and hold a thermometer with the bulb just under the water. Add boiling chips to prevent bumping. Bring the water to the boil with a very small flame and read the thermometer. Note any change in the reading if the thermometer touches the bottom of the test-tube. 5. Boiling point of inflammable liquids (i) Use a different method of heating inflammable liquids, e.g. ethanol, bp 78.4oC and acetone, bp 56oC. Put 2 cm of the inflammable liquid in a test-tube. Put the test-tube in an empty beaker. Put a thermometer into the test-tube with the bulb in the liquid. Boil water in an electric jug or on an electrical hot plate. Pour the hot water into the beaker so that the level is higher than the inflammable liquid in the test-tube. Stir the inflammable liquid gently with the thermometer and read the thermometer when the inflammable liquid boils. (ii) Use a very small test-tube or seal one end of a piece of glass tubing, 8 cm length and 3 cm external diameter. Put the inflammable liquid into this tube. Put a capillary tube, sealed at one end, into the inflammable liquid with the sealed end up and the open end down in the inflammable liquid. Use a rubber band to attach the tube containing inflammable and capillary tube to the bulb of a thermometer. Hold the apparatus in a beaker of water and heat gently with a Bunsen burner flame. When the temperature rises, bubbles slowly come out of the capillary tube. At the boiling point the bubbles suddenly come out as a steady stream. Read the temperature. Let the water cool and read the temperature again when the steady stream of bubbles ceases. Calculate the boiling point as the average of the two readings. 6. Pressure affects the boiling point Put water in a sidearm test-tube or in a round-bottom flask with a 2-hole stopper. Insert a thermometer through a hole in the stopper so that the bulb of the thermometer reaches, but does not touch, the bottom of the test-tube or flask. Add boiling chips to prevent bumping. Boil the water and read the temperature on the thermometer. Stop heating. Connect a water pump to the sidearm or to the second hole of the 2-hole stopper. When the water stops boiling, turn on the water pump to reduce the pressure.
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Stop heating. Connect a water pump to the sidearm or to the second hole of the 2-hole stopper. When the water stops boiling, turn on the water pump to reduce the pressure. Read the temperature, heat to boiling and read the temperature again. 7(a). Solubility in water Test different salts taken to show that each has a different solubility in water. Take 5 g samples and try to dissolve each in 15 mL of water in a test-tube. Attach the stopper then shake each test-tube vigorously for the same time to show that solubility is a characteristic of a particular substance, e.g. sugar, common salt, potassium nitrate, calcium sulphate. The solubility of a salt is usually expressed as the number of grams able to dissolve in 100 g water at 20oC, e.g. ammonium chloride 37.2 g, barium chloride 35.7 g, calcium chloride 42.7 g, copper sulphate 20.7 g, lead nitrate 54.4 g, magnesium sulphate 25.2 g, potassium chloride 34.0 g, potassium iodide 144.0 g, sodium bicarbonate 9.6 g, sodium chloride 36.0 g, sodium hydroxide 109.0 g, sodium nitrate 87.5 g. (b) Solubility of a substance in water at a given temperature Put about 50 cm3 of water in a beaker and add baking powder, sodium bicarbonate, gradually while stirring. Potassium sulphate is an alternative substance. Make a saturated solution by stirring until no more solute will dissolve. Read the temperature of the saturated solution. Weigh a clean evaporating dish, w1. Add some clear saturated solution and weigh again, w2. Carefully evaporate the solution in the evaporating dish to dryness and weigh again, w3. The mass of the sodium bicarbonate dissolved = w3 - w1. The mass of water = w2 - w1 - w3. Calculate the solubility of the sodium bicarbonate as g per 100 g water at room temperature.
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The mass of the sodium bicarbonate dissolved = w3 - w1. The mass of water = w2 - w1 - w3. Calculate the solubility of the sodium bicarbonate as g per 100 g water at room temperature. 8(a) Solubility and particle size To show that small particles dissolve faster than the large particles, add 4 g of coarse able salt to one test-tube half filled with water, and 4 g of fine table salt to a second test-tube containing the same amount of water or use whole and crushed copper sulphate crystals. Stir or shake both tubes equally and simultaneously. Pause after every few seconds to observe the amount of undissolved salt left in each tube. (b) Solubility and solvents (i) Fill two test-tubes one third full with (a) water and (b) methylated spirits. To each test-tube add 1 g sodium chloride, attach a stopper and shake. Sodium chloride dissolves readily in water but not so readily in methylated spirits. (ii) Fill two test-tubes one third full with (a) 88g water and (b) a solution of 1 g potassium iodide in 5 mL water. To each test-tube add iodine crystals, attach a stopper and shake. Sodium chloride dissolves readily in water but not so readily in alcohol
DETERMINATION OF DENSITY
9(a) Density of a solid The density of a solid is the ratio of mass to volume. We can find the density of a regular solid with a balance and ruler. We can find the volume of an insoluble irregular solid with a measuring cylinder. Half fill a graduated cylinder with water. Read the volume, immerse the solid in the water and read the volume again. The difference in the two readings is the volume of the solid. (b) Density of a liquid Weigh a small container, fill with liquid and weigh again. Transfer the liquid to a measuring cylinder. Find the density by dividing the mass of the liquid by the volume. Heat of fusion and vaporization
10(a) Separate by sublimation Separate iodine from a mixture of crystals of iodine and sodium chloride. Heat the mixture in an evaporating dish with a funnel placed over it. The iodine sublimes on to the cool sides of the funnel. SEPARATION OF MIXTURES
(b) Separate by distillation Put 10 mL ink in a flat bottom conical flask.
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Heat the mixture in an evaporating dish with a funnel placed over it. The iodine sublimes on to the cool sides of the funnel. SEPARATION OF MIXTURES
(b) Separate by distillation Put 10 mL ink in a flat bottom conical flask. Add boiling chips to prevent bumping. Fit a stopper with a delivery tube reaching half way down a collecting test-tube or an U-tube, in a beaker of water. Heat the ink with a Bunsen burner flame. Drops of a colourless liquid appear in the collecting tube. Identify the liquid as water by its action of turning white anhydrous copper sulphate to blue hydrated copper sulphate. Do not allow ink to froth up or splash into the delivery tube. (c) Separate salt and sand Prepare a mixture of salt and sand. Put about 2 mL of the mixture in a test-tube. Add 5 mL water and shake until all the salt has dissolved. Pour the contents of the tube into a filter paper in a funnel over an evaporating basin. Wash the test-tube with water and add this to the filter paper. The sand will remain on the filter paper and may be dried and collected. The salt can be recovered from the filtrate by warming the evaporation basin to drive off the water. (d) Solvent extraction of oil from nuts Put groundnuts (peanuts) or pieces of chopped coconut into a mortar. Add 20 mL of acetone or methylated spirits. Grind the nuts in the solvent as finely as possible. Pour off the liquid into a test-tube and filter into an evaporating basin. Warm the evaporating basin for 10 minutes. The solvent evaporates leaving the oil extracted from the nuts. (e) Separate two immiscible liquids of different density Use a separating funnel or make a separating funnel with a piece of wide tubing fitted with a stopper, tube and rubber tube with a clip. Shake the mixture thoroughly in a closed container then run it into the separating funnel. Wait until a clear boundary appears between the two liquids and then run off the denser layer into a beaker below. 11(b). HEATING AND BURNING DIFFERENT SUBSTANCES Substances that gain mass when heated. Clean 25 cm of magnesium ribbon and cut into 1 cm pieces. Weigh a crucible plus lid, put the pieces of magnesium ribbon in the crucible and weigh again.
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HEATING AND BURNING DIFFERENT SUBSTANCES Substances that gain mass when heated. Clean 25 cm of magnesium ribbon and cut into 1 cm pieces. Weigh a crucible plus lid, put the pieces of magnesium ribbon in the crucible and weigh again. Put the crucible on a pipe clay triangle supported on a tripod. Heat very gently then strongly. Hold the crucible lid in a pair of tongs close to the crucible. The magnesium ribbon darkens just before it begins to melt. At the first sign of burning, place the lid on the crucible and remove the Bunsen burner. Every 4 seconds, raise the lid to allow more air to enter but do not allow any white magnesium oxide smoke to escape. When the magnesium stops burning on raising the lid, remove the lid. Heat the crucible again strongly but still hold the crucible lid in a pair of tongs close to the crucible in case the magnesium starts to burn again. Leave the crucible to cool. When cool, weigh the crucible plus lid plus contents. Calculate the increase in mass of the magnesium. (b) Substances that lose mass when heated (i) Weigh a test-tube containing 1 cm potassium permanganate crystals and a 1 cm plug of cotton wool at the mouth to prevent loss of any solid during heating. Heat the test-tube and cotton wool and weigh it again. Note any change in the potassium permanganate crystals. Note any loss in mass due to the loss of water of crystallization. (ii) Weigh a test-tube containing 1 cm copper carbonate and a 1 cm plug of cotton wool at the mouth to prevent loss of any solid during heating. Heat the test-tube and cotton wool and weigh it again. Note any change in the copper carbonate. Note any loss in mass. (b) Substances that neither gain nor lose mass when heated Weigh a test-tube containing 1 cm dry zinc oxide and a 1 cm plug of cotton wool at the mouth to prevent loss of any solid during heating. Heat the test-tube and cotton wool and weigh it again. Note any change in the zinc oxide. Note any loss in mass. (c) Effect of heat on copper sulphate crystals Crush blue copper sulphate crystals and put them into a dry test-tube to a depth of 4 cm. Heat the tube gently.
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Note any loss in mass. (c) Effect of heat on copper sulphate crystals Crush blue copper sulphate crystals and put them into a dry test-tube to a depth of 4 cm. Heat the tube gently. Note whether vapour collects on the cooler parts, change of colour from blue to white, and any liquid collecting in the receiving tube. Identity of the liquid by measuring the boiling point. When all the copper sulphate crystals have changed to white and the tube has cooled, hold the tube in your hand and pour the liquid back on to the white crystals. Note whether the blue colour restored and if any heat is given back. Blue copper sulphate crystals + heat -> white, anhydrous copper sulphate + water. This is a reversible change. 12.PREPARING, COLLECTING AND TESTING GASES (b) Hydrogen Be careful! A dangerous explosion may occur if you use any vessel bigger than a small test-tube when igniting the gas, particularly if it is mixed with air. Never dry hydrogen gas with concentrated sulphuric acid! (i) Place a few pieces of granulated zinc or zinc foil from the casing of an old dry cell in a boiling tube; add 2 drops of copper sulphate solution. An alternative to the thistle funnel at A is a syringe as shown at B. Pour enough 1 M sulphuric acid down the thistle funnel on to the zinc to cover the bottom of the funnel tube. Discard the first two or three test tubes of hydrogen, as they will contain displaced air. Collect test tubes of the gas and stopper them. Test the third test-tube of gas by holding a lighted taper or splint over the mouth as soon as you take out the stopper. Pure hydrogen burns with a quiet "pop" sound. (ii) Alternatively, add sulphuric acid from a syringe. Gas cannot escape through the syringe so you do not need to cover the tube of the syringe with acid. (iii) Hydrogen burns in air to form water vapour. When hydrogen ignites in a dry test-tube, note any vapour or mist on the sides of the test-tube. (iv) Investigate whether hydrogen is lighter than air by "pouring" the gas into a test-tube held either above the first tube or below it. Use a lighted taper to investigate where the hydrogen has gone. (v) Blow soap bubbles by holding the delivery tube of the apparatus in detergent or soap solution.
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(iv) Investigate whether hydrogen is lighter than air by "pouring" the gas into a test-tube held either above the first tube or below it. Use a lighted taper to investigate where the hydrogen has gone. (v) Blow soap bubbles by holding the delivery tube of the apparatus in detergent or soap solution. The hydrogen bubbles will rise into the air, showing the low density of hydrogen gas. (b) Oxygen (i) Prepare oxygen safely by decomposition of hydrogen peroxide solution. Put 20 mL hydrogen peroxide into a 100 mL bottle. Fix a delivery tube to the bottle. Add two spatulas of manganese dioxide and oxygen bubbles off for collection. (ii) Oxygen is colourless and has no smell. To test whether the test-tube contains oxygen, light a splint of dry wood, blow out the flame leaving a glowing splint then put the glowing splint in a test-tube of oxygen. The glowing splint bursts into flame. This experiment is called the glowing splint test. (iii) Use an L-shape piece of nichrome wire with a shield to fit on the top to protect your hand. Fix steel wool into a loop in the lower end of the Nichrome wire. Heat the steel wool to red hot in a Bunsen burner flame then inserts it quickly into a test-tube of oxygen. (iv) Fix a small piece of charcoal into the loop in the lower end of the Nichrome wire. Ignite the charcoal in the Bunsen burner flame and then insert it quickly into another test-tube of oxygen. (v) Dip the loop in the lower end of the Nichrome wire into sulphur powder. Ignite the sulphur powder in a Bunsen burner flame and then insert it quickly into another test-tube of oxygen. (c) Hydrogen chloride Put rock salt, sodium chloride, into a 100 mL filtering flask. Coarse rock salt causes less frothing than the fine salt. Carefully add concentrated sulphuric acid down the thistle funnel. You can collect hydrogen chloride gas by upward displacement of air, as in the diagram. (i) Collect four test tubes of the gas and cork them. Remove the cork from one of these test tubes under water. Note the solubility of hydrogen chloride. (ii) Hold a piece of cotton wool soaked in ammonium hydroxide at the mouth of a test-tube of hydrogen chloride. (iii) Shake a test-tube of the gas with water to obtain a solution of hydrogen chloride.
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Note the solubility of hydrogen chloride. (ii) Hold a piece of cotton wool soaked in ammonium hydroxide at the mouth of a test-tube of hydrogen chloride. (iii) Shake a test-tube of the gas with water to obtain a solution of hydrogen chloride. Test the solution with an acid/base indicator. Put a piece of magnesium ribbon in the solution. Collect any gas formed and test for hydrogen with the glowing splint test. (d) Ammonia (i) Put a mixture of calcium hydroxide and ammonium chloride into a test-tube to a depth of 4 cm. Fill a U-tube with lumps of calcium oxide mixed with cotton wool. The cotton wool is to prevent blocking of the tube. Gently heat the test-tube. The calcium oxide dries the ammonia gas. Test whether the receiver test-tube is full by holding a piece of red litmus paper at the opening. Collect test tubes of ammonia and cork them. The method of collection illustrates that ammonia gas is lighter than air. (ii) Fill a flask with ammonia. Fit a cork and tube into the flask as shown. The tube should have been drawn out into a jet. Warm the flask gently to expand the gas and then hold the flask upside down with the tube in the water. Water will spray into the flask from the jet. (e) Carbon dioxide Many reactions can be used to produce carbon dioxide gas. Marble chips or other carbonate rock treated with dilute acid provides a good source. The gas is not too soluble to be collected by water displacement, as shown above for the preparation of hydrogen. Alternatively, carbon dioxide can be collected by displacing air from dry bottles. To test if the bottle is full, lower a lighted splint or taper into the top of the jar. If the flame is extinguished at the entrance as at (ii), then the jar is full. Put a cardboard cover over the top to prevent diffusion of the gas. Check the density of the carbon dioxide by pouring" the gas into another bottle either above or below the first bottle. Find where the gas is by testing with a lighted splint. Note: The presence of carbon dioxide can be confirmed by the fact that limewater becomes "milky" when the gas is Passed through it. (f) Cooking and carbon dioxide The purpose of baking powder or soda in cooking is to produce tiny bubbles of carbon dioxide. This expands the pastry, cake or dough, making it light and pleasant to eat.
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Note: The presence of carbon dioxide can be confirmed by the fact that limewater becomes "milky" when the gas is Passed through it. (f) Cooking and carbon dioxide The purpose of baking powder or soda in cooking is to produce tiny bubbles of carbon dioxide. This expands the pastry, cake or dough, making it light and pleasant to eat. Yeast cells do the same thing in bread making, though this takes longer. Baking powder, or sodium bicarbonate, NHCO3, reacts with an acid such as lactic acid from sour milk to produce carbon dioxide. Commercial "baking powders" often contain a solid acid, which only reacts with the sodium bicarbonate when moist. (i) Put baking powder into water. Note whether carbon dioxide gas forms. Note whether carbon dioxide forms when you put sodium bicarbonate into water. Add baking powder in a test-tube with vinegar or lemon juice (acetic acid). Note whether carbon dioxide forms. (ii) Make a sugar solution and half fill a jar with this solution. Add a spoonful of yeast and leave to stand for 2 days. Construct a bubbler to fit on the top of the jar. Note whether the yeast forms a gas. Note whether carbon dioxide gas collects in the upper part of the jar. 12(a) RUSTING Take 7 test tubes and 11 clean nails. Prepare the tubes as shown below: Tube 1: Put 2 clean nails in the test-tube and half cover them with distilled water. These nails are in contact with air and water and form the control experiment. Tube 2: Put a few pieces of anhydrous calcium chloride or silica gel in the bottom of a dry test-tube, and also two nails. Put a plug of cotton wool in the top of the tube. These nails are in contact with air, but not moisture. Tube 3: Boil water for several minutes to expel dissolved air and pour into the test-tube whilst hot. Put 2 nails in the water. Put a little Vaseline or a few drops of olive oil on the surface of the hot water. The Vaseline will melt and form an airtight layer, solidifying as the water-cools. These nails are in contact with water but not air. Tube 4: Half cover 2 nails with water containing a little common salt dissolved in it. These nails are in contact with air, water and salt. Tube 5: Wrap a piece of zinc foil round part of a nail.
