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      I am presented with the... I am presented with the problem of deriving a method to find out what products are formed in the thermal decomposition of copper carbonate CuCO3. The two possible equations for this reaction are: 1 2CuCO3s Cu2Os + 2CO2g + 1/2O2g 2 CuCO3s CuOs + CO2g I will use the volume of gas produced to check which equation is right this is explained further on. Testing the gases limewater or burning splint released may seem less cumbersome however there is a possibility of the gas produced being a mixture of carbon dioxide and oxygen hence testing the gas will be inaccurate and inconclusive. I think that the copper carbonate, which is a green powder, will decompose to produce a black powder, which is known as copper oxide. Gas will be given off during the experiment and the end of the experiment will be observed when there is no more production of gas. If the amount of gas released is close to 60.7cm3 volumes are derived from calculations below then equation one 1 is correct. If the amount of gas released is close to 97.2cm3 then equation two 2 is correct. Additionally to this it is known that when a metal carbonate thermally decomposes, it forms its oxide and carbon dioxide, hence I believe that equation two is right. The method I have devised to find out which equation is correct is written below with reference to the mole. If 0.5g of copper carbonate is used then the number of moles can be calculated using moles = mass/molecular mass Mr Mr of CuCO3 = 63.5 + 12 + 16*3 = 123.5 Therefore the number of moles of CuCO3 in equation 1 is: n = 0.5/2 * 123.5 = 2.02429*10-3 moles The ratio of CuCO3 to CO2 is; 2 : 2 1 : 1 Therefore the mole ratio is; 2.02429*10-3 : 2.02429*10-3 Hence the volume of carbon dioxide produced can be calculated: Moles = volume cm3/ 24000 Volume = moles * 24000 = 2.02429*10-3 * 24000 = 48.58cm3 The ratio of CuCO3 to O2 is; 2 : 1/2 4:1 Therefore the mole ratio is; 2.02429*10-3 : 5.06073*10-4 Hence the volume of oxygen gas produced can be calculated using same equation as above: Volume = 5.06073*10-4 * 24000 = 12.15cm3 So the total volume of gas produced would be : 12.15cm3 + 48.58cm3 = 60.73cm3 = 60.7cm3 The number of moles in 0.5g of copper carbonate in equation 2 is: Moles = mass/molecular mass Mr = 0.5/ 123.5 = 4.04858*10-3 moles The ratio of CuCO3 to CO2 is;1:1 Therefore the mole ratio is; 4.04858*10-3 : 4.04858*10-3 Hence the volume of carbon dioxide produced can be calculated as follows: Moles n = volume cm3/24000 Volume = n * 24000 Volume = 4.04858 *10-3 * 24000 = 97.17cm3 = 97.2cm3 Since no other gas is produced, the total volume of gas produced would be 97.2cm3. Equipment: · 0.5g of CuCO3 · 200cm3 measuring cylinder · Trough · Digital weighing balance · Bunsen burner · Tripod · Heat mat · Delivery tube · Boiling tube · Clamp and stand · Wire mesh · Funnel paper · Tray Method: Firstly, an accurate weight of copper carbonate is needed. Put one funnel paper onto the digital balance and 'tare' so that the reading becomes zero. Remove the funnel paper and place it on the worktop next to the balance then add copper carbonate onto it funnel paper and weigh it. If it is not yet the correct mass 0.5g then remove it and add some more copper carbonate. This should be done until the weight is 5g and adjustments made accordingly. The reason for removing the funnel paper from the balance before adding the solid onto it is because some of the solid may fall onto the weighing balance and hence will not be used in the experiment but will still be weighed hence used in possible calculations thus distorting the possible results. After this is done, put the copper carbonate from the funnel paper into the boiling tube. Place the stand and bunsen burner on top of the heat mat and connect the boiling tube to the clamp. Attach the delivery tube to the boiling tube making sure that there is no space for air to escape. Half fill the trough with water then place the tray in it and also fill the measuring cylinder this must be completely full. Then cover the measuring cylinder with cling film and invert it into the trough of water. Ensuring that the end of the measuring cylinder is in the water, remove the cling film and