NYJC Prelim CHEM P3 ANS

NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CHEMISTRY 9746/03 Paper 3 16 September 2009 2 hours Candidates answer on writing paper Additional Materials: Data Booklet Graph paper READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams, graphs or rough working. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer 4 out of 5 questions. Write your answers on the writing paper provided Begin each question on a fresh piece of paper. You are advised to show all working and calculations This document consists of 9 printed pages and 1 blank page

2 H2 Chemistry 9746/03/NYJC J2/09PX 1(a) (i) Element A The sharp drop in value from G to H i ndicates that G is from Group I where the second ionization energy involves t he removal of an electron from the inner shell . Hence Element A is in Group III and is Aluminium. [2] (b) (i) AB(CO3)2 → AO + BO + 2 CO2 [1] (ii) Assume : AB(CO3)2 → AO (s) + BO (aq) + 2 CO2 Masses given : 0.400 0.057 (0.275 – 0.057) x Therefore, x = 0.400 – 0.275 = 0.125 g [1] (iii) Soln : Let Ar of A be a and that of B be b AB(CO3)2 → AO (s) + BO (aq) + 2 CO2 Mass / g : 0.400 0.057 0.218 0.125 Determining amts and using mol ratios Amt of CO2 = 0.125/44 = 0.00284 mol Amt of AO = 0.057/(a + 16) = 0.00284 /2 mol Giving a = 24.1, Therefore A is Mg Amt of BO = 0.218/(b + 16) = 0.00284 /2 mol Giving b = 137.5, Therefore B is Ba (c) MgO > NaF > H2O > CH3NH2 > CO2  CO 2 have simple molecular structure consisting of CO2 molecules held together by van der waals forces.  CH 3NH2 and H2O have simple molecular structures consisting of CH3NH2 and H2O molecules held together by stronger hydrogen bonds.  H 2O has 2 lone pair hydrogen units while CH3NH2 has 1 thus H2O has higher boiling point.  NaF and MgO have ionic structures consis ting of oppositely charged ions held together by strong electrostatic forces. MgO has higher boiling point because it has higher ionic charges than NaF. [5]

3 H2 Chemistry 9746/03/NYJC J2/09PX (d) (i) For the amino acid resi due to be found on the outer surface of a water soluble globular protein, the R group must be a hydrophilic group.  Hence, Serine (ser) will be found on the outer surface The OH group on serine can form hydrogen bonds with water hence soluble.  Lysine (Lys) will also be found on the outer surface The NH2 group on lysine can form hydrogen bonds with water hence making it soluble. [2] (ii) For Phe, Pro, Leu : Hydrophobic or van der Waals forces For Lys, Ser: hydrogen bonding [1] (iii) Denaturation of proteins. When heated, the weaker R group interactions such as hydrogen bond and hydrophobic interactions stabilising the tertiary structure will be disrupted. Hence the polypeptide chain will uncoil itself and lose its shape. However, the primary structure will still be intact. [3] (iv) for Lys (the zwitterions for any other amino acid can also be drawn) [1] (v) [2] When a small amount of acid (H +) is added, CH C O O- H3N + (CH2)4 NH2 H+ + CH C O OH H3N + (CH2)4 NH2 When a small amount of base (OH) is added, CH C O O- H3N + (CH2)4 NH2 OH- + + H2O CH C O O - H2N (CH2)4 NH2

4 H2 Chemistry 9746/03/NYJC J2/09PX 2(a) (i) Energy kJ mol-1 0 2Fe(s) + 3/2 O2(g) Fe2O3 (s) Hf [Fe2O3 (s)] 2Fe(g) + 3/2 O2(g) Hatm[Fe (s)]2x 2x (1st IE + 2nd IE + 3rd IE) 2Fe3+(g) + 3/2 O2(g) + 6e 3/2 x BE (O=O) 2Fe3+(g) + 3O(g) + 6e 2Fe3+(g) + 3O-(g) + 3e 3x EA1[O] 2Fe3+(g) + 3O2-(g) L.E. 3C (s) + 2Fe2O3 (s) By Hess’ Law, ∆H f = 2(414) + 2(762 + 1560 + 2960) + 3/2 (496) + 3 (-141) + 3 (844) + L.E. L.E. = - 1.51 x 104 kJ mol-1 [4] (ii) O -(g) + e → O2-(g) 2nd E.A. involves an electron added to negatively charged O-. Energy needed to overcome repulsion between two ne gatively charged species, therefore ∆H is positive. [1] (iii) 4Fe (s) + 3 CO2 (g) 3CO2 (g) + 2Fe2O3 (s) Hr + 3O2 + 3O23(-394) (-3288/2) Hr = 3(-394) - (-3288/2) = +462 kJ mol-1 By Hess' Law [4] (b) (i) CH2CHO CONH2 A & B [2]

