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tigerclouds

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Re: VCE Chemistry Question Thread
« Reply #8685 on: August 22, 2020, 12:10:36 pm »
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Hi, could someone please help me with the following?

1. How do we predict half-equations for organic molecules and acid-base reactions in general? I know this is a vague question, but I don't understand how to predict reactions to perform calculations related to volumetric analysis

2. I have no idea what this question is asking me to do and the solutions don't make sense to me:
"The amount of vitamin C in fruit juice can be determined by titration with a standard 0.0100 M iodine solution:
                                      C6H8O6(aq) + I2(aq) → C6H6O6(aq) + 2H+(aq) + 2I–(aq)
If the maximum concentration of vitamin C is likely to be 0.00050 g mL–1, describe how you would perform the analysis. You should mention the volume of the fruit juice used and the maximum titre of iodine you would expect to obtain.

keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8686 on: August 22, 2020, 03:06:58 pm »
+4
Hello
would this be right ,sorry for asking i wasnt there when my teacher went over this

That's fine.

Also how do you know if the molecule is polar or non polar i am quite confused

So like, people will shy away from this method - mainly because it involves learning something new - but I personally think this is the easiest way to learn polarity. First, you need to learn about vectors and how to add vectors to each other geometrically:

https://www.mathsisfun.com/algebra/vectors.html

Read this - it's an amazing resource. All you need to understand is:

a) What a vector is (a direction with magnitude)
b) How to add one vector to another (head-to-tail)

If you don't understand the website, just ask - but it should be straightforward. You don't need to understand how subtraction, scalar multiplcation, notation, or how ANYTHING else works. Just how to add vectors together. Got it? Awesome.

If you look at the picture attached, you can actually draw vectors over a pair of atoms. You can think of these vectors as a "force" that the electrons exert when they're more attracted to one atom over the other. (it's slightly more complicated than that, but this intuition is enough to explain polarity on a basic level) You'll notice there's a plus on the not-pointy end - you can use this plus to remember which side is more positive. There are three rules you need to know about these arrows:

  • The arrow points towards the more electronegative atom.
  • The bigger the difference in electronegativity, the bigger the vector will be.
  • The arrows ONLY exist between two atoms - lone pairs don't count.

Using this, how can you tell if a molecule is polar or not? Well, if you draw all of these arrows between each pair of bonding atoms (we call these "dipole vectors"), and add them all up, then the molecule will be NON-POLAR is the arrows return to the starting point, and POLAR if they do not. If you look at my second picture, there's an example for ACl3 - in this case, since the arrows return to the starting point (in red), the molecule must be non-polar. However - if you were to do this with HCN (this molecule is linear, with a single bond connecting the H to the C and a triple bond to the C and N), then the two dipole vectors would be different sizes, and would point to somewhere that isn't the middle - try it out for yourself!

The only problem with this method, is you need to add the vectors in 3D - so it's harder to do for parent geometries with 4, 5, or 6 electron domains. However, something that is handy to know - in a tetrahedral atom, if all of the dipole vectors are the same size, then the molecule will be non-polar. The same for a trigonal bipyramidal molecule or an octahedral molecule. Hopefully knowing this will make it easier for you to try and imagine what's happening in 3D.

and thank you for your really detailed explanation above!

All good, man!

Hi, could someone please help me with the following?

1. How do we predict half-equations for organic molecules and acid-base reactions in general? I know this is a vague question, but I don't understand how to predict reactions to perform calculations related to volumetric analysis

Half-reactions is something we discuss for redox reactions, not acid-base reactions or typical organic reactions. The general trick is:

a) If it's an acid base reaction, you'll be making a salt + water
b) If it's an organic molecule, you need to understand its reactivity and possible reaction pathways
c) If you have no clue and the question hasn't told you what it should be, assume 1:1 stoichiometry. At the very least, you'll get consequential marks

2. I have no idea what this question is asking me to do and the solutions don't make sense to me:
"The amount of vitamin C in fruit juice can be determined by titration with a standard 0.0100 M iodine solution:
                                      C6H8O6(aq) + I2(aq) → C6H6O6(aq) + 2H+(aq) + 2I–(aq)
If the maximum concentration of vitamin C is likely to be 0.00050 g mL–1, describe how you would perform the analysis. You should mention the volume of the fruit juice used and the maximum titre of iodine you would expect to obtain.

lol they're not trying to make this simple before. I feel the thinking is outside the study design a bit, but the calculations certainly aren't.

