Post by: Mao on November 11, 2010, 01:16:04 pm
Big thanks to my high school chem teacher. And anyone who help me find wrong answers :P
Ammonia
a)i) 3H2 + N2 <--> 2NH3
ii) porous iron catalyst (Fe2O3)
iii) provides an alternate pathway with lower activation energy for the reaction to take place,
b)i)highly corrosive, toxic gas
ii) burns skin, lungs, etc...
iii) wearing protective equipment, ventilation
Post by: 10weid on November 11, 2010, 03:22:23 pm
Post by: minilunchbox on November 11, 2010, 03:24:13 pm
Post by: 98.40_for_sure on November 11, 2010, 03:24:41 pm
Oh god, to download or not to download.
DO IT. you wouldn't do it...
Post by: 10weid on November 11, 2010, 03:25:22 pm
Post by: z1m2v3 on November 11, 2010, 03:25:43 pm
Post by: Studyinghard on November 11, 2010, 03:26:04 pm
Post by: 10weid on November 11, 2010, 03:26:51 pm
Post by: 10weid on November 11, 2010, 03:29:01 pm
Post by: VeryCrazyEdu. on November 11, 2010, 03:29:28 pm
Post by: Souljette_93 on November 11, 2010, 03:29:36 pm
Also for the chosen chemical only sulfuric acid is given :( , wouldn't you know of Ammonia as well?
Post by: mustafa on November 11, 2010, 03:30:19 pm
also pretty sure 8 is A, dilution decreases ocncentration of all ions - just because hydronium ion conc decresases doesnt necessarily mean OH- increases...
dilution increases OH concentration as there is a higher concentration of OH in water than in acid. Therefore, OH increases. BTW, unless temperature is changed, ANY change in one ion concentration is iversely proportional to the change in the other ion concentration in order to satisfy the [OH][H3O]=10^-14 rule.
Post by: ckg93 on November 11, 2010, 03:30:45 pm
The oxygen's in D don't balance.
Post by: Souljette_93 on November 11, 2010, 03:30:56 pm
also pretty sure 8 is A, dilution decreases ocncentration of all ions - just because hydronium ion conc decresases doesnt necessarily mean OH- increases...
Why not? I always thought if one increases, the other decreases.
Post by: z1m2v3 on November 11, 2010, 03:31:22 pm
D is not balanced (there are 2 oxygen molecules on the left and 3 on the right)
Post by: 10weid on November 11, 2010, 03:31:59 pm
also pretty sure 8 is A, dilution decreases ocncentration of all ions - just because hydronium ion conc decresases doesnt necessarily mean OH- increases...
dilution increases OH concentration as there is a higher concentration of OH in water than in acid. Therefore, OH increases. BTW, unless temperature is changed, ANY change in one ion concentration is iversely proportional to the change in the other ion concentration in order to satisfy the [OH][H3O]=10^-14 rule.
but this is a solution of HCl, npt pure water, thus the Kw rule does not necessarily apply - i think... wtev, i know that if a base is diluted, OH- decreases, and if an acid is diltued, H+ is decreased.. so ill stick with my answer..
Post by: mustafa on November 11, 2010, 03:32:34 pm
And seriously there is no need to correct him by stating H2's enthalpy as 268 instead of 289 or watever. Concept is the same
Post by: mirra250 on November 11, 2010, 03:33:07 pm
Post by: 98.40_for_sure on November 11, 2010, 03:33:31 pm
also pretty sure 8 is A, dilution decreases ocncentration of all ions - just because hydronium ion conc decresases doesnt necessarily mean OH- increases...
Why not? I always thought if one increases, the other decreases.
Dilution decreases ALL concs... nuff said? you can't dilute parts of your solution. why is there even debate about this
Post by: Mao on November 11, 2010, 03:33:45 pm
yeah i think 4 is A
also isnt question 16 C instead of D?
D is not balanced (there are 2 oxygen molecules on the left and 3 on the right)
ur combustion reaction of benzoic is wrong as well..yes. yes. yes. Fixed. Still trawling the thread for other mistakes. That's what I get for not reading the question or writing anything down. :P
Post by: m@tty on November 11, 2010, 03:34:16 pm
also pretty sure 8 is A, dilution decreases ocncentration of all ions - just because hydronium ion conc decresases doesnt necessarily mean OH- increases...
The concentration oh
Also, Vanadium Oxide is fine for catalyst instead of Vanadium Pentoxide.
Post by: Souljette_93 on November 11, 2010, 03:34:39 pm
Mai is right, vanadium (v) oxide is the same as vanadium pentoxide.
And seriously there is no need to correct him by stating H2's enthalpy as 268 instead of 289 or watever. Concept is the same
Yeah it is, but he did ask to point out the mistakes. And plus it did stumped me at first..
Post by: Mao on November 11, 2010, 03:34:43 pm
also catalyst foro sulfuric is vanadium pentoxide..
That is the same as Vanadium (V) oxide
Post by: 10weid on November 11, 2010, 03:36:07 pm
also, u reckon they'll cut mc19 outta the exam?
Post by: akira88 on November 11, 2010, 03:36:47 pm
16 is C. D is not balanced (2:3 oxygen atoms)Yeah the equation you gave isn't balanced...
Post by: Souljette_93 on November 11, 2010, 03:37:09 pm
also pretty sure 8 is A, dilution decreases ocncentration of all ions - just because hydronium ion conc decresases doesnt necessarily mean OH- increases...
Why not? I always thought if one increases, the other decreases.
Dilution decreases ALL concs... nuff said? you can't dilute parts of your solution. why is there even debate about this
No debate, i want to know the right answer. And it also because that would mean my whole concept it wrong.
Post by: Mao on November 11, 2010, 03:37:12 pm
and H2 combusts in the data book at 286 kJ/mol not 289...
Mao, there is an error in Q's 2.1.b ii) The combustion is -286 Kj/mol, not 289
Also for the chosen chemical only sulfuric acid is given :( , wouldn't you know of Ammonia as well?
oops. fixing now.
I do know production of Ammonia, but not at the VCE level, so I'd have no idea what you guys know is what I think I know and what I think you know might be completely different to what you think I know about what I think you know. (Got it? :P)
Post by: mustafa on November 11, 2010, 03:37:54 pm
also pretty sure 8 is A, dilution decreases ocncentration of all ions - just because hydronium ion conc decresases doesnt necessarily mean OH- increases...
