September 23, 2020, 06:41:53 pm

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#### Corey King

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##### Re: VCE Chemistry Question Thread
« Reply #8775 on: September 14, 2020, 02:58:01 pm »
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I mean, what makes you think it only works for the cathode reaction?

Definitely not going to see the 6d electrons start getting added if the 6p sub-shell isn't filled. Remember that Bi is in the p-block and the 6th period, so 6p is going to be the last shell filled.

Oh okay. See, I thought that you basically counted from left to right on the periodic table when assigning electrons to their orbitals.
So Rb would be [Kr] 5s^1
And so then Zr would be [Kr] 5s^2 , 5d^2
Following this, would Bi not be something like [Xe] 6s^...... 6d^..... and then, 6p^.... (Bi is perhaps a bad choice for me since I know nothing about the Lanthanides )

#### keltingmeith

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##### Re: VCE Chemistry Question Thread
« Reply #8776 on: September 14, 2020, 03:29:02 pm »
+5
Hi guys I’m trying to improve my knowledge of experimental design before the exam by thinking of hypothetical scenarios does anyone know what systematic and random errors as well as how to improve precision and accuracy for this experiment?

To determine the effect quantity of current has on mass of copper deposited at cathode.

What would even be systematic? Mis  calibration? is the only one I can think of but that’ doesn’t work all the time And for random error I want to say the person might accidentally let the current pass through cell for too long, but my teacher says personal errors are the worse to use

I know improving validity would be to control stuff like time current passed through etc and reliability would be repeat to take an average.

Thanks for the help!! 😊

So what you've given here isn't an experiment - it's just an aim. To know what might induce an error, or how to improve precision/accuracy, you need to have an actual experiment to go with it. So maybe the experiment is:

Quote
An electrolytic cell is setup using inert platinum electrodes in a solution of copper sulphate. Each electrode is weighed before the experiment. A current of 5A is applied to the circuit for 5 seconds, after which the electrodes are then removed from solution, left to air-dry, and then weighed again. This process is repeated again for currents of 10A, 15A, and 20A.

So NOW we have something to work with. Firstly, potential systematic errors - the copper being left to air dry could actually oxidise, and form copper oxide, which would affect the mass. There's also miscalibration of the equipment (after all, we never calibrated our current source in this experiment, now did we? I don't like bringing up miscalibration if something HAS been calibrated, unless the method is described and the method sounds like a miscalibration). Random errors - 5 seconds is a very short time, and it's very likely likely some of the experiments were done for slightly less, some slightly more - this would be less an issue if you were doing this for a much longer time frame, but still possible on a larger time frame. To improve accuracy in this case, it would be better to let the electrodes dry in a nitrogen-rich environment, so that there will be little-to-no oxidation by the air, and to improve precision you could run the current over a much longer time-frame - even 5 minutes each would be a MASSIVE improvement, and being off by a second would mean an error of less than 0.34%.

Oh okay. See, I thought that you basically counted from left to right on the periodic table when assigning electrons to their orbitals.
So Rb would be [Kr] 5s^1
And so then Zr would be [Kr] 5s^2 , 5d^2
Following this, would Bi not be something like [Xe] 6s^...... 6d^..... and then, 6p^.... (Bi is perhaps a bad choice for me since I know nothing about the Lanthanides )

Rb is right, Zr is not, and Bi definitely not - sorry! Subshells are not filled up that easily, they are filled up in the following order:

Although you should be aware of the exceptions that are Copper and Chromium (both of these fill up the d subshell before the s subshell).

Basically, even though d starts in period 4, it starts from energy level 3 (or the 3d subshell). f is similar - it essentially starts in period 6, but starts from energy level 4 (or the 4f subshell). The reason for this is less complicated than you'd think, but it's because this convention is based on quantum mechanics, and you're learning an extremely simplified version of it.