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Tube 4: Half cover 2 nails with water containing a little common salt dissolved in it. These nails are in contact with air, water and salt. Tube 5: Wrap a piece of zinc foil round part of a nail. Put the nail in the test-tube and almost cover with tap water. Tube 6: Wrap a piece of tin foil round part of the nail. Put the nail in the test-tube and add tap water as you did for tube 5. Tube 7: Wrap a piece of copper wire round a nail and put it in the test-tube exactly like tubes 5 and 6. Stand these 7 test tubes in a rack and leave for several days. Note the conditions for rusting and which metal, (zinc, copper or tin), is best at preventing rusting. (b) INCREASE IN MASS OF IRON DURING RUSTING Counterbalance a piece of iron on a knife-edge, using a brass weight or stone. Leave in moist air or on a window ledge for a few days and note the effect of the rust on the longer arm of the lever. During rusting, metallic ion changes to Fe (OH) 3.xH2O. (c) WHAT COMBINES WITH IRON DURING RUSTING Moisten the inside of a test-tube with water, sprinkle into it a spatula measure of iron filings and rotate it horizontally so that the filings spread and adhere to the walls. Alternatively, push a small plug of moistened iron wool to the bottom of the tube. Invert the test-tube in a beaker about one third full of water. Use the beaker lip to support the tube. The water levels inside and outside the tube should be the same and the level should be marked on the tube. Leave the tube in this position for a few days. The iron will rust and the water level will rise up inside the tube, finally becoming steady. Again add water to the beaker until the levels inside and outside the tube are the same and mark the new level. It will be seen that one fifth of the air volume has been used up, suggesting that oxygen has been used up in the rusting of iron. The residual gas does not support combustion of a lighted splint.
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Again add water to the beaker until the levels inside and outside the tube are the same and mark the new level. It will be seen that one fifth of the air volume has been used up, suggesting that oxygen has been used up in the rusting of iron. The residual gas does not support combustion of a lighted splint. 13 COLOURED EXTRACTS FROM FLOWERS AS INDICATORS OF ACIDS AND BASES (a) Extract coloured substances from plants Select brightly coloured flowers, such as the purple and red bougainvillaea, or coloured leaves. Squeeze or grind one of the coloured flowers or leaves in a mortar with a mixture made of 2 mL acetone and 2 mL ethanol. By this means the colouring matter will be extracted into the solvent. Filter and collect the filtrate. Repeat this experiment with one or two other flower colours. Keep these coloured solutions for use as "indicators" in the next experiment. (b) Plant extracts to indicate whether a substance is acidic or basic (i) Put a spot of the coloured flower extract on to a filter paper and leave to dry. Put one drop of lemon juice on to the spot then note any change of colour. Repeat the experiment with other fruit juices and vinegar. They are acidic substances. Note any colour change with dilute hydrochloric acid. Different colours suggest that some substances are more acidic than others. (ii) Put some of the original filtrate on to another piece of filter paper. When dry, note the colours given by sodium bicarbonate solution, washing soda, limewater and a dilute solution of sodium hydroxide. These are alkaline, or basic, substances. Note whether they all give the same colour: Plant extracts can act as indicators to test whether a substance is acidic or basic. (iii) Add a few drops of sodium bicarbonate solution to 1 mL of flower extract indicator in a test-tube. Then add lemon juice and note any colour change. (iv) Repeat the experiment with limewater and indicator followed by dilute hydrochloric acid. Note any colour change. Note whether you can get back the original colour by adding more limewater. Note how many times you can change the indicator colour before the test-tube is full. (v) Litmus, an extract of lichens, is another plant indicator. An acidic solution turns blue litmus red. An alkaline solution turns red litmus blue.
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(v) Litmus, an extract of lichens, is another plant indicator. An acidic solution turns blue litmus red. An alkaline solution turns red litmus blue. (v) Universal Indicator can be in the form of a solution or dried on filter paper. Universal indicator not only indicates whether a substance is acidic or basic but also how acidic it is. Investigate the effect of Universal Indicator on the solutions above. To avoid using the name of a colour to indicate acidity, we use a scale of numbers from 0 to 14 called the pH scale. We can use the pH scale to express the degree of acidity. Acid solutions have a pH value less than 7. Alkaline or basic solutions have a pH value greater than 7. Solutions with a pH value of 7 are neither acidic nor basic, they are neutral. On the bottle or packet of Universal indicator, a colour chart shows the colour and the pH value associated with this colour as follows: (a) Colour (b) pH (c) Acid/Base (a) Red (b) 1-3 (c) very acidic (a) Orange (b) 4-5 (c) weak acid (a) Yellow (b) 6 (c) very weak acid (a) Green (b) 7 (c) neutral (a) Blue (b) 8 (c) very weak base (a) Indigo (b) 9-10 (c) weak base (a) Violet (b) 11- 14 (c) very basic Use 2 drops of Universal Indicator to 10 mL of solution to be tested. Test the pH value of lemon juice, vinegar, sodium bicarbonate solution, washing soda, lime water, sodium hydroxide solution, tap water, distilled water. 14. CRYSTALS(a) Crystal growth Sodium thiosulphate crystals grow rapidly from a super-saturated aqueous solution. The formula for the crystals is Na2S2O3.10H2O. On heating, these crystals dissolve in some of their water of crystallization. Put 3 cm of sodium thiosulphate crystals in a test-tube. Add 2 drops of water. Heat gently until all the crystals have dissolved. They appear to "melt". Leave to cool. Crystals may not form unless you drop a tiny seed crystal of sodium thiosulphate into the solution. Then crystal growth commences and spreads rapidly through the whole solution.
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Leave to cool. Crystals may not form unless you drop a tiny seed crystal of sodium thiosulphate into the solution. Then crystal growth commences and spreads rapidly through the whole solution. Watch the growth from one centre. Hold the tube in the hand while crystallization occurs. (b) Crystals of naphthalene grow from the melt Put a little naphthalene on a glass slide. Hold over a flame until the crystals melt. Put a cover slip over the liquid and allow to cool. Watch the crystals grow using a hand lens. Sometimes crystals will grow from several points simultaneously to make boundaries where they meet. Draw the shape of the boundary between the forming crystals and the melt. View the crystals through Polaroid filters. (c) Grow large crystals (i) Use a 0.5 - 0.8 cm long seed crystal to start growing large crystals. Make seed crystals by slow evaporation of 30 mL of saturated solution in a glass dish. Dry a selected seed crystal. Tie a piece of clean cotton around it without touching the seed crystal with your hands because impurities easily affect the size and shape of the crystal. Hang the seed crystal 5 cm from the base of the container with a bent wire. Fill a jar (with a screw-on lid) with a solution of the salt less than saturation strength before you put the seed crystal in position. If little crystals grow on the surface of the seed crystal, then screw on the lid of the jar to make the little crystals dissolve. Prevent crystals growing on the sides of a crystallizing dish by rubbing Vaseline round the upper inside rim. Evaporation may be increased by sitting the crystal growing jar on a tin with a 5 watt bulb mounted inside it. An air flow over the solution surface given by a fan will also hasten crystal growth. You can preserve large crystals by painting with a clear varnish. (ii) You can also support the seed crystal at the end of a glass tube. Heat a 3 mm bore piece of glass tubing in a flame until the end softens sufficiently to squeeze with pliers to make a smaller hole. When the glass tubing cools, drop seed crystals on its end until one catches in the smaller hole. Keep this crystal in place by dropping other crystals on it. Support the glass tubing vertically so that the seed crystal at the end is immersed in the solution of the salt. The seed crystal should then grow. Turn the crystal regularly so that growth on all faces is equal. 15.
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The seed crystal should then grow. Turn the crystal regularly so that growth on all faces is equal. 15. DIFFUSION
(a) Diffusion of heavy carbon dioxide gas upwards (i) Fill a jar with carbon dioxide and invert it over a similar jar full of air. After a few moments separate the jars, pour a little lime water in the lower one and shake it. The lime water will turn milky indicating that the carbon dioxide has fallen into the lower jar because it is the heavier gas. (ii) Repeat the experiment with the carbon dioxide in the lower jar and invert a jar of air on top of it. If the jars are left for about 5 minutes carbon dioxide will be carried into the upper jar by diffusion; in the same way air will be carried into the lower jar. The lime water test will show the presence of carbon dioxide in the upper jar. (b) Diffusion rates of ammonia and hydrogen chloride gases The long glass tube should be horizontal. Corks should fit at both ends. Using a pair of tongs or tweezers, dip a piece of cotton wool into concentrated hydrochloric acid and another piece into concentrated ammonium hydroxide. Drain off excess liquid. As nearly as possible at the same time, put the ammonia cotton wool at one end of the tube and the acid cotton wool at the other. Close the ends of the tube with corks. After a while, look carefully for a white ring which will form where the ammonia gas and the hydrogen chloride gas meet after diffusing through the air towards each other. Ammonia is the less dense gas and the white ring of ammonium chloride should form nearer to the hydrogen chloride end than from the ammonia end of the tube. (c) Diffusion of liquids (i) Place a crystal of potassium dichromate, potassium dichromate (VI), or ammonium dichromate at the bottom of a beaker of water. To do this, put a glass tube into the beaker of water so that it touches the bottom, then to drop the crystal down the tube. Close the top of the tube with your finger and remove the tube gently, leaving the crystal in the beaker. The colour of the dissolving crystal will spread throughout the water in quite a short time. (ii) Fill a very small open bottle with a strong solution of potassium permanganate, potassium manganate (VII). Place this in a larger jar.
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The colour of the dissolving crystal will spread throughout the water in quite a short time. (ii) Fill a very small open bottle with a strong solution of potassium permanganate, potassium manganate (VII). Place this in a larger jar. Fill the larger jar very carefully by pouring water down the side until the water level is above the top of the small bottle. Leave this for a few days. The potassium permanganate solution diffuses evenly through the water. (d) Size of a molecule We select an oil molecule because it has a density less than the density of water. Oil floats on the surface and does not dissolve in the water. If the water has a large enough surface area, we assumed that thin oil will spread out in a layer one molecule thick called a monomolecular layer and not form little "hills" of molecules. If we know the original volume of oil and the surface area that it forms, then we can calculate the thickness of a monomolecular layer dividing the volume by the area. Use a tray with area > 30 cm2 so as not to restrict the oil film. Sprinkle the surface of the water with a very fine light powder, e.g. talc powder. When you put oil on the water, it pushes the powder aside so you can easily see the area covered by the oil. Pour thin petroleum distillate oil into a burette. Find the volume of fifty drops by running oil from the burette drop by drop and counting the drops. Allow one more drop to fall on a piece of plastic. Touch the oil drop with the point of a glass rod and then touch the prepared water surface. The oil spreads out. Measure the approximate area over which it spreads. Estimate what fraction of oil was removed by the glass point by using the glass point to remove successive fractions from the drop until it has been used up. The volume of oil put on the water can be calculated and an estimate made of the thickness of the oil layer, about 10-6 mm. This is an approximate dimension of a single molecule of the oil. 16. ELECTRICITY
(a) Solids that conduct electricity The source of the DC supply can be dry cells in series giving 6 volts. The bulb, which should be low power, indicates when the current is flowing. The electrodes may be carbon or steel, perhaps mounted in a wooden support, cork or rubber stopper so as to keep the electrodes a constant distance apart.
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ELECTRICITY
(a) Solids that conduct electricity The source of the DC supply can be dry cells in series giving 6 volts. The bulb, which should be low power, indicates when the current is flowing. The electrodes may be carbon or steel, perhaps mounted in a wooden support, cork or rubber stopper so as to keep the electrodes a constant distance apart. (i) Test the conductivity of solids by making a good contact between the surface of the solid and the two electrodes. The surface of the solid must first be cleaned. All metals conduct electricity. Carbon conducts electricity. Note whether non-metallic solids, e.g. plastics, naphthalene, wax, sugar, sodium chloride and sulphur, conduct electricity. (ii) Glass can be a conductor. Heat a glass rod until it becomes very hot and begins to soften. Test the hot, soft part with the conductivity apparatus. When molten, glass is a good conductor of electricity. (b) LIQUIDS THAT CONDUCT ELECTRICITY (i) First test liquids obtained by melting substances. Melt the following substances, but heat very gently and cautiously because otherwise they may ignite and burn: sulphur, wax, naphthalene, polyethylene material, tin, lead and, if available, a low melting point salt such as lead bromide, m.p. 488oC, or potassium iodide, m.p. 682oC . Test the conductivity of the melt by dipping in the electrodes and waiting a few moments for the electrodes to reach the same temperature. This ensures that the electrodes are in contact with the liquid and not the solidified melt. Scrape and clean the electrodes between each test. (ii) Test ethanol, or methylated spirits, acetone, carbon (IV) chloride, vinegar, sugar solution, copper (II) sulphate solution, sodium chloride solution, and other substances dissolved in water. Clean and dry the electrodes between each test. (iii) Test pure distilled water for conductivity. Put the electrodes into a beaker of distilled water. The bulb does not light up so pure water does not conduct electricity. Very gradually stir small crystals of common salt into the water. Note any changes to the bulb as the salt dissolves.
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The bulb does not light up so pure water does not conduct electricity. Very gradually stir small crystals of common salt into the water. Note any changes to the bulb as the salt dissolves. Classify substances into the following groups: (a) those which conduct electricity in the solid state and those which do not; (b) those which conduct in the liquid state and those which do not; (c) those which conduct when dissolved in water and those which do not
(c) ELECTROLYSIS OF LEAD BROMIDE There are very few suitable low melting point salts. Lead bromide has a low melting point and makes an interesting electrolysis experiment. Potassium bromide may have too high a melting point, 682oC, to melt easily. The lead bromide is melted in a 100 mL hard glass beaker, or in a crucible. The carbon electrodes are supported by a strip of wood with two holes bored 2 cm apart for the electrodes. Connect crocodile clips to the rods and complete the circuit with a torch bulb, to indicate when a current is flowing, and a 12-volt torch battery or cells wired in series. The electrodes can be labelled positive and negative. The only ions present in this melt are the bromide and lead ions. Bromine is readily seen coming off at the positive electrode, which is the anode. The fact that bromine appears only at the positive electrode helps in the understanding of the existence of a negative bromide ion. Lead has both a lower melting point and a greater density than lead bromide and therefore appears as a melt at the bottom of the beaker. The small globule of lead, which accumulates at the negative electrode, the cathode, can be seen after about 10 minutes of electrolysis. Decant off the molten lead bromide carefully into another crucible. The electric current has split up crystalline lead bromide into bromine gas and lead metal. (d) ELECTROLYSIS In aqueous solutions there are usually four ions present, two from the water and two from the dissolved salt. The products will be gaseous, or metals which are deposited on the negative electrode. It is composed of an open cylinder of glass approximately 8 cm high and 2.5 cm in diameter. A small bottle of similar size with the bottom cut off would do just as well.
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The products will be gaseous, or metals which are deposited on the negative electrode. It is composed of an open cylinder of glass approximately 8 cm high and 2.5 cm in diameter. A small bottle of similar size with the bottom cut off would do just as well. The cylinder has a 2-hole rubber stopper carrying two carbon electrodes with connecting leads to a battery, or DC supply of 4 to 6 volts. If cork is used, this must be made leak proof by covering the whole of the bottom surface round the electrodes and the glass edge with Faraday's wax or a similar soft wax. The electrodes may be carbon rods from a dry cell or pencil leads. The alloy supports for the coiled filament in electric light bulbs have also been found suitable for electrodes. The electrodes should project about 2 cm into the cylinder and also 2 cm below for attaching the leads to the battery. Pencil leads are brittle, and if they are used it is better to fix the electrodes in the following way. Solder a piece of stout copper wire to a 4 cm length of braided copper screening wire. Drill two holes in the rubber stopper with a 1 mm drill. Insert the copper wire into the hole from above and pull it right through the stopper until the screening wire is also pulled a little way into the hole. Into the core of this screening wire insert the pencil lead securely. Then pull the screening wire with the lead further into the stopper so that the lead electrode is firmly held in the stopper. The excess copper wire is cut off. This procedure is repeated with the other electrode. The solution is placed in the glass cylinder. The two small tubes are then filled with the solution, and carefully inverted over the electrodes. The electrodes are connected to a safe DC supply with a small bulb in series. Increase the voltage until the bulb lights, showing that a current is flowing. When this happens, cut out the bulb from the circuit by closing the switch, as shown. This will allow a larger current to flow. The tubes collect any gas given off and the properties of the gas should be tested. Using carbon electrodes, the following results will be found. (d) ELECTROLYSIS OF WATER Pure water does not conduct electricity. For this reason 2 or 3 mL of dilute sulphuric acid or dilute sodium sulphate solution is added to the water in the electrolysis cell.
