JK-TK/Newdata huggingface dataset:
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. Another 10cm3 of water was added and the mixture heated to dissolve then cooled for the crystals to reappear .The table below shows the the results obtained Total volume of water added(cm3) 10.0 20.0 30.0 40.0 50.0 Mass of KClO3 5.0 5.0 5.0 5.0 5.0 Temperature at which crystals appear 80.0 65.0 55.0 45.0 30.0 Solubility of KclO3 50.0 25.0 16.6667 12.5 10.0 (a)Complete the table to show the solubility of KclO3 at different temperatures.
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. Another 10cm3 of water was added and the mixture heated to dissolve then cooled for the crystals to reappear .The table below shows the the results obtained Total volume of water added(cm3) 10.0 20.0 30.0 40.0 50.0 Mass of KClO3 5.0 5.0 5.0 5.0 5.0 Temperature at which crystals appear 80.0 65.0 55.0 45.0 30.0 Solubility of KclO3 50.0 25.0 16.6667 12.5 10.0 (a)Complete the table to show the solubility of KclO3 at different temperatures. (b)Plot a graph of mass of KClO3 per 100g water against temperature at which crystals form. (c)From the graph, show and determine ; (i)the solubility of KClO3 at I. 50oC From a well plotted graph = 14.5 g KClO3/100g water II. 35oC From a well plotted graph = 9.0 g KclO3/100g water (ii)the temperature at which the solubility is: I.10g/100g water From a well plotted graph = 38.0 oC II.45g/100g water From a well plotted graph = 77.5 oC (d)Explain the shape of the graph. Solubility of KClO3 increase with increase in temperature/more KclO3dissolve as temperature rises. (e)What happens when 100g per 100g water is cooled to 35.0 oC Solubility before heating = 100.0
(From the graph) Solubility after cooling = 9.0 Mass of salt precipitated/crystallization = 91.0 g 9. 25.0cm3 of water dissolved various masses of ammonium chloride crystals at different temperatures as shown in the table below.
Solubility of KClO3 increase with increase in temperature/more KclO3dissolve as temperature rises. (e)What happens when 100g per 100g water is cooled to 35.0 oC Solubility before heating = 100.0
(From the graph) Solubility after cooling = 9.0 Mass of salt precipitated/crystallization = 91.0 g 9. 25.0cm3 of water dissolved various masses of ammonium chloride crystals at different temperatures as shown in the table below. Mass of ammonium chloride(grams) 4.0 4.5 5.5 6.5 9.0 Temperature at which solid dissolved(oC) 30.0 50.0 70.0 90.0 120.0 Solubility of NH4Cl 16.0 18.0 22.0 26.0 36.0 (a)Complete the table (b)Plot a solubility curve (c)What happens when a saturated solution of ammonium chloride is cooled from 80oC to 40oC. (From the graph )Solubility at 80oC = 24.0 g Less (From the graph )Solubility at 40oC = 16.8 g Mass of crystallized/precipitated = 7.2 g 20. Solubility and solubility curves are therefore used (i) to know the effect of temperature on the solubility of a salt (ii)to fractional crystallize two soluble salts by applying their differences in solubility at different temperatures. (iii)determine the mass of crystal that is obtained from crystallization. 21.Natural fractional crystallization takes place in Kenya/East Africa at: (i) Lake Magadi during extraction of soda ash(Sodium carbonate) from Trona(sodium sesquicarbonate) (ii) Ngomeni near Malindi at the Indian Ocean Coastline during the extraction of common salt(sodium chloride). 22.Extraction of soda ash from Lake Magadi in Kenya Rain water drains underground in the great rift valley and percolate underground where it is heated geothermically. The hot water dissolves underground soluble sodium compounds and comes out on the surface as alkaline springs which are found around the edges of Lake Magadi in Kenya.