Also, lanthanoids can be thought of just like the transitional metals in how they fit into the periodic table, but starting from period 6 - and so you can use the same rules as you know for working around the transition metals. Here's a visualisation:

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#### Coolgalbornin03Lo

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##### Re: VCE Chemistry Question Thread
« Reply #8777 on: September 15, 2020, 08:40:03 am »
0
What is voltage in electrolysis?

Is that the Q which is measured in coulombs. All I know is Volts = J/C

Can too much voltage be a systematic error in electrolysis- and if so why?
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#### Corey King

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##### Re: VCE Chemistry Question Thread
« Reply #8778 on: September 15, 2020, 01:24:31 pm »
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So what you've given here isn't an experiment - it's just an aim. To know what might induce an error, or how to improve precision/accuracy, you need to have an actual experiment to go with it. So maybe the experiment is:

So NOW we have something to work with. Firstly, potential systematic errors - the copper being left to air dry could actually oxidise, and form copper oxide, which would affect the mass. There's also miscalibration of the equipment (after all, we never calibrated our current source in this experiment, now did we? I don't like bringing up miscalibration if something HAS been calibrated, unless the method is described and the method sounds like a miscalibration). Random errors - 5 seconds is a very short time, and it's very likely likely some of the experiments were done for slightly less, some slightly more - this would be less an issue if you were doing this for a much longer time frame, but still possible on a larger time frame. To improve accuracy in this case, it would be better to let the electrodes dry in a nitrogen-rich environment, so that there will be little-to-no oxidation by the air, and to improve precision you could run the current over a much longer time-frame - even 5 minutes each would be a MASSIVE improvement, and being off by a second would mean an error of less than 0.34%.

Rb is right, Zr is not, and Bi definitely not - sorry! Subshells are not filled up that easily, they are filled up in the following order:

(Image removed from quote.)

Although you should be aware of the exceptions that are Copper and Chromium (both of these fill up the d subshell before the s subshell).

Basically, even though d starts in period 4, it starts from energy level 3 (or the 3d subshell). f is similar - it essentially starts in period 6, but starts from energy level 4 (or the 4f subshell). The reason for this is less complicated than you'd think, but it's because this convention is based on quantum mechanics, and you're learning an extremely simplified version of it.

Also, lanthanoids can be thought of just like the transitional metals in how they fit into the periodic table, but starting from period 6 - and so you can use the same rules as you know for working around the transition metals. Here's a visualisation:

(Image removed from quote.)

Ah right. I did not know that the d shell begins to fill at energy level 3.
So then, why is the electron configuration of K = K:[Ar]4s^1
And not equal to K:[Ar]3d^1?

#### keltingmeith

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##### Re: VCE Chemistry Question Thread
« Reply #8779 on: September 15, 2020, 01:50:29 pm »
+4
What is voltage in electrolysis?

Is that the Q which is measured in coulombs. All I know is Volts = J/C

Can too much voltage be a systematic error in electrolysis- and if so why?

Remember that galvanic cells put out a certain voltage based on the Ecell value. For an electrolytic cell, instead the voltage is the amount of energy required to be put into the system for it to run. The two are mirror images of each. For this reason, do you think it might be a systematic error if you add too much voltage?

Ah right. I did not know that the d shell begins to fill at energy level 3.
So then, why is the electron configuration of K = K:[Ar]4s^1
And not equal to K:[Ar]3d^1?

The reason for this is because the energy of the 4s sub-shell is lower than the energy of the 3d sub-shell.

Remember how a little bit back I mentioned every new year in chemistry, you're told something isn't exactly true? We're back at it again. The idea that the 3rd shell is lower in energy than the 4th shell is true for all elements up to 20 - and this is why you were taught that back then. As I said, the model you were taught is useful, but it's not ENTIRELY correct. This breaks down once you get into sub-shells, and you learn that the 4th shell starts to have electrons in it BEFORE the 3rd shell is full.