place it on the tray. Connect the delivery tube from the boiling tube into the measuring cylinder ensuring that it remains completely full of water and that if a gas were to be passed through the tube then all of it would go into the measuring cylinder and not the trough. Place the tripod and wire mesh over the bunsen burner and make sure that the boiling tube is positioned over the tripod. When all the apparatus is set up, light the bunsen burner. When no more gas is being collected in the inverted measuring cylinder, the reaction is complete. Record the reading of the measuring cylinder and put off the bunsen burner. Compare this reading with the theoretical volumes previously calculated. The volumes calculated may not be exactly the same as recorded from the experiment but there should be a similarity. Whichever volume calculated is closest to the volume of gas released during the experiment relates to the correct equation. In order to get an accurate conclusion and to identify any anomalies, the experiment should be carried out three times, making sure that the mass of copper carbonate used and all other conditions remain the same so that a fair test is carried out. The value of gas recorded will differ slightly from the volume calculated previously because the theoretical yield is usually greater than the actual yield. This could be because of incomplete thermal decomposition or a variety of other factors. In the experiment a digital balance is used because it gives accurate readings which is important because the results depend greatly on the mass of copper carbonate used. Also the measuring cylinder 200cm3 used is suitable because it is accurate to the nearest 1cm3 with divisions of 2cm3, this ensures quite accurate readings. A gas syringe is not used because the gas released may be too much for the gas syringe to measure. The maximum amount of gas, which can be released is 97.2cm3 approximately "“ allow for experimental error and the gas syringe has a maximum capacity of 100cm3 which is too close to 97.2cm3 because it is possible that more than this amount will be released hence a gas syringe is not suitable for usage. The boiling tube is heated over the wire mesh because of safety reasons, for example to prevent the tube from possible breakage under the heat. Also precaution must be taken when working around the bunsen flame. Additionally to this, if any water is spilt when setting up the inverted measuring cylinder or at any other time, then it must be wiped and dried instantly in order to prevent any accidents or injuries for example slipping. Goggles must be worn at all times during the practical and long hair must be tied back to prevent any accidents.   

      I am presented with the problem of deriving a method to find out what products are formed in the thermal decomposition of copper carbonate CuCO3. The two possible equations for this reaction are: 1 2CuCO3s Cu2Os + 2CO2g + 1/2O2g 2 CuCO3s CuOs + CO2g ...

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      For this investigation I am... For this investigation I am reacting magnesium ribbon Mg with Hydrochloric acid HCl I am measuring the rate of reaction between the two, and to do this am measuring the hydrogen given off by the reaction. Magnesium is a shiny silver coloured metal, an element with an atomic number of 12, which belongs to the group 2 alkali metal group; it therefore has only two orbiting electrons and is very unstable and reactive. Hydrochloric acid is a solution of hydrogen and chloride a colourless acidic gas in water, Hydrochloric Acid has a high acidity and therefore will react with an alkali metal, the magnesium will displace the hydrogen and bond with the chlorine giving off hydrogen gas and producing the salt magnesium chloride. To keep the experiment fair I must look at all the factors in the reaction, firstly the concentration of the hydrochloric acid is important because it determines how fast the reaction will happen, more HCl molecules in the hydrochloric acid solution will produce more collisions with the magnesium and dissolve it quicker this is the only variable which I am planning to vary, I will keep the different variations of acid in different containers to prevent contamination and will