VCE Chemistry Question Thread

[deStudent]

http://m.imgur.com/a/6CLvf

For Q17)b) the answer is attached which I don't quite understand. Isn't their answer saying that S2O42- is under going reduction and oxidation at the same time? I know that reduction and oxidation occurs simultaneously, but it possible for it to occur on the same specie (S2O42-)? Been stuck on this since I couldn't find anything undergoing oxidation.

For the other part (b): is Fe2+ a solid or aqueous solution here? I wrote solid since its in a tablet, but the answer has it as aqueous?

Cheers all

[keltingmeith]

For Q17)b) the answer is attached which I don't quite understand. Isn't their answer saying that S2O42- is under going reduction and oxidation at the same time? I know that reduction and oxidation occurs simultaneously, but it possible for it to occur on the same specie (S2O42-)? Been stuck on this since I couldn't find anything undergoing oxidation.

Yeah, that's absolutely fine, you're just thinking about it weirdly. Notice that on the LHS of the equation, there's TWO molecules that contain sulphur, not just the one. In reality, one single molecule isn't both reduced and oxidised (which is how you seem to be picturing it), but rather two identical molecules react with each other, and one of them will be oxidised, and the other reduced.

For the other part (b): is Fe2+ a solid or aqueous solution here? I wrote solid since its in a tablet, but the answer has it as aqueous?

Cheers all

Key word you missed is solution. I see your logic, but an important note that is often overlooked (hell, I didn't actually realise this until second year uni!) - reacting something that's solid (such as our tablet) is really damned hard, and so most chemical reactions will happen in solution. That's not to say they don't happen, but if you're ever talking about reactions in VCE, it's safe to assume that they're either liquid or in solution unless they specifically state, "is reacted in the solid state" (and tbh, I doubt they ever will).

[homosapien]

What would be the best books for doing well in chem?
Apart from my textbook, I have purchased A+ notes and checkpoints
Are there any other books (more for questions) that I should get. I was thinking NEAP smartstudy questions but anything else?
Also id love to go to tsfx lectures but they cost a lot, especially if i was to go for a few subjects so is it worth it?
thanks

[Vaike]

Hey all,
Was just wondering if anyone would be able to explain these redox questions from the pearson textbook. In both cases, each question lists a species, hydrochloric acid and potassium dichromate, but then goes on to use only chlorine ions and dichromate respectively in the half and overall equations, seemingly dropping the hydrogen and potassium ions. I assume it is because they don't take part in the redox reaction similar to spectator ions in an ionic equation, is this correct? And if so, how do we identify them as such? Thanks    

[keltingmeith]

What would be the best books for doing well in chem?
Apart from my textbook, I have purchased A+ notes and checkpoints
Are there any other books (more for questions) that I should get. I was thinking NEAP smartstudy questions but anything else?

Tbh, I don't think more books = higher grades. What's more important is you have thorough understanding of the content and why you got questions wrong. If you do this, and still end up running out of questions, you should get together with your friends and make up questions for each other to try and answer. The better you know the content, the easier it'll be to come up with questions.

Also id love to go to tsfx lectures but they cost a lot, especially if i was to go for a few subjects so is it worth it?
thanks

Nah just go to the AN ones - they're free and serve the exact same function. Click the link at the top to go to the booking place.

Hey all,
Was just wondering if anyone would be able to explain these redox questions from the pearson textbook. In both cases, each question lists a species, hydrochloric acid and potassium dichromate, but then goes on to use only chlorine ions and dichromate respectively in the half and overall equations, seemingly dropping the hydrogen and potassium ions. I assume it is because they don't take part in the redox reaction similar to spectator ions in an ionic equation, is this correct? And if so, how do we identify them as such? Thanks    

You've got it in one - as for identifying as such, always break things up into singular ions. From there, it's always strongest reductant with strongest oxidant, and everything else is just a spectator.

[deStudent]

Yeah, that's absolutely fine, you're just thinking about it weirdly. Notice that on the LHS of the equation, there's TWO molecules that contain sulphur, not just the one. In reality, one single molecule isn't both reduced and oxidised (which is how you seem to be picturing it), but rather two identical molecules react with each other, and one of them will be oxidised, and the other reduced.

