# ATAR Notes: Forum

## VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE => VCE Chemistry => Topic started by: psyxwar on November 18, 2013, 05:43:57 pm

Title: VCE Chemistry Question Thread
Post by: psyxwar on November 18, 2013, 05:43:57 pm
VCE CHEMISTRY Q&A THREAD

To go straight to posts from 2018, click here.

What is this thread for?
If you have general questions about the VCE Chemistry course or how to improve in certain areas, this is the place to ask! 👌

Who can/will answer questions?
Everyone is welcome to contribute; even if you're unsure of yourself, providing different perspectives is incredibly valuable.

Please don't be dissuaded by the fact that you haven't finished Year 12, or didn't score as highly as others, or your advice contradicts something else you've seen on this thread, or whatever; none of this disqualifies you from helping others. And if you're worried you do have some sort of misconception, put it out there and someone else can clarify and modify your understanding!

There'll be a whole bunch of other high-scoring students with their own wealths of wisdom to share with you, including TuteSmart tutors! So you may even get multiple answers from different people offering their insights - very cool.

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OTHER CHEMISTRY RESOURCES

Original post.
Post all your chemistry questions here :) Please help each other out - that's how you all improve! Many high achieving past students will also be floating around to help you out.

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Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on November 18, 2013, 05:44:43 pm
Anyway, what is the best way to learn the different analysis techniques? For example, mass spec. Should I aim to develop a deep understanding of the process, or is that a waste of time?
Title: Re: VCE Chemistry Question Thread
Post by: Stick on November 18, 2013, 06:08:47 pm

No, "trendsetting 2013ers" is getting old and daft now. Why not this instead?

(http://i.imgur.com/zm7GcS8.jpg)

That's where it is now.
Title: Re: VCE Chemistry Question Thread
Post by: Scooby on November 18, 2013, 06:19:19 pm
Anyway, what is the best way to learn the different analysis techniques? For example, mass spec. Should I aim to develop a deep understanding of the process, or is that a waste of time?

Waste of time. It's important that you're able to interpret the results, but as for understanding the techniques themselves, you'll get by with a pretty basic level of knowledge
Title: Re: VCE Chemistry Question Thread
Post by: DJA on November 18, 2013, 06:37:30 pm
Alright, considering some of us will have started/ be starting the coursework around now

I am still embroiled in 1/2 exams. Arghh…..

Nevertheless, I am looking forward to Chemistry next year. It’s been a really fun subject this year. Anyone doing uni chem in '14?
Title: Re: VCE Chemistry Question Thread
Post by: clıppy on November 18, 2013, 06:43:16 pm
Look at that, I might actually be able to help here next year. Unless of course nliu is still around, I can't compete with that.
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on November 18, 2013, 06:44:18 pm
I am still embroiled in 1/2 exams. Arghh…..

Nevertheless, I am looking forward to Chemistry next year. It’s been a really fun subject this year. Anyone doing uni chem in '14?
Yeah me too :P

And yes, I'm doing MUEP, hopefully at Scotch! Looking forward to meeting Chris Commons :D
Title: Re: VCE Chemistry Question Thread
Post by: DJA on November 18, 2013, 07:05:25 pm
Yeah me too :P

And yes, I'm doing MUEP, hopefully at Scotch! Looking forward to meeting Chris Commons :D

Really??

Sweet! I’m doing uni chem too so I’ll probably get to meet you in person next year. Chris Commons is an absolutely stellar bloke! I’m sure we will have an awesome class. My friends who are doing uni chem are genii at chem-very smart guys.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on November 18, 2013, 07:26:31 pm
Should I spend the summer reading over area of study one of unit 3?? I'm probably going to spend a bit of time going over 1/2 content relevant to 3/4 chem.
Title: Re: VCE Chemistry Question Thread
Post by: clıppy on November 18, 2013, 07:47:17 pm
Should I spend the summer reading over area of study one of unit 3?? I'm probably going to spend a bit of time going over 1/2 content relevant to 3/4 chem.
The way it went at my school the only assumed things were knowing how to find mols, particles, concentration and using PV=nRT. Almost everything else that was needed we went through again.
If you feel like you're not confident with those things (and possibly redox too) you could revise them, but from a personal standpoint, I see no reason to revise much of 1/2.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on November 18, 2013, 08:02:06 pm
OK.
Personally, what I found was needed from 1/2 chemistry in 3/4:
Organic nomenclature
Really simple acid-base stuff
Redox reactions
Gases, concentrations and stoichiometry
Maybe solubility? Not too important in 3/4

As you can see, there's not too much that you DO need to know from 1/2

Anyway, what is the best way to learn the different analysis techniques? For example, mass spec. Should I aim to develop a deep understanding of the process, or is that a waste of time?

Personally, the way I learnt them was that I kept reading them until I could explain them to myself and that it made sense. A deep understanding isn't really needed; perhaps just learn what's in the textbook. I've found that's more than enough.

Look at that, I might actually be able to help here next year. Unless of course nliu is still around, I can't compete with that.

Hey, with that attitude, you really won't be able to help :P

Should I spend the summer reading over area of study one of unit 3?? I'm probably going to spend a bit of time going over 1/2 content relevant to 3/4 chem.

What I did in the holidays was that I read through the entire textbook and noted what I found easy and what wasn't so easy. For me, generally it was ok, aside from fractional distillation (like, the exact mechanisms at play, and guess what happened to this part of the course?), analytical techniques and biomolecules (I just hated biology in year 10). I did enough reading to have an idea of what they were, but maybe not a particularly in-depth knowledge. By the time class time came to look over those things, I had seen them before, I had thought about them, and remembering that stuff became easy.
Title: Re: VCE Chemistry Question Thread
Post by: Penguuu on December 28, 2013, 11:45:39 am

Please help!  :) (this thread kind of died, not sure where to post haha)
A solution of potassium permanganate can be standardised using pure iron wire. in a particular experiment 0.317 g of iron wire was dissolved such that Fe2+ ions were formed, and the resulting solution was made up to 250.0 mL.
20ml aliquots of this solution were then taken, with 11.72ml of the permanganate solution being required.
calculate the molarity of the permanganate solution.
Title: Re: VCE Chemistry Question Thread
Post by: DJA on December 28, 2013, 12:21:49 pm
Please help!  :) (this thread kind of died, not sure where to post haha)
A solution of potassium permanganate can be standardised using pure iron wire. in a particular experiment 0.317 g of iron wire was dissolved such that Fe2+ ions were formed, and the resulting solution was made up to 250.0 mL.
20ml aliquots of this solution were then taken, with 11.72ml of the permanganate solution being required.
calculate the molarity of the permanganate solution.

So a Volumetric analysis question:
Step 1: Redox reaction check lzxnl's post above

Looking at the question lets convert everything to moles first as it is easier!
n(Fe2+)= 0.317/55.8 = 0.005681mol total in 250.0 mL of solution.

Thus in a 20mL aliquot, we use a down-scaling factor to find out the amount of Fe2+ in 20 mL
ie. n(Fe2+) in a 20mL aliquot = 20/250 x 0.005681 = 0.0004545 mol

Now remember the mole ratio worked out earlier, we can see that we need 2 times more MnO4- than Fe2+ for the reaction to go to completion (the 1:5 ratio).
Thus we can conclude:
n(MnO4-) required = 5 x n(Fe2+)
= 1/5 x 0.0004545 = 9.09*10-5 mol

Now to find the concentration of the permanganate solution. We know that 11.72mL of the solution reacted to completion with the Fe2+ solution. We also know the mol which reacted from our working so far. So we are left with an arbitrary calculation:
C(permanganate solution) = 9.09*10-5/0.01172 = 7.76M*10-3

The concentration of the permaganate solution is 7.76M*10-3 rounded to 3 sf.

Cheers lzxnl for checking.
Title: Re: VCE Chemistry Question Thread
Post by: Penguuu on December 28, 2013, 01:08:02 pm
Thanks DJALogical, unfortunately I dont have the solutions, but it looks right.
Ive always been confused with the 'down scaling', didnt know you could do that!
Thank you very much :)
Title: Re: VCE Chemistry Question Thread
Post by: DJA on December 28, 2013, 03:14:58 pm
Thanks DJALogical, unfortunately I dont have the solutions, but it looks right.
Ive always been confused with the 'down scaling', didnt know you could do that!
Thank you very much :)

You're welcome!

On a side note: Should this question forum be stickied for ease of access (like in bio for example)? Mod anyone? :D
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on December 28, 2013, 03:46:31 pm
Please help!  :) (this thread kind of died, not sure where to post haha)
A solution of potassium permanganate can be standardised using pure iron wire. in a particular experiment 0.317 g of iron wire was dissolved such that Fe2+ ions were formed, and the resulting solution was made up to 250.0 mL.
20ml aliquots of this solution were then taken, with 11.72ml of the permanganate solution being required.
calculate the molarity of the permanganate solution.

So a Volumetric analysis question:
The reaction between the potassium permanganate and the iron wire can be written omitting spectator ions in the ionic form as follows (Due to the potassium permanganate being composed of K+ and MnO4-)

MnO4-  + Fe2+ ---> Fe(MnO4)2
If we balance this equation we put a 2 in front of the MnO4 and get a 2:1 mole ratio:
2MnO4-  + Fe2+ ---> Fe(MnO4)2
Keep this in mind for later.

Looking at the question lets convert everything to moles first as it is easier!
n(Fe2+)= 0.317/55.8 = 0.005681mol total in 250.0 mL of solution.

Thus in a 20mL aliquot, we use a down-scaling factor to find out the amount of Fe2+ in 20 mL
ie. n(Fe2+) in a 20mL aliquot = 20/250 x 0.005681 = 0.0004545 mol

Now remember the mole ratio worked out earlier, we can see that we need 2 times more MnO4- than Fe2+ for the reaction to go to completion (the 2:1 ratio).
Thus we can conclude:
n(MnO4-) required = 2 x n(Fe2+)
= 2 x 0.0004545 = 0.0009090 mol

Now to find the concentration of the permanganate solution. We know that 11.72mL of the solution reacted to completion with the Fe2+ solution. We also know the mol which reacted from our working so far. So we are left with an arbitrary calculation:
C(permanganate solution) = 0.0009090/0.01172 = 0.07756M

The concentration of the permaganate solution is 0.07756M rounded to 4 sf.
Can anyone else confirm my answer by checking the caluculations?

(This should be right, disclaimer I haven't done volumetric analysis for ages so I may have errors but hopefully not ;))

Firstly...I don't quite get your first step, DJALogical. The reaction between iron(II) and potassium permanganate is not a precipitation reaction but a redox reaction. Permanganate is a pretty powerful oxidant, enough to oxidise iron(II) to iron(III).

MnO4-(aq) + 8H+(aq) + 5 Fe2+(aq) => 5 Fe3+(aq) + Mn2+(aq) +4H2O(l)

I can just write this equation out because I've dealt with this equation too many times in trial exams, and because you can just replace any electrons in the permanganate reduction equation with Fe(II) ions :D

So...0.317 g iron wire eh? Molar mass of Fe = 55.8 g/mol => n(Fe2+) = 0.317/55.8 mol = 5.681 mmol
As you make a 250 mL solution of this, and then take 20 mL of it, you really just have 20mL/250 mL*5.681 mmol = 0.45448 mmol
Mol ratio suggests that n(Fe2+)=5n(MnO4-) => n(MnO4-)=9.0896*10-5 mol

Concentration = n/V = 9.0896*10-5 mol / 0.01172 L = 7.76*10-3 M
Note that as the mass of the wire is to 3 s.f., the answer is only to 3 s.f.

Interesting. My answer differs from yours by a factor of ten exactly.
Title: Re: VCE Chemistry Question Thread
Post by: DJA on December 28, 2013, 03:52:23 pm
Interesting. My answer differs from yours by a factor of ten exactly.

:) lol this is what happens when I haven't done redox titrations since like midyear chem olympiad-totally missed on on the possibility of it being a redox reaction.
Yep since the mole ratio if you realise it is redox is 5:1 whereas if you incorrectly assume precip, you get a mole ratio of 1:2.
Otherwise theoretically the steps following the redox reaction are the same.

Thanks for the pick up!! I'm modifying my post so I dont screw over the next poor sod who comes along.
Title: Re: VCE Chemistry Question Thread
Post by: DJA on December 28, 2013, 03:59:53 pm
lzxnl just for my own understanding could you write out the two separate ionic equations for the permaganate and Iron (II) so I can see what they look like and balance out? I've gotten it written down on paper I just want to check if its right with you. :)
Its been ages since I did redox.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on December 28, 2013, 04:03:41 pm
lzxnl just for my own understanding could you write out the two separate ionic equations for the permaganate and Iron (II) so I can see what they look like and balance out? I've gotten it written down on paper I just want to check if its right with you. :)
Its been ages since I did redox.

MnO4-(aq) + 8H+(aq) + 5e- => Mn2+(aq) +4H2O(l)

5Fe2+(aq) => 5Fe3+(aq) + 5e-
Title: Re: VCE Chemistry Question Thread
Post by: DJA on December 28, 2013, 04:06:07 pm
MnO4-(aq) + 8H+(aq) + 5e- => Mn2+(aq) +4H2O(l)

5Fe2+(aq) => 5Fe3+(aq) + 5e-

Cheers!

Also I wanted to ask, is there any 'atmospheric chemistry' in Chemistry 3/4? Because I hated that stuff in year 11 and am sincerely hoping I don't have to do it again.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on December 28, 2013, 04:15:02 pm
Nope. None of that rubbish at all. I hated it too.
If you're keen, look through last year's exam for a guideline on what's on the course now.
Title: Re: VCE Chemistry Question Thread
Post by: Penguuu on December 28, 2013, 05:04:32 pm
Ohhhh, I clearly need to practice more haha. Thanks both of youse ^^
Title: Re: VCE Chemistry Question Thread
Post by: Penguuu on December 29, 2013, 10:56:56 am
Hi guys, I'm stuck with this question.
A solution of hydrochloric acid of concentration 0.1M has a pH of 1. Ethanoic acid of the same concentration has a pH of approximately 3. Explain how this difference arises.
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on December 29, 2013, 11:22:46 am
Hi guys, I'm stuck with this question.
A solution of hydrochloric acid of concentration 0.1M has a pH of 1. Ethanoic acid of the same concentration has a pH of approximately 3. Explain how this difference arises.
Ethanoic acid is a weak acid and unlike hydrochloric acid (a strong acid), does not completely ionise in water. Ionising less = less H+ in solution = higher pH. In a solution of HCl, most of the molecules are Cl- and H+, whereas in a solution of ethanoic acid, most of the molecules are ethanoic acid molecules (CH3COOH) that have not ionised.
Title: Re: VCE Chemistry Question Thread
Post by: DJA on December 29, 2013, 11:24:50 am
Hi guys, I'm stuck with this question.
A solution of hydrochloric acid of concentration 0.1M has a pH of 1. Ethanoic acid of the same concentration has a pH of approximately 3. Explain how this difference arises.

Hydrochloric acid is a strong acid. A strong acid is one which ionises completely in water. We use a straight arrow to denote this
HCl + H2O -> H3O+ + Cl-

Ethanoic acid is a weak acid. A weak acid does not completely ionise in water, we use a reversible arrow to denote it.
CH3COOH + H2O ↔ H3O+ + CH3COO-

The H+ or H3O+ species is formed when an acid reacts with water. pH for practical purposes is the measure of the concentration of the hydronium ion H3O+ (the negative logarithm of the hydronium ion activity). Since ethanoic acid is not completely ionized, there is less than the maximum possible number of hydronium ions in the solution (there are un-ionized ethanoic acid molecules floating around) hence making it a higher pH than hydrochloric acid which is fully ionized and has more hydronium ions  in the solution than ethanoic acid giving it a lower pH.

(Edit:psyxwar beat me to it but ill post anyway for the sake of completeness)
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on December 29, 2013, 05:02:46 pm
Hi, can you guys help me with this question:
Write a balanced ionic equation for the reaction between iodine and sodium thisulfate (Na2S2O3)
I2 (aq) + 2Na2S2O3 (aq) -> 2NaI (aq) + Na2S4O6 (aq)
Thanks for help!!
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on December 29, 2013, 05:44:37 pm
In this reaction, the sodium ions don't really do anything so your ionic reaction is I2(aq)+2S2O32-(aq) => 2I-(aq) + S4O62-(aq)
Title: Re: VCE Chemistry Question Thread
Post by: Penguuu on January 02, 2014, 05:44:22 pm
Hi guys, I would love some help with this question. Ive already got the worked out solutions but i still dont understand it.
25.00ml of a 0.100 M solution of HCl is added to 25.00ml of a 0.180 M solution of naOH.
The concentration of OH-(aq) remaining in the solution, in M is

A. 0.0400

Thanks!  :)
Title: Re: VCE Chemistry Question Thread
Post by: brightsky on January 02, 2014, 06:04:38 pm
25.00mL of a 0.100 M solution of HCl is added to 25.00mL of a 0.180 M solution of NaOH. Find the concentration of OH-(aq) remaining in the solution in M.

The ionic equation for this reaction is:

H+(aq) + OH- (aq) --> H2O(l)

Note that Na+(aq) and Cl-(aq) are merely spectator ions.

n(H+) = n(HCl) = c*V = 0.02500*0.100 = 0.00250 mol
n(OH-) = n(NaOH) = c*V = 0.02500*0.180 = 0.00450 mol

It is clear that H+ is the limiting reagent. Hence, n(OH-) remaining after the reaction goes to completion is 0.00450 - 0.00250 = 0.00200.

This means that c(OH-) = n/V = 0.00200/0.05000 = 0.0400 M, as required.

Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 04, 2014, 10:35:39 am
When a certain non-metal whose formula is X8 burns in air XO3 forms. Write a balanced equation for this reaction. If 120.0g of oxygen gas is consumed completely, along with 80.0g of X8, identify element X.

Is this balanced equation correct: X8 (g) + 12 O2 (g) -> 8XO3 (g)
I'm not sure how to proceed with this question....

Your help is appreciated   :D
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on January 04, 2014, 10:42:40 am
When a certain non-metal whose formula is X8 burns in air XO3 forms. Write a balanced equation for this reaction. If 120.0g of oxygen gas is consumed completely, along with 80.0g of X8, identify element X.

Is this balanced equation correct: X8 (g) + 12 O2 (g) -> 8XO3 (g)
I'm not sure how to proceed with this question....

Your help is appreciated   :D

Sorry. The chemical formula makes it look slightly obvious as to what it is :P

Just saying, X8 looks like a solid to me. But you're not meant to know that.

Your equation is balanced correctly though. Now, you know that 120.0 g of oxygen gas is used up. This is 120/32=15/4 moles of oxygen gas (divide by 32 not 16 remember). This must have reacted with exactly 15/48 moles of X8 by mole ratios. If this mass of X8 is 80 grams, one mole of your unknown compound is 80*48/15=80*16/5=256 grams
This is for X8, so the molar mass of X is 32 grams
AKA sulfur

How many non-metals are there anyway that exist as X8 and form trioxides? :P
Title: Re: VCE Chemistry Question Thread
Post by: brightsky on January 04, 2014, 10:46:36 am
Sulfur fits the bill quite nicely. Let see if it is indeed sulfur...

n(O2) = m/M = 120.0/(2*16.0) = 120.0/32.0 = 3.75 mol
n(X8) = m/M = 80.0/M(X8) = 3.75/12 = 0.3125 mol (since the n(X8) : n(O2) = 1 : 12)
M(X8) = 80.0/0.3125 = 256 g/mol
M(X) = 256/8 = 32 g/mol

The element X is indeed sulfur.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 04, 2014, 10:55:23 am
X8 reacts completely with oxygen; a combustion reaction.

X8(s) + 12O2(g) ---> 8XO3(s)

Now that we have our balanced equation, we use stoichiometry.

We know that we've reacted 120.0g of O2 with 80g of X8.

n(O2): n(X8)
12: 1
120/32: n
Cross multiply
n(X8) = 0.3125 mol

We want to find the molecular mass of this metal, so.

0.3125 = 80 / mr
Mr(X8) = 256
Mr(X) = 256 / 8
Mr(X) = 32

Thus, by consulting our periodic table, we realise the metal is Sulphur.

Title: Re: VCE Chemistry Question Thread
Post by: DJA on January 04, 2014, 11:11:52 am
Sorry. The chemical formula makes it look slightly obvious as to what it is :P

AKA sulfur

How many non-metals are there anyway that exist as X8 and form trioxides? :P

lol

Nliu--> Looks at the question...A few seconds later knows what the answer is.
Who bothers to calculate these days anyway?
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 04, 2014, 11:22:49 am
Haha thanks for all your help and different ways to approaching the question.

I have another question:

A 10.0 mL sample of HNO3 was diluted to a volume of 100.0 mL. 25 mL of the dilute solution was needed to neutralise 50.0 mL of a 0.60 M KOH solution. What was the concentration of the original nitric acid?

Title: Re: VCE Chemistry Question Thread
Post by: DJA on January 04, 2014, 12:30:12 pm
Haha thanks for all your help and different ways to approaching the question.
I have another question:
A 10.0 mL sample of HNO3 was diluted to a volume of 100.0 mL. 25 mL of the dilute solution was needed to neutralise 50.0 mL of a 0.60 M KOH solution. What was the concentration of the original nitric acid?

The equation for the neutralisation reaction from the info is this:
HNO3 + KOH ---> KNO3 +H2O

The mole ratio between HNO3 and KOH is 1:1 meaning that basically 1 mol of HNO3 will react with 1 mol of KOH. Keep this ratio in mind.

Transform everything to moles for calculation:
n(KOH) in moles = 0.60 * 0.050 = 0.030mol KOH reacted

Remember the mole ratio? We now know that:
n(HNO3) = n(KOH) = 0.030mol  due to the 1:1 ratio

Now this 0.030mol is from 0.025L of solution. Hence in 0.100 L of solution we use a multiplication factor of 4 to get the amount in mol in the 100mL.
n(HNO3) in 100.0mL = 4 * 0.030mol = 0.120 mol

This amount is the total mol of HNO3 from the very start (which was originally in only 10mL of solution). Therefore we now can easily calculate the concentration of the original HNO3.
C(HNO3) original = 0.120 / 0.010 = 12M concentration for the nitric acid.

(sheesh that's some strong nitric acid...now I am less sure my answer is right)
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 04, 2014, 01:04:24 pm
Yep that's the correct answer thank you!

(sheesh that's some strong nitric acid...now I am less sure my answer is right)
Title: Re: VCE Chemistry Question Thread
Post by: DJA on January 04, 2014, 01:09:16 pm
Yep that's the correct answer thank you!

You're very welcome.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 04, 2014, 01:55:52 pm
How would you draw compound C3H3O6?

Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on January 04, 2014, 02:05:59 pm
How would you draw compound C3H3O6?

That's not a stable molecule; odd number of electrons and organic molecules like that aren't stable
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 04, 2014, 06:01:56 pm
Thanks! But it's the answer I have for this question:

Compound Z consists of 26.7% carbon, 2.2% hydrogen and 71.1% oxygen.
Its molecular mass is 90.0g mol-1.

Can someone please clarify?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 04, 2014, 06:19:04 pm
Thanks! But it's the answer I have for this question:

Compound Z consists of 26.7% carbon, 2.2% hydrogen and 71.1% oxygen.
Its molecular mass is 90.0g mol-1.

Can someone please clarify?

That compound's empirical formula is CHO2

The compound's molecular formula is C2H2O4
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 04, 2014, 06:24:25 pm
Thanks! It makes more sense now
Title: Re: VCE Chemistry Question Thread
Post by: Professor_Oak on January 05, 2014, 12:24:15 pm
Hi guys can somebody explain the logic behind 25c in a relatively detailed manner? I can sort of see how it may work but I'm still a bit confused with this sort of question.
(http://i.imgur.com/2kkVhPd.png)
Title: Re: VCE Chemistry Question Thread
Post by: clıppy on January 05, 2014, 12:55:05 pm
We know in the 20mL of CH3CH2OH we had 0.000690 mol
We want to figure out how many mols of CH3CH2OH was in the initial 250mL used.

The way I did these questions was setting it up like a sort of ratio.

0.000690 mol -> 20mL
x mol -> 250mL

x =0.000690 * (250/20)
Title: Re: VCE Chemistry Question Thread
Post by: brightsky on January 05, 2014, 12:55:39 pm
I have no idea what the actual question is, but I'm assuming that you had a 10 mL beer sample, which you placed in a 250.0 mL volumetric flask, which you dilated to 250.0 mL, from which you then withdrew a 20.00 mL aliquot, with which you then performed some sort of reaction. The question, I'm assuming, is calculate the amount of ethanol in the original 10 mL beer sample.

In part b) of the question, you found that the amount of ethanol in the 20.00 mL aliquot is 0.00068925 mol. Now, you withdrew this 20.00 mL aliquot from the 250.0 mL volumetric flask. Logically, then, the amount of ethanol in the 250.0 mL volumetric flask is 250.0/20.00 multiplied by the amount of ethanol in the 20.00 mL aliquot, which works out to be 0.008615 mol. Now we have the amount of ethanol in the 250.0 mL volumetric flask. Now, all the ethanol in the volumetric flask was sourced from the beer sample; you placed the beer sample into the volumetric flask and then added distilled/deionised water until the volumetric flask was filled up to the calibration mark. Logically, then, the amount of ethanol in the beer sample would equal to the amount of ethanol in the 250.0 mL volumetric flask. Hence, the amount of ethanol in the beer sample is 0.008615 mol, which, to three sig figs, is 0.00862 mol.

Let me know if any of the assumptions I made above are incorrect.
Title: Re: VCE Chemistry Question Thread
Post by: DJA on January 05, 2014, 12:58:43 pm
Hi guys can somebody explain the logic behind 25c in a relatively detailed manner? I can sort of see how it may work but I'm still a bit confused with this sort of question.
Image removed)

Hey mate. So basically what 25 c) is explaining is what I call a scaling factor.
Note that so far from the calculations, the n(CH3CH2OH)in 20mL=0.000690mol. This amount is the amount in only 20mL of solution which was removed from the 250.0mL flask for the purpose of the titration.

In order to find the total amount of CH3CH2OH which came from the beer sample, we multiply by 250.0/20.0 which is just a scaling factor which calculates how much mol of CH3CH2OH was there originally.

After that multiplication we get the answer 0.008615mol which is the amount orginally in the 250.0mL flask which incendentally is also the amount from the original beer sample (10mL) which was transferred into the 250mL flask at the beginning of the titration.
Then the answer is just rounded to 3 sf to get the final solution 0.00862mol.

Hope that made sense. Post if you need clarification  :)

brightsky beat me to it but i'll post anyway
Title: Re: VCE Chemistry Question Thread
Post by: Professor_Oak on January 05, 2014, 01:47:42 pm
Thanks guys! I think I get it now. And DJA logical I'll see you next year for MUEP chem!

We know in the 20mL of CH3CH2OH we had 0.000690 mol
We want to figure out how many mols of CH3CH2OH was in the initial 250mL used.

The way I did these questions was setting it up like a sort of ratio.

0.000690 mol -> 20mL
x mol -> 250mL

x =0.000690 * (250/20)
I think this explanation is the best way to put it, it reminds me a bit of the solubility stuff from 1/2.
Title: Re: VCE Chemistry Question Thread
Post by: DJA on January 05, 2014, 04:12:09 pm
Thanks guys! I think I get it now. And DJA logical I'll see you next year for MUEP chem!

Make sure you wear a name tag called 'Professor Oak' so I know who you are ;)

I'll write DJALogical on my forehead in black texta for the first session ;D

Also I believe psyxwar MAY be doing MUEP chem as well.
Title: Re: VCE Chemistry Question Thread
Post by: Only Cheating Yourself on January 05, 2014, 05:05:07 pm
How many hours a night do you guys plan on studying?
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on January 06, 2014, 01:45:47 pm
Make sure you wear a name tag called 'Professor Oak' so I know who you are ;)

I'll write DJALogical on my forehead in black texta for the first session ;D

Also I believe psyxwar MAY be doing MUEP chem as well.
Yeah I am.

Fractional distillation was taken off the course right? The 2013 SD has "production of fuels from fractional distillation of crude oil" in a table of key ideas under the "Advice for Teachers" bit, but in the summary of changes it seems to have been deleted.
Title: Re: VCE Chemistry Question Thread
Post by: brightsky on January 06, 2014, 01:53:48 pm
You need to know what fractional distillation is (i.e. a technique that separates a mixture of liquids based on boiling point differences). You do not, however, need to know anything about how fractional distillation works (e.g. fractionating column, temperature gradient, etc.).
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on January 06, 2014, 01:56:23 pm
You need to know what fractional distillation is (i.e. a technique that separates a mixture of liquids based on boiling point differences). You do not, however, need to know anything about how fractional distillation works (e.g. fractionating column, temperature gradient, etc.).
But what's the point of knowing what it is without knowing how it works? o.o
Title: Re: VCE Chemistry Question Thread
Post by: DJA on January 06, 2014, 02:11:52 pm
But what's the point of knowing what it is without knowing how it works? o.o

None. You just learn it for the sake of the course.  ::)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on January 06, 2014, 03:18:30 pm
But what's the point of knowing what it is without knowing how it works? o.o

psyxwar, you have made the common and forgivable mistake of assuming that the VCE course's main aim is to teach students stuff.
Title: Re: VCE Chemistry Question Thread
Post by: PsychoT on January 07, 2014, 12:56:35 pm
Hi, i've got two questions, if anyone could help me out with them.

1. A precipitate of Chromium (III) oxide is formed from the reaction between potassium hydroxide and Chromium (III) Chloride. What mass of precipitate would be formed if 500ml of 0.25M potassium hydroxide was reacted with excess chromium (III) chloride?

2. Leaded Petrol contains the additive tetraethyl lead Pb (C2H5)4 which reduces engine knock when the petrol is combusted.
The reaction is represented as follows:
4C2H5Cl (l) + 4 Na (s) + Pb (s) --> Pb(C2H5)4 (l) + 4 NaCl (s)

What mass of chloroethane, C2H4Cl would be required to produced 40.4g of tetraethyl lead, Pb (C2H5)4?

Cheers guys, read through this place for a while. Reckon i'll start to get involved.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on January 07, 2014, 01:14:31 pm
Hi, i've got two questions, if anyone could help me out with them.

1. A precipitate of Chromium (II) oxide is formed from the reaction between potassium hydroxide and Chromium (III) Chloride. What mass of precipitate would be formed if 500ml of 0.25M potassium hydroxide was reacted with excess chromium (III) chloride?

2. Leaded Petrol contains the additive tetraethyl lead Pb (C2H5)4 which reduces engine knock when the petrol is combusted.
The reaction is represented as follows:
4C2H5Cl (l) + 4 Na (s) + Pb (s) --> Pb(C2H5)4 (l) + 4 NaCl (s)

What mass of chloroethane, C2H4Cl would be required to produced 40.4g of tetraethyl lead, Pb (C2H5)4?

Cheers guys, read through this place for a while. Reckon i'll start to get involved.

For the first part, let's break this question down.

The question says excess CrCl3 so the amount of chromium oxide you get is directly determined by the amount of potassium hydroxide there is.

Now, you need a chemical equation for this. Are you sure the chromium (III) chloride is converted into chromium (II) oxide? From my own knowledge, the chromium shouldn't be reduced. Otherwise, this question makes no sense chemically.

Second part.

Work out the number of moles of tetraethyl lead. Then, use the mole ratios to see that the mass of chloroethane required is four times as great as the number of moles of tetraethyl lead. Use this amount of chloroethane to work out the mass of chloroethane.
Title: Re: VCE Chemistry Question Thread
Post by: PsychoT on January 07, 2014, 01:18:53 pm
For the first part, let's break this question down.

The question says excess CrCl3 so the amount of chromium oxide you get is directly determined by the amount of potassium hydroxide there is.

Now, you need a chemical equation for this. Are you sure the chromium (III) chloride is converted into chromium (II) oxide? From my own knowledge, the chromium shouldn't be reduced. Otherwise, this question makes no sense chemically.

Second part.

Work out the number of moles of tetraethyl lead. Then, use the mole ratios to see that the mass of chloroethane required is four times as great as the number of moles of tetraethyl lead. Use this amount of chloroethane to work out the mass of chloroethane.

Cheers. On the first part, that's a typo on my behalf. Should be Chromium (III) oxide.

Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on January 07, 2014, 01:35:51 pm
1. A precipitate of Chromium (III) oxide is formed from the reaction between potassium hydroxide and Chromium (III) Chloride. What mass of precipitate would be formed if 500ml of 0.25M potassium hydroxide was reacted with excess chromium (III) chloride?

OK.
Let's work out the reaction stepwise.
Working out an equation to get chromium (III) hydroxide is easy. You just replace the chlorides in the chromium chloride with hydroxides from KOH. That reaction is  3KOH + CrCl3 => Cr(OH)3 + 3 KCl
Now we want to write a reaction to convert chromium hydroxide into chromium oxide. Note that we can write OH- + OH- => H2O + O2-. This reaction doesn't really happen; it's just to allow us to work out how to balance the equation.
So we have 6OH- => 3H2O + 3 O2-. Adding chromium (III) ions yields 2Cr3+ + 6OH- => 2Cr3+ + 3H2O + 3 O2-. In other words, we can replace 2Cr(OH)3 with Cr2O3 + 3 H2O
Doubling the coefficients of our above equation:
6KOH + 2CrCl3 => 2Cr(OH)3 + 6 KCl
Which can now be rewritten as
6KOH + 2CrCl3 => Cr2O3 + 3 H2O + 6 KCl

Now 500 mL 0.25 M KOH is the limiting reagent, so the mole ratios use n(KOH). We have 0.125 moles of KOH, so from mole ratios we have a sixth of this as chromium (III) oxide aka 1/48 moles. Use molar mass of chromium oxide. That would be your precipitate; KCl dissolves.
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on January 07, 2014, 01:59:31 pm

Now we want to write a reaction to convert chromium hydroxide into chromium oxide. Note that we can write OH- + OH- => H2O + O2-. This reaction doesn't really happen; it's just to allow us to work out how to balance the equation.
So we have 6OH- => 3H2O + 3 O2-. Adding chromium (III) ions yields 2Cr3+ + 6OH- => 2Cr3+ + 3H2O + 3 O2-. In other words, we can replace 2Cr(OH)3 with Cr2O3 + 3 H2O

Is that method in the textbook?
Or is that just a way to break it down (showing the steps)

Because I just go, "Yeah, Cr2O3 is the product, KCl another (by simply looking at what's left) and then adding H2O for the remaining H atoms etc"

But i'd like to know your process :D
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on January 07, 2014, 02:05:07 pm
I would probably work it out something like that as well, because for me, the first step, the conversion to chromium hydroxide, is relatively simple to see.

That's not in the textbook though; I do whatever method suits the question.
Title: Re: VCE Chemistry Question Thread
Post by: SunnyB on January 07, 2014, 03:39:13 pm
what are you guys doing for holiday hw?
Title: Re: VCE Chemistry Question Thread
Post by: PsychoT on January 07, 2014, 04:46:09 pm
I would probably work it out something like that as well, because for me, the first step, the conversion to chromium hydroxide, is relatively simple to see.

That's not in the textbook though; I do whatever method suits the question.

Cheers mate. Got that downpat now. Appreciate it alot.
Title: Re: VCE Chemistry Question Thread
Post by: Strawberrry on January 09, 2014, 11:06:40 am
How come this acid + base equation doesnt have water as a product?

HCl (aq) + NH3 (aq) -----> NH4Cl (aq)

Thank you in advance :)
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 09, 2014, 11:13:25 am
How come this acid + base equation doesnt have water as a product?

HCl (aq) + NH3 (aq) -----> NH4Cl (aq)

Thank you in advance :)

HCl dissociates to H+ and Cl-. Your product is NH4Cl; so, added to NH3 are the H+ and the Cl-, meaning there isn't any by-product formed. Plus oxygen isn't involved in the reaction, so even if excess H+ existed, water cannot be formed as a by-product.

Hope this helped.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on January 09, 2014, 11:23:53 am
How come this acid + base equation doesnt have water as a product?

HCl (aq) + NH3 (aq) -----> NH4Cl (aq)

Thank you in advance :)

Not all bases directly contain OH-. Only those bases which contain a hydroxide, oxide or carbonate ion (for VCE purposes) form water when neutralised by an acid. An acid is merely a species that donates protons and a base is merely a species that accepts protons. An acid base reaction is thus just a proton transfer reaction; water can only be formed in specific circumstances.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 09, 2014, 11:29:27 am
Not all bases directly contain OH-. Only those bases which contain a hydroxide, oxide or carbonate ion (for VCE purposes) form water when neutralised by an acid. An acid is merely a species that donates protons and a base is merely a species that accepts protons. An acid base reaction is thus just a proton transfer reaction; water can only be formed in specific circumstances.

Is that simply because there is actually oxygen present in the base or acid? So for instance, sodium hydroxide's hydroxyl group reacts with dissociated H+ ions from the acid to form water as the by-product of the reaction. That's the logic I'm following so just clarifying to check whether that's right or wrong.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on January 09, 2014, 11:50:54 am
Is that simply because there is actually oxygen present in the base or acid? So for instance, sodium hydroxide's hydroxyl group reacts with dissociated H+ ions from the acid to form water as the by-product of the reaction. That's the logic I'm following so just clarifying to check whether that's right or wrong.

The presence of oxygen is one thing, but in something like the phosphate ion, which also has oxygen, that doesn't form water when reacted with an acid. The oxide ion forms water because it is the conjugate base of the hydroxide ion (I know right?) and accepts two hydrogens to form water. Carbonate ions form water because carbonic acid is unstable.

