Login

Welcome, Guest. Please login or register.

March 29, 2024, 12:01:36 am

Author Topic: VCE Chemistry Question Thread  (Read 2313126 times)  Share 

0 Members and 9 Guests are viewing this topic.

Calebark

  • biscuits of disappointment
  • National Moderator
  • ATAR Notes Superstar
  • *****
  • Posts: 2670
  • Respect: +2741
Re: VCE Chemistry Question Thread
« Reply #9075 on: April 08, 2021, 12:47:05 pm »
+1
As you go down a group does the effective nuclear charge increase slightly or does it remain relatively constant 
If iodine has a higher effective nuclear charge then wouldn’t it have a smaller atomic radius

Effective nuclear charge (Zeff) decreases down a group. However you are correct that the Zeff increases as atomic radius decreases. As you go down, remember there are more shells of electron. Each inner shell can shield the valence shell from the nucleus' positive attraction, allowing the valence electrons to be further out increasing atomic radii. This same effect decreases the effective nuclear charge.

I'll also add that Zeff increases across a period. Can you figure why? It's similar to the above.
Spoiler
The number of protons/valence increase but number of shells remain the same, meaning the level of shielding is the same despite the additional attraction, so Zeff increases → radii decreases
🐢A turtle has flippers and a tortoise has clubs🐢

shahifa

  • Adventurer
  • *
  • Posts: 24
  • Respect: 0
Re: VCE Chemistry Question Thread
« Reply #9076 on: April 24, 2021, 02:20:33 pm »
0
Hey Guys!

I was wondering if anyone had any practice tests or quizzes for the topics of Metals and Ionic Compounds. I have a SAC coming up, and I've done all my revision and memorisation and some short quizzes, but if anyone had anything extra they could send, that would be great! If not, thats fine! (btw this is Unit 1 Chemistry)

Thanks
Shahifa

Owlbird83

  • BLAA 2020
  • Victorian Moderator
  • Forum Leader
  • *****
  • Posts: 553
  • Respect: +785
Re: VCE Chemistry Question Thread
« Reply #9077 on: April 24, 2021, 02:55:44 pm »
+4
Hey Guys!

I was wondering if anyone had any practice tests or quizzes for the topics of Metals and Ionic Compounds. I have a SAC coming up, and I've done all my revision and memorisation and some short quizzes, but if anyone had anything extra they could send, that would be great! If not, thats fine! (btw this is Unit 1 Chemistry)

Thanks
Shahifa

Hi Shahifa,
Sorry it's against forum rules for people to share/distribute any copyright practice sacs/etc.
Perhaps you could try see if your local library might have a copy of chem 1/2 checkpoints that you could borrow? (i found this helpful sometimes to get extra revision during yr11). Also, maybe you could try seeing if your teacher/school has extra revision materials?
2018: Biology
2019: Chemistry, Physics, Math Methods, English, Japanese
2020: Bachelor of Psychology (Monash)

shahifa

  • Adventurer
  • *
  • Posts: 24
  • Respect: 0
Re: VCE Chemistry Question Thread
« Reply #9078 on: April 24, 2021, 06:23:34 pm »
+3
Hi Shahifa,
Sorry it's against forum rules for people to share/distribute any copyright practice sacs/etc.
Perhaps you could try see if your local library might have a copy of chem 1/2 checkpoints that you could borrow? (i found this helpful sometimes to get extra revision during yr11). Also, maybe you could try seeing if your teacher/school has extra revision materials?

Oh sorry I didn't know, but thanks for the tips!

vehura

  • Trailblazer
  • *
  • Posts: 43
  • she/her
  • Respect: +33
Re: VCE Chemistry Question Thread
« Reply #9079 on: April 26, 2021, 12:11:27 pm »
0
Hi everyone!

I'm having some trouble getting my head around Le Chatelier's principle, specifically how changes to the system favours certain reactions. I was given a question today:

Equation 1: Hb(aq) + 4O2(aq) ⇌ Hb(O2)4(aq)
Equation 2: Hb(aq) + 4CO(aq) ⇌ Hb(CO)4

Patients suffering from carbon monoxide poisoning are treated with pure oxygen (hyperbaric) to remove CO from the Hb. With reference to Le Chatelier's theory, explain how this treatment works.


I wrote this:

Pure oxygen will be added to the system. Le Chatelier's theory states that the system will attempt to favour a reaction in order to counteract a change to the system. Therefore, to counteract an increase of pure oxygen, the system will favour the reverse reaction in order to use up the CO bound to the Hb.

But this doesn't really make sense... I know that you will have to favour the reverse reaction in order to remove the CO, but if the oxygen was added to the reactants side then you'd have to favour the forward reaction to remove it but then you'd just make more CO! If someone could explain this I would be really grateful.  :)

Thanks!
class of 2021
2020: psych (50)
2021: eng (50) lit (47)

Bri MT

  • VIC MVP - 2018
  • Administrator
  • ATAR Notes Legend
  • *****
  • Posts: 4719
  • invest in wellbeing so it can invest in you
  • Respect: +3677
Re: VCE Chemistry Question Thread
« Reply #9080 on: April 26, 2021, 12:52:51 pm »
+7
Hi everyone!