wash out containers before I use them too, the heat of the reaction will also determine how fast it takes place, more heat means more movement of molecules and more collisions between the HCl and the Mg particles, the amount of magnesium will also effect the reaction rate so I will use measured amounts that are all the same, and the surface area of the magnesium must be kept constant too because more surface area means more collisions and effects reaction speed, the apparatus used will be kept the same to give the reaction the same amount of room to take place in, keeping the equipment still which will reduce vibrations which in turn effect the reaction rate, the amount of hydrochloric acid used will be kept constant too because more HCl acid means more HCl molecules to collide with the magnesium, there is also the pressure at which the reaction takes place which will cause more reaction because the particles would be pushed closer together, although the gas given off would have to be measured at room pressure or the volume would be changed by the pressure, The issue of room temperature and the starting temperature of the solution will also be kept constant by keeping the acid bottles in a constant temperature and measuring the temperature at the beginning of the experiment to make sure that this temperature stays the same and therefore there will be a constant amount of activation energy and the reactions will start with the same amount of heat present and therefore will start at the same rate of reaction if no variables were changed. I am not planning to vary the pressure because it brings up to many complications and is not needed, I will start the stop watch as soon as the magnesium ribbon hits the acid and close the bung instantly, also to do this someone else will time for me while I handle the magnesium and close the bung up, I will also clean the jar in which the reaction was carried out to prevent inconsistencies in the strength of acid and possible side effects caused by residues left from the reaction. To carry out the experiment we will be using a conical flask which the reaction will take place in, a bung with tubes coming out of it for the gas to flow down, 4 small measuring cylinders to measure the amount of acid used the measuring cylinders measure in ml, which is equivalent to cm3 in a ration of 1:1 so we can use them to measure the volume of the gas, a margarine tub filled with water and a large measuring cylinder filled with water which will be put upside down in the water to measure gas produced by the reaction, varied strengths of hydrochloric Acid in 10ml quantities, Magnesium ribbon in4cm lengths, a stop watch to time the reaction, and safety goggles for protection from acid splashes. The hydrochloric acid will be put in the conical flask, the magnesium will be dropped in to start the reaction, the bung will be promptly placed on top of the conical flask, from the bung tubes run in to the margarine tub full of water and under the measuring tube so that when the reaction takes place gas will be pushed through the tube and collect in the measuring cylinder. There will be 4 variations of the acid strength, 2 molar, 1.5 molar, 1 molar, and 0.5 molar. I predict that the higher the concentration of the acid the quicker the reaction, but there will be a point where all the magnesium is depleted and the reaction rate will level out, some of the weaker concentrations will not reach that level, but some of the stronger ones should and there will be a point where there is no more magnesium to react and the gas is no longer produced. Before I did the actual experiments I tried some preliminary tests with some 1m acid and some 2m acid, the hydrogen was produced as soon as I dropped the magnesium ribbon in the acid, and the 2m acid reaction finished quicker than the 1m acid reaction, this determines that my original assumption was correct and the magnesium was dissolved quicker in the 2m acid, although both reactions produced the same amount of gas because I used the same amount of magnesium and therefore the reaction was limited to the amount of magnesium to react with, I tested the gas produced under a flame and it produced a high pitched squeak which indicated the presence of hydrogen