2010 HSC Chemistry Part B
Needs to be completed and checked.
Question 21 (3 marks)
Hydrochloric acid is a strong acid and completely dissociates in water.
HCl(aq) → H+(aq) + Cl-(aq) 100% dissociation
0.001 mol L-1 of HCl produces 0.001 mol L-1 of H+ ions which has a pH of 3.
Ethanoic acid acid is a weak acid and only partially dissociates in water.
CH3COOH(aq) ↔ CH3COO-(aq) + H+(aq) 1.8% dissociation
0.056 mol L-1 of CH3COOH only produces 0.001 mol L-1 of H+ ions which has a pH of 3.
Since both solution produce equal concentrations of H+ ions they produce the same pH.
Question 22 (6 marks)
(a) Short chained esters like methyl propanoate are commonly used as solvents.
(b) Refluxing raises the reaction temperature increasing the rate of reactions and prevents the loss of volatile materials.
(c) Two reactants used to prepare methyl propanoate are methanol and propanoic acid.
Drawings needed
Question 23 (3 marks)
(a) CH3(CH2)3OH(l) + 6O2(g) → 4CO2(g) + 5H2O(l)
(b) The molecular formula for butan-1-ol (1-butanol old terminology) is C4H10O.
Butan-1-ol has a molar mass of 74.12 g mol-1
ΔH kJ g-1 = ΔH kJ mol-1 / Molar mass g mol-1
ΔH kJ g-1 = 2676 / 74.12
ΔHcombustion = 36.1 kJ g-1
The heat of combustion cannot be greater than the theoretical value so Fuel A and B must be ruled out.
Assuming heat loss and incomplete combustion from the experiment, Fuel C is most likely to be butano-1-ol.
Question 24 (4 marks)
(a) CH2CH2 + H2 → CH3CH3
The type of reaction is an addition reaction. H2 is added across the double bond.
(b) Bubble gas through a solution of aqueous bromine water in the absence of UV light. Alkenes rapidly decolourlise bromine water and alkanes don't.
If the water does not change from orange/ brown to colourless then all the ethene has been converted to ethane.
Question 25 (5 marks)
Dissolved oxygen (DO) is a measured of the dissolved oxygen in a sample of water. Biochemical oxygen demand (BOD5) measures the difference between the initial dissolved oxygen content and the dissolved oxygen that is left after being placed 5 days in the dark at 20°C
A high DO indicates that the natural water is 'healthy' and is able to support a wide range of active organisms. A low DO indicates stagnant like water which would support a smaller variety of organisms.
A high BOD5 indicates that the natural water contains a high level of organic pollution and is unhealthy as the oxygen is used up by microorganisms as they digest the oragnic matter. A low BOD5 indicates that the water is fresher and contains very little organic matter.
It is important to meaure both DO and BOD5 in water as they both measure different things. DO measures the of the initial oxygen content of the water and BOD5 measures the amount of 'organic pollution' in the water.
Question 26 (4 marks)
I would have to say that the markers would accept both reactions as the equation 2 is outside the current syllabus.
Step 1: Write an equation for the reaction.
Equation 1: Zn(s) + 2 HNO3(aq) → Zn(NO3)2(aq) + H2(g)
Technically incorrect, as dilute nitric acid is still a strong oxidising agent and slowly reacts with zinc to produce colourless nitrigen monoxide according to the second equation.
Equation 2: 3 Zn(s) + 8 HNO3(aq) → 3 Zn(NO3)2(aq) + 2 NO(g) + 4 H2O(l)
Step 2: Determine the number of moles of zinc.
nZn = mZn / MZn
nZn = 10.0 g /65.41
nZn = 0.15288.. mol (leave in calculator memory)
Step 3: Determine the moles of acid.
n(HNO3) = c.v
n(HNO3) = 0.50 x 0.20 mol
n(HNO3) = 0.10 mol
Step 4: Determine the limiting reagent.
Using equation one as most students would,
1 mole Zn reacts with 2 moles of HNO3
Therefore, 0.05 moles Zn reacts with 0.10 moles of HNO3
There is excess zinc. Nitric acid is the limiting reagent.
Step 5: Determine the moles of gas produced
From equation one, 0.10 moles of HNO3 produces 0.05 moles of H2(g)
Questions 27 to 31 should be completed before 30th March 2011