Marking_Guidelines_Chem_Prelim_2008_Exam_Choice

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Exam Choice 2008 Chemistry Preliminary Examination . Marking Guidelines and model Answers. Part A Multiple Choice 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 C D D B A C A B C A B B A B D Part B Question 16 Marking Guidelines Marks identifies the electrical conductivity of each allotrope and explains this in terms of the structure and bonding in each 4 identifies the electrical conductivity of each allotrope and correctly explains this in terms of the structure and bonding for two of the three 3 identifies the electrical conductivity of two allotropes and correctly explains this in terms of the structure and bonding for one 2 identifies one aspect of the structure, bonding or electrical conductivity of one allotrope 1 For a substance to conduct electricity it must have delocalised charged particles that can travel through the sample. Diamond is composed of a 3D lattice of atoms, each covalently bonded via all four valence electrons, to neighbouring carbon atoms. There are no delocalised electrons hence diamond does not conduct electricity. Fullerenes are discrete molecules, for example C 60 , where carbon atoms are also bonded to neighbouring carbons by covalent bonds. There are also no mobile charged particles to conduct electricity through the sample, so neither do fullerenes conduct electricity. Graphite consists of layers of 2D lattices. Bonding within layers is covalent, but using only 3 of carbon’s 4 valence electrons. The fourth electron is delocalised between layers, and as a result, graphite is a good electrical conductor. Question 17 Marking Guidelines Marks explains the use of metals throughout history, with reference to extraction of the metals from their ores, including a balanced chemical equation. 4 explains the use of metals throughout history, with reference to extraction of the metals from their ores 3 outlines the use of metals over time, with limited reference to ease of extraction 2 lists copper, bronze and iron in their correct chronological order 1 Different metals have been used over time because of the ease of their extraction from their ores. Unreactive metals were the first to be extensively used, such as gold and silver, because they are found uncombined in nature. Copper, which can be extracted from its oxide simply by using carbon in the heat of a fire, was the first metal to be extracted. Bronze, an alloy of copper and tin, can be similarly extracted from a mixture of their ores. This required little or no expertise and very little energy. The use of iron however, required much hotter temperatures, and thus more advanced technology, and occurred later. Later still aluminium use began, because of the requirement of electrolysis to extract it from its ore. The overall reaction to produce aluminium via electrolysis is: 2Al 2 O 3 (l)  4Al (l) + 3O 2 (g) Question 18 (a) Marking Guidelines Marks correctly identifies the separation process as not physical and gives a valid reason 2 correctly identifies the separation process as not solely physical 1 The process is based on a chemical reaction to convert the Fe into FeCl 2 . Therefore the process is not a physical separation. Question 18 (b) Marking Guidelines Marks writes a correct, balanced chemical equation 1 Fe(s) + 2HCl(aq) -> FeCl 2 (aq) + H 2 (g) Question 18 (c) Marking Guidelines Marks correctly calculates the % iron in the mixture and demonstrates that it is not in a 2:1 sulfur to iron ratio 3 calculates % iron in the mixture 2 calculates the mass of sulfur or iron present in the mixture 1 mass of residue (sulfur): 4.25-1.86 = 2.39g mass of mixture: 360.18-356.98 = 3.20g mass of iron = 3.2- 2.39 = 0.81 g %Fe = 0.81 / 3.20 x 100 = 25.3 % Therefore the % S is approximately 74.7% This means the mixture does not meet the teacher's requirements as it would need to be approx 67% S and 33% Fe to be 2 parts S : 1 part Fe. 1

Question 19 Marking Guidelines Marks explains the trend in first ionisation energy and atomic radius, across a period and down a group, using effective nuclear charge, and provides an example of each trend in the P.T. 5 identifies the trends in first IE and atomic radius in the periodic table, and provides a limited explanation in terms of effective nuclear charge 3-4 identifies one or two trends in first IE and/or atomic radius in the periodic table 1-2 Moving across a period in the periodic table, the number of protons in the nucleus increases, and the number of electrons in the valence shell increases. Because this valence shell is not increasingly shielded from the nucleus, effective nuclear charge increases across a period, and the valence shell is pulled more closely to the nucleus. This results in an increased first ionisation energy moving across a period (eg the first IE of Na is much lower than that of Ar), and the atomic radius decreases across a period (eg Li atoms are much larger than Ne). Moving