Why not? I always thought if one increases, the other decreases.
Dilution decreases ALL concs... nuff said? you can't dilute parts of your solution. why is there even debate about this
Jess Christ.....
Pure water has a higher concerration on OH ions than an acid. So when mixed, the end solution will have an overall increased concentration of OH.
Surely ppl dd not forget about the OH concentration in pure water.
Besides the concentration of H+ ions and the concentration of OH ions needs to multiply to give A fixed value at a fixed temperatre. Temp was fixed in question....
Therfore is concentration of H+ ions decreases concentration of OH ions must increase if temp is kept constant.
Post by: VeryCrazyEdu. on November 11, 2010, 03:39:02 pm
Post by: m@tty on November 11, 2010, 03:39:17 pm
cool, sorry i just never eard it called oxide b4 :p
also, u reckon they'll cut mc19 outta the exam?
Why would they do that?
The question was fine?..
Post by: ghadz7 on November 11, 2010, 03:40:55 pm
cool, sorry i just never eard it called oxide b4 :p
also, u reckon they'll cut mc19 outta the exam?
Why would they do that?
The question was fine?..
How'd you go m@tty? Full marks?
Post by: mirra250 on November 11, 2010, 03:41:30 pm
Finally someone gives a proper explanation!also pretty sure 8 is A, dilution decreases ocncentration of all ions - just because hydronium ion conc decresases doesnt necessarily mean OH- increases...
Why not? I always thought if one increases, the other decreases.
Dilution decreases ALL concs... nuff said? you can't dilute parts of your solution. why is there even debate about this
Jess Christ.....
Pure water has a higher concerration on OH ions than an acid. So when mixed, the end solution will have an overall increased concentration of OH.
Surely ppl dd not forget about the OH concentration in pure water.
Besides the concentration of H+ ions and the concentration of OH ions needs to multiply to give A fixed value at a fixed temperatre. Temp was fixed in question....
Therfore is concentration of H+ ions decreases concentration of OH ions must increase if temp is kept constant.
Post by: m@tty on November 11, 2010, 03:42:01 pm
Lol man.. I haven't spotted any errors yet.. :buck2: I think I did decently :Pcool, sorry i just never eard it called oxide b4 :p
also, u reckon they'll cut mc19 outta the exam?
Why would they do that?
The question was fine?..
How'd you go m@tty? Full marks?
EDIT: I got one MC wrong so far...
Post by: Mao on November 11, 2010, 03:42:15 pm
Why doesn't the Li+ move to the left?? I thought it would finish the 'circuit' by doing so....feel free to correct me :S
As the cell discharges, the reaction at anode is LiC6 --> C6 + Li+(electrolyte) + e, so Li+ is being discharged from LiC6 into the electrolyte, movement to the right.
To 'complete' the circuit, you need a NEGATIVE charge to 'complete' the circuit in the same direction, or a positive going in the opposite direction.
Post by: vexx on November 11, 2010, 03:42:27 pm
a)i) 3H2 + N2 <--> 2NH3
ii) porous iron catalyst (Fe2O3)
iii) provides an alternative pathway with lower activation energy for the reaction to take place.
b)i)highly corrosive, toxic gas
ii) burns skin, lungs, etc...
iii) wearing protective equipment, ventilation
Post by: mustafa on November 11, 2010, 03:42:49 pm
I acknowledge Mao's argument that electrons flow from the reductant molecule to te anode electrode, but in this case, both are Fe(s) ...,
electrons don't flow from Fe(s) to Fe(s) in a logical VCAA question...
Although Tyne logic of that question is somewhat stupid.
Post by: bomb on November 11, 2010, 03:43:06 pm
and H2 combusts in the data book at 286 kJ/mol not 289...Mao, there is an error in Q's 2.1.b ii) The combustion is -286 Kj/mol, not 289
Also for the chosen chemical only sulfuric acid is given :( , wouldn't you know of Ammonia as well?
oops. fixing now.
I do know production of Ammonia, but not at the VCE level, so I'd have no idea what you guys know is what I think I know and what I think you know might be completely different to what you think I know about what I think you know. (Got it? :P)
7b i put lithium has a low Mr and is lightweight...acceptable answer?
Post by: Mao on November 11, 2010, 03:43:59 pm
electrons don't flow from Fe(s) to Fe(s) in a logical VCAA question...
And fuel cells don't recharge, in VCELand, or real world :P
Post by: mirra250 on November 11, 2010, 03:44:23 pm
it clearly wasnt and you're deluded if you believe C is correct. It is completely ambiguous and unacceptable for a VCE exam and hence will either be removed or everyone who answered will receive a mark. If you think it's another answer than C, then you're totally wrong :)cool, sorry i just never eard it called oxide b4 :p
also, u reckon they'll cut mc19 outta the exam?
Why would they do that?
The question was fine?..
Post by: m@tty on November 11, 2010, 03:44:48 pm
I disAgree with 19 MC. The anode and reductand are the same.
I acknowledge Mao's argument that electrons flow from the reductant molecule to te anode electrode, but in this case, both are Fe(s) ...,
electrons don't flow from Fe(s) to Fe(s) in a logical VCAA question...
Although Tyne logic of that question is somewhat stupid.
No they're not. One is a reactant, the other is an electrode.
Post by: Souljette_93 on November 11, 2010, 03:44:56 pm
Ammonia
a)i) 3H2 + N2 <--> 2NH3
ii) porous iron catalyst (Fe2O3)
Isn't it fe3O4? lol i wrote that and in brackets i wrote "porous iron"
Post by: Mao on November 11, 2010, 03:45:12 pm
and H2 combusts in the data book at 286 kJ/mol not 289...Mao, there is an error in Q's 2.1.b ii) The combustion is -286 Kj/mol, not 289
Also for the chosen chemical only sulfuric acid is given :( , wouldn't you know of Ammonia as well?
oops. fixing now.
I do know production of Ammonia, but not at the VCE level, so I'd have no idea what you guys know is what I think I know and what I think you know might be completely different to what you think I know about what I think you know. (Got it? :P)
7b i put lithium has a low Mr and is lightweight...acceptable answer?
Potentially, yeah. Light weight and packs charge is generally a good thing. I'd accept that.