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Using carbon electrodes, the following results will be found. (d) ELECTROLYSIS OF WATER Pure water does not conduct electricity. For this reason 2 or 3 mL of dilute sulphuric acid or dilute sodium sulphate solution is added to the water in the electrolysis cell. Connect the cell to the DC supply and watch for bubbles of gas at both electrodes. If none appear, add a little more acid or sodium sulphate solution. After 5 to 10 minutes there should be enough hydrogen and oxygen gas for testing. Predict at which electrode each gas will appear. (e) Electrolysis of solutions of ionic salts Most ionic salts can be used satisfactorily in electrolysis. Concentrations of 1M or less are suitable. Potassium iodide gives iodine at the anode and hydrogen gas at the cathode. Zinc sulphate gives a spongy mass of zinc at the cathode, oxygen gas at the anode. Lead acetate deposits lead on the cathode and oxygen gas is produced at the anode. If the lead acetate solution is cloudy, add a few drops of acetic acid. Sodium chloride gives hydrogen gas at the cathode and chlorine gas at the anode. Copper sulphate deposits copper at the cathode and oxygen gas is produced at the anode. 17. COMMON REACTIONS
(a) DISPLACEMENT OF COPPER FROM AQUEOUS SOLUTION OF COPPER IONS (i) A metal higher in the activity order can displace copper metal from a solution of copper ions. Put 10 mL of molar copper sulphate solution in a small beaker. Clean magnesium ribbon and cut into 0.5 cm pieces. Add these pieces to the copper sulphate solution one at a time. The reaction can be vigorous! Copper metal deposits and the blue colour gradually disappear as the magnesium displaces the copper ion. Note any heat given out by the reaction. When the solution is colourless, decant the solution from the red copper powder at the bottom of the beaker. 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.
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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.
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(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.
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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.
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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!
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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. The list is not intended to be exhaustive: in particular, items (such as Bunsen burners, tripods) that are commonly regarded as standard equipment in a chemical laboratory are not included in this list. One burette, 50 cm3
A measuring cylinder, 50 cm³ or 25 cm³
A filter funnel
One pipette, 25cm3
Two conical flasks within the range 150cm3 0r 250cm3
A beaker, squat form with lip: 250 cm³
A thermometer, -10°C to + 110°C at 1 °C
A polystyrene or other plastic beaker of approximate capacity 150 cm³
Clocks (or wall-clock) to measure to an accuracy of about 1s. (Where clocks are specified, candidates may use their own wristwatch if they prefer)
Wash bottle
Test tubes (some of which should be Pyrex or hard glass). Approximately
125 mm x 16 mm
Boiling tubes, approximately 150 mm x 25 mm
Stirring rod
ACCURACY
Unless a question instructs candidates differently they should assume that readings from equipment and apparatus ought to be made with the following precision:
Burette readings should be to the nearest 0.05 cm3
Weighings should be made to 0.01 g or 0.001 g depending on the precision of the balance
0 to 1000C thermometers should be read to the nearest 0.50C and 0 to 500C thermometers to the nearest 0.20C
Timers will normally be read to the nearest second. It is important that when candidates record reading they include the appropriate number of decimal places. For example a burette reading of exactly 24.7 cm3 should be recorded in a results table as 24.70cm3. A temperature of reading of exactly 350C should be recorded as 35.50C.
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It is important that when candidates record reading they include the appropriate number of decimal places. For example a burette reading of exactly 24.7 cm3 should be recorded in a results table as 24.70cm3. A temperature of reading of exactly 350C should be recorded as 35.50C. When titres have to be averaged, it is important that the mean is expressed to either the nearest 0.05cm3 or to the second decimal point. For example, if a candidate records four different titres as listed below, the mean can be worked out. 26.50cm3 26.25cm3 26.60cm3 26.65cm3
The candidate is expected to ignore the second titre and average the remaining three
26.50 + 26.60 + 26.65 = 26.583 this should be recorded as 26.60cm3
3
(to the nearest 0.05cm3)
In general, a final should always be given to the same number of significant figures as is suggested in the exercise. A significant proportion of marks for quantitative exercise will be awarded for accuracy. The marks will be awarded by comparing the candidate’s results with the teacher’s reports. CALCULATIONS
Usually calculations will be structured. Candidates will be led through a series of steps leading to a final value. Since most of the marks for these steps will be for a correct method rather than the numerical answer, it is important that candidates include their working even if this seems to be trivial. No marks can be awarded for an incorrect answer without working but a correct method followed by an incorrect answer will receive credit. GRAPHS
Some exercises in practical chemistry will require candidates to treat their readings graphically. The question will however instruct the candidate which axes to use for each quantity being plotted. Some useful points for candidates to keep in mind when constructing graphs are listed below. Candidates will normally be instructed to put the dependent variable, the quantity being measured e.g. temperature on the y- axis. The predetermined quantity e.g. volume will be on the x-axis. The scales should be chosen so that the results are spread out as far apart as the size of the grid allows but not at the expense of using a sensible scale. For example using 1cm to represent 3 units might spread the readings better than using 1 cm to represent 4 units but the scale may be difficult to read. It is always advisable ton use even scales.
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The scales should be chosen so that the results are spread out as far apart as the size of the grid allows but not at the expense of using a sensible scale. For example using 1cm to represent 3 units might spread the readings better than using 1 cm to represent 4 units but the scale may be difficult to read. It is always advisable ton use even scales. The origin (0,0) need not necessarily be included on either scale if it is not relevant. For example if temperature readings between 210C and 280C are plotted, there is no need to begin the axes at zero. The axes must be clearly labelled with the quantity being plotted e.g. mass and its units e.g. kilograms
The points plotted may be joined with a straight line or a smooth curve. Since readings are all subject to experimental error, the line drawn may not necessarily pass through every point. Points should never be joined by a series of short straight lines. Doing a Titration
Begin by preparing your burette, as described on the burette page. Your burette should be conditioned and filled with titrant solution. You should check for air bubbles and leaks, before proceeding with the titration.Take an initial volume reading and record it in your notebook. Before beginning a titration, you should always calculate the expected endpoint volume. Prepare the solution to be analysed by placing it in a clean Erlenmeyer flask or beaker. If your sample is a solid, make sure it is completely dissolved. Put a magnetic stirrer in the flask and add indicator. Use the burette to deliver a stream of titrant to within a couple of mL of your expected endpoint. You will see the indicator change colour when the titrant hits the solution in the flask, but the colour change disappears upon stirring.Approach the endpoint more slowly and watch the colour of your flask carefully. Use a wash bottle to rinse the sides of the flask and the tip of the burette; to be sure all titrant is mixed in the flask. As you approach the endpoint, you may need to add a partial drop of titrant. You can do this with a rapid spin of a Teflon stopcock or by partially opening the stopcock and rinsing the partial drop into the flask with a wash bottle. Ask your TA to demonstrate these techniques for you, in the lab.Make sure you know what the endpoint should look like. For phenolphthalein, the endpoint is the first permanent pale pink.
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You can do this with a rapid spin of a Teflon stopcock or by partially opening the stopcock and rinsing the partial drop into the flask with a wash bottle. Ask your TA to demonstrate these techniques for you, in the lab.Make sure you know what the endpoint should look like. For phenolphthalein, the endpoint is the first permanent pale pink. The pale pink fades in 10 to 20 minutes.If you think you might have reached the endpoint, you can record the volume reading and add another partial drop. Sometimes it is easier to tell when you have gone past the endpoint. When you have reached the endpoint, read the final volume in the burette and record it in your notebook. Subtract the initial volume to determine the amount of titrant delivered. Use this, the concentration of the titrant, and the stoichiometry of the titration reaction to calculate the number of moles of reactant in your analyte solution. Indicators
An indicator is a substance used in titrations, which has one colour in the presence of an excess of one reagent and a different colour in the presence of an excess of the other. Examples of indicators include;
Methyl orange
Phenolphthalein
Methyl red
Bromothymol blue
The pH scale
A much more useful measure of the strength of an acid solution was worked out by the Danish biochemist S. Sorensen. He worked in the laboratories of the Carlsberg breweries and was interested in checking the acidity of beer. The scale he introduced was the pH scale. The scale runs from 1 to 14, and the following general rules apply. Rules for the pH scale:
Acid have a pH less than 7,
The more acidic a solution, the lower the pH. Neutral substances, such as pure water, have a pH of 7,
Alkalis have a pH greater than 7,
The more alkaline a solution, the higher the pH. The pH of a solution can be measured in several ways. Universal indicator papers that are sensitive over the full range of values can be used. Some colour changes of common indicators are shown below. Indicator colour changes
Candidates may be asked to carry out exercises involving:
Simple quantitative experiments involving the measurement of volumes:
Speeds of reaction. Measurement of temperature based on a thermometer with 10C graduations. Problems of an investigatory nature, possibly including suitable organic compounds. Filtration.
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Measurement of temperature based on a thermometer with 10C graduations. Problems of an investigatory nature, possibly including suitable organic compounds. Filtration. Identification of ions and gasses as specified in the curriculum. HEALTH AND SAFETY
Candidates must follow the health and Safety policy normally operates in their laboratories when carrying out the practical Examination. Eye protection must always be worn. Laboratory overalls are recommended. All substances should be regarded as being potentially toxic and hazardous. Hazard labels (e.g. flammable) should be read and appropriate precautions (e.g. keep liquid away from flame) taken. All substances spilled on the skin should be rinsed off immediately. Chemicals must never be tasted Gases and vapours should never be smelt unless the question instructs the candidates to do so and then only with great care. 2.0 QUANTITATIVE ANALYSIS
2.1 Molar solutions and volumetric analysis. Amount of substance
In quantitative analysis, it is important to measure the amounts or moles of reacting substances accurately. To do that, we must use values of relative atomic masses expressed on a periodic table. If we are given the mass of a compound, we can determine the number of moles. The first step is to calculate the mass of one mole of the compound by summing up the relative atomic masses of the constituent atoms. E.g. 1 mole of calcium carbonate (CaCO3) will have a mass of 40 + 12 + 48 = 100g
10g of CaCO3 will contain 10/100 = 0.1moles
concentration of a solution
It is very useful to be know exactly how much of a dissolved substance is present in a solution of particular concentration or volume of a solution. So we need a standard way of comparing the concentrations of solutions. The concentration of a solution is determined from the number of moles of solute dissolved in one litre (1dm3) of solution. If one mole of a solute is dissolved in water and the volume of the solution made up to 1litre (1dm3), this solution in known as a molar solution or 1M solution. If two moles are dissolved in 1litre (1dm3), the resulting solution is 2M and so on.
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The concentration of a solution is determined from the number of moles of solute dissolved in one litre (1dm3) of solution. If one mole of a solute is dissolved in water and the volume of the solution made up to 1litre (1dm3), this solution in known as a molar solution or 1M solution. If two moles are dissolved in 1litre (1dm3), the resulting solution is 2M and so on. You need to be able to calculate
(i) The number of moles or mass of substance in an aqueous solution of given volume and concentration
(ii) The concentration of an aqueous solution given the amount of substance and volume of water. (iii) Use the equation: molarity of Z = moles of Z / volume in dm3
Remember: moles Z = mass Z / formula mass of Z
Concentration can also be expressed directly in grams per litre (1dm3). 9.8g of H2SO4 can be expressed either as 0.1M or 9.8g/dm3
Molarity (M) = g/dm3 / formula mass
Example 1: 5.95g of potassium bromide were dissolved in 400cm3 of water. Calculate its molarity. [Ar values: K = 39, Br = 80]
Moles = mass / formula mass, (KBr = 39 + 80 = 119)
mol KBr = 5.95/119 = 0.05 mol
400 cm3 = 400/1000 = 0.4 dm3
molarity = moles of solute / volume of solution
molarity of KBr solution = 0.05/0.4 = 0.125M
2.3 Volumetric calculations (acid-alkali titrations)
Chemical Equations
These balancing numbers have an additional meaning where these amount calculations are concerned. The balancing numbers give the numbers of moles present for each chemical involved in the reaction. For example, sodium chloride may be prepared by the reaction of sodium hydroxide and hydrochloric acid, according to the following equation -
NaOH (aq) + HCl (aq) NaCl (aq) + H2O (l)
The equation specifies what amounts of sodium hydroxide and hydrochloric acid will react together and what amounts of products are produced.
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[Ar values: K = 39, Br = 80]
Moles = mass / formula mass, (KBr = 39 + 80 = 119)
mol KBr = 5.95/119 = 0.05 mol
400 cm3 = 400/1000 = 0.4 dm3
molarity = moles of solute / volume of solution
molarity of KBr solution = 0.05/0.4 = 0.125M
2.3 Volumetric calculations (acid-alkali titrations)
Chemical Equations
These balancing numbers have an additional meaning where these amount calculations are concerned. The balancing numbers give the numbers of moles present for each chemical involved in the reaction. For example, sodium chloride may be prepared by the reaction of sodium hydroxide and hydrochloric acid, according to the following equation -
NaOH (aq) + HCl (aq) NaCl (aq) + H2O (l)
The equation specifies what amounts of sodium hydroxide and hydrochloric acid will react together and what amounts of products are produced. 1NaOH (aq) + 1HCl (aq) 1NaCl (aq) + 1H2O (l)
This equation states that 1 mol of sodium hydroxide and 1 mol of hydrochloric acid will react together to give of 1 mol sodium chloride and of 1 mol water. The ratio of NaOH: HCl: NaCl: H2O is 1: 1: 1: 1. What this means is that if the amount of any one of the components in the above reaction is known then the others may be worked out from the ratio above. Or, if the masses of chemicals reacting together are known then amounts can be calculated and the balancing numbers deduced from the amounts reacting together. 2.4 Titration Experiment
Titrations can be used to find the concentration of an acid or alkali from the relative volumes used and the concentration of one of the two reactants. Titration involves the neutralisation of an acid with an alkali or a soluble carbonate. Since both reactants and products are colourless, an indicator is used to find the neutralisation point or end point i.e. the point at which the acid has been neutralised. This process is done in three stages. The acid is poured into the burette.
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the point at which the acid has been neutralised. This process is done in three stages. The acid is poured into the burette. A known volume of the alkali is transferred using a pipette into four separate conical flasks. Two or three drops of suitable indicator are added to each of the conical flasks. The acid is run into the flask until the indicator just changes colour. The difference between the initial and final burette readings gives the volume of the acid used commonly known as the titre. The first titration usually gives and approximate end point and is treated as the trial. The experiment is repeated for each of the other conical flask, to try to obtain the end point accurately. This is usually done by running out the acid to a point one unit away from the trial, the going drop-by-drop until an accurate end point is obtained. The volume of the acid required for each change is read off and recorded in a table similar to the one below. PROCEDURE DURING TITRATION
Titrations require continuous shaking of the conical flask and its contents. Check the quantity of the pipette as indicated on the bulb of the pipette and remember to record it in the appropriate place. As you release the liquid from a pipette into a conical flask, one should not blow out the last drop remaining in the jet. Before you take any reading from the burette after filling it, first allow the solution to run out to fill the tap and jet of the burette, and then you begin taking your readings. Always write the reading immediately you take them. When completing the table of results, you will be expected to complete all columns as accurately as the as the limits of the apparatus can allow e.g. burette used is usually read to the nearest read to the nearest 0.05 cm 3, pipette is accurate to 0.05 cm 3 (1 drop). Use minimum amount of indicator possible (2 – 3 drops) and recognize the end point has been reached e.g. when the colour just changes. Show the values that can be averaged to obtain an acceptable value for use in calculations (only those values within 0.2 cm 3 should be averaged). Attempt to work out the questions from the first principles and not use the formula method, which has its own limitations.
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when the colour just changes. Show the values that can be averaged to obtain an acceptable value for use in calculations (only those values within 0.2 cm 3 should be averaged). Attempt to work out the questions from the first principles and not use the formula method, which has its own limitations. GENERAL NOTES FOR QUALITATIVE AND QUANTITIVE ANALYSIS
Indicators
An indicator is a substance used in titrations, which has one colour in the presence of an excess of one reagent and a different colour in the presence of an excess of the other. Examples of indicators include;
Methyl orange
Phenolphthalein
Methyl red
Bromothymol blue
Indicator colour changes
Candidates may be asked to carry out exercises involving:
Simple quantitative experiments involving the measurement of volumes:
Speeds of reaction. Measurement of temperature based on a thermometer with 10C graduations. Problems of an investigatory nature, possibly including suitable organic compounds. Filtration. Identification of ions and gasses as specified in the curriculum. QUALITATIVE ANALYSIS
Qualitative analysis is mainly about identification of substances. No emphasis is laid on amount or quantity. In qualitative analysis exercises, candidates should use approximately 1cm depth of a solution (1-2cm3) for each test and add reagents slowly, ensuring good mixing, until no further change is seen. Candidates should indicate at what stage a change occurs, writing any deductions alongside the observations on which they are based. Answers should include details of colour changes and precipitates formed and the names and chemical tests for any gases evolved. Marks for deductions or conclusions can only be gained if the appropriate observations are recorded. QUALITATIVE ANALYSIS NOTES
TABLE 1:TESTING FOR GASES
TABLE 2: TEST FOR ANIONS
TABLE 3: TESTS FOR CATIONS
TABLE 4: ORGANIC TESTS
TABLE 5: MISCELLANEOUS TESTS
TABLE 6. FLAME TESTS
A tungsten wire loop is first dipped into some concentrated hydrochloric acid to dissolve any oxides and hence clean the wire. It is then dipped into salt powder and introduced into a colourless Bunsen burner flame. The colour is then observed and inferences made.
