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#### redpanda83

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##### Re: VCE Chemistry Question Thread
« Reply #8205 on: October 18, 2019, 09:05:48 pm »
0
Oooh. That makes more sense. Thank you!
if you read the data booklet, it literally explains it
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#### dev_xy

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##### Re: VCE Chemistry Question Thread
« Reply #8206 on: October 18, 2019, 09:56:03 pm »
0
From what I remember haven’t done chem in a while lol, what you described is the enthalpy of the reaction, which can be both either positively (endothermic reaction) or negative (exothermic reaction). However, the molar heat of combustion is different, and is - by convention - expressed as an absolute value.

what do you mean by 'absolute value"?

#### angrybiscuit

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##### Re: VCE Chemistry Question Thread
« Reply #8207 on: October 18, 2019, 10:23:08 pm »
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Your understanding is correct, it is a mistake on VCAA's part. Delta Hc is always a positive value.

This is interesting. I've had a loong argument with my teachers regarding the fact that they would mark down students for not putting a negative value for Delta Hc. Still angry
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#### hums_student

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##### Re: VCE Chemistry Question Thread
« Reply #8208 on: October 18, 2019, 10:39:10 pm »
+3
what do you mean by 'absolute value"?

An absolute value means it's always positive.
Basically | -x | = x
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#### EllingtonFeint

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##### Re: VCE Chemistry Question Thread
« Reply #8209 on: October 19, 2019, 05:42:50 pm »
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Hey hey,
-Do we need to know about Normal Phase HPLC vs Reverse Phase HPLC??
-What exactly is an analyte? Is it, like, synonymous with molecule?

-I do not understand Q 5 C i on the 2018 exam. Could somebody please talk me through it? (I don't really understand the formulas used in the examiner's report)

-Q 8. Woah. I have NO IDEA what is going on there. That's a fuel cell right?? Could somebody please direct me to like a super basic youtube video or something cos my knowledge is SERIOUSLY lacking here
I have no idea what the polarity is at Electrode W or Y, what happens in an alkaline environment (and for that matter an acidic environ) and ahh I am so sorry for probably annoying everybody on here with my incessant questions.
« Last Edit: October 19, 2019, 06:27:19 pm by EllingtonFeint »
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#### turtlebanana

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##### Re: VCE Chemistry Question Thread
« Reply #8210 on: October 19, 2019, 06:19:42 pm »
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**CALORIMETRY QUESTION**

Does anyone have an explanation for why: the energy from a snack bar (take this as an example since its an exam question from neap) available for the human body to use is more than 10% lower than the amount determined by calorimetry?

I understand that during an actual experiment, practically speaking, heat is lost to the SURROUNDINGS which results in the energy transferred to the water being less than if theoretically calculated. But is energy lost when a person eats a snack bar? (like how is it lower than determined by calorimetry?) The only reason i can think of is if the person doesn't consume all of it (i.e. some crumbs or something fall off), if it melts as well then some of it can like drip away? Maybe i'm totally off haha

Appreciate any help!
« Last Edit: October 19, 2019, 06:21:16 pm by turtlebanana »

#### jollyboat

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##### Re: VCE Chemistry Question Thread
« Reply #8211 on: October 19, 2019, 06:41:55 pm »
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Hey I was wondering if anybody knows how strict VCAA marking is with states, (aq) vs (l)? For example I was doing a question where you were required to write an equation, and I wrote CH3COOH(aq) + H2O(l) ––> CH3COO–(aq) + H3O+(aq). However the examination report said to write the reactant CH3COOH as a liquid, but didn't indicate if (aq) was accepted too.

#### jkay__

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##### Re: VCE Chemistry Question Thread
« Reply #8212 on: October 19, 2019, 06:45:01 pm »
+1
**CALORIMETRY QUESTION**

Does anyone have an explanation for why: the energy from a snack bar (take this as an example since its an exam question from neap) available for the human body to use is more than 10% lower than the amount determined by calorimetry?

I understand that during an actual experiment, practically speaking, heat is lost to the SURROUNDINGS which results in the energy transferred to the water being less than if theoretically calculated. But is energy lost when a person eats a snack bar? (like how is it lower than determined by calorimetry?) The only reason i can think of is if the person doesn't consume all of it (i.e. some crumbs or something fall off), if it melts as well then some of it can like drip away? Maybe i'm totally off haha

Appreciate any help!

Not all of an energy bar might be able to be digested (cellulose, dietary fibre), and not all of the carbs/proteins/fats might be digested with 100% efficiency (not all of it is taken in by body)
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#### Erutepa

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##### Re: VCE Chemistry Question Thread
« Reply #8213 on: October 19, 2019, 09:34:15 pm »
+3
Hey hey,
-Do we need to know about Normal Phase HPLC vs Reverse Phase HPLC??
-What exactly is an analyte? Is it, like, synonymous with molecule?