down a group in the periodic table, although the number of protons is increasing, the number of full electron shells between the nucleus and the valence shell is increasing, the valence shell is increasingly shielded from the nucleus moving down a group. As a result, first IE decreases down a group and atomic radius increases. For example, Li atoms are much smaller than Cs, whereas Cs is much more reactive, because of its lower first ionisation energy. Question 20 (a) Marking Guidelines Marks explains how the addition of a catalyst affects the rate of reaction using the particle theory 2 identifies how the addition of a catalyst affects the rate of reaction 1 To react, particles must collide with sufficient energy to overcome the activation energy. Addition of a catalyst increases reaction rate by providing an alternative mechanism with a lower activation energy, so that more collisions between particles result in a reaction. Question 20 (b) Marking Guidelines Marks explains how an increase in T affects the rate of reaction using the particle theory 2 identifies how an increase in T affects the rate of reaction 1 To react, particles must collide, and they must collide with sufficient energy to overcome the activation energy. Increasing the temperature increases the rate because particles have more kinetic energy at increased temperature. Thus, they collide more frequently, and with more energy, both of which increase the rate. Question 21 (a) Marking Guidelines Marks writes two correct half equations 2 writes one correct half equation OR writes two half-equations based on an incorrect valency of X 1 X(s) -> X 2+ + 2e - 2H + + 2e - -> H 2 (g) Question 21 (b) Marking Guidelines Marks outlines an appropriate method to determine the reactivity of Y and Z 2 outlines an appropriate but limited or partially correct method to determine the reactivity of Y and Z 1 A solid sample of metal Y can be placed into a solution of metal Z nitrate in a test tube. If metal Y displaces metal Z from its solution, then Y is more reactive than Z. Question 22 Marking Guidelines Marks provides two comprehensive methods for producing the desired solution, using the solution and the solid, and includes equipment to be used and necessary calculations 7 provides two methods for producing the desired solution, using the solution and the solid, and includes most of the equipment to be used and most steps in the calculations 5-6 outlines two methods for producing the desired solution, using the solution and the solid, with limited equipment included, and attempts to include calculations 3-4 response contains one or two correct steps 1-2 Using the solution, the student could obtain 25.0 mL of 0.01 M Na 2 CO 3 (aq): - transfer 20.0 mL of the 0.05 M solution to a 100 mL volumetric flask using a clean 20.0 mL volumetric pipette - fill up to the mark with distilled water and mix thoroughly by inversion - transfer 25.0 mL of the solution to the reaction vessel being used using a 25.0 mL volumetric pipette - the solution produced would be 0.01 M because: o C 1 V 1 = C 2 V 2 o 0.05 x 20 = C 2 x 100 o C 2 = 0.01 M 2

Using the solid, the student could prepare the solution as follows: - weigh a 25 mL beaker using an electronic balance - transfer solid Na 2 CO 3 to the beaker and reweigh - dissolve the solid in 20 mL of distilled water and quantitatively transfer this to a 100 mL volumetric flask (using a small funnel, and by washing the beaker 3x with 5 mL of distilled water and transferring the washings to the volumetric flask) - fill the volumetric flask up to the mark using distilled water and mix thoroughly by inversion - obtain 25.0 mL of the solution using a 25.0 mL volumetric pipette - * the mass of Na 2 CO 3 (s) needed can be calculated in the following way: o n(Na 2 CO 3 ) =CV = 0.1 x 0.01 = 0.001 mol o m(Na 2 CO 3 ) = n x MM = 0.001 x 105.99 = 0.11g. Question 23 (a) Marking Guidelines Marks draws the structure of butane 1 C C C C H H H H H H H H H H Question 23 (b) Marking Guidelines Marks identifies the trend shown in the table and explains it in terms of intermolecular forces 3 identifies the trend and gives a partial explanation 2 identifies the trend shown 1 The data provided show that as the number of carbons in the alkane increases, so does the boiling point. This is because as the number of carbons increases, so does the molecular mass of the alkane. Alkanes are non-polar molecules and as such the only intermolecular forces present are dispersion forces. These increase with increasing molecular mass because the number of electrons in the molecule to create temporary dipoles increases. As a result, as molecular mass increases the energy required to separate molecules from each other increases, and thus the boiling point increases, since boiling a sample separates molecules as they enter the gas phase. Question 23 (c) Marking Guidelines Marks identifies two similarities and/or differences in the safe storage of butane and octane 