and proves more of my hypothesis correct and I can determine that when the magnesium ribbon reacts with the hydrochloric acid, magnesium chloride is formed. Here is a graph to show my preliminary test results. Time 1m HCl 2m HCl 30 "“ 0:30 22 cm3 43 cm3 60 "“ 1:00 33 cm3 45 cm3 90 "“ 1:30 43 cm3 46 cm3 120 "“ 2:00 45 cm3 46 cm3 150 "“ 2:30 46 cm3 46 cm3 180 "“ 3:00 46 cm3 46 cm3 210 "“ 3:30 46 cm3 46 cm3 240 "“ 4:00 46 cm3 46 cm3 It would appear that the reaction levels out at the point where 46 cm3 of hydrogen gas is produced when using a 4cm long piece of magnesium. I wrote down the equation to show the reaction between the 2 reactents: Magnesium + Hydrochloric acid = Magnesium Chloride + Hydrogen Mgs + 2HClaq = MgClaq + Hg I will repeat each experiment 4 times to smooth out inconsistencies and be able to produce an average result. To ensure that the experiment is carried out safely I will wear protective goggles at all times when handling acid, stand up while doing experiments to get out of the way quickly in the case of acid spills, and keep a clear and tidy workspace around me to prevent things getting in the way and being damaged by acid. With the equipment set up I would drop the magnesium In to the acid, then begin timing for a set amount of time even if the reaction had finished, and measure the gas produced in the large measuring cylinder and note the volume every 30 seconds for 4 minutes. Here are my 4 results tables Time 0.5m HCl 1m HCl 1.5m HCl 2m HCl 30 "“ 0:30 7 cm3 23 cm3 39 cm3 43 cm3 60 "“ 1:00 12 cm3 34 cm3 42 cm3 45 cm3 90 "“ 1:30 19 cm3 42 cm3 45 cm3 47 cm3 120 "“ 2:00 26 cm3 46 cm3 47 cm3 47 cm3 150 "“ 2:30 30 cm3 46 cm3 47 cm3 47 cm3 180 "“ 3:00 34 cm3 46 cm3 47 cm3 47 cm3 210 "“ 3:30 37 cm3 46 cm3 47 cm3 47 cm3 240 "“ 4:00 39 cm3 46 cm3 47 cm3 47 cm3 Time 0.5m HCl 1m HCl 1.5m HCl 2m HCl 30 "“ 0:30 7 cm3 21 cm3 38 cm3 44 cm3 60 "“ 1:00 11 cm3 35 cm3 43 cm3 46 cm3 90 "“ 1:30 16 cm3 42 cm3 45 cm3 47 cm3 120 "“ 2:00 27 cm3 47 cm3 48 cm3 47 cm3 150 "“ 2:30 32 cm3 48 cm3 48 cm3 47 cm3 180 "“ 3:00 35 cm3 48 cm3 49 cm3 47 cm3 210 "“ 3:30 37 cm3 48 cm3 49 cm3 47 cm3 240 "“ 4:00 40 cm3 48 cm3 49 cm3 47 cm3 Time 0.5m HCl 1m HCl 1.5m HCl 2m HCl 30 "“ 0:30 6 cm3 20 cm3 38 cm3 44 cm3 60 "“ 1:00 12 cm3 33 cm3 43 cm3 47 cm3 90 "“ 1:30 17 cm3 43 cm3 46 cm3 48 cm3 120 "“ 2:00 26 cm3 48 cm3 49 cm3 48 cm3 150 "“ 2:30 29 cm3 49 cm3 49 cm3 48 cm3 180 "“ 3:00 34 cm3 49 cm3 49 cm3 48 cm3 210 "“ 3:30 36 cm3 49 cm3 49 cm3 48 cm3 240 "“ 4:00 39 cm3 49 cm3 49 cm3 48 cm3 Time 0.5m HCl 1m HCl 1.5m HCl 2m HCl 30 "“ 0:30 8 cm3 25 cm3 40 cm3 45 cm3 60 "“ 1:00 13 cm3 35 cm3 47 cm3 47 cm3 90 "“ 1:30 19 cm3 42 cm3 49 cm3 49 cm3 120 "“ 2:00 26 cm3 49 cm3 49 cm3 49 cm3 150 "“ 2:30 32 cm3 50 cm3 49 cm3 49 cm3 180 "“ 3:00 36 cm3 50 cm3 49 cm3 49 cm3 210 "“ 3:30 39 cm3 50 cm3 49 cm3 49 cm3 240 "“ 4:00 42 cm3 50 cm3 49 cm3 49 cm3 These are the 4 sets of results, I recoded 2 each lesson, the last results seem to be react slightly quicker than the others, this could be due to temperature of the room or contamination of the equipment, the reactions seem to have happened quicker, although they don't seem to be too random and I will use them in my average table set of results. Time 0.5mHCl 1mHCl 1.5mHCl 2mHCl 30 "“ 0:30 7 cm3 22.25 cm3 38.75 cm3 44 cm3 60 "“ 1:00 12 cm3 34.25 cm3 43.75 cm3 46.25 cm3 90 "“ 1:30 17.75 cm3 42.25 cm3 46.5 cm3 47.25 cm3 120 "“ 2:00 26.25 cm3 47.5 cm3 48.25 cm3 47.25 cm3 150 "“ 2:30 30.75 cm3 49 cm3 48.25 cm3 47.25 cm3 180 "“ 3:00 34.75 cm3 49 cm3 48.5 cm3 47.25 cm3 210 "“ 3:30 37.25 cm3 49 cm3 48.5 cm3 47.25 cm3 240 "“ 4:00 40 cm3 49 cm3 48.5 cm3 47.25 cm3 It was noticeable, when looking at the results table, that the more concentrated acid had a faster rate of reaction than the less concentrated acid. This was probably because there are more particles in a concentrated acid and therefore more collisions will occur, for example; the 0.5 molar acid reactions produced on average 7 cm3 of hydrogen gas in the first 30 seconds, whereas the 1.5 molar acid reactions produced on average 38.75 cm3 of hydrogen gas in the first 30 seconds. The results appear to level out at around about 48.4cm3; the concentration of the 0.5 acid reactions does not level out because we stopped the timer before