I'm having some trouble getting my head around Le Chatelier's principle, specifically how changes to the system favours certain reactions. I was given a question today:

Equation 1: Hb(aq) + 4O2(aq) ⇌ Hb(O2)4(aq)
Equation 2: Hb(aq) + 4CO(aq) ⇌ Hb(CO)4

Patients suffering from carbon monoxide poisoning are treated with pure oxygen (hyperbaric) to remove CO from the Hb. With reference to Le Chatelier's theory, explain how this treatment works.


I wrote this:

Pure oxygen will be added to the system. Le Chatelier's theory states that the system will attempt to favour a reaction in order to counteract a change to the system. Therefore, to counteract an increase of pure oxygen, the system will favour the reverse reaction in order to use up the CO bound to the Hb.

But this doesn't really make sense... I know that you will have to favour the reverse reaction in order to remove the CO, but if the oxygen was added to the reactants side then you'd have to favour the forward reaction to remove it but then you'd just make more CO! If someone could explain this I would be really grateful.  :)

Thanks!


Hi!

So what we want to do here is uncouple the CO from the Hb to make the Hb available for oxygen. If we externally added CO that would be a problem because it would drive reaction 2 forwards, but really what we care about is reducing Hb(CO4) - we don't really care about the CO just being there if it isn't taking up our Hb.

Not sure if I've interpreted your confusion correctly, please speak up if you're still confused :)

Owlbird83

  • BLAA 2020
  • Victorian Moderator
  • Forum Leader
  • *****
  • Posts: 553
  • Respect: +785
Re: VCE Chemistry Question Thread
« Reply #9081 on: April 26, 2021, 01:20:36 pm »
+6
Hi everyone!

I'm having some trouble getting my head around Le Chatelier's principle, specifically how changes to the system favours certain reactions. I was given a question today:

Equation 1: Hb(aq) + 4O2(aq) ⇌ Hb(O2)4(aq)
Equation 2: Hb(aq) + 4CO(aq) ⇌ Hb(CO)4

Patients suffering from carbon monoxide poisoning are treated with pure oxygen (hyperbaric) to remove CO from the Hb. With reference to Le Chatelier's theory, explain how this treatment works.


I wrote this:

Pure oxygen will be added to the system. Le Chatelier's theory states that the system will attempt to favour a reaction in order to counteract a change to the system. Therefore, to counteract an increase of pure oxygen, the system will favour the reverse reaction in order to use up the CO bound to the Hb.

But this doesn't really make sense... I know that you will have to favour the reverse reaction in order to remove the CO, but if the oxygen was added to the reactants side then you'd have to favour the forward reaction to remove it but then you'd just make more CO! If someone could explain this I would be really grateful.  :)

Thanks!

Hey vehura, you might find this thread on carbon monoxide poisoning helpful.
I feel like the combined equation makes it a lot easier to understand what's happening. You are adding more on the right side, so in order to counteract the change, the equilibrium shifts towards the left, the side with more O2 bonded to the haemoglobin.
Hb(O2)4(aq)+4CO(aq)⇌Hb(CO)4(aq)+4O2(aq)

I think instead of thinking about it as 'using up' the HbCO, it would be better to think about it like how the Hb much rather prefers bonding to CO  (so there is a much larger amount of HbCO than HbO) but when you flood the system with oxygen, some of the Hb bonds with bonds oxygen instead of carbon monoxide. The increased amount of reactants on the left makes the reverse reaction become favoured.
My chem teacher used an analogy iirc with Haemoglobin being a person who has a much higher chance of picking toxic people that don't treat it well (CO) to hang with, despite there being nice, caring people (O2). We if we flood their life with the nice O2 people then it kind of increases the likelihood that they will hang with more nice people than toxic people.
I hope this isn't too confusing i'm not sure if i've phrased it clear enough, it's difficult to get straight away, i recommend that thread though it has some really good explanations!

I personally think your answer would be fine, but maybe remove the 'in order to use up the CO part'? Also maybe specify that you mean equation2 when you are talking about it favouring the reverse reaction.

edit: Bri beat me hehe, i'll leave in case it helps having it explained different ways
« Last Edit: April 26, 2021, 01:24:01 pm by Owlbird83 »
2018: Biology
2019: Chemistry, Physics, Math Methods, English, Japanese
2020: Bachelor of Psychology (Monash)

vehura

  • Trailblazer
  • *
  • Posts: 43
  • she/her
  • Respect: +33
Re: VCE Chemistry Question Thread
« Reply #9082 on: April 26, 2021, 02:08:19 pm »
+1
Hi!
Hey vehura,

