2010 exam choice s - Exam Choice 2010 Chemistry Trial HSC...

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1 Exam Choice 2010 Chemistry Trial HSC examination . Marking Guidelines and model Answers. Section I A Multiple Choice 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 A A C C D C A C D C A A D B B B D B C B Section I B 21.a . Marking Guidelines Marks Explains the different reactivities including a correct chemical equation. 2 Explains the different reactivities. 1 Butane contains only a carbon-carbon single bond, whereas butene contains a carbon-carbon double bond. The double bond is much reactive than the single bond. This is illustrated by the fact that 1-butene undergoes an addition reaction with Br 2 (aq) (reaction 1), whereas butane undergoes substitution, and only in the presence of UV light (reaction 2). Reaction 1 21.b. Marking Guidelines Marks Gives the correct name for the product of the reaction identified in (a) 1 1.2 dibromobutane 22.a. Marking Guidelines Marks Gives the systematic name of one monomer. 1 Chloroethene 22.b Marking Guidelines Marks Draws the structure of a segment of the polymer chain. 1 22.c. Marking Guidelines Marks Accounts for a use of the polymer in terms of its properties. 2 Identifies a use OR a property of the polymer. 1 Poly(vinyl chloride) commonly used for waterproof clothing and shower curtains because it is water resistant and has a good tensile strength. 23.a. Marking Guidelines Marks Writes a balanced equation for the reaction. 1 Cu (s) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2Ag (s) 23.b. Marking Guidelines Marks Correctly determines the concentration of both species, with sufficient working. 3 Correctly determines the concentration of the nitrate ion AND Determines the moles of Ag + consumed in the reaction. 2 Correctly determines the concentration of the NO 3 - Calculates the initial moles of Cu or AgNO 3 1 (i) n (NO 3 - ) remains unchanged as it is a spectator ion.
2 c(NO 3 - ) = 0.50M (2 sig figs) (ii) n (Cu) = m/MM = 1.15 / 63.55 = 0.0181 mol n (AgNO 3 ) = c x V = 0.5 x 0.1 = 0.05 mol 1 mole Cu requires 2 moles of AgNO 3 0.0181 mol Cu requires 0.03619 mol AgNO 3 . Since mol of Ag we have exceeds that required, AgNO 3 is in excess and copper limits reaction and is totally consumed. 0.03619 moles of Ag + are consumed. n(Ag + ) remaining = n initial – n consumed = 0.05 – 0.03619 = 0.01381 mol c(Ag + ) in remaining solution = n/V = 0.01381/0.1 = 0.14M (2 sig figs) 24.a. Marking Guidelines Marks Identifies an appropriate instrument to detect the radiation. 1 Geiger tube 24.b. Marking Guidelines Marks Correctly determines the concentration of both species, with sufficient working. 2 Correctly determines the concentration of the nitrate ion AND Determines the moles of Ag + consumed in the reaction. 1 In this application, some, but not all of the radiation must be able to pass through the foil so that the levels passing through can be detected on the other side. This allows for the thickness to be monitored.

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