the reaction had time to complete. I have made a graph of the average reaction rate for this experiment. The numbers along the bottom indicate time; the numbers along the side indicate cm3 of gas produced. The graph supports my original prediction, it shows that the higher the concentration of the acid in molars the faster the reaction occurs and hydrogen is produced quicker, therefore I can deduce that In a higher concentration there are more acid particles to react with the magnesium ribbon and therefore it is dissolved faster. Therefore if you increase the concentration of the acid you are introducing more particles into the reaction which will in turn produce a faster reaction because there will be more collisions between the particles which is what increases the reaction rate. If we would have carried on the practical for a longer time the 0.5 molar reactions would have eventually levelled out at about 48.4cm3. While performing the experiment I had to ensure that the temperature was kept constant throughout, because varying temeperature will vary the results, if the temperature increases from the start time to the finish time then the reaction speeds will get quicker at a slightly greater rate, there was also the issue of room temperature which we measured but could not do much about because the room is a large environment and has many sources of heat. The reaction could have been sped up or slowed down in many ways but the amount of hydrogen produced remains constant. There are always ways to improve an experiment like this, I could have measured the temperature of the acid to make sure that it all started at the same temperature, and could have recorded temperature results while doing the practical too, so that I have 2 sets of results for each experiment, and could compare these and analyse how they are relevant to the experiment. Also the measuring of the acid could have been improved using small measuring syringes, and the measuring of the hydrogen produced could also be improved using the gas measuring syringe which would have produced much more accurate results because getting the upturned measuring cylinder in water without letting air in was difficult and the reading of the measuring cylinder could have been improved using the gas measuring syringe because the results would have been more accurate. Also weighing the magnesium ribbon, and cutting it more precisely would have helped get more accurate results, the magnesium was also covered in a white powder, some pieces more so than others, this is magnesium oxide, produced where the magnesium has been exposed to air, the pieces with more magnesium oxide on them would have less magnesium to react with the acid and the oxide may slow the reaction by getting in the way, or reacting with the acid and producing water. I could have cleaned each piece of magnesium with some emery cloth to reduce the magnesium oxide. I could have also tried varying other constants in my experiment, beginning of the year I temperature, the presence of a catalyst, the surface area of the magnesium or the pressure of the reaction chamber. These differentials in the variables would affect the reaction rates in different ways, but the tests all followed the same predicted pattern and shows that there is a level where all the magnesium is depleted, if we had used more magnesium and less hydrochloric acid we could have found a point where the amount of hydrochloric acid levels out before the magnesium, but we would have to use a lot of magnesium because using a small amount of hydrochloric acid would make it much harder to measure the results with current equipment. I also did some extra tests using 3 cm pieces of magnesium, and measuring how much gas was produced, to do these measurements we used the same equipment apart from the measuring cylinder which was replaced with the gas measuring syringe, the measuring syringe was much more accurate than the cylinder, and gave us better readings. The amount of gas produced from the 3cm piece of magnesium levels out at a point of 32 cm3 if we put this in a line graph with the maximum amount of hydrogen produced from the 5.5 cm pieces of magnesium we can predict how much the most amount of hydrogen that can be produced by