Essentially, the question wants you to describe an experiment that will tell you what the concentration of vitamin C is. They've even given away half the answer - the experiment you'll do is a titration! So you just need to explain how you would do that titration. You don't need to go step-by-step (although that would be one way to answer it - it wouldn't be wrong, certainly), but you need to explain roughly how you'd go about it. Eg, "I would pour the fruit juice all over the bench, then add iodine until there's a colour change and I reach the end point". Things to think about:

-This is a titration. So, you're going to have something in a burette, and something in a conical flask. What do you think would make more sense to put where, and how much of each are you going to put in where?
-If the amount of vitamin C they've given is correct, how much of it will react with the iodine? What volume would that correspond to?

tigerclouds

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Re: VCE Chemistry Question Thread
« Reply #8687 on: August 22, 2020, 04:14:24 pm »
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Half-reactions is something we discuss for redox reactions, not acid-base reactions or typical organic reactions. The general trick is:

a) If it's an acid base reaction, you'll be making a salt + water
b) If it's an organic molecule, you need to understand its reactivity and possible reaction pathways
c) If you have no clue and the question hasn't told you what it should be, assume 1:1 stoichiometry. At the very least, you'll get consequential marks
Thank you for your reply. I know half-reactions are typically used for redox equations but I asked this because one question in my book asked me to find the half-equation of propan-2-ol to from propanone and whilst this was relatively straightforward, I was wondering how I would know how to form half-equations for more complex organic molecules. Also for acid base reactions, why is it that sometimes water isn't formed or it is by ions are also present as by-products? How can we predict this?


Essentially, the question wants you to describe an experiment that will tell you what the concentration of vitamin C is. They've even given away half the answer - the experiment you'll do is a titration! So you just need to explain how you would do that titration. You don't need to go step-by-step (although that would be one way to answer it - it wouldn't be wrong, certainly), but you need to explain roughly how you'd go about it. Eg, "I would pour the fruit juice all over the bench, then add iodine until there's a colour change and I reach the end point". Things to think about:

-This is a titration. So, you're going to have something in a burette, and something in a conical flask. What do you think would make more sense to put where, and how much of each are you going to put in where?
-If the amount of vitamin C they've given is correct, how much of it will react with the iodine? What volume would that correspond to?
Listing the steps make sense but I don't know how to derive the volume from the information they've given us. The solutions say: "If the maximum concentration of vitamin C in fruit juice were 0.000 50 g mL–1, titration of an aliquot of 50.00 mL of fruit juice with 0.0100 M iodine solution would give a maximum titre of 14.20 mL." but I don't understand where these values have come from. Also, I read your earlier post about the unknown volume always being in the burette but since the volume for neither of the reactants is known, how do we determine this? Is the solution with a known concentration usually placed in the burette?
Thank you for your time.

keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8688 on: August 22, 2020, 04:48:11 pm »
+3
Thank you for your reply. I know half-reactions are typically used for redox equations but I asked this because one question in my book asked me to find the half-equation of propan-2-ol to from propanone and whilst this was relatively straightforward, I was wondering how I would know how to form half-equations for more complex organic molecules. Also for acid base reactions, why is it that sometimes water isn't formed or it is by ions are also present as by-products? How can we predict this?

I'm happy to try and explain, but I'm honestly unsure what these half-equations would look like - especially considering propanol to propanone IS a redox reaction... Can you give me what the "half-equation" in question looked like?