5 H2 Chemistry 9746/03/NYJC J2/09PX reagents and conditions: A dd Fehling’s reagent to A and B seperately, and warm. Observations: For A, reddish brown prec ipitate formed. For B, no precipitate formed. Other possible answers: (Write their observations) Tollen’s reagent, warm K 2Cr2O7 with dil H2SO4 and heat 2,4-DNPH, warm (ii) O O C H3 C H3 O O CH3 CH3 X & Y [2] reagents and conditions: Step 1: Heat both X and Y separately with dil H2SO4 . Step 2: To the resulting mixture from X and Y separately, add K2Cr2O7 with dil H2SO4 and heat. Observations: For X, orange K 2Cr2O7 turns green. For Y, orange K 2Cr2O7 remains. Other possible answers: KMnO4 with dil H2SO4 and heat Or just use the oxidizing agent, with dil H 2SO4, heat, since the acid present will hydrolyse the ester then immediately followed by oxidation of the alcohol (c) (i) N N CH3 .. : The nitrogen atom (circled) is attached to the three alkyl groups which are electro-donating, making the lone pair more available for bonding to H +/ hence stabilizing the conjugate acid by dispersing the positive charge on the nitrogen atom. [3] (ii) N N CH3 .. : H O H .. .. H O H .... ||||||||||||||| |||||||||||||||         [2]

6 H2 Chemistry 9746/03/NYJC J2/09PX (d)(i) CH 3CH(NH2)CH2CONH2 [1] (ii) Cl - NH3 +CH2CH2CH2COOH [1] 3(a) (i) N 2O (g)  N2 (g) + ½ O2 (g) Initial P/kPa 25.0 0 0 Change /kPa -x +x + ½ x Final P / kPa 25.0-x x ½ x PT = 25.0 - x + x + ½ x = 25.0 + ½ x x = 2(PT - 25.0) ( where x = amt of N2O decomposed at any given time ) PN2O = 25.0 - x = 75.0 - 2PT At 230 s, PT = 30.0, PN2O = 15.0 kPa At 590 s, PT = 34.0, PN2O = 7.0 kPa (ii) [4] (iii) At any time , PN2 = x = 2(PT - 25.0) = 25.0 - PN2O 1m [3]

7 H2 Chemistry 9746/03/NYJC J2/09PX No. of half-lives 0 1 2 PN2O / kPa 25.0 12.5 6.25 PN2 / kPa 0 12.5 18.75 setting up table (such as the one above) and some attempt to complete it… Ans : 2 half lives (b) (i) e- config of cation of Y : 1s2 2s2 2p6 3s2 3p6 3d9 Identity of X : NH3 or ammonia [2] (ii) [4] ( Y represents Cu) Cu(NO 3)2 + 6 H2O  [Cu(H2O)6] 2+ + 2 NO3 – Blue [Cu(H2O)6] 2+ + 2 OH –  Cu(OH)2(s) + 6 H2O light blue ppt. [Cu(H2O)6] 2+ + 4 NH3  [Cu(NH3)4(H2O)2] 2+ + 4 H2O excess dark blue filtrate Mg(H2O)6 2+ + 2 OH -  Mg(OH)2 + 6H2O white residue, insoluble in excess NH 3 (c) (i) C C C C OH H H H H H N H [1] (ii) [3]

8 H2 Chemistry 9746/03/NYJC J2/09PX C CH3 CH3 OHNC __ H+ (aq), 2H2O reflux C CH3 CH3 OHHOOC C C H2 CH3 COOH con