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FLAME TESTS
A tungsten wire loop is first dipped into some concentrated hydrochloric acid to dissolve any oxides and hence clean the wire. It is then dipped into salt powder and introduced into a colourless Bunsen burner flame. The colour is then observed and inferences made. TABLE 7: HEAT ANALYSIS
TABLE 8: SOLUBLE AND INSOLUBLE SALTS
TABLE 9: GENERAL PRELIMINARY TESTS
TABLE 10 : IGNITION ANALYSIS
Candidates may be asked to heat an unknown alone in an ignition tube. The colour changes or the identity of any gases evolved may provide evidence as to the identity of the unknown. TABLE 10: RESIDUE WHEN HOT AND COLD
Residue remaining after ignition may have a different colour when hot and cold
TABLE 11: HEATING A SOLID
Gases or vapours may be evolved during heating of the solid
TABLE 12: SOLUBILITY
The patterns of solubility for various types of salts
KEY
Ppt = precipitate
Sol = solution
Insol = insoluble
Xs = excess
Dil = dilute
1.HYDROGEN CHLORIDE
Physical Properties
Chemical Properties
Footnotes:
Dilute hydrochloric acid is one of the three common dilute acids used in the laboratory. Concentrated hydrochloric acid is used in the manufacture of many chemicals. 2.CARBON DIOXIDE
Physical Properties
Chemical Properties
Footnotes:
Makes up 0.04% of the gases in the air. Rainwater is slightly acidic because carbon dioxide dissolves in it. Needed by green plants during photosynthesis. Product of respiration in living things. Product of combustion
Is one of the "greenhouse" gases present in the atmosphere. Used in fire extinguishers since it is heavier than air and forms a "blanket" around the fire. This prevents combustion since carbon dioxide does not support it and the presence of carbon dioxide stops oxygen reaching the combustible material. 3.CHLORINE GAS
Physical Properties
Chemical Properties
Footnotes:
One of the "family" of halogen gases (iodine and bromine are in the same family)
Used in water purification. Used to make bleaching powder, disinfectants and antiseptics
Also used to make some explosives, poison gases and pesticides.
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This prevents combustion since carbon dioxide does not support it and the presence of carbon dioxide stops oxygen reaching the combustible material.
3.CHLORINE GAS
Physical Properties
Chemical Properties
Footnotes:
One of the "family" of halogen gases (iodine and bromine are in the same family)
Used in water purification.
Used to make bleaching powder, disinfectants and antiseptics
Also used to make some explosives, poison gases and pesticides.
4.HYDROGEN CHLORIDE GAS
Physical Properties
Chemical Properties
Footnotes:
Dilute hydrochloric acid is one of the three common dilute acids used in the laboratory.
Concentrated hydrochloric acid is used in the manufacture of many chemicals.
5.HYDROGEN GAS
Physical Properties
Chemical Properties
Footnotes:
The lightest gas known.
Once used in airships but replaced by helium which is not explosive.
Used to make ammonia which is needed in the manufacture of fertilizers and explosives.
6.NITROGEN
Physical Properties
Chemical Properties
Footnotes:
Makes up around 79% of the gases in the air.
Used to make ammonia gas, which in turn is used to make explosives and fertilizers.
Used in light bulbs and thermometers because it is not reactive.
Also used as the atmosphere in rooms where explosives are stored.
7.OXYGEN
Physical Properties
Chemical Properties
Footnotes:
Makes up 20% of the gases in the air.
Needed by the majority of living organisms for respiration
Is produced by green plants as a by-product of photosynthesis.
8.SULPHUR DIOXIDE GAS
Physical Properties
Chemical Properties
Footnotes:
It is used as a bleaching agent.
It is one of the gases responsible for air pollution.
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18.0.0 ACIDS, BASES AND SALTS (25 LESSONS) A.ACIDS AND BASES At a school laboratory: (i)An acid may be defined as a substance that turn litmus red. (ii)A base may be defined as a substance that turn litmus blue. Litmus is a lichen found mainly in West Africa. It changes its colour depending on whether the solution it is in, is basic/alkaline or acidic.It is thus able to identify/show whether 1. An acid is a substance that dissolves in water to form H+/H3O+ as the only positive ion/cation. This is called the Arrhenius definition of an acid. From this definition, an acid dissociate/ionize in water releasing H+ thus: HCl(aq) -> H+ (aq) + Cl- (aq) HNO3(aq) -> H+ (aq) + NO3- (aq) CH3COOH(aq) -> H+ (aq) + CH3COO-(aq) H2SO4(aq) -> 2H+ (aq) + SO42-(aq) H2CO3(aq) -> 2H+ (aq) + CO32-(aq) H3PO4(aq) -> 3H+ (aq) + PO43-(aq) 2.A base is a substance which dissolves in water to form OH- as the only negatively charged ion/anion. This is called Arrhenius definition of a base. From this definition, a base dissociate/ionize in water releasing OH- thus: KOH(aq) -> K+(aq) + OH-(aq) NaOH(aq) -> Na+(aq) + OH-(aq) NH4OH(aq) -> NH4+(aq) + OH-(aq) Ca(OH)2(aq) -> Ca2+(aq) + 2OH-(aq) Mg(OH)2(aq) -> Mg2+(aq) + 2OH-(aq) 3. An acid is a proton donor. A base is a proton acceptor. This is called Bronsted-Lowry definition of acids and bases. From this definition, an acid donates H+ . H+ has no electrons and neutrons .It contains only a proton. Examples I.
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From this definition, an acid donates H+ . H+ has no electrons and neutrons .It contains only a proton. Examples I. From the equation: HCl(aq) + H2O(l) === H3O+(aq) + Cl- (aq) (a)(i)For the forward reaction from left to right, H2O gains a proton to form H3O+ and thus H2O is a proton acceptor .It is a Bronsted-Lowry base (ii) For the backward reaction from right to left, H3O+ donates a proton to form H2O and thus H3O+ is an ‘opposite’ proton donor. It is a BronstedLowry conjugate acid (b)(i)For the forward reaction from left to right, HCl donates a proton to form Cl- and thus HCl is a proton donor . It is a Bronsted-Lowry acid (ii) For the backward reaction from right to left, Cl- gains a proton to form HCl and thus Cl- is an ‘opposite’ proton acceptor. It is a Bronsted-Lowry conjugate base. Every base /acid from Bronsted-Lowry definition thus must have a conjugate product/reactant. II. From the equation: HCl(aq) + NH3(aq) === NH4+(aq) + Cl- (aq) (a)(i)For the forward reaction from left to right, NH3 gains a proton to form NH4+ and thus NH3 is a proton acceptor . It is a Bronsted-Lowry base (ii) For the backward reaction from right to left, NH4+ donates a proton to form NH3 and thus NH4+ is an ‘opposite’ proton donor. It is a Bronsted-Lowry conjugate acid (b)(i)For the forward reaction from left to right, HCl donates a proton to form Cl- and thus HCl is a proton donor . It is a Bronsted-Lowry acid (ii) For the backward reaction from right to left, Cl- gains a proton to form HCl and thus Cl- is an ‘opposite’ proton acceptor. It is a Bronsted-Lowry conjugate base. 4. Acids and bases show acidic and alkaline properties/characteristics only in water but not in other solvents e.g.
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It is a Bronsted-Lowry conjugate base. 4. Acids and bases show acidic and alkaline properties/characteristics only in water but not in other solvents e.g. (a)Hydrogen chloride gas dissolves in water to form hydrochloric acid Hydrochloric acid dissociates/ionizes in water to free H+(aq)/H3O+(aq) ions. The free H3O+(aq) / H+(aq) ions are responsible for: (i)turning blue litmus paper/solution red. (ii)show pH value 1/2/3/4/5/6 (iii)are good electrolytes/conductors of electricity/undergo electrolysis. (iv)react with metals to produce /evolve hydrogen gas and a salt. i.e. Ionically: -For a monovalent metal: 2M(s) + 2H+(aq) -> 2M+(aq) + H2(g) -For a divalent metal: M(s) + 2H+(aq) -> M2+(aq) + H2(g) -For a trivalent metal: 2M(s) + 6H+(aq) -> 2M3+(aq) + 3H2(g) Examples: -For a monovalent metal: 2Na(s) + 2H+(aq) -> 2Na+(aq) + H2(g) -For a divalent metal: Ca(s) + 2H+(aq) -> Ca2+(aq) + H2(g) -For a trivalent metal: 2Al(s) + 6H+(aq) -> 2Al3+(aq) + 3H2(g) (v)react with metal carbonates and hhydrogen carbonates to produce /evolve carbon(IV)oxide gas ,water and a salt. i.e.
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i.e. Ionically: -For a monovalent metal: M2O(s) + 2H+(aq) -> 2M+(aq) + H2O (l) MOH(aq) + H+(aq) -> M+(aq) + H2O (l)
-For a divalent metal: MO(s) + 2H+(aq) -> M2+(aq) + H2O (l) M(OH) 2(s) + 2H+(aq) -> M2+(aq) + 2H2O(l) -For a trivalent metal: M2O3(s) + 6H+(aq) -> 2M3+(aq) + 3H2O (l) M(OH) 3(s) + 3H+(aq) -> M3+(aq) + 3H2O(l) Examples: -For a monovalent metal: K2O(s) + 2H+(aq) -> 2K+(aq) + H2O (l) NH4OH(aq) + H+(aq) -> NH4+(aq) + H2O (l) -For a divalent metal: ZnO (s) + 2H+(aq) -> Zn2+(aq) + H2O (l) Pb(OH) 2(s) + 2H+(aq) -> Pb2+(aq) + 2H2O(l) (b)Hydrogen chloride gas dissolves in methylbenzene /benzene but does not dissociate /ionize into free ions. It exists in molecular state showing none of the above properties. (c)Ammonia gas dissolves in water to form aqueous ammonia which dissociate/ionize to free NH4+ (aq) and OH-(aq) ions. This dissociation/ionization makes aqueous ammonia to: (i)turn litmus paper/solution blue. (ii)have pH 8/9/10/11 (iii)be a good electrical conductor (iv)react with acids to form ammonium salt and water only. NH4OH(aq) + HCl(aq) -> NH4Cl(aq) + H2O(l) (d)Ammonia gas dissolves in methylbenzene/benzene /kerosene but does not dissociate into free ions therefore existing as molecules 6.
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This dissociation/ionization makes aqueous ammonia to: (i)turn litmus paper/solution blue. (ii)have pH 8/9/10/11 (iii)be a good electrical conductor (iv)react with acids to form ammonium salt and water only. NH4OH(aq) + HCl(aq) -> NH4Cl(aq) + H2O(l) (d)Ammonia gas dissolves in methylbenzene/benzene /kerosene but does not dissociate into free ions therefore existing as molecules 6. Solvents are either polar or non-polar. A polar solvent is one which dissolves ionic compounds and other polar solvents. Water is polar solvent that dissolves ionic and polar substance by surrounding the free ions as below: H ð+ H ð+ O ð- H ð+ H ð+ H ð+ O ð- H ð+ H ð+ H ð+
O ð- H+ O ð- O ð- Cl - O ð- H ð+ Hð+ H ð+ O ð- H ð+ H+ H ð+ H ð+ H ð+ O ð- Beaker Cl- Cl- H+ water H+ Cl- H+ Free ions Note:Water is polar .It is made up of : Oxygen atom is partially negative and two hydrogen atoms which are partially positive. They surround the free H+ and Cl- ions. A non polar solvent is one which dissolved non-polar substances and covalent compounds. If a polar ionic compound is dissolved in non-polar solvent ,it does not ionize/dissociate into free ions as below: H-Cl H-Cl methyl benzene H-Cl H-Cl Covalent bond 7. Some acids and bases are strong while others are weak. (a)A strong acid/base is one which is fully/wholly/completely dissociated / ionized into many free H+ /OH- ions i.e. I. Strong acids exists more as free H+ ions than molecules. e.g.
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I. Strong acids exists more as free H+ ions than molecules. e.g. HCl(aq) H+(aq) + Cl- (aq) (molecules) (cation) (anion) HNO3(aq) H+(aq) + NO3-(aq) (molecules) (cation) (anion) H2SO4(aq) 2H+(aq) + SO42-(aq) (molecules) (cation) (anion) II. Strong bases/alkalis exists more as free OH- ions than molecules. e.g. KOH(aq) K+(aq) + OH- (aq) (molecules) (cation) (anion) NaOH(aq) Na+(aq) + OH-(aq) (molecules) (cation) (anion) (b) A weak base/acid is one which is partially /partly dissociated /ionized in water into free OH- (aq) and H+(aq) ions. I. Weak acids exists more as molecules than as free H+ ions. e.g. CH3COOH(aq) H+(aq) + CH3COO- (aq) (molecules) (cation) (anion) H3PO4(aq) 3H+(aq) + PO43-(aq) (molecules) (cation) (anion) H2CO3(aq) 2H+(aq) + CO32-(aq) (molecules) (cation) (anion) II. Weak bases/alkalis exists more as molecules than free OH- ions. e.g. NH4OH(aq) NH4+(aq) + OH- (aq) (molecules) (cation) (anion) Ca(OH)2(aq) Ca2+(aq) + 2OH-(aq) (molecules) (cation) (anion) Mg(OH)2(aq) Mg2+(aq) + 2OH-(aq) (molecules) (cation) (anion)
8. The concentration of an acid/base/alkali is based on the number of moles of acid/bases dissolved in a decimeter(litre)of the solution. An acid/base/alkali with more acid/base/alkali in a decimeter(litre) of solution is said to be concentrated while that with less is said to be dilute. 9.
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The concentration of an acid/base/alkali is based on the number of moles of acid/bases dissolved in a decimeter(litre)of the solution. An acid/base/alkali with more acid/base/alkali in a decimeter(litre) of solution is said to be concentrated while that with less is said to be dilute. 9. (a) (i)strong acids have pH 1/2/3 while weak acids have high pH 4/5/6. (ii)a neutral solution have pH 7 (iii)strong alkalis/bases have pH 12/13/14 while weak bases/alkalis have pH 11/10 /9 / 8. (b) pH is a measure of H+(aq) concentration in a solution. The higher the H+(aq)ions concentration ; -the higher the acidity -the lower the pH -the lower the concentration of OH-(aq) -the lower the alkalinity At pH 7 , a solution has equal concentration of H+(aq) and OH-(aq). Beyond pH 7,the concentration of the OH-(aq) increases as the H+(aq) ions decreases. 10.(a) When acids /bases dissolve in water, the ions present in the solution conduct electricity. The more the dissociation the higher the yield of ions and the greater the electrical conductivity of the solution. A compound that conducts electricity in an electrolyte and thus a compound showing high electrical conductivity is a strong electrolyte while a compound showing low electrical conductivity is a weak electrolyte. (b) Practically, a bright light on a bulb ,a high voltage reading from a voltmeter high ammeter reading from an ammeter, a big deflection on a galvanometer is an indicator of strong electrolyte(acid/base) and the opposite for weak electrolytes(acids/base) 11. Some compounds exhibit/show both properties of acids and bases/alkalis. A substance that reacts with both acids and bases is said to be amphotellic. The examples below show the amphotellic properties of: (a) Zinc (II)oxide(ZnO) and Zinc hydroxide(Zn(OH)2) (i)When ½ spatula full of Zinc(II)oxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution.
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Some compounds exhibit/show both properties of acids and bases/alkalis. A substance that reacts with both acids and bases is said to be amphotellic. The examples below show the amphotellic properties of: (a) Zinc (II)oxide(ZnO) and Zinc hydroxide(Zn(OH)2) (i)When ½ spatula full of Zinc(II)oxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e. (i) when reacting with nitric(V)acid, the oxide shows basic properties by reacting with an acid to form a simple salt and water only. Basic oxide + Acid -> salt + water Examples: Chemical equation ZnO(s) + 2HNO3(aq) -> Zn(NO3) 2 (aq) + H2O(l) ZnO(s) + 2HCl(aq) -> ZnCl2 (aq) + H2O(l) ZnO(s) + H2SO4(aq) -> ZnSO4 (aq) + H2O(l) Ionic equation ZnO(s) + 2H+ (aq) -> Zn 2+ (aq) + H2O(l) (ii) when reacting with sodium hydroxide, the oxide shows acidic properties by reacting with a base to form a complex salt. Basic oxide + Base/alkali + Water -> Complex salt Examples: Chemical equation 1.When Zinc oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxozincate(II) complex salt. ZnO(s) + 2NaOH(aq) + H2O(l) -> Na2Zn(OH) 4(aq) 2.When Zinc oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxozincate(II) complex salt.
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Basic oxide + Acid -> salt + water Examples: Chemical equation ZnO(s) + 2HNO3(aq) -> Zn(NO3) 2 (aq) + H2O(l) ZnO(s) + 2HCl(aq) -> ZnCl2 (aq) + H2O(l) ZnO(s) + H2SO4(aq) -> ZnSO4 (aq) + H2O(l) Ionic equation ZnO(s) + 2H+ (aq) -> Zn 2+ (aq) + H2O(l) (ii) when reacting with sodium hydroxide, the oxide shows acidic properties by reacting with a base to form a complex salt. Basic oxide + Base/alkali + Water -> Complex salt Examples: Chemical equation 1.When Zinc oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxozincate(II) complex salt. ZnO(s) + 2NaOH(aq) + H2O(l) -> Na2Zn(OH) 4(aq) 2.When Zinc oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxozincate(II) complex salt. ZnO(s) + 2KOH(aq) + H2O(l) -> K2Zn(OH) 4(aq) Ionic equation ZnO(s) + 2OH-(aq) + H2O(l) -> 2[Zn(OH) 4]2- (aq) (ii)When Zinc(II)hydroxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e. (i) when reacting with nitric(V)acid, the hydroxide shows basic properties. It reacts with an acid to form a simple salt and water only.