You can always go by first principles - remember, all an acid really is, is a proton donor, and a base a proton acceptor. For example, with HCl and NaOH:

HCl + NaOH --->H+ + Cl- + Na+ + OH-

And we know that that proton must go to something negative, so it's going to attach to the OH. The sodium is then left with the chloride, and we get:

HCl + NaOH ---> NaCl + H2O

Or maybe something slightly more complicated:

NH3 + HCl ---> NH3 + H+ + Cl-

In this case, the base is NH3. The negative bit the H is going to attach to is the N, which gives us:

NH3 + HCl ---> NH4++ Cl-

You might also write:

NH3 + HCl ---> NH4Cl

Listing the steps make sense but I don't know how to derive the volume from the information they've given us. The solutions say: "If the maximum concentration of vitamin C in fruit juice were 0.000 50 g mL–1, titration of an aliquot of 50.00 mL of fruit juice with 0.0100 M iodine solution would give a maximum titre of 14.20 mL." but I don't understand where these values have come from. Also, I read your earlier post about the unknown volume always being in the burette but since the volume for neither of the reactants is known, how do we determine this? Is the solution with a known concentration usually placed in the burette?
Thank you for your time.

The reason you don't know volumes is because they're something that are decided during the experiment planning process. That's what you're doing now - planning an experiment, so you need to choose the volumes! The basis of a titration experiment is:

You have two solutions - one of a known concentration, one of an unknown concentration.

There's no such things as "volumes" or "burettes" or "indicators" in the premise - just that you know one concentration, but not the other. However, before you do an experiment, you need to decide what volumes you use, what indicators you use, and if you're going to use a burette (note: you don't always have to, but it's /usually/ recommended that you do). That's what this question is all about - deciding those volumes.

And you want to know the best part? There's not actually a wrong answer! You'll need to decide what volume of which chemical goes into the conical flask, but it doesn't matter which volume you use or what chemical you put into the flask - all that will change is what chemical you put into the burette, and how much of it you'll need to add in the titration. In this example, they've used 50 mL of juice in the conical, and that will require 14.20 mL of iodine. But you could've done this the other way round - you could've put 14.20 mL of iodine in the conical, and then required a titre of 50.00 mL of juice. Or you could've used 5 mL of juice in the conical, and that would've required 1.42 mL of iodine from the burette. It doesn't matter what volumes you decide - you just need to decide them.

In proper experiment design (which is why I think this might be slightly out of the curriculum), you typically want to put the thing of known concentration in the conical because it makes the maths more intuitive - but you don't HAVE to do that (and it honestly doesn't even make the maths easier or harder). You also want to choose an aliquot size such that the titre is closeish to being within 40-60% of the burette's total volume (10-15mL for a 25mL burette), to minimise uncertainties. You'll also typically use a round number for the aliquot, because pipettes typically only come in those round sizes. But you also wouldn't necessarily be wrong if you picked weird numbers - it's just that you'd find it frustrating to physically do the actual experiment, but that wouldn't make your answer wrong.

zhouzhennan

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Re: VCE Chemistry Question Thread
« Reply #8689 on: August 22, 2020, 08:14:55 pm »
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For Unit 4 AOS1, is it necessary to memorise the bond strength sequence of different carbon bonds? Like C- F, C-H, C-O, C-C etc. Do we need to know which goes first?


tigerclouds

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Re: VCE Chemistry Question Thread
« Reply #8690 on: August 22, 2020, 09:36:38 pm »
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I'm happy to try and explain, but I'm honestly unsure what these half-equations would look like - especially considering propanol to propanone IS a redox reaction... Can you give me what the "half-equation" in question looked like?
So this was the question: "Some alcohols can be converted into ketones by oxidising agents. Write a balanced half-equation for the oxidation of propan-2-ol (CH3CHOHCH3) to form propanone (CH3COCH3)." and the answer is "CH3CHOHCH3(aq) → CH3COCH3(aq) + 2H+(aq) + 2e–" and I get that but if we were asked to write the half-equation for the oxidation of ethanol for example, I wouldn't know what to do.


NH3 + HCl ---> NH3 + H+ + Cl-
In this case, the base is NH3. The negative bit the H is going to attach to is the N, which gives us:

NH3 + HCl ---> NH4++ Cl-

You might also write:

NH3 + HCl ---> NH4Cl
Ok so I know this probably sounds very dumb, but how do we know that it's only the HCl that dissociates into separate ions, why can't NH3 do that and produce NCl and H4? Sorry I think I'm lacking the basic foundations for acid-base reactions and I don't know what to do to relearn it.