21.Natural fractional crystallization takes place in Kenya/East Africa at: (i) Lake Magadi during extraction of soda ash(Sodium carbonate) from Trona(sodium sesquicarbonate) (ii) Ngomeni near Malindi at the Indian Ocean Coastline during the extraction of common salt(sodium chloride). 22.Extraction of soda ash from Lake Magadi in Kenya Rain water drains underground in the great rift valley and percolate underground where it is heated geothermically. The hot water dissolves underground soluble sodium compounds and comes out on the surface as alkaline springs which are found around the edges of Lake Magadi in Kenya. Temperatures around the lake are very high (30-40oC) during the day. The solubility of trona decrease with increase in temperature therefore solid crystals of trona grows on top of the lake (upto or more than 30metres thick) A bucket dredger mines the trona which is then crushed ,mixed with lake liquor and pumped to washery plant where it is further refined to a green granular product called CRS. The CRS is then heated to chemically decompose trona to soda ash(Sodium carbonate) Chemical equation 2Na2CO3.NaHCO3.2H2O(s) -> 3Na2CO3 (s) + CO2(g) + 5H2O(l) Soda ash(Sodium carbonate) is then stored .It is called Magadi Soda. Magadi Soda is used : (i) make glass (ii) for making soapless detergents (iii) softening hard water. (iv) Common salt is colledcted at night because its solubility decreases with decrease in temperature. It is used as salt lick/feed for animals. Summary flow diagram showing the extraction of Soda ash from Trona Sodium chloride and Trona dissolved in the sea Natural fractional crystallization Crystals of Trona (Day time) Crystals of sodium chloride(At night) Dredging /scooping/ digging Crushing Furnace (Heating) Carbon(IV) oxide Soda ash
23.Extraction of common salt from Indian Ocean at Ngomeni in Kenya Oceans are salty.They contain a variety of dissolved salts (about 77% being sodium chloride). During high tide ,water is collected into shallow pods and allowed to crystallize as evaporation takes place.The pods are constructed in series to increase the rate of evaporation.
It is used as salt lick/feed for animals. Summary flow diagram showing the extraction of Soda ash from Trona Sodium chloride and Trona dissolved in the sea Natural fractional crystallization Crystals of Trona (Day time) Crystals of sodium chloride(At night) Dredging /scooping/ digging Crushing Furnace (Heating) Carbon(IV) oxide Soda ash
23.Extraction of common salt from Indian Ocean at Ngomeni in Kenya Oceans are salty.They contain a variety of dissolved salts (about 77% being sodium chloride). During high tide ,water is collected into shallow pods and allowed to crystallize as evaporation takes place.The pods are constructed in series to increase the rate of evaporation. At the final pod ,the crystals are scapped together,piled in a heap and washed with brine (concentrated sodium chloride). It contains MgCl2 and CaCl2 . MgCl2 and CaCl2are hygroscopic. They absorb water from the atmosphere and form a solution. This makes table salt damp/wet on exposure to the atmosphere. 24.Some water form lather easily with soap while others do not. Water which form lather easily with soap is said to be “soft” Water which do not form lather easily with soap is said to be “hard” Hardness of water is caused by the presence of Ca2+ and Mg2+ ions. Ca2+ and Mg2+ ions react with soap to form an insoluble grey /white suspension/precipitate called Scum/ curd. Ca2+ and Mg2+ ions in water come from the water sources passing through rocks containing soluble salts of Ca2+ and Mg2+ e.g. Limestone or gypsum There are two types of water hardness: (a)temporary hardness of water (b)permanent hardness of water (a)temporary hardness of water
Temporary hardness of water is caused by the presence of dissolved calcium hydrogen carbonate/Ca(HCO3)2 and magnesium hydrogen carbonate/Mg(HCO3)2 When rain water dissolve carbon(IV) oxide from the air it forms waek carbonic(IV) acid i.e. CO2(g) + H2O(l) -> H2CO3(aq) When carbonic(IV) acid passes through limestone/dolomite rocks it reacts to form soluble salts i.e.