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i.e. (i) when reacting with nitric(V)acid, the hydroxide shows basic properties. It reacts with an acid to form a simple salt and water only. Basic hydroxide + Acid -> salt + water Examples: Chemical equation Zn(OH) 2 (s) + 2HNO3(aq) -> Zn(NO3) 2 (aq) + 2H2O(l)
Zn(OH) 2 (s) + 2HCl(aq) -> ZnCl2 (aq) + 2H2O(l) Zn(OH) 2 (s) + H2SO4(aq) -> ZnSO4 (aq) + 2H2O(l) Ionic equation Zn(OH) 2 (s) + 2H+ (aq) -> Zn 2+ (aq) + 2H2O(l) (ii) when reacting with sodium hydroxide, the hydroxide shows acidic properties by reacting with a base to form a complex salt. Basic hydroxide + Base/alkali -> Complex salt Examples: Chemical equation 1.When Zinc hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxozincate(II) complex salt. Zn(OH) 2 (s) + 2NaOH(aq) -> Na2Zn(OH) 4(aq) 2.When Zinc hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxozincate(II) complex salt. Zn(OH) 2 (s) + 2KOH(aq) -> K2Zn(OH) 4(aq) Ionic equation Zn(OH) 2 (s) + 2OH-(aq) -> 2[Zn(OH) 4]2- (aq) (b) Lead (II)oxide(PbO) and Lead(II) hydroxide (Pb(OH)2) (i)When ½ spatula full of Lead(II)oxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e.
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Zn(OH) 2 (s) + 2NaOH(aq) -> Na2Zn(OH) 4(aq) 2.When Zinc hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxozincate(II) complex salt. Zn(OH) 2 (s) + 2KOH(aq) -> K2Zn(OH) 4(aq) Ionic equation Zn(OH) 2 (s) + 2OH-(aq) -> 2[Zn(OH) 4]2- (aq) (b) Lead (II)oxide(PbO) and Lead(II) hydroxide (Pb(OH)2) (i)When ½ spatula full of Lead(II)oxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e. (i) when reacting with nitric(V)acid, the oxide shows basic properties by reacting with an acid to form a simple salt and water only. All other Lead salts are insoluble. Chemical equation PbO(s) + 2HNO3(aq) -> Pb(NO3) 2 (aq) + H2O(l) Ionic equation PbO(s) + 2H+ (aq) -> Pb 2+ (aq) + H2O(l) (ii) when reacting with sodium hydroxide, the oxide shows acidic properties by reacting with a base to form a complex salt. Chemical equation 1.When Lead(II) oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoplumbate(II) complex salt. PbO(s) + 2NaOH(aq) + H2O(l) -> Na2Pb(OH) 4(aq) 2.When Lead(II) oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoplumbate(II) complex salt.
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Chemical equation PbO(s) + 2HNO3(aq) -> Pb(NO3) 2 (aq) + H2O(l) Ionic equation PbO(s) + 2H+ (aq) -> Pb 2+ (aq) + H2O(l) (ii) when reacting with sodium hydroxide, the oxide shows acidic properties by reacting with a base to form a complex salt. Chemical equation 1.When Lead(II) oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoplumbate(II) complex salt. PbO(s) + 2NaOH(aq) + H2O(l) -> Na2Pb(OH) 4(aq) 2.When Lead(II) oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoplumbate(II) complex salt. PbO(s) + 2KOH(aq) + H2O(l) -> K2Pb(OH) 4(aq) Ionic equation PbO(s) + 2OH-(aq) + H2O(l) -> 2[Pb(OH) 4]2- (aq) (ii)When Lead(II)hydroxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e. (i) when reacting with nitric(V)acid, the hydroxide shows basic properties. It reacts with the acid to form a simple salt and water only. Chemical equation Pb(OH) 2 (s) + 2HNO3(aq) -> Pb(NO3) 2 (aq) + 2H2O(l) Ionic equation Pb(OH) 2 (s) + 2H+ (aq) -> Pb 2+ (aq) + 2H2O(l) (ii) when reacting with sodium hydroxide, the hydroxide shows acidic properties. It reacts with a base to form a complex salt. Chemical equation 1.When Lead(II) hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoplumbate(II) complex salt.
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Chemical equation Pb(OH) 2 (s) + 2HNO3(aq) -> Pb(NO3) 2 (aq) + 2H2O(l) Ionic equation Pb(OH) 2 (s) + 2H+ (aq) -> Pb 2+ (aq) + 2H2O(l) (ii) when reacting with sodium hydroxide, the hydroxide shows acidic properties. It reacts with a base to form a complex salt. Chemical equation 1.When Lead(II) hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoplumbate(II) complex salt. Pb(OH) 2 (s) + 2NaOH(aq) -> Na2Pb(OH) 4(aq) 2.When Lead(II) hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoplumbate(II) complex salt. Pb(OH) 2 (s) + 2KOH(aq) -> K2Pb(OH) 4(aq) Ionic equation Pb(OH) 2 (s) + 2OH-(aq) -> 2[Pb(OH) 4]2- (aq)
(c)Aluminium(III)oxide(Al2O3) and Aluminium(III)hydroxide(Al(OH)3) (i)When ½ spatula full of Aluminium(III)oxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e. (i) when reacting with nitric(V)acid, the oxide shows basic properties by reacting with an acid to form a simple salt and water only.
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Pb(OH) 2 (s) + 2KOH(aq) -> K2Pb(OH) 4(aq) Ionic equation Pb(OH) 2 (s) + 2OH-(aq) -> 2[Pb(OH) 4]2- (aq)
(c)Aluminium(III)oxide(Al2O3) and Aluminium(III)hydroxide(Al(OH)3) (i)When ½ spatula full of Aluminium(III)oxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e. (i) when reacting with nitric(V)acid, the oxide shows basic properties by reacting with an acid to form a simple salt and water only. Chemical equation Al2O3 (s) + 6HNO3(aq) -> Al(NO3)3 (aq) + 3H2O(l) Al2O3 (s) + 6HCl(aq) -> AlCl3 (aq) + 3H2O(l) Al2O3 (s) + 3H2SO4(aq) -> Al2(SO4)3 (aq) + 3H2O(l) Ionic equation Al2O3 (s) + 3H+ (aq) -> Al 3+ (aq) + 3H2O(l) (ii) when reacting with sodium hydroxide, the oxide shows acidic properties by reacting with a base to form a complex salt. Chemical equation 1.When Aluminium(III) oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoaluminate(III) complex salt. Al2O3 (s) + 2NaOH(aq) + 3H2O(l) -> 2NaAl(OH) 4(aq) 2.When Aluminium(III) oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoaluminate(II) complex salt.
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Chemical equation Al2O3 (s) + 6HNO3(aq) -> Al(NO3)3 (aq) + 3H2O(l) Al2O3 (s) + 6HCl(aq) -> AlCl3 (aq) + 3H2O(l) Al2O3 (s) + 3H2SO4(aq) -> Al2(SO4)3 (aq) + 3H2O(l) Ionic equation Al2O3 (s) + 3H+ (aq) -> Al 3+ (aq) + 3H2O(l) (ii) when reacting with sodium hydroxide, the oxide shows acidic properties by reacting with a base to form a complex salt. Chemical equation 1.When Aluminium(III) oxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoaluminate(III) complex salt. Al2O3 (s) + 2NaOH(aq) + 3H2O(l) -> 2NaAl(OH) 4(aq) 2.When Aluminium(III) oxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoaluminate(II) complex salt. Al2O3 (s) + 2KOH(aq) + 3H2O(l) -> 2NaAl(OH) 4(aq) Ionic equation Al2O3 (s) + 2OH-(aq) + 3H2O(l) -> 2[Al(OH) 4]- (aq) (ii)When Aluminium(III)hydroxide is placed in a boiling tube containing 10cm3 of either 2M nitric(V)acid or 2M sodium hydroxide hydroxide solution, it dissolves on both the acid and the alkali/base to form a colourless solution. i.e. (i) when reacting with nitric(V)acid, the hydroxide shows basic properties. It reacts with the acid to form a simple salt and water only.
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i.e. (i) when reacting with nitric(V)acid, the hydroxide shows basic properties. It reacts with the acid to form a simple salt and water only. Chemical equation Al(OH) 3 (s) + 3HNO3(aq) -> Al(NO3)3 (aq) + 3H2O(l)
Al(OH)3 (s) + 3HCl(aq) -> AlCl3 (aq) + 3H2O(l) 2Al(OH)3 (s) + 3H2SO4(aq) -> Al2(SO4)3 (aq) + 3H2O(l) Ionic equation Al(OH)3 (s) + 3H+ (aq) -> Al 3+ (aq) + 3H2O(l) (ii) when reacting with sodium hydroxide, the hydroxide shows acidic properties. It reacts with a base to form a complex salt. Chemical equation 1.When aluminium(III) hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoaluminate(III) complex salt. Al(OH) 3 (s) + NaOH(aq) -> NaAl(OH) 4(aq) 2.When aluminium(III) hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoaluminate(III) complex salt.
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It reacts with a base to form a complex salt. Chemical equation 1.When aluminium(III) hydroxide is reacted with sodium hydroxide the complex salt is sodium tetrahydroxoaluminate(III) complex salt. Al(OH) 3 (s) + NaOH(aq) -> NaAl(OH) 4(aq) 2.When aluminium(III) hydroxide is reacted with potassium hydroxide the complex salt is potassium tetrahydroxoaluminate(III) complex salt. Al(OH) 3 (s) + KOH(aq) -> KAl(OH) 4(aq) Ionic equation Al(OH) 3 (s) + OH-(aq) -> [Al(OH) 4]- (aq) Summary of amphotellic oxides/hydroxides Oxide Hydroxide Formula of simple salt from nitric (V)acid Formula of complex salt from sodium hydroxide ZnO Zn(OH)2 Zn(NO3)2 Na2Zn(OH)4 [Zn(OH)4]2-(aq) Sodium tetrahydroxozincate(II) PbO Pb(OH)2 Pb(NO3)2 Na2Pb(OH)4 [Pb(OH)4]2-(aq) Sodium tetrahydroxoplumbate(II) Al2O3 Al(OH) 3 Al(NO3)3 NaAl(OH)4 [Al(OH)4]-(aq) Sodium tetrahydroxoaluminate(II) 12.(a) A salt is an ionic compound formed when the cation from a base combine with the anion derived from an acid. A salt is therefore formed when the hydrogen ions in an acid are replaced wholly/fully or partially/partly ,directly or indirectly by a metal or ammonium radical. (b) The number of ionizable/replaceable hydrogen in an acid is called basicity of an acid. Some acids are therefore: (i)monobasic acids generally denoted HX e.g. HCl, HNO3,HCOOH,CH3COOH. (ii)dibasic acids ; generally denoted H2X e.g. H2SO4, H2SO3, H2CO3,HOOCOOH. (iii)tribasic acids ; generally denoted H3X e.g. H3PO4.
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H2SO4, H2SO3, H2CO3,HOOCOOH. (iii)tribasic acids ; generally denoted H3X e.g. H3PO4. (c) Some salts are normal salts while other are acid salts. (i)A normal salt is formed when all the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical. (ii)An acid salt is formed when part/portion the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical. Table showing normal and acid salts derived from common acids Acid name Chemical formula Basicity Normal salt Acid salt Hydrochloric acid HCl Monobasic Chloride(Cl-) None Nitric(V)acid HNO3 Monobasic Nitrate(V)(NO3-) None Nitric(III)acid HNO2 Monobasic Nitrate(III)(NO2-) None Sulphuric(VI)acid H2SO4 Dibasic Sulphate(VI) (SO42-) Hydrogen sulphate(VI) (HSO4-) Sulphuric(IV)acid H2SO3 Dibasic Sulphate(IV) (SO32-) Hydrogen sulphate(IV) (HSO3-) Carbonic(IV)acid H2CO3 Dibasic Carbonate(IV)(CO32-) Hydrogen carbonate(IV) (HCO3-) Phosphoric(V) H3PO4 Tribasic Phosphate(V)(PO43-) Dihydrogen
acid phosphate(V) (H2PO42-) Hydrogen diphosphate(V) (HP2O42-) The table below show shows some examples of salts.
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(i)A normal salt is formed when all the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical. (ii)An acid salt is formed when part/portion the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical. Table showing normal and acid salts derived from common acids Acid name Chemical formula Basicity Normal salt Acid salt Hydrochloric acid HCl Monobasic Chloride(Cl-) None Nitric(V)acid HNO3 Monobasic Nitrate(V)(NO3-) None Nitric(III)acid HNO2 Monobasic Nitrate(III)(NO2-) None Sulphuric(VI)acid H2SO4 Dibasic Sulphate(VI) (SO42-) Hydrogen sulphate(VI) (HSO4-) Sulphuric(IV)acid H2SO3 Dibasic Sulphate(IV) (SO32-) Hydrogen sulphate(IV) (HSO3-) Carbonic(IV)acid H2CO3 Dibasic Carbonate(IV)(CO32-) Hydrogen carbonate(IV) (HCO3-) Phosphoric(V) H3PO4 Tribasic Phosphate(V)(PO43-) Dihydrogen
acid phosphate(V) (H2PO42-) Hydrogen diphosphate(V) (HP2O42-) The table below show shows some examples of salts. Base/alkali Cation Acid Anion Salt Chemical name of salts NaOH Na+ HCl Cl NaCl Sodium(I)chloride Mg(OH)2 Mg2+ H2SO4 SO42 MgSO4 Mg(HSO4)2 Magnesium sulphate(VI) Magnesium hydrogen sulphate(VI) Pb(OH)2 Pb2+ HNO3 NO3 Pb(NO3)2 Lead(II)nitrate(V) Ba(OH)2 Ba2+ HNO3 NO3 Ba(NO3)2 Barium(II)nitrate(V) Ca(OH)2 Ba2+ H2SO4 SO42 MgSO4 Calcium sulphate(VI) NH4OH NH4+ H3PO4 PO43 (NH4 )3PO4 (NH4 )2HPO4 NH4 H2PO4 Ammonium phosphate(V) Diammonium phosphate(V) Ammonium diphosphate(V) KOH K+ H3PO4 PO43 K3PO4 Potassium phosphate(V) Al(OH)3 Al3+ H2SO4 SO42 Al2(SO4)2 Aluminium(III)sulphate(VI) Fe(OH)2 Fe2+ H2SO4 SO42 FeSO4 Iron(II)sulphate(VI) Fe(OH)3 Fe3+ H2SO4 SO42 Fe2(SO4)2 Iron(III)sulphate(VI) (d) Some salts undergo hygroscopy, deliquescence and efflorescence.
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(ii)An acid salt is formed when part/portion the ionizable /replaceable hydrogen in an acid is replaced by a metal or metallic /ammonium radical. Table showing normal and acid salts derived from common acids Acid name Chemical formula Basicity Normal salt Acid salt Hydrochloric acid HCl Monobasic Chloride(Cl-) None Nitric(V)acid HNO3 Monobasic Nitrate(V)(NO3-) None Nitric(III)acid HNO2 Monobasic Nitrate(III)(NO2-) None Sulphuric(VI)acid H2SO4 Dibasic Sulphate(VI) (SO42-) Hydrogen sulphate(VI) (HSO4-) Sulphuric(IV)acid H2SO3 Dibasic Sulphate(IV) (SO32-) Hydrogen sulphate(IV) (HSO3-) Carbonic(IV)acid H2CO3 Dibasic Carbonate(IV)(CO32-) Hydrogen carbonate(IV) (HCO3-) Phosphoric(V) H3PO4 Tribasic Phosphate(V)(PO43-) Dihydrogen
acid phosphate(V) (H2PO42-) Hydrogen diphosphate(V) (HP2O42-) The table below show shows some examples of salts. Base/alkali Cation Acid Anion Salt Chemical name of salts NaOH Na+ HCl Cl NaCl Sodium(I)chloride Mg(OH)2 Mg2+ H2SO4 SO42 MgSO4 Mg(HSO4)2 Magnesium sulphate(VI) Magnesium hydrogen sulphate(VI) Pb(OH)2 Pb2+ HNO3 NO3 Pb(NO3)2 Lead(II)nitrate(V) Ba(OH)2 Ba2+ HNO3 NO3 Ba(NO3)2 Barium(II)nitrate(V) Ca(OH)2 Ba2+ H2SO4 SO42 MgSO4 Calcium sulphate(VI) NH4OH NH4+ H3PO4 PO43 (NH4 )3PO4 (NH4 )2HPO4 NH4 H2PO4 Ammonium phosphate(V) Diammonium phosphate(V) Ammonium diphosphate(V) KOH K+ H3PO4 PO43 K3PO4 Potassium phosphate(V) Al(OH)3 Al3+ H2SO4 SO42 Al2(SO4)2 Aluminium(III)sulphate(VI) Fe(OH)2 Fe2+ H2SO4 SO42 FeSO4 Iron(II)sulphate(VI) Fe(OH)3 Fe3+ H2SO4 SO42 Fe2(SO4)2 Iron(III)sulphate(VI) (d) Some salts undergo hygroscopy, deliquescence and efflorescence. (i) Hygroscopic salts /compounds are those that absorb water from the atmosphere but do not form a solution.
