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March 29, 2024, 04:11:21 pm

Author Topic: VCE Chemistry Question Thread  (Read 2313589 times)  Share 

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keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8910 on: November 16, 2020, 12:45:35 am »
+7
Thank you for that! Because I'm referring to individual parts to questions and their answers, they won't make sense if I only take a picture of that part of the question so I think it'll be easier if I give you the link to the exam:
https://www.vcaa.vic.edu.au/Documents/exams/chemistry/2019/NHT/2019chem-nht-w.pdf

and this is the examiner's report: https://www.vcaa.vic.edu.au/Documents/exams/chemistry/2019/NHT/chemistrynht_examrep19.pdf

Sorry - I'm currently in the middle of working 12 hour days, I saw NHT and read NEAP . Let's get to work!

Hey guys, can someone please help me with the following questions I have from the 2019 NHT exam, I'd greatly appreciate it! (Sorry they're so many)

- question 6f: I don't understand the solutions from the examiner's report especially with regards to why a narrower tube or meniscus of a pipette increases accuracy, what do they mean by that? And on that note, does anyone know of a way I can improve my understanding of titration errors - like any good videos that explain it because I've tried to look online and I can't find anything.

- question 7a: why do we not include H2O in the Kc equation

- question 7bi: I interpreted it as the addition of H2O will cause the system to favour the reverse reaction so that it is consumed but the answers refer to opposing the dilution by favouring the side with more ions so is my justification incorrect?

- question 7bii: why do the answers talk about the rate of reaction when the question asks for the effect on the position of equilbirum? I spoke about LCP.

- question 8d: How did they use the chemical shift values to determine that there is an ethyl group? It doesn't match up with the values from the data book
 
- question 3cii and 9b: how do we know how reactive a metal is and if it's going to react violently and why is it bad for something to react with oxygen? Also for question 9b, why can't the electrolyte react with Na, isn't that the point?


6f. Yeah, that's a very niche point. Basically, the meniscus is the drop at the top of the pipette/burette. You'd think to yourself that a narrower tube means flatter of a drop, but it's actually the opposite - the wider the tube, the flatter the drop. The thing is, these glassware are calibrated around the bottom of that drop, so you want the bottom of it to be as obvious as possible - hence why you want a smaller tube, to see the drop better. As for titration errors - the big ones are inappropriate washing, but other ones include the ability of the acid or base to absorb moisture over time, the fact that the end point is not the same as the equivalence point

Tbh, I would've just said because the pipette has a higher precision than a measuring cylinder. It's not even about the size of the tube - it's just that pipettes are designed and calibrated to be really good, whereas measuring cylinders are designed to be good enough. It's a precision vs. cost thing, that's all

7a. Oh man, I love this experiment!!! So much better to see than just look at equations for. But, I digress. The reason we don't include water is because the reaction is happening in water, so we say it has an effective concentration of 1.0 M. The actual reason has to do with the fact that equilibrium expressions actually have nothing to do with concentrations, only this weird physical measurement called activity - but at low concentrations, activity can be approximated with concentration. For higher concentrations, like the solvent, the effective activity is 1 and can be ignored.

7bi. Not at all - your justification is honestly more correct, because a dilution would shift to the right, not the left - and having done this experiment I can promise you, it goes pink when you add water.

7bii. This is a matter of reading the question - not that is specifically says to answer in terms of collision theory. Collision theory is all about rate, NOT about equilibrium, hence why they discuss rate.

8d. The NHT group may have a different data booklet, so double check you have the right one. The more important bit is you know it's an ethyl group if there's a \(CH_3\) at around 1ppm, and a \(CH_2\) at around 2ppm, with the first being a triplet and the second a quartet. Doesn't matter if it's off by even as much as you see here. The ppms can move around a bit, so don't think they have to be exactly where the book says - that's why using integration, splitting, etc. is way more important than shift.

3cii/9b. The lower the E0 value, the more reactive the forward reaction is. This isn't always true, mind, but it's an incredibly good rule of thumb, and enough for VCE. Even better, for what they've done - transition metals, not very reactive, all other metals, very reactive. And no - the point isn't for the electrolyte to react with the sodium. The electrolyte is only meant to carry charge around. A similar example - let's say you're eating a sandwich while playing with an animal. If the animal ate your sandwich, this would be a sad thing, because you wanted to eat that sandwich. Same thing here - we want the sodium electrode to react in the electrochemical reaction, not in a different reaction that doesn't produce electrochemical energy.

tigerclouds

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Re: VCE Chemistry Question Thread
« Reply #8911 on: November 16, 2020, 09:40:30 am »
0
Thank you so so much for taking the time to explain that, it really helped!
7a. Oh man, I love this experiment!!! So much better to see than just look at equations for. But, I digress. The reason we don't include water is because the reaction is happening in water, so we say it has an effective concentration of 1.0 M. The actual reason has to do with the fact that equilibrium expressions actually have nothing to do with concentrations, only this weird physical measurement called activity - but at low concentrations, activity can be approximated with concentration. For higher concentrations, like the solvent, the effective activity is 1 and can be ignored.
Just on this point, does that mean that we never include H2O in the expression if either of the reactants and products are aqueous?

keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8912 on: November 16, 2020, 09:44:11 am »
+4
Thank you so so much for taking the time to explain that, it really helped!Just on this point, does that mean that we never include H2O in the expression if either of the reactants and products are aqueous?