And the reason for this is because the energy level of a shell isn't constant - it actually depends on the sub-shell in question (and when you hit university, you'll learn that even that isn't constant, and that the energy level of each sub-shell changes for some atoms depending on what might be bonding to it. Ooft!). As it turns out, the 3d sub-shell - despite being in the 3rd shell - is HIGHER in energy than the 4s sub-shell - despite being in the 4th shell. That's where that arrow diagram above comes into play - it tells you what the order of the energy levels roughly are: 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s... and you get the idea.

So, when it comes to potassium putting electrons in its energy levels, it has 19 electrons and asks: what's my lowest energy unfilled sub-shell? Well, 1s, of course, so now its electron configuration and electron count is:

1s2, 17

And if you keep asking this question, you'll fill up the sub-shells like so:

1s2 2s2, 15
1s2 2s2 2p6, 9
1s2 2s2 2p6 3s2, 7
1s2 2s2 2p6 3s2 3p6, 1
1s2 2s2 2p6 3s2 3p6 4s1, 0

Which is long and tedious, so surely there's a better way? And that way is you can combine the information of that arrow-filling diagram (which tells you the order of the energy levels) and the periodic table (see below for one which tells you which block corresponds to which sub-shell):

So, we know that K is in the s-block and the 4th period, which corresponds to the 4s block. This means that every sub-shell below 4s will be filled:

1s2 2s2 2p6 3s2 3p6

Since it is in the first group of the s-block, then that means there's only 1 leftover electron to add to the 4s sub-shell, giving us an electronic configuration of:

1s2 2s2 2p6 3s2 3p6 4s1

If I were to do this for tungsten (W - group 6, period 6), I'd note that it's in the third row of the d-block, which corresponds to sub-shell 3(because third row)+2(because the d-block starts at energy level 3)=5d. So now I need to fill up all the energy levels below the 5d sub-shell - using the diagram above, that would be the following:

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14

Next, tungsten is the 4th element in the d-block, so it has 4 electrons in its valence sub-shell, so the final configuration is:

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d4

And if you want to double-check that's correct, if you add up all the electrons, you should get 74 (the atomic number of W): 2+2+6+2+6+2+10+6+2+10+6+2+14+4=74

But also, it's worth noting that the study design only requires you to know how to do this up to 36 - so feel free to test with monstrously big ones, like tungsten, but know that you'll never have to go past Krypton. This also technically means all you REALLY need to know is that the 3d electrons are higher in energy than 4s, but not than 4d, and that's why you fill up 4s before you fill up 4d.
« Last Edit: September 15, 2020, 01:52:57 pm by keltingmeith »
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#### ArtyDreams

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##### Re: VCE Chemistry Question Thread
« Reply #8780 on: September 16, 2020, 03:14:32 pm »
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I just had a quick question regarding all this zwitterion stuff because I keep getting confused.

So at a NEUTRAL pH, an amino acid is a zwitterion, with an NH3+ and a COO- group.
If it is a high pH, we have NH2 still and COO- (anion)
If its at a low pH, we have NH3+ and COOH (cation)

So, when we are drawing a zwitterion, of an amino acid that has an acid or base in the side chain, this is NOT affected right? It just remains the same. But if it is at a high pH, will an acidic side chain become COO-? Will a basic chain remian the same?
If it as a low pH, will an acidic side chain remain? etc

Pretty much, I'm just curious if the side chains have to be drawn differently at diff pHs when it contains an acidic or basic group.
An exaple would be lysine. Would the Nh2 Side chain become NH3+ at a low pH?

I hope this question makes sense.
Thank you!
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#### Coolgalbornin03Lo

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##### Re: VCE Chemistry Question Thread
« Reply #8781 on: September 16, 2020, 03:33:19 pm »
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I just had a quick question regarding all this zwitterion stuff because I keep getting confused.