Also, if I have missed the karma to anyone who helped out on making the solution set better, pm me and I will karma.
Post by: 10weid on November 11, 2010, 03:45:17 pm
Post by: m@tty on November 11, 2010, 03:45:57 pm
ok, just asked my dad, who has a medical degree with an additional year of study in chemistry, and topped chemistry at melb university 2 years in a row, he said fuel cells ARE rechargeable,.
Last years examiner's report clearly said they were not. On top of that I've never read that they are. And seriously? How can you get the reactions to go backward in there? Split the H2O up?..
Post by: andy456 on November 11, 2010, 03:46:15 pm
would it be 1 mark for correct equation
1 mark for correct co-efficients
1 mark for correct delta H
Post by: cama23 on November 11, 2010, 03:46:30 pm
shit yeahhhhh
Post by: Mao on November 11, 2010, 03:47:30 pm
ok, just asked my dad, who has a medical degree with an additional year of study in chemistry, and topped chemistry at melb university 2 years in a row, he said fuel cells ARE rechargeable,.
I challenge this. Source? (journal article, textbook reference, anything?)
Post by: 10weid on November 11, 2010, 03:47:34 pm
ok, just asked my dad, who has a medical degree with an additional year of study in chemistry, and topped chemistry at melb university 2 years in a row, he said fuel cells ARE rechargeable,.
Last years examiner's report clearly said they were not.
whatever the answer is, (and this isnt me admitting defeat:P), ill take my 46-4 in chem, it'll be in my bottom 2, and prepare for my interview:)
as long as they dont ask me about the rechargeable nature of fuel cells in relation to medicine! :p
Post by: mirra250 on November 11, 2010, 03:49:15 pm
LMAO.. yes.. it's called electrolysis of waterok, just asked my dad, who has a medical degree with an additional year of study in chemistry, and topped chemistry at melb university 2 years in a row, he said fuel cells ARE rechargeable,.
Last years examiner's report clearly said they were not. On top of that I've never read that they are. And seriously? How can you get the reactions to go backward in there? Split the H2O up?..
Post by: Souljette_93 on November 11, 2010, 03:50:21 pm
ok, just asked my dad, who has a medical degree with an additional year of study in chemistry, and topped chemistry at melb university 2 years in a row, he said fuel cells ARE rechargeable,.
Regardless if they are or not, it's all going to depend on what VCAA accepts. They are the one's who is going to give the mark.
Post by: Mao on November 11, 2010, 03:50:46 pm
ok, just asked my dad, who has a medical degree with an additional year of study in chemistry, and topped chemistry at melb university 2 years in a row, he said fuel cells ARE rechargeable,.
Last years examiner's report clearly said they were not.
whatever the answer is, (and this isnt me admitting defeat:P), ill take my 46-4 in chem, it'll be in my bottom 2, and prepare for my interview:)
as long as they dont ask me about the rechargeable nature of fuel cells in relation to medicine! :p
GL man :) I still want you to prove to me I am wrong. I firmly believe there's no such thing as 'rechargeable fuel cell', that'll be like a 'two-directional hose' for watering your garden..
Well, looks like the discussion have calmed down a bit. I will go get out of bed now and do some boring stuff. I'll be back later tonight, don't hesitate to email. :)
Post by: Mao on November 11, 2010, 03:51:26 pm
LMAO.. yes.. it's called electrolysis of waterok, just asked my dad, who has a medical degree with an additional year of study in chemistry, and topped chemistry at melb university 2 years in a row, he said fuel cells ARE rechargeable,.
Last years examiner's report clearly said they were not. On top of that I've never read that they are. And seriously? How can you get the reactions to go backward in there? Split the H2O up?..
ELECTROLYSIS OF WATER DOESN'T USE A BLOODY POROUS PLATINUM ELECTRODE!!!@!@!@ (sorry :P)
Post by: m@tty on November 11, 2010, 03:51:33 pm
LMAO.. yes.. it's called electrolysis of waterok, just asked my dad, who has a medical degree with an additional year of study in chemistry, and topped chemistry at melb university 2 years in a row, he said fuel cells ARE rechargeable,.
Last years examiner's report clearly said they were not. On top of that I've never read that they are. And seriously? How can you get the reactions to go backward in there? Split the H2O up?..
How would you do this in the fuel cell? The H2O is released into the atmosphere.. Clearly you can't electrolyse it..
EDIT: what Mao said..
Post by: mustafa on November 11, 2010, 03:52:04 pm
sure fuel cells are rechargeable.
Last year the assessors said try ACCEPTED responses saying fuel cells were unrechargeable. Which means they can be recharged, but VCAA was just being nice.
Post by: m@tty on November 11, 2010, 03:53:51 pm
I think you missed that a current was being supplied?
Post by: mirra250 on November 11, 2010, 03:54:22 pm
LOL how are they not electrolysed with a porous platinum electrode?LMAO.. yes.. it's called electrolysis of waterok, just asked my dad, who has a medical degree with an additional year of study in chemistry, and topped chemistry at melb university 2 years in a row, he said fuel cells ARE rechargeable,.
Last years examiner's report clearly said they were not. On top of that I've never read that they are. And seriously? How can you get the reactions to go backward in there? Split the H2O up?..
How would you do this in the fuel cell? The H2O is released into the atmosphere.. Clearly you can't electrolyse it..
EDIT: what Mao said..
and clearly the water can be harnessed and pumped back into the fuel cell just as the O2 and H2 are and then applied electricity to it. Clearly it would never happen but does not make it impossible
Post by: Mao on November 11, 2010, 03:54:33 pm
I'm
sure fuel cells are rechargeable.
Last year the assessors said try ACCEPTED responses saying fuel cells were unrechargeable. Which means they can be recharged, but VCAA was just being nice.
Your tap in the sink can also suck up water, but at no point in their operation will they return water to the dam against the main flow. Same principle. :)
Post by: 98.40_for_sure on November 11, 2010, 03:55:36 pm
I don't know if it has been contested yet.. But I believe you've got 18 wrong. It was being electrolysed. So the strongest oxidant is oxidised.. I believe that the correct answer is B. The anode reaction is the oxidation of the sulfate ion and the cathode is the reduction of water..
I think you missed that a current was being supplied?