Good question - my gut says no, but I can think of weird situations where it might be relevant still, which is the problem with trying to explain such complicated things using approximations and simplifications (it's actually a similar problem with electrochem). I would say that for the purposes of VCE, you will never be in a situation where you include the concentration of water - and definitely, you will never need to include it if either the products or the reactants are in an aqueous state. Some company exams might not behave as nicely, though - so if they have a situation where you include the concentration of water, you're likely better off ignoring that example if you can't figure out why they did include it.

Coolgalbornin03Lo

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Re: VCE Chemistry Question Thread
« Reply #8913 on: November 17, 2020, 05:45:53 pm »
0
Hey guys when discussing why certain polysaccharides of glucose are less soluble is it better to say hydroxyl groups exposed or glycosidic links exposed? Also is the bonding between respective molecules hydrogen or dispersion? Like for amylopectin would I say due to its branched structure it undergoes long range dispersion forces or long range hydrogen bonding which causes more hydroxyl groups to be exposed?

Thanks guys!
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keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8914 on: November 17, 2020, 07:50:06 pm »
+7
Hey guys when discussing why certain polysaccharides of glucose are less soluble is it better to say hydroxyl groups exposed or glycosidic links exposed? Also is the bonding between respective molecules hydrogen or dispersion? Like for amylopectin would I say due to its branched structure it undergoes long range dispersion forces or long range hydrogen bonding which causes more hydroxyl groups to be exposed?

Thanks guys!

I'm going to start with the second question because I think there might be some confusion around intermolecular forces. Both dispersion forces and hydrogen bonding are examples of intermolecular forces  - and the third one you should be familiar with are dipole-dipole interactions. There are more, but in VCE, we're just going to think of these 3. They can also be examples of intramolecular forces, but we're just going to focus on them being intermolecular for now.

In order of strength, these forces go hydrogen bonding >> dipole-dipole interactions >> dispersion forces, and molecules are always going to preference the strongest interaction that they can. They might undergo multiple (for example, sodium dodecyl sulphate is a molecule that undergoes both hydrogen bonding, but also dispersion forces), but they will always preference the strongest available to them. Just because two molecules are the same, you cannot say that they will undergo hydrogen bonding or dispersion forces just because they're the same, because it depends on the molecule.

Some examples - hexane can only undergo dispersion forces. It cannot form hydrogen bonds or dipole-dipole interactions. So, hexane will form dispersion forces with other hexane molecules. Water can do all three, but it will favour hydrogen bonds with itself, because it is both a hydrogen-bond donor and a hydrogen bond acceptor. A weirder example is acetone. Acetone is a hydrogen bond acceptor, but not a donor - it is also soluble in both hexane and water. So in hexane, acetone will only form dispersion forces with hexane molecules. In water, acetone will form hydrogen bonds with other water molecules. However, since acetone is polar, it can form dipole-dipole interactions as well. So, acetone molecules will form dipole-dipole interactions with itself - and it will do that whether it's dissolved in hexane or with water, because that's the strongest force it can form with itself. So, if you imagine a giant pot filled with hexane, water, and acetone (note: this is a pretend pot, because in a real pot, the water and hexane will not mix), then you will see all forces happen at once. Whenever acetone gets close to water, it will start to form hydrogen bonds. However, when the acetone moves away from the water, it will start to lose those hydrogen bonds and form different interactions depending on what it finds while moving around - if it's another acetone molecule, that'll be dipole-dipole, and if it's hexane, it'll just be dispersion forces. I've attached a picture to hopefully get this idea across. Dispersion forces are not shown, but everywhere you see an acetone near a hexane molecule, you can assume there are dispersion forces there. Hydrogen bonding and dipole-dipole interactions are the red dotted lines

As for the first question - there's two contributing reasons that causes them to be less soluble. The first is the fact that it gets bigger - the bigger something is, the more bonds it can make with other molecules that are the same. It's the same reason that as you increase carbon chain length in methane, ethane, propane, etc., that the state of matter at room temperature changes from gas, to solid, to liquid. The other one is that you lose hydrogen bonding groups. Remember - water wants to form hydrogen bonds, because it can form them, so those are all that water will be looking for. As the chain gets bigger, and hydroxy groups turn into glycosidic links, you start to lose available hydrogen bonding sites, and so solubility in water will decrease. The key point, though, is that it's not one or the other - it's both, and both are equally correct.

