Login

Welcome, Guest. Please login or register.

November 06, 2025, 02:31:50 am

Author Topic: Chemistry 3/4 2013 Thread  (Read 449330 times)  Share 

0 Members and 1 Guest are viewing this topic.

xenial

  • Guest
Re: Chemistry 3/4 2013 Thread
« Reply #1215 on: August 21, 2013, 07:37:08 am »
0
Okay thankyou!

So basically, it's like methods in that you shouldn't waste time making notes on the subject, you should just start doing prac exams and that'll cover any holes you have in your knowledge?

Edward21

  • Victorian
  • Forum Leader
  • ****
  • Posts: 625
  • Don't ask me, all I do is calculate pH.
  • Respect: +27
Re: Chemistry 3/4 2013 Thread
« Reply #1216 on: August 23, 2013, 10:30:15 pm »
0
With AAS, does it use both the Visible AND UV regions of the EM spectrum??
2012 Biology [44] 2013 Chemistry [50] Italian [38] English [48] Health & HD [45] Methods [34] ATAR: 99.10

2014-2016 Bachelor of Biomedicine - The University of Melbourne


Alwin

  • Victorian
  • Forum Leader
  • ****
  • Posts: 838
  • Respect: +241
Re: Chemistry 3/4 2013 Thread
« Reply #1217 on: August 24, 2013, 12:04:15 am »
0
With AAS, does it use both the Visible AND UV regions of the EM spectrum??

Remember that AAS uses a cathode lamp, with the metal specifically chosen :) So, if the metal absorbs strongly in the UV region of the EM spec then the lamp will also emit this specific wavelength.
However, the (monochrome) detector is usually only set to one wavelength that the cathode lamp emits strongly. So in theory, yes, AAS can use UV regions of the EM spec, but most metals (not all) emit a wavelength in the visible region, so the detector is set to detect this.

Hope it helps ^^
2012:  Methods [48] Physics [49]
2013:  English [40] (oops) Chemistry [46] Spesh [42] Indo SL [34] Uni Maths: Melb UMEP [4.5] Monash MUEP [just for a bit of fun]
2014:  BAeroEng/BComm

A pessimist says a glass is half empty, an optimist says a glass is half full.
An engineer says the glass has a safety factor of 2.0

Edward21

  • Victorian
  • Forum Leader
  • ****
  • Posts: 625
  • Don't ask me, all I do is calculate pH.
  • Respect: +27
Re: Chemistry 3/4 2013 Thread
« Reply #1218 on: August 24, 2013, 07:56:48 am »
0
Remember that AAS uses a cathode lamp, with the metal specifically chosen :) So, if the metal absorbs strongly in the UV region of the EM spec then the lamp will also emit this specific wavelength.
However, the (monochrome) detector is usually only set to one wavelength that the cathode lamp emits strongly. So in theory, yes, AAS can use UV regions of the EM spec, but most metals (not all) emit a wavelength in the visible region, so the detector is set to detect this.

Hope it helps ^^
Thanks, damn TSFX practice paper -_-
2012 Biology [44] 2013 Chemistry [50] Italian [38] English [48] Health & HD [45] Methods [34] ATAR: 99.10

2014-2016 Bachelor of Biomedicine - The University of Melbourne


teletubbies_95

  • Victorian
  • Forum Leader
  • ****
  • Posts: 691
  • heartbreaker <3 JB
  • Respect: +24
  • School: Mac.Rob
  • School Grad Year: 2013
Re: Chemistry 3/4 2013 Thread
« Reply #1219 on: August 24, 2013, 10:41:22 am »
0
Hey guys !
I was doing vcaa 2012 , I was having trouble on determining what to include in my answer for question 3 (c) in the short answer?
2012: Psychology(46) Biology (44)
2013: Chem(41)---EngLang(44)--HealthnHuman(47)---Methods(41)--DEAKIN PSYCH(4.5)
ATAR=99.10! :) umat=94ile
i liek lala :) arre bhaiya aal izz well :) <3

xenial

  • Guest
Re: Chemistry 3/4 2013 Thread
« Reply #1220 on: August 24, 2013, 10:48:34 am »
0
Thanks!