You don't call hydroxide ions hydroxyl groups; that term is reserved for OH groups that aren't ions.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 09, 2014, 12:38:53 pm
The presence of oxygen is one thing, but in something like the phosphate ion, which also has oxygen, that doesn't form water when reacted with an acid. The oxide ion forms water because it is the conjugate base of the hydroxide ion (I know right?) and accepts two hydrogens to form water. Carbonate ions form water because carbonic acid is unstable.

You don't call hydroxide ions hydroxyl groups; that term is reserved for OH groups that aren't ions.

Ah ok thanks for the heads up :)
Title: Re: VCE Chemistry Question Thread
Post by: Strawberrry on January 09, 2014, 04:21:24 pm
The first dot point of the study design refers to 'titration curves'... What does it actually mean?
Thanks  :D
Title: Re: VCE Chemistry Question Thread
Post by: clıppy on January 09, 2014, 04:23:36 pm
Graphs that look something like this (http://www.chemguide.co.uk/physical/acidbaseeqia/wasb2.gif)

You mostly use them in questions asking what type of indicator to use by looking at the equivalence point and comparing to your data book.
Title: Re: VCE Chemistry Question Thread
Post by: Tomas Sadauskas on January 09, 2014, 04:39:31 pm
Just wondering, with the titration curve above, why is the graph so steep/vertical around the equivalence point (ph 8-11ish)?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 09, 2014, 04:49:04 pm
Just wondering, with the titration curve above, why is the graph so steep/vertical around the equivalence point (ph 8-11ish)?

Because the end point (the point at which the inducator changes colour, and theoretically, this occurs at the same time as the equivalence point, where reactants are in stoichiometrically equal proportions) is sharp. That is, there is a very precise and quick colour change, represented by the steepness.

When looking at back titrations, you notice that they're essentially done because the end point is broad; that is, there is no real quick & precise colour change of the indicator. In the circumstance of a broad end point, there is a nore diagonal slope which indicates the end point, and so to obtain accurate results (this is obtained with a sharp end point), you carry out a back titration. Hope this helped :)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on January 09, 2014, 04:56:21 pm
Just wondering, with the titration curve above, why is the graph so steep/vertical around the equivalence point (ph 8-11ish)?

Because the end point (the point at which the inducator changes colour, and theoretically, this occurs at the same time as the equivalence point, where reactants are in stoichiometrically equal proportions) is sharp. That is, there is a very precise and quick colour change, represented by the steepness.

When looking at back titrations, you notice that they're essentially done because the end point is broad; that is, there is no real quick & precise colour change of the indicator. In the circumstance of a broad end point, there is a nore diagonal slope which indicates the end point, and so to obtain accurate results (this is obtained with a sharp end point), you carry out a back titration. Hope this helped :)

It's not really that. Let's say we're adding an acid to a base. Near the end point, the amount of base is almost zero. Adding the same amount of acid then leads to a larger drop in pH because less of it is neutralised by the base. This is why the curve would drop more. And vice-versa if we were adding base to acid.

The reason why weak acid and weak base titrations don't have clear endpoints is because when adding the weak acid to the weak base, the concentrations of H+ and OH- are quite low. Therefore, the reaction doesn't take place as readily and adding bits of the weak acid doesn't lead to as great a change in the concentration of H+.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 09, 2014, 05:00:52 pm
What is the most VCAA could ask us to so with a titration curve? I know how to read off whether the aliquot is an acid/base and determine its strength, I know how to identify equivalence point and which indicator would be best to observe the end point. Is this sufficient? Is there any more? I mean, concentrations could be read off and then apply values to stoichiometry, but thats about all I found.

It's not really that. Let's say we're adding an acid to a base. Near the end point, the amount of base is almost zero. Adding the same amount of acid then leads to a larger drop in pH because less of it is neutralised by the base. This is why the curve would drop more. And vice-versa if we were adding base to acid.

The reason why weak acid and weak base titrations don't have clear endpoints is because when adding the weak acid to the weak base, the concentrations of H+ and OH- are quite low. Therefore, the reaction doesn't take place as readily and adding bits of the weak acid doesn't lead to as great a change in the concentration of H+.

Thanks :)
Title: Re: VCE Chemistry Question Thread
Post by: Edward21 on January 09, 2014, 08:55:23 pm
Just a little heads up, in the new 2013 study design, titration curves are mentioned explicitly in a new dot-point, I think you can interpret that as: they're trying to make it clear that titration curves need to be understood well, and you could potentially be asked to graph one. They'd be a little tricky though, they'd use a weak acid or base. Ie. Titration curve of NaOH and CH3COOH. Where at the equivalence point you have water and the CH3COO- ethanoate ion, this is the conjugate base of the acid CH3COOH which accepts H+ from H2O forming OH- ions (at least that's what I understood from the textbook..)

That's why the pH isn't 7 or "neutral" at 25 degrees, but above 7 at 25 degrees with more [OH-]>[H+] in the solution. It could be a 3 mark question where 1 is for the correct shape --> added base (NaOH) and the line keeps on rising or pH rises with more basic titre volume added, 2nd mark for the equivalence point being over 7 due to the reason I stated, and the 3rd mark for the correct aligning of the equivalence point (to neutralise the acid) with the volume of NaOH as per calculation.

I expected this to come up in the 2013 exam, so just familiarise yourself with doing this just in case for the 2014 exam :)
Title: Re: VCE Chemistry Question Thread
Post by: Strawberrry on January 11, 2014, 06:17:10 pm
Thank you everyone!!

When you dilute something, does the amount, in mol, stay the same?
Title: Re: VCE Chemistry Question Thread
Post by: DJA on January 11, 2014, 06:57:59 pm
Thank you everyone!!

When you dilute something, does the amount, in mol, stay the same?

Yes.
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on January 11, 2014, 07:15:45 pm
On that note, what's the purpose of diluting a solution before like a titration?
Is it just to get enough volume?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 11, 2014, 07:20:05 pm
On that note, what's the purpose of diluting a solution before like a titration?
Is it just to get enough volume?

Well you don't have to dilute something. But the reason dilutions are made is usually because you have a stock of 12M HCl for example, and you want about 0.5M only. Therefore, you dilute! :)

Also, the smaller the concentration, the more credible your experimental results.
Title: Re: VCE Chemistry Question Thread
Post by: brightsky on January 11, 2014, 09:06:45 pm
On that note, what's the purpose of diluting a solution before like a titration?
Is it just to get enough volume?

If the concentration of the titrant is too high, then the titre would, in all likelihood, be too small, since then a smaller volume of the titrant is needed for the equivalence point to be reached. We usually want the titre to be of moderate measure (not too small and not too big), so in order to achieve this, we dilute before the titration.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 11, 2014, 09:09:51 pm
If the concentration of the titrant is too high, then the titre would, in all likelihood, be too small, since then a smaller volume of the titrant is needed for the equivalence point to be reached. We usually want the titre to be of moderate measure (not too small and not too big), so in order to achieve this, we dilute before the titration.

So does that mean that if the concentration is too high, the titre value will be very low, and so the accuracy and credibility of results is less than if the titre value was higher, achieved by diluting your titrant??

Just need a clarification! Thanks brightsky!
Title: Re: VCE Chemistry Question Thread
Post by: brightsky on January 11, 2014, 09:15:57 pm
So does that mean that if the concentration is too high, the titre value will be very low, and so the accuracy and credibility of results is less than if the titre value was higher, achieved by diluting your titrant??

Just need a clarification! Thanks brightsky!

Pretty much! We wouldn't want the concentration of the titrant to be too low either, because then the titre would be too large. Consider what would happen if you had a titre of 60 mL and a burette with a maximum capacity of 50 mL...
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 11, 2014, 09:18:03 pm
Pretty much! We wouldn't want the concentration of the titrant to be too low either, because then the titre would be too large. Consider what would happen if you had a titre of 60 mL and a burette with a maximum capacity of 50 mL...

LOL!

Thanks brightsky.

Title: Re: VCE Chemistry Question Thread
Post by: PsychoT on January 12, 2014, 06:09:34 pm
I'm baaaccck.

Anyone care to help me out with this one?

If 50cm3 of 0.1M HNO3 is mixed with 60cm3 of 0.1M Ca(OH)2, what volume of 0.050M H2SO4 would be required to neutralize the solution?

Cheers.
Title: Re: VCE Chemistry Question Thread
Post by: PsychoT on January 12, 2014, 06:19:52 pm
Got one more, the answers in this little booklet I have are pissing me off, I reckon most have mistakes.

What is the mass of the Phosphorus Oxide P4O10 that is produced from the reaction between 4g of P406 and 4g of I2?

5P406 + 8I2 --> 4P2I4 + 3P4O10
Title: Re: VCE Chemistry Question Thread
Post by: SocialRhubarb on January 12, 2014, 06:31:40 pm
If 50cm3 of 0.1M HNO3 is mixed with 60cm3 of 0.1M Ca(OH)2, what volume of 0.050M H2SO4 would be required to neutralize the solution?

50 cm3 is equivalent to 0.05 L. 0.05 L of 0.1M HNO3 contains 0.005 mol of H+.

60 cm3 is equivalent to 0.06 L. 0.06 L of 0.1M Ca(OH)2 contains 0.012 mol of OH-, as each mole of calcium hydroxide produces two moles of hydroxide ion.

Hence, there will be 0.007 mol of hydroxide ions left in solution.

0.050 M H2SO4 has a H+ concentration of 0.10 M. Hence, 0.07 L of solution is required, which is equivalent to 70 cm3.
Title: Re: VCE Chemistry Question Thread
Post by: PsychoT on January 12, 2014, 07:05:54 pm
50 cm3 is equivalent to 0.05 L. 0.05 L of 0.1M HNO3 contains 0.005 mol of H+.

60 cm3 is equivalent to 0.06 L. 0.06 L of 0.1M Ca(OH)2 contains 0.012 mol of OH-, as each mole of calcium hydroxide produces two moles of hydroxide ion.

Hence, there will be 0.007 mol of hydroxide ions left in solution.

0.050 M H2SO4 has a H+ concentration of 0.10 M. Hence, 0.07 L of solution is required, which is equivalent to 70 cm3.

Thankyou.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 12, 2014, 07:28:32 pm
In the Heinemann 2 textbook, pg32, there is a table of sample titration results.
I'm not sure why the initial burette reading changes for each titration number.

Can someone please explain? Thank you  :D
Title: Re: VCE Chemistry Question Thread
Post by: brightsky on January 12, 2014, 07:32:49 pm
We expect the initial burette readings to change, because during each titration, a certain volume of solution is dispensed from the burette. For this reason, it is natural that the initial burette reading for the next titration is lower.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 12, 2014, 07:42:49 pm
Thanks I understand this concept however in the table it moves from 0.00 to 21.00 to 1.00 to 22.00 to 2.00 in each titration number. Can you account for this difference?
Title: Re: VCE Chemistry Question Thread
Post by: Edward21 on January 12, 2014, 07:47:12 pm
Thanks I understand this concept however in the table it moves from 0.00 to 21.00 to 1.00 to 22.00 to 2.00 in each titration number. Can you account for this difference?
You're reading the table wrong (sorry if that's harsh!) it's the difference between the start and end that gives you the difference, the ACTUAL titre volume. You can't keep on dispensing liquid from a burette, it runs out, you just make note of where it starts and where it ends, and the difference is the titre :) If you look down the bottom the titres are all fairly similar volumes. When you do titrations in class it will make a lot more sense  :D
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 12, 2014, 08:08:07 pm
Ok thanks! Just to clarify, does that mean after titre 1, more standard solution is poured into the burette, and hence the increase in brunette reading number 2?
Title: Re: VCE Chemistry Question Thread
Post by: Edward21 on January 12, 2014, 08:40:00 pm
Ok thanks! Just to clarify, does that mean after titre 1, more standard solution is poured into the burette, and hence the increase in brunette reading number 2?
Probably, I mean if the end titre of #1 doesn't match the starting of #2 then you can assume something's happened :) It's actually strange that it doesn't follow on logically.. in practice exams I've done it has the final reading of one titration matching the starting of the next?? But as I said when you do this in class you'll see what I mean
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on January 12, 2014, 08:48:29 pm
Ok thanks! Just to clarify, does that mean after titre 1, more standard solution is poured into the burette, and hence the increase in brunette reading number 2?

The increase in burette reading would have been because it (the solution in the burette) had been dispensed (probably for convenience).

if MORE solution was poured in, that would have resulted in a decrease in burette reading.

Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 12, 2014, 08:55:45 pm
Thanks for the explanation snorlax + edward21!
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 13, 2014, 09:13:33 am
Convert this molar concentration to units of ppm.
0.0036 M Ca(OH)2 solution.

These are my steps:
0.0036mol / 1000 mL
m(Ca(OH)2) = 0.0036 x (40.1+2x17) = 0.26676g

0.26676g / 1000 g = 266.76g / 1,000,000g

Answer: 2.7 x 102 ppm

Textbook answer: 270 ppm

I want to ask why is it to 3 significant figures? Thank you!
Title: Re: VCE Chemistry Question Thread
Post by: brightsky on January 13, 2014, 09:21:22 am
The textbook is wrong. Your working is perfect except make sure you write 17.0 and not just 17, because all relative atomic masses are given in the data book to 1 decimal place.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 13, 2014, 09:27:29 am
Thanks for clearing it up!
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 13, 2014, 01:09:58 pm
Assuming complete disassociation occurs, calculate the pH of the following solution at 25o C : 1M HClO4.

pH = -log10(1) = 0

Can an acid ever have a pH of zero?

Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: Edward21 on January 13, 2014, 01:23:20 pm
Assuming complete disassociation occurs, calculate the pH of the following solution at 25o C : 1M HClO4.

pH = -log10(1) = 0

Can an acid ever have a pH of zero?

Thanks!
Yes! In fact the standard half-cell (look at your electrochemical series, see the equation with 0.00 Eo value) has a pH of 0 too! You can get negative pH and pH over 14, they just generally come between 0-14 :)
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 13, 2014, 01:30:31 pm
That's interesting, thanks!
Title: Re: VCE Chemistry Question Thread
Post by: Edward21 on January 13, 2014, 01:43:48 pm
That's interesting, thanks!
I remember when I got my first pH of -0.30 (2M acid) and 14.7 NaOH haha :P you learn 0-14 is just a loose guide for most solutions.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 13, 2014, 03:18:25 pm
Just as sharp end points = strong acids and bases and
broad end points = strong acid/weak base or strong base/weak acid;
I was wondering whether weak acid and weak base would also mean a sharp end point?
Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: Stick on January 13, 2014, 03:22:10 pm
Unfortunately I think you've gotten yourself a bit confused there. :S Sharp end points occur whenever a strong acid or base is involved - regardless of whether the other reactant is strong or weak. Broad end points only occur when a weak acid reacts with a weak base. Hopefully that's cleared it up for you. :)
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 13, 2014, 03:27:02 pm
Hahaha thanks a lot, now I'll remember my silly mistake!
Title: Re: VCE Chemistry Question Thread
Post by: Edward21 on January 13, 2014, 05:00:37 pm
Hahaha thanks a lot, now I'll remember my silly mistake!
Just google some titration curves and get a taste of what gives what :) there's a few in the textbook too but look online for more.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 13, 2014, 05:37:33 pm
Quickly to clarify this: also part of my summary!

Weak Acid + Weak base (need back titration)
Strong Acid + Strong base (no need for back titration)
Strong Acid + weak base (no need for back titration)
Weak Acid + Strong base (no need for back titration).

Essentially, is it accurate to say that we carry out back titrations because the end point is very broad when we react weak acid with a weak base? We actually increase the accuracy of our results by carrying out back titrations.

Thanks for the clarification! :)
Title: Re: VCE Chemistry Question Thread
Post by: SocialRhubarb on January 13, 2014, 05:43:25 pm
Yes, that seems accurate, but bear in mind there are also other reasons for conducting back titrations, apart from the strength of your titrant and analyte.
Title: Re: VCE Chemistry Question Thread
Post by: Spxtcs on January 13, 2014, 05:45:24 pm
This has probably already been asked, but what exactly is a sharp end point and a broad end point? I get the idea of end points/equivalence points but titration curves are confusing the hell out of me!
And how do you know what indicators to use just by looking at titration curves/end points?
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on January 13, 2014, 05:48:49 pm
And how do you know what indicators to use just by looking at titration curves/end points?
To know which indicator to use, you check out the pH of the end point.
I think the list of indicators and the pH they best work in are in the data book?

Yeah - page 11 of the data book
http://www.vcaa.vic.edu.au/Documents/exams/chemistry/chemdata_2012-w.pdf
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 13, 2014, 05:50:53 pm
This has probably already been asked, but what exactly is a sharp end point and a broad end point? I get the idea of end points/equivalence points but titration curves are confusing the hell out of me!
And how do you know what indicators to use just by looking at titration curves/end points?

The end point of a titration is the point at which the indicator changes colour; theoretically speaking, the same as the equivalence point, at which the reactants are in stoichiometrically equal proportions, and neutralisation has taken place. A sharp end point refers to a very precise, quick change in the colour of the indicator, that of which occurs in a reaction between a strong base/acid and another reactant of any strength (i.e. Strong acid + weak base, Strong base + weak acid or Strong acid + strong base). Sometimes, the colour change of the indicator isn't very quick and precise. As a result, the end point is very broad. This is essentially what happens when you react a weak acid with a weak base.

Q2. When using titration curves, look at your end point. This is basically the close-to-vertical gradient of your titration curve. Go to the middle of that curve and look at the corresponding pH. Then, look at the different indicators and pick the one which the corresponding pH of the end point (as read from your curve) is most suitable to. E.g. an end point with pH 7 would mean you'd use an indicator with a range where 7 falls into (e.g. an indicator with a pH range 6-8).
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 13, 2014, 06:00:56 pm
Just google some titration curves and get a taste of what gives what :) there's a few in the textbook too but look online for more.

Thanks for the advice! Also, about the chemical formula of sodium acetate:

Which one is better: NaCH3COO or CH3COONa? Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 13, 2014, 06:01:49 pm
Thanks for the advice! Also, about the chemical formula of sodium acetate:

Which one is better: NaCH3COO or CH3COONa? Thanks!

I go with the second one :) I think the conventional way is that of the latter formula for sodium acetate (for some off reason LOL).
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 13, 2014, 06:04:36 pm
That's also what I was thinking. Thanks for verifying!
Title: Re: VCE Chemistry Question Thread
Post by: Spxtcs on January 13, 2014, 06:07:37 pm
The end point of a titration is the point at which the indicator changes colour; theoretically speaking, the same as the equivalence point, at which the reactants are in stoichiometrically equal proportions, and neutralisation has taken place. A sharp end point refers to a very precise, quick change in the colour of the indicator, that of which occurs in a reaction between a strong base/acid and another reactant of any strength (i.e. Strong acid + weak base, Strong base + weak acid or Strong acid + strong base). Sometimes, the colour change of the indicator isn't very quick and precise. As a result, the end point is very broad. This is essentially what happens when you react a weak acid with a weak base.

Q2. When using titration curves, look at your end point. This is basically the close-to-vertical gradient of your titration curve. Go to the middle of that curve and look at the corresponding pH. Then, look at the different indicators and pick the one which the corresponding pH of the end point (as read from your curve) is most suitable to. E.g. an end point with pH 7 would mean you'd use an indicator with a range where 7 falls into (e.g. an indicator with a pH range 6-8).

Thank you so much!
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on January 13, 2014, 07:42:09 pm
Slight correction. As the titration curves are so steep near the equivalence point, it is ok if you choose an indicator that has a pH range slightly away from the equivalence point pH. For a sharp enough titration curve, the difference between the endpoint and the equivalence point ends up being tiny and negligible anyway. For this reason, phenolphthalein is a valid indicator for use in sodium hydroxide/nitric acid titrations, for instance.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 14, 2014, 11:57:43 am
If writing an ionic equation from a full equation including ethanoic acid, should we dissociate the H+ and CH3COO? And also, please explain? Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: brightsky on January 14, 2014, 12:02:45 pm
No. Ethanoic acid is a weak acid, so it only partially ionizes in water. If you pour ethanoic acid into water, most of the ethanoic acid molecules will retain the proton.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 14, 2014, 12:05:07 pm
Thank you!  :)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on January 14, 2014, 01:53:03 pm
Thanks for the advice! Also, about the chemical formula of sodium acetate:

Which one is better: NaCH3COO or CH3COONa? Thanks!

The secondone is preferred because the sodium ion is a cation and the negative charge in the acetate ion is on tje oxygen. Thus the plus and minus charges are put together in the formula
Title: Re: VCE Chemistry Question Thread
Post by: PsychoT on January 14, 2014, 03:30:57 pm
Anyone got a good online resource to learn about the work covered in AoS 1 (Analytical Chemistry) and AoS2 2 (Organic Reaction Pathways)?

Don't get my books till the 23rd, want to get a little headstart.

Cheers.
Title: Re: VCE Chemistry Question Thread
Post by: BLACKCATT on January 14, 2014, 03:34:57 pm
How do you work out the reaction between potassium dichromate (k2cr2o7) and tin chloride(sncl2)?
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 14, 2014, 03:48:14 pm
K2Cr2O7 (aq) + SnCl2 (aq) -> 2KCl (aq) + SnCr2O7 (s)

1. "Swap" the partners of each reactant with the other.
2. Balance equation.
3. Include states.

**I'm not sure about the solubility of tin (II) dichromate, if anyone can confirm it'll be great!
Title: Re: VCE Chemistry Question Thread
Post by: BLACKCATT on January 14, 2014, 03:56:29 pm
Ohhh right thanks.
What if the potassium was omitted? Whats the reaction between the dichromate ion(cr2o7^-2, thats a negative charge) and sncl2?
Title: Re: VCE Chemistry Question Thread
Post by: Daenerys Targaryen on January 14, 2014, 04:11:12 pm
You would have 2Cl- on the right hand side. (the K also omitted)
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 15, 2014, 08:10:16 am
Washing soda is added to hard water to allow soap to lather.
A certain brand of washing soda contains partially hydrated sodium carbonate solid.
A 0.300 g sample completely reacts with 20.0 mL of 0.250 M hydrochloride acid.

Na2CO3 (s) + 2HCl (aq) -> 2NaCl (aq) + H2O (l) + CO2 (g)

Why does the reaction not take into account the "partially hydrated" term, in that the chemical formula for sodium carbonate be Na2CO3.xH2O (ss) thanks!
Title: Re: VCE Chemistry Question Thread
Post by: Daenerys Targaryen on January 15, 2014, 10:59:32 am
Washing soda is added to hard water to allow soap to lather.
A certain brand of washing soda contains partially hydrated sodium carbonate solid.
A 0.300 g sample completely reacts with 20.0 mL of 0.250 M hydrochloride acid.

Na2CO3 (s) + 2HCl (aq) -> 2NaCl (aq) + H2O (l) + CO2 (g)

Why does the reaction not take into account the "partially hydrated" term, in that the chemical formula for sodium carbonate be Na2CO3.xH2O (ss) thanks!

I think it is because the water on the carbonate wouldn't react with HCl and thus just appear as water on the right hand side. Thus not really needing to take it into account
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on January 15, 2014, 12:10:43 pm
Washing soda is added to hard water to allow soap to lather.
A certain brand of washing soda contains partially hydrated sodium carbonate solid.
A 0.300 g sample completely reacts with 20.0 mL of 0.250 M hydrochloride acid.

Na2CO3 (s) + 2HCl (aq) -> 2NaCl (aq) + H2O (l) + CO2 (g)

Why does the reaction not take into account the "partially hydrated" term, in that the chemical formula for sodium carbonate be Na2CO3.xH2O (ss) thanks!

I think it is because the water on the carbonate wouldn't react with HCl and thus just appear as water on the right hand side. Thus not really needing to take it into account

I was initially thinking that too, but then, having not done the calculations, I thought that perhaps the question wanted us to find the water of crystallisation (the x term in the formula given). Then I saw that the numbers were too nice for that to happen.

What's the question though?
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 15, 2014, 01:25:31 pm
Thanks for your responses!
The question is to write a balanced equation.
Title: Re: VCE Chemistry Question Thread
Post by: Edward Elric on January 15, 2014, 01:55:55 pm
Hey Chemistry folks
Can someone help me out with the following Q
1) write the half equation(partial ionic equation) for the oxidation of chlorine gas (Cl2) to Hypochlorous acid (HOCL)

2) the oxidant and the reactant of : 2KMnO4(aq) + 5H2S(aq) + 6HCL(aq) -----> 2MnCl2(aq) + 5S(s) + 2KCl(aq) + 8H2O(l)

3)  write balanced equations for the following redox reactions: Under certain conditions, zinc can react with concentrated nitric acid (HNO3) to form the zinc ion (Zn^2+) the ammonium ion (NH4^+) and water.
Any solutions to the Q, will be much appreciated!
Title: Re: VCE Chemistry Question Thread
Post by: Edward21 on January 15, 2014, 07:36:24 pm
Hey Chemistry folks
Can someone help me out with the following Q
1) write the half equation(partial ionic equation) for the oxidation of chlorine gas (Cl2) to Hypochlorous acid (HOCL)

2) the oxidant and the reactant of : 2KMnO4(aq) + 5H2S(aq) + 6HCL(aq) -----> 2MnCl2(aq) + 5S(s) + 2KCl(aq) + 8H2O(l)

3)  write balanced equations for the following redox reactions: Under certain conditions, zinc can react with concentrated nitric acid (HNO3) to form the zinc ion (Zn^2+) the ammonium ion (NH4^+) and water.
Any solutions to the Q, will be much appreciated!

1. 2H2O(l) + Cl2(g) --> 2HOCl(aq) + 2H+(aq) + 2e-

2. oxidant and reductant? The oxidant gets reduced, and oxidises something else by taking electrons: [KMnO4] (+7 to +2) (Mn)
Reductant will give away electrons to reduce something and is oxidised: [H2S/S2-] (-2 to 0) (S) I've separated S2- from H2S as it's in the aqueous state..

3. (Zn(s) --> Zn2+(aq) + 2e-)*4
8e- + 9H+(aq) + HNO3(aq) --> NH4+(aq) +3H2O(l)

4Zn(s) + 9H+(aq) + HNO3(aq) --> 4Zn2+(aq) + NH4+(aq) + 3H2O(l)
Title: Re: VCE Chemistry Question Thread
Post by: Edward Elric on January 15, 2014, 09:26:06 pm
1. 2H2O(l) + Cl2(g) --> 2HOCl(aq) + 2H+(aq) + 2e-

2. oxidant and reductant? The oxidant gets reduced, and oxidises something else by taking electrons: [KMnO4] (+7 to +2) (Mn)
Reductant will give away electrons to reduce something and is oxidised: [H2S/S2-] (-2 to 0) (S) I've separated S2- from H2S as it's in the aqueous state..

3. (Zn(s) --> Zn2+(aq) + 2e-)*4
8e- + 9H+(aq) + HNO3(aq) --> NH4+(aq) +3H2O(l)

4Zn(s) + 9H+(aq) + HNO3(aq) --> 4Zn2+(aq) + NH4+(aq) + 3H2O(l)

Thank you so much! Legend! But i Think the oxidation no of 2KMnO4 to 2MnCl2 is from +12 to +2 instead of +7 to +2 :P but either way it doesn't make a difference.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on January 16, 2014, 12:24:29 am
The manganese in permanganate has oxidation number +7, not +12. There is no substance in existence that has anything in such a high oxidation number. I can't imagine anything losing twelve electrons.
Title: Re: VCE Chemistry Question Thread
Post by: Edward Elric on January 16, 2014, 03:10:56 am
The manganese in permanganate has oxidation number +7, not +12. There is no substance in existence that has anything in such a high oxidation number. I can't imagine anything losing twelve electrons.

Ohh okay, sorry i must have done my oxidation nos incorrectly, cheers :) I think I used the coefficient to find the oxidation no instead of the subscript lol
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 16, 2014, 10:38:30 am
When copper (II) sulfate solution is stored in an iron container, the container gradually corrodes.
Write a balanced ionic equation to represent the reaction if the products of the reaction are copper and iron (II) sulfate solution.

Answer: Cu2+ (aq) Fe (s) -> Cu (s) + Fe2+ (aq)

I would just to like to check whether we are allowed to dissociate ions in redox reactions?
And if so, is it because redox reactions are essentially ionic equations due to the transfer of electrons?

Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 16, 2014, 11:25:18 am
When copper (II) sulfate solution is stored in an iron container, the container gradually corrodes.
Write a balanced ionic equation to represent the reaction if the products of the reaction are copper and iron (II) sulfate solution.

Answer: Cu2+ (aq) Fe (s) -> Cu (s) + Fe2+ (aq)

I would just to like to check whether we are allowed to dissociate ions in redox reactions?
And if so, is it because redox reactions are essentially ionic equations due to the transfer of electrons?

Thanks!

Your half equations (i.e. reduction and oxidation half-equations) should be based on an ionic equation. Write your balanced equation, then write your ionic equation, and then, deduce your oxidation half equation and reduction half-equation.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 16, 2014, 08:54:29 pm
Thanks for your help!
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 17, 2014, 12:32:06 pm
Just a question about sig figs:

A cylinder has a volume of 39.0 L filled with nitrogen gas to a pressure of 3.00 atmospheres at 23oC. What mass of nitrogen does the cylinder contain?

Should the answer be to 2 or 3 sig figs? As there is a conversion of the 23 -> 296, do we now refer to the 296 as the given value and ignore the 23? Thanks!

Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 17, 2014, 12:36:10 pm
Just a question about sig figs:

A cylinder has a volume of 39.0 L filled with nitrogen gas to a pressure of 3.00 atmospheres at 23oC. What mass of nitrogen does the cylinder contain?

Should the answer be to 2 or 3 sig figs? As there is a conversion of the 23 -> 296, do we now refer to the 296 as the given value and ignore the 23? Thanks!

2 sig figs. Its what you start with!
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 17, 2014, 12:38:37 pm
Cool thanks for checking (back of book has 3)
Title: Re: VCE Chemistry Question Thread
Post by: thushan on January 17, 2014, 12:39:06 pm
Nup, 3 sig figs.

Temperature in K = 23 + 273 (273 is the figure given in the Data Book as opposed to 273.0 etc) = 296 K.

When adding, its decimal places that matter, hence 296 K is the value to correct accuracy.

Since 296 happens to also have 3 sig figs, and so do all the other data that you use to multiply, the answer is to 3 sig figs.

In addition/subtraction you consider decimal places. In multiplication/division you consider sig figs.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 17, 2014, 12:39:54 pm
Nup, 3 sig figs.

Temperature in K = 23 + 273 (273 is the figure given in the Data Book as opposed to 273.0 etc) = 296 K.

When adding, its decimal places that matter, hence 296 K is the value to correct accuracy.

Since 296 happens to also have 3 sig figs, and so do all the other data that you use to multiply, the answer is to 3 sig figs.

In addition/subtraction you consider decimal places. In multiplication/division you consider sig figs.

Oh sorry eagles! Thanks for that thushan!
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 17, 2014, 12:43:12 pm
Dw yacoubb

Thanks for both of your responses!
Title: Re: VCE Chemistry Question Thread
Post by: Rod on January 17, 2014, 12:44:46 pm
Nup, 3 sig figs.

Temperature in K = 23 + 273 (273 is the figure given in the Data Book as opposed to 273.0 etc) = 296 K.

When adding, its decimal places that matter, hence 296 K is the value to correct accuracy.

Since 296 happens to also have 3 sig figs, and so do all the other data that you use to multiply, the answer is to 3 sig figs.

In addition/subtraction you consider decimal places. In multiplication/division you consider sig figs.
Hi Thushan,

Our teacher has told us to always use 3 sig figs when answering questions. But don't we look at the values given in the question, and then use the sigfigs from the question for our answer? Example the highest number of sig figs used in the question is 2, so then we have our answer in 2 sig figs. Or is my teacher right and and do we only use 3?

Thanks
Title: Re: VCE Chemistry Question Thread
Post by: thushan on January 17, 2014, 01:07:30 pm
Keep intermediate steps to correct sig figs - don't rely on the start of the question. However, use your calculator value for subsequent steps.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 18, 2014, 05:51:45 pm
4.15 g tungsten is burned in chlorine and 8.95 g tungsten chloride (WCl6) was formed.
Find the relative atomic mass of tungsten.

Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 18, 2014, 05:56:47 pm
4.15 g tungsten is burned in chlorine and 8.95 g tungsten chloride (WCl6) was formed.
Find the relative atomic mass of tungsten.

We have 8.95g of WCl6.
n(WCl6) = 8.95/(6x35.5 + 183.8) =0.022555443 mol

There is one tungsten cation in tungsten chloride. Therefore n(WCl6) = n(W)

Therefore, Mr of W = 4.15 / 0.022555443 mol = 183.9910615, to 3 sig figs, is 184.

Your answer is 184. But, why would it ask you to find the atomic mass of tungsten when you already have it on your periodic table?
Title: Re: VCE Chemistry Question Thread
Post by: SocialRhubarb on January 18, 2014, 06:05:54 pm
4.15 g tungsten is burned in chlorine and 8.95 g tungsten chloride (WCl6) was formed.
Find the relative atomic mass of tungsten.

We begin with 4.15 g of pure tungsten, and finish with 8.95 g of tungsten chloride. All of the tungsten in the original sample is present in the tungsten chloride, so there must be 4.8 g of chlorine in the sample, as 4.15 g of the 8.95 g sample is tungsten.

4.8 g of chlorine is equivalent to 0.135 mol of chlorine. The ratio of tungsten to chlorine is 1 : 6, so the amount of tungsten is one sixth the amount of chlorine. Therefore, the amount of tungsten is 0.0225 mol.

We know that the amount of tungsten in the tungsten chloride is the same as the original amount of tungsten. Therefore, 0.0225 mol of tungsten has a mass of 4.15 g, and hence, by ratios, 1 mol of tungsten has a mass of 184.16 g, approximately 184.

Finally, the question asks for a relative atomic mass, so the answer must be in atomic mass units, so the relative atomic mass of tungsten is 184 atomic mass units.

We have 8.95g of WCl6.
n(WCl6) = 8.95/(6x35.5 + 183.8) =0.022555443 mol

There is one tungsten cation in tungsten chloride. Therefore n(WCl6) = n(W)

Therefore, Mr of W = 4.15 / 0.022555443 mol = 183.9910615, to 3 sig figs, is 184.

Your answer is 184. But, why would it ask you to find the atomic mass of tungsten when you already have it on your periodic table?

You can't use the value for the molar mass of tungsten in the calculation of the relative atomic mass of tungsten. It's similar to mathematics - if a question asks you to prove a relationship, you cannot assume the relationship is true.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 18, 2014, 06:09:53 pm
Awesome! Thank you  :)
Title: Re: VCE Chemistry Question Thread
Post by: survivor on January 19, 2014, 08:11:02 pm
A sample of Na2CO3 contains 1.50 mole of sodium ions. Calculate the total mass of Na2CO3 in the sample.

Thanks
Title: Re: VCE Chemistry Question Thread
Post by: LOLs99 on January 19, 2014, 08:19:22 pm
A sample of Na2CO3 contains 1.50 mole of sodium ions. Calculate the total mass of Na2CO3 in the sample.

Thanks
n (Na2CO3)= 1/2 of n(Na+)
So n(Na2CO3)= 1.50mol x 0.5=0.75mol
m(Na2CO3)= 0.75mol x (23+23+12+18+18+18)= 84g

Hopefully my molar mass is correct.
Title: Re: VCE Chemistry Question Thread
Post by: scribble on January 19, 2014, 08:23:12 pm
n(Na+) = m/M = 1.50
n(Na2CO3) = 1/2*n(Na+) = 0.750 mol
m(Na2CO3) = nM = 0.75*(23.0*2+12+16*3) = 79.5g to 3sigfigs
Title: Re: VCE Chemistry Question Thread
Post by: LOLs99 on January 19, 2014, 08:35:18 pm
n(Na+) = m/M = 1.50
n(Na2CO3) = 1/2*n(Na+) = 0.750 mol
m(Na2CO3) = nM = 0.75*(23.0*2+12+16*3) = 79.5g to 3sigfigs

Opps Haha thanks for reminding me my molar mass :)  oxygen is 16.
Title: Re: VCE Chemistry Question Thread
Post by: scribble on January 19, 2014, 08:37:13 pm
hahah water is 18 :P
Title: Re: VCE Chemistry Question Thread
Post by: Edward Elric on January 21, 2014, 02:20:45 am
Oxalic acid, C2H2O4 may be reacted according to the following equation.
Cr2o7^2-(aq) +8H^+(aq) + 3C2H2O4(aq) -------> 2Cr^3+(aq)+7H2O(l) + 6CO2(g)
a) write balanced half equations for the reactions occurring,
b) 20.0g of an impure sample of oxalic acid was dissolved in water to make 500.0 mL of solution. 23.0 mL of this solution was acidified with sulfuric acid, and 28.5 mL of 0.0880 M standard potassium dichromate solution was required to reach the endpoint.
i) Determine the amount (in mol) of oxalic acid in the 23.0 mL sample.
ii) Determine the percentage by mass of oxalic acid in the impure sample.

can someone do these Q and see whether i got it right as i don't have the solutions (please show all working), my answers are as follow:

a) Cr2O7^2-(aq) + 14H^+(aq) + 6e^-  ---------> 2Cr^3+(aq) + 7H2O(l)  (1)
3C2H2O4(aq) -------> 6CO2(g) + 6H^+ + 6e^-                                     (2)
bi) 0.164 mol
ii) 73.6 %
Thanks in Advance :)
Title: Re: VCE Chemistry Question Thread
Post by: scribble on January 21, 2014, 03:38:53 am
a) is fine
b)i) this part asks you for the amount in mol in the 23.0 mL sample, not in the entire sample, so all you need to do is this:
n(Cr2O7^2-) = cV = 0.0880*28.5*10^-3=0.002508mol
n(C2H2O4)=3*n(Cr2O7^2-) = 3*0.002508 = 0.00752 mol (correct to 3sigfigs)

ii) n(C2H2O4)total= 0.007524 * 100/23 = 0.1636 mol
m(C2H2O4)=nM=0.1636*(12*2+1*2+16*4)=14.7g
% mass = 14.7/20.0 * 100 = 73.6% (correct to 3 sigfigs)
Title: Re: VCE Chemistry Question Thread
Post by: BLACKCATT on January 21, 2014, 09:42:00 am
ii) n(C2H2O4)total= 0.007524 * 500/23 = 0.1636 mol

just a correction, sorry
Title: Re: VCE Chemistry Question Thread
Post by: Edward Elric on January 21, 2014, 02:08:06 pm
a) is fine
b)i) this part asks you for the amount in mol in the 23.0 mL sample, not in the entire sample, so all you need to do is this:
n(Cr2O7^2-) = cV = 0.0880*28.5*10^-3=0.002508mol
n(C2H2O4)=3*n(Cr2O7^2-) = 3*0.002508 = 0.00752 mol (correct to 3sigfigs)

ii) n(C2H2O4)total= 0.007524 * 100/23 = 0.1636 mol
m(C2H2O4)=nM=0.1636*(12*2+1*2+16*4)=14.7g
% mass = 14.7/20.0 * 100 = 73.6% (correct to 3 sigfigs)

Thank you very much Scribble :D
Title: Re: VCE Chemistry Question Thread
Post by: ETTH96 on January 21, 2014, 02:16:23 pm
How do I find the new pH?