Ca2+ and Mg2+ ions in water come from the water sources passing through rocks containing soluble salts of Ca2+ and Mg2+ e.g. Limestone or gypsum There are two types of water hardness: (a)temporary hardness of water (b)permanent hardness of water (a)temporary hardness of water
Temporary hardness of water is caused by the presence of dissolved calcium hydrogen carbonate/Ca(HCO3)2 and magnesium hydrogen carbonate/Mg(HCO3)2 When rain water dissolve carbon(IV) oxide from the air it forms waek carbonic(IV) acid i.e. CO2(g) + H2O(l) -> H2CO3(aq) When carbonic(IV) acid passes through limestone/dolomite rocks it reacts to form soluble salts i.e. In limestone areas; H2CO3(aq) + CaCO3(s) -> Ca(HCO3)2 (aq) In dolomite areas; H2CO3(aq) + MgCO3(s) -> Mg(HCO3)2 (aq) (b)permanent hardness of water Permanent hardness of water is caused by the presence of dissolved calcium sulphate(VI)/CaSO4 and magnesium sulphate(VI)/Mg SO4 Permanent hardness of water is caused by water dissolving CaSO4 and MgSO4 from ground rocks. Hardness of water can be removed by the following methods: (a)Removing temporary hardness of water (i)Boiling/heating. Boiling decomposes insoluble calcium hydrogen carbonate/Ca(HCO3)2 and magnesium hydrogen carbonate/Mg(HCO3)2 to insoluble CaCO3 and MgCO3 that precipitate away. i.e Chemical equation Ca(HCO3)2(aq) -> CaCO3 (s) + CO2(g) + H2O(l) Mg(HCO3)2(aq) -> MgCO3 (s) + CO2(g) + H2O(l) (ii)Adding sodium carbonate (IV) /Washing soda. Since boiling is expensive on a large scale ,a calculated amount of sodium carbonate decahydrate /Na2CO3.10H2O precipitates insoluble Ca2+(aq) and Mg2+(aq) ions as carbonates to remove both temporary and permanent hardness of water .This a double decomposition reaction where two soluble salts form an insoluble and soluble salt. i.e.
i.e Chemical equation Ca(HCO3)2(aq) -> CaCO3 (s) + CO2(g) + H2O(l) Mg(HCO3)2(aq) -> MgCO3 (s) + CO2(g) + H2O(l) (ii)Adding sodium carbonate (IV) /Washing soda. Since boiling is expensive on a large scale ,a calculated amount of sodium carbonate decahydrate /Na2CO3.10H2O precipitates insoluble Ca2+(aq) and Mg2+(aq) ions as carbonates to remove both temporary and permanent hardness of water .This a double decomposition reaction where two soluble salts form an insoluble and soluble salt. i.e. (i)with temporary hard water Chemical equation Na2CO3 (aq) + Ca(HCO3) 2 (aq) -> NaHCO3(aq) + CaCO3 (s) Na2CO3 (aq) + Mg(HCO3) 2 (aq) -> NaHCO3(aq) + MgCO3 (s) Ionic equation CO32 (aq) + Ca2+ (aq) -> CaCO3 (s)
CO32 (aq) + Mg2+ (aq) -> MgCO3 (s) (ii)with permanent hard water Chemical equation Na2CO3 (aq) + MgSO4 (aq) -> Na2SO4 (aq) + MgCO3 (s) Na2CO3 (aq) + CaSO4 (aq) -> Na2SO4 (aq) + MgCO3 (s) Ionic equation CO32 (aq) + Ca2+ (aq) -> CaCO3 (s) CO32 (aq) + Mg2+ (aq) -> MgCO3 (s) (iii)Adding calcium (II)hydroxide/Lime water Lime water/calcium hydroxide removes only temporary hardness of water from by precipitating insoluble calcium carbonate(IV). Chemical equation Ca(OH)2 (aq) + Ca(HCO3) 2 (aq) -> 2H2O(l) + 2CaCO3 (s) Excess of Lime water/calcium hydroxide should not be used because it dissolves again to form soluble calcium hydrogen carbonate(IV) causing the hardness again.
