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How did you go? (2 Viewers)

P!xel

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That's right

carbon monoxide was removed from the environment , causing the sharp drop in CO making the equilibrium shift further to the right
Yes, that's right.

And the third change in graph involved removal of both products and reactants (mostly products), causing the equilibrium to shift to the right, so reactants dropped and products increased.
 

jeshkaaa

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I said something similar. Does anyone remember the actual question for it? I can't seem to remember ><"
 

The Nomad

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GUYS! I have a good feeling 90% of state wrote C for the last MC question but i know taht cant be write. DId anyone get A or andd B and can explain how?
I got A.

Concentration of oxygen was 8mg/L, but there was a 10L sample, so 80mg/10L.

n = mass/molar mass = volume/molar volume.
Volume = molar volume x mass/molar mass = 24.79 x 0.08/32 = 0.062L = 62 mL.

Answer is A. I hope.
 

akira-sb

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GUYS! I have a good feeling 90% of state wrote C for the last MC question but i know taht cant be write. DId anyone get A or andd B and can explain how?
I got A .. cause
In 10L:
Concentration 02 = 10 x 8
= 80mg
80/1000 = 0.08g
moles of O2 = 0.08/32 (molar mass of O2)
=2.5 x 10^-3
Litres = (2.5 x 10^-3) X 24.79
= 0.061975L
Therefore 61.975ml
Thus A for most correct.
 

akira-sb

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Yes, that's right.

And the third change in graph involved removal of both products and reactants (mostly products), causing the equilibrium to shift to the right, so reactants dropped and products increased.
The third one I think was a decrease in pressure, which would account for the decrease in ALL 3 of the gases. Then due to Le Chatliers, it shifted to the right i think to counter the disturbance
 

eriito

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I got A .. cause
In 10L:
Concentration 02 = 10 x 8
= 80mg
80/1000 = 0.08g
moles of O2 = 0.08/32 (molar mass of O2)
=2.5 x 10^-3
Litres = (2.5 x 10^-3) X 24.79
= 0.061975L
Therefore 61.975ml
Thus A for most correct.
:party:
 

norelle

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Apr 30, 2009
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hey guys
I found this paper fairly straight forward
UNTIL
I found out that I did a mistake on the last question
the Fe2+, Fe3+ and Pb galvanic cell question,,
I put Pb2+ reduced to Pb, and Fe oxidised to Fe2+
Which are WRONG.

there are 3 marks for writing half-equations
1mark for calculating potential electrode
3marks for filling the blanks of the cell diagram

so if I did the first question wrong.
WHOLE thing will be marked wrong also??
7marks deducted straight away from me?
or will i still get marks for error-carrying??
gosh im so upset now :(
 

Tumnus

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I thought it wasn't tooo bad, Industrial was great! though there were some difficult core ones, e.g bread nitrogen and that galvanic cell was confusing. Should have spent more time on the graph thing since i didn't have the time to finish it...wasted it trying to figure the other ones out :( stupid time management! lost me marks again....:mad1:
 

cely

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yes i agree i struggled with time to i missed out a three mark question in the shipwrecks which i didnt see and then saw it at the last minute :( and i knew exactly how to answer it. i didnt really get the titration with the two acids? the ph of one went up and leveled out again and then went up? anyone no why? was it a buffer?
 

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