Also how much do we need to know about batteries? Would we be able to need to know the differences between them - what different ones are used for? Be able to draw them etc.?

Homer

  • Victorian
  • Forum Obsessive
  • ***
  • Posts: 431
  • Respect: +10
Re: Chemistry 3/4 2013 Thread
« Reply #1221 on: August 25, 2013, 02:34:22 pm »
0
The following equations form part of the electrochemical series. They are ranked in the order shown.
Ag+(aq) + e– (Forward and back arrow) Ag(s)
Pb2+(aq) + 2e– (Forward and back arrow) Pb(s)
Fe2+(aq) + 2e– (Forward and back arrow) Fe(s)
Zn2+(aq) + 2e– (Forward and back arrow) Zn(s)
Mg2+(aq) + 2e– (Forward and back arrow) Mg(s)

Lead rods are placed in solutions of silver nitrate, iron(lI) sulfate and magnesium chloride. In which solutions would you expect to see a coating of another metal form on the lead rod? Explain

Coating? Isn't that similar to electrolysis? So for Lead, shouldn't something below it be used? Fe2+? since is is the stongest oxidant below solid lead?

ANS: Ag+ ions :s
Bachelor of Laws/Engineering

2013 ATAR: 98.65

Specialist Maths [53.06] Maths Methods [48.83] Physics [48.22]

Donuts. Is there anything they can't do?

thushan

  • ATAR Notes Lecturer
  • Honorary Moderator
  • ATAR Notes Legend
  • *******
  • Posts: 4959
  • Respect: +626
Re: Chemistry 3/4 2013 Thread
« Reply #1222 on: August 25, 2013, 02:40:41 pm »
+1
Not on this occasion. This is a simple spontaneous redox reaction between a metal and a metal ion - there is no external current added here.

Ag+ is the only oxidant with the potential to oxidise the Pb (s) in the lead. In doing so, the Ag+ is reduced and you get an Ag deposit.
Managing Director  and Senior Content Developer - Decode Publishing (2020+)
http://www.decodeguides.com.au

Basic Physician Trainee - Monash Health (2019-)
Medical Intern - Alfred Hospital (2018)
MBBS (Hons.) - Monash Uni
BMedSci (Hons.) - Monash Uni

Former ATARNotes Lecturer for Chemistry, Biology

Homer

  • Victorian
  • Forum Obsessive
  • ***
  • Posts: 431
  • Respect: +10
Re: Chemistry 3/4 2013 Thread
« Reply #1223 on: August 25, 2013, 02:56:52 pm »
0
Not on this occasion. This is a simple spontaneous redox reaction between a metal and a metal ion - there is no external current added here.

Ag+ is the only oxidant with the potential to oxidise the Pb (s) in the lead. In doing so, the Ag+ is reduced and you get an Ag deposit.

but in the cell formed from Ag+/Ag and Pb/Pb2+ half cells, wouldn't the electron flow from Pb electrode to the Ag+/Ag half cells? So shouldn't the Pb electrode disintegrate over time instead of being coated by another metal?

Another question:
"lead would be able to reduce which ions from the electrochemical series.." What do they mean when they say lead would be able to reduce?"
« Last Edit: August 25, 2013, 03:12:10 pm by Homer »
Bachelor of Laws/Engineering

2013 ATAR: 98.65

Specialist Maths [53.06] Maths Methods [48.83] Physics [48.22]

Donuts. Is there anything they can't do?

thushan

  • ATAR Notes Lecturer
  • Honorary Moderator
  • ATAR Notes Legend
  • *******
  • Posts: 4959
  • Respect: +626
Re: Chemistry 3/4 2013 Thread
« Reply #1224 on: August 25, 2013, 03:12:13 pm »
0
but in the cell formed from Ag+/Ag and Pb/Pb2+ half cells, wouldn't the electron flow from Pb electrode to the Ag+/Ag half cells? So shouldn't the Pb electrode disintegrate over time instead of being coated by another metal?