a reaction between a different length of magnesium with hydrochloric acid. Here is a graph of the readings I got using the 4 cm piece of magnesium. Time 0.5m 1m 1.5m 2m 30 - 0:30 8 12 27 29 60 - 1:00 12 17 33 31 90 - 1:30 17 25 34 31 120 - 2:00 19 28 34 31 150 - 2:30 23 33 34 31 180 - 3:00 26 35 34 31 210 - 3:30 28 35 34 31 240 - 4:00 30 34 34 31 the results are much more varied in these tests, this could be due to greater accuracy and being able to note these variances, or from some sort of contaminant, the average amount of gas produced for a 5.5 cm piece of magnesium was 48.4 cm3 the average amount of gas produced for a 3 cm piece of magnesium was 42 cm3 the more magnesium means there are more magnesium atoms to react with the hydrochloric acid molecules and therefore more hydrogen is produced, if we put these in to a line graph we can use it to estimate how much hydrogen would be produced by other lengths of magnesium. I have drawn a line of best fit between the two points, with this graph we can estimate how much hydrogen would be produced if we reacted 1cm of magnesium. The graph gives a reading of around about 18 cm3 of hydrogen gas produced, if I had extra time we could test this theory, but unfortunately we do not. I also only did one set of results for the tests with the gas syringe, if I would have been able to continue this experiment further I could have produced more average results and seen if my predictions for the 1 cm piece of magnesium was correct. We could have also varied the concentration of the acid more so, and used less or more acid to get more accurate results or results for different test situations, instead of changing the strength we could have changed the amount of acid, or the temperature of the acid, or try varying these together and see how they effect each other.   

      For this investigation I am reacting magnesium ribbon Mg with Hydrochloric acid HCl I am measuring the rate of reaction between the two, and to do this am measuring the hydrogen given off by the reaction. Magnesium is a shiny silver coloured metal, an element with an atomic...

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      Determination of the relative atomic mass...Determination of the relative atomic mass of Lithium Implementing For all my results I have decided to use three significant figures. This is because the apparatus I was told to use for the experiment gave me results to three significant figures. There was, therefor, no point in calculate to any further accuracy as this would just define a figure that has already been calculated to a specific depth of accuracy. Results Method 1 Mass of Lithium used 1.09 g Volume of gas produced 175 cm3 Method 2 1 2 3 Final Burette Reading 42.1 42.3 42.3 Initial Reading 0 0 0 Volume added cm3 42.1 42.3 42.3 Average 42.2 cm3 Analysing Method 1 Calculate the number of moles of hydrogen. 1 175 = 0.00729 Number of moles = Amount of gas produced 2400 2400 This was calculated to find out the number of moles in the Hydrogen gas that had produced. Deduce from this the number of moles of Lithium. 2 The number of moles of Lithium = 0.00729 x 2= 0.0146 If you look at the chemical equation¡V 2Lis + 2H2Ol ------"žÂ³ 2LiOHaq + H2g You can see that the molar ratio of number of moles of hydrogen to number of moles of lithium is 1:2. As I had already calculated the number of moles for Hydrogen, I can multiply it by two and this will give me the number of moles of Lithium. Using your values above and the original mass of Lithium, calculate the relative atomic mass of lithium. 3 1.09 = 7.13 Relative atomic mass= Mass used in grams 0.146 Number of moles From the amount of lithium used and the number of moles of lithium. I used the above equation to calculate the relative atomic mass of lithium. Method 2 Calculate the number of moles of HCl used in the titration. Average volume 1 42.7 x 0.1= 0.00422 Number of moles of HCl = concentration x volume 1000 1000 The formula I used is the one shown above. This enabled me to calculate the number of moles of HCl used in the titration experiment, as the HCl was a solution. Deduce the number of moles of LiOH used in the titration. 