JK-TK/Newdata huggingface dataset:
Identify the particle: (i)with the highest mass. Alpha/ α (ii) almost equal to an electron Beta/ β 1.a)State two differences between chemical and nuclear reactions(2mks) (i) Nuclear reactions mainly involve protons and neutrons in the nucleus of an atom.Chemical reactions mainly involve outer electrons in the energy levels an atom. (ii) Nuclear reactions form a new element. Chemical reactions do not form new elements (iii) Nuclear reactions mainly involve evolution/production of large quantity of heat/energy.Chemical reactions produce or absorb smaller quantity of heat/energy. (iv)Nuclear reactions are accompanied by a loss in mass /mass defect. Chemical reactions are not accompanied by a loss in mass. (v)Rate of decay/ disintegration of nuclide is independent of physical conditionsThe rate of a chemical reaction is dependent on physical conditions of temperature/pressure/purity/particle size/ surface area b)Below is a radioactive decay series starting from 21483 Bi and ending at 20682 Pb. Study it and answer the question that follows. 20 Identify the particles emitted in steps I and III (2mks) I - α-particle III - β-ray ii)Write the nuclear equation for the reaction which takes place in (a) step I 21483Bi -> 21081Bi + 4 2 He (b) step 1 to 3 21483Bi -> 21081Bi + 4 2 He + 2 0 -1 e (c) step 3 to 5 21082Pb -> 20682Pb + 4 2 He + 2 0 -1 e (c) step 1 to 5 21483Bi -> 20682Pb + 2 4 2 He + 3 0 -1 e The table below give the percentages of a radioactive isotope of Bismuth that remains after decaying at different times. Time (min) 0 6 12 22 38 62 100 Percentage of Bismuth 100 81 65 46 29 12 3 i)On the grid below , plot a graph of the percentage of Bismuth remaining(Vertical axis) against time. 21 ii)Using the graph, determine the: I. Half – life of the Bismuth isotope II.
Time (min) 0 6 12 22 38 62 100 Percentage of Bismuth 100 81 65 46 29 12 3 i)On the grid below , plot a graph of the percentage of Bismuth remaining(Vertical axis) against time. 21 ii)Using the graph, determine the: I. Half – life of the Bismuth isotope II. Original mass of the Bismuth isotope given that the mass that remained after 70 minutes was 0.16g (2mks) d) Give one use of radioactive isotopes in medicine (1mk) 14.a)Distinguish between nuclear fission and nuclear fusion. (2mks) Describe how solid wastes containing radioactive substances should be disposed of. (1mk) b)(i)Find the values of Z1 and Z2 in the nuclear equation below Z1 1 94 140 1 U + n -> Sr + Xe + 2 n 92 0 38 Z2 0 iii)What type of nuclear reaction is represented in b (i) above? A radioactive cobalt 6128Co undergoes decay by emitting a beta particle and forming Nickel atom, Write a balanced decay equation for the above change 1 mark If a sample of the cobalt has an activity of 1000 counts per minute, determine the time it would take for its activity to decrease to 62.50 if the half-life of the element is 30years 2 marks Define the term half-life. The diagram below shows the rays emitted by a radioactive sample
22 a) Identify the rays S,R and Q S- Beta ( β )particle/ray R- Alpha (α )particle/ray Q- Gamma (y )particle/ray b) State what would happen if an aluminium plate is placed in the path of ray R,S and Q: R-is blocked/stopped/do not pass through Q-is not blocked/pass through S-is blocked/stopped/do not pass through (c)The diagram bellow is the radioactive decay series of nuclide A which is 24194Pu.Use it to answer the questions that follow. The letters are not the actual symbols of the elements. (a)Which letter represent the : Explain.
The diagram below shows the rays emitted by a radioactive sample
22 a) Identify the rays S,R and Q S- Beta ( β )particle/ray R- Alpha (α )particle/ray Q- Gamma (y )particle/ray b) State what would happen if an aluminium plate is placed in the path of ray R,S and Q: R-is blocked/stopped/do not pass through Q-is not blocked/pass through S-is blocked/stopped/do not pass through (c)The diagram bellow is the radioactive decay series of nuclide A which is 24194Pu.Use it to answer the questions that follow.

The letters are not the actual symbols of the elements.

(a)Which letter represent the : Explain.

