• YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page

Predictions for Chemistry 2014 HSC? (7 Viewers)

SuchSmallHands

Well-Known Member
Joined
Nov 13, 2012
Messages
1,391
Gender
Female
HSC
2014
B
B
C
A
D
D
A
A
B
B
D
C
B
C
C
B
C
B
B
C

LOL same as SSH
Since all three of us got pretty much the same answers these are probably correct (unless all three of us overlooked something) for anyone looking for the answers later
 

SuchSmallHands

Well-Known Member
Joined
Nov 13, 2012
Messages
1,391
Gender
Female
HSC
2014
Okay 2011 is opening now. I'll do those 3:
6. D
7. C
20. B (better would be hexan-3-ol but whatever)
 

QZP

Well-Known Member
Joined
Oct 7, 2013
Messages
839
Gender
Undisclosed
HSC
2014
Explain why ozone gas is more reactive than oxygen gas (2)
 
Joined
Feb 16, 2014
Messages
2,258
Gender
Male
HSC
2014
Since all three of us got pretty much the same answers these are probably correct (unless all three of us overlooked something) for anyone looking for the answers later
Nope we are all correct, i just realised i had the 2010 paper along with the solutions.
 

SuchSmallHands

Well-Known Member
Joined
Nov 13, 2012
Messages
1,391
Gender
Female
HSC
2014
how did u get 6d?
can u please explain
Each O carries a charge of 2-. Therefore the two together have a net negative charge of 4-. V must be in the 5+ state to give the ion an overall charge of 1+. It becomes VO2+. The O has a charge of 2-. To give net charge of 2+ V must be in 4+ oxidation state.
 

timeflies

Active Member
Joined
Nov 12, 2013
Messages
211
Gender
Male
HSC
2014
JRAHS 2011 Trials MCQ answers (my own and not confirmed):
1C 2C 3B 4C 5A 6D 7C 8D 9C 10B 11A 12C 13D 14D 15B 16C 17C 18A 19B 20B

Correct answers to incorrect answers above:
2A 8C
I think you got trolled in question 5. A says that the MASS number is greater than 83, it would be right if it said atomic number. So I think the best answer here would be D haha?
 
  • Like
Reactions: QZP

QZP

Well-Known Member
Joined
Oct 7, 2013
Messages
839
Gender
Undisclosed
HSC
2014
Life is pointless :'(

EDIT: Well, looking at that question again it's quite stupid because n: p ratio is irrelevant once Z>=83.
 
Last edited:

enigma_1

~~~~ Miss Cricket ~~~~
Joined
Feb 27, 2013
Messages
4,281
Location
Lords
Gender
Female
HSC
2014
guys help how do you upload photos here?

The thing isn't working for me :(
 

timeflies

Active Member
Joined
Nov 12, 2013
Messages
211
Gender
Male
HSC
2014
Life is pointless :'(

EDIT: Well, looking at that question again it's quite stupid because n: p ratio is irrelevant once Z>=83.
Yeah all the answers were quite bad for that one tbh but D was probably the "best" one
 

QZP

Well-Known Member
Joined
Oct 7, 2013
Messages
839
Gender
Undisclosed
HSC
2014
Yeah all the answers were quite bad for that one tbh but D was probably the "best" one
Really? I would have gone with B. Calculated n: p ratio comes out to something like 1.6:1 which is not a low ratio :S

Edit: actually idk anymore I wouldn't have gone with any of them lol -_-
 
Joined
Feb 16, 2014
Messages
2,258
Gender
Male
HSC
2014
wait nevermind

Need help with this question. The solution is beside it, but what exactly do you do for the dilute part?

https://www.dropbox.com/sh/u9krve2zulw3eky/AABoea9WYPUOnyp5revb0ZZia?dl=0
ez

Okay so you know you have your equation blah blah, and you work out the concentration of the acid to be 0.33015. HOWEVER, this is the concentration of the dilute acid, we want the concentration of undilute acid. NOW, this is the best way to remember how to go from here, so the acid went from 10 to 200mL so 20x dilution right? The concentration of the dilute solution is going to be significantly lower than the undilute, so in order to get undilute, you multiply the concentration by 20.
Concentration of undilute sample is now 6.603mol/L. But it wants it in g/L.
So you do 6.603 x molecular mass of the acid/L
= ~650g/L
 
Joined
Feb 16, 2014
Messages
2,258
Gender
Male
HSC
2014
Describe how ethanol may be prepared from renewable and non-renewable resources. Include relevant equations and state any reaction conditions required. 3 marks.

Quick question guys, would you go all the way back to how ethylene is fractionally distilled from petroleum, or it is produced from catalytic cracking, OR because it's 3 marks would you just go from ethanol is produced from hydration of ethylene. (non-renewable resources way)
 

QZP

Well-Known Member
Joined
Oct 7, 2013
Messages
839
Gender
Undisclosed
HSC
2014
Renewable: Fermentation of glucose from crops
Non-renewable: Hydration of ethylene obtained from petroleum fractions or cracking

Quick question back @ you, how come you specifically mentioned catalytic cracking? :S Isn't ethylene also from thermal
 

SuchSmallHands

Well-Known Member
Joined
Nov 13, 2012
Messages
1,391
Gender
Female
HSC
2014
Describe how ethanol may be prepared from renewable and non-renewable resources. Include relevant equations and state any reaction conditions required. 3 marks.

Quick question guys, would you go all the way back to how ethylene is fractionally distilled from petroleum, or it is produced from catalytic cracking, OR because it's 3 marks would you just go from ethanol is produced from hydration of ethylene. (non-renewable resources way)
I wouldn't personally, this question already asks for so much for only three marks.

Edit: actually yeah I would. It says from non-renewable resources in particular so I'd give a quick sentence to that and move on.
 

enigma_1

~~~~ Miss Cricket ~~~~
Joined
Feb 27, 2013
Messages
4,281
Location
Lords
Gender
Female
HSC
2014
ez

Okay so you know you have your equation blah blah, and you work out the concentration of the acid to be 0.33015. HOWEVER, this is the concentration of the dilute acid, we want the concentration of undilute acid. NOW, this is the best way to remember how to go from here, so the acid went from 10 to 200mL so 20x dilution right? The concentration of the dilute solution is going to be significantly lower than the undilute, so in order to get undilute, you multiply the concentration by 20.
Concentration of undilute sample is now 6.603mol/L. But it wants it in g/L.
So you do 6.603 x molecular mass of the acid/L
= ~650g/L
yay thankss! :)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 7)

Top