0.100 M NaOH solution is added to a 10.0 mL aliquot of 0.100 M HCl solution

Thanks
Title: Re: VCE Chemistry Question Thread
Post by: emilyhobbes on January 21, 2014, 02:18:59 pm
does it say how much NaOH solution has been added?
Title: Re: VCE Chemistry Question Thread
Post by: ETTH96 on January 21, 2014, 02:22:36 pm
Nope, I've been given 4 different graphs and it asks which best represents the change in pH... How do I know which one it is?

Ive summed it down to two graphs, since its a base added to a acid the graph should go upwards.. right?

Graph 1:
EP is at 5
Highest pH is 9
Lowest pH is 1

Graph 2:
EP is at 7
Highest pH is 13
Lowest pH is 1

Sorry, if this is a bit hard to imagine..
Title: Re: VCE Chemistry Question Thread
Post by: ETTH96 on January 21, 2014, 02:24:51 pm
Oh wait, sorry it does say on the graph: 0.010 L NaOH added. SORRY
Title: Re: VCE Chemistry Question Thread
Post by: emilyhobbes on January 21, 2014, 02:43:23 pm
hahah, no worries!
Okay, so HCl is a strong acid and NaOH is a strong base, so this means that the end point is going to happen around ph7 and it'll be sharp

and yes, the graph should go upwards, so it sounds like graph 2 to me :)
Title: Re: VCE Chemistry Question Thread
Post by: ETTH96 on January 21, 2014, 05:01:59 pm
Thanks emilyhobbes, really helpful! Can anyone help me with this question? Its pretty long :(

Hydrogen peroxide added with potassium permanganate...
Equation:    2MnO4(-) +5H2O2 +6H+ ==> 2Mn(2+) +8H2O +5O2

a)  The label on the bottle of the solution of hydrogen peroxide says it is 10.0%, by mass, hydrogen peroxide, and has a density of 1.19 gmL-1. Show that this data is consistent with a concentration of 3.50 M

^I've just completed this question, I did this:

1000 x 1.19 = 1190g
therefore 10% of 1190 is 119g

n(H2O2)= 119/34 = 3.50 mol
c(H2O2)=3.50/1 = 3.50 M

I STILL NEED HELP WITH THIS, PLEASE:
b)  20.00 ml of this solution is pipetted into a 250ml volumetric flask, and the volume is made up to the mark with distilled water. 25.00 ml of this solution is titrated, in the presence of dilute sulphuric acid, with a 0.129 M solution of potassium permanganate, KMnO4.
What volume of the potassium permanganate solution is required for complete reaction of 25.00 ml of the diluted sample?
Title: Re: VCE Chemistry Question Thread
Post by: clıppy on January 21, 2014, 06:49:08 pm
Thanks emilyhobbes, really helpful! Can anyone help me with this question? Its pretty long :(

Hydrogen peroxide added with potassium permanganate...
Equation:    2MnO4(-) +5H2O2 +6H+ ==> 2Mn(2+) +8H2O +5O2

I STILL NEED HELP WITH THIS, PLEASE:
b)  20.00 ml of this solution is pipetted into a 250ml volumetric flask, and the volume is made up to the mark with distilled water. 25.00 ml of this solution is titrated, in the presence of dilute sulphuric acid, with a 0.129 M solution of potassium permanganate, KMnO4.
What volume of the potassium permanganate solution is required for complete reaction of 25.00 ml of the diluted sample?

So we've got 20mL of 3.5M H2O2
n(H2O2) initial = CV = 3.5 * 0.02 = 0.07mol in 250mL (we took the 20mL and diluted with water up to the mark, but the number of mols is still the same)

So now how if we took 25mL of this new solution out, how many mols would we have?
0.07 mol in 250mL
x mol in 25mL
x = (25 * 0.07) / 250
x = 0.007 mol in 25mL

So now we know how many mols of H2O2 are going to react with the MnO4, we can use stoichiometry to find out how many mols of MnO4 we need.
n(MnO4) = 2/5 * n(H2O2)
n(MnO4) = 0.0028 mol
Volume of MnO4 = n/C
Volume = 0.0028/0.129
Volume = 0.0217L
Volume = 21.71mL

My chemistry is pretty rusty so if I've done something horribly wrong, please let me know.
Title: Re: VCE Chemistry Question Thread
Post by: Irving4Prez on January 21, 2014, 08:58:20 pm
What is the difference between an end point and an equivalence point?
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on January 21, 2014, 09:13:40 pm
What is the difference between an end point and an equivalence point?
Equivalence point is the point when the amounts of acid and based have reacted are stoichiometrically equivalents
End points are the point at which the indicator permanently changes colour
Hope this helps!
Title: Re: VCE Chemistry Question Thread
Post by: ETTH96 on January 21, 2014, 11:18:36 pm
Hey guys, can someone let me know if I'm doing this right?

2MnO4 + 16H+ +5(C2O4)2 --> 2Mn(2+) +8H2O +10CO2

use this equation to calculate the volume of a manganate(VII) solution of concentration 0.0200 M, required for complete reaction with 0.2000g of iron (II) oxalate.

This is what I did:
n(FeC2O4)= 0.2000/143.8 = 0.00139 mol

n(MnO4) = 5/2 x 0.00139 = 0.00348 mol

therefore, v(MnO4)= 0.00348/0.0200 = 0.174 L

???
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on January 21, 2014, 11:48:36 pm
Equivalence point is the point when the amounts of acid and based have reacted are stoichiometrically equivalents
End points are the point at which the indicator permanently changes colour
Hope this helps!
When the indicator permanently changes colour, doesn't that mean the acid and base that react are stoichiometrically equivalent?
I always thought the end point was the point the indicator first changes colour? (not permanent)
And the equivalence point where the indicator changes colour permanently. :/

Someone help distinguish the two please :)
Title: Re: VCE Chemistry Question Thread
Post by: emilyhobbes on January 22, 2014, 12:14:15 am
the endpoint is the point where the indicator changes colour
BUT the acid and base, in all probability, aren't going to be in exact stoich proportions at this point, because the endpoint is kind of just like an approximation. In theory, the endpoint SHOULD be as close to the equivalence point (when acid and base are in stoich proportions) as possible, but say you picked a super unsuitable indicator, it's endpoint isn't going to be near the equiv point and won't be an accurate representation of where the titration (etc) is complete

hopefully that makes some sense
Title: Re: VCE Chemistry Question Thread
Post by: Rod on January 22, 2014, 01:31:13 am
When the indicator permanently changes colour, doesn't that mean the acid and base that react are stoichiometrically equivalent?
I always thought the end point was the point the indicator first changes colour? (not permanent)
And the equivalence point where the indicator changes colour permanently. :/

Someone help distinguish the two please :)
No, the end point is just a point in which the indicator changes colour. It is never dependent on the acid and base reaction. The equivalence point, however is the exact point at which the acid and base are in the correct stoich proportions. We need to match the equivalence point with the end point to accurate find out the titre. This is why you must always pick an indicator who's end point lies within the pH of the equivalence point. If we pick the wrong indicator, it will indicate a random pH!

We will be doing a major SAC on this so most school's will be spending some good time going over this.

Hope that helps,

Rod
Title: Re: VCE Chemistry Question Thread
Post by: Rod on January 22, 2014, 01:39:37 am
Hey guys, can someone let me know if I'm doing this right?

2MnO4 + 16H+ +5(C2O4)2 --> 2Mn(2+) +8H2O +10CO2

use this equation to calculate the volume of a manganate(VII) solution of concentration 0.0200 M, required for complete reaction with 0.2000g of iron (II) oxalate.

This is what I did:
n(FeC2O4)= 0.2000/143.8 = 0.00139 mol
It looks okay but I'm having trouble understanding your equation. You do not have iron (II) oxalate in your equation. If I assume that (C2O4) is iron oxalate then;
n(f2c2o4) = 0.2/143.8
= 0.00139 mol

n2(mn) /5f2c204 =n(mn04) = 2/5 x 0.00139 = 0.000556 mol
v (mn04) = 0.00556/0.0200 = 27.8 mL
I get a completely different answer to you.

n(MnO4) = 5/2 x 0.00139 = 0.00348 mol

therefore, v(MnO4)= 0.00348/0.0200 = 0.174 L

???
Title: Re: VCE Chemistry Question Thread
Post by: Rod on January 22, 2014, 01:40:19 am
Hey guys, can someone let me know if I'm doing this right?

2MnO4 + 16H+ +5(C2O4)2 --> 2Mn(2+) +8H2O +10CO2

use this equation to calculate the volume of a manganate(VII) solution of concentration 0.0200 M, required for complete reaction with 0.2000g of iron (II) oxalate.

This is what I did:
n(FeC2O4)= 0.2000/143.8 = 0.00139 mol

n(MnO4) = 5/2 x 0.00139 = 0.00348 mol

therefore, v(MnO4)= 0.00348/0.0200 = 0.174 L

???
It looks okay but I'm having trouble understanding your equation. You do not have iron (II) oxalate in your equation. If I assume that (C2O4) is iron oxalate then;
n(f2c2o4) = 0.2/143.8
= 0.00139 mol

n2(mn) /5f2c204 =n(mn04) = 2/5 x 0.00139 = 0.000556 mol
v (mn04) = 0.00556/0.0200 = 27.8 mL
I get a completely different answer to you.
Title: Re: VCE Chemistry Question Thread
Post by: Rod on January 22, 2014, 01:42:11 am
Okay! I see the difference between my answer and yours.

I did 2/5 x the mole of f2c204 while you did 5/2.

It should be 2/5, not 5/2 because n(2mn04)/n(5f2c204) = 2/5.

Hope that helps,

Rod
Title: Re: VCE Chemistry Question Thread
Post by: brightsky on January 22, 2014, 01:42:25 am
Hey guys, can someone let me know if I'm doing this right?

2MnO4 + 16H+ +5(C2O4)2 --> 2Mn(2+) +8H2O +10CO2

use this equation to calculate the volume of a manganate(VII) solution of concentration 0.0200 M, required for complete reaction with 0.2000g of iron (II) oxalate.

This is what I did:
n(FeC2O4)= 0.2000/143.8 = 0.00139 mol

n(MnO4) = 5/2 x 0.00139 = 0.00348 mol

therefore, v(MnO4)= 0.00348/0.0200 = 0.174 L

???

looks good, except watch the sig figs. the answer to the first equation should be to 4 sig figs. i'd also add more working. for example after i calculate n(FeC2O4) i'd state explicitly that the answer to that equals to n(C2O4). notice that iron(II) oxalate isn't actually present in the equation provided. i would also watch the use of superscripts. permanganate ion has a charge of 1-, so you should include that in your working.

EDIT: and yes, the 2/5 thing as pointed out by rod. :P

When the indicator permanently changes colour, doesn't that mean the acid and base that react are stoichiometrically equivalent?
I always thought the end point was the point the indicator first changes colour? (not permanent)
And the equivalence point where the indicator changes colour permanently. :/

Someone help distinguish the two please :)

you'll find that sometimes the solution in the conical flask into which you are dispensing the solution in the burette will change colour at the first drop. the colour will of course quickly vanish, but a colour changes occurs nevertheless. it is a permanent colour change that we are looking for. a pH indicator is a weak acid/base whose conjugate base/acid is of a different colour. we exploit this property during titration to determine roughly where the equivalence point lies. envisage this. say you have a solution of HCl in the conical flask and a solution of NaOH in the burette, and you add a few drops of phenolphthalein into the HCl. phenolphthalein is itself a weak acid, which has the formula C20H14O4, but for the purposes of this explanation, we shall suppose that the formula is simply HIn instead. now the moment you add the phenolphthalein into the HCl solution (which, remember, is basically just a bunch of H+ and Cl- ions floating around amidst a sea of H2O molecules), the phenolphthalein molecules will react with the water according to the equation:

HIn(aq) + H2O(l) <--> In-(aq) + H3O+(aq)
no colour                     pink

the concentration fraction for this reaction is therefore:

CF = [In-][H3O+]/[HIn]

when the reaction reaches equilibrium, the concentration fraction becomes the equilibrium constant, which for phenolphthalein is 7.9*10^(-12) according to google.

Ka = [In-][H3O+]/[HIn] = 7.9*10^(-12)

now you can see that once the reaction reaches equilibrium, there is A LOT more HIn floating around than there are In-. this is why HIn is the dominant species, and the overall solution appears colourless initially. but as we add more and more NaOH into the solution, there will be less and less H3O+. so to regain equilibrium, the system will favour the forward reaction so as to replenish the lost H3O+ ions. this means less HIn and more In-. eventually, we'll get to a stage where In- becomes the dominant species, at which point the solution will appear pink.

but when exactly does the transition from colourless to pink occur, you may ask? i said earlier that the first colour changes is often experienced at the first drop. this is mainly because when the first drop hits the solution, in the region where the drop hits, there will be a sudden deficiency of H3O+ ions, and the system will immediately respond by favouring the forward reaction, producing more In- at the expense of HIn. but all of this happens locally and almost instantaneously. when the OH- disperse, the deficiency becomes less pronounced, and so the solution quickly returns to colourless. but there is a point where the overall deficiency of H3O+ is so high that the colour change is more or less permanent (permanent is a dodgy word to use here because you'll find that sometimes if you leave the solution out for a few hours it will go back to being colourless). when does exactly does this occur? to appreciate this, we must first do a little maths.

let us return to the equilibrium law expression and do a little bit of rearranging:

Ka = [In-][H3O+]/[HIn]
pKa = -log_10([In-][H3O+]/[HIn])
pKa = -log_10[H3O+] - log_10([In-]/[HIn]) (yay log laws from methods)
pKa = pH - log_10([In-]/[HIn])
pH = pKa + log_10([In-]/[HIn])

this last equation is what we call the henderson-hasselbalch equation. fancy name for a not-so-fancy equation. it so happens that the human eye (which btw is a very bad detector) can detect a colour change only when the ratio [In-]/[HIn] is between 0.1 and 100. this means that the human eye can only detect a colour change when the pH of the solution is:

pH = pKa +- 1 (where +- is the plus/minus symol)

now the Ka of an indicator varies according to how weak an acid/base it is, but we know that for phenolphthalein, Ka = 7.9*10^(-12) as stated above. this means that pKa = 9.7. so this means that the human eye can detect a colour change only when the pH of the solution is between 8.7 and 10.7. this is basically how they determine the pH range of a particular indicator. everything ties back to Ka values.

how do we choose which indicator to use? well, we know for instance that if we react a strong base with a strong acid, the reaction will reach completion (not equilibrium because the species are strong) roughly when pH = 7 (and you can prove why this is the case). so we want an indicator that will tell us when the pH of the solution is roughly 7; the pH curve is very steep around the pH = 7 region, so if we are off by a bit it doesn't really matter, and it so happens that phenolphthalein can do this because the human eye detects a colour change in phenolphthalein when pH is between 8.7 - 10.7, as per all the calculations above (you might think that this is way off pH = 7, but if you look at the standard pH curve for strong acid/strong base reaction, you'll find that it is not really).

man didn't envisage this post to be so long but hopefully i've shed some light on how indicators work...
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 22, 2014, 09:49:54 am
What is the difference between an end point and an equivalence point?

Theoretically, its meant to be the same thing. An end point is the point at which the indicator changes colour, and the equivalence point is the point at which the reactants are in stoichiometrically equal mole ratios.
Title: Re: VCE Chemistry Question Thread
Post by: thushan on January 22, 2014, 10:03:39 am
Equivalence point is practically where the two reactants (titrant and analyte) have been added in their stoichiometric ratios such that neither reagent is in excess.

End point is the point at which an indicator changes colour.

In a particular acid-base titration, you could choose from a wide variety of indicators, each which would have their own endpoint in the titration that you are doing (they would change colour at different times, depending on the pH at which they change colour). Your aim is to choose the indicator whose endpoint coincides with the equivalence point of the titration.

You don't see an equivalence point in an acid base titration since the reagents and products are colourless. You see an endpoint (you see indicators changing colours). You use the endpoint as a marker for the equivalence point of the titration.
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on January 22, 2014, 01:10:00 pm
Cheers everyone

@ ~3:45
He says the complete colour change is where the end point is - and where the equivalence point is IF choosing an appropriate indicator.

I understand what the equivalence point is, but not quite sure what you guys mean by "end point being the point at which an indicator changes colour"

In this video, is he correct in saying it's the end point? Is it also the equivalence point? (because he chose an appropriate indicator?)

Title: Re: VCE Chemistry Question Thread
Post by: Rod on January 22, 2014, 01:17:56 pm
Cheers everyone

@ ~3:45
He says the complete colour change is where the end point is - and where the equivalence point is IF choosing an appropriate indicator.

I understand what the equivalence point is, but not quite sure what you guys mean by "end point being the point at which an indicator changes colour"

In this video, is he correct in saying it's the end point? Is it also the equivalence point? (because he chose an appropriate indicator?)
'The point at which an indicator changes colour' - Just a specific pH range in which the indicator HAS to change colour. The pH range is THAT POINT. Different indicators have different pH ranges and therefore different POINTS at which they change colour. For example, the Ph range for methyl orange is from 3 to 4.4, therefore the point at which they change colour can ONLY be from 3 to 4.4 pH. On the other hand, the indicator cresol red changes colour when the pH is within 7 to 8.2, SO that is also the point at which the indicator changes colour.

I'm trying to keep it as simple as I can, hope that helps. If you are still confused ask!

Rod
Title: Re: VCE Chemistry Question Thread
Post by: Rod on January 22, 2014, 01:22:15 pm
I got around concepts like endpoint, equivalence point, average titre, aliquots, primary standards, standard solution, the types of cuves in pH graphs and all the equipment used during a titration when I did a practical on titration in year 11. So I think it is essential that you ask your teacher to do a practise prac on titrations and even back titrations, it will help a lot.
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on January 22, 2014, 01:35:13 pm
I think I understand now..
I think I was assuming that when the colour changes (permanently), that was the equivalence point...but that's the end point right?
Doing a titration alone can't determine the precise/true equivalence point, but only a close indication of the equivalence point?

In other words, the end point shows us the physical event (colour change), while the true equivalence point is determined based on the end point?

PLS SAY I'M RIGHT
omg. this seems like a straightforward concept, but I seem to be unsure...

EDIT: Just read thushans post again..What he means is what I said in this post, right? :S
Title: Re: VCE Chemistry Question Thread
Post by: Rod on January 22, 2014, 01:43:51 pm
I think I understand now..
I think I was assuming that when the colour changes (permanently), that was the equivalence point...but that's the end point right?
Doing a titration alone can't determine the precise/true equivalence point, but only a close indication of the equivalence point?

In other words, the end point shows us the physical event (colour change), while the true equivalence point is determined based on the end point?

PLS SAY I'M RIGHT
omg. this seems like a straightforward concept, but I seem to be unsure...

EDIT: Just read thushans post again..What he means is what I said in this post, right? :S
Yes you've nearly got it! Good job!

There is just one thing that you said that I feel isn't correct. The equivalence point NEVER makes the colour change. It's just a specific pH at which the standard and uknown both come to the appropriate stoic proportions. Scientists know at what pH the equivalence point is, so by knowing this we can use an indicator (this makes the colour change) which changes the colour when it is at the correct pH. So for example, it's like we react two solutions that are at correct stoic proportions at a pH of 4 (so the equivalence point is 4). We would then use an indicator that changes colour at a pH close to 4 (EXAMPLE - Methyl orange which ranges from 3 to 4.4). Thus, it will indicate a close approximation of the equivalence point. In other words, the end point is used to determine the equivalence point.

Other then that, everything else you have said is great. Good job bud!

Rod
Title: Re: VCE Chemistry Question Thread
Post by: ETTH96 on January 22, 2014, 03:07:45 pm
Thanks Rod and Brightsky, should I just remember when I ratio: put the coefficient of the mol of what I am looking for on top of the fraction and what I'm ratio-ing it against at the bottom :P Thanks heaps!!

Could anybody help me out with this question?

Ammonia is produced from its elements on a large scale using the Haber process (what on earth is this..?!). For reasons of environmental safety the concentration of ammonia in the air downwind of an ammonia production plant was mreasured by the following procedure:

a 20000 litre at stp sample of the air was slowly bubbled through an excess of dilute hydrochloric acid. the resulting solution was made alkaline and heated, the ammonia liberated being dissolved in exactly 50ml of 0.10 M HCl acid, which is a large excess. 40.00 mL of 0.10 M sodium hydroxide of 0.10 M Solution were required to neutralise the excess of acid.

Calculate the concentration of ammonia in the air in units of moles of ammonia per litre of air.

Is this some type of back titration?! Could someone give me some hints im so lost.. What do I do with the whole STP thing? What should I be thinking as soon as I read STP or SLC in a question? Thanks
Title: Re: VCE Chemistry Question Thread
Post by: TimewaveZero on January 22, 2014, 05:22:20 pm
Just after some advice on how to word a discussion question.

Compare your calculated result with the manufacturers analysis.

The prac was to find by means of redox titration, the concentration of iron in a fertiliser. I calculated it to be 18.58%, with the manufacturers analysis saying 20%. How should I word the question?

Should I just say something along the lines of: The calculated result compares relatively well (does it?) with the concentration of iron supplied on the label by the manufacturer, considering human error in the process of our analysis.
Title: Re: VCE Chemistry Question Thread
Post by: brightsky on January 22, 2014, 08:15:54 pm
Just after some advice on how to word a discussion question.

Compare your calculated result with the manufacturers analysis.

The prac was to find by means of redox titration, the concentration of iron in a fertiliser. I calculated it to be 18.58%, with the manufacturers analysis saying 20%. How should I word the question?

Should I just say something along the lines of: The calculated result compares relatively well (does it?) with the concentration of iron supplied on the label by the manufacturer, considering human error in the process of our analysis.

no, scientists hate vague adverbs like 'relatively', 'fairly', 'rather', etc.. state clearly whether the calculated concentration of iron is higher or lower than the quoted concentration of iron, and give the stats.
Title: Re: VCE Chemistry Question Thread
Post by: TimewaveZero on January 22, 2014, 08:18:34 pm
Oh okay, my Chem teacher loves the word relatively :/ haha

I'll reword it then.
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on January 23, 2014, 04:57:18 pm
When balancing equations does 6HNO3 mean that there are 6 hydrogen, 6 nitrogens and 18 oxygen? I really should know this but I've forgotten
Title: Re: VCE Chemistry Question Thread
Post by: brightsky on January 23, 2014, 04:59:34 pm
indeed. in its pure form, HNO3 exists as a molecule. think of 6HNO3 as six individual HNO3 molecules. if you tally the number of hydrogen, nitrogen and oxygen atoms respectively, you'd get what you got.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 24, 2014, 09:05:06 am
Regarding the experimental set-up of evaporation (gravimetric analysis), is it necessary to write where to heat the sample?

For instance, "take the sample and heat in a desiccator until constant mass" or is it better to leave the desiccator bit out?

Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: brightsky on January 24, 2014, 09:13:11 am
yeah i've never seen the term 'desiccator' included in the description. you can heat it up in an oven if you really wanted to. just write 'heat sample to constant mass'.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on January 24, 2014, 09:22:50 am
Ok thanks!
Title: Re: VCE Chemistry Question Thread
Post by: T-Infinite on January 24, 2014, 09:57:23 pm
Hey guys, quick question. How do you find the amount in mol of the ions, when you're only given the concentration and the molar mass? I may have forgotten some steps. Question is,

For a 0.20M solution of potassium sulfate, K2SO4, calculate the amount, in mol of:
a) potasium ions, K+
b) sulfate ions SO4^2-
c) oxygen atoms

Title: Re: VCE Chemistry Question Thread
Post by: T-Infinite on January 24, 2014, 10:09:07 pm
When you have 0.20M of K2S04, consider the fact that all the K+, SO4 2- etc are all in the same solution i.e they all have the same volume. Now, let's bring back to the mole equation n= C x V. Obviously, if everything has the same volume now, so n = C. Hence, 0.20M of K2S04 will have 0.40M of K+ i.e 0.40 mol of K+ as well. Apply the same procedure for the others :D!
Hope this helps.
How come the answer for a) says it's 0.10mol ? o-o
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on January 24, 2014, 10:18:58 pm
How come the answer for a) says it's 0.10mol ? o-o
If it's a textbook question then I'm probably wrong at the finding the mole point. But I'm sure that if you have for e.g. 2M of H2SO4, there will be 4M of H+ if complete dissociation occurs. Sorry if this doesn't help.
Title: Re: VCE Chemistry Question Thread
Post by: T-Infinite on January 24, 2014, 10:28:01 pm
If it's a textbook question then I'm probably wrong at the finding the mole point. But I'm sure that if you have for e.g. 2M of H2SO4, there will be 4M of H+ if complete dissociation occurs. Sorry if this doesn't help.
Yeah, I just saw that example from the TSFX notes booklet. But I'm still unsure how to find the amounts in mol T_T.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on January 25, 2014, 03:33:52 am
If it's a textbook question then I'm probably wrong at the finding the mole point. But I'm sure that if you have for e.g. 2M of H2SO4, there will be 4M of H+ if complete dissociation occurs. Sorry if this doesn't help.

Strictly speaking, 2M of sulfuric acid does not mean 4M of H+ as sulfuric acid only completely ionises the first time to form HSO4 -. This acid doesn't completely dissociate (around 10% dissociation here) so the concentration of H+ is closer to 2M than 4M. If you're interested and want to do the calculations, the Ka of HSO4 - is around 0.01. Try a calculation where [H+] = 2M and [HSO4-] = 2M initially and you'll find that the concentration of H+ at equilibrium is still very close to 2M.

Hey guys, quick question. How do you find the amount in mol of the ions, when you're only given the concentration and the molar mass? I may have forgotten some steps. Question is,

For a 0.20M solution of potassium sulfate, K2SO4, calculate the amount, in mol of:
a) potasium ions, K+
b) sulfate ions SO4^2-
c) oxygen atoms

Erm...you can't...because you need a volume. You would have 0.40 M K+ ions but you can't find the number of moles. n=cV remember
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on January 25, 2014, 11:27:02 am
Yeah this is a textbook missprint.  see this thread for others who fell over the same question:

Missprint Chapter 3 Question 6

It should have given you a volume of 250ml.

If it's a textbook question then I'm probably wrong at the finding the mole point. But I'm sure that if you have for e.g. 2M of H2SO4, there will be 4M of H+ if complete dissociation occurs. Sorry if this doesn't help.
Title: Re: VCE Chemistry Question Thread
Post by: T-Infinite on January 25, 2014, 11:27:39 am
Erm...you can't...because you need a volume. You would have 0.40 M K+ ions but you can't find the number of moles. n=cV remember
Really? Why is this question in my textbook? I thought the same thing when I first saw the question, they didn't give us volume... so I'd assumed there was another way.
Title: Re: VCE Chemistry Question Thread
Post by: T-Infinite on January 25, 2014, 11:29:03 am
Yeah this is a textbook missprint.  see this thread for others who fell over the same question:

Missprint Chapter 3 Question 6

It should have given you a volume of 250ml.
Ohhh, thank you for telling me this! :D I suspected there was something odd about it.

*soz for double post.
Title: Re: VCE Chemistry Question Thread
Post by: survivor on January 25, 2014, 01:24:35 pm
3 mol of Fe2O3 (s) and 6 mol of CO (g) react according to the equation:  Fe2O3 (s) + 3CO (g)  ->  2Fe (l) + 3CO2 (g)

c) Determine the the amount, in mol, of substance present when the reaction is complete.

The answer is 11 mol, but i don't understand how they get there. Thanks
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 25, 2014, 01:27:27 pm
3 mol of Fe2O3 (s) and 6 mol of CO (g) react according to the equation:  Fe2O3 (s) + 3CO (g)  ->  2Fe (l) + 3CO2 (g)

c) Determine the the amount, in mol, of substance present when the reaction is complete.

The answer is 11 mol, but i don't understand how they get there. Thanks

Ok maybe its just me, but what substance is it referring to. I feel like this question is awfully ambiguous.
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on January 25, 2014, 01:31:02 pm
3 mol of Fe2O3 (s) and 6 mol of CO (g) react according to the equation:  Fe2O3 (s) + 3CO (g)  ->  2Fe (l) + 3CO2 (g)

c) Determine the the amount, in mol, of substance present when the reaction is complete.

The answer is 11 mol, but i don't understand how they get there. Thanks
Limiting reagent is CO (6 mol CO reacts with 2mol Fe2O3). Hence, we have 1 mol of unreacted Fe2O3
2 mol Fe2O3 reacting with 6 mol of CO yields 4 mol Fe (2*2) and 6mol CO2 (2*3)

4+6+1=11 mol
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 25, 2014, 01:37:13 pm
Limiting reagent is CO (6 mol CO reacts with 2mol Fe2O3). Hence, we have 1 mol of unreacted Fe2O3
2 mol Fe2O3 reacting with 6 mol of CO yields 4 mol Fe (2*2) and 6mol CO2 (2*3)

4+6+1=11 mol

Thanks for that psyxwar :)

But could you please show how you got the steps? Thanks
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on January 25, 2014, 01:38:18 pm
Thanks for that psyxwar :)

But could you please show how you got the steps? Thanks
What do you mean? Isn't it just stoich?
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on January 25, 2014, 02:00:59 pm
If I spilled HCl on the floor, how would I safely clean it?
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on January 25, 2014, 02:15:28 pm
Tell your teacher.    :P

If they are not there, neutralise it with a weak base (sodium carbonate) then mop it up.  Then tell your teacher.

If I spilled HCl on the floor, how would I safely clean it?
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on January 25, 2014, 04:15:13 pm
Find the amount in mol of

H2O molecules in 20.0 g of CuSO4.5H2O

Cl− ions in 34 g of FeCl3
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 25, 2014, 04:29:32 pm
Find the amount in mol of

H2O molecules in 20.0 g of CuSO4.5H2O

Cl− ions in 34 g of FeCl3

1. n(CuSO4.5H2O) = 20/249.6 = 0.08012821mol
n(H2O) = n(CuSO4.5H2O) x 5 = 0.401mol

2. n(FeCl3) = 34/162.3 = 0.2094886 mol
n(Cl-) = n(FeCl3) x 3 = 0.63 mol
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on January 25, 2014, 04:41:06 pm
Quick Q about pH curves

In the image i've attached, the pH curve on the right is a titration between a strong acid and strong base. Why is phenophthalein used as the indicator? I thought it would be bromothymol blue? There's still a sharp end point between 6.0 - 7.6
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 25, 2014, 04:49:31 pm
Quick Q about pH curves

In the image i've attached, the pH curve on the right is a titration between a strong acid and strong base. Why is phenophthalein used as the indicator? I thought it would be bromothymol blue? There's still a sharp end point between 6.0 - 7.6

To determine which indicator is best to use, look at the centre of the almost vertical drop (i.e the end point) for the equivalence point pH. Using that value, determine which is the BEST indicator.
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on January 25, 2014, 05:00:43 pm
In a chemical reaction how can you tell which species is acting an acid?

Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 25, 2014, 05:02:17 pm
In a chemical reaction how can you tell which species is acting an acid?

The species that donates a proton/s (i.e. H+ ion).
Title: Re: VCE Chemistry Question Thread
Post by: Bluegirl on January 25, 2014, 05:06:14 pm
how do i write and equation for the oxidation of nitrogen(II) oxide in air?
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on January 25, 2014, 05:07:07 pm
The species that donates a proton/s (i.e. H+ ion).

so if there is NH4+ on the reactants side and the product was NH3 does this mean it donated the H+?
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on January 25, 2014, 05:22:07 pm
so if there is NH4+ on the reactants side and the product was NH3 does this mean it donated the H+?

Yep! NH4+ would be the acid and NH3 would be it's conjugate base.

The NH4+ would've transferred or "donated" it's proton (H+ ion), causing it to become NH3
Title: Re: VCE Chemistry Question Thread
Post by: MagicGecko on January 25, 2014, 05:25:41 pm
how do i write and equation for the oxidation of nitrogen(II) oxide in air?
2NO(g) + O2(g) ----> 2NO2(g)

so if there is NH4+ on the reactants side and the product was NH3 does this mean it donated the H+?
Yes since NH4+ lost a H+ it means it donated that H+ to whatever else was on the reactant side to become NH3. So NH4+ acted as an acid.
Title: Re: VCE Chemistry Question Thread
Post by: Bluegirl on January 25, 2014, 05:26:30 pm
2NO(g) + O2(g) ----> 2NO2(g)

Thanks
Title: Re: VCE Chemistry Question Thread
Post by: T-Infinite on January 25, 2014, 07:20:50 pm

Assign oxidation numbers to each element in this equation, and hence identify the oxidant and reductant

Fe2O3 (s) + 3CO (g) -----> 2Fe (s) + 3CO2 (g)

I get so confused with the whole oxidant/reductant/oxidation/reduction thing. @[email protected]
Title: Re: VCE Chemistry Question Thread
Post by: tinkerbell on January 25, 2014, 07:35:51 pm

Assign oxidation numbers to each element in this equation, and hence identify the oxidant and reductant

Fe2O3 (s) + 3CO (g) -----> 2Fe (s) + 3CO2 (g)

I get so confused with the whole oxidant/reductant/oxidation/reduction thing. @[email protected]

To assign oxidation numbers, you have to follow some rules, one of them being that a free element such as iron, which is a product in this reaction, has an oxidation number of 0.

Another important rule is that a compound has a net oxidation number of 0-so adding up the oxidation numbers of the individual elements making up the compound. So,  Fe203 is a compound and it must have a net oxidation charge of 0. Valence of oxygen is 2 and we know oxygen is electronegative (has an affinity for electrons), so the oxidation number for oxygen is -2 times 3 (because there are 3 oxygen atoms and we are accounting for all of them).

So, this means that the Fe2 component must have a positive charge of +6. This charge is distributed equally between the 2 Fe atoms.

Let's look at CO-oxygen -2, Carbon must have +2 oxidation number.

CO2-oxygen negative 4, so carbon positive 4.

Now that we have assigned our oxidation numbers, our next step is to work out what is the oxidising agent and what is the reducing agent.

Oxidising agent-what oxidises something and itself gets reduced. To be reduced the oxidation number needs to decrease. We see this with Fe203, which is a reactant and leads to the free element Fe. So, this is a oxidising agent/reductant.

CO (carbon monoxide) is the reducing agent because it gets oxidised-the oxidation number on the carbon atom increases.

Increase in oxidation number means-become oxidised and therefore reducing agent. reducing agent reduces Fe2O3 (because you get free iron).

Hope this makes sense,

Good luck.
Title: Re: VCE Chemistry Question Thread
Post by: T-Infinite on January 25, 2014, 07:47:13 pm
To assign oxidation numbers, you have to follow some rules, one of them being that a free element such as iron, which is a product in this reaction, has an oxidation number of 0.

Another important rule is that a compound has a net oxidation number of 0-so adding up the oxidation numbers of the individual elements making up the compound. So,  Fe203 is a compound and it must have a net oxidation charge of 0. Valence of oxygen is 2 and we know oxygen is electronegative (has an affinity for electrons), so the oxidation number for oxygen is -2 times 3 (because there are 3 oxygen atoms and we are accounting for all of them).

So, this means that the Fe2 component must have a positive charge of +6. This charge is distributed equally between the 2 Fe atoms.

Let's look at CO-oxygen -2, Carbon must have +2 oxidation number.

CO2-oxygen negative 4, so carbon positive 4.

Now that we have assigned our oxidation numbers, our next step is to work out what is the oxidising agent and what is the reducing agent.

Oxidising agent-what oxidises something and itself gets reduced. To be reduced the oxidation number needs to decrease. We see this with Fe203, which is a reactant and leads to the free element Fe. So, this is a oxidising agent/reductant.

CO (carbon monoxide) is the reducing agent because it gets oxidised-the oxidation number on the carbon atom increases.