So at a NEUTRAL pH, an amino acid is a zwitterion, with an NH3+ and a COO- group.
If it is a high pH, we have NH2 still and COO- (anion)
If its at a low pH, we have NH3+ and COOH (cation)

So, when we are drawing a zwitterion, of an amino acid that has an acid or base in the side chain, this is NOT affected right? It just remains the same. But if it is at a high pH, will an acidic side chain become COO-? Will a basic chain remian the same?
If it as a low pH, will an acidic side chain remain? etc

Pretty much, I'm just curious if the side chains have to be drawn differently at diff pHs when it contains an acidic or basic group.
An exaple would be lysine. Would the Nh2 Side chain become NH3+ at a low pH?

I hope this question makes sense.
Thank you!

Ionic Side chains are affected At high and low pH’s my question is in zwitterions at normal pH are side chains affected 🤔
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#### Sine

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##### Re: VCE Chemistry Question Thread
« Reply #8782 on: September 16, 2020, 03:36:14 pm »
+4
-snip-
Side chains are definitely impacted by pH.

The pH in which the majority of the side chain is either protonated or deprotonated depends on the pKa of the side chain of the amino acid.

The same principles apply though. E.g. for Lysine the pKa of the side chain is 10.5. So below a pH of 10.5 the majority of species will be protonated with a NH3+ side chain and above a pH of 10.5 the majority of species will be NH2.

#### keltingmeith

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##### Re: VCE Chemistry Question Thread
« Reply #8783 on: September 16, 2020, 03:38:55 pm »
+3
I just had a quick question regarding all this zwitterion stuff because I keep getting confused.

So at a NEUTRAL pH, an amino acid is a zwitterion, with an NH3+ and a COO- group.
If it is a high pH, we have NH2 still and COO- (anion)
If its at a low pH, we have NH3+ and COOH (cation)

So, when we are drawing a zwitterion, of an amino acid that has an acid or base in the side chain, this is NOT affected right? It just remains the same. But if it is at a high pH, will an acidic side chain become COO-? Will a basic chain remian the same?
If it as a low pH, will an acidic side chain remain? etc

Pretty much, I'm just curious if the side chains have to be drawn differently at diff pHs when it contains an acidic or basic group.
An exaple would be lysine. Would the Nh2 Side chain become NH3+ at a low pH?

I hope this question makes sense.
Thank you!

Absolutely pH will affect the side-chain! Smart thinking. The answer to this question is: beyond VCE. lel.

Whether or not an amino acid will be protonated on a specific point or not depends on the pKa of each side-chain, and it's not always true that it goes in the order of the side-chain being the last effected - in fact, lysine is one example of the side-chain staying protonated while the amine on the alpha carbon isn't! So, what's the pKa? just like pH=-log([H]), pKa=-log(Ka). Cool, so what's Ka? It's just a very specific example of an equilibrium constant. For any acid A, it can undergo the reaction:

$HA+H_2O\leftrightharpoons A^-+H_3O^+$

And Ka is the equilibrium constant for this reaction. Each different functional group has a different Ka value, so lysine actually has THREE of them - one for the carboxylic acid, one for the amine on the alpha carbon, and one for the amine on the side-chain.

So, for the question: how would you know what bits are protonated in a VCE exam? Well, according to the examination report for last year's exam (section B, Q1b(iii)), you should just assume the side-chains ARE affected by whatever the pH is. I.e., if they say high pH, just assume that the side-chains are also deprotonated, and if they're low pH, assume the side-chains are also protonated. I can't find a question ever asked by VCAA for more middling pHs, but based on the 2013 exam, I would assume they would HAVE to give you more information in the question so that you could make an answer. Eg,

Quote from: Question made up by meltingkeith
Lysine forms an ion with both two positive charges and a negative charge at a pH around 4. At a pH of 9, it forms a species with one positive charge and a negative charge. Draw two possible structures for lysine at a pH of 9.

If they were to ask you which of those structures is the correct one, they'd have to give you more information about which of the two is the more acidic.

