SRJC H2 Chem 2013 Prelim P1 Soln

SRJC A 1 Th W wa A B C D Answer One 1o Four 2o 5 mol o 2 Cl ex pr de W Answer Numbe Numbe Answer he diagram When glucos ater are form 4 5 6 7 r: B alcohol con alcohols co of H2O forme luster dec a xperimental rotons that ecay of radi Which row in A B C D r: B r of protons r of neutron below show se reacts w med? Assu nverted to c onverted to ed in total. ay is on e discovery, are heavier um-223. the table c Number of 74 82 74 82 s in A = 88 ns in 223-ra ws the struc ith hot aci d me the cyc carboxylic a ketones: R e rare ins t where a pa r than an  - correctly des protons – 6 = 82 dium = 223 1 ctural formu dified potass lic ring rem cid: RCH2O RR’CHOH + tance of arent atomi -particle. O Ra଼଼ ଶଶଷ  ଵ scribes the Numb 3 – 88 = 135 ula of glucos sium dichro ains intact a OH + 2[O]  [O]  RR’C scientific p ic nucleus e ne of the fi C଺ ଵସ +A nuclear ma ber of neutro 119 127 209 217 5 se. omate(VI), after the rea  RCOOH + C=O + H2O phenomena emits a clus rst predictio ake-up of ele ons Tur how many action. + H2O O a predicte d ster of neut ons was the ement A ? rn Over] moles of d before trons and e nuclear 1222

2   Turn Over] Number of neutrons in 14-carbon = 14 – 6 = 8 Number of neutrons in X = 135 – 8 = 127 3 A given mass of ideal gas occupies a volume V and exerts a pressure p at 30 oC. At which temperature will the same mass of the ideal gas occupy a volume ௏ ଷ and exert a pressure 2p? A 20 oC B 20 K C 202 oC D 202 K Answer: D pV = nRT (2p)( ௏ ଷ) = nR(xT) ଶ ଷ(pV) = x(nRT) x = ଶ ଷ Original T = 30 + 273 = 303 K New T = ଶ ଷ × 303 = 202 K 4 The boiling point of water (100 oC) is greater than that of ammonia (–33 oC). Which statement is a correct explanation of this? A Ammonia has intramolecular hydrogen bonds, which water does not have. B The Mr of water is greater than that in ammonia, so van der Waals’ forces are stronger in water. C There are, on average, more hydrogen bonds between water molecules than there are between ammonia molecules. D The O–H bond requires 460 kJ mol –1 to overcome, while the N–H bond only requires 390 kJ mol–1 to overcome. Answer: C 5 The standard enthalpy change of formation of hydrazine, N2H4(g), is x kJ mol–1. The bond energy of the NN bond is y kJ mol–1. The bond energy of the H–H bond is z kJ mol–1. 1223

3   Turn Over] What is the standard enthalpy change of atomisation of hydrazine? A (x + y + 2z) kJ mol–1 B (y + 2z – x) kJ mol–1 C (x + 2y + 4z) kJ mol–1 D (2y + 4z – x) kJ mol–1 Answer: B By Hess’ Law, Ha = –Hf + BE(NN) + 2BE(H–H) = (y + 2z – x) kJ mol–1 6 The Gsolution Ɵ and Ssolution Ɵ for silver chloride are +55.6 kJ mol –1 and +33.2 J mol –1 K–1 respectively. What is the enthalpy change when 287 g of silver chloride is precipitated under the same conditions? A +65.5 kJ