Increase in oxidation number means-become oxidised and therefore reducing agent. reducing agent reduces Fe2O3 (because you get free iron).

Hope this makes sense,

Good luck.
Thank you! Your step-by-step explanation really helps :) !
Title: Re: VCE Chemistry Question Thread
Post by: PsychoT on January 25, 2014, 10:20:31 pm
Can anyone help me out with this?

The percentage of iron in steel wool was analysed by dissolving 0.1674g in 50ml  of sulfuric acid to produce Fe2+. These ions were then titrated against 0.0213M Mn04- solution, producing Fe3+ and Mn2+ ions. The volume of permanganate solution required was 26.46 ml. Calculate the percentage of iron in the steel wool.
Title: Re: VCE Chemistry Question Thread
Post by: LOLs99 on January 25, 2014, 11:19:25 pm
Lets write down some equations first regarding this reaction.
Oxidation: $Fe^{2+}$ --->$Fe^{3+}$ + $e^-$ (multiply by 5 and add oxidation and reduction equation together)
Reduction: $MnO_4^- + 8H^+ + 5e^- ---> Mn^{2+} + 4H_2O$
Overall: $5Fe^{2+} + MnO_4^- + 8H^+ ---> 5Fe^{3+} + Mn^{2+} + 4H_2O$

So now lets start some calculations : $
n(MnO_4^-) = 0.02646 L$
x $0.0213M = 0.000564mol$

$n(Fe^{2+})= 0.000564mol$ x $5= 0.00282mol$

$n(Fe^{2+})= n(Fe)_{original} = 0.00282 mol$

$m(Fe) = 0.00282 mol$ x $55.9 g mol^{-1} = 0.158 g$

Therefore :   % Fe =$\frac{0.158}{0.1674}$ x $100$ = 94.1 % (3 sig fig)

-----

First time using LATEX. Sorry if it is messy. TOOK ME AGES TO USE LATEX!!
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on January 25, 2014, 11:34:10 pm
what is the difference between the relative molecular mass, Mr of water and the molar mass of water?
Title: Re: VCE Chemistry Question Thread
Post by: brightsky on January 25, 2014, 11:53:06 pm
not much, except relative molecular mass has no units but molar mass has units g/mol.
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on January 25, 2014, 11:56:36 pm
A haemoglobin molecule contains four iron atoms and when analysed is found to contain 0.33% iron by mass. What is the molecular mass, Mr of haemoglobin?

Title: Re: VCE Chemistry Question Thread
Post by: brightsky on January 26, 2014, 12:13:26 am
iron has an atomic mass of around 55.8. so four iron atoms would have combined mass of 223.2. if we denote the molecular mass of haemoglobin as x, then 0.33/100 * x = 223.2. so x = 67636.4 g/mol. pretty big molecule.
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on January 26, 2014, 01:02:45 am
iron has an atomic mass of around 55.8. so four iron atoms would have combined mass of 223.2. if we denote the molecular mass of haemoglobin as x, then 0.33/100 * x = 223.2. so x = 67636.4 g/mol. pretty big molecule.

thanks but i'm having trouble trying to figure out the steps you took. I realise that the mass is 223.2g and the mol is 0.33/100? so would you use Mr=m/n?
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on January 26, 2014, 01:20:12 am
A solution was prepared by dissolving 4.680g of sodium chloride and 1.705g of sodium sulfate in water, so the total volume of the solution was 500ml. what is the sodium ion concentration.

so far I have done $n(NaCl) = \frac{4.680}{58.44}=0.0801 mol$ and $n(Na2SO4) = \frac{1.705}{166.97} = 0.0102 mol$

so what do I do from here? Am I able to add the 2 mols together then divide by 0.5L even though the substances were different?
Title: Re: VCE Chemistry Question Thread
Post by: LOLs99 on January 26, 2014, 05:25:23 am
A solution was prepared by dissolving 4.680g of sodium chloride and 1.705g of sodium sulfate in water, so the total volume of the solution was 500ml. what is the sodium ion concentration.

so far I have done $n(NaCl) = \frac{4.680}{58.44}=0.0801 mol$ and $n(Na2SO4) = \frac{1.705}{166.97} = 0.0102 mol$

so what do I do from here? Am I able to add the 2 mols together then divide by 0.5L even though the substances were different?
I will continue from your step since it looks good. Just make sure u use the VCAA data booklet (download from vcaa website or ask your teacher for a nice copy).

From Na2SO4: n(Na+) = n(Na2SO4) x 2 (as there a 2 sodium atoms present in each molecule)
= 0.0204 mol
From NaCl: n(Na+)= n(NaCl)= 0.0801 mol

So now we have all the number of moles of sodium ion from 2 solution.
Add them up. n(Na+)total= 0.0801mol+0.0204mol= 0.1005mol
Finally divide by the total volume. c(Na+)= 0.1005mol/0.500L = 0.201 M

Hope it is right. ** Yawn **
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on January 26, 2014, 02:29:13 pm
Write balanced equations for each of the following:

Magnesium ribbon is burnt in air
Potassium metal dissolves violently in water
An aqueous sodium hydroxide solution is neutralised with  dilute aqueous sulfuric acid

Some explanation would be helpful as well for example would burnt in air mean I should add O2 to the reactants?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 26, 2014, 02:31:39 pm
Write balanced equations for each of the following:

Magnesium ribbon is burnt in air
Potassium metal dissolves violently in water
An aqueous sodium hydroxide solution is neutralised with  dilute aqueous sulfuric acid

(a) 2Mg(s) + O2(g) ---> 2MgO(s)

(b) K(s) + H2O(l) ---> KO(aq) + H2(g)

(c) 2NaOH(aq) + H2SO4(aq) ---> Na2SO4(aq) + 2H2O(l)

I could be wrong, but that's my try ^^. Clarification would be great :)
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on January 26, 2014, 02:48:58 pm
And also an aqueous dilute nitric acid solution reacts with solid zinc carbonate
Title: Re: VCE Chemistry Question Thread
Post by: MagicGecko on January 26, 2014, 03:08:47 pm
2HNO3(aq) + ZnCO3(s) --->  Zn(NO3)2 + H2O(l) + CO2(g)
Title: Re: VCE Chemistry Question Thread
Post by: thushan on January 26, 2014, 04:28:36 pm
(a) 2Mg(s) + O2(g) ---> 2MgO(s)

(b) K(s) + H2O(l) ---> KO(aq) + H2(g)

(c) 2NaOH(aq) + H2SO4(aq) ---> Na2SO4(aq) + 2H2O(l)

I could be wrong, but that's my try ^^. Clarification would be great :)

2 K(s) + 2 H2O(l) ---> 2 KOH (aq) + H2 (g)

is what you mean.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 26, 2014, 05:19:31 pm
2 K(s) + 2 H2O(l) ---> 2 KOH (aq) + H2 (g)

is what you mean.

I see :) Thanks thushan.
Title: Re: VCE Chemistry Question Thread
Post by: T-Infinite on January 27, 2014, 12:33:39 am
Someone please clarify this question for me, I've been doing every other questions correct so far until I got to this part where everything looked so odd. I'm not sure if the textbook answer at the back is wrong or not..

Question is...

The following equations are not balanced
(i) Identify the speciies that has been reduced and the species that has been oxidised
(ii) Write balanced half equations for the oxidation and reduction reactions
(iii) Combine the half equations to write a balanced equation.

d) MnO2 (s) + H+ (aq) + S (s) ---> Mn 2+ (aq) + H2O (l) + SO2 (g)

So I got part (i) correct but part (ii) and (iii) slightly off.

The answer at the back says that
(ii): 2MnO2 (s) + 4H+ (aq) + 2e ---> Mn2+ +2H2O (l)
S (s) + 2H2O (l) ---> SO2 (g) + 4H+ (aq) + 4e-
(iii) MnO2 (s) + 4H+ (aq) + S (s) ---> 2Mn2+ + 2H2O (l) +SO2 (g)

I may have done something wrong, who knows.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on January 27, 2014, 12:50:25 am
Someone please clarify this question for me, I've been doing every other questions correct so far until I got to this part where everything looked so odd. I'm not sure if the textbook answer at the back is wrong or not..

Question is...

The following equations are not balanced
(i) Identify the speciies that has been reduced and the species that has been oxidised
(ii) Write balanced half equations for the oxidation and reduction reactions
(iii) Combine the half equations to write a balanced equation.

d) MnO2 (s) + H+ (aq) + S (s) ---> Mn 2+ (aq) + H2O (l) + SO2 (g)

So I got part (i) correct but part (ii) and (iii) slightly off.

The answer at the back says that
(ii): 2MnO2 (s) + 4H+ (aq) + 2e ---> Mn2+ +2H2O (l)
S (s) + 2H2O (l) ---> SO2 (g) + 4H+ (aq) + 4e-
(iii) MnO2 (s) + 4H+ (aq) + S (s) ---> 2Mn2+ + 2H2O (l) +SO2 (g)

I may have done something wrong, who knows.

(ii) should have a coefficient of 1 for manganese dioxide
(iii) should therefore be slightly different
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 27, 2014, 10:08:57 am
Someone please clarify this question for me, I've been doing every other questions correct so far until I got to this part where everything looked so odd. I'm not sure if the textbook answer at the back is wrong or not..

Question is...

The following equations are not balanced
(i) Identify the speciies that has been reduced and the species that has been oxidised
(ii) Write balanced half equations for the oxidation and reduction reactions
(iii) Combine the half equations to write a balanced equation.

d) MnO2 (s) + H+ (aq) + S (s) ---> Mn 2+ (aq) + H2O (l) + SO2 (g)

So I got part (i) correct but part (ii) and (iii) slightly off.

The answer at the back says that
(ii): 2MnO2 (s) + 4H+ (aq) + 2e ---> Mn2+ +2H2O (l)
S (s) + 2H2O (l) ---> SO2 (g) + 4H+ (aq) + 4e-
(iii) MnO2 (s) + 4H+ (aq) + S (s) ---> 2Mn2+ + 2H2O (l) +SO2 (g)

I may have done something wrong, who knows.

8H+(aq) + 2MnO2(s) +4e  ---> 2Mn2+(aq) + 4H2O(l)

S(s) +2H2O(l) ---> SO2(g) + 4H+(aq) + 4e

Balanced equation:
4H+(aq) + 2MnO2(s)  ---> 2Mn2+(aq) + 2H2O(l)

S(s)  ---> SO2(g)

Balanced equation:
S(s) + 2MnO2(s) + 4H+(aq) ---> 2Mn2+(aq) + 2H2O(l) + SO2(g)

lznxl- why is the coefficient of manganese dioxide 1? :/
Title: Re: VCE Chemistry Question Thread
Post by: T-Infinite on January 27, 2014, 10:17:01 am

8H+(aq) + 2MnO2(s) +4e  ---> 2Mn2+(aq) + 4H2O(l)

S(s) +2H2O(l) ---> SO2(g) + 4H+(aq) + 4e

Balanced equation:
4H+(aq) + 2MnO2(s)  ---> 2Mn2+(aq) + 2H2O(l)

S(s)  ---> SO2(g)

Balanced equation:
S(s) + 2MnO2(s) + 4H+(aq) ---> 2Mn2+(aq) + 2H2O(l) + SO2(g)

lznxl- why is the coefficient of manganese dioxide 1? :/
Just wondering, what happened to the 8H+ on the left and the 4H+ on the right? Did they cancel out leaving only 4H+ on the left for the final balanced equation? o-o
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on January 27, 2014, 10:28:12 am
Just wondering, what happened to the 8H+ on the left and the 4H+ on the right? Did they cancel out leaving only 4H+ on the left for the final balanced equation? o-o

Correct; the H+ ions cancel out, as well as waters and electrons.
Title: Re: VCE Chemistry Question Thread
Post by: T-Infinite on January 27, 2014, 10:32:13 am
Correct; the H+ ions cancel out, as well as waters and electrons.
By the way, in the final equation (iii) , the textbook says there is only 1MnO2 ? I'm so confused, in the half equation theres 2MnO2 but in full equation theres only 1?
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on January 27, 2014, 01:30:15 pm
In gravimetric analysis, what's the point of rinsing the precipitate containing the substance being analysed with small amounts of solvent?
Title: Re: VCE Chemistry Question Thread
Post by: clıppy on January 27, 2014, 01:48:41 pm
Let's say that I'm precipitating PbSO4 from an FeSO4 solution. I've precipitated the PbSO4 and I've taken it out of the solution. To make sure there is no extra weight that isn't PbSO4, I'll rinse my precipitate with water several times to wash out everything that I don't want and then I'll dry my precipitate in an oven several times to make sure that all the water evaporates.
It's basically to get rid of anything that isn't the precipitate that could alter your final measured weight
Title: Re: VCE Chemistry Question Thread
Post by: PsychoT on January 27, 2014, 10:26:00 pm
How do I work out how many Cl- ions there are in 13.4g of Nickel Chloride (NiCl2)?

Cheers.
Title: Re: VCE Chemistry Question Thread
Post by: thushan on January 27, 2014, 10:36:06 pm
You're given a mass of NiCl2. The question asks you to count the number of Cl- ions. Since you need to count an amount, yet you're given a mass, the first thing to do is to determine the amount, in mol, of NiCl2.

Once you've got that, you know that for every NiCl2 formula unit you have, you have 2 Cl- ions. Use this fact to determine the amount of Cl- ions, given you know the amount of NiCl2.
Title: Re: VCE Chemistry Question Thread
Post by: PsychoT on January 27, 2014, 10:48:17 pm
You're given a mass of NiCl2. The question asks you to count the number of Cl- ions. Since you need to count an amount, yet you're given a mass, the first thing to do is to determine the amount, in mol, of NiCl2.

Once you've got that, you know that for every NiCl2 formula unit you have, you have 2 Cl- ions. Use this fact to determine the amount of Cl- ions, given you know the amount of NiCl2.

Cheers. Worked it out. Trying to get back into Chemistry is tough.
Title: Re: VCE Chemistry Question Thread
Post by: TimewaveZero on January 29, 2014, 07:30:00 pm
I've probably made a silly mistake here but I just can't see it haha

Serotonin (C10H12N2O molecular mass = 176g mol -1) is a compound that conducts nerve impulses in the brain and muscles. A sample of spinal fluid from a volunteer in a study was found to contain a serotonin concentration of 1.5 ng L-1 (1.5 nano grams per litre)
How many molecules of of serotonin are there in one millilitre of the fluid?

my calculations:

1 millilitre = 0.001 L
=> 1.5 * 0.001 = 0.0015 nanograms in 1 millilitre

= 0.0000000015 grams

n(c10h12n2o) = 0.0000000015 / 176

= 8.52 * 10^-12

n of molecules should equal (8.52 * 10^-12) * (6.02 * 10^23) = 5.13 * 10^12

the MC answers are:

A. 5.13 * 10^9
B. 9.03 * 10^11
C. 5.13 * 10^27
D. 9.03 * 10^29

Why is A) to the power of 9? Did I do something wrong with the dividing?
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on January 29, 2014, 07:48:48 pm
I've probably made a silly mistake here but I just can't see it haha

Serotonin (C10H12N2O molecular mass = 176g mol -1) is a compound that conducts nerve impulses in the brain and muscles. A sample of spinal fluid from a volunteer in a study was found to contain a serotonin concentration of 1.5 ng L-1 (1.5 nano grams per litre)
How many molecules of of serotonin are there in one millilitre of the fluid?

my calculations:

1 millilitre = 0.001 L
=> 1.5 * 0.001 = 0.0015 nanograms in 1 millilitre

= 0.0000000015 grams

n(c10h12n2o) = 0.0000000015 / 176

= 8.52 * 10^-12

n of molecules should equal (8.52 * 10^-12) * (6.02 * 10^23) = 5.13 * 10^12

the MC answers are:

A. 5.13 * 10^9
B. 9.03 * 10^11
C. 5.13 * 10^27
D. 9.03 * 10^29

Why is A) to the power of 9? Did I do something wrong with the dividing?
I think you got that bold step wrong, 0.0015 nanogram = 0.0015 x 10 ^-9 g= 1.5 x 10^-12.
Hence, if you fix that number and sub it into the other steps, everything would be fine!
Title: Re: VCE Chemistry Question Thread
Post by: TimewaveZero on January 29, 2014, 07:57:06 pm
I think you got that bold step wrong, 0.0015 nanogram = 0.0015 x 10 ^-9 g= 1.5 x 10^-12.
Hence, if you fix that number and sub it into the other steps, everything would be fine!

Ah yes, that is right. Although, I subbed the new value in, and am now presented with 5.13 * 10^24. ???? hahahaha
Title: Re: VCE Chemistry Question Thread
Post by: darklight on January 29, 2014, 07:59:53 pm
Ah yes, that is right. Although, I subbed the new value in, and am now presented with 5.13 * 10^24. ???? hahahaha

I just did it with that value and it did come to A, so maybe try again.

Also: why does one need a H+ catalyst when reacting an alkene with water? Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: TimewaveZero on January 29, 2014, 08:05:35 pm
Ahh yes I did too. I mistakingly left out the power of -15 in the final calculation. My error.
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on January 29, 2014, 08:17:33 pm
VCE don't really go into the reasons for the catalyst, only which one is needed for each reaction.  Have a look at this page for the reason for the H+ in addition reaction.  From what i can gather it has the ability to break the double bond thus allow the reaction to proceed.  H+ is left over from the breaking of water into the OH- function group attached to the hydrocarbon.  So it is a catalyst in that it is present at the start and finish of the reaction, however it is actually involved in the reaction.

Hope this clears it up a little.

I just did it with that value and it did come to A, so maybe try again.

Also: why does one need a H+ catalyst when reacting an alkene with water? Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on January 29, 2014, 11:33:34 pm
VCE don't really go into the reasons for the catalyst, only which one is needed for each reaction.  Have a look at this page for the reason for the H+ in addition reaction.  From what i can gather it has the ability to break the double bond thus allow the reaction to proceed.  H+ is left over from the breaking of water into the OH- function group attached to the hydrocarbon.  So it is a catalyst in that it is present at the start and finish of the reaction, however it is actually involved in the reaction.

Hope this clears it up a little.

Here's my take on it. I haven't read the link so I may well be repeating what the link says.
In an alkene, the carbon-carbon double bond contains a rather large amount of electrons (relative to other bonds). Therefore, anything with a positive charge would be attracted to this double bond. The H+ ion, a proton, is well suited to attacking the double bond as it is small. The proton then drags some electron density from the double bond, forming a carbon-hydrogen single bond. This creates a positive charge on the other carbon. Now, as this carbon has a positive charge, it attracts the negative oxygen in water much more readily and the reaction proceeds faster.

In contrast, without the acid catalyst, you're trying to react a water molecule with an alkene. Now, the double bond has lots of electrons, which repels the water molecule. This increases the activation energy of the reaction and thus slows it down.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on January 31, 2014, 07:54:42 am
Can someone please clarify that i've understood this correctly:

For the equation:
Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g),
we know that n(Mg)=0.0551 and n(HCl)=0.0600.

Since the Mg is in excess, the amount reacting would follow the mole ratio. This means, 0.03mol of Mg would react with 0.0600 mol of HCl. This would produce 0.09mol of MgCl2 and H2 gas.
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on January 31, 2014, 08:11:25 am
Can someone please clarify that i've understood this correctly:

For the equation:
Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g),
we know that n(Mg)=0.0551 and n(HCl)=0.0600.

Since the Mg is in excess, the amount reacting would follow the mole ratio. This means, 0.03mol of Mg would react with 0.0600 mol of HCl. This would produce 0.09mol of MgCl2 and H2 gas.
Looking at the mole ratio: you have 1:2:1:1
You are right at the point Mg is the excess reactant, hence the calculation will base on the mole of HCl. According to the mole ratio above, if you have 0.0600 mole of HCl, Mg will be halved that mole amount and equal 0.0300 mole. Other products also have the same mole ratio as Mg (1:1:1), therefore, MgCl2 and H2 will have the 0.0300 as well.
Remember that determining mole is based on the mole ratio on the equation you got. I think you are confused with the Conservation of Mass theory, which means the MASS of reactants must equal the mass of products. Note that this is mole, not mass.
Hope this helps!
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on February 01, 2014, 10:11:53 am
Can someone please clarify that i've understood this correctly:

For the equation:
Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g),
we know that n(Mg)=0.0551 and n(HCl)=0.0600.

Since the Mg is in excess, the amount reacting would follow the mole ratio. This means, 0.03mol of Mg would react with 0.0600 mol of HCl. This would produce 0.09mol of MgCl2 and H2 gas.
correct, except your sig figs are off.
Title: Re: VCE Chemistry Question Thread
Post by: Sanguinne on February 01, 2014, 11:57:03 am
What are the states for organic compound such as esters?
Would they be aqueous or solid or gas or liquid?

For example, if i had a question which asked me to write the equation of the formation of ethyl butanoate, i get
H2SO4
CH3CH2OH + CH3CH2CH2COOH ------> CH3CH2CH2COOCH2CH3 + H20(l)

what are the states for this organic compounds?
Title: Re: VCE Chemistry Question Thread
Post by: brightsky on February 01, 2014, 11:59:20 am
Usually liquid.
Title: Re: VCE Chemistry Question Thread
Post by: darklight on February 01, 2014, 01:01:30 pm
I've heard that states aren't that important in organic chem. Is this true?
Title: Re: VCE Chemistry Question Thread
Post by: Edward21 on February 01, 2014, 02:07:32 pm
I've heard that states aren't that important in organic chem. Is this true?
IF you've got a combustion equation, you'd put what you're oxidising in the (l) state. States still need to be there though! Compare the states that you've put down, with what's in the answer. It's confusing when we ferment glucose and get CO2 and ethanol, because some reason that both the glucose and ethanol are (l), the ethanol only is (aq) or none are (l).. :\ i've seen this vary from resource to resource unfortunately..
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 01, 2014, 03:19:30 pm
Can someone explain how to:

Calculate the mass, in mg, of manganese in the steel sample

in the attached question?

This is SA Q. 2bii) of the 2008 Chemistry Exam (if the image doesn't work)
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on February 01, 2014, 04:37:06 pm
From the graph:
[Mn] = 35 (however no units are given)

Because no units are given we cant really complete the whole question but lets do what we can.

This [Mn] being 35 is in a diluted sample.  So we need to find the concentration of Mn in the undiluted sample.

C1V1=C2V2
C1= ?
V1= 25ml
C2= 35
V2= 100ml

C1=35*100/25
= 140
So we have a concentration of the undiluted sample in the 1L being 140.  My guess is you cut off the x-axis label with the units of concentration. We need this to complete the question.

Can someone explain how to:

Calculate the mass, in mg, of manganese in the steel sample

in the attached question?

This is SA Q. 2bii) of the 2008 Chemistry Exam (if the image doesn't work)
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 01, 2014, 06:05:03 pm
My guess is you cut off the x-axis label with the units of concentration. We need this to complete the question.
I'm so sorry!! Yes I did. The unit of measure is the concentration of MnO4- (aq) (mgL-1)
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on February 01, 2014, 10:01:59 pm
Okay, so where were we...  (please note i have changed my tune a little from my last post, the graph tells us [MnO4-] not [Mn]

140mg/L  this is the mass of MnO4- in the undiluted solution

You are looking for only the Mn not the MnO4-.  so: %Mn in  MnO4- = 46.17%.

mass of Mn in 140mg of    MnO4-    = 140*0.4617
= 64.64mg
= 0.06464g

%Mn in sample.     = Mass of Mn/Mass of sample*100
= 0.06464/13.936*100
= 0.46%

I'm so sorry!! Yes I did. The unit of measure is the concentration of MnO4- (aq) (mgL-1)
Title: Re: VCE Chemistry Question Thread
Post by: EFPBH on February 02, 2014, 12:17:13 am
Hello, need help with this question. ( could you explain each step )

The active ingredient of an antacid is magnesium carbonate. When treated with excess hydrochloric acid, a 3.50 g of the antacid produced 714 mL of carbon dioxide, measured at 22.0 C and 101.3 kPa pressure. Calculate the percentage of the magnesium carbonate in the powder.

thanks :)
Title: Re: VCE Chemistry Question Thread
Post by: Edward21 on February 02, 2014, 12:36:27 am
Hello, need help with this question. ( could you explain each step )

The active ingredient of an antacid is magnesium carbonate. When treated with excess hydrochloric acid, a 3.50 g of the antacid produced 714 mL of carbon dioxide, measured at 22.0 C and 101.3 kPa pressure. Calculate the percentage of the magnesium carbonate in the powder.

thanks :)

MgCO3(s) + 2HCl(aq) --> CO2(g) + H2O(l) + MgCl2(aq)

If pV=nRT

n(CO2)=(101.3kPa x 0.714L)/(8.31 x (22+273)K)=0.0296mol
n(MgCO3)=n(CO2)=0.0295mol
m(MgCO3)=nM=0.0295mol x 84.3gmol=2.49g

%MgCO3=(2.49g)/(3.50g) x 100% = 71.1% purity  :)
Title: Re: VCE Chemistry Question Thread
Post by: bucklr on February 02, 2014, 12:42:14 am
Hello, need help with this question. ( could you explain each step )

The active ingredient of an antacid is magnesium carbonate. When treated with excess hydrochloric acid, a 3.50 g of the antacid produced 714 mL of carbon dioxide, measured at 22.0 C and 101.3 kPa pressure. Calculate the percentage of the magnesium carbonate in the powder.

thanks :)
using - MgCO3(s)+2HCL(aq) -> MgCl(aq)+H2O(l)+CO2(g)

0.714L of CO(2) produced
22.0oC -> convert to Kelvin by adding 273 -> 295oK

use gas equation to find mol of CO2

1=(0.714x101.3)/(8.31xnx295)
n=0.0295 mol of CO2

use balanced chemical equation to find mol of MgCO2

therefore... n(MgCO3)=0.0295mol
find mass of MgCO3 (using periodic table molar mass of Mg=24.3, C=12.0, O=32.0)
therefore... 0.0295x(24.3+12.0+48.0)=2.49g
find percentage by mass
100x2.49/3.50=71.1%(w/w)
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 02, 2014, 08:52:35 pm
If I have 1.305g of Na2CO3 and put this in a 250.0mL flask (and filled it up to the mark w/ deionised water) does this mean there is 1.305g of NaCO3 per 250mL?

Seems pretty straight forward but I just want to make sure :p
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on February 02, 2014, 09:00:36 pm
Yep, and you would have 5.22g per 1L,  Provided that is is pure Na2CO3

If I have 1.305g of Na2CO3 and put this in a 250.0mL flask (and filled it up to the mark w/ deionised water) does this mean there is 1.305g of NaCO3 per 250mL?

Seems pretty straight forward but I just want to make sure :p
Title: Re: VCE Chemistry Question Thread
Post by: eagles on February 03, 2014, 10:30:35 am
If 3.0 g of NaOH is added to 500mL of 0.10 M HCl, what's the resulting pH?
Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on February 03, 2014, 05:11:54 pm
Here is a hint:

Step 1) Write an equation.
Step 2) Find mole of each reactant.
Step 3) This is limiting reagent, so work out which one is left over. (we know its limiting regent as we have knowledge of both reactants).
Step 4) Work out how much is left over.
Step 5) Work out pH based on what is left over.

If 3.0 g of NaOH is added to 500mL of 0.10 M HCl, what's the resulting pH?
Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 03, 2014, 06:56:50 pm
Is the mole of the original solution = to the mole of the diluted solution?
The q. I just did is implying that it is
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on February 03, 2014, 07:14:03 pm
Is the mole of the original solution = to the mole of the diluted solution?
The q. I just did is implying that it is
Absolutely right. When you dilute something, you are adding water to it, hence, the amount of solute will not change, the only thing that is changing is the concentration.
You should recall the equation from unit 1/2 that:
c1 x V1 = c2 x V2
( this is because n1 = n2)
Hope this helps!
Title: Re: VCE Chemistry Question Thread
Post by: eagles on February 03, 2014, 07:17:41 pm
Thank you!

Here is a hint:

Step 1) Write an equation.
Step 2) Find mole of each reactant.
Step 3) This is limiting reagent, so work out which one is left over. (we know its limiting regent as we have knowledge of both reactants).
Step 4) Work out how much is left over.
Step 5) Work out pH based on what is left over.
Title: Re: VCE Chemistry Question Thread
Post by: fadzsta1 on February 03, 2014, 08:17:49 pm
Hey guys,
From Chem Heinemann 3/4 book - pg 59 Q22a:

The thermite process can be used to weld lengths of railway track together. A mould
is placed over the two ends of rails to be joined and it is filled with a ‘charge’ of
aluminium powder and iron(III) oxide. When the mixture is ignited, a redox reaction
occurs to form molten iron which joins the rails together.
a) Write a half equation for the conversion of iron(III) oxide to metallic iron.

My answer:        Fe2O3(s) + 6H+(aq) + 6e- ----> 2Fe(s) + 3H2O(l)
Actual answer:   Fe2O3(s) + 6e- ----> 2Fe(l) + 3O2-(s)

Where the heck did that oxygen ion come from? Isn't it supposed to be water to balance the hydrogen ions? How is that ion solid? So many questions... any help would be appreciated! Thanks!!
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 03, 2014, 08:21:34 pm
Hey guys,
From Chem Heinemann 3/4 book - pg 59 Q22a:

The thermite process can be used to weld lengths of railway track together. A mould
is placed over the two ends of rails to be joined and it is filled with a ‘charge’ of
aluminium powder and iron(III) oxide. When the mixture is ignited, a redox reaction
occurs to form molten iron which joins the rails together.
a) Write a half equation for the conversion of iron(III) oxide to metallic iron.

My answer:        Fe2O3(s) + 6H+(aq) + 6e- ----> 2Fe(s) + 3H2O(l)
Actual answer:   Fe2O3(s) + 6e- ----> 2Fe(l) + 3O2-(s)

Where the heck did that oxygen ion come from? Isn't it supposed to be water to balance the hydrogen ions? How is that ion solid? So many questions... any help would be appreciated! Thanks!!

The end of q22 a) says 'Write a half equation for the conversion of iron(III) oxide to,metallic iron and oxide ions
Title: Re: VCE Chemistry Question Thread
Post by: fadzsta1 on February 03, 2014, 08:29:10 pm
.... Perhaps the older version of the book has a typo there. Thanks for your help :)
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 03, 2014, 08:38:08 pm
.... Perhaps the older version of the book has a typo there. Thanks for your help :)

Ohhh I guess it does!  Do you have the heinemann chemistry student workbook? There's a similar Q. to this (Page 22, c))

Have I figured the half equation correctly?

Fe2O3(s) --> 2Fe(l) + 3H2O(l)
Fe2O3(s) +6H+ + 6e --> 2Fe(l) + 3H2O
Fe2O3(s) +6H+ + 6e --> 2Fe(l) + 6H+ + 3O2-
(now the 6H+ cancel out)
Fe2O3(s) + 6e --> 2Fe(l) + 3O2-

would 3O2- be in an (aq) state since it is now an ion? In the book it says that it is (s)?
Title: Re: VCE Chemistry Question Thread
Post by: fadzsta1 on February 03, 2014, 08:46:50 pm
I don't understand why the H2O dissociates, then cancels out the hydrogen ions... I also don't understand why the O2- ions are in the solid state....
I'm just gonna wait here till one of the experts on here delivers...
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 03, 2014, 08:53:54 pm
I don't understand why the H2O dissociates, then cancels out the hydrogen ions...

I did that so that that it is an ionic equation and then cancelled out the hydrogen ions since they are spectator ions *NOT SURE IF CORRECT* (probably isn't)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on February 03, 2014, 09:08:56 pm
Ohhh I guess it does!  Do you have the heinemann chemistry student workbook? There's a similar Q. to this (Page 22, c))

Have I figured the half equation correctly?

Fe2O3(s) --> 2Fe(l) + 3H2O(l)
Fe2O3(s) +6H+ + 6e --> 2Fe(l) + 3H2O
Fe2O3(s) +6H+ + 6e --> 2Fe(l) + 6H+ + 3O2-
(now the 6H+ cancel out)
Fe2O3(s) + 6e --> 2Fe(l) + 3O2-

would 3O2- be in an (aq) state since it is now an ion? In the book it says that it is (s)?

I think there is a serious misunderstanding here. The thermite reaction does not involve water here. It is just aluminium powder and iron (III) oxide. The overall reaction is Fe2O3(s) + 2Al(s) => 2 Fe(l) + Al2O3(s)
Redox reactions don't always have to proceed in aqueous solution!

Also, even if this reaction did occur in water, aluminium oxide is insoluble in water.
EVEN IF aluminium oxide dissolved in water, the oxide ion cannot exist in water in aqueous form. Think about it: the oxide ion's conjugate acid is the hydroxide ion, which is a strong base. Therefore, the oxide ion must be a ridiculously powerful base capable of immediately deprotonating most things around it (like water). Aqueous oxide ions quickly become hydroxide ions.

I don't understand why the H2O dissociates, then cancels out the hydrogen ions... I also don't understand why the O2- ions are in the solid state....
I'm just gonna wait here till one of the experts on here delivers...

The water doesn't dissociate. Oxide ions react with water by O2-(aq) + H2O(l) => 2OH-(aq)
Now if this hydroxide encounters hydrogen ions...you know what happens.
Title: Re: VCE Chemistry Question Thread
Post by: fadzsta1 on February 03, 2014, 09:29:45 pm
You are a god i love you
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 03, 2014, 09:43:57 pm
Ohhh I understand. Thanks so much :-)

What would the oxidation half equation be?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on February 03, 2014, 10:16:35 pm
Ohhh I understand. Thanks so much :-)

What would the oxidation half equation be?

Well...given the question...I'd say it would be Fe2O3(s)+ 6e- => 2Fe(l) + 3O2- (s)

Even though I dislike the practice of writing an ion as a solid. And yes the metallic iron is molten as per the question.

Edit: I've read the question incorrectly; this is the reduction half equation.
Edit again: how did I put the electrons on the wrong side of the equation.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 03, 2014, 10:20:39 pm
Thanks :) what would the reduction half equation be?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on February 04, 2014, 02:27:39 pm
Thanks :) what would the reduction half equation be?

Somehow I managed to misread oxidation as reduction. Sigh. Holiday mode does stuff to your mind.

I gave you the reduction equation above, where iron ions are reduced to iron metal.
The oxidation equation would be simply Al(s) => Al3+(s) + 3e-

There we go. I finally demonstrate that I know what oxidation and reduction are -.-
Title: Re: VCE Chemistry Question Thread
Post by: ealam2 on February 04, 2014, 03:36:41 pm
Hi! Could someone please explain HPLC (High Performance Liquid Chromatography) in detail? Having an excursion tomorrow and will be doing a prac on this concept. Thanks!  :)
Title: Re: VCE Chemistry Question Thread
Post by: Edward21 on February 04, 2014, 07:18:56 pm
Hi! Could someone please explain HPLC (High Performance Liquid Chromatography) in detail? Having an excursion tomorrow and will be doing a prac on this concept. Thanks!  :)
This is a form of column chromatography, that involves separation of compounds (as in all types of chromatography) but generally we're dealing with compounds whose molar masses are > 300gmol that cannot be vaporised, if they could be then we'd use Gas Chromatography. Our stationary phase is heaps of these tiny little balls in the column (silica), we have small sized balls to increase the surface area, to ensure maximal resolution (ie. better separation) of our sample. The mobile phase is a non-polar solvent (often hexane) which is passed through the column at High Pressure *hint hint* because of all these small little balls with so much surface area, we need high pressure to be applied to our mobile phase to move the compounds through :) the rest is essentially the same as basic chromatography principles, the higher the temp. of the column the faster the compounds will be eluted, the more soluble a component is in the mobile phase, the quicker it will move through the column just like in Gas-Column Chromatography. Chemguide.co.uk would have a more detailed explanation :)
Title: Re: VCE Chemistry Question Thread
Post by: scribble on February 04, 2014, 09:36:13 pm
o.O i always thought that hplc used a solvent as the mobile phase, not a gas.
Title: Re: VCE Chemistry Question Thread
Post by: ealam2 on February 04, 2014, 09:45:40 pm
o.O i always thought that hplc used a solvent as the mobile phase, not a gas.

Yeah I read that in chemguide, unless gas is a solvent? Is it? Sorry I feel so dumb right now.  :-[
Title: Re: VCE Chemistry Question Thread
Post by: scribble on February 04, 2014, 10:03:12 pm
:S ive been under the impression that with hplc, the stationary phase was the packing inside of the column (eg silica) and then the solvent was the mobile phase. Then there's a pump attached to the front of the column that constantly pumps more of the solvent/mobile phase through the column, sweeping the sample along and with it and separating out the components.
in normal phase chromatography, the stationary phase is more polar than the mobile phase and so components that are highly polar have a higher affinity to the stationary phase than a component that is less polar. hence, polar components spend more time adsorbed onto the stationary phase and so would stay in the column for longer/have a higher retention time than non polar components. in reverse phase chromatography, the stationary phase is less polar than the mobile phase so the opposite occurs; polar components are eluted from the column faster than non polar components and have a lower retention time.