(i)shortest lived nuclide L-has the shortest half life (ii)longest lived nuclide
23 P-Is stable (iii) nuclide with highest n/p ratio L-has the shortest half life thus most unstable thus easily/quickly decay/disintegrate (iv) nuclide with lowest n/p ratio P-is stable thus do not decay/disintegrate (b)How long would it take for the following: (i)Nuclide A to change to B 10 years (half life of A) (ii) Nuclide D to change to H 27days +162000years+70000years+16days 232000 years and 43 days (iii) Nuclide A to change to P 27days +162000years+70000years+16days 232000 years and 43 days Study THE END
12.0.0 GAS LAWS (15 LESSONS) (a)Gas laws 1. Matter is made up of small particle in accordance to Kinetic Theory of matter: Naturally, there are basically three states of matter: Solid, Liquid and gas: (i)A solid is made up of particles which are very closely packed with a definite/fixed shape and fixed/definite volume /occupies definite space. It has a very high density. (ii) A liquid is made up of particles which have some degree of freedom. It thus has no definite/fixed shape. It takes the shape of the container it is put. A liquid has fixed/definite volume/occupies definite space. (iii)A gas is made up of particles free from each other. It thus has no definite /fixed shape. It takes the shape of the container it is put. It has no fixed/definite volume/occupies every space in a container. 2.Gases are affected by physical conditions. There are two physical conditions: (i)Temperature (ii)Pressure 3. The SI unit of temperature is Kelvin(K). Degrees Celsius/Centigrade(oC) are also used. The two units can be interconverted from the relationship: oC + 273= K K -273 = oC Practice examples 1. Convert the following into Kelvin. (i) O oC oC + 273 = K substituting : O oC + 273 = 273 K (ii) -273 oC oC + 273 = K substituting : -273oC + 273 = 0 K (iii) 25 oC oC + 273 = K substituting : 25 oC + 273 = 298 K (iv) 100 oC oC + 273 = K substituting : 100 oC + 273 = 373 K
2. Convert the following into degrees Celsius/Centigrade(oC).
Convert the following into Kelvin. (i) O oC oC + 273 = K substituting : O oC + 273 = 273 K (ii) -273 oC oC + 273 = K substituting : -273oC + 273 = 0 K (iii) 25 oC oC + 273 = K substituting : 25 oC + 273 = 298 K (iv) 100 oC oC + 273 = K substituting : 100 oC + 273 = 373 K
2. Convert the following into degrees Celsius/Centigrade(oC). (i) 10 K K -273 = oC substituting: 10 – 273 = -263 oC (ii) (i) 1 K K -273 = oC substituting: 1 – 273 = -272 oC (iii) 110 K K -273 = oC substituting: 110 – 273 = -163 oC (iv) -24 K K -273 = oC substituting: -24 – 273 = -297 oC The standard temperature is 273K = 0 oC. The room temperature is assumed to be 298K = 25oC 4. The SI unit of pressure is Pascal(Pa) / Newton per metre squared (Nm-2) . Millimeters’ of mercury(mmHg) ,centimeters of mercury(cmHg) and atmospheres are also commonly used. The units are not interconvertible but Pascals(Pa) are equal to Newton per metre squared(Nm-2). The standard pressure is the atmospheric pressure. Atmospheric pressure is equal to about: (i)101325 Pa (ii)101325 Nm-2 (iii)760 mmHg (iv)76 cmHg (v)one atmosphere. 5. Molecules of gases are always in continuous random motion at high speed. This motion is affected by the physical conditions of temperature and pressure. Physical conditions change the volume occupied by gases in a closed system. The effect of physical conditions of temperature and pressure was investigated and expressed in both Boyles and Charles laws. 6.
Physical conditions change the volume occupied by gases in a closed system. The effect of physical conditions of temperature and pressure was investigated and expressed in both Boyles and Charles laws. 6. Boyles law states that “the volume of a fixed mass of a gas is inversely proportional to the pressure at constant/fixed temperature ” Mathematically:
Volume α 1 (Fixed /constant Temperature) Pressure V α 1 (Fixed /constant T) ie PV = Constant(k) P From Boyles law , an increase in pressure of a gas cause a decrease in volume. i.e doubling the pressure cause the volume to be halved. Graphically therefore a plot of volume(V) against pressure (P) produces a curve. V P Graphically a plot of volume(V) against inverse/reciprocal of pressure (1/p) produces a straight line V 1/P For two gases then P1 V1 = P2 V2 P1 = Pressure of gas 1 V1 = Volume of gas 1 P2 = Pressure of gas 2 V2 = Volume of gas 2
Practice examples: 1. A fixed mass of gas at 102300Pa pressure has a volume of 25cm3.Calculate its volume if the pressure is doubled. Working P1 V1 = P2 V2 Substituting :102300 x 25 = (102300 x 2) x V2 V2 = 102300 x 25 = 12.5cm3 (102300 x 2) 2. Calculate the pressure which must be applied to a fixed mass of 100cm3 of Oxygen for its volume to triple at 100000Nm-2. P1 V1 = P2 V2 Substituting :100000 x 100 = P2 x (100 x 3) V2 = 100000 x 100 = 33333.3333 Nm-2 (100 x 3) 3.A 60cm3 weather ballon full of Hydrogen at atmospheric pressure of 101325Pa was released into the atmosphere. Will the ballon reach stratosphere where the pressure is 90000Pa?