Download the updated one :P
Post by: Mao on November 11, 2010, 03:55:50 pm
I don't know if it has been contested yet.. But I believe you've got 18 wrong. It was being electrolysed. So the strongest oxidant is oxidised.. I believe that the correct answer is B. The anode reaction is the oxidation of the sulfate ion and the cathode is the reduction of water..
I think you missed that a current was being supplied?
My 18 is B?
Post by: mirra250 on November 11, 2010, 03:56:04 pm
I don't know if it has been contested yet.. But I believe you've got 18 wrong. It was being electrolysed. So the strongest oxidant is oxidised.. I believe that the correct answer is B. The anode reaction is the oxidation of the sulfate ion and the cathode is the reduction of water..that was incoherent but nonetheless incorrect. an oxidant is REDUCED, a reductant is OXIDISED. In this reaction, water is the strongest reductant hence is oxidised
I think you missed that a current was being supplied?
Post by: mirra250 on November 11, 2010, 03:56:44 pm
cool story. clearly completely differentI'm
sure fuel cells are rechargeable.
Last year the assessors said try ACCEPTED responses saying fuel cells were unrechargeable. Which means they can be recharged, but VCAA was just being nice.
Your tap in the sink can also suck up water, but at no point in their operation will they return water to the dam against the main flow. Same principle. :)
Post by: _henwee on November 11, 2010, 03:56:59 pm
I MISSED THE LABELLING ARROW QUESTIONS!!! FML!!!
Post by: ckg93 on November 11, 2010, 03:57:23 pm
According to the to my chem teacher, it MUST be PENToxide to get the full mark, however, the PROPER name is Di-vanadium Pentoxide.also catalyst foro sulfuric is vanadium pentoxide..
That is the same as Vanadium (V) oxide
I assumed that the dilution by water would decrease both, however was the increase in [OH-] (by adding water) significantly more than the decrease by dilution??? Taking into account 10 ml of 0.01M Hcl WILL have OH- ions already.
Post by: m@tty on November 11, 2010, 03:57:58 pm
I don't know if it has been contested yet.. But I believe you've got 18 wrong. It was being electrolysed. So the strongest oxidant is oxidised.. I believe that the correct answer is B. The anode reaction is the oxidation of the sulfate ion and the cathode is the reduction of water..
I think you missed that a current was being supplied?
My 18 is B?
I meant A.
Don't you take the highest in the series for oxidation in electrolosis? Thus the sulfate ion?
If it wasn't for the current I agree it would be water oxidised..
Post by: jasoN- on November 11, 2010, 03:58:04 pm
Post by: andy456 on November 11, 2010, 03:59:46 pm
I'mAre you serious...... read what you wrote...
sure fuel cells are rechargeable.
Last year the assessors said try ACCEPTED responses saying fuel cells were unrechargeable. Which means they can be recharged, but VCAA was just being nice.
Accepted responses saying that fuel cells were UNRECHARGEABLE.........that means cannot be recharged
From VCAA's assessment report last year (page 11, answer to 5d.)
Acceptable responses included:
fuel cells are not rechargeable
fuel cells needs continuous supply of reactants
fuels in fuel cells are molecular, not ionic
products of fuel cells are not recycled
some fuel cells produce CO2
fuel cell electrodes are porous.
VCAA is not going to go back on their word
Post by: bomb on November 11, 2010, 04:00:10 pm
If it's rechargeable then a CONTINUOUS FLOW OF REACTANTS is not possible, and the REACTANTS WOULD BE STORED, which would make the other answers right.
Post by: Souljette_93 on November 11, 2010, 04:00:21 pm
I don't know if it has been contested yet.. But I believe you've got 18 wrong. It was being electrolysed. So the strongest oxidant is oxidised.. I believe that the correct answer is B. The anode reaction is the oxidation of the sulfate ion and the cathode is the reduction of water..
I think you missed that a current was being supplied?
My 18 is B?
I meant A.
Don't you take the highest in the series for oxidation in electrolosis? Thus the sulfate ion?
I got A, and i also didn't understand your explanation. ( in the solutions )
Post by: m@tty on November 11, 2010, 04:01:59 pm
I don't know if it has been contested yet.. But I believe you've got 18 wrong. It was being electrolysed. So the strongest oxidant is oxidised.. I believe that the correct answer is B. The anode reaction is the oxidation of the sulfate ion and the cathode is the reduction of water..that was incoherent but nonetheless incorrect. an oxidant is REDUCED, a reductant is OXIDISED. In this reaction, water is the strongest reductant hence is oxidised
I think you missed that a current was being supplied?
All half equations in the ECS can go either backward or forward. I know that what I said is technically incorrect, but the species in the half equations higher in the series are often termed as stronger oxidant. The species on both sides. So I guess what I was saying was the weakest reductant will be oxidised.
Looks like you missed that a current was being supplied too :buck2:
Post by: jasoN- on November 11, 2010, 04:03:14 pm
Post by: stonecold on November 11, 2010, 04:04:38 pm
I don't know if it has been contested yet.. But I believe you've got 18 wrong. It was being electrolysed. So the strongest oxidant is oxidised.. I believe that the correct answer is B. The anode reaction is the oxidation of the sulfate ion and the cathode is the reduction of water..
I think you missed that a current was being supplied?
My 18 is B?
I meant A.
Don't you take the highest in the series for oxidation in electrolosis? Thus the sulfate ion?
I got A, and i also didn't understand your explanation. ( in the solutions )
Water is a stronger reductant than what SO42- is...
Post by: m@tty on November 11, 2010, 04:06:49 pm
Post by: stonecold on November 11, 2010, 04:07:42 pm
But a current was supplied so it was electrolosis. Can someone at least acknowledge that they read this?
Yeah, and 2.03 volts is higher than 1.23 volts or whatever water is, so water is the strongest reductant, and will be oxidised...
Post by: m@tty on November 11, 2010, 04:09:23 pm
Sure, that would happen in a galvanic cell. But it wouldn't also happen during electrolosis, would it?
Post by: vexx on November 11, 2010, 04:12:17 pm
Ammonia
a)i) 3H2 + N2 <--> 2NH3
ii) porous iron catalyst (Fe2O3)
Isn't it fe3O4? lol i wrote that and in brackets i wrote "porous iron"
its an iron oxide, the most common one is Fe2O3, but Fe3O4 can be used too just rarer.