EFPBH, you pretty much need to work out the mole for any chemistry question. to find the empirical formula, you're trying to find the atom ratio of each element in the hydrocarbon. moles are just a convenient way of counting atoms.
Title: Re: VCE Chemistry Question Thread
Post by: ealam2 on February 04, 2014, 10:09:12 pm
Thank you scribble and Edward21!  :) Hope I understand it more after my excursion and that it all sinks into my brain.   :)
Title: Re: VCE Chemistry Question Thread
Post by: EFPBH on February 04, 2014, 10:22:19 pm
Permethrin, C21H20Cl2O3, is one of the pyrethroid insecticides that are replacing organochlorine and organophosphate compunds. if a sample of permethrin contains 0.480g of carbon then the mass of chlorine present in the sample is ?

could you please explain each step, much appreciated :)
Title: Re: VCE Chemistry Question Thread
Post by: scribble on February 04, 2014, 10:28:22 pm
work out how much carbon you have in mol. then, as you know that the carbon, chlorine ratio is 21:2, you can work out how much chlorine you have in mol. then you can convert that to mass.
n(C)=0.480/12.0 = 0.0400mol
n(Cl)=2/21 * n(C) = 2/21*0.0400=0.0038095mol
m(Cl)=nM=0.0038095*35.5=0.135g to 3sigfigs
Title: Re: VCE Chemistry Question Thread
Post by: Edward21 on February 04, 2014, 10:55:13 pm
Yeah I read that in chemguide, unless gas is a solvent? Is it? Sorry I feel so dumb right now.  :-[
I fixed it up now, my mistake!
Title: Re: VCE Chemistry Question Thread
Post by: ealam2 on February 05, 2014, 03:58:06 pm
Thank you very much, Edward21! :)
Title: Re: VCE Chemistry Question Thread
Post by: eagles on February 05, 2014, 06:30:04 pm
There is a mixture of aluminium powder and iron (III) oxide.
When it is ignited, a redox reaction occurs to form molten iron.

How would we write the overall equation for this process?

Cheers!
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on February 05, 2014, 10:07:00 pm
You'll find the answer is on this/last page. Including a very stupid mistake by me in misreading oxidation as reduction.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 06, 2014, 05:28:01 am
Well...given the question...I'd say it would be Fe2O3(s) => 2Fe(l) + 3O2- (s) + 6e-

Edit: I've read the question incorrectly; this is the reduction half equation.

I asked my teacher about this and he said that the 3O2 isn't supposed to be there and now I'm pretty confused. I was wondering if someone could show me step by step to find this half equation.

Btw, if this is a reduction half equation, why aren't the electrons on the left side?

Thanks :-)
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on February 06, 2014, 07:14:01 am
I asked my teacher about this and he said that the 3O2 isn't supposed to be there and now I'm pretty confused. I was wondering if someone could show me step by step to find this half equation.

Btw, if this is a reduction half equation, why aren't the electrons on the left side?

Thanks :-)
Here is my answer, step by step, ( I'm too lazy to include states in each step, but I will at the end of my answer)
Step 1: you balance the KEY elements
Fe2O3 -> 2Fe
Step 2: balancing the O by adding water ( as you can see, there are 3O on the LHS, hence, you must add 3H2O to the RHS)
Fe2O3 -> 2Fe + 3H2O
Step 3: balance the H+ , on the  RHS, you see that there are 6H+, hence, to balance this, add 6H+ on the LHS
Fe2O3 + 6H+ -> 2Fe + 3H2O
Step 4: balancing all the charges by adding electrons. As you can see, on the LHS:
Fe2O3 + 6H+ -> 2Fe + 3H2O
(   0     +   6   ->  0    +  0     )      ( This step is actually finding the Oxidation number of each compound, add them together and find out where you can put the electrons to balance both side of the equation)
Hence, I see that there are 6+ on the LHS and 0 on the RHS, I would add 6e- to the LHS to balance the charges since 6 + (-6) = 0 ( = RHS)

Fe2O3 + 6H+ 6e- -> 2Fe + 3H2O
Step 5: include stages
Fe2O3(s) + 6H+(aq) + 6e- -> 2Fe(l) + 3H2O(l)
Hope this helps, let me know if there is any errors!
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on February 06, 2014, 10:38:27 am
Here is my answer, step by step, ( I'm too lazy to include states in each step, but I will at the end of my answer)
Step 1: you balance the KEY elements
Fe2O3 -> 2Fe
Step 2: balancing the O by adding water ( as you can see, there are 3O on the LHS, hence, you must add 3H2O to the RHS)
Fe2O3 -> 2Fe + 3H2O
Step 3: balance the H+ , on the  RHS, you see that there are 6H+, hence, to balance this, add 6H+ on the LHS
Fe2O3 + 6H+ -> 2Fe + 3H2O
Step 4: balancing all the charges by adding electrons. As you can see, on the LHS:
Fe2O3 + 6H+ -> 2Fe + 3H2O
(   0     +   6   ->  0    +  0     )      ( This step is actually finding the Oxidation number of each compound, add them together and find out where you can put the electrons to balance both side of the equation)
Hence, I see that there are 6+ on the LHS and 0 on the RHS, I would add 6e- to the LHS to balance the charges since 6 + (-6) = 0 ( = RHS)

Fe2O3 + 6H+ 6e- -> 2Fe + 3H2O
Step 5: include stages
Fe2O3(s) + 6H+(aq) + 6e- -> 2Fe(l) + 3H2O(l)
Hope this helps, let me know if there is any errors!

Firstly, where does the H+ come from? In the given question you just have aluminium and iron oxide. That's why my answer has an oxide in it.

I asked my teacher about this and he said that the 3O2 isn't supposed to be there and now I'm pretty confused. I was wondering if someone could show me step by step to find this half equation.

Btw, if this is a reduction half equation, why aren't the electrons on the left side?

Thanks :-)

What does your teacher say? The products of the reaction should be aluminium oxide and molten iron metal.
And yeah it's a reduction except I'm stupid and I put the electrons on the wrong side of the equation. ALWAYS balance your charges. Not like what I'm doing here. I'm getting so careless now.
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on February 06, 2014, 10:55:29 am
Hmm I get it now, just find out the question is 2 pages away. Thanks anw!

Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 06, 2014, 09:13:38 pm

What does your teacher say?

He said the answer should be what nhmn0301 stated  ???
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on February 06, 2014, 09:25:57 pm
He said the answer should be what nhmn0301 stated  ???

With waters? Ask him where the hydrogen atoms come from. The reaction involves aluminium metal with iron oxide, all in solid form.
Title: Re: VCE Chemistry Question Thread
Post by: darklight on February 06, 2014, 10:47:57 pm
Chloroform's molecules are made up of C, H and Cl. In an experiment, all of the chlorine present in a 3.59 gram sample of chloroform was converted into 12.92 gram of solid silver chloride, while the carbon content  of chloroform sample was converted into methanoic acid of mass 1.38g. 6.0 *10^23 chloroform molecules have the same mass as 6.0 *10^24 carbon atoms. Find the molecular formula of chloroform.

I know using the information that molar mass is 120 and the moles of C and Cl. But how do you find mole of H? Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on February 06, 2014, 11:41:01 pm
You can work out the mass of chlorine and carbon in the sample of chloroform given. Subtract this from the mass of chloroform given, and you should have the mass of hydrogen in chloroform.
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on February 07, 2014, 11:06:15 am
He said the answer should be what nhmn0301 stated  ???
My teacher said what lzxnl stated. It actually makes sense to me now :D!
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 07, 2014, 04:28:58 pm
My teacher said what lzxnl stated. It actually makes sense to me now :D!

Do you mind explaining it to me? I still don't understand it lol
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on February 07, 2014, 05:48:57 pm
This is the main confusing part here:
m(C+H+Cl) = 3.59 (chloroform contains carbon hydrogen and chlorine)
m(H) = 3.95 - m(C + Cl)     (hydrogen is chloroform minus carbon and chlorine)

------ From the start:

The idea is to find the mole of chlorine from the silver chloride (AgCl) and carbon from methanol acid (CHOOH).

Convert this mole of chlorine and carbon to mass and work out the hydrogen via the following:

m(C+H+Cl) = 3.59 (chloroform contains carbon hydrogen and chlorine)
m(H) = 3.95 - m(C + Cl)     (hydrogen is chloroform minus carbon and chlorine)

Convert the m(H) into mole and then do you normal empirical formula of n(C:H:Cl)
Then find the molecular formula.

Hope this helps out.

Do you mind explaining it to me? I still don't understand it lol
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on February 07, 2014, 05:54:47 pm
Do you mind explaining it to me? I still don't understand it lol
Sure, but since Im still trying to learn the concept, so please excuse me if I make any mistakes or get you confused:
I think the first thing that makes both of us make mistakes in this question is because we assume that, ALL redox reaction equations can be derived by adding water, H+ and stuff to both side without knowing when and where can we apply this concept.
Firstly, it is good to have a general idea of what is happening in a thermite process, the Fe2O3 and Al reacts together and generates extreme heat, this process is a redox reaction.
Before looking into the question, I guess you can write an overall equation for this:
Fe2O3 + Al -> Al203 + Fe  (sorry I'm really lazy to include states)
You can see that: Fe goes from +3 to 0 and Al goes from 0 to +3. Hence, Fe undergoes reduction and Al undergoes oxidation. Oxygen is actually just a spectator ions, do not even involve much in the redox process.
In Fe2O3, you know that you have 2Fe(3+) and 3O (-2), hence, the decomposition reaction comes as:
Fe2O3 -> 2Fe + 3O(-2)         If you question why at this stage, "2Fe" in the product has a charge of 0 but not +3, is because Fe has already gained 3 electrons from Aluminium, hence, it has successfully changed its charge to 0.
But we won't stop at that, since the equation is not balancing between the LHS and the RHS, since 3O (2-) carries a charge of -6, we need to add 6e- to the LHS as well. ( We are not adding H+ to balance because  in this reaction, there's no way water can present, we are talking about solid substances producing extreme heat). So our final answer should be:
Fe2O3 + 6e -> 2Fe + 3O(-2)
I know my explanation is not good enough and might still contain some errors, but it's my best, hope this helps!

Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on February 07, 2014, 06:14:20 pm
Yeah, the chemistry is all there
Title: Re: VCE Chemistry Question Thread
Post by: noah the lettuce on February 07, 2014, 08:33:07 pm
I have this chemistry question that I'm not too sure about the answer, if anyone could help me out?
A certain mineral sample was found to contain some aluminium oxide, Al2O3. The Al3+ ions in this mineral can be selectively extracted by precipitating it as Al(C9H6NO)3. When a 2.600g sample of the mineral was treated this way, the precipitate weighed 0.1747g. What was the percentage by mass of the Al in the original ore?
Mineral=2.600g
m (Al(C9H6NO)3)= 0.1747g
M=459g/mol
n= 0.00038061mol=n (Al)

Al2O3--> 2Al
n (Al2)=1/2*n (Al)
n (Al2)= 0.000190305mol
M (Al2)= 54g/mol
m (Al2)=0.01027647g

% Al=(0.01027647/2.600)*100
=0.3952%
Title: Re: VCE Chemistry Question Thread
Post by: scribble on February 07, 2014, 08:43:13 pm
what you've done is right, but the second part is a tad unecessry.
once you have n(Al)= 0.00038061mol
you can just go m(Al)=nM= 0.00038061*27.0 = 0.01027647g
and then workout the percentage instead of working out the mol of Al2 and then the molar mass of Al2 :)
Title: Re: VCE Chemistry Question Thread
Post by: noah the lettuce on February 07, 2014, 08:53:08 pm
what you've done is right, but the second part is a tad unecessry.
once you have n(Al)= 0.00038061mol
you can just go m(Al)=nM= 0.00038061*27.0 = 0.01027647g
and then workout the percentage instead of working out the mol of Al2 and then the molar mass of Al2 :)
Awesome, thank you  :D didn't realise I could skip that Al2 part, that cuts down the working out, but it's going to be a habit I'll have to cut out :'(
Title: Re: VCE Chemistry Question Thread
Post by: PsychoT on February 08, 2014, 10:50:35 am
Hi, I did the Volumetric determination of Nitrogen content in Lawn fertilser back titration prac yesterday, someone care to help me out with a few of these questions?

Average titre of HCl = 5.5mL (concordant titre)
c(HCl) = 0.1406
n(HCl) = 0.0007733 mol
v(NaOH) = 0.02L
c(NaOH) = 0.097M
n(NaOH) = 0.00194 mol

Equations:
NH4+ + OH- ---> NH3 + H20 (Ammonium compounding in the fertiliser)
HCl + NaOH ---> NaCl + H20

I'm now stuck when I get to here.

2d. Find the amount, in mol of Sodium Hydroxide that did not react with the fertiliser solution in each conical flask (is this just 0.00194 - 0.0007733)?
3. Then I need to calculate the amount of mol of Sodium Hydroxide that did react with the fertiliser solution (seems pretty easy once I know if my 2d is correct)
4. Find the amount of NH4+ ions present in each 20mL aliqout
5. Calculate the amount, in mol of NH4+ ions that were originally present in the 250mL volumetric flask
6. I should be about to find out the mass of nitrogen and the percentage by mass once I know this.

Will supply more info if needed, cheers guys!
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on February 08, 2014, 01:07:34 pm
Hi, I did the Volumetric determination of Nitrogen content in Lawn fertilser back titration prac yesterday, someone care to help me out with a few of these questions?

Average titre of HCl = 5.5mL (concordant titre)
c(HCl) = 0.1406
n(HCl) = 0.0007733 mol
v(NaOH) = 0.02L
c(NaOH) = 0.097M
n(NaOH) = 0.00194 mol

Equations:
NH4+ + OH- ---> NH3 + H20 (Ammonium compounding in the fertiliser)
HCl + NaOH ---> NaCl + H20

I'm now stuck when I get to here.

2d. Find the amount, in mol of Sodium Hydroxide that did not react with the fertiliser solution in each conical flask (is this just 0.00194 - 0.0007733)?
3. Then I need to calculate the amount of mol of Sodium Hydroxide that did react with the fertiliser solution (seems pretty easy once I know if my 2d is correct)
4. Find the amount of NH4+ ions present in each 20mL aliqout
5. Calculate the amount, in mol of NH4+ ions that were originally present in the 250mL volumetric flask
6. I should be about to find out the mass of nitrogen and the percentage by mass once I know this.

Will supply more info if needed, cheers guys!

The amount of sodium hydroxide that did not react with the ammonium ions is simply the amount that reacted with the HCl, which is 0.0007733 mol.
The amount that DID react with the fertiliser is 0.00194 - 0.0007733 mol.
Then you can work out n(NH4+) etc.
Title: Re: VCE Chemistry Question Thread
Post by: PsychoT on February 08, 2014, 01:24:06 pm
The amount of sodium hydroxide that did not react with the ammonium ions is simply the amount that reacted with the HCl, which is 0.0007733 mol.
The amount that DID react with the fertiliser is 0.00194 - 0.0007733 mol.
Then you can work out n(NH4+) etc.

That's perfect mate. Exactly what I was stuck on, cheers.
Title: Re: VCE Chemistry Question Thread
Post by: Sanguinne on February 08, 2014, 03:04:40 pm
Having a bit of trouble with the theory behind atomic absorption spectroscopy.

What does the monochromator actually do to the light after some of it has been absorbed by the metal atoms?
Title: Re: VCE Chemistry Question Thread
Post by: Ancora_Imparo on February 08, 2014, 05:06:18 pm
The monochromator essentially filters out all the unwanted wavelengths of light, such as those given off by the flame, leaving only a specific wavelength or a very narrow band of wavelengths to be passed to the recorder.
Title: Re: VCE Chemistry Question Thread
Post by: Sanguinne on February 08, 2014, 05:21:00 pm
The monochromator essentially filters out all the unwanted wavelengths of light, such as those given off by the flame, leaving only a specific wavelength or a very narrow band of wavelengths to be passed to the recorder.

I thought that the reason the cathode lamp was used was to produce wavelengths of light corresponding to the metal being analysed. If this is true, what is the difference between the lamp and the monochromator?
Title: Re: VCE Chemistry Question Thread
Post by: Ancora_Imparo on February 08, 2014, 05:45:09 pm
I thought that the reason the cathode lamp was used was to produce wavelengths of light corresponding to the metal being analysed. If this is true, what is the difference between the lamp and the monochromator?

That is indeed true. However, after the light has been passed through the flame where the metal atoms are being atomised, there will be interference from the flame itself. Thus the monochromator is used to ensure that these unwanted wavelengths are not passed through to the recorder.
Title: Re: VCE Chemistry Question Thread
Post by: Sanguinne on February 08, 2014, 06:33:32 pm
That is indeed true. However, after the light has been passed through the flame where the metal atoms are being atomised, there will be interference from the flame itself. Thus the monochromator is used to ensure that these unwanted wavelengths are not passed through to the recorder.

oh ok, thanks  :)
Title: Re: VCE Chemistry Question Thread
Post by: darklight on February 08, 2014, 09:56:17 pm
Hey guys,

In the Heinemann textbook, when showing how a chloroalkane can become an alkanol, it demonstrates that H+ (aq) and Cl-(aq) are separate in the solution. Why doesn't they join together to from HCl (aq). P. 149 for anyone who has the book :)
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 09, 2014, 10:16:08 am
For gas chromatography, is the stationary phase already inside the column? ..
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on February 09, 2014, 10:18:29 am
What is a dessicator and its function exactly?
Title: Re: VCE Chemistry Question Thread
Post by: Ancora_Imparo on February 09, 2014, 10:51:38 am
For gas chromatography, is the stationary phase already inside the column? ..

Yes.

What is a dessicator and its function exactly?

A desiccator is a container that is used to keep substances dry. It contains compounds known as desiccants which absorb water, thus keeping the substance in concern dry.
Title: Re: VCE Chemistry Question Thread
Post by: DJA on February 09, 2014, 11:16:46 am
What is a dessicator and its function exactly?

Just to add to that definition: A desiccator is used to dehydrate – to remove water from substances – for example in gravimetric analysis where the water needs to be removed from the precipitate. Some common dehydrating agents that are placed in the bottom of the desiccator to remove the water are fused calcium chloride, concentrated H2SO4 or anhydrous copper sulphate.
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on February 09, 2014, 01:03:57 pm
thanks guys.

tsfx states that typical uncertainties in volumetric analysis are:
* 20ml pipette: +/- 0.05ml
* burette: +/- 0.02ml

I thought the pipette was more accurate than the burette? o.o
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on February 09, 2014, 01:17:03 pm
So ionic compounds, such as HCl, dissociate into their separate ions in aqueous solution. This is why solutions with ionic compounds are able to conduct electricity - there are always ions moving about, free to conduct electrical charge.

Thus, HCl actually exists as H+ and Cl- in solution and NaOH actually exists as Na+ and OH-, but for convenience, when writing chemical equations, we group them together.

Slight correction. HCl is a polar covalent compound. In its pure gaseous form, it is a covalent molecule. When dissolved in water, HCl reacts with water in an acid base reaction to form H3O+ (or H+(aq)) and Cl-. HCl is a strong acid as chloride ion is quite stable (full octet of electrons) so you only get H+ and Cl- in an aqueous solution of HCl.
If dissolved in liquid HF, however, HCl is protonated by the stronger acid (liquid HF is a strong acid unlike aqueous HF) and acts as a base. The ionization of HCl is a result of its reaction with water, not because it is ionic.

NaOH is ionic though so when that dissolves it really does split apart into ions.

Title: Re: VCE Chemistry Question Thread
Post by: Ancora_Imparo on February 09, 2014, 02:32:34 pm
Slight correction. HCl is a polar covalent compound. In its pure gaseous form, it is a covalent molecule. When dissolved in water, HCl reacts with water in an acid base reaction to form H3O+ (or H+(aq)) and Cl-. HCl is a strong acid as chloride ion is quite stable (full octet of electrons) so you only get H+ and Cl- in an aqueous solution of HCl.
If dissolved in liquid HF, however, HCl is protonated by the stronger acid (liquid HF is a strong acid unlike aqueous HF) and acts as a base. The ionization of HCl is a result of its reaction with water, not because it is ionic.

NaOH is ionic though so when that dissolves it really does split apart into ions.

Whoops! Don't know what I was thinking when I said that. Message to take home: HCl is covalent, not ionic!
Title: Re: VCE Chemistry Question Thread
Post by: DJA on February 09, 2014, 04:17:02 pm
thanks guys.

tsfx states that typical uncertainties in volumetric analysis are:
* 20ml pipette: +/- 0.05ml
* burette: +/- 0.02ml

I thought the pipette was more accurate than the burette? o.o

from what I know that seems correct psy. I have always been taught that 20 ml pipette is +/- 0.05ml and burette is +/- 0.02ml.
Seems legit- please correct me anyone if its wrong

The only thin I know is that the 250mL volumetric flask is the one which is +/- 0.3ml. So far more inaccurate

Title: Re: VCE Chemistry Question Thread
Post by: Sanguinne on February 09, 2014, 08:34:12 pm
q60 in checkpoints in chemical analysis

The amount of copper in a solution of copper (II) sulfate  can be determined using atomic absorption spectroscopy. When a blue copper(II) sulfate solution is introduced into an AAS, a green flame is observed.
Consider the following statements.

I) A copper (II) sulfate solution appears blue because it absorbs red light.
II) The metal species undergoes oxidation in the flame.
III) The flame is green due to electron transitions from a higher energy state to lower energy state.
Which of the above statements are true?
I understand why II is not true and why III is true, but I dont understand why I is true. When would it absorb red light and why would it appear blue?
Thanks
Title: Re: VCE Chemistry Question Thread
Post by: scribble on February 10, 2014, 01:14:41 am
it absorbs the light that is complementary to the colour that it appears. that is the colour that is opposite to it on a colour wheel.
the idea is, if we have "white" light, which is a mixture of all colours, and the copper absorbs the red light, then what we see is a mixture of all the colours of light except for red. this appears blue. (or more correctly, cyan)

and heres a picture of a colour wheel:
http://www.d.umn.edu/~mharvey/colorwheel.jpg
Title: Re: VCE Chemistry Question Thread
Post by: Sanguinne on February 10, 2014, 11:27:39 am
it absorbs the light that is complementary to the colour that it appears. that is the colour that is opposite to it on a colour wheel.
the idea is, if we have "white" light, which is a mixture of all colours, and the copper absorbs the red light, then what we see is a mixture of all the colours of light except for red. this appears blue. (or more correctly, cyan)

and heres a picture of a colour wheel:
http://www.d.umn.edu/~mharvey/colorwheel.jpg

that makes sense, thanks.
do i have to know the colour wheel?
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 10, 2014, 06:12:08 pm
For short answer Q. 2 part b - http://www.vcaa.vic.edu.au/Documents/exams/chemistry/2012/2012chem1-w.pdf (Sorry I couldn't printscreen the question - it wouldn't fit!)
How do we find the answer and differentiate between the bottom two spots?

I knew it was one of them but didn't know which :P!
Title: Re: VCE Chemistry Question Thread
Post by: Ancora_Imparo on February 10, 2014, 06:56:52 pm
So using the Rf values from solvent G, we know that alanine is one of the middle two dots (going from left to right), as its Rf value is also in the middle.

Using the Rf values from solvent F, we know that alanine had to have travelled the furthest downwards as it has the lowest Rf value. So, out of the two middle dots we were considering before, alanine would correspond to the lower of those two.
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on February 10, 2014, 09:46:01 pm
"The amount of additional potassium permangenate soln required to reach end point is minute. This excess will notionally increase the calculated amount of iron present but, in practice, the extra has no effect on the calculated amount. Why not?"

I actually have no idea. Only thing that comes to mind is the uncertainties of the burette.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 11, 2014, 06:18:28 am
So using the Rf values from solvent G, we know that alanine is one of the middle two dots (going from left to right),

How do we know this? Sorry I don't really understand the question. What is happening as it is turned over and why are the dots represented differently? .... It's like on the first one there is one point of origin and on the second one there are 3 .. If that makes sense.. :|
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on February 11, 2014, 10:46:39 am
This is two stage TLC.  First a chromatogram was ran in direction of solvent G, there two substances with the same Rf value, 0.51, one happens to be alanine the other threonine.  We don't know which is which so we need to seperate these two by using another solvent (solvent F).  The below diagram shows the end result.

You can see the first origin for solvent G is the red line, the origin for solvent F is the green line.  Here we can measure the Rf values for each solvent and thus identify the compounds.

Hope this makes a bit of sense.  In terms of why you get three dots and then four.  It is due to the Alanine and Threonine having a very similar affinity for solvent G (meaning they have the same Rf), where as these two must have a different affinity for solvent F (meaning they have the different Rf).

How do we know this? Sorry I don't really understand the question. What is happening as it is turned over and why are the dots represented differently? .... It's like on the first one there is one point of origin and on the second one there are 3 .. If that makes sense.. :|
Title: Re: VCE Chemistry Question Thread
Post by: eagles on February 11, 2014, 10:32:28 pm
Explain why both factors of low temperature and high pressure will cause a real gas to deviate from ideal gas behaviour.

That low temperature should decrease the random movement of gas particles, as there is less kinetic energy generated from random collisions, and thereby reduce overall pressure.
When low temperature of a real gas induces high pressure, the real gas deviates from ideal gas behaviour.

Can you help me improve this?  :)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on February 11, 2014, 11:25:32 pm

Explain why low temperature and high pressure will cause a real gas to deviate from ideal gas behaviour.

A detailed explanation is appreciated. Cheers!

At low temperatures, gas molecules have less kinetic energy on average, so intermolecular forces have a stronger effect overall on molecular motions. As the intermolecular forces can no longer be neglected, the gases deviate from ideal gas behaviour.

At high pressures (i.e small space, or really high temperatures), gas molecules interact with each other a lot more as the space is relatively small for the amount of energy they have (end up getting closer to each other). Again, intermolecular forces have a greater effect overall.

Or, you can think about it this way. Reducing the temperature and pressurising a gas are the only ways of changing the state of a gas (unless you want to consider supercritical fluids and plasmas as different states of matter...but let's not go there). This implies that in reality, intermolecular forces must be able to hold a gas together if we keep reducing the temperature or pressurise the gas.

Explain why both factors of low temperature and high pressure will cause a real gas to deviate from ideal gas behaviour.

That low temperature should decrease the random movement of gas particles, as there is less kinetic energy generated from random collisions, and thereby reduce overall pressure.
When low temperature of a real gas induces high pressure, the real gas deviates from ideal gas behaviour.

Can you help me improve this?  :)

(just because this is the updated version of the quote)

Low temperatures do not induce high pressure. Just saying. You've said that yourself, so you've contradicted yourself.
There is a difference between the two cases. Also, low temperatures do not "generate" less kinetic energy; the particles just have less kinetic energy overall. Your answer doesn't really tell me much.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on February 12, 2014, 05:31:57 pm
At low temperatures, gas molecules have less kinetic energy on average, so intermolecular forces have a stronger effect overall on molecular motions. As the intermolecular forces can no longer be neglected, the gases deviate from ideal gas behaviour.

At high pressures (i.e small space, or really high temperatures), gas molecules interact with each other a lot more as the space is relatively small for the amount of energy they have (end up getting closer to each other). Again, intermolecular forces have a greater effect overall.

Low temperatures do not induce high pressure. Just saying. You've said that yourself, so you've contradicted yourself.
There is a difference between the two cases. Also, low temperatures do not "generate" less kinetic energy; the particles just have less kinetic energy overall. Your answer doesn't really tell me much.

Thank you! However, I've bolded the bits which I don't understand clearly from the explanation. I don't see the intended emphasis of the intermolecular forces.
Do you mind elaborating? Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on February 12, 2014, 06:03:46 pm
The ideal gas model makes the assumption that intermolecular forces are negligible. Under the ideal gas model, gases cannot change state, for those require intermolecular forces. The fact that increasing pressures or decreasing temperatures allow gases to change state shows that the ideal gas model begins to break down there.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on February 12, 2014, 07:18:38 pm
Yep, that makes sense now. Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: Sanguinne on February 12, 2014, 08:59:09 pm
q12 of the extended response in chemical analysis in checkpoints

Copper metal does dot dissolve in HCL, but can be dissolved by reaction with concentrated nitric acid, HNO3. The products are, Cu2+, NO2, and water. Give balanced ionic equations for

b) the reduction process
How would i know if its HNO3 or NO3- that goes under reduction. I thought it was HNO3 but the answer is NO3-

thanks
Title: Re: VCE Chemistry Question Thread
Post by: scribble on February 12, 2014, 09:04:38 pm
HNO3 is a very strong acid and usually ionises completely into H+ and NO3-.
I wouldn't say it matters hugely if you used HNO3 in your equation over NO3- though.
Title: Re: VCE Chemistry Question Thread
Post by: PsychoT on February 13, 2014, 05:05:18 pm
I need an easy way to remember what Polar and Non-Polar molecules are. Can anyone explain it in a way that I might remember.
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on February 13, 2014, 08:00:22 pm
Generally non-polar molecules are symmetrical and polar molecules are asymmetrical.  In reality we it is not a black and white thing as in polar and non-polar, we deal with varying degrees polarity.  Here are few key things to remember:

If the molecule is only carbon it is highly non-polar.
If a molecule contains OH,NH2,NH,COOH then the degree of polarity depends on the length of the carbon chains and the amount of OH.  the longer the carbon chain the less polar it is, the more functional groups the more polar it is.

Hope this helps out a little.

I need an easy way to remember what Polar and Non-Polar molecules are. Can anyone explain it in a way that I might remember.
Title: Re: VCE Chemistry Question Thread
Post by: PsychoT on February 13, 2014, 08:31:29 pm
Generally non-polar molecules are symmetrical and polar molecules are asymmetrical.  In reality we it is not a black and white thing as in polar and non-polar, we deal with varying degrees polarity.  Here are few key things to remember:

If the molecule is only carbon it is highly non-polar.
If a molecule contains OH,NH2,NH,COOH then the degree of polarity depends on the length of the carbon chains and the amount of OH.  the longer the carbon chain the less polar it is, the more functional groups the more polar it is.

Hope this helps out a little.

Cheers Jason! Helps alot.

Spent time this afternoon going through you Chromatography YouTube videos, they were great! Feel like i'm ahead of the rest of my class now. Definitely going to keep on doing it to consolidate knowledge.
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on February 13, 2014, 08:34:46 pm
A 150mL aqueous NaCl solution contains 0.0045g NaCl.
Calculate the concentration in ppm

easy way to do this?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on February 13, 2014, 09:42:30 pm
A 150mL aqueous NaCl solution contains 0.0045g NaCl.
Calculate the concentration in ppm

easy way to do this?

Yep. Very easy way, actually.
Parts per million is a mass fraction. Originally, it means 1 g solute per million grams solvent, but given that 1 L water is pretty close to 1000 g, we can say 1 ppm is really 1g/1000L. Or, rephrased, 1 ug/mL
So, we have 0.0045g/150 mL = 4500 ug/150 mL = 30ug/mL = 30 ppm
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on February 14, 2014, 12:01:42 am
So we don't need to calculate the concentration?
While I was doing the question, it felt like it took longer than It needed. :/
Title: Re: VCE Chemistry Question Thread
Post by: Edward21 on February 14, 2014, 12:20:22 am
So we don't need to calculate the concentration?
While I was doing the question, it felt like it took longer than It needed. :/
Ppm is a particular unit ratio for concentration! For ease of use, we simply use either mg/L and ug/mL as they will both give you the ppm. But essentially it's telling you how many parts of what you're looking at, per 10^6 (or a million) parts of solvent, so this is used for small concentrations.

0.0045g of NaCl in 150mL? Let's convert!

Ppm=mg/L=(0.0045X10^3 mg/0.150L)=30mg/L or 30ppm.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 14, 2014, 05:20:58 pm
I don't understand this question at alllllllllllllllllllllll

1.5 mol of zinc metal was requiored to reduce 1 mol of vanadium compound. The redox half equation for the reduction of vanadium would be:

a) VO2+(aq) + 2H+(aq) + e -> V3+ + H2O (l)

b) VO2+ (aq) + 4H+ (aq) + 3e -> V2+ (aq) + 2H2O (l)

c) V3+ + e -> V2+(aq)

d) VO2+ (aq) + 2H+ (aq) + 2e -> V2+ (aq) + H2O (l)
Title: Re: VCE Chemistry Question Thread
Post by: thushan on February 14, 2014, 06:13:41 pm
Think about how many moles of electrons 1.5 mol of zinc gives off when it is oxidised to Zn2+. That number of electrons would be mopped up by 1 mol of vanadium.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 14, 2014, 06:41:26 pm
Think about how many moles of electrons 1.5 mol of zinc gives off when it is oxidised to Zn2+. That number of electrons would be mopped up by 1 mol of vanadium.

I'm so sorry but I still don't understand, what do you mean by 'how many moles of electrons'? I didn't know electrons have a mole number?

Is it because 2e are needed for the oxidation of Zn and since there is .5 more mole of zinc metal than of the vanadium compound that the amount of electrons are 1? ...? O.ooo
Title: Re: VCE Chemistry Question Thread
Post by: Ancora_Imparo on February 15, 2014, 12:09:46 pm
I'm so sorry but I still don't understand, what do you mean by 'how many moles of electrons'? I didn't know electrons have a mole number?

Is it because 2e are needed for the oxidation of Zn and since there is .5 more mole of zinc metal than of the vanadium compound that the amount of electrons are 1? ...? O.ooo

The word 'mole' just refers to Avogadro's number, which is approximately 6.02 * 1023. You can apply 'mole' to anything, not just atoms or molecules - the number of apples I have, the number of times I've failed at Chemistry etc. So if I have one mole of electrons, it just means that I have 6.02 * 1023 electrons.

So if we have 1.5 mol of zinc, and each zinc atom gives off 2 electrons when it is oxidised to Zn2+, we would have 1.5 * 2 = 3.0 mol of electrons.

Now we know that 3.0 mol of electrons are required to reduce 1 mol of the vanadium compound. That means the ratio between the electrons and the vanadium compound in the equation must be 3:1. Hence, the answer is equation b).
Title: Re: VCE Chemistry Question Thread
Post by: eagles on February 16, 2014, 06:39:45 pm
Why do larger alcohol molecules become progressively less soluble?

That the influence of the OH- group and its ability to make hydrogen bonds with water is diminished when the chain increases, in that the straight and unreactive C-C chain dominates. The increase in the alkanol homologous series leads to a decreasing solubility ability.

Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on February 16, 2014, 09:57:04 pm
Why do larger alcohol molecules become progressively less soluble?

That the influence of the OH- group and its ability to make hydrogen bonds with water is diminished when the chain increases, in that the straight and unreactive C-C chain dominates. The increase in the alkanol homologous series leads to a decreasing solubility ability.

The hydroxyl group is neutral. It is best to just refer to it as a -OH group. Larger alcohol molecules have a longer non-polar end, the carbon chain, and the longer the carbon chain the more dominant the non-polar, and thus insoluble, part of the molecule becomes. Eventually, it is no longer favourable for the water molecules to bond to the alkanol; they'd rather bond to themselves as they're much more polar than the alkanol.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on February 17, 2014, 05:58:50 am
Thanks!

I also read somewhere that the solubility of larger alcohols decreases rapidly with size because the longer hydrocarbon tails of the molecules get between the water molecules and break hydrogen bonds. These broken hydrogen bonds are only replaced by much weaker dispersion forces. Hence the less soluble a larger molecule.

Is this also a valid explanation?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on February 17, 2014, 11:18:44 am
Something like that, yes. The important bit is that water molecules can only bond a certain number of times. If they form bonds to non-polar molecules, the forces of attraction aren't that great and the broken hydrogen bonds aren't compensated for. The larger the alkanol is, the larger the non-polar end.

Example: something like methanoic acid is miscible in water (it will mix with water in all proportions), but huge acids like decanoic acid instead form dimers in water, where the hydrophilic acid functional groups bond to each other and the hydrophobic ends point in opposite directions. This way, the acid functional group is unable to bond to water and is thus insoluble.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on February 17, 2014, 12:35:41 pm
Righto. Thanks :)
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on February 17, 2014, 05:12:33 pm
What is atomic absorption spectroscopy?
Easy explanation pls ;)
Title: Re: VCE Chemistry Question Thread
Post by: thushan on February 17, 2014, 05:19:15 pm
What is atomic absorption spectroscopy?
Easy explanation pls ;)

A technique where you determine the concentration of a metal ion by looking at how much light of a wavelength it absorbs. The more light it absorbs, the more concentrated the metal ion in solution.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on February 17, 2014, 06:34:09 pm
Thanks thushan

That is an awesome explanation. It has helped me so much. ;D ;D ;D
Title: Re: VCE Chemistry Question Thread
Post by: eagles on February 17, 2014, 08:14:15 pm
Can C5H10 have isomers such as 3-methyl,but-1-ene, and 3-methyl,but-2-ene?

Also, I have a question about significant figures. Do we need to consider the values from the periodic table? If all other values of a question use 4 significant figures but the elements we refer on the periodic table show 3, will this affect the significant figures of the final answer?

Cheers! xD
Title: Re: VCE Chemistry Question Thread
Post by: Edward21 on February 17, 2014, 08:48:26 pm
Can C5H10 have isomers such as 3-methyl,but-1-ene, and 3-methyl,but-2-ene?

Also, I have a question about significant figures. Do we need to consider the values from the periodic table? If all other values of a question use 4 significant figures but the elements we refer on the periodic table show 3, will this affect the significant figures of the final answer?

Cheers! xD
Yup 3-methylbut-1-ene and 2-methylbut-2-ene (you wouldn't use 3-methyl as you can flip this molecule the other way and realise 2-methyl will work as the C=C bond is symmetrical).

I'm quite certain that your answer should be to 3, but some may disagree, I'd like to hear another opinion on this, I never really had this clarified for me..
Title: Re: VCE Chemistry Question Thread
Post by: RKTR on February 17, 2014, 09:20:44 pm

Also, I have a question about significant figures. Do we need to consider the values from the periodic table? If all other values of a question use 4 significant figures but the elements we refer on the periodic table show 3, will this affect the significant figures of the final answer?