-I do not understand Q 5 C i on the 2018 exam. Could somebody please talk me through it? (I don't really understand the formulas used in the examiner's report)

-Q 8. Woah. I have NO IDEA what is going on there. That's a fuel cell right?? Could somebody please direct me to like a super basic youtube video or something cos my knowledge is SERIOUSLY lacking here
I have no idea what the polarity is at Electrode W or Y, what happens in an alkaline environment (and for that matter an acidic environ) and ahh I am so sorry for probably annoying everybody on here with my incessant questions.

1) You don't need to know about the normal/reverse phase HPLC

2) An analyte is a substance which is being analysed. If you are analysing a particular molecule, then you can call it the analyte since you are analysing it.

3) sorry, I can't help with the 2018 chem exam questions as I am saving that exam to do a bit later and don't want to read through it yet haha.
However, I can say that when a question asks for the polarity of an electrode, it is asking if it is of positive or negative polarity.
To determine this, you first want to identify if it's discharging/producing voltage or recharging/receiving voltage.
- If it's discharging, the negative electrode will be where the oxidation reaction occurs (loss of electrons) and is the anode, and the positive electrode will be where the reduction reaction occurs (gain of electrons) and is the cathode.
- If its recharging/consuming voltage, the negative electrode will be where the reduction reaction occurs and will the cathode, and the positive electrode will be where the oxidation reaction occurs and will the anode

To summarise:
Discharge(producing voltage):
- Anode=Oxidation=Negative
- Cathode=Reduction=Positive
Recharge(consuming voltage):
- Anode=Oxidation=Positive
- Cathode=Reduction=Negative

In terms of Balancing in acidic and alkaline environments:
- In acidic environments, when balancing the hydrogens in a half cell reaction, you add H+ ions to the relevant side
- In alkaline environments, when balancing the hydrogens in a half cell reaction, you would add OH- ions to the relevant side. In order to work this out, you would first balance by adding H+ ions to the relevant side, but would then add a necessary amount of OH- ions to both sides such to neutralise the H+. After this, your balanced half cell equation should involve water on one side and OH- on the other.
This is probably a confusing explanation, so I will put an example here:
Write the half-equation for the oxidation of MnO2 to MnO4- in an Acidic environment
1. Balance main/key atoms
- here this is Mn, and is already balanced so:
$MnO_2 \rightarrow MnO^-_4$
2. Balance Oxygen by adding H2O
$MnO_2 + 2H_2O \rightarrow MnO^-_4$
$MnO_2 +2H_2O \rightarrow MnO^-_4+4H^+$
4. Add electrons (in order to balance out the oxidation numbers of both sides of the reaction)
$MnO_2 +2H_2O \rightarrow MnO^-_4+4H+ +3e^-$
$MnO_2(s) +2H_2O(l) \rightarrow MnO^-_4(aq)+4H^+(aq) +3e^-$
Write the half-equation for the oxidation of MnO2 to MnO4- in an alkaline environment
6. Add OH- to both sides to neutralise the H+
$MnO_2(s) +2H_2O(l) +4OH^-(a)\rightarrow MnO^-_4(aq)+4H^+(aq)+4OH^-(aq) +3e^-$
and then combine the H+ and OH- to get:
$MnO_2(s) +2H_2O(l) +4OH^-(aq)\rightarrow MnO^-_4(aq)+4H_2O(l) +3e^-$
and finally, cancel out any water on both sides
$MnO_2(s) +4OH^-(aq)\rightarrow MnO^-_4(aq)+2H_2O(l) +3e^-$

I didn't really use any videos for chem, so I can't really recommend any. I can recommend looking at the chemistry 3/4 lecture slides here: https://atarnotes.com/note/lecture-slides-september-2019-42/ as they do a good job at summarising the course, but are perhaps not the best if you are lost on a topic.

And as a side note, don't be sorry about asking so many questions. People here are happy to answer questions especially since you have a go at them and explain what you're thought processes are.
« Last Edit: October 19, 2019, 10:10:06 pm by Erutepa »
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#### colline

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##### Re: VCE Chemistry Question Thread
« Reply #8214 on: October 20, 2019, 08:59:10 pm »
+3
Hey I was wondering if anybody knows how strict VCAA marking is with states, (aq) vs (l)? For example I was doing a question where you were required to write an equation, and I wrote CH3COOH(aq) + H2O(l) ––> CH3COO–(aq) + H3O+(aq). However the examination report said to write the reactant CH3COOH as a liquid, but didn't indicate if (aq) was accepted too.

I think the only type of question where VCAA do not mark on states is electrolysis, so I would assume that only CH3COOH(l) would have been accepted.
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#### EllingtonFeint

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##### Re: VCE Chemistry Question Thread
« Reply #8215 on: October 22, 2019, 11:42:57 am »
0
I don't understand what is going on here ...