Not exactly the half cell, but that's essentially what happens. The surface Pb is replaced with Ag. So it takes on the appearance of a coat.
Managing Director  and Senior Content Developer - Decode Publishing (2020+)
http://www.decodeguides.com.au

Basic Physician Trainee - Monash Health (2019-)
Medical Intern - Alfred Hospital (2018)
MBBS (Hons.) - Monash Uni
BMedSci (Hons.) - Monash Uni

Former ATARNotes Lecturer for Chemistry, Biology

Jeggz

  • Victorian
  • Forum Obsessive
  • ***
  • Posts: 493
  • Respect: +42
  • School: Presbyterian Ladies' College
  • School Grad Year: 2013
Re: Chemistry 3/4 2013 Thread
« Reply #1225 on: August 25, 2013, 03:43:24 pm »
0
For equilibrium, do we need to know what the concentration fraction or Qc is?
Melbourne University - Commerce; Actuarial Studies.

Tutoring 2015 - Email/PM for places!

lzxnl

  • Victorian
  • ATAR Notes Legend
  • *******
  • Posts: 3432
  • Respect: +215
Re: Chemistry 3/4 2013 Thread
« Reply #1226 on: August 25, 2013, 04:15:22 pm »
0
For equilibrium, do we need to know what the concentration fraction or Qc is?

What do you mean? You need to know how to work it out, that's for sure.
2012
Mathematical Methods (50) Chinese SL (45~52)

2013
English Language (50) Chemistry (50) Specialist Mathematics (49~54.9) Physics (49) UMEP Physics (96%) ATAR 99.95

2014-2016: University of Melbourne, Bachelor of Science, Diploma in Mathematical Sciences (Applied Maths)

2017-2018: Master of Science (Applied Mathematics)

2019-2024: PhD, MIT (Applied Mathematics)

Accepting students for VCE tutoring in Maths Methods, Specialist Maths and Physics! (and university maths/physics too) PM for more details

Jeggz

  • Victorian
  • Forum Obsessive
  • ***
  • Posts: 493
  • Respect: +42
  • School: Presbyterian Ladies' College
  • School Grad Year: 2013
Re: Chemistry 3/4 2013 Thread
« Reply #1227 on: August 25, 2013, 04:25:00 pm »
0
What's the difference between Qc and Kc though? I'm kinda confused :\
Melbourne University - Commerce; Actuarial Studies.

Tutoring 2015 - Email/PM for places!

thushan

  • ATAR Notes Lecturer
  • Honorary Moderator
  • ATAR Notes Legend
  • *******
  • Posts: 4959
  • Respect: +626
Re: Chemistry 3/4 2013 Thread
« Reply #1228 on: August 25, 2013, 04:32:28 pm »
+1
What's the difference between Qc and Kc though? I'm kinda confused :\

Qc is the concentration fraction at any point in time, Kc is the value that the concentration fraction would be if and only if the system were in equilibrium.
Managing Director  and Senior Content Developer - Decode Publishing (2020+)
http://www.decodeguides.com.au

Basic Physician Trainee - Monash Health (2019-)
Medical Intern - Alfred Hospital (2018)
MBBS (Hons.) - Monash Uni
BMedSci (Hons.) - Monash Uni

Former ATARNotes Lecturer for Chemistry, Biology

Alwin

  • Victorian
  • Forum Leader
  • ****
  • Posts: 838
  • Respect: +241
Re: Chemistry 3/4 2013 Thread
« Reply #1229 on: August 25, 2013, 06:15:25 pm »
+2
You also can use Qc (sometimes also called CF, concentration fraction which is pretty dumb since my teacher uses CF to mean calibration factor too) to determine if the system is at equilibrium or not + which way it will shift. Also, some ppl write Keq for some reason unbeknownst to me. anways:

eg, Qc = Kc its at equilibrium. Qc > Kc, POE will move to the left and if Qc < Kc then POE moves to the right

:)
2012:  Methods [48] Physics [49]
2013:  English [40] (oops) Chemistry [46] Spesh [42] Indo SL [34] Uni Maths: Melb UMEP [4.5] Monash MUEP [just for a bit of fun]
2014:  BAeroEng/BComm

A pessimist says a glass is half empty, an optimist says a glass is half full.
An engineer says the glass has a safety factor of 2.0