Cheers! xD

Help With Sig Figs
THE FOOLPROOF GUIDE TO SIGNIFICANT FIGURES
Title: Re: VCE Chemistry Question Thread
Post by: Limista on February 17, 2014, 09:28:07 pm

Also, I have a question about significant figures. Do we need to consider the values from the periodic table? If all other values of a question use 4 significant figures but the elements we refer on the periodic table show 3, will this affect the significant figures of the final answer?

yeah you do need to consider the values from the periodic table.

when doing sig figs, you need to consider:
* what numerical values you are using from the worded question, and seeing how many sig figs they have. Let's assume in your case it is 4.
* very likely you are going to need molar masses from the periodic table. So now you're going to be dealing with numerical values that have 3 sig figs.
* if adding/subtracting a numerical value with 4 sig figs to a numerical value with 3 sig figs, the answer you get must be to 4 sig figs.
* if multiplying/dividing a numerical value with 4 sig figs to a numerical value with 3 sig figs, the answer you get must be to 3 sig figs.

EDIT: wanted to add some more... if the least number of sig figs in the worded question is 2, then your final answer must be to 2 sig figs. If the least number of sig figs in the worded question is 3, then your final answer must be to 3 sig figs. If the least number of sig figs in the worded question is 4, then your answer must still be to 3 sig figs PROVIDED that you used numerical values with 3 sig figs in your calculation (which would occur if you used molar masses from the periodic table).

hope that makes sense  :)
Title: Re: VCE Chemistry Question Thread
Post by: eagles on February 18, 2014, 06:09:01 pm
Thanks for all your help  ;D
Title: Re: VCE Chemistry Question Thread
Post by: eagles on February 20, 2014, 11:02:24 pm
Why does the equivalence point of a titration not necessarily going to be value 7?

Assuming that it is a neutralisation reaction, why, when the mole ratios of the 2 reactants are the same, may have a pH deviating from 7?

Thanks.
Title: Re: VCE Chemistry Question Thread
Post by: Edward21 on February 20, 2014, 11:26:35 pm
Why does the equivalence point of a titration not necessarily going to be value 7?

Assuming that it is a neutralisation reaction, why, when the mole ratios of the 2 reactants are the same, may have a pH deviating from 7?

Thanks.
Well.. (this could take a while, hang in there!) at the equivalence point.. we have water being produced and a salt (if you react and acid and base) no acid nor base is in excess at this point (this is the point at which the reactants are present in their stoichiometric ratios, and are producing the products)

1st example. Strong acid + strong base. Let's try NaOH and HCl. We get out H2O and NaCl (this is a neutral salt) hence why at 25 degrees C, the pH is approximately 7.

2nd example. Strong acid + weak base. Let's try H2SO4 and NH3. We get out H2O and a salt of ammonium sulfate (NH4)2SO4. The NH4+ ion is a weak acid it has the ability to donate H+ and it does just that!! it increases the [H+] concentration of the solution at the equivalence point.. this is why the pH is lower than 7 at the equivalence point not because we've got a strong acid, but rather we're producing a conjugate acid, which is making the pH lower.

3rd example. Weak acid + strong base. Let's try CH3COOH and NaOH. We get H2O and CH3COONa, the CH3COO- (ethanoate ion) is the conjugate base of the ethanoic acid. This base can accept H+ from water, increasing the [OH-] concentration thus raising the pH. This is why the pH is above 7 at the equivalence point not because we've got a strong base, but rather we're producing a conjugate base from our weak acid, thus the pH is above 7 (when at 25 degrees celsius).

4th example. Weak acid and weak base. Let's try HOCl (hypochlorous acid) and NH3. We get H2O and ammonium hypochlorite. NH4OCl. The NH4+ is a weak acid, it donates H+ to the solution, but since we've got HOCl and NH3 in a 1:1 mol ratio, we also produce equal amounts of the hypochlorite ion OCl- which is a weak acid and will accept H+ from water producing OH- in solution, or taking up H+ already there. The result is a slightly lower pH from the NH4+ and a slightly higher pH from the OCl- base.. this produces an equivalence point that is relatively neutral at 25 degrees (7) because you're producing a conjugate acid and base at the same time.

The message of the story? The salt produced determines the pH :)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on February 21, 2014, 08:49:03 am
I have to disagree slightly. The acid strength does actually matter. With a strong acid weak base titration, as the acid is so much stronger, it protonates the weak base leaving a weak acid at equilibrium. However, with a weak acid and weak base titration, you really have to check the acidity constants of the weak acid and the conjugate acid of the weak base. Let's take sodium hydrogen sulfate and ammonia. Ammonia's conjugate acid, ammonium ion has Ka of around 5.6 times 10^-10, while sodium hydrogen sulfate itself has a Ka of around 0.01. This means that after reaction  we're left with ammonium sulfate. As sulfate is the conjugate of a stronger acid than ammonium is, sulfate ion is a far weaker base than ammonia. Similarly, ammonium ions are weaker acids than the hydrogen sulfate ion. Therefore, at equilibrium we will have mostly ammonium sulfate. As ammonium ions are more acidic than sulfate ions are basic, the solution would be slightly acidic at equivalence.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on February 22, 2014, 01:50:29 pm
Which of the following best describes what occurs when a substance absorbs infrared radiation?
a) some of its bonds begin to vibrate
b) Vibrating bonds increase the intensity of their vibration

I have already eliminated c and d but I cant decide which one is right from a and b.
Pls provide a brief explanation. Thanks
Title: Re: VCE Chemistry Question Thread
Post by: Stick on February 22, 2014, 04:04:43 pm
The answer is B because bonds are always vibrating - however infrared radiation can alter their degree of vibration. Option A implies that bonds are static unless infrared radiation is applied.
Title: Re: VCE Chemistry Question Thread
Post by: Bronzebottom64 on February 22, 2014, 05:47:23 pm
Hi all, I was doing some preparations for a SAC on Monday about Redox titrations. I was wondering if anyone could just give me run through of any example involving a dilution. Thanks
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on February 22, 2014, 05:55:01 pm
Hi all, I was doing some preparations for a SAC on Monday about Redox titrations. I was wondering if anyone could just give me run through of any example involving a dilution. Thanks

The thing to remember is to find number of moles of X in conical flask containing 20mL aliquot. Then, to find n(X) in say the 250mL volumetric flask, you multiply n(X) by (250/20). Work from there :)
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on February 22, 2014, 06:47:41 pm
The answer is B because bonds are always vibrating - however infrared radiation can alter their degree of vibration. Option A implies that bonds are static unless infrared radiation is applied.

That makes sense. Thanks for that Stick :)
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on February 22, 2014, 07:34:21 pm
Can someone pls explain the uses of each spectroscopy technique briefly?
eg: uv spectroscopy is used to determine the concentration of a substance

THANKS!!!!!!!!!!!!
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on February 22, 2014, 07:41:51 pm
Hey guys

How much do we need to know about Biochemical fuels?

Thanks
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on February 23, 2014, 02:07:07 pm
pls exlain briefly the chemical principles of AAS. What is it and how does it work?
My teacher has explained it a number of times but I still dont get it :(
Title: Re: VCE Chemistry Question Thread
Post by: thushan on February 23, 2014, 02:35:58 pm
pls exlain briefly the chemical principles of AAS. What is it and how does it work?
My teacher has explained it a number of times but I still dont get it :(

Electrons in atoms are arranged in shells. To move an electron from one shell to a outer shell, you need to input a specific amount of energy (exactly X joules, anything just under or over it will not work), since it takes energy to pull an electron away from the positively charged nucleus.

Different elements have different distances between their shells. This is because every element has a different nuclear charge (eg. sodium has a nuclear charge of +11), which tends to pull all the electrons towards the nucleus. Also, different elements have a different electron configuration (eg. sodium has a 2, 8, 2 configuration). The more electrons on the outer shell, the further from the nucleus the shell tends to go, because the electrons repel one another.

If different elements have different distances between their shells, the energy required to move electrons from one shell to a higher shell will be different for different elements. Hence, different elements can absorb only specific energies.

In AAS, the energy is delivered as a light, which is basically a bunch of photons (light particles). Each photon has an energy, which is proportional to its frequency (E = hv, where E = energy, v = frequency and h is a constant). Different elements therefore absorb different specific frequencies of light - which is how one can differentiate between elements in AAS.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on February 23, 2014, 02:48:00 pm
Electrons in atoms are arranged in shells. To move an electron from one shell to a outer shell, you need to input a specific amount of energy (exactly X joules, anything just under or over it will not work), since it takes energy to pull an electron away from the positively charged nucleus.

Different elements have different distances between their shells. This is because every element has a different nuclear charge (eg. sodium has a nuclear charge of +11), which tends to pull all the electrons towards the nucleus. Also, different elements have a different electron configuration (eg. sodium has a 2, 8, 2 configuration). The more electrons on the outer shell, the further from the nucleus the shell tends to go, because the electrons repel one another.

If different elements have different distances between their shells, the energy required to move electrons from one shell to a higher shell will be different for different elements. Hence, different elements can absorb only specific energies.

In AAS, the energy is delivered as a light, which is basically a bunch of photons (light particles). Each photon has an energy, which is proportional to its frequency (E = hv, where E = energy, v = frequency and h is a constant). Different elements therefore absorb different specific frequencies of light - which is how one can differentiate between elements in AAS.

Thanks for that. I get it but I dont understand how the process occurs. What is the need for the flame and the other instruments such as a monochromator?
Title: Re: VCE Chemistry Question Thread
Post by: Limista on February 23, 2014, 03:33:42 pm
Thanks for that. I get it but I dont understand how the process occurs. What is the need for the flame and the other instruments such as a monochromator?

Purpose of these instruments = to carry out Atomic Absorption Spectroscopy.

The only things you need to know about these instruments in 3/4 chemistry, is a brief definition of how they function and the order in which they function.

(http://www.mercury-instruments.com/images_technology/AAS-scetch.gif)

By 'order', I also mean that you should be asking yourself the steps involved in AAS to determine the concentration of the sample, and then how you would obtain the concentration of an unknown sample (in this case, you would plot a graph with concentrations of different known samples on y axis, absorbance readings on x axis, then draw a line of best fit. Use the absorbance reading of the unknown sample and apply to graph to determine concentration of unknown sample).
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on February 24, 2014, 07:44:22 pm
Help pls:)

A group of students dissolved a 1.028g sample of a Group 2 metal ion chloride anhydrous solid in water so that the total volume was 250.0mL. They then titrated 20.00ml aliquots of this solution with a 0.1074 M aqueous silver nitrate solution to precipitate the silver chloride. The average titre was found to be 13.78mL. The reaction between silver and chloride ions can be represented by:
Ag++Cl---> AgCl(s). Both solutions are aqueous.
The group 2 metal ion present in the sample would be:
a) Mg2+
b) Ba2+
c) Sr2+
d) Ca2+
Title: Re: VCE Chemistry Question Thread
Post by: RKTR on February 24, 2014, 08:36:36 pm
Find no of mol of AgNO3
n=cv
=0.1074(0.01378)
=0.00148mol
n of Ag+=0.00148mol
n of Cl-in aliquot =0.00148mol
n of Cl- in250ml =250/20 (0.00148)
=0.0185mol
n of MCl2 = 1/2 x 0.0185mol
=0.00925mol
Mass/ no of mol = molar mass
Molar mass= 1.028/0.00925=111.1g/mol
111.1-2(35.5)=40.1= molar mass of the metal ion which is Ca2+

Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on February 24, 2014, 09:02:59 pm
What is co-precipitation exactly?
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 25, 2014, 07:04:07 am
In infrared spectrosopy, how do you distinquish the different bonds when the peaks in an infrared spectrum are really similar?

For example, I was asked to identify what bond is undergoing a vibrational change in ethanol at 2950cm-1cm
So
The O-H bond occurs at a wavenumber of 2500-3300
The C-H bond occurs at 2850 - 3300

I know both the bonds exist in ethanol and the wave numbers fit for both.. so why is it C-H and also how do they know that it's "streching"?
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on February 25, 2014, 07:45:39 am
This is just means that you don't get the one precipitate that you are looking for.  You might get other precipitates forming that will add to the mass on the filter paper.

Things like Carbonates, Hydroxides, Phosphates can all create precipitates, by the addition of an acid this lowers their likely hood of forming and thus messing up your analysis.

What is co-precipitation exactly?
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on February 25, 2014, 07:48:53 am
Once you get more into IR and do a lot of problems you will see that the O-H peak is much more broad.  The CH peaks are a lot more defined and sharp (normally jagged as there are way more than one CH in most molecules) these also normally appear bang on 3000.
--> walk through on IR: http://www.youtube.com/watch?v=uDBWtDASYn4

In infrared spectrosopy, how do you distinquish the different bonds when the peaks in an infrared spectrum are really similar?

For example, I was asked to identify what bond is undergoing a vibrational change in ethanol at 2950cm-1cm
So
The O-H bond occurs at a wavenumber of 2500-3300
The C-H bond occurs at 2850 - 3300

I know both the bonds exist in ethanol and the wave numbers fit for both.. so why is it C-H and also how do they know that it's "streching"?
Title: Re: VCE Chemistry Question Thread
Post by: Einstein on February 25, 2014, 06:17:58 pm
im stuck on what i do for 3c, i know 3d is just 3c x 6.02.

the mol for acetic acid molecules if 0.27. And the answer the says 0.540 moles for 3c, why do they do 0.27 x 2, i know theirs 2 oxygen molecules but still?

Thanks
Title: Re: VCE Chemistry Question Thread
Post by: Einstein on February 25, 2014, 06:30:17 pm
nvm i figured it out haha, stupid question
Title: Re: VCE Chemistry Question Thread
Post by: Lizzy7 on February 26, 2014, 04:28:22 pm

When doing a gravimetric analysis prac, students were asked to collect 1g of Fertiliser and grind it. After grinding it, it was dissolved with water. This solution was filtered and HCl was added to the filtrate. Then, it was heated and after it boiled, BaCl2 was added to form a precipitate BaSO4.

My question is, why was HCl added? Is it because it would cancel out the other ions in the fertiliser so that it wouldn't form precipitates with Barium? or is it because the fertiliser solution was alkaline (since fertilisers contain NH4 and are generally more basic) and so adding an acid would neutralise it?

It would be great if someone helped me out :)

Title: Re: VCE Chemistry Question Thread
Post by: RKTR on February 26, 2014, 04:39:05 pm
Adding HCl in gravimetric analysis

Title: Re: VCE Chemistry Question Thread
Post by: Lizzy7 on February 26, 2014, 07:11:40 pm
Adding HCl in gravimetric analysis

Thank you RKTR !!
Title: Re: VCE Chemistry Question Thread
Post by: JadedBlack on February 26, 2014, 07:14:12 pm
A copper ore contains 3.0% by mass of the metal. If the copper present as Cu2S, calculate the percentage of Cu2S in the ore

answer supposed to be 3.8%, but I've fiddled around a bit and haven't gotten it :/
Title: Re: VCE Chemistry Question Thread
Post by: RKTR on February 26, 2014, 08:22:28 pm
A copper ore contains 3.0% by mass of the metal. If the copper present as Cu2S, calculate the percentage of Cu2S in the ore

answer supposed to be 3.8%, but I've fiddled around a bit and haven't gotten it :/
mass of Cu=2(63.5)=127 g
mass of Cu/ mass of metal x 100=3.0
mass of Cu = 3.0/100 x mass of metal
127 x 100/3.0 =mass of metal =4233.33 g
% of Cu2S = [2(63.5)+32.0 ]/4233.33 x 100 =3.8%
Title: Re: VCE Chemistry Question Thread
Post by: JadedBlack on February 26, 2014, 08:30:33 pm
thank you, can't believe how simple that was. Shows how much I need some sleep lol
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on February 26, 2014, 09:07:49 pm
A copper ore contains 3.0% by mass of the metal. If the copper present as Cu2S, calculate the percentage of Cu2S in the ore

answer supposed to be 3.8%, but I've fiddled around a bit and haven't gotten it :/

Couldn't you also just say that copper metal itself is 2*63.6/(2*63.6+32) of the copper(I) sulfide present by mass, or 79.9% of the Cu2S, so the % by mass of the copper sulfide must be 3.%/0.799 = 3.75% => 3.8%?
Title: Re: VCE Chemistry Question Thread
Post by: Lizzy7 on February 27, 2014, 06:20:57 pm
Hello everyone it'd be great if someone were to help me with this question:

C14H18Cl6N2 is a drug available for the treatment of hypertension. A 1.356g sample of tablets containing this drug was treated to release all the Cl as Cl- ion. The other constituents were filtered off. An excess of silver nitrate solution was then added to the filtrate, forming an AgCl precipitate which was filtered, dried and weighed. Its mass was 1.125 g. Assuming that the drug was the only constituent of the table containing Cl, what was the percentage by mass of the drug in the tablets.

So here is what i am up to:

n(AgCl)= 1.125/143.4 =0.007845 mol since mole ratios are 1:1 I derived that n(Cl)=0.007845 mol

n(Cl) in tablet= 0.007845*6 = 0.04707 mol  [Was this what I was supposed to do? since there were 6 Cl in the drug]

How do I then proceed from here?

Thank You in advance !
Title: Re: VCE Chemistry Question Thread
Post by: RKTR on February 27, 2014, 07:03:41 pm
1 mol of the drug gives you 6 mol of Cl-
so u have no. of mol of Cl- ,you divide by 6 to get no.of mol of the drug
then multiply by molar mass to get mass of drug
mass of drug/ 1.356 x100= your answer
Title: Re: VCE Chemistry Question Thread
Post by: Lizzy7 on February 27, 2014, 07:22:40 pm
1 mol of the drug gives you 6 mol of Cl-
so u have no. of mol of Cl- ,you divide by 6 to get no.of mol of the drug
then multiply by molar mass to get mass of drug
mass of drug/ 1.356 x100= your answer

Thank you RKTR !  :)
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on February 27, 2014, 09:07:44 pm
Find no of mol of AgNO3
n=cv
=0.1074(0.01378)
=0.00148mol
n of Ag+=0.00148mol
n of Cl-in aliquot =0.00148mol
n of Cl- in250ml =250/20 (0.00148)
=0.0185mol
n of MCl2 = 1/2 x 0.0185mol
=0.00925mol

Mass/ no of mol = molar mass
Molar mass= 1.028/0.00925=111.1g/mol
111.1-2(35.5)=40.1= molar mass of the metal ion which is Ca2+

why is this multiplied by a half? I thought it would be 2 because of required/given?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on February 27, 2014, 09:22:58 pm
Don't rote learn formulas. Think about where they come from.
Pretend magnesium chloride is a box containing two chloride ions. Then, if you had ten chloride ions, you could put them in five boxes. Hence, the number of magnesium chloride units is half the number of chloride ions.
Title: Re: VCE Chemistry Question Thread
Post by: Lizzy7 on February 27, 2014, 09:50:57 pm
Don't rote learn formulas. Think about where they come from.
Pretend magnesium chloride is a box containing two chloride ions. Then, if you had ten chloride ions, you could put them in five boxes. Hence, the number of magnesium chloride units is half the number of chloride ions.

Mind explosion right now  ;D This was the same principle to my question before, but now your analogy helped me understand it better !

^_^ Thank you *cyber high five*

Title: Re: VCE Chemistry Question Thread
Post by: ETTH96 on February 28, 2014, 02:56:45 pm
Hey guys, can someone help me with this question?

All the iron present in a 1.048g sample of a breakfast food was extracted and the total volume of the solution formed was diluted to give a total volume of 100.0 mL. When analysed in the spectrophotometer, an absorbance of 0.23 was obtained. the numerical value for the ion content of this food sample, expressed in mg per 100g of food is...  (A DIAGRAM IS GIVEN, AND AT AN ABSORBANCE OF 0.23, THE CONCENTRATION OF FE(IRON) IS 1.8x10^5 M.

a) 12
b) 1.7
c) 9.6
d) 96
Title: Re: VCE Chemistry Question Thread
Post by: Robert123 on February 28, 2014, 06:21:09 pm
Hey guys, can someone help me with this question?

All the iron present in a 1.048g sample of a breakfast food was extracted and the total volume of the solution formed was diluted to give a total volume of 100.0 mL. When analysed in the spectrophotometer, an absorbance of 0.23 was obtained. the numerical value for the ion content of this food sample, expressed in mg per 100g of food is...  (A DIAGRAM IS GIVEN, AND AT AN ABSORBANCE OF 0.23, THE CONCENTRATION OF FE(IRON) IS 1.8x10^5 M.

a) 12
b) 1.7
c) 9.6
d) 96
I don't have a calculator  on me at the moment but the method to solve it will be
n(Fe)=cv
=1.8E-5*0.1=1.8*10^-6 mols
m(fe)=n*M
=1.8*10^-6*55.8 (not 100% sure of iron's molar mass off the top of my head)

From there we know the amount of iron in g within that sample. To calculate how much is in 100g, multiply it by 100/1.048 then convert it into mg ( multiply solution by 10^3)
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on March 01, 2014, 05:27:43 pm
How is the gas chromatography instrument calibrated?
Thanks :)
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on March 01, 2014, 08:30:04 pm
In Gas Chromatography you need to run a series of pure standard solutions of varying concentrations.

Pure solutions: you can see the Rt value of the species in question

Varying concentration: You can create a calibration curve and determine the concentration.

How is the gas chromatography instrument calibrated?
Thanks :)
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on March 01, 2014, 09:13:25 pm
if the average volume of three concordant titres is 22.23333 mL, how do I express it?
22.23mL or 22.20 mL?
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on March 01, 2014, 11:08:27 pm
a question asked me to work out molarity which is concentration. They then used that value as mol or n to find mass.

how is concentration/molarity the same as n or mol?
Title: Re: VCE Chemistry Question Thread
Post by: Limista on March 02, 2014, 04:12:44 am
a question asked me to work out molarity which is concentration. They then used that value as mol or n to find mass.

how is concentration/molarity the same as n or mol?

concentration and no. mole are two different things; they are not the same.

With this question, let's say the concentration value is 5M.

You know that C = n/V
So n = C x V
If C = 5 mol/L, then V = 1L  and n = 5 mol

Applying it to this second formula where   m = n x M
You would substitute n with 5, so m = 5 x M, where M = molar mass
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on March 02, 2014, 10:02:00 am
if the average volume of three concordant titres is 22.23333 mL, how do I express it?
22.23mL or 22.20 mL?

I don't think it matters too much since you will have to convert it to litres anyway.
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on March 02, 2014, 10:10:12 am
I don't think it matters too much since you will have to convert it to litres anyway.
It does matter, that's why I asked
Title: Re: VCE Chemistry Question Thread
Post by: saba.ay on March 02, 2014, 11:08:17 am
if the average volume of three concordant titres is 22.23333 mL, how do I express it?
22.23mL or 22.20 mL?

Well I'd recommend you use 22.23333 mL in your calculations but if you absolutely have to write it and the number of sig figs is 4, then you'd write 22.23mL . I don't know why you'd round to 22.20 mL ? :/
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on March 02, 2014, 12:05:44 pm
Well I'd recommend you use 22.23333 mL in your calculations but if you absolutely have to write it and the number of sig figs is 4, then you'd write 22.23mL . I don't know why you'd round to 22.20 mL ? :/
isn't it important to keep your titre to 2 decimal places, rather than 4 sig figs? I mean if it was like 9.00 mL, you wouldn't go to 4 sig figs?
I dunno, my teacher was saying something about rounding your three concordant titres to 0.1 mL - I actually have no idea.
I'll go with 22.23, thanks!
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on March 02, 2014, 12:09:54 pm
The only reason for this i could see is not to go from concentration in to grams, but to go from concentration in mol/L into concentration in g/L.

You are not changing concentration into "mol" into "grams", you are just converting the "mol" in "mol/L" into "grams" in "g/L".

Hope this makes sense.

a question asked me to work out molarity which is concentration. They then used that value as mol or n to find mass.

how is concentration/molarity the same as n or mol?
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on March 02, 2014, 01:14:11 pm
I don't get back titration worded questions. Can anyone tell me the steps required to do the question?

Title: Re: VCE Chemistry Question Thread
Post by: RKTR on March 02, 2014, 01:32:20 pm
I don't get back titration worded questions. Can anyone tell me the steps required to do the question?
2HNO3 (aq) +Ca(OH)2 (aq)-->Ca(NO3)2 (aq)+2H2O（l)
n of HNO3=cv=0.050(0.10)=0.0050mol
n of Ca(OH)2=cv=0.060(0.10)=0.0060 mol
from the equation we know that calcium hydroxide is in excess as only 0.0025 of it will react with nitric acid
calcium hydroxide left will be 0.0035 mol
H2SO4(aq)+Ca(OH)2(aq)-->CaSO4(aq)+2H2O(l)
n of sulfuric acid needed = n of calcium hydroxide left=0.0035 mol
v=n/c = 0.0035/0.050 = 0.070L
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on March 02, 2014, 05:40:14 pm
AN unkown organic compound was analysed to determine its indentity. The compound contained only carbon, hydrogen and oxygen and did not contain any double carbon-carbon bonds. When completely oxideised in air, 8.20g of this compound produced 8.19g of water, 20.2g of carbon dioxide.

Determine the empirical formula of this compound

(Thanks for the help! - you don't need to write out all the working out if it takes too long! I just need the answer.. thanks :))

I think i'm just unsure about the hydrogen.. I found the mole to be 0.455mol and therefore the mass 0.910g.. is this correct?
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on March 02, 2014, 05:47:45 pm
AN unkown organic compound was analysed to determine its indentity. The compound contained only carbon, hydrogen and oxygen and did not contain any double carbon-carbon bonds. When completely oxideised in air, 8.20g of this compound produced 8.19g of water, 20.2g of carbon dioxide.

Determine the empirical formula of this compound

(Thanks for the help! - you don't need to write out all the working out if it takes too long! I just need the answer.. thanks :))

I think i'm just unsure about the hydrogen.. I found the mole to be 0.455mol and therefore the mass 0.910g.. is this correct?
I got C4H2O3
H2O = 8.19g
n=m/M
n=0.455
n(H2O) x 2 = n(H)
n(H) = 0.91

hope this clears it up, if not i'll post my complete working out!
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on March 02, 2014, 05:58:06 pm
I'm skeptical solely based on the ridiculous number of double bonds that must be present in a molecule like that...

So, the number of moles of water gives the number of moles of hydrogen atoms in the organic compound. Similarly, the number of moles of carbon dioxide is the number of moles of carbon atoms in the original compound.

8.19 g of water has 8.19/18*2 moles of hydrogen atoms => 0.91 mol hydrogen, which is 0.91 g hydrogen (these are hydrogen atoms, not hydrogen molecules!). Why? 8.19/18 is the number of moles of water molecules and each water molecule has two hydrogens.

Number of moles of CO2 is then 20.2/44 mol = 0.459 moles of CO2 and hence carbon atoms. The mass of carbon atoms in the original compound is then 5.51 g. In total, we've accounted for 6.42 g of stuff in the organic compound. This leaves 8.20g - 6.42g = 1.78g of oxygen atoms, which is 1.78/16 = around 0.11 mol oxygen atoms.

Putting this altogether, we have 0.11 mol O, 0.91 mol H and 0.45 mol C. Ratio looks like 1 O to 4 C to 8 H to me i.e. C4H8O, which could be anything from a cyclobutanol (not very stable those things) to tetrahydrofuran (very stable solvent; five membered ring) or a ketone or even an aldehyde. None of these contain carbon-carbon double bonds.
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on March 02, 2014, 10:08:07 pm
in a redox full equation I have 10H+ and 5H2O on the reactants side and 16H+ and 8H20 on the products side.
The answer gives 6H+, no H20 on reactants and 6H+ and 3H20 on products. Does this mean we are allowed to just cancel the h20 and H+ by subtracting the smaller from the larger? please explain
Title: Re: VCE Chemistry Question Thread
Post by: mad_maxine on March 02, 2014, 10:14:30 pm
in a redox full equation I have 10H+ and 5H2O on the reactants side and 16H+ and 8H20 on the products side.
The answer gives 6H+, no H20 on reactants and 6H+ and 3H20 on products. Does this mean we are allowed to just cancel the h20 and H+ by subtracting the smaller from the larger? please explain

Yep, you just subtract the smaller number of H+s and H2Os from the larger amounts, and whatever the difference is is your number.
So 6H+ on the products side, and 3H2O on the products side

:)
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on March 02, 2014, 10:44:37 pm
Yep, you just subtract the smaller number of H+s and H2Os from the larger amounts, and whatever the difference is is your number.
So 6H+ on the products side, and 3H2O on the products side

:)

thanks and also this question

Fe2O3(s) + 3CO(g) -> 2Fe(l) +3CO2 (g)

during this reaction the oxidation number of iron changes from

a. +3 to 0 and CO is the reductant
b. +6 to 0 and CO is the reductant
c. +3 to 0 and CO is the oxidant
d. +6 to 0 and CO is the oxidant

cant seem to figure out how CO oxidises or reduces
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on March 02, 2014, 11:20:53 pm
Check the oxidation number of the carbon atom in CO and what happens to it when it becomes CO2
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on March 03, 2014, 09:46:58 pm
Anyone???
For the first question, use C1V1=C2V2
is the answer 240mL?
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on March 03, 2014, 09:50:25 pm
For gravimetric analysis, back titration and redox titration, what are some common errors from these experiments and how does rinsing cause errors in these experiments?
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on March 04, 2014, 06:21:18 am
Hey guys I know this may be simple but it's so important for calculations and I don't completely understand the concept

Can someone please explain the basics of converting ppm to the desired unit?

Like I understand that ppm = mg/L but when changing the units, what do you do? If I wanted this value in grams, could I just multiply the mg amount by 1000?

Can ppm be anything as long as it is one unit to a million? Eg. so 1ppm can be 1g per 106g yet can also be 1ml per 106ml?

How do I convert ppm to M? If the value is in g per L, can I just multiply it by the M?

Sorry if some of that didn't make sense.. this is just how I've understood it :| Thanks for the help :-)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on March 04, 2014, 08:43:05 am
Hey guys I know this may be simple but it's so important for calculations and I don't completely understand the concept

Can someone please explain the basics of converting ppm to the desired unit?

Like I understand that ppm = mg/L but when changing the units, what do you do? If I wanted this value in grams, could I just multiply the mg amount by 1000?

Can ppm be anything as long as it is one unit to a million? Eg. so 1ppm can be 1g per 106g yet can also be 1ml per 106ml?

How do I convert ppm to M? If the value is in g per L, can I just multiply it by the M?

Sorry if some of that didn't make sense.. this is just how I've understood it :| Thanks for the help :-)

Ppm just means one part in a million. As you've said, it can mean 1 gram in a million grams, which is its common meaning. If you have mg/L, you would actually divide by 1000. You always less grams than mg.

Mg/l is only ppm because we take the mass of 1 ml of water to be 1 g. Thus, 1g/million grams = 1g/1000 L = 1 mg/L

To go from mg/L to M, you would convert to g/L then divide by the molar mass in g/mol. Try dividing the units g/L by g/mol. Yes you can do this to check your units.
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on March 04, 2014, 10:24:52 am
Yo team I'm taking an intro Chem subject at uni but since not many people take it I think it's best to ask in the VCE forums.

Homogenous vs heterogenous mixtures, is the different literally defined by what you can see with your own eyes?  That seems like a pretty vague line to me but everything I read seems to define the difference as being whether or not you can 'see' the individual components of the mixture.
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on March 04, 2014, 10:00:00 pm
In a redox reaction, can the hydrogens in water cancel out hydrogen ions?
So I have Iron3 oxide that converts into metallic iron
the half equation would like something like this:
Fe2O3 + 6H+ + 6e- -------> 2Fe + 3H2O              (ceebs with states)
Could Hydrogens cancel out to:
Fe2O3 + 6e- -------> 2Fe + 3O2-

Has that always been the case? Is it necessary? Need some confirmation

thanks.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on March 04, 2014, 10:07:45 pm
In a redox reaction, can the hydrogens in water cancel out hydrogen ions?
So I have Iron3 oxide that converts into metallic iron
the half equation would like something like this:
Fe2O3 + 6H+ + 6e- -------> 2Fe + 3H2O              (ceebs with states)
Could Hydrogens cancel out to:
Fe2O3 + 6e- -------> 2Fe + 3O2-

Has that always been the case? Is it necessary? Need some confirmation

thanks.

Ok, so the easiest thing to do is actually break it down to the ionic equation.

The iron ion in iron (III) oxide is Fe3+(aq).

We know that this is converted to metallic iron which is Fe(s) .

Fe3+(aq) + 3e ----> Fe(s)

Answering your question, no, the hydrogen atoms in water don't cancel out H+ ions.

Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on March 04, 2014, 10:35:24 pm
Chapter 5, Q22 a) of the heinamann textbook.
It seems like that's what they've done.
Anyone else confirm that I can cancel the Hydrogens like this?

Cheers
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on March 04, 2014, 10:41:26 pm
Chapter 5, Q22 a) of the heinamann textbook.
It seems like that's what they've done.
Anyone else confirm that I can cancel the Hydrogens like this?

Cheers

Ok. Its just you said that Fe solid was produced. You never mentioned oxide ions.

So

Fe2O3(s) + 3H2O (l) + 3e ----> 2Fe(l) + 3O2-(aq) + 6H+(aq)

Now I've seen the textbook answer, but I'm stumped. Why would they cancel out hydrogens? I think I need an explanation for that too.
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on March 04, 2014, 10:47:57 pm
Ok. Its just you said that Fe solid was produced. You never mentioned oxide ions.

So

Fe2O3(s) + 3H2O (l) + 3e ----> 2Fe(l) + 3O2-(aq) + 6H+(aq)

Now I've seen the textbook answer, but I'm stumped. Why would they cancel out hydrogens? I think I need an explanation for that too.

I'm not sure what you're trying to do..LOL.
Why did you...wut... your charges don't even balance..?

Pretty sure my equation was right. I just need to know how (if there's a rule or whatnot) hydrogens are allowed to cancel

Also, I did state Iron3 oxide converts into metallic iron

Title: Re: VCE Chemistry Question Thread
Post by: Capristo on March 04, 2014, 10:48:05 pm
Hey, guys..Just wondering if anyone could help me with this. UVvis-Spectroscopy

A 2.5g sample of fertilizer was weighed out and the iron II content was dissolved by stirring it with 25ml of 2M sulphuric acid. The solution was filtered into a 250ml volumetric flask. 0.01M KMno4 was added to the solution untill the first appearance of pink to ensure all of the Fe2+ were oxidised to Fe3+ ions. Finally, deionised water was added to make the solution up to a volume of 250ml.

50ml of the fertilizer solution was then transferred to a flask and 25ml of KSCN and HCL was added. Deionisated water was then added to bring the solution up to 250ml.

The solution was analysed using a spectrophotometer.

After sketching a calibration graph using the absorbance for the known concentration Fe3+ solutions, the concentration of the sample was determined to be 1.8ppm.

Detirmine the percentage by mass of fe2+ in the fertilizer.

Any help would be really appreciated, thanks!
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on March 04, 2014, 10:58:12 pm
I'm not sure what you're trying to do..LOL.
Why did you...wut... your charges don't even balance..?

Pretty sure my equation was right. I just need to know how (if there's a rule or whatnot) hydrogens are allowed to cancel

Also, I did state Iron3 oxide converts into metallic iron
Remember that redox does not always need water to happen. Especially in a thermite reaction one, in an extreme heat condition, you can not get water from anywhere.
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on March 04, 2014, 11:05:41 pm
Remember that redox does not always need water to happen. Especially in a thermite reaction one, in an extreme heat condition, you can not get water from anywhere.
Oh so because it was a thermite process, water wasn't produced to begin with?
Simply the Iron3 oxide converted into it's Iron and oxgygen?
That makes more sense
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on March 04, 2014, 11:31:43 pm
Oh so because it was a thermite process, water wasn't produced to begin with?
Simply the Iron3 oxide converted into it's Iron and oxgygen?
That makes more sense

I thought thermite reaction was between aluminium and iron oxide. All that happens is the aluminium is oxidised to a trivalent cation and the iron cations are reduced.

Hey, guys..Just wondering if anyone could help me with this. UVvis-Spectroscopy

A 2.5g sample of fertilizer was weighed out and the iron II content was dissolved by stirring it with 25ml of 2M sulphuric acid. The solution was filtered into a 250ml volumetric flask. 0.01M KMno4 was added to the solution untill the first appearance of pink to ensure all of the Fe2+ were oxidised to Fe3+ ions. Finally, deionised water was added to make the solution up to a volume of 250ml.

50ml of the fertilizer solution was then transferred to a flask and 25ml of KSCN and HCL was added. Deionisated water was then added to bring the solution up to 250ml.

The solution was analysed using a spectrophotometer.

After sketching a calibration graph using the absorbance for the known concentration Fe3+ solutions, the concentration of the sample was determined to be 1.8ppm.

Detirmine the percentage by mass of fe2+ in the fertilizer.

Any help would be really appreciated, thanks!

1.8 ppm => 1.8 mg/l
In 250 ml this means we had 0.45 mg of Fe3+
These ions were initially in 50 ml of fertiliser solution that was taken from 250 ml in total of Fe2+ solution, so in the first 250 ml flask we had five times 0.45 mg = 2.25 mg of Fe2+. This is in 2.5 g so % by mass is 2.25 mg / 2.5 g = 0.09%

I'm typing on my phone and I'm tired so if there's a mistake do tell me.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on March 05, 2014, 08:37:04 pm
If there is 0.075M of Cr2O7 2- , what is the concentration in a 150mL sample?