i.e. (i)with temporary hard water Chemical equation Na2CO3 (aq) + Ca(HCO3) 2 (aq) -> NaHCO3(aq) + CaCO3 (s) Na2CO3 (aq) + Mg(HCO3) 2 (aq) -> NaHCO3(aq) + MgCO3 (s) Ionic equation CO32 (aq) + Ca2+ (aq) -> CaCO3 (s)
CO32 (aq) + Mg2+ (aq) -> MgCO3 (s) (ii)with permanent hard water Chemical equation Na2CO3 (aq) + MgSO4 (aq) -> Na2SO4 (aq) + MgCO3 (s) Na2CO3 (aq) + CaSO4 (aq) -> Na2SO4 (aq) + MgCO3 (s) Ionic equation CO32 (aq) + Ca2+ (aq) -> CaCO3 (s) CO32 (aq) + Mg2+ (aq) -> MgCO3 (s) (iii)Adding calcium (II)hydroxide/Lime water Lime water/calcium hydroxide removes only temporary hardness of water from by precipitating insoluble calcium carbonate(IV). Chemical equation Ca(OH)2 (aq) + Ca(HCO3) 2 (aq) -> 2H2O(l) + 2CaCO3 (s) Excess of Lime water/calcium hydroxide should not be used because it dissolves again to form soluble calcium hydrogen carbonate(IV) causing the hardness again. (iv)Adding aqueous ammonia Aqueous ammonia removes temporary hardness of water by precipitating insoluble calcium carbonate(IV) and magnesium carbonate(IV) Chemical equation 2NH3 (aq) + Ca(HCO3) 2 (aq) -> (NH4) 2CO3(aq) + CaCO3 (s) 2NH3 (aq) + Mg(HCO3) 2 (aq) -> (NH4) 2CO3(aq) + MgCO3 (s) (v)Use of ion-exchange permutit This method involves packing a chamber with a resin made of insoluble complex of sodium salt called sodium permutit. The sodium permutit releases sodium ions that are exchanged with Mg2+ and Ca2+ ions in hard water making the water to be soft. i.e.
(iv)Adding aqueous ammonia Aqueous ammonia removes temporary hardness of water by precipitating insoluble calcium carbonate(IV) and magnesium carbonate(IV) Chemical equation 2NH3 (aq) + Ca(HCO3) 2 (aq) -> (NH4) 2CO3(aq) + CaCO3 (s) 2NH3 (aq) + Mg(HCO3) 2 (aq) -> (NH4) 2CO3(aq) + MgCO3 (s) (v)Use of ion-exchange permutit This method involves packing a chamber with a resin made of insoluble complex of sodium salt called sodium permutit. The sodium permutit releases sodium ions that are exchanged with Mg2+ and Ca2+ ions in hard water making the water to be soft. i.e. Na2X(aq) + Ca2+ (aq) -> Na+ (aq) + CaX(s) Na2X(aq) + Mg2+ (aq) -> Na+ (aq) + MgX(s) Hard water containing Mg2+ and Ca2+
Ion exchange resin as Sodium permutit ------- Na+ ions replace Mg2+ and Ca2+ to make the water soft. When all the Na+ ions in the resin is fully exchanged with Ca2+ and Ng2+ ions in the permutit column ,it is said to be exhausted. Brine /concentrated sodium chloride solution is passed through the permutit column to regenerated /recharge the column again. Hard water containing Mg2+ and Ca2+ Ion exchange resin as Sodium permutit
------- Na+ ions replace Mg2+ and Ca2+ to make the water soft. (vi)Deionization /demineralization This is an advanced ion exchange method of producing deionized water .Deionized water is extremely pure water made only of hydrogen and oxygen only without any dissolved substances. Deionization involve using the resins that remove all the cations by using: (i)A cation exchanger which remove /absorb all the cations present in water and leave only H+ ions. (ii)An anion exchanger which remove /absorb all the anions present in water and leave only OH- ions. The H+(aq) and OH- (aq) neutralize each other to form pure water.