Calculate the pressure which must be applied to a fixed mass of 100cm3 of Oxygen for its volume to triple at 100000Nm-2. P1 V1 = P2 V2 Substituting :100000 x 100 = P2 x (100 x 3) V2 = 100000 x 100 = 33333.3333 Nm-2 (100 x 3) 3.A 60cm3 weather ballon full of Hydrogen at atmospheric pressure of 101325Pa was released into the atmosphere. Will the ballon reach stratosphere where the pressure is 90000Pa? P1 V1 = P2 V2 Substituting :101325 x 60 = 90000 x V2 V2 = 101325 x 60 = 67.55 cm3 90000 The new volume at 67.55 cm3 exceed ballon capacity of 60.00 cm3.It will burst before reaching destination. 7.Charles law states that“the volume of a fixed mass of a gas is directly proportional to the absolute temperature at constant/fixed pressure ” Mathematically: Volume α Pressure (Fixed /constant pressure) V α T (Fixed /constant P) ie V = Constant(k) T From Charles law , an increase in temperature of a gas cause an increase in volume. i.e doubling the temperature cause the volume to be doubled. Gases expand/increase by 1/273 by volume on heating.Gases contact/decrease by 1/273 by volume on cooling at constant/fixed pressure. The volume of a gas continue decreasing with decrease in temperature until at -273oC /0 K the volume is zero. i.e. there is no gas. This temperature is called absolute zero. It is the lowest temperature at which a gas can exist. Graphically therefore a plot of volume(V) against Temperature(T) in: (i)oC produces a straight line that is extrapolated to the absolute zero of -273oC . V -273oC 0oC T(oC) (ii)Kelvin/K produces a straight line from absolute zero of O Kelvin V 0 T(Kelvin)
For two gases then V1 = V2 T1 T2 T1 = Temperature in Kelvin of gas 1 V1 = Volume of gas 1 T2 = Temperature in Kelvin of gas 2 V2 = Volume of gas 2 Practice examples: 1.
It is the lowest temperature at which a gas can exist. Graphically therefore a plot of volume(V) against Temperature(T) in: (i)oC produces a straight line that is extrapolated to the absolute zero of -273oC . V -273oC 0oC T(oC) (ii)Kelvin/K produces a straight line from absolute zero of O Kelvin V 0 T(Kelvin)
For two gases then V1 = V2 T1 T2 T1 = Temperature in Kelvin of gas 1 V1 = Volume of gas 1 T2 = Temperature in Kelvin of gas 2 V2 = Volume of gas 2 Practice examples: 1. 500cm3 of carbon(IV)oxide at 0oC was transfered into a cylinder at -4oC. If the capacity of the cylinder is 450 cm3,explain what happened. V1 = V2 substituting 500 = V2 T1 T2 (0 +273) (-4 +273) = 500 x (-4 x 273) = 492.674cm3 (0 + 273) The capacity of cylinder (500cm3) is less than new volume(492.674cm3). 7.326cm3(500-492.674cm3)of carbon(IV)oxide gas did not fit into the cylinder. 2. A mechanic was filling a deflated tyre with air in his closed garage using a hand pump. The capacity of the tyre was 40,000cm3 at room temperature. He rolled the tyre into the car outside. The temperature outside was 30oC.Explain what happens. V1 = V2 substituting 40000 = V2 T1 T2 (25 +273) (30 +273) = 40000 x (30 x 273) = 40671.1409cm3 (25 + 273) The capacity of a tyre (40000cm3) is less than new volume(40671.1409cm3). The tyre thus bursts. 3. A hydrogen gas balloon with 80cm3 was released from a research station at room temperature. If the temperature of the highest point it rose is -30oC , explain what happened.