I multiplied 0.075 but it's molar mass and then multiplied this value by 150/1000 to find that there are 2.43g per 150mL. Is this correct?? Is g per mL a unit for concentration?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on March 05, 2014, 09:44:24 pm
If there is 0.075M of Cr2O7 2- , what is the concentration in a 150mL sample?

I multiplied 0.075 but it's molar mass and then multiplied this value by 150/1000 to find that there are 2.43g per 150mL. Is this correct?? Is g per mL a unit for concentration?

The concentration is just 0.075 M. M is the unit for molar concentration. The volume is actually irrelevant.
Title: Re: VCE Chemistry Question Thread
Post by: Sense on March 05, 2014, 10:06:11 pm
I keep getting the complete wrong answer for every part of this practice question ... Could someone help D:

An impure sample of limestone, mainly calcium carbonate, was analysed by using a back titration. Approximately 1 g of the finely powdered limestone was weighed accurately into a conical flask. An excess of hydrochloric acid, exactly 50.00 mL, was added to the limestone. The mixture was stirred for 15 min with a magnetic stirrer to allow the reaction to be completed.
The mixture was then titrated with a standard solution of sodium hydroxide and the following results were obtained:
- Mass of watch glass = 8.7954 g
- Mass of anhydrous Calcium carbonate and watch glass = 9.8460 g
- Concentration of standard sodium hydroxide solution = 0.0489 M
- Titration value of sodium hydroxide obtained = 22.32 mL
- Concentration of hydrochloric acid = 0.395 M

a   Write balance equations for the two reactions that occur.
b   Determine the moles of hydrochloric acid in excess after the reaction with the limestone.
c   Calculate the total moles of hydrochloric acid added to the limestone.
d   How many moles of hydrochloric acid reacted with the limestone?
e   Calculate the number of moles of calcium carbonate in the limestone.
f   What is the percentage of calcium carbonate in the limestone?

Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on March 05, 2014, 11:00:17 pm
I keep getting the complete wrong answer for every part of this practice question ... Could someone help D:

An impure sample of limestone, mainly calcium carbonate, was analysed by using a back titration. Approximately 1 g of the finely powdered limestone was weighed accurately into a conical flask. An excess of hydrochloric acid, exactly 50.00 mL, was added to the limestone. The mixture was stirred for 15 min with a magnetic stirrer to allow the reaction to be completed.
The mixture was then titrated with a standard solution of sodium hydroxide and the following results were obtained:
- Mass of watch glass = 8.7954 g
- Mass of anhydrous Calcium carbonate and watch glass = 9.8460 g
- Concentration of standard sodium hydroxide solution = 0.0489 M
- Titration value of sodium hydroxide obtained = 22.32 mL
- Concentration of hydrochloric acid = 0.395 M

a   Write balance equations for the two reactions that occur.
b   Determine the moles of hydrochloric acid in excess after the reaction with the limestone.
c   Calculate the total moles of hydrochloric acid added to the limestone.
d   How many moles of hydrochloric acid reacted with the limestone?
e   Calculate the number of moles of calcium carbonate in the limestone.
f   What is the percentage of calcium carbonate in the limestone?

a) First is the calcium carbonate with the hydrochloric acid, the other is the hydrochloric acid with the sodium hydroxide
b)n(HCl)excess = n(NaOH) = cV = 0.0489*0.02232
n(HCl)reacted = (0.395*0.05) - (0.0489*0.02232)
c) & d) from above.
e) n(CaCO2) = 1/2 * n(HCl)reacted
f) m(CaCO2) = n(CaCO2) * Mr
then that mass over mass of initial sample, multiply by 100.
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on March 06, 2014, 07:45:02 am
Looks like the excess and reacted are around the wrong way:

"b)n(HCl)xs = n(NaOH) = cV = 0.0489*0.02232
n(HCl)reacted = (0.395*0.05) - (0.0489*0.02232)
c) & d) from above."

a) First is the calcium carbonate with the hydrochloric acid, the other is the hydrochloric acid with the sodium hydroxide
b)n(HCl)reacted = n(NaOH) = cV = 0.0489*0.02232
n(HCl)excess = (0.395*0.05) - (0.0489*0.02232)
c) & d) from above.
e) n(CaCO2) = 1/2 * n(HCl)reacted
f) m(CaCO2) = n(CaCO2) * Mr
then that mass over mass of initial sample, multiply by 100.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on March 06, 2014, 07:25:00 pm
Does anyone know any health risks associated with Atomic absorption spectroscopy, IR spectroscopy and UV-Vis spectroscopy?
Will be very helpful :D
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on March 06, 2014, 09:17:52 pm
A question from the Heinamann book:
In order to standardise a solution of hydrochloric acid, a
student titrated the solution against 20.00 mL aliquots of
a standard solution of sodium carbonate. Methyl orange
indicator was used to identify the end point of the reaction:
2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
The sodium carbonate solution had been prepared by
dissolving 1.236 g of anhydrous Na2CO3 in water and making
the solution up to 250.0 mL in a volumetric fl ask. The titres
recorded were 21.56 mL, 20.98 mL, 20.96 mL
and 21.03 mL.
a What value for the titre of hydrochloric acid solution
should the student use in the calculation of the acid
b What is the molarity of the sodium carbonate solution?
c Calculate the concentration of the HC l, in mol L−1.

Having trouble with c). I'm getting 1.111M while the answer suggests 0.08892 M

Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: Gutthi on March 06, 2014, 09:36:55 pm
While titrating Iodine against Wine (to determine the wine's Sulfur Dioxide content) why is the Iodine titrated QUICKLY?
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on March 06, 2014, 11:13:07 pm
Looks like there might be a couple of things you are looking over here:

1) The Sodium Carbonate was added to a 250ml volumetric flask.
2) The ratio of HCl to Na2CO3 is not 1:1

A question from the Heinamann book:
In order to standardise a solution of hydrochloric acid, a
student titrated the solution against 20.00 mL aliquots of
a standard solution of sodium carbonate. Methyl orange
indicator was used to identify the end point of the reaction:
2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
The sodium carbonate solution had been prepared by
dissolving 1.236 g of anhydrous Na2CO3 in water and making
the solution up to 250.0 mL in a volumetric fl ask. The titres
recorded were 21.56 mL, 20.98 mL, 20.96 mL
and 21.03 mL.
a What value for the titre of hydrochloric acid solution
should the student use in the calculation of the acid
b What is the molarity of the sodium carbonate solution?
c Calculate the concentration of the HC l, in mol L−1.

Having trouble with c). I'm getting 1.111M while the answer suggests 0.08892 M

Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on March 07, 2014, 07:38:42 pm
Looks like there might be a couple of things you are looking over here:

1) The Sodium Carbonate was added to a 250ml volumetric flask.
2) The ratio of HCl to Na2CO3 is not 1:1
b) n(Na2CO3) = 1.236/106.0 = 0.01166mol
c(Na2CO3) = 0.01166/0.250 = 0.04664 M
*I check if b) was right*

c)n(HCl) = 2 x n(Na2CO3) = 0.02332mol
c(HCl) = 0.02332/0.02099 = 1.111 M

It's c) that doesn't come off right. Am I doing a step incorrectly?

Thanks
Title: Re: VCE Chemistry Question Thread
Post by: thushan on March 07, 2014, 08:59:32 pm
b) n(Na2CO3) = 1.236/106.0 = 0.01166mol
c(Na2CO3) = 0.01166/0.250 = 0.04664 M
*I check if b) was right*

c)n(HCl) = 2 x n(Na2CO3) = 0.02332mol
c(HCl) = 0.02332/0.02099 = 1.111 M

It's c) that doesn't come off right. Am I doing a step incorrectly?

Thanks

"c)n(HCl) = 2 x n(Na2CO3) = 0.02332mol"

Your issue is here :( the n(HCl) in the titre is indeed twice the amount of Na2CO3 in the aliquot. However, the amount of Na2CO3 in the aliquot is not 0.01166 mol, this is the amount of Na2CO3 in the 250.0 mL volumetric flask. Remember you are taking 20.00 mL aliquots of the Na2CO3 from the 250.0 mL sample.

Alternative solution:
n(Na2CO3)250.0 mL = 1.236/106.0 = 0.01166mol
n(Na2CO3)20.00 mL aliquot = 0.01166 x 20.00/250.0 = 0.0009328 mol
n(HCl)titre = n(Na2CO3)20.00 mL aliquot x 2 = 0.001866 mol
[HCl] = 0.001866/0.02099 = 0.08888 M

I bet you anything the 0.08892 M answer is because they rounded off too early.
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on March 07, 2014, 10:41:10 pm
"c)n(HCl) = 2 x n(Na2CO3) = 0.02332mol"

Your issue is here :( the n(HCl) in the titre is indeed twice the amount of Na2CO3 in the aliquot. However, the amount of Na2CO3 in the aliquot is not 0.01166 mol, this is the amount of Na2CO3 in the 250.0 mL volumetric flask. Remember you are taking 20.00 mL aliquots of the Na2CO3 from the 250.0 mL sample.

Alternative solution:
n(Na2CO3)250.0 mL = 1.236/106.0 = 0.01166mol
n(Na2CO3)20.00 mL aliquot = 0.01166 x 20.00/250.0 = 0.0009328 mol
n(HCl)titre = n(Na2CO3)20.00 mL aliquot x 2 = 0.001866 mol
[HCl] = 0.001866/0.02099 = 0.08888 M

I bet you anything the 0.08892 M answer is because they rounded off too early.

Ahh yess, I was using the amount in the 250.0mL flask than the 20.0mL titre!
Thanks for that!
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on March 07, 2014, 11:13:54 pm
While titrating Iodine against Wine (to determine the wine's Sulfur Dioxide content) why is the Iodine titrated QUICKLY?

Because sulfur dioxide (SO2) is produced during the titration, and you want to carry out the titration BEFORE the gas leaves the conical flask.
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on March 08, 2014, 08:10:53 pm
how do I split this overall ionic equation into reduction and oxidation half equations
2 S2O42-(aq) + H2O (l)  ->  S2O32-(aq) + 2 HSO3-(aq)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on March 08, 2014, 09:41:48 pm
how do I split this overall ionic equation into reduction and oxidation half equations
2 S2O42-(aq) + H2O (l)  ->  S2O32-(aq) + 2 HSO3-(aq)

Firstly, what's being oxidised? Check oxidation numbers first.
In the left hand side, the sulfur has an oxidation number of +3, in thiosulfate it has +2 while in hydrogen sulfite it has +4. This is a bit of a trippy question because S2O42- is disproportionating; i.e. it is oxidising AND reducing itself. Try balancing S2O42- => S2O32-
and S2O42- => HSO3-
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on March 09, 2014, 09:42:43 pm
In a typical chromatogram for HPLC/GLC how do I read it? What does it actually mean?

What does the relative peak height tell you about a component?
What does the area under the peak tell you about the component?

Why doesn't the slowest component (one with the highest retention time) have the most area all the time? - (so I'm wondering about the relationship between retention time and each component on a chromatogram)

ALSO:
http://puu.sh/7oHzs.png
Is eluent the same as waste?

Thanks
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on March 09, 2014, 09:53:13 pm
In a typical chromatogram for HPLC/GLC how do I read it? What does it actually mean?

What does the relative peak height tell you about a component?
What does the area under the peak tell you about the component?

Why doesn't the slowest component (one with the highest retention time) have the most area all the time? - (so I'm wondering about the relationship between retention time and each component on a chromatogram)

Thanks

Ok:

(1.) Area under peak = Concentration of that component in the sample.
(2.) Relative peak height I'm pretty sure means the same as area under peak, and thus means concentration. Could someone please clarify?!
(3.) Retention time depends upon how strongly adsorbed the component is to the stationary phase, and how soluble the component is in the mobile phase. If a component has a high Rt, this means it is strongly adsorbed to the stationary phase. Just because the component is strongly adsorbed does not mean it is highly concentrated in the sample.
Title: Re: VCE Chemistry Question Thread
Post by: Scooby on March 10, 2014, 02:59:55 am
Ok:

(1.) Area under peak = Concentration of that component in the sample.
(2.) Relative peak height I'm pretty sure means the same as area under peak, and thus means concentration. Could someone please clarify?!
(3.) Retention time depends upon how strongly adsorbed the component is to the stationary phase, and how soluble the component is in the mobile phase. If a component has a high Rt, this means it is strongly adsorbed to the stationary phase. Just because the component is strongly adsorbed does not mean it is highly concentrated in the sample.

The peak height is the peak height and the peak area is the peak area. The peak height is sometimes used to determine concentration, but obviously that's a lot less accurate than using the peak area, especially if the peak is really wide
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on March 10, 2014, 03:44:18 am
The peak height is the peak height and the peak area is the peak area. The peak height is sometimes used to determine concentration, but obviously that's a lot less accurate than using the peak area, especially if the peak is really wide

Thanks for clarifying that Scooby. :)
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on March 10, 2014, 10:34:41 am
The peak height is the peak height and the peak area is the peak area. The peak height is sometimes used to determine concentration, but obviously that's a lot less accurate than using the peak area, especially if the peak is really wide
Uh, does the peak height mean anything? I don't think stating that a peak height is the peak height and peak area is the pear area is going to help me in any way..
There has to be a reason why different components have differing peak heights
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on March 10, 2014, 10:58:53 am
For Q 4biv of Exam 2004 (Exam 1.. Short Answer ),  - Sorry guys couldn't printscreen the whole q!

But why do we use the absorbance of 7.5mg/L? I used 1.5mg/L and although is is wrong, it was still awarded 1 mark.. for some reason...?

It says 'Q mol In was used up at the point the absorbance reached zero' but I don't understand why this value is used.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on March 10, 2014, 11:08:20 am
For Q 4biv of Exam 2004 (Exam 1.. Short Answer ),  - Sorry guys couldn't printscreen the whole q!

But why do we use the absorbance of 7.5mg/L? I used 1.5mg/L and although is is wrong, it was still awarded 1 mark.. for some reason...?

It says 'Q mol In was used up at the point the absorbance reached zero' but I don't understand why this value is used.

Have u drawn the calibration curve for it?
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on March 10, 2014, 11:12:08 am
Uh, does the peak height mean anything? I don't think stating that a peak height is the peak height and peak area is the pear area is going to help me in any way..
There has to be a reason why different components have differing peak heights

The only questions that you're going to get relating to peak height is to construct a callibration curve based on this data. The peak height basically tells you the absorbance/concentration of a certain compound in the sample. To find what these compounds are, the peak times are compared to a different graph where the Rt value of compounds are known.

So if a compound is known to show a peak at 10 mins at certain conditions and there is a peak at 10 mins in the unkown compound in the same conditions we can assume that this peak is of the same compound.

Have u drawn the calibration curve for it?

Yes :)
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on March 10, 2014, 11:19:25 am
Help pls :)
1-propylamine may be produced by a reaction between 1-chloropropane and one other molecular compound. When 2.17g of 1-chloropropane is completely converted to 1-propylamine, what mass of the other compound would be expected to react?
Title: Re: VCE Chemistry Question Thread
Post by: Scooby on March 10, 2014, 12:56:11 pm
Thanks for clarifying that Scooby. :)

No problem :)

Help pls :)
1-propylamine may be produced by a reaction between 1-chloropropane and one other molecular compound. When 2.17g of 1-chloropropane is completely converted to 1-propylamine, what mass of the other compound would be expected to react?

To produce propan-1-amine from 1-chloropropane we need to react it with ammonia. Use the mass of 1-chloropropane given and its molar mass to find the number of moles of it that reacted. From that determine the number of moles of ammonia that reacted. Since you have the number of moles of ammonia reacted and its molar mass, you can work out the mass that reacted as well
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on March 10, 2014, 03:27:22 pm
is 0.025mg/L of Ca2+ equal to 6.25x10-7mol L
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on March 10, 2014, 03:45:45 pm
is 0.025mg/L of Ca2+ equal to 6.25x10-7mol L

Yes.

0.025 mg/L (or ppm) / 1000 = 0.000025 g/L
0.000025 g/L divided by 40.1 (Mr of Ca) = 6.25 x 10-7 mol/L or M.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on March 10, 2014, 03:52:33 pm
Yes.

0.025 mg/L (or ppm) / 1000 = 0.000025 g/L
0.000025 g/L divided by 40.1 (Mr of Ca) = 6.25 x 10-7 mol/L or M.

Wooo!! Thanks
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on March 11, 2014, 07:39:47 pm
Can anyone name and similarities or differences between IR spec, AAS and UV-visible spectroscopy?
Title: Re: VCE Chemistry Question Thread
Post by: bonappler on March 11, 2014, 10:12:49 pm
Yeah I need help with this
Grains of fine California beach sand are approximately spheres with an average radius of 50 μm and are made of silicon dioxide, which has a density of 2.8 × 103 kg/m3. What mass of sand grains would have a total surface area (the total area of all the individual spheres) equal to the surface area of a cube 0.9 m on an edge?

Title: Re: VCE Chemistry Question Thread
Post by: Aurelian on March 12, 2014, 06:32:03 pm
Yeah I need help with this
Grains of fine California beach sand are approximately spheres with an average radius of 50 μm and are made of silicon dioxide, which has a density of 2.8 × 103 kg/m3. What mass of sand grains would have a total surface area (the total area of all the individual spheres) equal to the surface area of a cube 0.9 m on an edge?

Lol what a bizarre question. Okay so, firstly think about what the surface area of that cube is going to be. Then think about what the surface area of an individual grain of sand is going to be. Then figure out how many grains of sand you're going to need.

Then think about what the mass of an individual grain of sand is going to be (hint: use the density). Then, the mass of sand you'll need will just be the number of grains required multiplied by the mass of each grain...!

Does that help? =)
Title: Re: VCE Chemistry Question Thread
Post by: bonappler on March 13, 2014, 11:41:48 am
Actually, I got an answer but I'm not sure if it is correct
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on March 13, 2014, 07:01:29 pm
Has anyone had a chemistry sac on spectroscopy?
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on March 15, 2014, 11:21:06 am
this may sound stupid but whats the difference between Mol/L and Molarity (concentration)

just say you had a substance and it was 6.5mol/L and you wanted to find the mass...
would you go m(substance)=6.5 x (molar mass of substance)   ?

Title: Re: VCE Chemistry Question Thread
Post by: RKTR on March 15, 2014, 12:03:07 pm
For your first question,those two are the same.

Concentration(g/L)=molarity (mol/L) x molar mass (g/mol)

Then you need the volume to find the mass

Sometimes if you can't remember the formula, try using the units to help you
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on March 15, 2014, 12:31:01 pm
Can someone give me the half equation of the oxidation of ethanol to ethanal?
Or, if that's not possible, the full equation of ethanol being oxidised to ethanal using an excess of potassium dichromate, please.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on March 15, 2014, 02:00:44 pm
Can someone give me the half equation of the oxidation of ethanol to ethanal?
Or, if that's not possible, the full equation of ethanol being oxidised to ethanal using an excess of potassium dichromate, please.

Ethanol is C2H5OH
Ethanal is C2H4O

So really it's just C2H5OH => C2H4O + 2H+ + 2e-
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on March 15, 2014, 04:25:13 pm
Why is a fine layer of powder (Alumina) spread on a glass/plastic plate in Thin-Layer Chromatography?
Is it to increase surface area for components to be adsorbed into the stationary phase more effectively?

Also in chromatography, when it says you need an appropriate solvent/mobile phase what does that mean? - So, say my sample is a polar compound, would I want a polar solvent, or what? (polar-attracts-polar rule?)

Also! Does the degree of adsorption to the stationary phase depend on the: chemical structure of the stationary phase? or is it the chemical structure of the component?

Thanks,
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on March 15, 2014, 08:06:04 pm
In chromatography, are the Rf values and the Rt value, inversely proportional?  ???
Title: Re: VCE Chemistry Question Thread
Post by: RKTR on March 15, 2014, 10:03:25 pm
In chromatography, are the Rf values and the Rt value, inversely proportional?  ???
Yup
Title: Re: VCE Chemistry Question Thread
Post by: grannysmith on March 16, 2014, 03:49:46 pm
Do we need to show working out in chemistry?
Title: Re: VCE Chemistry Question Thread
Post by: HEN_iP on March 16, 2014, 05:16:52 pm
Do we need to show working out in chemistry?

Always should. Shows the examiners you know what you're doing and how you got there, obviously. In my opinion, also helps checking your answer because you'll have all your steps laid out.
Title: Re: VCE Chemistry Question Thread
Post by: ETTH96 on March 16, 2014, 09:06:13 pm
Hi Guys, I'm having a little bit of trouble understanding organic chemical reactions.. Like substitution, addition and condensation, I believe?

I don't understand how it works, when do i know what to use/identify what has been used etc... and how are they different? I've tried youtubing some videos but i just don't understand? could someone please explain
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on March 16, 2014, 09:26:25 pm
Hi Guys, I'm having a little bit of trouble understanding organic chemical reactions.. Like substitution, addition and condensation, I believe?

I don't understand how it works, when do i know what to use/identify what has been used etc... and how are they different? I've tried youtubing some videos but i just don't understand? could someone please explain

I'm going to go overkill here because I can.

For substitution, it occurs when carbon is bonded to something more electronegative than itself. Like, you'll never have normal substitution of a hydrogen atom via a polar reaction; halogenation occurs because the X-X halogen dissociates into VERY reactive halogen radicals.
Let's take methyl bromide. Now, bromine is more electronegative than carbon, so the carbon is positively charged (slightly). This means that the carbon will attract anything negatively charged, like hydroxide ion. When a hydroxide ion meets this methyl bromide molecule, the OH- is attracted to the positive carbon. As it begins to form a bond with the carbon (i.e. donates a pair of electrons to form a covalent bond), there are three things that can happen. 1, the carbon kicks out a hydrogen atom while giving all of its electrons to the hydrogen forming H-. That's not likely as H- is way too unstable. 2, the carbon kicks out the hydroxide ion again and no reaction happens. 3, the carbon kicks out the bromine giving its electrons to the bromine in order to form a bond with hydroxide.

Which of 2/3 is going to happen? Well, HBr is a MUCH stronger acid than HOH (also known as water :P). This suggests that Br- is more stable than OH-, so we can reasonably expect that Br- will be kicked off preferentially to OH-. This is indeed what happens and OH- does normally substitute for bromine. Use similar reasoning for substituting OH- for Cl.

What about ammonia? Well, HBr is still a much stronger acid than NH4+, so Br- is more stable than ammonia. That's why ammonia will also substitute for bromine atoms. Look up SN2 (substitution nucleophilic bimolecular) if you want more information on this type of reaction.

Now let's look at addition. Generally addition means addition over a double bond. In the VCE course, addition is generally of a hydrogen halide or of water across a double bond (do remind me if there are any others). Let's see what happens with the addition of HX where X is a halogen.

The double bond is quite electron rich, so it is a nice target for something like a H+ cation. What happens is that in addition, one carbon forms a bond with the proton, leaving the other carbon positively charged (carbon has no lone pairs, so to form a bond with another atom, it must drag electrons from elsewhere; in this case, one carbon breaks the C=C double bond and converts it into a single bond). This positively charged carbon then attracts the X- ion and you've completed the addition of HX. Something similar happens with X2, although I won't go into the details.

Now condensation? This is a VEEEERRRRYYY broad class of reactions; I wouldn't even call it a class of reactions. In esterification, the simplified answer is, the alcohol oxygen in ROH bonds with the carboxylic acid carbon in R'COOH. The alcohol hydrogen migrates to the acid OH group, that group leaves as water and you're left with R'COOR (yes this is overly simplified but I don't want to complicate things any further xP)
Why does this happen? Well, the oxygen in ROH is quite negative and the carbon in RCOOH is VERY positive due to its C=O and its C-OH bonds. That's why the reaction would occur.

As for amide formation, directly mixing a carboxylic acid and an amine won't give you an amide (even with acid catalyst) because protonated amines and carboxylates are both not very reactive. However, for the sake of VCE, imagine the amine nitrogen (with its negative charge) attacking or forming a bond with the positive carbon in the RCOOH, and then imagine the amine nitrogen replacing the OH group. You'll need to give me more specific reactions here I'm afraid; I can come up with a dozen condensation reactions that are probably all irrelevant.

How to identify what's used? Substitution, just remember that alcohols can be prepared from alkyl halides (slightly more complicated than this but oh well) and that amines can be prepared from alcohols and alkyl halides. Condensation, well in VCE there's only one way of making esters and amides. For addition, you're generally told to start with an addition or from something like ethene. You literally break the C=C bond and add atoms to each side.
Title: Re: VCE Chemistry Question Thread
Post by: rhinwarr on March 17, 2014, 09:30:47 pm
1) A sample of limestone is analysed for its calcium carbonate content as follows. A 25.00g sample is crushed and heated to a high temperature. It is cooled, weighed and reheated until a constant mass is obtained. The mass remaining at the end of the process if 11.64g. Find the percentage purity of CaCO3 in the sample.

CO2 is the gas released so the change in mass must be the CO2 right? But then if I use CO2 to find the mass of CaCO3, I get a mass that is greater than 25g. Can someone help please? The answer is supposed to be 95%.

2) A gravimetric analysis on fertiliser is carried out to find the mass of phosphorus. A 14.298g sample of fertiliser yielded 4.107g of Mg2P2O7. Determine the mass of phosphorus.

What I did was m(P)=4.107 x 62/222.6=1.144g
But the answer is supposed to be 0.8967g
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on March 18, 2014, 02:30:26 pm
1) A sample of limestone is analysed for its calcium carbonate content as follows. A 25.00g sample is crushed and heated to a high temperature. It is cooled, weighed and reheated until a constant mass is obtained. The mass remaining at the end of the process if 11.64g. Find the percentage purity of CaCO3 in the sample.

CO2 is the gas released so the change in mass must be the CO2 right? But then if I use CO2 to find the mass of CaCO3, I get a mass that is greater than 25g. Can someone help please? The answer is supposed to be 95%.

2) A gravimetric analysis on fertiliser is carried out to find the mass of phosphorus. A 14.298g sample of fertiliser yielded 4.107g of Mg2P2O7. Determine the mass of phosphorus.

What I did was m(P)=4.107 x 62/222.6=1.144g
But the answer is supposed to be 0.8967g

determine the mol of Mg2P207 using n=4.107/M. From that you know there are 2 x n(mg2p2o7) of phosphorus
using that mol value, you can calculate the mass using m=n x M
Title: Re: VCE Chemistry Question Thread
Post by: ETTH96 on March 18, 2014, 09:25:01 pm
Thanks lzxnl!!! Really helpful :)

Can someone help me out with this question?

1-chlorobutane can be hydrolysed to 1-butanol. 1-butanol can then be oxidised to a carboxylic acid of empirical formula C4H8O2.
i) give the name or formula of a suitable laboratory oxidising agent for the reaction of 1-butanol to a carboxylic acid.

How do I do this..? How do I know what to use? I'm not too sure how to do this

ii) give the systematic name for the carboxylic acid.

Thanks guys
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on March 18, 2014, 09:30:22 pm
Thanks lzxnl!!! Really helpful :)

Can someone help me out with this question?

1-chlorobutane can be hydrolysed to 1-butanol. 1-butanol can then be oxidised to a carboxylic acid of empirical formula C4H8O2.
i) give the name or formula of a suitable laboratory oxidising agent for the reaction of 1-butanol to a carboxylic acid.

How do I do this..? How do I know what to use? I'm not too sure how to do this

ii) give the systematic name for the carboxylic acid.

Thanks guys

This is just stuff you have to know. Primary alcohols are oxidised to carboxylic acids with the same number of carbons by KMnO4 or K2Cr2O7/Na2Cr2O7
The carboxylic acid would then be butanoic acid
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on March 21, 2014, 05:29:13 pm
Is

CH3CH2CH(NH2)(CH2)2COOH

4-amino penatnoic acid?
Title: Re: VCE Chemistry Question Thread
Post by: Aurelian on March 21, 2014, 05:37:35 pm
Is

CH3CH2CH(NH2)(CH2)2COOH

4-amino penatnoic acid?

No, it's 4-aminohexanoic acid - there are six carbons in the chain.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on March 21, 2014, 06:27:38 pm
The concentration of potassium carbonate in mineral water from a country springs was found by titrating 20.00cm3 aliquots of the mineral water against 0.150 M hydrochloric acid using methyl orange as the indicator. The average titre of several titration was 12.30cm3.

a) write an equation for the reaction formed.
(How do we know when to write ionic or full chemical equations?)

b) find the molarity of the potassium carbonate in the mineral water.

Can I have help for these questions? :)
Title: Re: VCE Chemistry Question Thread
Post by: PB on March 21, 2014, 07:11:04 pm
a) if they simply said write an equation, I am pretty sure you can just write the ionic. However, you might want to err on the side of caution and simply write out the full equation! It is easier to comprehend anyway.

So first up, lets try and remember - what do you get when you mix a metal carbonate with an acid?   1. a salt  2. water  3. CO2

NOW BEFORE LOOKING BELOW, TRY TO FORMULATE THE EQUATION FIRST FOR YOURSELF!

Ans
general unbalanced equation
(1)HCl + K2CO3   ->  CO2  + H2O + KCl
and balancing it..
(2) K2CO3(aq) +2HCl (aq)-> H2O (l)  +CO2 (g) + 2KCl(aq)

wah lah
Title: Re: VCE Chemistry Question Thread
Post by: PB on March 21, 2014, 07:25:56 pm
b) now try not to get confused with the cm3 units.   Remember that 1x1x1cm will simply make a tiny, little box that can only hold 1ml of water at room temperature.   In short, 1cm3 = 1ml
ok so we found out that 12.30ml of .150M HCl was required to eat up all that potassium carbonate in the mineral water. so lets try and find how many mol of HCl was in that 12.30ml.

n(HCl)=cv = .150 *.01230  = 1.845 x 10 ^-3 mols

finding n(K2CO3)...
n(K2CO3): n(HCl)
1:2
9.225 x10 ^-4  : 1.845 x 10 ^-3

therefore, the molarity/concentration of potassium carbonate in 20mls of mineral water is

c=n/v
= 9.225 x10 ^-4 divide by .02
=.04612M (4 s.f.)

hope that helped :P

Title: Re: VCE Chemistry Question Thread
Post by: eagles on March 21, 2014, 08:38:22 pm
Thanks, I had the same answers as you did but the answer for question b) is 0.092M.
The worked solutions show that:
c1v1 = c2 = v2
c1 = 0.15 x 0.01230/0.0200 = 0.092M.

Can you help me explain their working?

Cheers!
Title: Re: VCE Chemistry Question Thread
Post by: PB on March 21, 2014, 11:35:32 pm
lol what the... I think they have used the wrong formula.  c1v1=c2v2   should only be used when you are trying to find the concentration of a solution after is has been diluted with water.

I am pretty the answers are wrong man :P because what their working out would be finding out is the new concentration of HCl in the mineral water after it has been diluted to 20mls from 12.30mls.

and mind you, it is not even the molarity of HCl we are after...it is K2CO3.  Pretty sure the solutions are wrong.

P.S. how about some +1's if I have helped you hey? ;)    Help me up to +5 respect on my 300th post!

Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on March 22, 2014, 03:06:09 pm
Can someone pls help me with MC q9 in  chem exam 1 2012?
Its naming compounds and I really suck at it :(
Title: Re: VCE Chemistry Question Thread
Post by: PB on March 22, 2014, 05:57:26 pm
Ok well, first up - we know that this is an ester due to the COO- group trapped in the middle of the carbon chain.
and how do we name esters? its easy dw.
Lets take a look at how esters are formed first. It is formed from a carboxylic acid joining together with an alcohol.

Looking at the diagram, we can see that what was once the carboxylic acid in the ester has three carbons (CH3CH2COO...) so it would form the SECOND part of the name - propanOATE.

The first part of the ester's name would simply be related to the other carbon chain hanging off on the other side of the ester. In the diagram, we can see that this carbon chain is only 2 carbons long.  Hence the FIRST part of the name would be - ethYL.

Put the two parts together and you get Ethyl Propanoate.

So it goes that the general 'formula' of ester nomenclature is   -   (prefix 1)yl   (prefix 2)oate.

where:
(prefix 1) =  the number of carbons in what was once the alcohol
(prefix 2) =  the number of carbons in what was once the carboxylic acid
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on March 22, 2014, 06:38:21 pm
OMG THANK YOU SOOOO MUCH!!!!
U r so good at explaining things. I would give u more than +1 if I could.
Kind request, whoever benefitted from PB's explanation, pls +1
Title: Re: VCE Chemistry Question Thread
Post by: PB on March 22, 2014, 07:13:00 pm
Thanks rishi :) really appreciate it!
Title: Re: VCE Chemistry Question Thread
Post by: eagles on March 23, 2014, 08:45:50 am
Hi! Please see attached the problem I'm working on.
Should the answer to question b) be component B as it is more strongly absorbed to the stationary phase than component a? Thus it will travel more slowly and move least distance on the paper. Hence the lower RF value.

Thank you!
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on March 23, 2014, 11:10:06 am
The only problem I can see is that the question said "absorbed" instead of "adsorbed".  In chromatography compounds are adsorbed (bonded to the stationary phase) not absorbed (taken into).  Could be a typo, but you do need to be careful not to confuse the two.

Your logic is correct in regards to Rf values though.

Hi! Please see attached the problem I'm working on.
Should the answer to question b) be component B as it is more strongly absorbed to the stationary phase than component a? Thus it will travel more slowly and move least distance on the paper. Hence the lower RF value.

Thank you!
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on March 23, 2014, 04:44:55 pm
"Write a sequence of reactions that shows the production of ethanoic acid to ethane"

The answers state that

First Cl is used as a catalyst to subtitute H

Then OH is used to substitute the Cl

Then it is oxidised and becomes ethanoic acid

But why? Why is ethane reacted with Cl first? Why can't ethane react with OH first? I'm confused about the whole process. Can someone please explain what has happened? Thanks :)
Title: Re: VCE Chemistry Question Thread
Post by: PB on March 23, 2014, 04:58:50 pm
lol, this question has been asked soooo many times, even in the years before you XD
In the end everyone just takes it as a fact and moves on because the actual mechanics of it is way out of the course. :P

Simply put, it is easier to substitute a Cl with OH than it is to substitute a H with OH (you also get higher yield of Ethanol this way). VCAA is NEVER going to test you on why that is - just that it is that way.

One more thing, Chlorine is NOT the catalyst as it takes part in the reaction.  UV light is the catalayst.

Hope that helps :P
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on March 23, 2014, 05:13:22 pm
lol, this question has been asked soooo many times, even in the years before you XD
In the end everyone just takes it as a fact and moves on because the actual mechanics of it is way out of the course. :P

Simply put, it is easier to substitute a Cl than with OH than it is to substitute a H with OH (you also get higher yield of Ethanol this way). VCAA is NEVER going to test you on why that is - just that it is that way.

One more thing, Chlorine is NOT the catalyst as it takes part in the reaction.  UV light is the catalayst.

Hope that helps :P

ohh okay thanks :)

So do alkanes always have to react with Cl? My friend said that they can react with any of the halogens?
Title: Re: VCE Chemistry Question Thread
Post by: PB on March 23, 2014, 05:16:44 pm
Yeah for sure, you could do it with Bromine as well.  Just as long as the halogen is electronegative enough :)
Title: Re: VCE Chemistry Question Thread
Post by: thushan on March 23, 2014, 05:22:04 pm
Yeah for sure, you could do it with Bromine as well.  Just as long as the halogen is electronegative enough :)

Iodine works too. Fluorine doesn't, though. If you mix methane and fluorine, fluorine just rips all the H out and you get C and HF.
Title: Re: VCE Chemistry Question Thread
Post by: PB on March 23, 2014, 05:24:46 pm
Iodine works too. Fluorine doesn't, though. If you mix methane and fluorine, fluorine just rips all the H out and you get C and HF.

What he said ^
Title: Re: VCE Chemistry Question Thread
Post by: Jawnle on March 23, 2014, 05:49:37 pm
Chem Klub or Khem Club?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on March 23, 2014, 06:01:58 pm
Iodine works too. Fluorine doesn't, though. If you mix methane and fluorine, fluorine just rips all the H out and you get C and HF.

What about excess fluorine? :D
Title: Re: VCE Chemistry Question Thread
Post by: eagles on March 23, 2014, 08:46:58 pm
I was just wondering about the reaction of water and carbon dioxide gas which gives us carbonic acid.
Would we use a complete reaction arrow or an equilibrium arrow?

Thank you.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on March 23, 2014, 09:24:33 pm
If you were to draw in arrow sizes to reflect the proportion of completion of the reaction...you wouldn't be able to see much of the forward arrow :P
It's not a very favourable reaction. Carbon dioxide doesn't dissolve in water to any significant extent, nor does it form much of an acid. That's why when we write acid-base equations with carbonate, we neglect the formation of carbonic acid and go straight to H2O and CO2.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on March 23, 2014, 10:36:44 pm
Haha, thanks.

Given a class prac, we need to find the titratable acidity content in wine, as expressed if all is tartaric acid, which is a diprotic acid with molar mass 150g mol-1. I'm having trouble understanding what this means and would really appreciate an explanation.

Also, I'm unsure about the purpose of step 4 in our experiment of determining acidity in wine:

1. Fill the burette with NaOH.
2. Add approx 100mL of deionised water to a 250mL conical flask.
3. Add 3 or 4 drops of phenolphthalein indicator to the flask and mix well.
4. Add 0.1M NaOH from the burette to the flask until the solution turns a pale pink colour that persists for at least 30 seconds. This takes only a few drops and there is no need to record this volume.
5. Pipette 10.0mL of wine into conical flask and record initial burette reading.
6. Titrate the solution in the conical flask with NaOH until the pink colour again persists for at least 30 seconds.
7. Record final burette reading and repeat steps to obtain concordant titres.