Deionization involve using the resins that remove all the cations by using: (i)A cation exchanger which remove /absorb all the cations present in water and leave only H+ ions. (ii)An anion exchanger which remove /absorb all the anions present in water and leave only OH- ions. The H+(aq) and OH- (aq) neutralize each other to form pure water. Chemical equation H+(aq) + OH- (aq) -> H2O(l) When exhausted the cation exchanger is regenerated by adding H+(aq) from sulphuric(VI)acid/hydrochloric acid. When exhausted the anion exchanger is regenerated by adding OH-(aq) from sodium hydroxide. Advantages of hard water Hard water has the following advantages: (i)Ca2+(aq) in hard water are useful in bone and teeth formation (ii) is good for brewing beer (iii)contains minerals that cause it to have better /sweet taste (iv)animals like snails and coral polyps use calcium to make their shells and coral reefs respectively. (v)processing mineral water Disadvantages of hard water
Hardness of water: (i)waste a lot of soap during washing before lather is formed. (ii)causes stains/blemishes/marks on clothes/garments (iii)causes fur on electric appliances like kettle ,boilers and pipes form decomposition of carbonates on heating .This reduces their efficiency hence more/higher cost of power/electricity. Sample revision questions In an experiment, soap solution was added to three separate samples of water. The table below shows the volumes of soap solution required to form lather with 1000cm3 of each sample of water before and after boiling. Sample I Sample II Sample III Volume of soap before water is boiled (cm3) 27.0 3.0 10.0 Volume of soap after water is boiled(cm3) 27.0 3.0 3.0 a) Which water sample is likely to be soft? Explain. (2mks) Sample II: Uses little sample of soap . c) Name the change in the volume of soap solution used in sample III (1mk) On heating the sample water become soft bcause it is temporary hard.
Explain. (2mks) Sample II: Uses little sample of soap . c) Name the change in the volume of soap solution used in sample III (1mk) On heating the sample water become soft bcause it is temporary hard. 2.Study the scheme below and use it to aanswer the questions that follow:
(a)Write the formula of: (i)Cation in solution K Al3+ (ii)white ppt L Al(OH)3 (iii) colourless solution M [Al(OH)4]- (iv) colourless solution N AlCl3 (v)white ppt P Al(OH)3 (b)Write the ionic equation for the reaction for the formation of: (i)white ppt L Al3+(aq) + 3OH- (aq) -> Al(OH)3(s) (v)white ppt P Al3+(aq) + 3OH- (aq) -> Al(OH)3(s) (c)What property is illustrated in the formation of colourless solution M and N. Amphotellic
19.0.0 ENERGY CHANGES IN CHEMICAL AND PHYSICAL PROCESSES (25 LESSONS) 1.Introduction to Energy changes Energy is the capacity to do work. There are many/various forms of energy like heat, electric, mechanical, and/ or chemical energy.There are two types of energy: (i)Kinetic Energy(KE) ;the energy in motion. (ii)Potential Energy(PE); the stored/internal energy. Energy like matter , is neither created nor destroyed but can be transformed /changed from one form to the other/ is interconvertible. This is the principle of conservation of energy. e.g. Electrical energy into heat through a filament in bulb. Chemical and physical processes take place with absorption or evolution/production of energy mainly in form of heat The study of energy changes that accompany physical/chemical reaction/changes is called Thermochemistry. Physical/chemical reaction/changes that involve energy changes are called thermochemical reactions. The SI unit of energy is the Joule(J).Kilo Joules(kJ)and megaJoules(MJ) are also used. The Joule(J) is defined as the: (i) quantity of energy transferred when a force of one newton acts through a distance of one metre. 2 (ii) quantity of energy transferred when one coulomb of electric charge is passed through a potential difference of one volt.