Post by: monicapham93 on November 11, 2010, 04:12:58 pm
about the ammonia
is it wrong to say that if theres an excess there will be no change?? cause i thought it meant, what effect does it have on the relitive proportions of the product gasses in the atmosphere (according to the eqn)
argh. i dunno i probs read it wrong
Post by: stonecold on November 11, 2010, 04:13:28 pm
But doesn't the current force a reaction which will not normally happen?
Sure, that would happen in a galvanic cell. But it wouldn't also happen during electrolosis, would it?
I think I get where you're coming from, but water will always be reduced before a sulfate ion.
Think of the hundreds of Q's you've done. Spectator ions never get oxidised, because they are too weak of a reductant.
Post by: m@tty on November 11, 2010, 04:14:05 pm
Post by: stonecold on November 11, 2010, 04:15:00 pm
Wait, I see the error... The cathode reaction is the reduction of water .. which is -0.83 V and then you take the closest one above.. being the other water.. :buck2: HOW COULD I MISS THAT, ARGH.
DW, it ranks up there with my misreading of mol ratio error in MC Q7 lol...
Post by: ckg93 on November 11, 2010, 04:25:17 pm
But doesn't the current force a reaction which will not normally happen?Only after the first reaction has been completed. In other words, the water would have to be used up before the second reaction could occur.
Sure, that would happen in a galvanic cell. But it wouldn't also happen during electrolosis, would it?
Post by: Chavi on November 11, 2010, 04:26:42 pm
ok, just asked my dad, who has a medical degree with an additional year of study in chemistry, and topped chemistry at melb university 2 years in a row, he said fuel cells ARE rechargeable,.Ahh yes Dyeni. Echoes of Yr 9 science when you convinced the teacher to give you that 1/2 mark advantage due to some obscure finding in an ancient medicine manual. .
Post by: becca92 on November 11, 2010, 04:26:48 pm
I wrote "It lowers the activation energy required for the reaction to proceed but did not mention alternative pathway.. will i still get the mark?
Post by: stonecold on November 11, 2010, 04:28:35 pm
ok, just asked my dad, who has a medical degree with an additional year of study in chemistry, and topped chemistry at melb university 2 years in a row, he said fuel cells ARE rechargeable,.Ahh yes Dyeni. Echoes of Yr 9 science when you convinced the teacher to give you that 1/2 mark advantage due to some obscure finding in an ancient medicine manual. .
VCAA makes up their own rules. As has been seen in past examiners reports, 'fuel cells are non rechargeable.'
Post by: Souljette_93 on November 11, 2010, 04:29:28 pm
For 2.5a "iii, Provide alternative reaction pathways with lowe ractivation energy"
I wrote "It lowers the activation energy required for the reaction to proceed but did not mention alternative pathway.. will i still get the mark?
Yeah of course. I think the major thing you had to mention was "lowering the activation energy". There are multiple answers, so don't worry, if you have a look at the examiners report, you will see that they accept a variety of answers depending on the the question.
Post by: m@tty on November 11, 2010, 04:31:58 pm
Post by: physics on November 11, 2010, 04:33:49 pm
and the concentration of H...i gpot 1.6 or something :S
Post by: m@tty on November 11, 2010, 04:35:20 pm
And crap that's a high concentration... didn't you get suspicious when a weak acid produced a solution of such high pH?
Post by: Souljette_93 on November 11, 2010, 04:36:39 pm
FOR first short answer...the K value doesnt include H2o does it?
and the concentration of H...i gpot 1.6 or something :S
As far as i know, you include water, this is not a Ka expression.
Also in the examiner's report (2009), they said:
Quote
The most common error was leaving [H2O] out of the equilibrium expression.
Post by: ckg93 on November 11, 2010, 04:38:27 pm
for Q2 3ii
about the ammonia
is it wrong to say that if theres an excess there will be no change?? cause i thought it meant, what effect does it have on the relitive proportions of the product gasses in the atmosphere (according to the eqn)
Hey Monica, Chris from NYSF here. You were referring to section B - Question 3 - Part a ii right??? I put down what I thought would be produced, considering the yields would be different between the two parts. I also said that NH4OH would be produced in excess ammonia.
Post by: physics on November 11, 2010, 04:41:47 pm
FOR first short answer...the K value doesnt include H2o does it?consequential marks atleast ? :(
and the concentration of H...i gpot 1.6 or something :S
As far as i know, you include water, this is not a Ka expression.
Also in the examiner's report (2009), they said:QuoteThe most common error was leaving [H2O] out of the equilibrium expression.
Post by: monicapham93 on November 11, 2010, 04:43:26 pm
for Q2 3ii
about the ammonia
is it wrong to say that if theres an excess there will be no change?? cause i thought it meant, what effect does it have on the relitive proportions of the product gasses in the atmosphere (according to the eqn)
Hey Monica, Chris from NYSF here. You were referring to section B - Question 3 - Part a ii right??? I put down what I thought would be produced, considering the yields would be different between the two parts. I also said that NH4OH would be produced in excess ammonia.
ooh chris, as in from camb boys???
heyyyaaa!
yeah ... im probs all wrong, i didnt really get the question so i tried to answer it the only way that made sense to me
... lucky its one mark though!
Post by: ckg93 on November 11, 2010, 04:50:17 pm
For 2.5a "iii, Provide alternative reaction pathways with lowe ractivation energy"By adsorbing the reactants to the surface of the catalyst, the catalyst allows a greater percentage of collisions to react. Hence the activation energy is reduced by the catalyst adsorbing the reactants to the catalyst surface. I'm not sure that you will get the mark, but be hopeful (it was a relatively tricky exam)
I wrote "It lowers the activation energy required for the reaction to proceed but did not mention alternative pathway.. will i still get the mark?
Post by: Mao on November 11, 2010, 04:56:43 pm
According to the to my chem teacher, it MUST be PENToxide to get the full mark, however, the PROPER name is Di-vanadium Pentoxide.also catalyst foro sulfuric is vanadium pentoxide..
That is the same as Vanadium (V) oxide
yes, that is one of its names. I assure you that other name names (Vanadium pentoxide, vanadium (V) oxide, divanadium pentaoxide, etc) will also be accepted.
Post by: ckg93 on November 11, 2010, 05:02:00 pm
And yeah mon, it is me. I bet you'll still beat me lol.
