(Go here instead!
http://vcenotes.com/forum/index.php/topic,7618.0.html)
Encased my solutions below anyway
==========================================================================================================
Ignore red font, these answers are incorrect according to above.Possibly correct...EXTENDED RESPONSE
1)a) Mg + 2HCl -> MgCl2 + H2
b) Powdered => greater surface area => more frequent collisions => increased rate of reaction
Steeper graph attaining maximum V(H2) over less time
2)a) Calibration Factor = 1.87 kJ/°C
b) Final Temperature = 69.8 °C
3)a) Acid II, because Ka is proportional to 10^(-pH); substituting values shows that II has the lowest [H+]
b) Acid III, since pH < 1;
Quote from Pandemonium: [
http://vcenotes.com/forum/index.php/topic,7616.msg94599.html#msg94599 ]
We know that the [acid] was all 0.1M
therefore, assuming complete ionisation of the acid, we'd have [H+] = 0.1M
pH would therefore be equal to 1 at the minimum if it were monoprotic.
acid III i think had pH 0.7, therefore it had to be at least diprotic.
c) Percentage Ionisation = 7.9%
d) Ratio of [OH-] values = 100
e) Same change in pH, +1 (tick the third box)
Since dilution by a factor of 10 <=> divide [H+] by 10,
new pH = -log10{[H+]/10} = old pH - log10(.1) = old pH 1 + 1 => same change in pH for both acids (+1)f)i) [HCOOH] = 7.9 x 10^-3 Mf)ii) Exothermic, heated => K decreases => [H+] decreases => pH increases
4)a) Kc = 0.48 M
We are told 0.45mol N2O4 is initially placed in the container, and 0.36mol NO2 forms;
Thus due to the 1:2 stoichiometry, 0.18mol N2O4 reacts to form 0.36mol NO2; and 0.27mol N2O4 remains.
Thus Kc = (0.36)^2 / (0.27) = 0.48 M
b) Endothermic reaction, since the lower temperature results in a lower Kc value.
5)
a) Sulfuric Acid: circle O2 [I would like to make this blue...but I suspect VCAA do want more than one reactant circled...]
b) 2SO2 + O2 <-> 2SO3
c) In the converter, heated gas is recycled between runs through catalyst beds, hence reducing heating (energy) costs
d)i) Superphosphate [fertiliser]
d)ii) Did not expect this to be assumed knowledge...alternately you can write the equation for production of ammonium sulfate, also a fertiliser.
Superphosphate: Ca3(PO4)2 + 2H2SO4 + 4H2O -> Ca(H2PO4)2 + 2CaSO4.2H2O ... lol
( I was stymied and wrote the equation for dehydration of glucose
)
6)a)i)2CH3OH + 3O2 -> 2CO2 + 4H2O
a)ii) deltaH = -1450 kJ/mol
b)i) 2.20 x 10^6 C
b)ii) 122 g (3 s.f.)
b)iii) Gaseous CO2 escapes into the air thereby lowering expected yield of methanol
c) Overall neutrality - CO2 is produced by methanol combustion but consumed by electrolysis
7)a)i) electron flow direction: <----
a)ii) Cd -> Cd2+ + 2e-
a)iii) Inert / does not form precipitate with electrolytes
b)i) - (upper circle), + (lower circle)
b)ii) Products of discharge reaction remain in contact with electrodes in a convertible form.
b)iii) Anode equation [discharge]: Cd + 2OH- -> Cd(OH)2 + 2e-
b)iv) Cathode equation [recharge]: Cd(OH)2 + 2e- -> Cd + 2OH-
a)i) H2 + 2OH- -> 2H2O + 2e- I can't believe I did that...
a)ii) O2 + 4H+ + 4e- -> 2H2O
b) H2PO4- flow direction: <-------
c)i) 1.8 x 10^2 kJ
c)ii) Efficiency = 62%
d) Advantage: Fuel cells are more efficient due to their facility of directly converting chemical energy into electrical energy.
Disadvantage: Because fuel cell technology is still developing, they are more expensive than internal combustion engines.
9)a) m(CO2) = 8.24 x 10^15 kg
b) Energy produced = 3.69 x 10^19 kJ
^ [3 significant figures instead of what it should be (2, from 0.42% etc.), but VCAA allows one decimal place either way]
==========================================================================================================
Home
Help
Search
Login