3. A hydrogen gas balloon with 80cm3 was released from a research station at room temperature. If the temperature of the highest point it rose is -30oC , explain what happened. V1 = V2 substituting 80 = V2 T1 T2 (25 +273) (-30 +273) = 80 x (-30 x 273) = 65.2349cm3
(25 + 273) The capacity of balloon (80cm3) is more than new volume (65.2349cm3). The balloon thus remained intact. 8. The continuous random motion of gases differ from gas to the other.The movement of molecules (of a gas) from region of high concentration to a region of low concentration is called diffusion. The rate of diffusion of a gas depends on its density. i.e. The higher the rate of diffusion, the less dense the gas. The density of a gas depends on its molar mass/relative molecular mass. i.e. The higher the density the higher the molar mass/relative atomic mass and thus the lower the rate of diffusion. Examples 1.Carbon (IV)oxide(CO2) has a molar mass of 44g.Nitrogen(N2)has a molar mass of 28g. (N2)is thus lighter/less dense than Carbon (IV)oxide(CO2). N2 diffuses faster than CO2. 2.Ammonia(NH3) has a molar mass of 17g.Nitrogen(N2)has a molar mass of 28g. (N2)is thus about twice lighter/less dense than Ammonia(NH3). Ammonia(NH3) diffuses twice faster than N2. 3. Ammonia(NH3) has a molar mass of 17g.Hydrogen chloride gas has a molar mass of 36.5g.Both gases on contact react to form white fumes of ammonium chloride .When a glass/cotton wool dipped in ammonia and another glass/cotton wool dipped in hydrochloric acid are placed at opposite ends of a glass tube, both gases diffuse towards each other. A white disk appears near to glass/cotton wool dipped in hydrochloric acid. This is because hydrogen chloride is heavier/denser than Ammonia and thus its rate of diffusion is lower . The rate of diffusion of a gas is in accordance to Grahams law of diffusion.
A white disk appears near to glass/cotton wool dipped in hydrochloric acid. This is because hydrogen chloride is heavier/denser than Ammonia and thus its rate of diffusion is lower . The rate of diffusion of a gas is in accordance to Grahams law of diffusion. Grahams law states that: “the rate of diffusion of a gas is inversely proportional to the square root of its density, at the same/constant/fixed temperature and pressure” Mathematically R α 1 and since density is proportional to mass then R α 1 √ p √ m For two gases then: R1 = R2 where: R1 and R2 is the rate of diffusion of 1st and 2nd gas. √M2 √M1 M1 and M2 is the molar mass of 1st and 2nd gas. Since rate is inverse of time. i.e. the higher the rate the less the time: For two gases then: T1 = T2 where: T1 and T2 is the time taken for 1st and 2nd gas to diffuse. √M1 √M2 M1 and M2 is the molar mass of 1st and 2nd gas. Practice examples: 1. It takes 30 seconds for 100cm3 of carbon(IV)oxide to diffuse across a porous plate. How long will it take 150cm3 of nitrogen(IV)oxide to diffuse across the same plate under the same conditions of temperature and pressure.
Molar mass CO2 = 44g Molar mass HCl = 36.5g T CO2 = √ molar mass CO2 => 200 seconds = √ 44.0 T HCl √ molar mass HCl T HCl √ 36.5 T HCl = 200seconds x √ 36.5 = 182.1588 seconds √ 44.0 3. Oxygen gas takes 250 seconds to diffuse through a porous diaphragm. Calculate the molar mass of gas Z which takes 227 second to diffuse. Molar mass O2 = 32g Molar mass Z = x g T O2 = √ molar mass O2 => 250 seconds = √ 32.0 T Z √ molar mass Z 227seconds √ x √ x = 227seconds x √ 32 = 26.3828 grams 250 4. 25cm3 of carbon(II)oxide diffuses across a porous plate in 25seconds. How long will it take 75cm3 of Carbon(IV)oxide to diffuse across the same plate under the same conditions of temperature and pressure.