Also Mao, for Section B - Question 2 - Part a iv; can you also say that "The experiment would been conducted under standard conditions"? I said this because the K value for lactic acid was used, and that value in the data booklet was based on standard conditions.
Post by: darkphoenix on November 11, 2010, 05:12:37 pm
Post by: ckg93 on November 11, 2010, 05:23:01 pm
Also, your equation is not balanced properly, it should be 17(oxygen gas) and also 14(Carbon Dioxide) There are 7 Carbons in benzoic acid.
Post by: Aden on November 11, 2010, 05:33:39 pm
Basically, his rationale is that if you were to make every species in each reaction 1mol in a 1L container (yes, it is impossible given the nature of the reversible reaction, but for simplicity's sake lets just go with it), then it goes like this:
Initial CF/K:
A.
B. Not applicable at all
C.
D.
The volume is halved, so the concentration of each species will double, therefore CF immediately after:
A.
B. Not applicable
C.
D.
So the percentage change in each case is:
A. 100%
B. Whatever
C. 100%
D. 75%
So... to prove him wrong... we...?
Post by: stonecold on November 11, 2010, 05:36:03 pm
A friend of mine is arguing that the answer to multiple choice Question 7 is either A or C, and not D.
Basically, his rationale is that if you were to make every species in each reaction 1mol in a 1L container (yes, it is impossible given the nature of the reversible reaction, but for simplicity's sake lets just go with it), then it goes like this:
Initial CF/K:
A.
B. Not applicable at all
C.
D.
The volume is halved, so the concentration of each species will double, therefore CF immediately after:
A.
B. Not applicable
C.
D.
So the percentage change in each case is:
A. 100%
B. Whatever
C. 100%
D. 75%
So... to prove him wrong... we...?
I stuffed this question and chose A. But looking at it again, I can see why it is D....
Post by: Aden on November 11, 2010, 05:39:29 pm
I stuffed this question and chose A. But looking at it again, I can see why it is D....
My friend is arguing that A or C are both correct, since they have the largest percentage change as shown by my previous post =\
Post by: stonecold on November 11, 2010, 05:42:47 pm
I stuffed this question and chose A. But looking at it again, I can see why it is D....
My friend is arguing that A or C are both correct, since they have the largest percentage change as shown by my previous post =\
lol i dunno. for my sake i hope he's right but i doubt it.
vcaa don't seem to reward the smart ass approach to questions...
Post by: m@tty on November 11, 2010, 05:48:32 pm
The was I looked at it was A-factor of 2; B-factor of 2; C-no change; D - reduce by a factor of 4. Clearly the one which has undergone the biggest 'change' is D.
Post by: TehGullz on November 11, 2010, 05:52:30 pm
Also Mao, for Section B - Question 2 - Part a iv; can you also say that "The experiment would been conducted under standard conditions"? I said this because the K value for lactic acid was used, and that value in the data booklet was based on standard conditions.
I said this too, anyone think it'd get marks?
Post by: stonecold on November 11, 2010, 05:58:26 pm
Also Mao, for Section B - Question 2 - Part a iv; can you also say that "The experiment would been conducted under standard conditions"? I said this because the K value for lactic acid was used, and that value in the data booklet was based on standard conditions.
I said this too, anyone think it'd get marks?
This will certainly get marks. It is a valid point.
Post by: m@tty on November 11, 2010, 06:17:36 pm
Also Mao, for Section B - Question 2 - Part a iv; can you also say that "The experiment would been conducted under standard conditions"? I said this because the K value for lactic acid was used, and that value in the data booklet was based on standard conditions.
I said this too, anyone think it'd get marks?
This will certainly get marks. It is a valid point.
Yeah that's what I said. Though I do suspect that they were looking for the answer which Mao gave. Certainly should get the marks though.
Post by: happyhappyland on November 11, 2010, 06:20:00 pm
Post by: stonecold on November 11, 2010, 07:18:15 pm
Also Mao, for Section B - Question 2 - Part a iv; can you also say that "The experiment would been conducted under standard conditions"? I said this because the K value for lactic acid was used, and that value in the data booklet was based on standard conditions.
I said this too, anyone think it'd get marks?
This will certainly get marks. It is a valid point.
Yeah that's what I said. Though I do suspect that they were looking for the answer which Mao gave. Certainly should get the marks though.
I put Mao's answer but I like this better :P
FUELS CELLS ARE NOT F******* RECHARGABLE... YOU PUT REACTANTS IN TO KEEP IT RUNNING... YOU DONT CONVERT THE PRODUCTS BACK INTO THE REACTIONS SO ITS NOT RECHARGABLE..................... if your dad is a doctor please tell me where he works cos im not going to see him
I agree. If VCAA dog us out of a mark on this Q then I'll be kinda pissed...especially after they've made it explicitly clear in examiners reports that fuel cells are not rechargeable.
Post by: teags92 on November 11, 2010, 08:02:51 pm
Post by: 10weid on November 11, 2010, 08:04:33 pm
catalyst for H2so4 is vanadium oxide NOT vanadium pentoxide....
theyre all acceptable.. if u have ever heard of the prefix 'pent' it denotes the number 5 - since 5 oxygen atoms are presnt in one mole of v2o5, pentoxide is technically correct.,
Post by: 10weid on November 11, 2010, 08:05:54 pm
Post by: qshyrn on November 11, 2010, 08:08:01 pm
Post by: qshyrn on November 11, 2010, 08:09:08 pm
umm the assumption with regard to the lactic acid, is it acceptable to say that it is assumed that all of the lactic acid dissolves in the water? coz if it doesnt all dissolve, the calculation is incorrect..nah dont think so. they asked u to find the percentage of ionization later on, so clearly theyre not gonna make u assume its 100 %
Post by: mirra250 on November 11, 2010, 08:12:34 pm
that doesnt make sense qshryn..?umm the assumption with regard to the lactic acid, is it acceptable to say that it is assumed that all of the lactic acid dissolves in the water? coz if it doesnt all dissolve, the calculation is incorrect..nah dont think so. they asked u to find the percentage of ionization later on, so clearly theyre not gonna make u assume its 100 %
yes i think they will accept it 10weird
Post by: 10weid on November 11, 2010, 08:13:01 pm
Post by: 10weid on November 11, 2010, 08:13:20 pm
umm the assumption with regard to the lactic acid, is it acceptable to say that it is assumed that all of the lactic acid dissolves in the water? coz if it doesnt all dissolve, the calculation is incorrect..nah dont think so. they asked u to find the percentage of ionization later on, so clearly theyre not gonna make u assume its 100 %
10weid* but thanks
that doesnt make sense qshryn..?