Key word you missed is solution. I see your logic, but an important note that is often overlooked (hell, I didn't actually realise this until second year uni!) - reacting something that's solid (such as our tablet) is really damned hard, and so most chemical reactions will happen in solution. That's not to say they don't happen, but if you're ever talking about reactions in VCE, it's safe to assume that they're either liquid or in solution unless they specifically state, "is reacted in the solid state" (and tbh, I doubt they ever will).

Thanks EulerFan.

One more thing about the first question. If the equation was set up in a way such that there was only one of those sulphur molecules in the equation, this means that it can't undergo both reduction and oxidation right? Because there's only one molecule of that type, instead of 2?

----
One question here if anyone is keen on reading my working, http://m.imgur.com/a/lh829
The answer wasn't provided and my answer doesn't sound that good..

[keltingmeith]

Thanks EulerFan.

One more thing about the first question. If the equation was set up in a way such that there was only one of those sulphur molecules in the equation, this means that it can't undergo both reduction and oxidation right? Because there's only one molecule of that type, instead of 2?

Firstly, I think you're missing the link between the theory and the practical. You're trying to describe a chemical reaction in which there's only one sulphur containing molecule. As someone who regularly images single molecules, good luck trying to do a reaction with only one of the sulphur containing molecules. Imagine this - you have your beakers set up, one is just water, the other is a salt of S2O4^2-. Now, your mission is this - take our your tweezers, pick up only a single molecule of S2O4^2-, and place it into the water. You can't tip the second beaker into the first - even if you only had a gram of your salt, you'd still have 0.01 moles of the anion, and that's 6.02*10^21 molecules! It's absolutely insane, and not really feasible.
tl;dr - your question isn't really appropriate. Try and look past the theory and into the practical that it represents. Remember - chemistry is a PHYSICAL science, so you should be able to picture some physicality to it at all times.

Secondly, I don't think it's impossible that a single molecule could be reduced and oxidised at the same time (essentially, for the molecule to oxidise itself, such that one functional group gets oxidised by another, with the oxidising group thus getting reduced). Probably beyond VCE level, can't find an example, but in my mind I don't see why such a molecule can't exist.

One question here if anyone is keen on reading my working, http://m.imgur.com/a/lh829
The answer wasn't provided and my answer doesn't sound that good..

There's not really much to say to this type of a question - nobody's going to ask this. It doesn't really test chemistry, it only tests if you kind of understand what's going on. Tbh, if VCAA wanted to ask something like this, they're more likely to ask something like, "what would happen if sodium chloride was used instead of hydrochloric acid? Would the same stoichiometric amounts be required?"

Your answer suggests you've got the basic gist of why things need to be balanced, so I wouldn't worry about it.

[Gogo14]

1. When water is release in a combustion reaction is it a gas or liquid?
2. What is the definition of oxidation number? I know that its used in redox but don't really know what it actually measures

Edit:

Firstly, I think you're missing the link between the theory and the practical. You're trying to describe a chemical reaction in which there's only one sulphur containing molecule. As someone who regularly images single molecules, good luck trying to do a reaction with only one of the sulphur containing molecules. Imagine this - you have your beakers set up, one is just water, the other is a salt of S2O4^2-. Now, your mission is this - take our your tweezers, pick up only a single molecule of S2O4^2-, and place it into the water. You can't tip the second beaker into the first - even if you only had a gram of your salt, you'd still have 0.01 moles of the anion, and that's 6.02*10^21 molecules! It's absolutely insane, and not really feasible.
tl;dr - your question isn't really appropriate. Try and look past the theory and into the practical that it represents. Remember - chemistry is a PHYSICAL science, so you should be able to picture some physicality to it at all times.

Secondly, I don't think it's impossible that a single molecule could be reduced and oxidised at the same time (essentially, for the molecule to oxidise itself, such that one functional group gets oxidised by another, with the oxidising group thus getting reduced). Probably beyond VCE level, can't find an example, but in my mind I don't see why such a molecule can't exist.

There's not really much to say to this type of a question - nobody's going to ask this. It doesn't really test chemistry, it only tests if you kind of understand what's going on. Tbh, if VCAA wanted to ask something like this, they're more likely to ask something like, "what would happen if sodium chloride was used instead of hydrochloric acid? Would the same stoichiometric amounts be required?"

Your answer suggests you've got the basic gist of why things need to be balanced, so I wouldn't worry about it.

not sure but could peroxides undergo both reduction and oxidation when it gets broken down?

