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what you said first is correct

it doesn't matter about the feasability of it. that's what they said happened. and if you calculate it correctly it is only about 3.something grams.. i hope you converted to KJ/mol not J/mol

12 or 13/15 stupid, stupid mistakes.

+1 , Fe3+ has a higher reduction potential than Fe2+ so just use that.

no, you're not meant to round until you get to your final answer, where you put it in the correct amount of sig figs. perhaps, as it said 50% is lost to the environment. keep reading the question buddy, don't cut corners :P

Cathode's are electrodes, as such they have to be connected to something... ions are not connected to anything. perhaps, but it was not electrolysis, regardless. There was no external power source.

you should never ever round. you probably won't get full marks. and also, it may have been 2s.f. not 3 (i dont remember though)

i don't know how you did it... but, /\H = -mc/\t where m was 210, c is 4.18, t was 65 /\H = 57.057 KJ/g turn this into /mol 57.057 x mm(ethanol) / m(ethanol) = 1376/2 (50% heat lost to environment) rearrange to find m(ethanol) m(ethanol) = 2624.622/688 =3.8 grams (2 sf)

silver ions have a higher reduction potential than water

You will get 0 for part a, unless it was out of 2 , you may get 1 any other parts (if correct) will get 100% so yes carry on errors apply

lol, Oxidation occurs at the Anode. platinum is inert meaning it doesn't oxidise or reduce. look at the standard potentials table, Pb -> Pb2+ + 2e- oxidation is loss. therefore the anode was lead.

i got 15 B but i dont know why, 1.8% nitrogen because it was 2s.f. i think... and if not i got like 1.77%. your ethanol is incorrect though, sorry dude

For nitrogen i think i got 1.77% as well. does anyone remember amount of significant figures though? and that was really quite easy, i'm thinking around 85+ for band 6

Man i pumped this section. i loved the 7 marker on the scientists haha. although i agree, a part II was dodgy for me, but the rest of it was totally fine.

lol oh well right? um ammonia bread wasnt too bad when i got my head around it ended up getting about 1.2% nitrogen.

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

That was a rather simple exam compared to past years, i loved the shipwrecks option. pumped that booklet front to back. however yes, that 6 mark equilibrium question was tricky. i think i said , incorrectly, what the last change was, but oh well. I'd say at least 80% raw

After that exam, i'd say you're about right anna band 6, 85-90% raw.

You speak the truth. What i meant to say is that everything is being based off of assumptions. If 50/84 is enough for an E4, Then 60/84 will definitely be an E4, Else It will probably be an E4 if you get 60, but not if you get 50. Is that a little better? :)

My math teacher, HSC marker, says that on average you need to get around 50/84 to get an E4, if you get 60 it's probably going to definitely be an E4