yes i think they will accept it 10weird
Post by: qshyrn on November 11, 2010, 08:15:21 pm
nooo, as in the 4.50grams of lactic acid all dissolved...ooo ok. are the Ka Values in the data book at 25degrees??? i wrote that i assumed the experiment was done at 25 degreres
Post by: crayolé on November 11, 2010, 09:29:54 pm
Post by: kakar0t on November 11, 2010, 09:31:13 pm
Post by: m@tty on November 11, 2010, 09:31:53 pm
Post by: kakar0t on November 11, 2010, 09:37:26 pm
Yep.
http://www.youtube.com/watch?v=GANY0T6Xkzk
Post by: Potter on November 11, 2010, 11:09:26 pm
Post by: Lighties on November 11, 2010, 11:26:51 pm
They obviously weren't looking for that approach as it yields two answers..
The was I looked at it was A-factor of 2; B-factor of 2; C-no change; D - reduce by a factor of 4. Clearly the one which has undergone the biggest 'change' is D.
I don't really get how you looked at it =| But they asked for the 'percentage change in the concentration fraction', so wouldn't a change in the denominator of Q (as in D) affect it less than the same change in the numerator of Q (such as A)?
Post by: a.zirek on November 12, 2010, 01:08:34 am
Post by: coletrain on November 12, 2010, 10:13:53 am
Post by: year12 on November 12, 2010, 11:18:50 am
Post by: d-ea-6 on November 12, 2010, 11:23:23 am
Because I assumed it was so I wrote down that
CH4 + 2 O2 → CO2 + 2 H2O and
2 H2 + O2 → 2 H2O
Therefore, the answer for the volume of hydrogen needed would be double the volume that Mao has given (12.4 L)
Post by: year12 on November 12, 2010, 11:32:12 am
With question 1.b.iii where methane and hydrogen are "burned", is that the same as combustion?
Because I assumed it was so I wrote down that
CH4 + 2 O2 → CO2 + 2 H2O and
2 H2 + O2 → 2 H2O
Therefore, the answer for the volume of hydrogen needed would be double the volume that Mao has given (12.4 L)
Yeah, burning = combustion but I don't get your logic. Are you saying that the volume needs to double because there is a coefficient of 2 in front of hydrogen...?
Post by: d-ea-6 on November 12, 2010, 11:35:14 am
Post by: andy456 on November 12, 2010, 11:39:39 am
for the lithium + water equation i mind blanked and wrote this
is this wrong??
Post by: year12 on November 12, 2010, 11:48:49 am
Yes :-\
Oh, okay. Now I get you. But you don't need to worry about that. All you need to do is get the required energy that needs to be produced by burning hydrogen. Then, divide that by 286 which gives you the no. of mole of H2 you need which can be used to find the required volume (Just like Mao has shown in his solution) because 286 is for 1 mole of H2 burnt. Hope I make some sense!
Post by: year12 on November 12, 2010, 11:53:18 am
I have a question too
for the lithium + water equation i mind blanked and wrote this
is this wrong??
well.. it definitely isn't 'wrong'. It just isn't simplified as such. You just have the two ions combined to get 2LiOH! Doubt you will lost marks, the significance of this Q was to show that H2 is formed -> high pressure -> explosion.
Post by: andy456 on November 12, 2010, 11:55:32 am
Ok so if my reaction is 'technically' right i should get the marks??I have a question too
for the lithium + water equation i mind blanked and wrote this
is this wrong??
well.. it definitely isn't 'wrong'. It just isn't simplified as such. You just have the two ions combined to get 2LiOH! Doubt you will lost marks, the significance of this Q was to show that H2 is formed -> high pressure -> explosion.
Post by: Greggler on November 12, 2010, 01:15:44 pm
Post by: andy456 on November 12, 2010, 01:43:39 pm
Post by: m@tty on November 12, 2010, 01:46:35 pm
would that question be 2 for the equation and 1 for why it exploded, or the other way around
Yeah, that sounds right. I dunno how they could give two for explosion..
Post by: stonecold on November 12, 2010, 03:43:05 pm
would that question be 2 for the equation and 1 for why it exploded, or the other way around
Yeah, that sounds right. I dunno how they could give two for explosion..
I said both of hydrogen gas is highly flammable/explosive and high pressure due to build up of hydrogen gas could lead to an explosion.
Post by: Mao on November 12, 2010, 04:13:56 pm
That is a VERY good point. As soon as the seal is ruptured it will combust. Didn't think of this at the time.would that question be 2 for the equation and 1 for why it exploded, or the other way around
Yeah, that sounds right. I dunno how they could give two for explosion..
I said both of hydrogen gas is highly flammable/explosive and high pressure due to build up of hydrogen gas could lead to an explosion.
Post by: Mao on November 12, 2010, 04:15:38 pm
With question 1.b.iii where methane and hydrogen are "burned", is that the same as combustion?
Because I assumed it was so I wrote down that
CH4 + 2 O2 → CO2 + 2 H2O and
2 H2 + O2 → 2 H2O
Therefore, the answer for the volume of hydrogen needed would be double the volume that Mao has given (12.4 L)
Yes, it's just combustion and your equation is correct.
But the value from the data booklet is for this equation: H2 + 0.5O2 --> H2O
The value from the data booklet (or any standard tables) is always per 1 mol of the reactant in question. :)
Post by: d-ea-6 on November 12, 2010, 04:26:32 pm
With question 1.b.iii where methane and hydrogen are "burned", is that the same as combustion?
Because I assumed it was so I wrote down that
CH4 + 2 O2 → CO2 + 2 H2O and
2 H2 + O2 → 2 H2O
Therefore, the answer for the volume of hydrogen needed would be double the volume that Mao has given (12.4 L)
Yes, it's just combustion and your equation is correct.