2 The number of moles of LiOH used in the titration = 0.00422 LiOHaq + HClaq -----"žÂ³ LiCl + H2g This calculation was much the same as calculation 2 of the last experiment but this time the molar ratio was 1:1 between LiOH and HCl, as shown in the equation above¡V Calculate the number of moles of LiOH present in 100cm3 of the solution from method 1. 3 The number of moles of LiOH present in 100 cm3 is = 0.00422 x 4 = 0.0169 The first equation shows that in 25 cm3 of LiOH there are 0.00422 moles. 100 cm3 is 4 times as much as 25cm3, therefor if I multiply 0.00422 by 4 then this would give me the number of moles on 100 cm3 Use the result and the original mass of lithium to calculate the relative atomic mass of Lithium. 4 0.104 = 6.15 Relative atomic mass = Mass used in g 0.0169 Number of moles I used the formula above to calculate the relative atomic mass. My average atomic mass for Lithium = 6.64 The actual relative atomic mass for Lithium = 6.90 Difference = 0.26 % difference = 3.77 Risk assessment The first safety hazard I encountered in my experiment was the Li that I placed into the waster in the first method. I was told it reacted violently with water to form lithium hydroxide. The Li was contained at almost all times throughout the experiment but as there was a chance of me being in proximity of the reaction then I wore goggles. Also I was shown that I was not to use a cube that has sides of more than 3mm. So I used a cube as near this number as possible but I ensured that I did not exceed this 3mm constraint at all times. The second safety hazard I encountered was the hydrogen gas. This is an extremely flammable gas and so I made sure that there were no naked flames around the experiment at all times. Thirdly I encountered lithium hydroxide this is only a problem when it exceeds 6.5 molar and as I used no where near this molarity all that this substance was, was a mild irritant. Therefor all I needed was eye protection and in the case of it getting on my skin I was to wash it off with water. HCl was the last harmful substance that I used in my experiment and the procedure for this was exactly the same as lithium hydroxide. Evaluation There are two ways that there could have been inaccuracies in my experiment, the first of which could have been my experimental procedure and the second could have been problems with the apparatus I used to implement my experiment. The first inaccuracy that I encountered really cancels itself out in some respects. There is oil used to stop the Li reacting with the air and this is helpful in some respects but not in others. This creates a multiple number of inaccuracies, as you have to remove the oil from the Li so that you weight the Li by itself to stop inaccuracies in later calculations. This though will then let the Li oxidise with the oxygen in the atmosphere, therefor I will, if the experiment is to be repeated use an inert atmosphere to stop any oxidisation from taking place. Also the only effective way to remove the oil from the Li is to cut the outside layer off the Li and this will ensure there is no oil left on the Li when it is weighed. Also this will stop the oil from being in the Li hydroxide and creating any inaccuracies in the second method. Hydrogen is another source for an inaccuracy. This is because in can be effected and lost. It can be lost in two ways, either when the bung is replaced onto the conical flask when the Li starts to react or when the hydrogen goes through the rubber tubing it can be lost. There are no ways to improve the experimental procedure for this and so they are inaccuracies that I cannot change. The third inaccuracy is the fact the gas expanded when it came from the Li and water as it was hot so the final gas reading would be greater than it should be and so it effected my results. I would of used a percentage error type of evaluation but as I don¡¦t have specific measurements that could of shown me how inaccurate the data I collected was I cannot use percentage errors for some data¡¦s and not for others.   

      Determination of the relative atomic mass of Lithium Implementing For all my results I have decided to use three significant figures. This is because the apparatus I was told to use for the experiment gave me results to three significant figures. There was, therefor, no point in calculate...