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#### EllingtonFeint

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##### Re: VCE Chemistry Question Thread
« Reply #8216 on: October 22, 2019, 02:23:23 pm »
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I think this is just looking at the change of the oxidation number in the second half equation with vanadium. We can see the oxidation number of vanadium on the left side of the second half equation is +4. The oxidation number of vanadium on the right side of the second half equation is +3.

Therefore, it is B as VO22+ has been reduced to V3+

Correct me, anyone, if I'm wrong but I feel the question was a just a little weird?

Yeah, so that's what I was thinking, but the answer is actually C (I probably should have said that at the start)...
This is what the examiner's report said:

O2(g) + 4H+(aq) + 4e– → 2 H2O(l) E0 = +1.23 V
VO2+(aq) + 2H+(aq) + e– → VO2+(aq) + H2O(l) E0 = +1.00 V
VO2+(aq) + 2H+(aq) + e– → V3+(aq) + H2O(l) E0 = +0.34 V
V3+(aq) + e– → V2+(aq) E0 = –0.26 V
The strongest reductant, V2+(aq), is oxidised to V3+(aq), which is both an oxidant and a reductant.
Subsequently as the [V2+(aq)] decreases, the next strongest reductant, V3+(aq), is oxidised to VO2+(aq) by O2.
If the reaction is allowed to continue, since VO2+(aq) is also both an oxidant and a reductant, VO2+(aq) can be oxidised to VO2+(aq) by O2(g).
So V2+(aq) can be oxidised to V3+(aq), VO2+(aq) and VO2+(aq).

I could not make sense of any of that

(I definitely think it's a weird question  )
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#### sweetcheeks

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##### Re: VCE Chemistry Question Thread
« Reply #8217 on: October 22, 2019, 03:15:16 pm »
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Yeah, so that's what I was thinking, but the answer is actually C (I probably should have said that at the start)...
This is what the examiner's report said:

O2(g) + 4H+(aq) + 4e– → 2 H2O(l) E0 = +1.23 V
VO2+(aq) + 2H+(aq) + e– → VO2+(aq) + H2O(l) E0 = +1.00 V
VO2+(aq) + 2H+(aq) + e– → V3+(aq) + H2O(l) E0 = +0.34 V
V3+(aq) + e– → V2+(aq) E0 = –0.26 V
The strongest reductant, V2+(aq), is oxidised to V3+(aq), which is both an oxidant and a reductant.
Subsequently as the [V2+(aq)] decreases, the next strongest reductant, V3+(aq), is oxidised to VO2+(aq) by O2.
If the reaction is allowed to continue, since VO2+(aq) is also both an oxidant and a reductant, VO2+(aq) can be oxidised to VO2+(aq) by O2(g).
So V2+(aq) can be oxidised to V3+(aq), VO2+(aq) and VO2+(aq).

I could not make sense of any of that

(I definitely think it's a weird question  )

Did you read the first part of the question?

#### EllingtonFeint

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##### Re: VCE Chemistry Question Thread
« Reply #8218 on: October 22, 2019, 03:18:07 pm »
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Did you read the first part of the question?

Yeah, but I still don't understand it!
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#### Erutepa

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##### Re: VCE Chemistry Question Thread
« Reply #8219 on: October 24, 2019, 07:51:05 am »
+4

Yeah, so that's what I was thinking, but the answer is actually C (I probably should have said that at the start)...
This is what the examiner's report said:

O2(g) + 4H+(aq) + 4e– → 2 H2O(l) E0 = +1.23 V
VO2+(aq) + 2H+(aq) + e– → VO2+(aq) + H2O(l) E0 = +1.00 V
VO2+(aq) + 2H+(aq) + e– → V3+(aq) + H2O(l) E0 = +0.34 V
V3+(aq) + e– → V2+(aq) E0 = –0.26 V
The strongest reductant, V2+(aq), is oxidised to V3+(aq), which is both an oxidant and a reductant.
Subsequently as the [V2+(aq)] decreases, the next strongest reductant, V3+(aq), is oxidised to VO2+(aq) by O2.
If the reaction is allowed to continue, since VO2+(aq) is also both an oxidant and a reductant, VO2+(aq) can be oxidised to VO2+(aq) by O2(g).
So V2+(aq) can be oxidised to V3+(aq), VO2+(aq) and VO2+(aq).

I could not make sense of any of that

(I definitely think it's a weird question  )
This answer is a bit odd, but
What this answer is saying is that V2+ is the strongest reductant which means it will be oxidised first. After all the V2+ is oxidised, the next strongest reductant will be oxidised (in this case V3+).
You can determine the strongest reductant by constructing an electrochemical series by listing all the possible half cell reactions from the greatest resultant voltage to the lowest. Species on the right side of the equation are reductants and increase in strength as you move further down the electrochemical series. Species on the left side of the equation are oxidants and increase in strength as you move further up the electrochemical series
I assume there was more information given in a question stem above that applied to the question, hence the 4 half cell equations provided in the answer guide.
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