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Table showing normal and acid salts derived from common acids Acid name Chemical formula Basicity Normal salt Acid salt Hydrochloric acid HCl Monobasic Chloride(Cl-) None Nitric(V)acid HNO3 Monobasic Nitrate(V)(NO3-) None Nitric(III)acid HNO2 Monobasic Nitrate(III)(NO2-) None Sulphuric(VI)acid H2SO4 Dibasic Sulphate(VI) (SO42-) Hydrogen sulphate(VI) (HSO4-) Sulphuric(IV)acid H2SO3 Dibasic Sulphate(IV) (SO32-) Hydrogen sulphate(IV) (HSO3-) Carbonic(IV)acid H2CO3 Dibasic Carbonate(IV)(CO32-) Hydrogen carbonate(IV) (HCO3-) Phosphoric(V) H3PO4 Tribasic Phosphate(V)(PO43-) Dihydrogen
acid phosphate(V) (H2PO42-) Hydrogen diphosphate(V) (HP2O42-) The table below show shows some examples of salts. Base/alkali Cation Acid Anion Salt Chemical name of salts NaOH Na+ HCl Cl NaCl Sodium(I)chloride Mg(OH)2 Mg2+ H2SO4 SO42 MgSO4 Mg(HSO4)2 Magnesium sulphate(VI) Magnesium hydrogen sulphate(VI) Pb(OH)2 Pb2+ HNO3 NO3 Pb(NO3)2 Lead(II)nitrate(V) Ba(OH)2 Ba2+ HNO3 NO3 Ba(NO3)2 Barium(II)nitrate(V) Ca(OH)2 Ba2+ H2SO4 SO42 MgSO4 Calcium sulphate(VI) NH4OH NH4+ H3PO4 PO43 (NH4 )3PO4 (NH4 )2HPO4 NH4 H2PO4 Ammonium phosphate(V) Diammonium phosphate(V) Ammonium diphosphate(V) KOH K+ H3PO4 PO43 K3PO4 Potassium phosphate(V) Al(OH)3 Al3+ H2SO4 SO42 Al2(SO4)2 Aluminium(III)sulphate(VI) Fe(OH)2 Fe2+ H2SO4 SO42 FeSO4 Iron(II)sulphate(VI) Fe(OH)3 Fe3+ H2SO4 SO42 Fe2(SO4)2 Iron(III)sulphate(VI) (d) Some salts undergo hygroscopy, deliquescence and efflorescence. (i) Hygroscopic salts /compounds are those that absorb water from the atmosphere but do not form a solution. Some salts which are hygroscopic include anhydrous copper(II)sulphate(VI), anhydrous cobalt(II)chloride, potassium nitrate(V) common table salt.
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e.g. CuCO3 (s) -> CuO(s) + CO2(g) CaCO3 (s) -> CaO(s) + CO2(g) PbCO3 (s) -> PbO(s) + CO2(g) FeCO3 (s) -> FeO(s) + CO2(g) ZnCO3 (s) -> ZnO(s) + CO2(g) (iii)Sodium hydrogen carbonate(IV) and Potassium hydrogen carbonate(IV)decompose on heating to give the corresponding carbonate (IV) and form water and carbon(IV)oxide gas. i.e. 2NaHCO 3(s) -> Na2CO3(s) + CO2(g) + H2O(l) 2KHCO 3(s) -> K2CO3(s) + CO2(g) + H2O(l) (iii) Calcium hydrogen carbonate (IV) and Magnesium hydrogen carbonate(IV) decompose on heating to give the corresponding carbonate (IV) and form water and carbon(IV)oxide gas. i. e. Ca(HCO3) 2(aq) -> CaCO3(s) + CO2(g) + H2O(l) Mg(HCO3) 2(aq) -> MgCO3(s) + CO2(g) + H2O(l) 15. Salts contain cation(positively charged ion) and anions(negatively charged ion).When dissolved in polar solvents/water. The cation and anion in a salt is determined/known usually by precipitation of the salt using a precipitating reagent. The colour of the precipitate is a basis of qualitative analysis of a compound. 16.Qualitative analysis is the process of identifying an unknown compound /salt by identifying the unique qualities of the salt/compound. It involves some of the following processes. (a)Reaction of cation with sodium/potassium hydroxide solution. Both sodium/potassium hydroxide solutions are precipitating reagents. The alkalis produce unique colour of a precipitate/suspension when a few/three drops is added and then excess alkali is added to unknown salt/compound solution. NB: Potassium hydroxide is not commonly used because it is more expensive than sodium hydroxide.
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Both sodium/potassium hydroxide solutions are precipitating reagents. The alkalis produce unique colour of a precipitate/suspension when a few/three drops is added and then excess alkali is added to unknown salt/compound solution. NB: Potassium hydroxide is not commonly used because it is more expensive than sodium hydroxide. The table below shows the observations, inferences / deductions and explanations from the following test tube experiments: Procedure Put about 2cm3 of MgCl2, CaCl2, AlCl3, NaCl, KCl, FeSO4, Fe2(SO4) 3, CuSO4, ZnSO4NH4NO3, Pb(NO3) 2, Ba(NO3) 2 each into separate test tubes. Add three
drops of 2M sodium hydroxide solution then excess (2/3 the length of a standard test tube). Observation Inference Explanation No white precipitate Na+ and K+ Both Na+ and K+ ions react with OH- from 2M sodium hydroxide solution to form soluble colourless solutions Na+(aq) + OH-(aq) -> NaOH(aq) K+(aq) + OH-(aq) -> KOH(aq) No white precipitate then pungent smell of ammonia /urine NH4+ ions NH4+ ions react with 2M sodium hydroxide solution to produce pungent smelling ammonia gas NH4+ (aq) + OH-(aq) -> NH3 (g) + H2O(l) White precipitate insoluble in excess Ba2+ ,Ca2+, Mg2+ ions Ba2+ ,Ca2+ and Mg2+ ions react with OH- from 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides. Ba2+(aq) + 2OH-(aq) -> Ba(OH) 2(s) Ca2+(aq) + 2OH-(aq) -> Ca(OH) 2(s) Mg2+(aq) + 2OH-(aq) -> Mg(OH) 2(s) White precipitate soluble in excess Zn2+ ,Pb2+, Al3+ ions Pb2+ ,Zn2+ and Al3+ ions react with OH- from 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides.
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Add three
drops of 2M sodium hydroxide solution then excess (2/3 the length of a standard test tube). Observation Inference Explanation No white precipitate Na+ and K+ Both Na+ and K+ ions react with OH- from 2M sodium hydroxide solution to form soluble colourless solutions Na+(aq) + OH-(aq) -> NaOH(aq) K+(aq) + OH-(aq) -> KOH(aq) No white precipitate then pungent smell of ammonia /urine NH4+ ions NH4+ ions react with 2M sodium hydroxide solution to produce pungent smelling ammonia gas NH4+ (aq) + OH-(aq) -> NH3 (g) + H2O(l) White precipitate insoluble in excess Ba2+ ,Ca2+, Mg2+ ions Ba2+ ,Ca2+ and Mg2+ ions react with OH- from 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides. Ba2+(aq) + 2OH-(aq) -> Ba(OH) 2(s) Ca2+(aq) + 2OH-(aq) -> Ca(OH) 2(s) Mg2+(aq) + 2OH-(aq) -> Mg(OH) 2(s) White precipitate soluble in excess Zn2+ ,Pb2+, Al3+ ions Pb2+ ,Zn2+ and Al3+ ions react with OH- from 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides. Zn2+(aq) + 2OH-(aq) -> Zn(OH) 2(s) Pb2+(aq) + 2OH-(aq) -> Pb(OH) 2(s) Al3+(aq) + 3OH-(aq) -> Al(OH) 3(s)
The hydroxides formed react with more OH- ions to form complex salts/ions.
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Observation Inference Explanation No white precipitate Na+ and K+ Both Na+ and K+ ions react with OH- from 2M sodium hydroxide solution to form soluble colourless solutions Na+(aq) + OH-(aq) -> NaOH(aq) K+(aq) + OH-(aq) -> KOH(aq) No white precipitate then pungent smell of ammonia /urine NH4+ ions NH4+ ions react with 2M sodium hydroxide solution to produce pungent smelling ammonia gas NH4+ (aq) + OH-(aq) -> NH3 (g) + H2O(l) White precipitate insoluble in excess Ba2+ ,Ca2+, Mg2+ ions Ba2+ ,Ca2+ and Mg2+ ions react with OH- from 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides. Ba2+(aq) + 2OH-(aq) -> Ba(OH) 2(s) Ca2+(aq) + 2OH-(aq) -> Ca(OH) 2(s) Mg2+(aq) + 2OH-(aq) -> Mg(OH) 2(s) White precipitate soluble in excess Zn2+ ,Pb2+, Al3+ ions Pb2+ ,Zn2+ and Al3+ ions react with OH- from 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides. Zn2+(aq) + 2OH-(aq) -> Zn(OH) 2(s) Pb2+(aq) + 2OH-(aq) -> Pb(OH) 2(s) Al3+(aq) + 3OH-(aq) -> Al(OH) 3(s)
The hydroxides formed react with more OH- ions to form complex salts/ions. Zn(OH) 2(s) + 2OH(aq) -> [ Zn(OH) 4]2-(aq) Pb(OH) 2(s) + 2OH(aq) -> [ Pb(OH) 4]2-(aq) Al(OH) 3(s) + OH(aq) -> [ Al(OH) 4]-(aq) Blue precipitate insoluble in excess Cu2+ Cu2+ ions react with OH- from 2M sodium hydroxide solution to form insoluble blue precipitate of copper(II) hydroxide.
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Ba2+(aq) + 2OH-(aq) -> Ba(OH) 2(s) Ca2+(aq) + 2OH-(aq) -> Ca(OH) 2(s) Mg2+(aq) + 2OH-(aq) -> Mg(OH) 2(s) White precipitate soluble in excess Zn2+ ,Pb2+, Al3+ ions Pb2+ ,Zn2+ and Al3+ ions react with OH- from 2M sodium hydroxide solution to form insoluble white precipitate of their hydroxides. Zn2+(aq) + 2OH-(aq) -> Zn(OH) 2(s) Pb2+(aq) + 2OH-(aq) -> Pb(OH) 2(s) Al3+(aq) + 3OH-(aq) -> Al(OH) 3(s)
The hydroxides formed react with more OH- ions to form complex salts/ions. Zn(OH) 2(s) + 2OH(aq) -> [ Zn(OH) 4]2-(aq) Pb(OH) 2(s) + 2OH(aq) -> [ Pb(OH) 4]2-(aq) Al(OH) 3(s) + OH(aq) -> [ Al(OH) 4]-(aq) Blue precipitate insoluble in excess Cu2+ Cu2+ ions react with OH- from 2M sodium hydroxide solution to form insoluble blue precipitate of copper(II) hydroxide. Cu2+(aq) + 2OH-(aq) -> Cu(OH) 2(s) Green precipitate insoluble in excess On adding 3cm3 of hydrogen peroxide, brown/yellow solution formed Fe2+ Fe2+ oxidized to Fe3+ Fe2+ ions react with OH- from 2M sodium hydroxide solution to form insoluble green precipitate of Iron(II) hydroxide.
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Zn2+(aq) + 2OH-(aq) -> Zn(OH) 2(s) Pb2+(aq) + 2OH-(aq) -> Pb(OH) 2(s) Al3+(aq) + 3OH-(aq) -> Al(OH) 3(s)
The hydroxides formed react with more OH- ions to form complex salts/ions. Zn(OH) 2(s) + 2OH(aq) -> [ Zn(OH) 4]2-(aq) Pb(OH) 2(s) + 2OH(aq) -> [ Pb(OH) 4]2-(aq) Al(OH) 3(s) + OH(aq) -> [ Al(OH) 4]-(aq) Blue precipitate insoluble in excess Cu2+ Cu2+ ions react with OH- from 2M sodium hydroxide solution to form insoluble blue precipitate of copper(II) hydroxide. Cu2+(aq) + 2OH-(aq) -> Cu(OH) 2(s) Green precipitate insoluble in excess On adding 3cm3 of hydrogen peroxide, brown/yellow solution formed Fe2+ Fe2+ oxidized to Fe3+ Fe2+ ions react with OH- from 2M sodium hydroxide solution to form insoluble green precipitate of Iron(II) hydroxide. Fe2+(aq) + 2OH-(aq) -> Fe(OH) 2(s) Hydrogen peroxide is an oxidizing agent that oxidizes green Fe2+ oxidized to brown Fe3+ Fe(OH) 2(s) + 2H+ -> Fe(OH) 3(aq) Brown precipitate insoluble in excess Fe3+ Fe3+ ions react with OH- from 2M sodium hydroxide solution to form insoluble brown precipitate of Iron(II) hydroxide. Fe3+(aq) + 3OH-(aq) -> Fe(OH) 3(s) (b)Reaction of cation with aqueous ammonia Aqueous ammonia precipitating reagent that can be used to identify the cations present in a salt. Like NaOH/KOH the OH- ion in NH4OH react with the cation to form a characteristic hydroxide .
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Fe2+(aq) + 2OH-(aq) -> Fe(OH) 2(s) Hydrogen peroxide is an oxidizing agent that oxidizes green Fe2+ oxidized to brown Fe3+ Fe(OH) 2(s) + 2H+ -> Fe(OH) 3(aq) Brown precipitate insoluble in excess Fe3+ Fe3+ ions react with OH- from 2M sodium hydroxide solution to form insoluble brown precipitate of Iron(II) hydroxide. Fe3+(aq) + 3OH-(aq) -> Fe(OH) 3(s) (b)Reaction of cation with aqueous ammonia Aqueous ammonia precipitating reagent that can be used to identify the cations present in a salt. Like NaOH/KOH the OH- ion in NH4OH react with the cation to form a characteristic hydroxide . Below are the observations ,inferences and explanations of the reactions of aqueous ammonia with salts from the following test tube reactions. Procedure Put about 2cm3 of MgCl2, CaCl2, AlCl3, NaCl, KCl, FeSO4, Fe2(SO4) 3, CuSO4, ZnSO4NH4NO3, Pb(NO3) 2, Ba(NO3) 2 each into separate test tubes. Add three drops of 2M aqueous ammonia then excess (2/3 the length of a standard test tube). Observation Inference Explanation No white precipitate Na+ and K+ NH4+,Na+ and K+ ions react with OH- from 2M aqueous ammonia to form soluble colourless solutions NH4+ (aq) + OH-(aq) -> NH4+OH(aq) Na+(aq) + OH-(aq) -> NaOH(aq) K+(aq) + OH-(aq) -> KOH(aq) White precipitate insoluble in excess Ba2+ ,Ca2+, Mg2+ ,Pb2+, Al3+, ions Ba2+ ,Ca2+,Mg2+ ,Pb2+ and Al3+, ions react with OH- from 2M aqueous ammonia to form insoluble white precipitate of their hydroxides.
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Procedure Put about 2cm3 of MgCl2, CaCl2, AlCl3, NaCl, KCl, FeSO4, Fe2(SO4) 3, CuSO4, ZnSO4NH4NO3, Pb(NO3) 2, Ba(NO3) 2 each into separate test tubes. Add three drops of 2M aqueous ammonia then excess (2/3 the length of a standard test tube). Observation Inference Explanation No white precipitate Na+ and K+ NH4+,Na+ and K+ ions react with OH- from 2M aqueous ammonia to form soluble colourless solutions NH4+ (aq) + OH-(aq) -> NH4+OH(aq) Na+(aq) + OH-(aq) -> NaOH(aq) K+(aq) + OH-(aq) -> KOH(aq) White precipitate insoluble in excess Ba2+ ,Ca2+, Mg2+ ,Pb2+, Al3+, ions Ba2+ ,Ca2+,Mg2+ ,Pb2+ and Al3+, ions react with OH- from 2M aqueous ammonia to form insoluble white precipitate of their hydroxides. Pb2+ (aq) + 2OH-(aq) -> Pb(OH) 2(s) Al3+ (aq) + 3OH-(aq) -> Al(OH) 3(s) Ba2+ (aq) + 2OH-(aq) -> Ba(OH) 2(s) Ca2+ (aq) + 2OH-(aq) -> Ca(OH) 2(s) Mg2+ (aq) + 2OH-(aq) -> Mg(OH) 2(s) White precipitate soluble in excess Zn2+ ions Zn2+ ions react with OH- from 2M aqueous ammonia to form insoluble white precipitate of Zinc hydroxide. Zn2+(aq) + 2OH-(aq) -> Zn(OH) 2(s) The Zinc hydroxides formed react NH3(aq) to form a complex salts/ions.