Thank you so much Bri MT! Your explanation was really helpful - I realise I was not thinking of the equations as part of the same reaction. And thanks so much for the detailed explanation and link to the thread Owlbird - I really appreciate it. The analogy is really useful also <3
class of 2021
2020: psych (50)
2021: eng (50) lit (47)

amanaazim

  • Forum Regular
  • **
  • Posts: 80
  • Respect: 0
Re: VCE Chemistry Question Thread
« Reply #9083 on: April 26, 2021, 07:54:53 pm »
0
hey so can someone explain the pH in fuel cells . if its a hydrogen oxygen fuel cell and the electrolyte is h+ ions then is the pH Increasing or decreasing.

same with a potassium hydroxide been an electrolyte of a methanol fuel cell would the pH near the cathode increase or decrease.

is it true overall pH stays the same but at anode and cathodes its different can someone explain please.
thank you.

thenuttyprofessor

  • Trailblazer
  • *
  • Posts: 48
  • Respect: 0
Re: VCE Chemistry Question Thread
« Reply #9084 on: April 27, 2021, 04:59:11 pm »
0
What is the mark breakdown per topic for the exam in chem generally?

Corey King

  • Trendsetter
  • **
  • Posts: 133
  • Respect: +3
Re: VCE Chemistry Question Thread
« Reply #9085 on: April 28, 2021, 06:06:52 pm »
0
Hey guys,
I'm wondering why Hydrogen and Oxygen spontaneously react in a fuel cell, but out in the the atmosphere, there is no reaction.
What causes this reaction to occur in the fuel cell and not occur in the air?

Sine

  • Werewolf
  • National Moderator
  • Great Wonder of ATAR Notes
  • *****
  • Posts: 5135
  • Respect: +2103
Re: VCE Chemistry Question Thread
« Reply #9086 on: April 28, 2021, 06:18:02 pm »
+3
What is the mark breakdown per topic for the exam in chem generally?
Hope this helps :)

https://jameskennedymonash.wordpress.com/2020/02/08/vce-chemistry-written-examination-paper-analysis-of-marks-allocated-per-chapter/

Red topics are high-yield, however this is just based on past exams

Corey King

  • Trendsetter
  • **
  • Posts: 133
  • Respect: +3
Re: VCE Chemistry Question Thread
« Reply #9087 on: April 28, 2021, 06:38:32 pm »
0
In a fuel cell,
What causes the fuel to be oxidised?
Is it simple diffusion across a "membrane", or are the electrons just stripped from some redox force across the electrode wire?

Erutepa

  • VIC MVP - 2019
  • Forum Leader
  • ****
  • Posts: 721
  • evenin'
  • Respect: +775
Re: VCE Chemistry Question Thread
« Reply #9088 on: May 02, 2021, 10:22:30 am »
+5
In a fuel cell,
What causes the fuel to be oxidised?
Is it simple diffusion across a "membrane", or are the electrons just stripped from some redox force across the electrode wire?
fuel cells work essentially the same way as your standard galvanic cells - after all they are a galvanic cell. You have a separation of reactants at two electrodes such that (given the reactants react spontaneously) one reactant will be oxidized at the anode which provides electrons to the cathode to reduce the other reactant. So In a fuel cell its a spontaneous redox reaction occuring that results in oxidation at the anode as well as reduction at the cathode. Hopefully this helps :)
Qualifications
 > Have counted to 227
 > Can draw really good spiders
 > 2 Poet points
 > 6.5 insanipi points
 > 1 Bri MT point

Corey King

  • Trendsetter
  • **
  • Posts: 133
  • Respect: +3
Re: VCE Chemistry Question Thread
« Reply #9089 on: May 02, 2021, 12:23:54 pm »
0
fuel cells work essentially the same way as your standard galvanic cells - after all they are a galvanic cell. You have a separation of reactants at two electrodes such that (given the reactants react spontaneously) one reactant will be oxidized at the anode which provides electrons to the cathode to reduce the other reactant. So In a fuel cell its a spontaneous redox reaction occuring that results in oxidation at the anode as well as reduction at the cathode. Hopefully this helps :)

Thanks Erutepa,

I remember when I asked this question that I read in the textbook that a catalyst is present that helps the oxidation occur in the anode. I used to just think that the electro-something forces caused the electrons to leave through the wire, directly from the reductant, like being pulled by a magnet. But this reading made it sound like the electrons are stripped, and they kind of pool, and this higher rate of reaction toward that side of the equilibrium leads to an imbalance in the number of electrons at each half cell, and then at some point the electrons will begin moving.

So, if that's true, then what is causing the oxidation exactly?

Hey guys,

I received this question from my teacher and gave this answer: https://snipboard.io/CU8IiZ.jpg
The correct answer was this: https://snipboard.io/FroaA7.jpg
My intuition told me that this would be the case, but I couldn't explain to myself why.

Do all lone atoms of an element group together in twos? Why is this pattern important and what makes it occur?

Many thanks,
Corey

Mod Edit: Merged double post
« Last Edit: May 02, 2021, 09:33:37 pm by Erutepa »