2 identifies one difference or similarity in the safe storage of butane and octane 1 Both hydrocarbons are flammable and must be stored away from naked flames and sources of ignition. However, butane is a gas at room temperature and therefore must be stored in a high pressure cylinder with a narrow neck and tightly fitting valve. Octane can be stored in metal drums because it is much less volatile than butane. Question 24 (a) Marking Guidelines Marks calculates the energy required to break the bonds, showing working 1 energy required = 4x414 + 2x498 = 2652 kJ Question 24 (b) Marking Guidelines Marks calculates the energy released showing working 1 energy released = 2x804 + 4x463 = 3460 kJ Question 24 (c) Marking Guidelines Marks calculates  H for the reaction using the results from (a) and (b) above 1 energy released – energy required = 3460-2652 = 808kJ because more energy is released than absorbed, the reaction is exothermic and so  H = -808kJ/mol Question 24 (d) Marking Guidelines Marks draws an energy profile diagram for the reaction that includes: - the correct shape for an exothermic reaction - labelled activation energy - reactants and products labelled - the value of  H from the calculation above 3 draws an energy profile diagram that shows most of the features above 2 draws an energy profile diagram with the correct shape for an exothermic reaction 1 3

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Question 25 (a) Marking Guidelines Marks uses the graph to estimate the boiling point of ethyne as -85 o C 1 Question 25 (b) Marking Guidelines Marks accounts for the higher BP of the species with MM 17, 18 and 20 in terms of H-bonding and compares the strength of these forces with those in the other molecules. 3 accounts for the higher BP of these species in terms of identified intermolecular forces 2 Identifies the intermolecular forces in species with MM17, 18 or 19 are hydrogen bonds OR Identifies the intermolecular forces other species are weak dispersion forces or dipole dipole interactions. 1 The species with MM = 17, 18 and 20 are NH 3 , H 2 O and HF. These species have intermolecular hydrogen bonding (whereas the other species shown do not). This is the strongest intermolecular force. Hence more energy is required to separate these molecules from each other, and hence their BPs are the highest shown. Question 25 (c) Marking Guidelines Marks explains why molar mass needs to be kept constant in terms of dispersion forces 2 provides a limited or partially correct explanation of why molar mass needs to be kept constant 1 All molecules have intermolecular dispersion forces, and these arise from the random motion of electrons creating temporary dipoles. The more electrons the stronger the dispersion forces, and the higher the molar mass of a species, the more electrons it contains. Therefore to keep the strength of dispersion forces constant so that only the effects of permanent dipoles are considered, molecular mass of the species being considered needs to be similar. Question 26 (a) Marking Guidelines Marks writes a balanced chemical equation including states 1 Cu(NO 3 ) 2 (aq) + Na 2 CO 3 (aq) -> CuCO 3 (s) + 2NaNO 3 (aq) Question 26 (b) Marking Guidelines Marks correctly calculates the mass of CuCO 3 produced, showing working and quoted to 3 significant figures 3 correctly calculates the mass of CuCO 3 , quoted to 3 sig. figs, but with one error OR correctly calculates the mass of CuCO 3 , quoted to the incorrect number of sig. figs. 2 response indicates an understanding that there is a limiting reagent present 1 n(Cu(NO 3 ) 2 ) = m/MM = 1.35/187.57 = 7.197x10 -3 n(Na 2 CO 3 ) = m/MM = 1.35/105.998 = 1.274x10 -2 therefore all of the Cu(NO 3 ) 2 will be used (limiting reagent) and hence n(CuCO 3 ) will be 7.197x10 -3 m(CuCO 3 ) = n x MM = 7.197x10 -3 x 123.55 = 0.889 g. Question 27 (a) Marking Guidelines Marks describes the basic procedure for the investigation, including: - basic equipment used - data to be measured (temp change, mass of chemical, mass of water used) - method of data collection 3 outlines the basic procedure with an error or omission of major step 2 describes one step in the procedure OR identifies TWO types of data that need to be collected in the investigation 1 1. Weigh out a small amount of solid NaOH using an electronic balance. 2. Measure 100.0mL of water using a measuring cylinder and pour into a polystyrofoam cup. 3. Measure the initial temperature of the water using a thermometer. 4. Add the NaOH and stir to dissolve. 5. Monitor the temperature of the solution and record the final temperature. Question 27 (b) Marking Guidelines Marks identifies one valid reason for a difference in the result gained experimentally and the theoretical result. 1 Some heat energy will be released into the surroundings other than water (eg the vessel will also gain heat) and this heat wasn’t measured. 4 energy (kJ/mol) reaction pathway reactants products activation energy  H = - 808kJ/mol