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Observation Inference Explanation No white precipitate Na+ and K+ NH4+,Na+ and K+ ions react with OH- from 2M aqueous ammonia to form soluble colourless solutions NH4+ (aq) + OH-(aq) -> NH4+OH(aq) Na+(aq) + OH-(aq) -> NaOH(aq) K+(aq) + OH-(aq) -> KOH(aq) White precipitate insoluble in excess Ba2+ ,Ca2+, Mg2+ ,Pb2+, Al3+, ions Ba2+ ,Ca2+,Mg2+ ,Pb2+ and Al3+, ions react with OH- from 2M aqueous ammonia to form insoluble white precipitate of their hydroxides. Pb2+ (aq) + 2OH-(aq) -> Pb(OH) 2(s) Al3+ (aq) + 3OH-(aq) -> Al(OH) 3(s) Ba2+ (aq) + 2OH-(aq) -> Ba(OH) 2(s) Ca2+ (aq) + 2OH-(aq) -> Ca(OH) 2(s) Mg2+ (aq) + 2OH-(aq) -> Mg(OH) 2(s) White precipitate soluble in excess Zn2+ ions Zn2+ ions react with OH- from 2M aqueous ammonia to form insoluble white precipitate of Zinc hydroxide. Zn2+(aq) + 2OH-(aq) -> Zn(OH) 2(s) The Zinc hydroxides formed react NH3(aq) to form a complex salts/ions. Zn(OH) 2(s) + 4NH3(aq) ->[ Zn(NH3) 4]2+(aq)+ 2OH-(aq) Blue precipitate Cu2+ Cu2+ ions react with OH- from 2M aqueous
that dissolves in excess ammonia solution to form a deep/royal blue solution ammonia to form blue precipitate of copper(II) hydroxide. Cu2+(aq) + 2OH-(aq) -> Cu(OH) 2(s) The copper(II) hydroxide formed react NH3(aq) to form a complex salts/ions.
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Zn2+(aq) + 2OH-(aq) -> Zn(OH) 2(s) The Zinc hydroxides formed react NH3(aq) to form a complex salts/ions. Zn(OH) 2(s) + 4NH3(aq) ->[ Zn(NH3) 4]2+(aq)+ 2OH-(aq) Blue precipitate Cu2+ Cu2+ ions react with OH- from 2M aqueous
that dissolves in excess ammonia solution to form a deep/royal blue solution ammonia to form blue precipitate of copper(II) hydroxide. Cu2+(aq) + 2OH-(aq) -> Cu(OH) 2(s) The copper(II) hydroxide formed react NH3(aq) to form a complex salts/ions. Cu(OH) 2 (s) + 4NH3(aq) ->[ Cu(NH3) 4]2+(aq)+ 2OH-(aq) Green precipitate insoluble in excess. On adding 3cm3 of hydrogen peroxide, brown/yellow solution formed Fe2+ Fe2+ oxidized to Fe3+ Fe2+ ions react with OH- from 2M aqueous ammonia to form insoluble green precipitate of Iron(II) hydroxide. Fe2+(aq) + 2OH-(aq) -> Fe(OH) 2(s) Hydrogen peroxide is an oxidizing agent that oxidizes green Fe2+ oxidized to brown Fe3+ Fe(OH) 2(s) + 2H+ -> Fe(OH) 3(aq) Brown precipitate insoluble in excess Fe3+ Fe3+ ions react with OH- from 2M aqueous ammonia to form insoluble brown precipitate of Iron(II) hydroxide. Fe3+(aq) + 3OH-(aq) -> Fe(OH) 3(s) Note (i) Only Zn2+ ions/salts form a white precipitate that dissolve in excess of both 2M sodium hydroxide and 2M aqueous ammonia. (ii) Pb2+ and Al3+ ions/salts form a white precipitate that dissolve in excess of 2M sodium hydroxide but not in 2M aqueous ammonia.
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Fe2+(aq) + 2OH-(aq) -> Fe(OH) 2(s) Hydrogen peroxide is an oxidizing agent that oxidizes green Fe2+ oxidized to brown Fe3+ Fe(OH) 2(s) + 2H+ -> Fe(OH) 3(aq) Brown precipitate insoluble in excess Fe3+ Fe3+ ions react with OH- from 2M aqueous ammonia to form insoluble brown precipitate of Iron(II) hydroxide. Fe3+(aq) + 3OH-(aq) -> Fe(OH) 3(s) Note (i) Only Zn2+ ions/salts form a white precipitate that dissolve in excess of both 2M sodium hydroxide and 2M aqueous ammonia. (ii) Pb2+ and Al3+ ions/salts form a white precipitate that dissolve in excess of 2M sodium hydroxide but not in 2M aqueous ammonia. (iii) Cu2+ ions/salts form a blue precipitate that dissolve to form a deep/royal blue solution in excess of 2M aqueous ammonia but only blue insoluble precipitate in 2M sodium hydroxide (c)Reaction of cation with Chloride (Cl-)ions All chlorides are soluble in water except Silver chloride and Lead (II)chloride (That dissolve in hot water).When a soluble chloride like NaCl, KCl, NH4Cl is
added to about 2cm3 of a salt containing Ag+ or Pb2+ions a white precipitate of AgCl or PbCl2 is formed. The following test tube reactions illustrate the above. Experiment Put about 2cm3 of silver nitrate(V) andLead(II)nitrate(V)solution into separate test tubes. Add five drops of NaCl /KCl / NH4Cl/HCl. Heat to boil. Observation Inference Explanation (i)White precipitate does not dissolve on heating Ag+ ions Ag+ ions reacts with Cl- ions from a soluble chloride salt to form a white precipitate of AgCl (ii)White precipitate dissolve on heating Pb2+ ions Pb2+ ions reacts with Cl- ions from a soluble chloride salt to form a white precipitate of PbCl2. PbCl2 dissolves on heating. Note Both Pb2+ and Al3+ ions forms an insoluble white precipitate in excess aqueous ammonia.
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Observation Inference Explanation (i)White precipitate does not dissolve on heating Ag+ ions Ag+ ions reacts with Cl- ions from a soluble chloride salt to form a white precipitate of AgCl (ii)White precipitate dissolve on heating Pb2+ ions Pb2+ ions reacts with Cl- ions from a soluble chloride salt to form a white precipitate of PbCl2. PbCl2 dissolves on heating. Note Both Pb2+ and Al3+ ions forms an insoluble white precipitate in excess aqueous ammonia. A white precipitate on adding Cl- ions/salts shows Pb2+. No white precipitate on adding Cl- ions/salts shows Al3+. Adding a chloride/ Cl- ions/salts can thus be used to separate the identity of Al3+ and Pb2+. (d)Reaction of cation with sulphate(VI)/SO42- and sulphate(IV)/SO32- ions All sulphate(VI) and sulphate(IV)/SO32- ions/salts are soluble/dissolve in water except Calcium sulphate(VI)/CaSO4, Calcium sulphate(IV)/CaSO3, Barium sulphate(VI)/BaSO4, Barium sulphate(IV)/BaSO3, Lead(II) sulphate(VI)/PbSO4 and Lead(II) sulphate(IV)/PbSO3.When a soluble sulphate(VI)/SO42- salt like Na2SO4, H2SO4, (NH4)2SO4 or Na2SO3 is added to a salt containing Ca2+, Pb2+, Ba2+ ions, a white precipitate is formed. The following test tube experiments illustrate the above. Procedure Place about 2cm3 of Ca(NO3)2, Ba(NO3)2, BaCl2 and Pb(NO3)2, in separate boiling tubes. Add six drops of sulphuric(VI)acid /sodium sulphate(VI)/ammonium sulphate(VI)solution. Repeat with six drops of sodium sulphate(IV). Observation Inference Explanation White Ca2+, Ba2+, CaSO3 and CaSO4 do not form a thick precipitate as
precipitate Pb2+ ions they are sparingly soluble.
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Add six drops of sulphuric(VI)acid /sodium sulphate(VI)/ammonium sulphate(VI)solution. Repeat with six drops of sodium sulphate(IV). Observation Inference Explanation White Ca2+, Ba2+, CaSO3 and CaSO4 do not form a thick precipitate as
precipitate Pb2+ ions they are sparingly soluble. Ca2+(aq)+ SO32-(aq) -> CaSO3(s) Ca2+(aq)+ SO42-(aq) -> CaSO4(s) Ba2+(aq)+ SO32-(aq) -> BaSO3(s) Ba2+(aq)+ SO42-(aq) -> BaSO4(s) Pb2+(aq)+ SO32-(aq) -> PbSO3(s) Pb2+(aq)+ SO42-(aq) -> PbSO4(s) (e)Reaction of cation with carbonate(IV)/CO32- ions All carbonate salts are insoluble except sodium/potassium carbonate(IV) and ammonium carbonate(IV). They dissociate /ionize to release CO32- ions. CO32- ions produce a white precipitate when the soluble carbonate salts is added to any metallic cation. Procedure Place about 2cm3 of Ca(NO3)2, Ba(NO3)2, MgCl2 ,Pb(NO3)2 andZnSO4 in separate boiling tubes. Add six drops of Potassium /sodium carbonate(IV)/ ammonium carbonate (IV)solution. Observation Inference Explanation Green precipitate Cu2+ ,Fe2+,ions CO32-(aq) Copper(II)carbonate(IV) and Iron(II) carbonate (IV) are precipitated as insoluble green precipitates. Cu2+(aq)+ CO32-(aq) -> CuCO3(s) Fe2+(aq)+ CO32-(aq) -> FeCO3(s) When sodium carbonate(IV)is added to CuCO3(s) the CO32-(aq) ions are first hydrolysed to produce CO2(g) and OH(aq)ions. CO32-(aq) + H2O (l) -> CO2 (g) + 2OH- (aq) The OH-(aq) ions further react to form basic copper(II) carbonate(IV). Basic copper(II) carbonate(IV) is the only green salt of copper.
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Cu2+(aq)+ CO32-(aq) -> CuCO3(s) Fe2+(aq)+ CO32-(aq) -> FeCO3(s) When sodium carbonate(IV)is added to CuCO3(s) the CO32-(aq) ions are first hydrolysed to produce CO2(g) and OH(aq)ions. CO32-(aq) + H2O (l) -> CO2 (g) + 2OH- (aq) The OH-(aq) ions further react to form basic copper(II) carbonate(IV). Basic copper(II) carbonate(IV) is the only green salt of copper. Cu2+(aq)+ CO32-(aq)+2OH- (aq)
->CuCO3.Cu(OH)2 (s) White precipitate CO32- White ppt of the carbonate(IV)salt is precipitated Ca2+(aq) + CO32- (aq) -> CaCO3(s) Mg2+(aq) + CO32- (aq) -> MgCO3(s) Pb2+(aq) + CO32- (aq) -> PbCO3(s) Zn2+(aq) + CO32- (aq) -> ZnCO3(s) Note (i)Iron(III)carbonate(IV) does not exist. (ii)Copper(II)Carbonate(IV) exist only as the basic CuCO3.Cu(OH) 2 (iii)Both BaCO3 and BaSO3 are insoluble white precipitate. If hydrochloric acid is added to the white precipitate; I. BaCO3 produces CO2 gas. When bubbled/directed into lime water solution,a white precipitate is formed. II. I. BaSO3 produces SO2 gas. When bubbled/directed into orange acidified potassium dichromate(VI) solution, it turns to green/decolorizes acidified potassium manganate(VII). (f) Reaction of cation with sulphide / S2- ions All sulphides are insoluble black solids/precipitates except sodium sulphide/ Na2S/ potassium sulphide/K2S.When a few/3drops of the soluble sulphide is added to a metal cation/salt, a black precipitate is formed. Procedure Place about 2cm3 of Cu(NO3)2, FeSO4, MgCl2,Pb(NO3)2 and ZnSO4 in separate boiling tubes.
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When bubbled/directed into orange acidified potassium dichromate(VI) solution, it turns to green/decolorizes acidified potassium manganate(VII). (f) Reaction of cation with sulphide / S2- ions All sulphides are insoluble black solids/precipitates except sodium sulphide/ Na2S/ potassium sulphide/K2S.When a few/3drops of the soluble sulphide is added to a metal cation/salt, a black precipitate is formed. Procedure Place about 2cm3 of Cu(NO3)2, FeSO4, MgCl2,Pb(NO3)2 and ZnSO4 in separate boiling tubes. Add six drops of Potassium /sodium sulphide solution. Observation Inference Explanation Black ppt S2- ions CuS, FeS,MgS,PbS, ZnS are black insoluble precipitates Cu2+(aq) + S2-(aq) -> CuS(s) Pb2+(aq) + S2-(aq) -> PbS(s) Fe2+(aq) + S2-(aq) -> FeS(s) Zn2+(aq) + S2-(aq) -> ZnS(s) Sample qualitative analysis guide
You are provided with solid Y(aluminium (III)sulphate(VI)hexahydrate).Carry out the following tests and record your observations and inferences in the space provided. 1(a) Appearance Observations inference (1mark) White crystalline solid Coloured ions Cu2+ , Fe2+ ,Fe3+ absent (b)Place about a half spatula full of the solid into a clean dry boiling tube. Heat gently then strongly. Observations inference (1mark) Colourless droplets formed on the cooler Hydrated compound/compound part of the test tube containing water of crystallization Solid remains a white residue (c)Place all the remaining portion of the solid in a test tube .Add about 10cm3 of distilled water. Shake thoroughly. Divide the mixture into five portions. Observation Inference (1mark) Solid dissolves to form Polar soluble compound a colourless solution Cu2+ , Fe2+ ,Fe3+ absent (i)To the first portion, add three drops of sodium hydroxide then add excess of the alkali.
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Shake thoroughly. Divide the mixture into five portions. Observation Inference (1mark) Solid dissolves to form Polar soluble compound a colourless solution Cu2+ , Fe2+ ,Fe3+ absent (i)To the first portion, add three drops of sodium hydroxide then add excess of the alkali. Observation Inference (1mark) White ppt, soluble in excess Zn2+ , Pb2+ , Al3+ (ii)To the second portion, add three drops of aqueous ammonia then add excess of the alkali. Observation Inference (1mark) White ppt, insoluble in excess Pb2+ , Al3+ (iii)To the third portion, add three drops of sodium sulphate(VI)solution. Observation Inference (1mark) No white ppt Al3+
(iv)I.To the fourth portion, add three drops of Lead(II)nitrate(IV)solution. Preserve Observation Inference (1mark) White ppt CO32-, SO42-, SO32-, Cl-, II.To the portion in (iv) I above , add five drops of dilute hydrochloric acid. Observation Inference (1mark) White ppt persist/remains SO42-, Cl-, III.To the portion in (iv) II above, heat to boil. Observation Inference (1mark) White ppt persist/remains SO42-, Note that: (i)From test above, it can be deduced that solid Y is hydrated aluminium(III)sulphate(VI) solid (ii)Any ion inferred from an observation below must be derived from previous correct observation and inferences above. e.g. Al3+ in c(iii) must be correctly inferred in either/or in c(ii) or c(i)above SO42- in c(iv)III must be correctly inferred in either/or in c(iv)II or c(iv)I above (iii)Contradiction in observations and inferences should be avoided.e.g. “White ppt soluble in excess” to infer presence of Al3+ ,Ba2+ ,Pb3+ (iv)Symbols of elements/ions should be correctly capitalized. e.g. “SO4-2” is wrong, “sO42-” is wrong, “cu2+” is wrong. Sample solutions of salt were labeled as I,II, III and IV.
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e.g. “SO4-2” is wrong, “sO42-” is wrong, “cu2+” is wrong. Sample solutions of salt were labeled as I,II, III and IV. The actual solutions, not in that order are lead nitrate, zinc sulphate potassium chloride and calcium chloride. a)When aqueous sodium carbonate was added to each sample separately, a white precipitate was formed in I, III and IV only. Identify solution II. b)When excess sodium hydroxide was added to each sample separately, a white precipitate was formed in solutions III and I only. Identify solution I 17.When solids/salts /solutes are added to a solvent ,some dissolve to form a solution. Solute + Solvent -> Solvent If a solution has a lot of solute dissolved in a solvent ,it is said to be concentrated. If a solution has little solute dissolved in a solvent ,it is said to be dilute. There is a limit to how much solute can dissolve in a given /specified amount of solvent/water at a given /specified temperature. The maximum mass of salt/solid/solute that dissolve in 100g of solvent/water at a specified temperature is called solubility of a salt. When no more solute can dissolve in a given amount of solvent at a specified temperature, a saturated solution is formed. For some salts, on heating, more of the salt/solid/solute dissolve in the saturated solution to form a super saturated solution. The solubility of a salt is thus calculated from the formula Solubility = Mass of solute/salt/solid x 100 Mass/volume of water/solvent Practice examples (a)Calculate the solubility of potassium nitrate(V) if 5.0 g of the salt is dissolved in 50.0cm3 of water. Solubility = Mass of solute/salt/solid x 100 =>( 5.0 x 100 ) = 10.0 g /100g H2O Mass/volume of water/solvent 50.0 (b)Calculate the solubility of potassium chlorate(V) if 50.0 g of the salt is dissolved in 250.0cm3 of water.