The reason you don't know volumes is because they're something that are decided during the experiment planning process. That's what you're doing now - planning an experiment, so you need to choose the volumes! The basis of a titration experiment is:

You have two solutions - one of a known concentration, one of an unknown concentration.

There's no such things as "volumes" or "burettes" or "indicators" in the premise - just that you know one concentration, but not the other. However, before you do an experiment, you need to decide what volumes you use, what indicators you use, and if you're going to use a burette (note: you don't always have to, but it's /usually/ recommended that you do). That's what this question is all about - deciding those volumes.

And you want to know the best part? There's not actually a wrong answer! You'll need to decide what volume of which chemical goes into the conical flask, but it doesn't matter which volume you use or what chemical you put into the flask - all that will change is what chemical you put into the burette, and how much of it you'll need to add in the titration. In this example, they've used 50 mL of juice in the conical, and that will require 14.20 mL of iodine. But you could've done this the other way round - you could've put 14.20 mL of iodine in the conical, and then required a titre of 50.00 mL of juice. Or you could've used 5 mL of juice in the conical, and that would've required 1.42 mL of iodine from the burette. It doesn't matter what volumes you decide - you just need to decide them.
Oh ok, I didn't realise it was an open-ended question.

In this example, they've used 50 mL of juice in the conical, and that will require 14.20 mL of iodine. But you could've done this the other way round - you could've put 14.20 mL of iodine in the conical, and then required a titre of 50.00 mL of juice. Or you could've used 5 mL of juice in the conical, and that would've required 1.42 mL of iodine from the burette. It doesn't matter what volumes you decide - you just need to decide them.
Could I just ask, so they randomly chose to use 50 mL of juice but how did they work out that 14.20 mL of iodine is required for that?

keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #8691 on: August 23, 2020, 11:04:36 am »
+3
For Unit 4 AOS1, is it necessary to memorise the bond strength sequence of different carbon bonds? Like C- F, C-H, C-O, C-C etc. Do we need to know which goes first?

Can't see it mentioned in the study design, and tbh can't even think of anything that the study design requires that would require you to know this in the first place. I think you can safely ignore this information.

So this was the question: "Some alcohols can be converted into ketones by oxidising agents. Write a balanced half-equation for the oxidation of propan-2-ol (CH3CHOHCH3) to form propanone (CH3COCH3)." and the answer is "CH3CHOHCH3(aq) → CH3COCH3(aq) + 2H+(aq) + 2e–" and I get that but if we were asked to write the half-equation for the oxidation of ethanol for example, I wouldn't know what to do.

Okay, cool - so to answer your original question: you don't have to worry about writing up half-reactions for acid-base reactions or organic reactions unless they're also a redox reaction. If they are a redox reaction, then they would have to tell you what the product is going to be - like they've done in this instance. For example, if I were to ask you, "write the oxidation reaction for solid iron" - you wouldn't know what to write, because there's two possible products that could be formed - either iron (II) or iron (III). Similar for the oxidation of ethanol - there's three possible products (that I can think of - there may be more) - ethan-1,1-diol, ethanal, and ethanoic acid. To write half-reactions for any of these, it would be the exact same process as you follow for inorganic redox reactions:

1. Balance oxygens using water
2. Balance hydrogens using protons
3. Balance charges using electrons

Ok so I know this probably sounds very dumb, but how do we know that it's only the HCl that dissociates into separate ions, why can't NH3 do that and produce NCl and H4? Sorry I think I'm lacking the basic foundations for acid-base reactions and I don't know what to do to relearn it.

So this comes down to knowing whether something ionises or not - and to know that, you'll need to know if something is a strong acid or base. Strong acids and bases will ionise, weak ones will not. For example, ethane (C2H6) is an extremely weak acid - it won't ionise, so you'll never form C2H5- and H+. Similar for NH3 - for any of those protons to come off the nitrogen, it would have to be a strong acid. But we know that it's not a strong acid, because it's a strong base and it's NOT amphiprotic. You can also gather this information from VSEPR - if something is still stable after taking a proton off, then it's acidic. If it doesn't look stable after taking a proton off, then it's not. If you were to draw NH2- using VSEPR, you'd find that the nitrogen would need to have two lone pairs - but that would put too much electron density nitrogen, and it wouldn't be happy, it would no longer be stable. This kind of thing just takes exposure to get used to.