The SI unit of energy is the Joule(J).Kilo Joules(kJ)and megaJoules(MJ) are also used. The Joule(J) is defined as the: (i) quantity of energy transferred when a force of one newton acts through a distance of one metre. 2 (ii) quantity of energy transferred when one coulomb of electric charge is passed through a potential difference of one volt. All thermochemical reactions should be carried out at standard conditions of: (i) 298K /25oC temperature (ii)101300Pa/101300N/m2 /760mmHg/1 atmosphere pressure. 2.Exothermic and endothermic processes/reactions Some reactions / processes take place with evolution/production of energy. They are said to be exothermic while others take place with absorption of energy. They are said to be endothermic. Practically exothermic reactions / processes cause a rise in temperature (by a rise in thermometer reading/mercury or alcohol level rise) Practically endothermic reactions / processes cause a fall in temperature (by a fall in thermometer reading/mercury or alcohol level decrease) To demonstrate/illustrate exothermic and endothermic processes/reactions a) Dissolving Potassium nitrate(V)/ammonium chloride crystals Procedure: Measure 20cm3 of water in a beaker. Determine and record its temperature T1.Put about 1.0g of Potassium nitrate(V) crystals into the beaker. Stir the mixture carefully and note the highest temperature rise /fall T2.Repeat the whole procedure by using ammonium chloride in place of Potassium nitrate (V) crystals. Sample results Temperture (oC) Using Potassium nitrate(V) crystals Using Ammonium chloride crystals T2(Final temperature) 21.0 23.0 T1 (Initial temperature) 25.0 26.0 Change in temperature(T2 –T1) 4.0 3.0 Note: (i)Initial(T1) temperature of dissolution of both potassium nitrate(V) crystals and ammonium chloride crystals is higher than the final temperature(T2) (ii) Change in temperature(T2 –T1) is not a mathematical “-4.0” or “-3.0”.
Determine and record its temperature T1.Put about 1.0g of Potassium nitrate(V) crystals into the beaker. Stir the mixture carefully and note the highest temperature rise /fall T2.Repeat the whole procedure by using ammonium chloride in place of Potassium nitrate (V) crystals. Sample results Temperture (oC) Using Potassium nitrate(V) crystals Using Ammonium chloride crystals T2(Final temperature) 21.0 23.0 T1 (Initial temperature) 25.0 26.0 Change in temperature(T2 –T1) 4.0 3.0 Note: (i)Initial(T1) temperature of dissolution of both potassium nitrate(V) crystals and ammonium chloride crystals is higher than the final temperature(T2) (ii) Change in temperature(T2 –T1) is not a mathematical “-4.0” or “-3.0”. (iii)Dissolution of both potassium nitrate(V) and ammonium chloride crystals is an endothermic process because initial(T1) temperature is higher than the final temperature(T2) thus causes a fall/drop in temperature. b) Dissolving concentrated sulphuric(VI) acid/sodium hydroxide crystals
3 Procedure: Measure 20cm3 of water in a beaker. Determine and record its temperature T1.Carefully put about 1.0g/four pellets of sodium hydroxide crystals into the beaker. Stir the mixture carefully and note the highest temperature rise /fall T2.Repeat the whole procedure by using 2cm3 of concentrated sulphuric(VI) acid in place of sodium hydroxide crystals. CAUTION: (i)Sodium hydroxide crystals are caustic and cause painful blisters on contact with skin. (ii) Concentrated sulphuric (VI) acid is corrosive and cause painful wounds on contact with skin. Sample results Temperture (oC) Using Sodium hydroxide pellets Using Concentrated sulphuric(VI) acid T2(Final temperature) 30.0 32.0 T1 (Initial temperature) 24.0 25.0 Change in temperature(T2 –T1) 6.0 7.0 Note: (i)Initial (T1) temperature of dissolution of both concentrated sulphuric (VI) acid and sodium hydroxide pellets is lower than the final temperature (T2).