[jh;)]

Hi, I just have a question here:

What is the resulting pH when 3.0g of NaOH is added to 500mL of 0.10M HCl? I've found the number of moles of both but I'm not sure what to do from here onwards.

Thanks

[Gogo14]

Hi, I just have a question here:

What is the resulting pH when 3.0g of NaOH is added to 500mL of 0.10M HCl? I've found the number of moles of both but I'm not sure what to do from here onwards.

Thanks

This is what I did:
1. find which compound is in excess. (NaOH). This is important as the rest of NaOH reacts with HCl to form water and Cl (both with neutral pH)
2. NaOH is in excess by 2.5*10^-2mol. Then calculate concentration.
3. Apply the pH formula to find the pH. (I got 13)

[keltingmeith]

1. When water is release in a combustion reaction is it a gas or liquid?

Good question! If something explodes, what temperature is it likely to be at/above? At what temperature does water become a gas?

2. What is the definition of oxidation number? I know that its used in redox but don't really know what it actually measures

It sort of measures how much "extra charge" an atom has in its "vicinity". For example - in water, O has a -2 oxidation number, and H has a +1 oxidation number. This means that O has an "extra" 2 electrons, whereas H is "missing" 1 electron. It's a bit of a weird way of thinking (particularly when you start to go further into how electrons work, where this model is overly simplistic...), but it's useful for predicting if a redox reaction has occurred. You can also generally use it to check to see if your molecule makes sense, but this is more synthetic stuff with metals, so not overly useful for VCE.

Edit:not sure but could peroxides undergo both reduction and oxidation when it gets broken down?

Yeah, probably the best example we'll be able to come up with.

[homosapien]

According to my textbook, petrodiesel is produced by fractional distillation of crude oil but it also says that a source of petrodiesel is petroleum. I'm not sure how these relate... is petroleum 'made up' of crude oil, are they the same thing or two different things? Could someone please describe the link between them thanks

[deStudent]

http://m.imgur.com/iFYelWA

For (a), the answer has 39mg/L. I get 0.032M by C = n/V, I'm a bit confused on what they did?

(b): I'm lost here, I tried going n = CV as a guess then solving for the mass. Then converting to m/m%.

[keltingmeith]

According to my textbook, petrodiesel is produced by fractional distillation of crude oil but it also says that a source of petrodiesel is petroleum. I'm not sure how these relate... is petroleum 'made up' of crude oil, are they the same thing or two different things? Could someone please describe the link between them thanks

Pretty sure they're different names for the same thing. BUT, it's really not that big a deal. At all. I would not worry about this at all.

http://m.imgur.com/iFYelWA
For (a), the answer has 39mg/L. I get 0.032M by C = n/V, I'm a bit confused on what they did?

Remember: M=mol/L=n/V. So, if you divide this by molar mass (Mr, so we don't get confused with M=molarity), you get n/VMr=(n/Mr)*(1/V)=m/V, which is in units of g/L. Changing this to mg/L should be easy enough, just multiply by 1000! They've done this because the units are easier for later on, but not necessary. If you convert your number by dividing by the molar mass, you should have something similar. (obviously not exact since getting the concentration at all is a bit of guesswork) Since they didn't specify what units, in an exam situation, you should still get the marks. (indeed, I would've done M myself, because that's what they usually want, and those are the units on the graph)

EDIT: Sorry, just took another look at what you said. You mentioned you used C=n/V to find the concentration. How did you do that? You should be reading this off the graph, not using a formula.

(b): I'm lost here, I tried going n = CV as a guess then solving for the mass. Then converting to m/m%.

So, you have the mol/L from part a (for ease, use their value, because that's what their answer will be based on. But of course, you should still get a simlar answer regardless), multiply by the volume of your solution (250 mL) to get this into mol. But you want g/g% - so, use m=nM to convert mol to g (remember, this is only for the phosphate in the tablet, not the whole tablet). After you've done this, you can divide by the mass of the full tablet (0.250 g) to get the percentage by mass.

[deStudent]

Thanks EulerFan,

I'm still a bit confused. What should I be reading off the graph with the method you suggested? So I went n/(Mr*V) = 0.250/31 /(31*250E-3), then *1000 which gave me 0.104? If I had to read off the graph, I got C = 0.1M -> g/L, multiply by 31. This wouldn't be correct either?