VCE Chemistry Question Thread

[paper-back]

For Question 3d, VCAA 2012 Exam 2 in the "Intuitive way" method they've said that the concentration of CH3(CH4)2COO- = CH3(CH4)COOH in step 5. How did they come to assume this?

[jyce]

For Question 3d, VCAA 2012 Exam 2 in the "Intuitive way" method they've said that the concentration of CH3(CH4)2COO- = CH3(CH4)COOH in step 5. How did they come to assume this?

K_a = ([sorbate ion]equ. x [H⁺]equ.) / [sorbic acid]equ.
--> K_a / [H⁺]equ. = [sorbate ion]equ. / [sorbic acid]equ.
and K_a / [H⁺]equ. = (1.73 x 10^-5) / 10^-4.76 = 1
--> [sorbate ion]equ. / [sorbic acid]equ. also = 1
--> [sorbate ion]equ. = [sorbic acid]equ.  (Here's what you're asking)
therefore, if at equilibrium the concentration of the conjugate base is the same as the concentration of the acid, the acid has ionised by 50%.

I'd be thoroughly impressed by anyone who comes across this line of thought, especially under the pressure of exam conditions and considering this is the only question of its kind.

[paper-back]

K_a = ([sorbate ion]equ. x [H⁺]equ.) / [sorbic acid]equ.
--> K_a / [H⁺]equ. = [sorbate ion]equ. / [sorbic acid]equ.
and K_a / [H⁺]equ. = (1.73 x 10^-5) / 10^-4.76 = 1
therefore, [sorbate ion]equ. / [sorbic acid]equ. also = 1
--> [sorbate ion]equ. = [sorbic acid]equ.  (Here's what you're asking)
therefore, if at equilibrium the concentration of the conjugate base is the same as the concentration of the acid, the acid has ionised by 50%.

I'd be thoroughly impressed by anyone who comes across this line of thought, especially under the pressure of exam conditions and considering this is the only question of its kind.

Oh wow, I usually just input 10^-4.76 into my calculator straight as it is and didn't relise it equals to 1.73 x 10^-5
Thanks!

[banaidil]

helloo, 2011 exam 1 Q4 has me confused
I dont under stand why we multiply the mole of percipitate by 1/2 and not by 2. can you please explain to me why?

[banaidil]

helloo, 2011 exam 1 Q4 has me confused
I dont under stand why we multiply the mole of percipitate by 1/2 and not by 2. can you please explain to me why?

[jyce]

helloo, 2011 exam 1 Q4 has me confused
I dont under stand why we multiply the mole of percipitate by 1/2 and not by 2. can you please explain to me why?

The precipitate (MgNH₄PO₄.6H₂O) has one phosphorous atom per molecule, while P₂O₅ has two phosphorous atoms per molecule. Therefore, one molecule of P₂O₅, with its two phosphorous atoms, will form two precipitate molecules, as each precipitate molecule only needs one phosphorous atom. In other words, for every two moles of precipitate that forms, only one mole of P₂O₅ has reacted.
So, the mole of P₂O₅ is half, NOT double, the mole of precipitate.

Hope this makes sense!

[bonjour-sarah]

Yes, my bet for the last question is either calorimetry or electrolysis. But don't hold me to these predictions! And you're welcome 

In terms of this question~ As it is a relatively new thing, I had just assumed it was like last year , analysing the errors and explaining the improvement. But when you say experimental design, could we be asked to design something from scratch? (Within reason of course ) . What is a good method of going about these types of questions, even the shorter ones, ie. for calorimetry or AAS in the past, as I never know what to/not to include / how specific to be? Thanks again!!!!

[jyce]

In terms of this question~ As it is a relatively new thing, I had just assumed it was like last year , analysing the errors and explaining the improvement. But when you say experimental design, could we be asked to design something from scratch? (Within reason of course ) . What is a good method of going about these types of questions, even the shorter ones, ie. for calorimetry or AAS in the past, as I never know what to/not to include / how specific to be? Thanks again!!!!

I highly doubt you'd be asked to come up with an entire experiment all on your own. As for what to write about, there seems to be two common mistakes to look out for: the experimenter writes a chemical reaction incorrectly (e.g. see the sample exam), and the experimenter wants to study the impact of a particular variable but manipulates multiple variables at once. For example, say we wanted to determine the impact of x, y and z on another variable a and we designed the following two conditions:
Condition 1: x = 1, y = 1, z = 1 and a = 5
Condition 2: x = 2, y = 2, z = 2 and a = 10
The student might then conclude that increasing z increases a. However, this might not be correct: because x, y and z were all changed together we can't determine whether the impact on a was due to one variable, two or a combination of x, y and z. You should be able to see parallels between this example and last year's exam. Also, when it comes to making improvements to the experiment, "conducting more trials" or "examining the impact of more variables" is always an easy suggestion to make. It's hard to give you any more specific advice as this is a relatively new topic to the exams and there's not many examples for me to draw upon.

Note that only the 2014 and sample VCAA exams have asked you to critique an experimental design as the last question of Section B. In the 2013 exam, you were instead asked to critique a set of student's notes about catalysts, so this year it could be something like this! Either way, 'critiquing' seems to be the theme of the last question.

[banaidil]

Hope this makes sense!

THANK YOUU, this helps alot

[banaidil]

Hope this makes sense!

THANK YOUU, this helps alot

[bonjour-sarah]

Which terminal do we connect our anode/ cathode to? Here's the question that confused me, because maybe I'm misinterpreting something. I had thought it was always negative to positive, hence anode to positive? But I don't think that's right

A student constructs a pile of discs by alternating ten copper discs and ten zinc discs separated with filter paper soaked in saline solution. Wires are connected to both ends of the arrangement and also attached to a voltmeter.
a) Which type of disc should be connected to the negative terminal of the voltmeter?
Answer:

Spoiler

: Zinc

[sunshine98]

Which terminal do we connect our anode/ cathode to? Here's the question that confused me, because maybe I'm misinterpreting something. I had thought it was always negative to positive, hence anode to positive? But I don't think that's right

A student constructs a pile of discs by alternating ten copper discs and ten zinc discs separated with filter paper soaked in saline solution. Wires are connected to both ends of the arrangement and also attached to a voltmeter.
a) Which type of disc should be connected to the negative terminal of the voltmeter?
Answer:

Spoiler

: Zinc

In electrolysis , you attach +ve to +ve and -ve to -ve.

Usually oxidation would take place with the Zn--> Zn2+ +2e , making it the anode and hence the negative electrode. 
But now you want to add electrons into this to drive reaction backwards. And hence you need the negative terminal to supply the electrons to the negative electrode.

[bonjour-sarah]

In electrolysis , you attach +ve to +ve and -ve to -ve.

Usually oxidation would take place with the Zn--> Zn2+ +2e , making it the anode and hence the negative electrode. 
But now you want to add electrons into this to drive reaction backwards. And hence you need the negative terminal to supply the electrons to the negative electrode.

But if there's a voltmeter and no battery pack, how can I assume electrolysis and not galvanic?

[alexathans10]

Does anyone have a copy of the neap 2015 trial exam units 3 & 4 answers?

[alexathans10]

Does anyone have a copy of the neap 2015 trial exam units 3 & 4 answers?