Also, hydrogen bonding HAS to be short ranged. If the molecules move far away, then the hydrogen bond becomes weaker, until eventually it's only as strong as a dispersion force.

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Re: VCE Chemistry Question Thread
« Reply #8915 on: November 19, 2020, 09:20:12 am »
0
Does anyone know how Edrolo works?

I want to get just the physical edrolo textbook for chem and not the online. Is this possible?

Rube

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Re: VCE Chemistry Question Thread
« Reply #8916 on: November 19, 2020, 03:16:16 pm »
0
When drawing organic chemistry structures do the bond angles matter in 3&4?

ArtyDreams

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Re: VCE Chemistry Question Thread
« Reply #8917 on: November 19, 2020, 03:48:51 pm »
+3
Does anyone know how Edrolo works?

I want to get just the physical edrolo textbook for chem and not the online. Is this possible?

We used Edrolo this year - I'm pretty sure the textbook and online comes as a bundle - but its worth asking your teachers, I think they'll know! Otherwise, I'm planning on giving my textbook away secondhand - you might be able to get one secondhand from someone!  ;)

When drawing organic chemistry structures do the bond angles matter in 3&4?

I don't think so!
« Last Edit: November 19, 2020, 04:28:41 pm by ArtyDreams »

Rube

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Re: VCE Chemistry Question Thread
« Reply #8918 on: November 19, 2020, 04:29:54 pm »
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Thanks!

Coolgalbornin03Lo

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Re: VCE Chemistry Question Thread
« Reply #8919 on: November 20, 2020, 08:46:52 am »
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In volumetric analysis what does “titration by solution x” mean? Is solution x the known solution being delivered from the Burette or the unknown solution being delivered into the aliquot?
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Re: VCE Chemistry Question Thread
« Reply #8920 on: November 20, 2020, 02:30:28 pm »
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Does anyone know how Edrolo works?

I want to get just the physical edrolo textbook for chem and not the online. Is this possible?
Also, is it possible for students to only get online, so they can purchase the textbooks secondhand?
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Re: VCE Chemistry Question Thread
« Reply #8921 on: November 20, 2020, 09:28:11 pm »
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Hi I was wondering if gravimetric analysis is still on the study design. I'm lowley panicking because I am seeing it in older vcaa exams idk if its still part of the course though.

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Re: VCE Chemistry Question Thread
« Reply #8922 on: November 20, 2020, 09:44:35 pm »
+2
Hi I was wondering if gravimetric analysis is still on the study design. I'm lowley panicking because I am seeing it in older vcaa exams idk if its still part of the course though.
Pretty sure it is a 1/2 concept, don't think it is part of the study design for 3/4.

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Re: VCE Chemistry Question Thread
« Reply #8923 on: November 21, 2020, 11:38:00 am »
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Can someone please explain the effects of washing different equipment in terms of titrations?
i.e. washing the burette/ pipette, conical flask/ volumetric flask with de-ionised water, etc. and how this effects the calculated value
It's hella confusing !

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Re: VCE Chemistry Question Thread
« Reply #8924 on: November 21, 2020, 12:22:03 pm »
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Can someone please explain the effects of washing different equipment in terms of titrations?
i.e. washing the burette/ pipette, conical flask/ volumetric flask with de-ionised water, etc. and how this effects the calculated value
It's hella confusing !

We know the number of mols by n = cv & we want to make sure we have accurate info on these so we can do our calculations with the right numbers.

Rinsing with distilled water means we don't add any extra solution but we might dilute the solution going in that glassware i.e. it could mean the real concentration is lower than we think it is.

Rinsing with the solution that will go in the glassware means we won't dilute it but we might add in some extra solution into the glassware.


In the burette you know what the volume you're giving is from reading the difference between your markings so it doesn't matter if there's a bit of extra solution in there - so long as you read the marks on the burette properly and stopped at the equivalence point you'd get the right volume. This means that if we rinse with the solution we're going to run through the burette that's not going to mess us our volume. We also know that rinsing with the solution won't mess up the concentration. Therefore, we should rinse the burette with the solution.

Ok, how about if we rinsed with water? We wouldn't be adding in any extra volume of our solution but we would be diluting it. This would mean that when we do the calculation n = cv, mols in a titre = average titre * concentration calculation we would overestimate the number of mols delivered in a titre compared to what it actually was. This is why we don't want to rinse the burette with water.


Ok, how about in the conical flask then? We already know the volume we'll use for our calculation from measuring the solution going into the flask.  What we really want to avoid is adding in any more mols. We don't care as much if we add in more water since we already have our volume measurement and the water won't add any more mols of our solution. This makes water a good choice for rinsing with. Again, we don't want to rinse with our solution here because we don't want to accidentally add any more mols of it. If we did that, we would need a larger titre from the burette than we should to reach the equivalence point.

Does this make sense so far?