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      Introduction Sulphate ions... Introduction Sulphate ions and iodine can be formed from reactions between peroxodisulphate ions and iodine ions. S2O82-aq + 2I-aq -----------> 2S042-aq + I2aq The reactants and the sulphate ions are colourless so the achievements of the experiment must be determined by the iodine produced. If starch is added to the reaction iodine is more clearly visible as a strong complex of blue-black colour is formed. For this experiment I will be measuring the time taken for the iodine to appear in the solution. To make this more clear and accurate thiosulphate ions are added to the mixture which will emphasize the changes. This chemical also returns iodine to iodine ions. 2S2O32-aq + I2aq -----------> S4062-aq + 2I-aq As the reactants of this reactions are colourless it is difficult to tell when the colour change will happen, and when it does it is quite sudden. This happens because the thiosulphate is being used up in the reactions, and when it is all used up the starch-iodine complex is formed resulting in the colour change. Equipment and apparatus needed: "¢ Test tubes "¢ Boiling tubes "¢ Thermometer 0-110°C "¢ Stopwatch "¢ Burettes "¢ 15cm3 Potassium iodide solution,KI, 1.00 mol dm-3 "¢ 10cm3 Potassium peroxodisulphate solution,K2S2O8, 0.0400 mol dm-3 "¢ 10cm3 Sodium thiosulphate solution, Na2S2O3, 0.0100 mol dm-3 "¢ 5cm3 fresh Starch solution "¢ Test tube rack Method "¢ Read and understand carefully the table below illustrating the different mixtures of concentrations for individual experiments. mixture Volume of KLaq/ cm3 Volume of water / cm3 Volume of Na2S2O3aq / cm3 Volume of starch solution/cm3 Volume of K2S2O8aq / cm3 1 5 0 2 1 2 2 4 1 2 1 2 3 3 2 2 1 2 4 2 3 2 1 2 5 1 4 2 1 2 "¢ When making the mixtures, make sure that all chemicals are added to a boiling tube, except the potassium peroxodisulphate which must be added separately as this is what will trigger the reaction. "¢ Make up mixture 1 and place a thermometer in the boiling tube. Add the potassium peroxodisulphate at begin the timing with the clock "¢ Stop timing when the solution turns blue-black "¢ Constantly stir the solution. "¢ Take a temperature reading "¢ Repeat procedure for each mixture of different concentrations. Results A Table mixture Concentration of I-aq / mol dm-3 Clock time / seconds Rate / mol dm-3 s-1 Temperature / °C 1 0.5 28.8 3.45 X10-7 23 2 0.4 40.9 2.44 X10-7 23 3 0.3 58.0 1.72 X10-7 26 4 0.2 136.7 7.32 X10-8 26 5 0.1 455.5 2.20 X10-8 24 Questions B Rate = concentration of I2 Time = [ I2 ] = mol dm-3 s-1 S [ I2 ] = 1.0 X10-5 Mixture: 1. = 1.0 X10-5 = 3.45 X10-7 28.8 2. = 1.0 X10-5 = 2.44 X10-7 40.9 3. = 1.0 X10-5 = 1.72 X10-7 58.0 4. = 1.0 X10-5 = 7.32 X10-8 136.7 5. = 1.0 X10-5 = 2.20 X10-8 455.5 C The Iodine ions are in excess in the reaction mixtures. The reactant that is not in excess in the reaction will be used up and it is this which will determine the total amount of iodine produced. D Moles = mol dm-3 x dm3 = 0.04 x 2/1000 = 8x10-5 mols of S2O8 2- E i 2cm3 of Na2S2O3 0.0100 mol dm-3 Moles of Na2S2O3 = 0.01 x 2/1000 = 2x10-5 moles of S2O3 ii Moles of = 1.0 X10-5 I2 iii Percentage % of reaction = 1.0 X10-5 x100 8.0 X10-5 = 12.5 %   

      Introduction Sulphate ions and iodine can be formed from reactions between peroxodisulphate ions and iodine ions. S2O82-aq + 2I-aq -----------> 2S042-aq + I2aq The reactants and the sulphate ions are colourless so the achievements of the experiment must be determined by the iodine produced. If...

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