SAMPLE CHEMISTRY QUESTIONS MIXTURE OF UNIT 3 & 4 MATERIALS O H

SAMPLE CHEMISTRY QUESTIONS
MIXTURE OF UNIT 3 & 4 MATERIALS
QUESTION 1
a.
(i)
NaOH ( aq )  HCl ( aq )  NaCl ( aq )  H 2 O( l )
n( NaOH )  n( HCl )  cV  1.00  0.0333  0.0333 mol
1 mark
(ii) CH 3 COOH ( aq )  NaOH ( aq )  CH 3COONa ( aq )  H 2O( l )
n(CH 3 COOH ) reacting  n( NaOH ) reacting
 n( NaOH ) initially added  n( NaOH ) in equation 3
 (1.00  0.050)  (0.0333)  0.0167 mol
2 marks
Catalyst
(iii) C 4 H 9 OH ( aq )  CH 3 COOH ( aq )  CH 3 COOC 4 H 9 ( aq )  H 2 O( l )
n(bu tan ol )  n(CH 3 COOH ) reacting in equation 1
 n(CH 3 COOH ) initially added  n(CH 3 COOH ) in equation 2
 0.05  0.0167  0.0333 mol
2 marks
(iv) Applying mole ratios:
Catalyst
C 4 H 9 OH ( aq )  CH 3 COOH ( aq )  CH 3 COOC 4 H 9 ( aq )  H 2 O( l )
1 mole
+ 1 mole
0.0333 mole + 0.0333 mole
 1 mol
 0.0333 mol
C 4 H 9 OH ( aq )
CH 3 COOH ( aq )
CH 3 COOC 4 H 9 ( aq )
Initial Mole
0.0500
0.0500
0
Final Mole
0.0500  0.0333  0.0167
0.0500  0.0333  0.0167
0  0.0333  0.0333
2 marks
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Unit 3 & 4 Chemistry - Mixed Questions
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(i)
b.
K
[CH 3 COOC 4 H 9 ( aq ) ][ H 2 O(l ) ]
[C 4 H 9 OH ( aq ) ][CH 3 COOH ( aq ) ]

[CH 3 COOC 4 H 9 ( aq ) ]
[C 4 H 9 OH ( aq ) ][CH 3 COOH ( aq ) ]
M 1
1 mark
(ii) As V  1 dm , C  n .
3
C 4 H 9 OH ( aq )
CH 3 COOH ( aq )
CH 3 COOC 4 H 9 ( aq )
Initial Mole
0.0500
0.0500
0
Final Mole
0.0500  0.0333  0.0167
0.0500  0.0333  0.0167
0  0.0333  0.0333
Eq Conc
0.0167 M
0.0167 M
0.0333 M
K
0.0333
 119.4019  119 M 1
2
(0.0167 )
1 mark
c.
(i)
Heating to 110 o C will evaporate water but not the organic molecules involved in
the reaction. The system will oppose this change by favouring the forward
reaction. Hence the yield of ester will increase.
2 marks
(ii) The catalyst used in this process is usually concentrated sulphuric acid as it is a
strong dehydrating agent (reaction requires the removal of water).
1 mark
(iii) Due to structural arrangement, esters cannot form hydrogen bonds with water which decreases their solubility.
Both alcohols and carboxylic acids display exposed polar covalent bonds that can
form hydrogen bonds with water, and hence are soluble to some extent in water.
2 marks
d.
The given molecules cannot be used to produce a polyester. Polyesters are produced
from the reaction of an alcohol that has two  OH functional groups with a carboxylic
acid that has two  COOH functional groups. The given structures do not possess the
functional groups for the polymerisation process to occur.
2 marks
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Unit 3 & 4 Chemistry - Mixed Questions
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QUESTION 2
(a)
(i)
(ii)
(b)
(i)
(ii)
3.125 =
3.125 =
[C6H6O6] = 0.315 M
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Unit 3 & 4 Chemistry - Mixed Questions
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(c)
(i)
(ii)
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QUESTION 3
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Unit 3 & 4 Chemistry - Mixed Questions
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QUESTION 4
a.
(i)
E = m × c × ∆T = 500 × 4.184 × 52.8 = 110457.6 J = 110 kJ
(ii) Energy is lost as heat to the surroundings.
(iii) mass(C2H5OH) = d × V = 0.789 × 5.00 = 3.945 g
n(C2H5OH) = mass/M = 3.945/46 = 0.08576 mol
0.08576 mol → 110 kJ then 1 mole → 110/0.08576 = 1282.6 kJ
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
b.
(i)
∆H = −1283 kJmol−1
Calibration factor = Energy/ ∆T = 110/52.8 = 2.08 kJ°C−1 = 2.08 kJ°K−1
(ii) Energy (kJ) = 505/15 × 2.50 = 84.2 kJ
∆T = Energy/CF = 84.2/2.08 = 40.5°C
c.
(i)
Monounsaturated.
(ii) Stearic acid is saturated. If it was the main component of olive oil then the
oil would be more solid.
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Unit 3 & 4 Chemistry - Mixed Questions
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QUESTION 5
a.
(i)
Hydrolysis reaction.
(ii) Fat formed from reaction of 3 fatty acid molecules and one glycerol
molecule.
(iii) 3 x CH3(CH2)14COOH + C3H8O3 - 3 (H2O)
Fat = 3 x (16C + 32H + 2O) + (3C + 8H + 3O) - 3 (2H + O)
(51C + 98H + 6O ) = M.F. = C51H98O6
b.
(i)
n (C6H12O6 ) = 0.20 / 180 = 0.0011  3114J
1n

x
x = 3114/ 0.0011 = 2831 kJ
H = - 2.83 x 10 3 kJmol -1
(ii) Oxidation in living cells occurs in series of steps that are catalysed by
enzymes. The reaction that occurs in the calorimeter is not catalysed by
enzymes and is spontaneous.
(Other possible answers ... air used in cells .. not pure O2)
(iii) To form glycogen .
(iv) The glucose molecule has hydroxyl (OH) functional groups present which are
polar. These OH groups hydrogen bond with water molecules through the
interaction:
Glucose - OH  + -------------  - OH2 *
where attraction of delta +ve H from -OH
constitutes a H bond.
to delta -ve O from H2O
(Or a diagram to show this interaction)
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Unit 3 & 4 Chemistry - Mixed Questions
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QUESTION 6
a.
(i)
(ii)
b.
(i)
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(ii)
c.
(i)
(ii)
(iii)
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Unit 3 & 4 Chemistry - Mixed Questions
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QUESTION 7
a.
(i)
(ii)
(iii)
(iv)
(v)
B, F and I
B and C
E
D and E
K and L
b.
Structure A
Structure B
Structure E
0.007 grams
4.1 grams
0.03 grams
Answer = 3  10 3 kJ
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