You should know some common acids/bases - but usually, you can discern from the material supplied in the question if something will ionise or not. You can also use solubility to work this out - if something is soluble, it'll ionise. This is why it's important to know common ions, too - that way you know if something will come off an atom or not. Remember - everything you learn is connected to everything else in chemistry, it's important to see and use those links.

Oh ok, I didn't realise it was an open-ended question.

I mean, something something that's your job as the student. Make sure you're reading the questions carefully - when you're asked how you would do an experiment, analysis, find something out, etc., and they haven't given you an experiment brief of some kind, then it should be ringing alarm bells in your head that the question is likely going to be a bit more open-ended.

Could I just ask, so they randomly chose to use 50 mL of juice but how did they work out that 14.20 mL of iodine is required for that?

The same way you would answer any titration question - you have a maximum concentration of vitamin c to expect, you have a chemical equation to work with, all you've gotta do now is the maths you'd normally do for a titration.

tigerclouds

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Re: VCE Chemistry Question Thread
« Reply #8692 on: August 23, 2020, 01:40:32 pm »
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Thank you for explaining that, it makes more sense.
So this comes down to knowing whether something ionises or not - and to know that, you'll need to know if something is a strong acid or base. Strong acids and bases will ionise, weak ones will not. For example, ethane (C2H6) is an extremely weak acid - it won't ionise, so you'll never form C2H5- and H+. Similar for NH3 - for any of those protons to come off the nitrogen, it would have to be a strong acid. But we know that it's not a strong acid, because it's a strong base and it's NOT amphiprotic. You can also gather this information from VSEPR - if something is still stable after taking a proton off, then it's acidic. If it doesn't look stable after taking a proton off, then it's not. If you were to draw NH2- using VSEPR, you'd find that the nitrogen would need to have two lone pairs - but that would put too much electron density nitrogen, and it wouldn't be happy, it would no longer be stable. This kind of thing just takes exposure to get used to.

You should know some common acids/bases - but usually, you can discern from the material supplied in the question if something will ionise or not. You can also use solubility to work this out - if something is soluble, it'll ionise. This is why it's important to know common ions, too - that way you know if something will come off an atom or not. Remember - everything you learn is connected to everything else in chemistry, it's important to see and use those links.
So are we going to have to memorise common examples of strong and weak acids and bases? Is there a faster and more logical way to determine whether a chemical is a strong acid or base (other than the OH group indicating it's a strong base)?

but usually, you can discern from the material supplied in the question if something will ionise or not.
Could you please give an example of this?
Thanks again.

Sine

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Re: VCE Chemistry Question Thread
« Reply #8693 on: August 23, 2020, 01:46:02 pm »
+4
Thank you for explaining that, it makes more sense.So are we going to have to memorise common examples of strong and weak acids and bases? Is there a faster and more logical way to determine whether a chemical is a strong acid or base (other than the OH group indicating it's a strong base)?
Thanks again.
Definitely have a few examples (I think 3 is good) for strong/weak acids and bases. VCAA tend to keep using the same sort of acids/bases in their questions so you will see the same things cropping up all the time. With more experience, you get better at thinking about whether something is likely to be a strong/weak acid or base.

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Re: VCE Chemistry Question Thread
« Reply #8694 on: August 24, 2020, 02:34:08 pm »
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Would it be incorrect to say polysaccharides are “non-crystalline and generally soluble in water” this was a test of one of my formative tests but my teacher is yet to mark it and the sac is soon. I chose this as the incorrect MC answer from the other 4.
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Re: VCE Chemistry Question Thread
« Reply #8695 on: August 24, 2020, 03:57:14 pm »
+1
Would it be incorrect to say polysaccharides are “non-crystalline and generally soluble in water” this was a test of one of my formative tests but my teacher is yet to mark it and the sac is soon. I chose this as the incorrect MC answer from the other 4.