Molar mass O2 = 32g Molar mass Z = x g T O2 = √ molar mass O2 => 250 seconds = √ 32.0 T Z √ molar mass Z 227seconds √ x √ x = 227seconds x √ 32 = 26.3828 grams 250 4. 25cm3 of carbon(II)oxide diffuses across a porous plate in 25seconds. How long will it take 75cm3 of Carbon(IV)oxide to diffuse across the same plate under the same conditions of temperature and pressure. (C=12.0,0=16.0) Molar mass CO2 = 44.0 Molar mass CO = 28.0 Method 1 25cm3 CO takes 25seconds 75cm3 takes 75 x25 = 75seconds 25
T CO2 = √ molar mass CO2 => T CO2seconds = √ 44.0 T CO √ molar mass CO 75 √ 28.0 T CO2 =75seconds x √ 44.0 = 94.0175 seconds √ 28.0 Method 2 25cm3 CO2 takes 25seconds 1cm3 takes 25 x1 = 1.0cm3sec-1 25 R CO2 = √ molar mass CO => x cm3sec-1 = √ 28.0 R CO √ molar mass CO2 1.0cm3sec-1 √ 44.0 R CO2 = 1.0cm3sec-1 x √ 28.0 = 0.7977cm3sec-1 √ 44.0 0.7977cm3 takes 1 seconds 75cm3 takes 75cm3 = 94.0203seconds 0.7977cm3
13.0.0 THE MOLE-FORMULAE AND CHEMICAL EQUATIONS (40 LESSONS) Introduction to the mole, molar masses and Relative atomic masses 1. The mole is the SI unit of the amount of substance. 2. The number of particles e.g. atoms, ions, molecules, electrons, cows, cars are all measured in terms of moles. 3. The number of particles in one mole is called the Avogadros Constant. It is denoted “L”.
3. The number of particles in one mole is called the Avogadros Constant. It is denoted “L”. The Avogadros Constant contain 6.023 x10 23 particles. i.e. 1mole = 6.023 x10 23 particles = 6.023 x10 23 2 moles = 2 x 6.023 x10 23 particles = 1.205 x10 24 0.2 moles = 0.2 x 6.023 x10 23 particles = 1.205 x10 22 0.0065 moles = 0.0065 x 6.023 x10 23 particles = 3.914 x10 21 3. The mass of one mole of a substance is called molar mass. The molar mass of: (i)an element has mass equal to relative atomic mass /RAM(in grams)of the element e.g. Molar mass of carbon(C)= relative atomic mass = 12.0g 6.023 x10 23 particles of carbon = 1 mole =12.0 g Molar mass of sodium(Na) = relative atomic mass = 23.0g 6.023 x10 23 particles of sodium = 1 mole =23.0 g Molar mass of Iron (Fe) = relative atomic mass = 56.0g 6.023 x10 23 particles of iron = 1 mole =56.0 g (ii)a molecule has mass equal to relative molecular mass /RMM (in grams)of the molecule. Relative molecular mass is the sum of the relative atomic masses of the elements making the molecule. The number of atoms making a molecule is called atomicity. Most gaseous molecules are diatomic (e.g. O2, H2, N2, F2, Cl2, Br2, I2)noble gases are monoatomic(e.g. He, Ar, Ne, Xe),Ozone gas(O3) is triatomic e.g.
Most gaseous molecules are diatomic (e.g. O2, H2, N2, F2, Cl2, Br2, I2)noble gases are monoatomic(e.g. He, Ar, Ne, Xe),Ozone gas(O3) is triatomic e.g. Molar mass Oxygen molecule(O2) =relative molecular mass =(16.0x 2)g =32.0g 6.023 x10 23 particles of Oxygen molecule = 1 mole = 32.0 g
2 2 Molar mass chlorine molecule(Cl2) =relative molecular mass =(35.5x 2)g =71.0g 6.023 x10 23 particles of chlorine molecule = 1 mole = 71.0 g Molar mass Nitrogen molecule(N2) =relative molecular mass =(14.0x 2)g =28.0g 6.023 x10 23 particles of Nitrogen molecule = 1 mole = 28.0 g (ii)a compound has mass equal to relative formular mass /RFM (in grams)of the molecule. Relative formular mass is the sum of the relative atomic masses of the elements making the compound. e.g.