EDIT: beaten by Sine, still posted because I also answered Coolgal's question in this one.
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#### ArtyDreams

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##### Re: VCE Chemistry Question Thread
« Reply #8784 on: September 16, 2020, 04:09:10 pm »
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Thank you so much for the help, sine and keltingmeith! Makes sense now
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#### Coolgalbornin03Lo

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##### Re: VCE Chemistry Question Thread
« Reply #8785 on: September 16, 2020, 10:00:22 pm »
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Hi guys! I have done the unit 3 2018 NEAP exam and have a few questions:
(which are attached below]
I have no idea why in this question they know that lead oxide is a product
I got solid lead + sulphate ions --------> Lead Sulphate + 2 electrons

for the second question 4b. ii I dont understand their reasoning at all.
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#### keltingmeith

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##### Re: VCE Chemistry Question Thread
« Reply #8786 on: September 16, 2020, 10:13:31 pm »
+5
Hi guys! I have done the unit 3 2018 NEAP exam and have a few questions:
(which are attached below]
I have no idea why in this question they know that lead oxide is a product
I got solid lead + sulphate ions --------> Lead Sulphate + 2 electrons

for the second question 4b. ii I dont understand their reasoning at all.

The language of the question is important - note that d (ii) is for when the battery is being RECHARGED, whereas the equation you're given is for when the battery is being DISCHARGED. This means the overall reaction that takes place is now:

2H2O(l) + 2PbSO4(s) ---> 2SO42-(aq) + 4H+(aq) + PbO2(s) + Pb(s)

and that the relevant oxidation and reduction equations are now reversed, as well.

As for 4b(ii), I don't think this is something VCAA can reasonably expect you to both know and be able to use. I'd ignore it, because tbh that's something I would believe upon hearing it, but not something I would've assumed if I'd had to come to the conclusion myself. Reminder that company exams are not infallible, it's somewhat common to come across questions every now and again that are a little out of the scope of the study design.
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#### Coolgalbornin03Lo

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##### Re: VCE Chemistry Question Thread
« Reply #8787 on: September 18, 2020, 08:53:06 pm »
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Does anyone have VCAAs expected definitions for fuels? For example I wrote bio fuels are derived from organic matter but this practise exam (VCAA 2008) says “biomass”. Is this the same thing?

Also in the last question of the VCAA exam it mentions the NAOH is neutralised with a certain amount of HCl. I used this to calculate the mole which I got correct as being the excess but why is the NaOH reacting found by minusing the amount which is required to neutralised?
« Last Edit: September 18, 2020, 09:02:29 pm by Coolgalbornin03Lo »
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#### Hello132

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##### Re: VCE Chemistry Question Thread
« Reply #8788 on: September 19, 2020, 10:11:42 am »
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Hi everyone,

I just had a few questions with the 2018 NHT Chemistry Exam

Q5aii)
It asked about predicting the gas formed at the anode in the electrolysis of concentrated potassium chloride. Whilst the examiner's report said the answer was O2, if the KCl solution was concentrated like it mentioned, would it not be Cl2 gas that is produced instead?

Q7a)
When writing oxidation and reduction half equations, why and when do we need to include SO4 2- ions

Thanks

#### zoharreznik

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##### Re: VCE Chemistry Question Thread
« Reply #8789 on: September 19, 2020, 07:07:31 pm »
+2
Hi everyone,

I just had a few questions with the 2018 NHT Chemistry Exam

Q5aii)
It asked about predicting the gas formed at the anode in the electrolysis of concentrated potassium chloride. Whilst the examiner's report said the answer was O2, if the KCl solution was concentrated like it mentioned, would it not be Cl2 gas that is produced instead?

Q7a)
When writing oxidation and reduction half equations, why and when do we need to include SO4 2- ions

Thanks

I would guess for 5aiii) that they've made a mistake. You would need Cl2 gas to produce the faint chlorine smell.

For 7a) you need to include SO4 2- ions as that is what is contained in the electrolyte, so that's what you're reacting and producing.
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