Before we answer this, what are your thoughts on why this statement is incorrect?

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Re: VCE Chemistry Question Thread
« Reply #8696 on: August 25, 2020, 10:48:54 am »
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1. (Image attached) Do we name this ethyl 3-methylpentanoate or just ethyl pentanoate?

2. For 1,1-dicholorobutane, how do we know which proton/hydrogen environment will have the largest downfield shift in H-NMR?
3. Does lower chemical shift means it has less attractions with its bonded atoms?

Thanks!

Chocolatepistachio

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Re: VCE Chemistry Question Thread
« Reply #8697 on: August 26, 2020, 07:09:21 pm »
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Hello
would this be right
also this question is a bit weird doesn't it already tell you the concentration of sodium hydroxide in the question

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Re: VCE Chemistry Question Thread
« Reply #8698 on: August 26, 2020, 07:27:12 pm »
+5
1. (Image attached) Do we name this ethyl 3-methylpentanoate or just ethyl pentanoate?

2. For 1,1-dicholorobutane, how do we know which proton/hydrogen environment will have the largest downfield shift in H-NMR?
3. Does lower chemical shift means it has less attractions with its bonded atoms?

Thanks!

I mean - why do you think that you wouldn't include the 3-methyl?

Chemical shift doesn't relate exactly to bond strength, but I think it can influence it? Chemical shift is a whole can of worms that you're learning about all the way up to after uni levels of chemistry. It's scary and I don't like it lol. As for dichlorobutane - the effect of the chlorine should be inductive, which means that the effect gets stronger the closer the proton environments are, and you can think of them as kinda happening "through bonds". This isn't technically correct, but is the easiest way to explain it - particularly at a VCE level. Chlorine, and halogens in general, are electron-withdrawing groups - this means that electrons are pulled closer to them. This also means that the electrons are pulled further away from the protons closer to those chlorine atoms. The further electrons are from those protons, the further downfield (ppm go up) their chemical shift will be. So, the proton attached to the first carbon will have the highest ppm, the next two (on the second carbon) will have the next highest, and so on and so forth. However, this effect will get smaller very quickly - and I wouldn't be surprised if the protons on carbon 2 and the protons on carbon 3 still overlap.

Also, you can use the NMR to confirm this. In 1,1-dichlorobutane, you expect to see four signals with different integrations - one for the first carbon (1H), one for the second carbon (2H), one for the third carbon (2H), and one for the fourth (3H). If your peak with the highest ppm has an integration of 1, and your lowest ppm has an integration of 3, then you know that the ones in the middle should follow the trend, with the left middle one being carbon 2, and the right middle one being carbon 3.

I do want to add - because chemical shift is so complicated, typically VCAA try not to ask you about assigning NMRs that require you to use it. This is also why you have some chemical shifts in your data booklet, so you don't have to do it from scratch.

Hello
would this be right
also this question is a bit weird doesn't it already tell you the concentration of sodium hydroxide in the question

Question isn't weird at all - it's saying, "we know the concentration is about 0.2, but we want to know it to more sig figs. Tell us what the more sig figs are". Think about it - if you want to use NaOH in a titration to HCl, then you're only going to know the HCl to however many sig figs you know the NaOH, so it's worth standardising the NaOH further so you can use it in better experiments.

However, what you've done is wrong, because you've essentially done the equation n=CV to find n, but then put that back into C=n/V using the same V number - it's basically the equivalent of doing x=3, adding one to both sides to get x+1=4, and then subtracting 1 from both sides to get x=3. What you need to do is INSTEAD use the amount of mole from the titration to get a BETTER concentration for NaOH.

This is the type of question that tests if you're thinking about the situation, or just plugging numbers into equations.

Coolgalbornin03Lo

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Re: VCE Chemistry Question Thread
« Reply #8699 on: August 26, 2020, 09:14:20 pm »
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Before we answer this, what are your thoughts on why this statement is incorrect?

I feel like I read somewhere although monosaccharides and soluble due to OH’s polysaccharides aren’t but I’m not sure why? Because they have polar glycosidic links but is it the packing?
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2020: English | Methods | Biology | Chemistry |              Psychology | ATAR: 0
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