It is the simplest whole number ratios in which atoms of elements combine to form the compound. 2.It is mathematically the lowest common multiple (LCM) of the atoms of the elements in the compound 3.Practically the empirical formula of a compound can be determined as in the following examples. To determine the empirical formula of copper oxide (a)Method 1:From copper to copper(II)oxide Procedure. Weigh a clean dry covered crucible(M1).Put two spatula full of copper powder into the crucible. Weigh again (M2).Heat the crucible on a strong Bunsen flame for five minutes. Lift the lid, and swirl the crucible carefully using a pair of tong. Cover the crucible and continue heating for another five minutes. Remove the lid and stop heating. Allow the crucible to cool. When cool replace the lid and weigh the contents again (M3). Sample results Mass of crucible(M1) 15.6g Mass of crucible + copper before heating(M2) 18.4 Mass of crucible + copper after heating(M3) 19.1 Sample questions 1. Calculate the mass of copper powder used. Mass of crucible + copper before heating(M2) = 18.4 Less Mass of crucible(M1) = - 15.6g Mass of copper 2.8 g 2. Calculate the mass of Oxygen used to react with copper. Method I Mass of crucible + copper after heating(M3) = 19.1g
7 7 Mass of crucible + copper before heating(M2) = - 18.4g Mass of Oxygen = 0.7 g Method II Mass of crucible + copper after heating(M3) = 19.1g Mass of crucible = - 15.6g Mass of copper(II)Oxide = 3.5 g Mass of copper(II)Oxide = 3.5 g Mass of copper = - 2.8 g Mass of Oxygen = 0.7 g 3.
Mass of crucible + copper before heating(M2) = 18.4 Less Mass of crucible(M1) = - 15.6g Mass of copper 2.8 g 2. Calculate the mass of Oxygen used to react with copper. Method I Mass of crucible + copper after heating(M3) = 19.1g
7 7 Mass of crucible + copper before heating(M2) = - 18.4g Mass of Oxygen = 0.7 g Method II Mass of crucible + copper after heating(M3) = 19.1g Mass of crucible = - 15.6g Mass of copper(II)Oxide = 3.5 g Mass of copper(II)Oxide = 3.5 g Mass of copper = - 2.8 g Mass of Oxygen = 0.7 g 3. Calculate the number of moles of: (i) copper used (Cu = 63.5) number of moles of copper = mass used => 2.8 = 0.0441moles Molar mass 63.5 (ii) Oxygen used (O = 16.0) number of moles of oxygen = mass used => 0.7 = 0.0441moles Molar mass 16.0 4. Determine the mole ratio of the reactants Moles of copper = 0.0441moles = 1 => Mole ratio Cu: O = 1:1 Moles of oxygen 0.0441moles 1 5.What is the empirical, formula of copper oxide formed. CuO (copper(II)oxide 6. State and explain the observations made during the experiment. Observation Colour change from brown to black Explanation Copper powder is brown. On heating it reacts with oxygen from the air to form black copper(II)oxide 7. Explain why magnesium ribbon/shavings would be unsuitable in a similar experiment as the one above. Hot magnesium generates enough heat energy to react with both Oxygen and Nitrogen in the air forming a white solid mixture of Magnesuin oxide and magnesium nitride. This causes experimental mass errors. (b)Method 2:From copper(II)oxide to copper
8 8 Procedure. Weigh a clean dry porcelain boat (M1). Put two spatula full of copper(II)oxide powder into the crucible.
(b)Method 2:From copper(II)oxide to copper
8 8 Procedure. Weigh a clean dry porcelain boat (M1). Put two spatula full of copper(II)oxide powder into the crucible. Reweigh the porcelain boat (M2).Put the porcelain boat in a glass tube and set up the apparatus as below; jgthungu@gmail.com52HEATCopper(II)oxideHydrogen /Laboratory gas /Ammonia gas/Carbon(II)Oxidegas from a generatorExcess hydrogenburningDetermining empirical formula from copper(II)oxide to copper Pass slowly(to prevent copper(II)oxide from being blown away)a stream of either dry Hydrogen /ammonia/laboratory gas/ carbon(II)oxide gas for about two minutes from a suitable generator. When all the in the apparatus set up is driven out ,heat the copper(II)oxide strongly for about five minutes until there is no further change. Stop heating. Continue passing the gases until the glass tube is cool. Turn off the gas generator. Carefully remove the porcelain boat form the combustion tube. Reweigh (M3). Sample results Mass of boat(M1) 15.6g Mass of boat before heating(M2) 19.1 Mass of boat after heating(M3) 18.4 Sample questions
9 9 1. Calculate the mass of copper(II)oxide used. Mass of boat before heating(M2) = 19.1 Mass of empty boat(M1) = - 15.6g Mass of copper(II)Oxide 3.5 g 2. Calculate the mass of (i) Oxygen. Mass of boat before heating(M2) = 19.1 Mass of boat after heating (M3) = - 18.4g Mass of oxygen = 0.7 g (ii)Copper Mass of copper(II)Oxide = 3.5 g Mass of oxygen = 0.7 g Mass of oxygen = 2.8 g 3.