But the value from the data booklet is for this equation: H2 + 0.5O2 --> H2O
The value from the data booklet (or any standard tables) is always per 1 mol of the reactant in question. :)
After I raised the query, I was afraid this was the case :P
This is what happens when you try a different study approach. I did the bulk of my chem study weeks ago, so when it came to the actual exam I think I forgot some things... (like this, for instance). Should've stuck with cramming!
Post by: D27RII on November 12, 2010, 04:32:49 pm
and D is wierd too, the question says "fuel cells and rechargeable cells" <----, doesnt that say taht fuel cells are not rechargeable??
still i think D is most right tho.
BTW, ONE important thing
for my lactic acid assumption i said the we assumed that H+ ion concentration was the same as the lactate ion concentration. is that a fair assumption?
and Hydrogen gas is not flammable, if u put down flammable u will get it wrong, it is combustible, DEFINITELY not flammable lol.
Post by: D27RII on November 12, 2010, 04:34:02 pm
Post by: D27RII on November 12, 2010, 05:09:58 pm
so u would need to mention that the reaction evolves IMMENSE heat
and that hydrogen gas is COMBUSTIBLE
Post by: Mao on November 12, 2010, 05:36:06 pm
also, its crucial to point out the basics...lithium is a very reactive element and the strongest reductant there is. Hence in its reaction with water, irrepsective of any Hydrogen gas is pretty darn violent. the hydrogen gas just adds the exploxive factor, but lithium and water otherwise relsesae a lot of heat anyway.
so u would need to mention that the reaction evolves IMMENSE heat
and that hydrogen gas is COMBUSTIBLE
Yes and no. The electrochemical series tell us nothing about the rate of reaction. We can only get this rate through experimentation. The reaction between lithium and water is not THAT energetic, I would contribute most of the explosive factor into the pressure build up. If you've done the experiment of Li in water vs Na in water, you can probably remember that Li was the boring one, just fizzing about and never caught on fire. The heat of reaction will predominantly go into increasing the pressure of H2 gas inside the battery by PV=nRT.
Post by: Mao on November 12, 2010, 05:38:35 pm
and Hydrogen gas is not flammable, if u put down flammable u will get it wrong, it is combustible, DEFINITELY not flammable lol.I disagree. Just about every MSDS on the planet will say H2 gas is an inorganic highly-flammable gas.
Post by: stonecold on November 12, 2010, 05:40:28 pm
and Hydrogen gas is not flammable, if u put down flammable u will get it wrong, it is combustible, DEFINITELY not flammable lol.I disagree. Just about every MSDS on the planet will say H2 gas is an inorganic highly-flammable gas.
I think the examiner will get the jist of what I meant, and I doubt they are that tight.
Post by: 8039 on November 13, 2010, 02:33:10 pm
Post by: D27RII on November 13, 2010, 03:26:29 pm
for my assumption question i said.
We assumed that the concentration of hydrogen ions was the same as the concentration of lactate ions as the two reactants are shown by the equation to exit in equimolar stoichiometric amounts.
is that a fair assumption considering that there are addition H+ ions in the neutral water itself?
Post by: m@tty on November 13, 2010, 03:30:05 pm
Reductant = A reducing agent
Anode = The positively charged electrode by which the electrons leave a device.
The reductant is the thing that releases electrons. At an atomic level this is the Fe atoms participating in the reaction (is this what you dispute? Dude, it's simple logic xD ). Then, by definition, the electrons produced must leave via the anode - ie. the electrons must "pass from the reductant to the anode as electrons are produced".
Therefore C.
QED.
EDIT:
besdies, electrons flowing from reductant to anode does not produce electricity, option C states "As electricity is being produced" - hence implies that electrons need to flow from reductant to CATHODE, not from reductant to anode.
Yes, that is true. But with intermediate and final steps shown it goes: reductant -> anode -> cathode -> oxidant.
PLLLSSS CHECK THIS ANSWER>
for my assumption question i said.
We assumed that the concentration of hydrogen ions was the same as the concentration of lactate ions as the two reactants are shown by the equation to exit in equimolar stoichiometric amounts.
is that a fair assumption considering that there are addition H+ ions in the neutral water itself?
I think that should be fine :)
Post by: DarkVisor on November 14, 2010, 05:17:07 pm
Post by: D27RII on November 16, 2010, 11:59:17 am
For the last question (The kid with the phosphoric acid fuel cell), I put "Non standard temp" for one point, but I also said that the NSC could cause the "Hydrogen peroxide reaction" to take place... Does anyone think they'll accept is? Wouldn't it technically be correct? :(
nope, H2O2 needs enzymes to be formed and enzymes to be decomposed. moreover in the different world of biology, two different catalysts are required for the forward and backward reactions respectively since the catalysts are organic and complex.
it shouldnt really be in the electrochemical series, because the reaction which produced H2O2 will never happen without enzymes and biological organelles, and the decomposing reaction will happen at such a slow temperature that it is completely negligible.
Post by: Shark 774 on November 16, 2010, 10:03:10 pm
also pretty sure 8 is A, dilution decreases ocncentration of all ions - just because hydronium ion conc decresases doesnt necessarily mean OH- increases...
The concentration ohis greater in the water, thus when it is added to the acidic solution the overall concentration of
Also, Vanadium Oxide is fine for catalyst instead of Vanadium Pentoxide.
I am not sure what the right answer is, but I don't agree with your reasoning. The OH- is greater in the water but as it is added to the HCl there is now a greater total volume and not to mention some of the OH- from the water will react with H+.
Post by: stonecold on November 16, 2010, 10:15:37 pm
also pretty sure 8 is A, dilution decreases ocncentration of all ions - just because hydronium ion conc decresases doesnt necessarily mean OH- increases...
The concentration oh
Also, Vanadium Oxide is fine for catalyst instead of Vanadium Pentoxide.
I am not sure what the right answer is, but I don't agree with your reasoning. The OH- is greater in the water but as it is added to the HCl there is now a greater total volume and not to mention some of the OH- from the water will react with H+.
Again agree with matty.
If H+ decreases, then OH- has to increase.
Post by: ckg93 on November 18, 2010, 12:01:06 pm
Hydrogen gas is not flammable, if u put down flammable u will get it wrong, it is combustible, DEFINITELY not flammable lol.
What about if we say it goes BOOM when around an open flame/ spark... I'm pretty sure the VCAA would not get bogged down in the definition difference between combustible and flammable...