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1.2 TEST MARK SCHEME
1. (a) H+(aq) + OH-(aq) à H2O(l) 1 (b) Ba2+(aq) + SO42-(aq) à BaSO4(s) 2
[3]
2.
(a) simplete ratio of atoms of each element in a compound (1) 1
(b) (1) × 12 (1)
(or definition in terms of moles) 2
(c) CHO = 29 = 6 (1)
 C6 H6 O6 (1) 2
(d) CH4O+ O2  CO2+2H2O (1) 1
[6]

3.
(a) (i) (1) = 1.64(1) allow 1.63 to 1.64
PV = nRT (1)
V = (1) if no × 3 CE allow use of p = 100 if answer in dm3 (1) allow 0.162 to 0.166 allow conseq on moles CH2NO2 5
(ii) V = V1 × (1) allow conseq on vol of gas products in (i)
= 0.410 (m3) (1) allow 0.4 to 0.42 or (1) or (for 1st mark)
Thus if ans = ans to (i) × allow 2 marks ignore units allow conseq on moles CH3NO2 and on omission of 3 2

(b) (i) (or pV = nRT) (1) (1) (1) = 1.99 × 107 Pa (1) (allow 1.99 to 2.00) 4
(ii) 1.99 × 107 × = 2.49 × 107 Pa (1) (allow conseq) 1
(iii) = 9.0 mol dm–3 (or M) (1) (1) 2
[14]

4.
(a) % 0 = 100 – 25.9 = 74.1 (1) wrong % AE –1
N : O = (1)
= 1 : 2.5
 N2O5 (1)
NO3 gains first two marks probably 3
(b) N2O5  2NO2 + O2 (1) allow × 2 if (d) is NO3 allow NO3  NO2 + O2 or × 2 1
[4]

5.
(a) ratio NaCl : Na2CO3 = 2 : 1 (1) mass ratio NaCl : Na2CO3 = 117 : 106 = 1 : 106/117 (1)
546 kg NaCl produces 546 × 106/117 (1)
= 495 kg Na2CO3 (unit required) (1) or moles NaCl = 546000/58.5 = 9333 (if 9.33, allow 3 max) (1) moles NaHCO3 = moles NaCl = 9333 (1) moles Na2CO3 = ½ moles NaHCO3 = ½ × 9333 = 4667 (1) mass Na2CO3 = 4667 × 106 = 495 kg (unit required) (1) or mass ratio NaCl: NaHCO3 = 58.5 : 84 (1) mass NaHCO3 = 546 × 84/58.5 = 784 kg (1) mass ratio NaHCO3 : Na2CO3 = 168 : 106 mass Na2CO3 = 784 × 106/168 (1)
= 495 kg (unit required) (1) 4

(b) 0.537 g in 100 cm3 scaled to 5.37 g dm–3(1) 5.37/106 = 0.0507 mol dm–3 allow 2-4 sig. figs, condone truncation (1) 2

(c) moles CO2 = moles Na2CO3 = (1) allow 0.045 to 0.046 moles in 1000