Chem_mock_ans

Chem_mock_ans - Chem Ans. 1a. i Relative isotopic masses™...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chem Ans. 1a. i Relative isotopic masses™ ¸ ª ¿Û 20 Y = 20 21 Y = 21 22 Y = 22 : (2) ii ‘ Relative atomic mass of Y is 20.86 ‘ means that the average mass of isotopes of atom Y based on 12 C scale is 20.86. (2) 20.86 h –ª –ª • ö ¿ YZ 4 – ª Y™ ¸ ¿Û ¾s ° Ü h -h 12 h iii. Let a% be the percentage of 22 Y, then a% , h ª ¿ ÛY ¸ ™ 20(2a ) + 21(100 − 3a) + 22(a ) = 20.86 100 ∴ a = 14 h • • 1b. : Isotopic mass v ì Relative atomic mass (2 ) (1) v¿ X t 12 – carbon scale h 12 – carbon scale h i. Ammonium sulphate can be manufactured by reacting ammonia with dilute sulphric acid. (1) E –™ªž ¾ x Ð }˜ 2 NH3 + H2SO4 (NH4)2SO4 (1) ii. Molecular mass of urea CON2H4 – ª ¿X Ûž = 12 + 16 + ( 14 x 2 ) + ( 1 x 4 ) = 60 g percentage by mass of N in urea (N = 14 x 2 x100% 60 ) (2) = 46.7% (1) Molecular mass of ammonium sulphate (NH4)2SO4 – ª¿4 Ûž = (14 x 4) x 2 + 32 + ( 16 x 4 ) = 132 g percentage by mass of N in ammonium sulphate (N ) íÈ = 14 x 2 x100% 132 = 21.2% (2) (1) iii. Nitrogen is used to manufacture protein which is important for plant growth. (1) iv. Two other factors: solubility and acidity of soil (2) €è E žš h • : Percentage by mass of nitrogen Z 4 –ª s¾ • –ö ª ° ¿ Ü percentage by mass of nitrogen 4 –ª s¾ • –ö ª ° ¿ Ü Z Ež š E žz › by mass of nitrogen percentage kF p •Îª ¬ kF – Õ¿ n E•ª ™ kξ – пF ž 2a. i. X E –ª ™ k(x ž пF ( h NO3- h ) ) (1) ii Fat and sodium hydroxide (1) –ª ¿ Û s š iii. 1. Very little foam formation and no scum or precipitate is obtained. (1) E ž –™ªk x Ð ¿F 2. Scum ( precipitate ) is formed. (1) ª ¿ Ûs š iv. (1) v – COO- (1) vi. Z (1) The highly branched hydrocarbon chain of Z makes it to be non-biodegradable. (1) h –ª™ k x E ž ¿Z ÐF vii The detergents emulsifies the oil ( dissolves the oil , or breaks down the oil to small droplets) which enables it to be dispersed or carried away water current. ( 2 ) šžEz viii. Detergent may be toxic and would kill the fish in the fish farms. (1) –ª ¾ s• –öª ¿ ° Ü 4 Z h • : Soapless detergent x –¿ Û • ª soapy detergent• ˜ –ª ¿ Û Soapy detergent ª¿Û•˜ : Soapless detergent –¿ Û • ˜ ª h h -COO- h h h fat h NaOH h h h petroleum h –ª ¿ Û • ˜ k˜ F –ª ¿ Û • ˜ h ª¿Û • ˜ h h E ž –Tñ¿Ð È ™ª 2b. i. Polythene or polyethene h HH H (1) H COO- ii. n C=C HH C H C n H (1) iii. 1. Thermoplastic (1) 2. Put the polymer in a beaker of boiling water or in an oven for some time. If the polymer is softened or melted and re-hardened by cooling , it is thermoplastic. (2) E ž ›z e Pz v h iv. h Plastic bags / squeeze bottle / plastic toy ( any two ) ( 2 ) /h /˜ h ™ : • – ¾ (•ZÜö¿° ª –ª • E –∪ ž b¿V ™ª Ð • E žP v z ›z 3. a) Gas syringe h ( 1 mark ) b) Zn(s) + 2HCl ( aq )→ ZnCl2 (aq ) + H2 (g ) ( 1 mark ) (C=C) c) Volume of hydrogen = 120 cm3 = 120 cm3 h No of moles of hydrogen produced = = ª¿Û› 120 = 0.005 24000 120 = 0.005 24000 ( 1 mark ) No of moles of HCl used = 2 x0.005 = 0.01 HCl› ˜ = 2 x0.005 = 0.01 ¿Û ( 1 mark ) Concentration of HCl = ¿Û›˜ = 1000 x0.01 = 0.2 mole dm-3 50 ( 1 mark ) 1000 x0.01 = 0.2 mole dm-3 50 d) The rate is slower as ethanoic acid is a weak acid and the hydrogen ion in the ethanoic acid is much smaller than that hydrochloric acid for the same concentration. ( 2 marks ) Ežš z Volume of carbon dioxide ( cm3¿ )• ¨ –Û ª ( cm3 ) 120 zinc + ethanoic acid h +h time ( min ) (h ) h ( 1 mark ) ª¿Û•¨ 1. E žz › e Pz v 2. • ¸Û¿ª– x (a) x (b) x (mole) = (c) x 3. (mole) = (mole) = x ìb+ a s x x È šÛ ¿ ª * (volume) (molarvolume) (molarity ) × x (volume)(cm3 ) 1000 Åá 4. a) mass of carbon = 3.52 × ¿Û•¸ (mass ) (molarmass ) = 3.52 × 12 12 = 3.52 × = 0.96 g 12 + 2(16) 44 12 12 = 3.52 × = 0.96 g 12 + 2(16) 44 ( 1 mark ) mass of hydrogen = 1.12 − 0.96 = 0.16 g = 1.12 − 0.96 = 0.16 g h ( 1 mark ) b) mass ratio h C:H= 0.96 0.16 : 12 1 ( 1 mark ) = 0.08 : 0.16 =1:2 Empirical formula is CH2 CH 2 h ( 1 mark ) c) Let the molecular formula of X be CnH2n CnH2n h ª ¿ ìX ø € 1 CnH2n + 1 n O2 → n CO2 + n H2O 2 volume of O2 ( O2 h ( 1 mark ) ) : volume of CnH2n ( CnH2n h ) 1 = 1 n:1 2 = 6:1 ∴ 6 n= 1 =4 1 2 ( 1 mark ) Therefore , the molecular formula is C4H8 C4H8 h d) CH3CH=CHCH3 and CH3CH2CH=CH2 ª¿ ìs • ( 2 marks ) .1 .2 h –‚ 쿪X Ch HX ‚쿪 (empirical formula)h C hª ¿ H X ì‚ ...
View Full Document

This note was uploaded on 09/02/2011 for the course CHINESE 12 taught by Professor Cn during the Spring '11 term at CNUAS.

Ask a homework question - tutors are online