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For some salts, on heating, more of the salt/solid/solute dissolve in the saturated solution to form a super saturated solution. The solubility of a salt is thus calculated from the formula Solubility = Mass of solute/salt/solid x 100 Mass/volume of water/solvent Practice examples (a)Calculate the solubility of potassium nitrate(V) if 5.0 g of the salt is dissolved in 50.0cm3 of water. Solubility = Mass of solute/salt/solid x 100 =>( 5.0 x 100 ) = 10.0 g /100g H2O Mass/volume of water/solvent 50.0 (b)Calculate the solubility of potassium chlorate(V) if 50.0 g of the salt is dissolved in 250.0cm3 of water. Solubility = Mass of solute/salt/solid x 100 =>( 50.0 x 100 ) = 20.0 g /100g H2O
Mass/volume of water/solvent 250.0 (c)If the solubility of potassium chlorate(V) is 5g/100g H2O at 80oC,how much can dissolve in 5cm3 of water at 80oC . Mass of solute/salt/solid = Solubility x Mass/volume of water/solvent 100 => 5 x 5 = 0.25g of KClO3 dissolve 100 (d)If the solubility of potassium chlorate(V) is 72g/100g H2O at 20oC,how much can saturate 25g of water at 20oC . Mass of solute/salt/solid = Solubility x Mass/volume of water/solvent 100 => 72 x 25 = 18.0g of KClO3 dissolve/saturate 100 (e) 22g of potassium nitrate(V) was dissolved in 40.0g of water at 10oC. Calculate the solubility of potassium nitrate(V) at 10oC. Solubility = Mass of solute/salt/solid x 100 =>( 22 x 100 ) = 55.0 g /100g H2O Mass/volume of water/solvent 40.0.
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Mass of solute/salt/solid = Solubility x Mass/volume of water/solvent 100 => 72 x 25 = 18.0g of KClO3 dissolve/saturate 100 (e) 22g of potassium nitrate(V) was dissolved in 40.0g of water at 10oC. Calculate the solubility of potassium nitrate(V) at 10oC. Solubility = Mass of solute/salt/solid x 100 =>( 22 x 100 ) = 55.0 g /100g H2O Mass/volume of water/solvent 40.0. (f)What volume of water should be added to 22.0g of water at 10oC if the solubility of KNO3 at 10oC is 5.0g/100g H2O? Solubility is mass/100g H2O => 22.0g + x = 100cm3/100g H2O X= 100 – 22 = 78 cm3 of H2O 18. A graph of solubility against temperature is called solubility curve. It shows the influence of temperature on solubility of different substances/solids/salts. Some substances dissolve more with increase in temperature while for others dissolve less with increase in temperature KNO3 KClO3 Saturated solution of KClO3 Solubility
Note: (i)solubility of KNO3 and KClO3 increase with increase in temperature. (ii)solubility of KNO3 is always higher than that of KClO3 at any specified temperature. (iii)solubility of NaCl decrease with increase in temperature. (iv)NaCl has the highest solubility at low temperature while KClO3 has the lowest solubility at low temperature. (v)At point A both NaCl and KNO3 are equally soluble. (vi)At point B both NaCl and KClO3 are equally soluble. (vii) An area above the solubility curve of the salt shows a saturated /supersaturated solution. (viii) An area below the solubility curve of the salt shows an unsaturated solution. 19.(a) For salts whose solubility increases with increase in temperature, crystals form when the salt solution at higher temperatures is cooled to a lower temperature.
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(vii) An area above the solubility curve of the salt shows a saturated /supersaturated solution. (viii) An area below the solubility curve of the salt shows an unsaturated solution. 19.(a) For salts whose solubility increases with increase in temperature, crystals form when the salt solution at higher temperatures is cooled to a lower temperature. (b) For salts whose solubility decreases with increase in temperature, crystals form when the salt solution at lower temperatures is heated to a higher temperature. The examples below shows determination of the mass of crystals deposited with changes in temperature. 1.The solubility of KClO3 at 100oC is 60g/100g water .What mass of KClO3 will be deposited at: (i)75 oC if the solubility is now 39g/100g water. At 100oC = 60.0g Less at 75oC = - 39.0g Mass of crystallized out 21.0g (i)35 oC if the solubility is now 28 g/100g water. At 100oC = 60.0g Less at 35oC = - 28.0.0g Mass of crystallized out 32.0g 2. KNO3 has a solubility of 42 g/100g water at 20oC.The salt was heated and added 38g more of the solute which dissolved at100oC. Calculate the solubility of KNO3 at 100oC. Solubility of KNO3 at 100oC = solubility at 20oC + mass of KNO3 added => 42g + 38g = 80g KNO3 /100g H2O 3. A salt solution has a mass of 65g containing 5g of solute. The solubility of this salt is 25g per 100g water at 20oC. 60g of the salt are added to the solution at 20oC.Calculate the mass of the solute that remain undissolved.
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A salt solution has a mass of 65g containing 5g of solute. The solubility of this salt is 25g per 100g water at 20oC. 60g of the salt are added to the solution at 20oC.Calculate the mass of the solute that remain undissolved. Mass of solvent at 20oC = mass of solution – mass of solute => 65 - 5 = 60g Solubility before adding salt = mass of solute x 100 Volume of solvent => 5 x 100 = 8.3333g/100g water 60 Mass of solute to equalize with solubility = 25 – 8.3333g = 16.6667g Mass of solute undissolved = 60.0 - 16.6667g = 43.3333 g 4. Study the table below Salt Solubility in gram at
50oC 20oC KNO3 90 30 KClO3 20 6 (i)What happens when the two salts are dissolved in water then cooled from 50oC to 20oC. (90 – 30) = 60.0 g of KNO3 crystals precipitate (20 – 6) = 14.0 g of KClO3 crystals precipitate (ii)State the assumption made in (i) above. Solubility of one salt has no effect on the solubility of the other. 5. 10.0 g of hydrated potassium carbonate (IV) K2CO3.xH2O on heating leave 7.93 of the hydrate. (a)Calculate the mass of anhydrous salt obtained. Hydrated on heating leave anhydrous = 7.93 g (b)Calculate the mass of water of crystallization in the hydrated salt Mass of water of crystallization = hydrated – anhydrous => 10.0 - 7.93 = 2.07 g (c)How many moles of anhydrous salt are there in 10of hydrate?
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10.0 g of hydrated potassium carbonate (IV) K2CO3.xH2O on heating leave 7.93 of the hydrate. (a)Calculate the mass of anhydrous salt obtained. Hydrated on heating leave anhydrous = 7.93 g (b)Calculate the mass of water of crystallization in the hydrated salt Mass of water of crystallization = hydrated – anhydrous => 10.0 - 7.93 = 2.07 g (c)How many moles of anhydrous salt are there in 10of hydrate? (K= 39.0,C=12.0.O= 16.0) Molar mass K2CO3= 138 Moles K2CO3 = mass of K2CO3 => 7.93 = 0.0515 moles Molar mass K2CO3 138 (d)How many moles of water are present in the hydrate for every one mole of K2CO3 ? (H=1.0.O= 16.0) Molar mass H2O = 18 Moles H2O = mass of H2O => 2.07 = 0.115 moles Molar mass H2O 18 Mole ratio H2O : K2CO3 = 0.115 moles 2 = 2 0.0515 moles 1 (e)What is the formula of the hydrated salt? K2CO3 .2 H2O 6. The table below shows the solubility of Potassium nitrate(V) at different temperatures. Temperature(oC) 5.0 10.0 15.0 30.0 40.0 50.0 60.0 mass KNO3/ 100g water 15.0 20.0 25.0 50.0 65.0 90.0 120.0 (a)Plot a graph of mass of in 100g water(y-axis) against temperature in oC
(b)From the graph show and determine (i)the mass of KNO3 dissolved at: I. 20oC From a correctly plotted graph = 32g II. 35oC From a correctly plotted graph = 57g III.
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Temperature(oC) 5.0 10.0 15.0 30.0 40.0 50.0 60.0 mass KNO3/ 100g water 15.0 20.0 25.0 50.0 65.0 90.0 120.0 (a)Plot a graph of mass of in 100g water(y-axis) against temperature in oC
(b)From the graph show and determine (i)the mass of KNO3 dissolved at: I. 20oC From a correctly plotted graph = 32g II. 35oC From a correctly plotted graph = 57g III. 55oC From a correctly plotted graph = 104g (ii)the temperature at which the following mass of KNO3 dissolved: I. 22g From a correctly plotted graph =13.0oC II. 30g From a correctly plotted graph =17.5oC III.100g From a correctly plotted graph =54.5oC (c)Explain the shape of your graph. Solubility of KNO3 increase with increase in temperature/More KNO3 dissolve as temperature rises. (d)Show on the graph the supersaturated and unsaturated solutions. Above the solubility curve write; “supersaturated” Below the solubility curve write; “unsaturated” (e)From your graph, calculate the amount of crystals obtained when a saturated solution of KNO3 containing 180g of the salt is cooled from 80oC to: I. 20oC Solubility before heating = 180 g Less Solubility after heating(from the graph) = 32 g Mass of KNO3crystals = 148 g II. 35oC Solubility before heating = 180 g Less Solubility after heating(from the graph) = 58 g Mass of KNO3crystals = 122 g III. 55oC
Solubility before heating = 180 g Less Solubility after heating(from the graph) = 102 g Mass of KNO3crystals = 78 g 7. The table below shows the solubility of salts A and B at various temperatures.
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35oC Solubility before heating = 180 g Less Solubility after heating(from the graph) = 58 g Mass of KNO3crystals = 122 g III. 55oC
Solubility before heating = 180 g Less Solubility after heating(from the graph) = 102 g Mass of KNO3crystals = 78 g 7. The table below shows the solubility of salts A and B at various temperatures. Temperature(oC) 0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 Solubility of A 28.0 31.0 34.0 37.0 40.0 43.0 45.0 48.0 51.0 Solubility of B 13.0 21.0 32.0 46.0 64.0 85.0 110.0 138.0 169.0 (a)On the same axis plot a graph of solubility (y-axis) against temperature for each salt. (b)At what temperature are the two salts equally soluble. The point of intersection of the two curves = 24oC (c)What happens when a mixture of 100g of salt B with 100g if water is heated to 80oC From the graph, the solubility of B at 80oC is 169g /100g water. All the 100g crystals of B dissolve. (d)What happens when the mixture in (c) above is then cooled from 50oC to 20oC. Method I. Total mass before cooling at 50oC = 100.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited 68.0 g Method II.
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(d)What happens when the mixture in (c) above is then cooled from 50oC to 20oC. Method I. Total mass before cooling at 50oC = 100.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited 68.0 g Method II. Mass of soluble salt crystals at 50oC added = 100 g (From graph)Solubility/mass before cooling at 50oC = 85.0 g Mass of crystals that cannot dissolve at 50oC 15.0 g (From graph) Solubility/mass before cooling at 50oC = 85.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited after cooling 53.0 g Total mass of crystals deposited = 15.0 + 53.0 = 68.0 g
(e)A mixture of 40g of A and 60g of B is added to 10g of water and heated to 70oC.The solution is then allowed to cool to 10oC.Describe clearly what happens.
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Method I. Total mass before cooling at 50oC = 100.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited 68.0 g Method II. Mass of soluble salt crystals at 50oC added = 100 g (From graph)Solubility/mass before cooling at 50oC = 85.0 g Mass of crystals that cannot dissolve at 50oC 15.0 g (From graph) Solubility/mass before cooling at 50oC = 85.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited after cooling 53.0 g Total mass of crystals deposited = 15.0 + 53.0 = 68.0 g
(e)A mixture of 40g of A and 60g of B is added to 10g of water and heated to 70oC.The solution is then allowed to cool to 10oC.Describe clearly what happens. I.For salt A Solubility of A before heating = mass of A x 100 Volume of water added => 40 x 100 = 400g/100g Water 10 (Theoretical)Solubility of A before heating = 400 g Less (From graph ) Solubility of A after heating at 70oC = 48g Mass of crystals that can not dissolve at70oC = 352 g (From graph ) Solubility of A after heating at 70oC = 48g Less (From graph ) Solubility of A after cooling to 10oC = 31g Mass of crystals that crystallize out on cooling to10oC = 17 g Mass of crystals that can not dissolve at70oC = 352 g Add Mass of crystals that crystallize out on cooling to10oC = 17 g Total mass of A that does not dissolve/crystallize/precipitate = 369 g I.For salt B Solubility of B before heating = mass of B x 100 Volume of water added => 60 x 100 = 600g/100g Water 10 (Theoretical)Solubility of B before heating = 600 g Less (From graph ) Solubility of B after heating at 70oC = 138g Mass of crystals that cannot dissolve at70oC = 462 g (From graph ) Solubility of B after heating at 70oC = 138g Less (From graph ) Solubility of B after cooling to 10oC = 21g Mass of crystals that crystallize out on cooling to10oC = 117 g Mass of crystals that cannot dissolve at70oC = 462 g Add Mass of crystals that crystallize out on cooling to10oC = 117 g Total mass of A that does not dissolve/crystallize/precipitate = 579 g
(f)State the assumption made in (e)above Solubility of one salt has no effect on the solubility of the other 8.
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Total mass before cooling at 50oC = 100.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited 68.0 g Method II. Mass of soluble salt crystals at 50oC added = 100 g (From graph)Solubility/mass before cooling at 50oC = 85.0 g Mass of crystals that cannot dissolve at 50oC 15.0 g (From graph) Solubility/mass before cooling at 50oC = 85.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited after cooling 53.0 g Total mass of crystals deposited = 15.0 + 53.0 = 68.0 g
(e)A mixture of 40g of A and 60g of B is added to 10g of water and heated to 70oC.The solution is then allowed to cool to 10oC.Describe clearly what happens. I.For salt A Solubility of A before heating = mass of A x 100 Volume of water added => 40 x 100 = 400g/100g Water 10 (Theoretical)Solubility of A before heating = 400 g Less (From graph ) Solubility of A after heating at 70oC = 48g Mass of crystals that can not dissolve at70oC = 352 g (From graph ) Solubility of A after heating at 70oC = 48g Less (From graph ) Solubility of A after cooling to 10oC = 31g Mass of crystals that crystallize out on cooling to10oC = 17 g Mass of crystals that can not dissolve at70oC = 352 g Add Mass of crystals that crystallize out on cooling to10oC = 17 g Total mass of A that does not dissolve/crystallize/precipitate = 369 g I.For salt B Solubility of B before heating = mass of B x 100 Volume of water added => 60 x 100 = 600g/100g Water 10 (Theoretical)Solubility of B before heating = 600 g Less (From graph ) Solubility of B after heating at 70oC = 138g Mass of crystals that cannot dissolve at70oC = 462 g (From graph ) Solubility of B after heating at 70oC = 138g Less (From graph ) Solubility of B after cooling to 10oC = 21g Mass of crystals that crystallize out on cooling to10oC = 117 g Mass of crystals that cannot dissolve at70oC = 462 g Add Mass of crystals that crystallize out on cooling to10oC = 117 g Total mass of A that does not dissolve/crystallize/precipitate = 579 g
(f)State the assumption made in (e)above Solubility of one salt has no effect on the solubility of the other 8. When 5.0 g of potassium chlorate (V) was put in 10cm3 of water and heated, the solid dissolves.
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Mass of soluble salt crystals at 50oC added = 100 g (From graph)Solubility/mass before cooling at 50oC = 85.0 g Mass of crystals that cannot dissolve at 50oC 15.0 g (From graph) Solubility/mass before cooling at 50oC = 85.0 g (From graph) Solubility/mass after cooling at 20oC = 32.0 g Mass of crystals deposited after cooling 53.0 g Total mass of crystals deposited = 15.0 + 53.0 = 68.0 g
(e)A mixture of 40g of A and 60g of B is added to 10g of water and heated to 70oC.The solution is then allowed to cool to 10oC.Describe clearly what happens. I.For salt A Solubility of A before heating = mass of A x 100 Volume of water added => 40 x 100 = 400g/100g Water 10 (Theoretical)Solubility of A before heating = 400 g Less (From graph ) Solubility of A after heating at 70oC = 48g Mass of crystals that can not dissolve at70oC = 352 g (From graph ) Solubility of A after heating at 70oC = 48g Less (From graph ) Solubility of A after cooling to 10oC = 31g Mass of crystals that crystallize out on cooling to10oC = 17 g Mass of crystals that can not dissolve at70oC = 352 g Add Mass of crystals that crystallize out on cooling to10oC = 17 g Total mass of A that does not dissolve/crystallize/precipitate = 369 g I.For salt B Solubility of B before heating = mass of B x 100 Volume of water added => 60 x 100 = 600g/100g Water 10 (Theoretical)Solubility of B before heating = 600 g Less (From graph ) Solubility of B after heating at 70oC = 138g Mass of crystals that cannot dissolve at70oC = 462 g (From graph ) Solubility of B after heating at 70oC = 138g Less (From graph ) Solubility of B after cooling to 10oC = 21g Mass of crystals that crystallize out on cooling to10oC = 117 g Mass of crystals that cannot dissolve at70oC = 462 g Add Mass of crystals that crystallize out on cooling to10oC = 117 g Total mass of A that does not dissolve/crystallize/precipitate = 579 g
(f)State the assumption made in (e)above Solubility of one salt has no effect on the solubility of the other 8. When 5.0 g of potassium chlorate (V) was put in 10cm3 of water and heated, the solid dissolves. When the solution was cooled , the temperature at which crystals reappear was noted.
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