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Unformatted text preview: on the right. Due to Le
Chatelier's principle an increase in pressure will shift the equlibrium to the right to minimise the
effect of the pressure increase. This decreases the concentration of H<sub>2</sub> and
N<sub>2</sub> and increases the concentration of NH<sub>3</sub>. The system then reaches
equlibrium and the concentrations will not change. _____________________________________________________________________________________________________
HSC CHEMISTRY PAST PAPER SOLUTIONS – ANDREW HARVEY 12 2000 HSC ––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––'', ''z''. Only one of these
options is on the list of choices so it must be ''x'', Pb, ''y'', ''z''.
Question 15:
(Here is how I would solve this question (there are probably better methods)):
Looking at Diagram A we can see that it is a dry cell. On a standard battery we know that the end
with the part raised is positive and the flat part is negative. So 3 must be negative terminal. So the
answer is either A or B. Now we know that electricity flows from cathode to anode, positive to
negative, therefore 1 must be the cathode. Hence the answer is B. SECTION I  Part B:
Question 16:
(a)
_____________________________________________________________________________________________________
HSC CHEMISTRY PAST PAPER SOLUTIONS – ANDREW HARVEY 16 2000 HSC –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
A mass of solid sodium hydrogen carbonate must be accurately weighted. This solid sodium
hydrogen carbonate must be transferred into a volumetric flask, which is then filled with water to the
calibration line. ''The moles of solid sodium hydrogen carbonate can be calculated (mass / molar
mass), and the volume of solution is known from the volumetric flask used. So concentration can be
calculated (concentration = number of moles of sodium hydrogen carbonate / total volume). As the
concentration is known accurately it is a standard solution.'' (The italics may not be required as it is
not part of outlining the procedure.)
(b)
<math>c= \frac {n}{v}</math>
<math>0.12 = \frac {n}{250 \times 10^{3}}</math>
''n'' = 0.03 moles
<math>n = \frac {m}{MM}</math>
<math>m = 0.03 \times \left ( 22.99 + 1.008 + 12.01 + 16.00 \times 3 \right )</math> (NB: this
assumes the equation is NaHCO<sub>3</sub>, which I am not sure of. Please check it.)
''m'' = 2.52 g
Question 17:
(a) The left one is "vinyl chloride" and the right is "styrene".
(b)
Polyvinylchloride (PVC) (made from the viny...
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This document was uploaded on 09/19/2013.
 Fall '13
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