Does this mean that the pink colour fades when we do step 5? Otherwise, how are we expected to see the pink colour change as mentioned in step 6? Had we omitted this step, that is pipetted the wine with the deionised water and phenolphthalein (without a drop of NaOH), would it have made any difference?

Can someone please also explain the need for adding the 100mL of water? I understand that it does not change the total amount of acid in the sample, so still unsure why step 2 is necessary.

All of your help is appreciated! Thank you  :D
Title: Re: VCE Chemistry Question Thread
Post by: ETTH96 on March 24, 2014, 06:51:52 pm
hey guys, sorry this may sound stupid but... in gravimetric analysis where a precipitate is formed, how do you know what to use to react with your substance to form a particular precipitate?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on March 24, 2014, 07:00:42 pm
Depends on what salt you want. If you want to precipitate silver chloride from silver ions, for instance, you need silver and chloride ions to be present in water. For instance, react silver nitrate with hydrochloric acid to form silver chloride precipitate and nitric acid.
Title: Re: VCE Chemistry Question Thread
Post by: eagles on March 24, 2014, 07:02:31 pm
Referring to the solubility table is quite helpful.

There are some precipitates which are good to remember that often appear in tests such as BaSO4 and AgCl.

For example, if given a reactant such as AgNO3, and in order to make a precipitate such as AgCl, we need a reactant with Cl in it. We can choose NaCl.

AgNO3 (aq) + NaCl (aq) --> AgCl (s) + NaNO3 (aq)

Title: Re: VCE Chemistry Question Thread
Post by: shaw on March 24, 2014, 11:35:50 pm
hi everyone I am new to this site and finally received my Extended Experimental Investigation: Analysis of fertiliser sac results back for which I only got 24/50 and after asking my teacher for reasons as to why I got this result he could not explain, would someone please help me out and post a previous sac which received the same or higher mark so I can see where I went wrong before I make the same mistakes on the next practical sac. I really want a good mark chemistry please help me out.

if anyone is on please help me out I'm that stressed I don't know what to do any more.  :'(
Title: Re: VCE Chemistry Question Thread
Post by: ETTH96 on March 25, 2014, 06:00:48 pm
Hey guys, I have a chemistry SAC tomorrow and it is based on the analysis of fertiliser; the sulfur, nitrogen and phosphorus contents. part of the SAC is that we have to draw up 1 table summarising all our results. The three methods of analysis we used were; gravimetric, volumetric(back titration) and colorimetric... so how can i go about setting up a table for my results? because i have no idea :(
Title: Re: VCE Chemistry Question Thread
Post by: mikehepro on March 25, 2014, 07:19:49 pm
Hey guys, just wondering how should I approach this question?
$0.470g$ of a compound containing carbon,hydrogen and oxygen produced $0.551g$ of water and $458 cm^3$ of carbon dioxide on being burnt at s.t.p. Calculate the empirical formula of the compound.
I got c:8 H:24 O:7 but that looks wrong to me.
Any help would be greatly appreciated.
Thanks.
Title: Re: VCE Chemistry Question Thread
Post by: Phenomenol on March 25, 2014, 07:26:52 pm
Hey guys, just wondering how should I approach this question?
$0.470g$ of a compound containing carbon,hydrogen and oxygen produced $0.551g$ of water and $458 cm^3$ of carbon dioxide on being burnt at s.t.p. Calculate the empirical formula of the compound.
I got c:8 H:24 O:7 but that looks wrong to me.
Any help would be greatly appreciated.
Thanks.

Hmm, try again. Remember, to find the mass of oxygen originally in the organic compound, subtract the mass of carbon and mass of hydrogen from 0.470g.
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on March 25, 2014, 07:48:40 pm
An alcoholic beer has a conc. of 2.5% v/v of ethanol. (density of ethanol @ 0.5g/mL)
Find the amount of ethanol consumed in a 375mL can. Mr of ethanol=46

Thanks
Title: Re: VCE Chemistry Question Thread
Post by: Phenomenol on March 25, 2014, 08:12:05 pm
An alcoholic beer has a conc. of 2.5% v/v of ethanol. (density of ethanol @ 0.5g/mL)
Find the amount of ethanol consumed in a 375mL can. Mr of ethanol=46

Thanks

Concentration m/v = concentration v/v * density of solute
Mass = volume * concentration m/v
Amount = mass/molar mass

Make sure your units are consistent as always!
Title: Re: VCE Chemistry Question Thread
Post by: mikehepro on March 25, 2014, 08:28:19 pm
Hmm, try again. Remember, to find the mass of oxygen originally in the organic compound, subtract the mass of carbon and mass of hydrogen from 0.470g.

Right,so I found out the how many mols of Co2 and H2O there is, which is roughly 0.03 and 0.02, however, how am I suppose to determine the mole ratio? Since there's o2 on the left side of the equation, unless I'm not doing it with the right method in the first place.
Thanks
Title: Re: VCE Chemistry Question Thread
Post by: Phenomenol on March 25, 2014, 08:35:47 pm
Right,so I found out the how many mols of Co2 and H2O there is, which is roughly 0.03 and 0.02, however, how am I suppose to determine the mole ratio? Since there's o2 on the left side of the equation, unless I'm not doing it with the right method in the first place.
Thanks

The trick here is to look at the masses of elements first, not amounts. Otherwise you're doing too much work.
Total mass(H) = 2.0/18.0 * 0.551g. This gives you the mass of hydrogen in the organic compound.

Total amount(CO2) = 0.458/22.4
Then find the mass of carbon in the CO2. This gives you the mass of carbon in the organic compound.

Simply subtract these masses from 0.470g to find the mass of oxygen in the compound. From there the routine method to find an empirical formula can be used.

So, the important thing to note is that all the carbon in the original compound is now in the form of CO2, and all the hydrogen in the original compound is now in the form of H2O.
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on March 25, 2014, 08:46:31 pm
There's a difference between de-ionised water and destilled water..But are they the same? Just difference processes?
If a question asked to write steps for like...cleaning equipment..which would I use?

Thanks
Title: Re: VCE Chemistry Question Thread
Post by: Phenomenol on March 25, 2014, 08:52:36 pm
There's a difference between de-ionised water and destilled water..But are they the same? Just difference processes?
If a question asked to write steps for like...cleaning equipment..which would I use?

Thanks

As far as I know, the purity of deionised water is higher than distilled water.
Also, you shouldn't necessarily clean equipment with either water. For example, a burette with which you will fill HCl solution in.
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on March 25, 2014, 09:10:21 pm
As far as I know, the purity of deionised water is higher than distilled water.
Also, you shouldn't necessarily clean equipment with either water. For example, a burette with which you will fill HCl solution in.
it's not like i'll be cleaning a concial flask with a solution?
Title: Re: VCE Chemistry Question Thread
Post by: Phenomenol on March 25, 2014, 09:18:14 pm
it's not like i'll be cleaning a concial flask with a solution?

If this is titration, and you're using the conical flask to hold your aliquot, correct. Clean it with water.
Title: Re: VCE Chemistry Question Thread
Post by: mikehepro on March 25, 2014, 10:04:38 pm
The trick here is to look at the masses of elements first, not amounts. Otherwise you're doing too much work.
Total mass(H) = 2.0/18.0 * 0.551g. This gives you the mass of hydrogen in the organic compound.

Total amount(CO2) = 0.458/22.4
Then find the mass of carbon in the CO2. This gives you the mass of carbon in the organic compound.

Simply subtract these masses from 0.470g to find the mass of oxygen in the compound. From there the routine method to find an empirical formula can be used.

So, the important thing to note is that all the carbon in the original compound is now in the form of CO2, and all the hydrogen in the original compound is now in the form of H2O.

Ah i get it now. Thank you for your help.
Really appreciate it.
Title: Re: VCE Chemistry Question Thread
Post by: Sanguinne on March 26, 2014, 05:51:37 pm
when turning a alkene into a chloroalkane, is AlCl3 necessary or can I use just use HCL as a catalyst?
Title: Re: VCE Chemistry Question Thread
Post by: darklight on March 26, 2014, 05:55:54 pm
Hi guys,

Say you are doing a volumetric analysis with the unknown solution being in the conical flask and the chemical with the known concentration is in the burette. What happens if the conical flask is rinsed by accident with the unknown solution?
Because you have more mol in the conical flask than expected, you'd need more titre from the burette, so you'd calculate a higher mol and hence higher concentration than expected. Is this right?

But what happens if you switch this around, and the known is in the conical and the unknown is in the burette, but you still wash the conical flask with the unknown? Because you have some of the unknown in the conical flask, it would already react with the known which would mean you would need less titre from the burette, which would mean that you would calculate a higher concentration right?

Is this reasoning right? Thanks :)
Title: Re: VCE Chemistry Question Thread
Post by: drake on March 26, 2014, 05:57:11 pm
when turning a alkene into a chloroalkane, is AlCl3 necessary or can I use just use HCL as a catalyst?

hey sanguinne, AlCl3 is not necessary. you just need HCl - but remember HCl is NOT a catalyst. it is simply a reactant!
Title: Re: VCE Chemistry Question Thread
Post by: Sanguinne on March 26, 2014, 06:26:45 pm
hey sanguinne, AlCl3 is not necessary. you just need HCl - but remember HCl is NOT a catalyst. it is simply a reactant!

oh ok
thanks   :D
Title: Re: VCE Chemistry Question Thread
Post by: Phenomenol on March 26, 2014, 07:10:54 pm
Hi guys,

Say you are doing a volumetric analysis with the unknown solution being in the conical flask and the chemical with the known concentration is in the burette. What happens if the conical flask is rinsed by accident with the unknown solution?
Because you have more mol in the conical flask than expected, you'd need more titre from the burette, so you'd calculate a higher mol and hence higher concentration than expected. Is this right?

But what happens if you switch this around, and the known is in the conical and the unknown is in the burette, but you still wash the conical flask with the unknown? Because you have some of the unknown in the conical flask, it would already react with the known which would mean you would need less titre from the burette, which would mean that you would calculate a higher concentration right?

Is this reasoning right? Thanks :)

Your reasoning is right. Good work.
Title: Re: VCE Chemistry Question Thread
Post by: darklight on March 26, 2014, 08:16:19 pm
Your reasoning is right. Good work.

Thank you! Another question:
If you rinsed the conical flask that contained the unknown solution with the known solution, this would cause a lower than expected conc. of the unknown solution because some of the known will react with the unknown prior to letting the titre run? Also, what is the effect when the burette which has a known solution is rinsed with the unknown solution?

thanks :)
Title: Re: VCE Chemistry Question Thread
Post by: Phenomenol on March 26, 2014, 09:08:10 pm
Thank you! Another question:
If you rinsed the conical flask that contained the unknown solution with the known solution, this would cause a lower than expected conc. of the unknown solution because some of the known will react with the unknown prior to letting the titre run? Also, what is the effect when the burette which has a known solution is rinsed with the unknown solution?

thanks :)

Some unknown will react prior to titration -> titre will be lower -> calculated amount of unknown is lower therefore lower than expected concentration of unknown. Yep.
Ughh, in the off chance that someone is stupid enough to do this (I doubt this scenario will ever come up), the concentration of the known solution will decrease -> titre will increase -> calculated amount of unknown is higher -> higher than expected concentration of unknown.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on March 26, 2014, 09:47:40 pm
Pentane boils at 36 degrees, while 2-methylbutane boils at 28 degrees. Explain why pentane has the higher boiling point.
Thanks :D
Title: Re: VCE Chemistry Question Thread
Post by: PB on March 26, 2014, 09:51:37 pm
2methyl butane has a methyl side group which prevents the molecules from getting close to each other. Pentane on the other hand, does not have this physical obstruction and hence, can get all nice and snuggly with each other.  The result is that the dispersion forces among the pentane would be a lot stronger because of the shorter distances between molecules.

Stronger dispersion forces = harder to evaporate using heat = higher boiling point.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on March 27, 2014, 05:08:20 pm
Devise reaction pathways for the following reactions:
a) ethylamine from ethene
Pls write why you did specific steps

Thanks ;D
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on March 28, 2014, 07:47:43 am
See the picture below.  Look at where you start. look at where you end up, thus you find the reactions to get you there.

Devise reaction pathways for the following reactions:
a) ethylamine from ethene
Pls write why you did specific steps

Thanks ;D
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on March 29, 2014, 09:50:00 pm
how many decimal places do we express Rf values to?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on March 30, 2014, 01:12:48 am
how many decimal places do we express Rf values to?

Usually 2.
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on March 30, 2014, 11:18:20 am
Usually 2.
so if get 8.777
then i'd express it as 8.78?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on March 30, 2014, 11:35:38 am
so if get 8.777
then i'd express it as 8.78?

Rf values are always below or equal to 1. But yes, if you got 0.777, that would be 0.78
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on March 30, 2014, 03:47:11 pm
Rf values are always below or equal to 1. But yes, if you got 0.777, that would be 0.78
LOL my bad wow hahahaha
thank you Yacoubb!!
Title: Re: VCE Chemistry Question Thread
Post by: maurlock on April 05, 2014, 09:41:15 pm
why will the ratio of n(base):n(acid) for a dicarboxylic acid reacting with NaOH be 2:1?
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on April 05, 2014, 10:03:34 pm
Acid:Base ratio is really a ratio of H+:OH-, A dicarboxylic acid has two COOH's that lose their hydrogens, sodium hydroxide only has one OH, so twice as much n(NaOH) is required.

thus the H+ : OH- is 1:2

why will the ratio of n(base):n(acid) for a dicarboxylic acid reacting with NaOH be 2:1?
Title: Re: VCE Chemistry Question Thread
Post by: maurlock on April 05, 2014, 10:05:46 pm
Thank you so much!
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 07, 2014, 08:45:11 pm
Can someone pls explain figure 12.13b from pg 185 of the heinemann textbook?
In the glycosidic linkage, where did the HOCH2 come from? I understand that there is a water molecule being released so that should just leave CH2 bonding with H shouldn't it? And oxygen for the glycosidic bond.
Hope you can make some sense of what I wrote here :P
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on April 08, 2014, 11:13:48 am
Can someone pls explain figure 12.13b from pg 185 of the heinemann textbook?
In the glycosidic linkage, where did the HOCH2 come from? I understand that there is a water molecule being released so that should just leave CH2 bonding with H shouldn't it? And oxygen for the glycosidic bond.
Hope you can make some sense of what I wrote here :P
I think the textbook got some mistakes with that fructose structure, as far as Im aware, isn't the OH supposed to be branching from the same carbon with an CH2OH ? thats why there an CH2OH in the next photo.
I think I figure out the problem here, there is a mistake in textbook, according to my 2010 version textbook, it looks exactly like your photo, however, according to another version of the textbook, it's CH2OH.
Hope this helps!
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 08, 2014, 11:21:46 am
I think the textbook got some mistakes with that fructose structure, as far as Im aware, isn't the OH supposed to be branching from the same carbon with an CH2OH ? thats why there an CH2OH in the next photo.
I think I figure out the problem here, there is a mistake in textbook, according to my 2010 version textbook, it looks exactly like your photo, however, according to another version of the textbook, it's CH2OH.
Hope this helps!

If that is the case, then the diagram makes sense. I think the diagram is wrong cause the CH2OH should be on the same branch as OH. Yeah, I'm pretty sure your're right. Thanks :D
Could someone pls confirm our theory
Title: Re: VCE Chemistry Question Thread
Post by: scribble on April 08, 2014, 11:34:12 am
i remember the diagram being wrong in my edition of the textbook. i wouldn't be surprised if they didnt correct it in the current edition.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 08, 2014, 05:03:20 pm
On a mass spectrum of ethanoic acid, is there no peak for OH- because it is an negative ion?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 08, 2014, 05:30:59 pm
I can't imagine how you'd form that negative ion. The electrons in the mass spectrometer are so energetic they can't be captured by molecules and radicals. That's why they knock out other electrons.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 08, 2014, 05:42:16 pm
I can't imagine how you'd form that negative ion. The electrons in the mass spectrometer are so energetic they can't be captured by molecules and radicals. That's why they knock out other electrons.

I'm a little confused

During the fragmentation process, there is a loss of CH3+ to form a COOH+ from ethanoic acid and both of these are on the mass spectrum

However, when there is a loss of OH (forming CH3CO+ then CH2CO+), the OH is not in the mass spectrum. Why?
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on April 08, 2014, 06:05:51 pm
How much of the theory behind instrumentation do we need to know? Or is it mainly application (reading spectrographs and interpreting them rather than explaining the theory behind it?)
Title: Re: VCE Chemistry Question Thread
Post by: Phenomenol on April 08, 2014, 06:23:25 pm
How much of the theory behind instrumentation do we need to know? Or is it mainly application (reading spectrographs and interpreting them rather than explaining the theory behind it?)

You need to have a rough idea of the principles, but not exactly how each apparatus works. For example, in mass spectroscopy, a magnetic field is used to differentiate between positively charged ions based on mass and charge. But, you won't need to know how the detector works, or how the electrons are fired to ionise molecules.
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on April 08, 2014, 06:34:28 pm
You need to have a rough idea of the principles, but not exactly how each apparatus works. For example, in mass spectroscopy, a magnetic field is used to differentiate between positively charged ions based on mass and charge. But, you won't need to know how the detector works, or how the electrons are fired to ionise molecules.
Thanks. How about things like degrees of freedom in IR?
Title: Re: VCE Chemistry Question Thread
Post by: Phenomenol on April 08, 2014, 06:38:13 pm
Thanks. How about things like degrees of freedom in IR?
I don't think so.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 08, 2014, 11:25:04 pm
How much of the theory behind instrumentation do we need to know? Or is it mainly application (reading spectrographs and interpreting them rather than explaining the theory behind it?)

I believe the study design states that you don't actually need to know anything about operations (e.g. how AAS works), but rather the interpretation of the qualitative and quantitative data produced by these instrumental analyses.
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on April 09, 2014, 04:24:29 pm
I saw an analogy being made between covalent bonds and springs, and thus stronger bonds and lighter atoms mean a higher frequency of vibration. Is it sufficient to say that say, a C=C bond vibrates at a higher frequency than say a C-C bond and therefore absorbs IR radiation of a higher freq/wavenumber?

Also is the reason carboxyllic acids dont exhibit the peak at roughly 2900 wavenumber due to the sp3 C-H bond because the broad and intense OH peak in that region masks it?
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 09, 2014, 04:33:51 pm
I saw an analogy being made between covalent bonds and springs, and thus stronger bonds and lighter atoms mean a higher frequency of vibration. Is it sufficient to say that say, a C=C bond vibrates at a higher frequency than say a C-C bond and therefore absorbs IR radiation of a higher freq/wavenumber?

Great question:
Basically, there are 2 factors affecting the frequency of radiation absorbed:
- Strength of the bond- as strength of bond increases (single --> double bond), frequency/energy increases. So the double bond does absorb more IR Radiation
- Mass of atoms in bond- As the masses increase, the freqency decreases (it will take a lot more energy to move bigger atoms so this makes sense)
Hope this helps :)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 09, 2014, 09:21:13 pm
I saw an analogy being made between covalent bonds and springs, and thus stronger bonds and lighter atoms mean a higher frequency of vibration. Is it sufficient to say that say, a C=C bond vibrates at a higher frequency than say a C-C bond and therefore absorbs IR radiation of a higher freq/wavenumber?

Also is the reason carboxyllic acids dont exhibit the peak at roughly 2900 wavenumber due to the sp3 C-H bond because the broad and intense OH peak in that region masks it?

Yeah. Think of absorption as a bit like springs. Essentially, the molecule bonds vibrate and they absorb radiation at the same frequency as the vibration frequency.

Yes. The OH peak masks it; if you have an oxygen atom in your compound but you don't see that masking but see a separate CH3 and OH absorption, you probably have an alcohol.
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on April 09, 2014, 09:37:57 pm
Thanks!

Can someone clarify the following (about NMR)? Correct to see if I've misunderstood anything etc. There probably is some irrelevant stuff; my teacher didn't teach it very well so I had to self learn a lot of it. Wall of text ahead.

General
• NMR spectroscopy uses an NMR instrument to induce a magnetic field.
• Protons, neutrons and electrons have spin, which can be +1/2 or -1/2. Spin is quantised; a nucleon or electron has to be one of the two aforementioned spin states.
• For a nucleus to be able to be analysed by NMR, it must be NMR active; that is, it must have a net nuclear spin: ie. it has an odd mass number. This is because if it had an even amount of nucleons, there would be no net nuclear spin as the spins of nucleons cancel out, leaving it overall neutral.
• NMR works by inducing resonance: nuclei in the lower energy state of being magnetically aligned with the magnetic field of the NMR instrument absorb energy and thus 'flip' and oppose the magnetic field. They then re-emit this energy and return to the lower energy state.
• Roughly 50% of nuclei are aligned both with and against the magnetic field. A slightly higher amount are aligned against the field, as it is a lower energy state
• No two NMR instruments (is this what they are called?) will have a magnetic field of exactly the same strength. Hence, we assign a chemical shift of 0 to tetramethylsilane (TMS) and measure it in concentration (ppm) rather than frequency as everything is standardised against TMS. TMS is symmetrical and thus produces only one peak, and its non-polar nature means it will appear on the far right (upfield/ on the shielded end) of the NMR spectrograph

Carbon 13 NMR
• Uses the carbon-13 isotope. Carbon-13 is pretty rare, it's abundance is something like 1%
• Spectrograph goes from 0-200ppm
• Number of peaks in carbon-13 NMR = number of different carbon environments
• Chemically equivalent carbon atoms are in the same carbon environment. This arises due to the symmetry of a molecule (CH3CH2CH3 has 2 carbon environments, one with 2 carbons and one with 1 carbon)
• Area under peaks is not proportional to the amount of carbon atoms in that particular environment

Proton NMR
• Uses the abundant hydrogen-1 isotope (something like 99% abundance).
• Spectrograph is usually 0-10ppm
• Number of peaks only indicative of number of hydrogen environments in low resolution proton NMR; in high resolution proton NMR peak splitting occurs, and a peak  may split into many subpeaks.
• Ratio of the area under the 'peaks'(where 1 peak may equal many subpeaks; one 'peak'=one hydrogen environment) is directly proportional to the ratio of hydrogen atoms in each environment.
• Chemical shift of protons determined by 3 effects:
• Magnetic arisotropy: this has the biggest affect on the chemical shift of a proton. Put simply, the magnetic field of nearby pi electrons in double bonds (ie. alkenes and arenes, but not alkynes(?)) acts in tandem with the NMR instrument's magnetic field, strengthening it. The effect of this is that more energy is required to flip the spin states of the protons, and hence radio waves of a higher frequency are required -> higher chemical shift. This effect diminishes the further away the proton is from the double bond.
• Inductive effects of nearby electronegative atoms: this has a smaller effect on the chemical shift of a proton. Electronegative elements hog electrons, and thus have a deshielding effect by diminishing the electron cloud around the proton. Hence, the proton is subject to more of the magnetic field's strength (no electrons to shield it) and thus more energy and thus a higher chemical shift is required for resonance
• Magnetic fields of nearby protons: this has the smallest effect on the chemical shift of a proton. Protons themselves act as magnets; they may have positive or negative spin (up or down spin?). Hence, a proton will be affected by the spin states of neighboring protons: if a proton neighboring a particular proton of interest, proton P, has a spin state is aligned with the magnetic field of the NMR instrument, then proton P will experience a slightly stronger magnetic field and thus require more energy to induce resonance, thus having a higher chemical shift. Conversely, if the neighboring proton's spin state is opposing the magnetic field of the NMR instrument, then proton P will experience a slightly weaker magnetic field and thus have a smaller chemical shift. This is the cause of peak splitting
• What is peak splitting? Peak splitting is when a peak is split into smaller subpeaks. Peak splitting occurs because protons themselves act as magnets. Hence, if a hydrogen nucleus has one neighboring hydrogen nucleus (in a different chemical environment), that neighboring hydrogen nucleus can either be aligned with or against the NMR's magnetic field. Hence, the peak splits into 2 subpeaks of roughly equal height (because the probability of that neighboring nucleus being aligned or against the NMR's magnetic field is approximately equal).
• 2 neighboring protons = 3 subpeaks in a 1:2:1 ratio. This is because the two neighboring protons can either both be against the field, both be aligned with the field, or one can be aligned and one can be against (for which there are 2 possible combinations, hence the 1:2:1 ratio)
• 3 neighboring protons = 4 subpeaks in a 1:3:3:1 ratio.
• In fact, there's a pattern. n neighboring protons will produce n+1 subpeaks: this is known as the n+1 rule. Note that chemically equivalent proteins don't cause splitting, nor do hydrogen bonding protons (eg, O-H, O-N). The ratio of the areas of these subpeaks can be modeled by Pascal's triangle.
• But why doesn't peak-splitting happen for carbon-13 NMR? The low abundance of carbon-13 means that it is very unlikely for carbon-13 atoms to be adjacent to eachother, hence they will not cause splitting.
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on April 10, 2014, 07:28:07 pm
to what extent do we need to know about how the equipment works in chromatography? e.g. how the columns work in gas chromatography, method of operation
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 10, 2014, 08:04:17 pm
to what extent do we need to know about how the equipment works in chromatography? e.g. how the columns work in gas chromatography, method of operation

Specifically mentions on the study design that you don't need to know this. :)
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 11, 2014, 09:49:26 am
In q 5c of 2012 chem exam 1, when drawing the structure, how can we determine on what carbon, the OH group is located?
Thanks :D
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on April 11, 2014, 10:43:19 am
A few ways to do this, one way, there are only 2 peaks in the carbon NMR, but 3 carbons in the molecule so there must be symmetry somewhere, only way symmetry can work with this molecule is 2-propanol.

Second way is using the 1H-NMR.  There is a septet with an area of 1, which means that there must be an environment with 1 hydrogen (CH) next to 6 neighbouring hydrogens (2x CH3).  The only way this can work is if it is 2-propanol.

CH3CH(OH)CH3

The H is the peak at 1.2ppm
The H is the peak at 3.6ppm

In q 5c of 2012 chem exam 1, when drawing the structure, how can we determine on what carbon, the OH group is located?
Thanks :D
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 11, 2014, 10:58:29 am
A few ways to do this, one way, there are only 2 peaks in the carbon NMR, but 3 carbons in the molecule so there must be symmetry somewhere, only way symmetry can work with this molecule is 2-propanol.

Second way is using the 1H-NMR.  There is a septet with an area of 1, which means that there must be an environment with 1 hydrogen (CH) next to 6 neighbouring hydrogens (2x CH3).  The only way this can work is if it is 2-propanol.

CH3CH(OH)CH3

The H is the peak at 1.2ppm

The H is the peak at 3.6ppm

Thanks heaps jgoudie...
btw, just wanted to say that all your videos are awesome and they have helped me sooo much. Just a quick question, on your biomolecules-fat video, I thought that the formula for an unsaturated fatty acid was CnH2n-2O2 not CnH2nO2. I'm sure that I'm making the mistake but could you please explain where.
Thanks
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 11, 2014, 11:21:12 am
For the synthesis of drugs using organic molecules, I'm aware that we need to know about the synthesis of acetyl-salicylic acid (aspirin) from salicylic acid. Do we also need to know about the production of penicillin? How about gene cloning?

Some clarification would be much appreciated!

Thank you!
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on April 11, 2014, 12:08:21 pm
Guys I'm doing uni Chem (but only fundamentals), can I ask a question here?

I'm really confused about the 'difference' between oxidation numbers and ionic charges.
It's clear to me what ionic charges are for standalone ions e.g. Cl-, K+, Ca2+ etc.
Now we have just learned oxidation numbers and their 'rules' which I understand is not the same as the ionic charge - fine.
BUT what seems to happen is that when figuring out the oxidation numbers for individual elements in a compound, sometimes they use the oxidation number 'rules' and sometimes they use the ionic charges.  Like my lecturer today said, for some compound that contained Cl, "The Chlorine has an oxidation number of -1 because it does in it's ionic state." but then thats not the case for other elements.
I'm REALLY confused about what to use when here, can anyone help?

Here's an example question that I found an answer to online.
Q: What is the oxidation number of iron in K4Fe(CN)6?
Well from the 'rules' we know that Group I elements will carry +1, so the K4 is +4.  That means the net charge of the Fe(CN)6 needs to be -4.
But then the answer goes on to say "CN has a -1 charge. It just does. Hence, CN is contributing -6. "

I know that the polyatomic ion CN has an ionic charge of -1 (CN-) but this just doesn't make sense to me, all of a sudden we are using the ionic charge for the polyatomic ion instead of the oxidation number?
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 11, 2014, 02:30:54 pm
Polyethenes are organic molecules that contain more than one carbon-to-carbon double bond. A particular polyene undergoes an addition reaction with bromine. The empirical formula of the product is C3H5Br2. The molecular formula of the polyene is likely to be:
a) C3H4
b) C3H5
c) C6H8
d) C6H10

I have no idea where to begin ???
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on April 11, 2014, 05:00:05 pm
Polyethenes are organic molecules that contain more than one carbon-to-carbon double bond. A particular polyene undergoes an addition reaction with bromine. The empirical formula of the product is C3H5Br2. The molecular formula of the polyene is likely to be:
a) C3H4
b) C3H5
c) C6H8
d) C6H10

I have no idea where to begin ???
The answer is D.
Now consider why we eliminate other solutions.
A and B, wrong since we can't see any double bonds there. C is still wrong eve though there is 3 double bonds, when C6H8 react with Br2, our product should be C6H8Br6 I.e C3H4B3.
So the answer is D, use the same method that we reason in answer C to prove that's it's true.
Hope this helps!
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 11, 2014, 07:09:21 pm
Guys I'm doing uni Chem (but only fundamentals), can I ask a question here? Sure! Especially as your question may be a concern to other VCE chemistry students

I'm really confused about the 'difference' between oxidation numbers and ionic charges.
It's clear to me what ionic charges are for standalone ions e.g. Cl-, K+, Ca2+ etc.
Now we have just learned oxidation numbers and their 'rules' which I understand is not the same as the ionic charge - fine.
BUT what seems to happen is that when figuring out the oxidation numbers for individual elements in a compound, sometimes they use the oxidation number 'rules' and sometimes they use the ionic charges.  Like my lecturer today said, for some compound that contained Cl, "The Chlorine has an oxidation number of -1 because it does in it's ionic state." but then thats not the case for other elements.
I'm REALLY confused about what to use when here, can anyone help?

Here's an example question that I found an answer to online.
Q: What is the oxidation number of iron in K4Fe(CN)6?
Well from the 'rules' we know that Group I elements will carry +1, so the K4 is +4.  That means the net charge of the Fe(CN)6 needs to be -4.
But then the answer goes on to say "CN has a -1 charge. It just does. Hence, CN is contributing -6. "

I know that the polyatomic ion CN has an ionic charge of -1 (CN-) but this just doesn't make sense to me, all of a sudden we are using the ionic charge for the polyatomic ion instead of the oxidation number?

I've decided to make a separate post in the VCE Chemistry forum about this question. Look for it there (:
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 12, 2014, 01:46:19 pm
Why is the following statement false ? :

Propene undergoes a polymerisation reaction expelling water in the process
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 12, 2014, 01:52:44 pm
Why is the following statement false ? :

Propene undergoes a polymerisation reaction expelling water in the process

Because in polymerisation of propene monomers, all that is occurring is that the double bond breaks to then join the monomers together. Water is not eliminated because there is no way for water to form (there is no hydroxyl, carboxyl, etc groups that react to form water). Thus propene polymerisation is not a condensation reaction.
Title: Re: VCE Chemistry Question Thread
Post by: Scooby on April 12, 2014, 01:59:14 pm
Why is the following statement false ? :

Propene undergoes a polymerisation reaction expelling water in the process

Addition polymerisation (eg. polymerisation of alkenes) doesn't liberate water, unlike condensation polymerisation
Title: Re: VCE Chemistry Question Thread
Post by: Jawnle on April 12, 2014, 04:48:29 pm
Hey guys, please clarify this for me. Thanks

Primary structure of DNA -  Is simply a straight chain of bases covalently bonded to deoxyribose molecules which are linked together by phosphate groups.
Secondary structure of DNA - When two base pairs go through a hydrogen bond with each other.
Tertiary structure of DNA - ?????????????????????????????????????????
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 12, 2014, 05:05:59 pm
Hey guys, please clarify this for me. Thanks

Primary structure of DNA -  Is simply a straight chain of bases covalently bonded to deoxyribose molecules which are linked together by phosphate groups.
Secondary structure of DNA - When two base pairs go through a hydrogen bond with each other.
Tertiary structure of DNA - ?????????????????????????????????????????

• Primary structure of DNA: the specific sequence of nucleotide bases of a polynucleotide strand. A nucleotide base consists of a nitrogenous base (A, T, G or C), linked covalently to a phosphate group and a deoxyribose (pentose sugar) part.

• Secondary structure of DNA: this refers to how two anti-parallel polynucleotide strands are linked by weak hydrogen bonding.

For VCE Chemistry, you only need to know about primary & secondary structures.
Title: Re: VCE Chemistry Question Thread
Post by: thushan on April 12, 2014, 05:19:36 pm
You must know about tertiary structures too. It is the overall 3-D structure of the protein due to interactions between Z groups in each of the amino acid residues. Such interactions include hydrogen bonding (eg. between OH groups between serine residues), ionic interactions between COO- and NH3+ groups in side chains and disulfide bridges between cysteine residues.
Title: Re: VCE Chemistry Question Thread
Post by: drake on April 12, 2014, 05:42:24 pm
i'm pretty sure the tertiary structure of DNA is the ionic interaction between DNA (the negatively charged phosphate groups) and histone molecules? i think thushan is talking about protein tertiary structure.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 12, 2014, 05:49:45 pm
i'm pretty sure the tertiary structure of DNA is the ionic interaction between DNA with histone molecules? i think thushan is talking about protein tertiary structure.

is that what the point in the study design: "the role of teritary structures of proteins in enzyme action" is referring to?

Also, do we need to know 'how DNA control biochmeical processes' and 'forensic applications'? I can't find it on the study design
Title: Re: VCE Chemistry Question Thread
Post by: vox nihili on April 12, 2014, 05:58:23 pm
i'm pretty sure the tertiary structure of DNA is the ionic interaction between DNA with histone molecules? i think thushan is talking about protein tertiary structure.
is that what the point in the study design: "the role of teritary structures of proteins in enzyme action" is referring to?

Also, do we need to know 'how DNA control biochmeical processes' and 'forensic applications'? I can't find it on the study design

Primary: sequence
Secondary: base pairing
Tertiary: helix

I would highly doubt that you do need to know that, Blondie21
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 12, 2014, 06:07:52 pm
Primary: sequence
Secondary: base pairing
Tertiary: helix

I would highly doubt that you do need to know that, Blondie21

Okay phew. Thanks :)))

Is the reaction between salicylic acid and ethanoic anhydride a substiution reaction?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 12, 2014, 07:48:36 pm
You must know about tertiary structures too. It is the overall 3-D structure of the protein due to interactions between Z groups in each of the amino acid residues. Such interactions include hydrogen bonding (eg. between OH groups between serine residues), ionic interactions between COO- and NH3+ groups in side chains and disulfide bridges between cysteine residues.

Yeah thushan DNA was being referred to, not proteins :)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 13, 2014, 01:40:49 am
Okay phew. Thanks :)))

Is the reaction between salicylic acid and ethanoic anhydride a substiution reaction?

Yes, it's a substitution. Esterification reactions are a special class of substitution in which the alcohol reacting with the acid substitutes for the OH group. The alcohol's original OH hydrogen is then transferred to the leaving OH group, giving the familiar condensation reaction.
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on April 13, 2014, 03:06:10 pm
Do dispersion forces exist between the compound and the mobile phase in gas chromatography?
Title: Re: VCE Chemistry Question Thread
Post by: vox nihili on April 13, 2014, 04:34:14 pm
Do dispersion forces exist between the compound and the mobile phase in gas chromatography?

If two things are coming into contact, odds on there are Van der Waals forces.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 13, 2014, 05:22:36 pm
Do we need to know the replication of DNA for chem? Its in ch 13 pg 209
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on April 13, 2014, 05:24:01 pm
If two things are coming into contact, odds on there are Van der Waals forces.
Yeah thats what I thought but the AN study guide for chem says they dont exist
Title: Re: VCE Chemistry Question Thread
Post by: thushan on April 13, 2014, 05:54:37 pm
Not in gas chromatography. The only reason that the sample is being swept along the column is because the mobile gas molecules are literally colliding with and knocking the sample molecules along the column. Theoretically, there would be van der Waals forces when the particles are close, but remember that the molecules would be passing one another at such high speed - we're talking gas here - that they would only be close to one another for a  fleeting amount of time.
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on April 13, 2014, 06:25:12 pm
Can someone tell me how CCl3CH2CCl3 is non polar, it looks like the molecule is pretty symmetrical
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on April 13, 2014, 06:33:54 pm
Not in gas chromatography. The only reason that the sample is being swept along the column is because the mobile gas molecules are literally colliding with and knocking the sample molecules along the column. Theoretically, there would be van der Waals forces when the particles are close, but remember that the molecules would be passing one another at such high speed - we're talking gas here - that they would only be close to one another for a  fleeting amount of time.
Ah fair enough, thanks
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 13, 2014, 06:35:26 pm
Can someone tell me how CCl3CH2CCl3 is non polar, it looks like the molecule is pretty symmetrical

If a molecule is symmetrical, it's non polar. I think you got a bit confused...
Remember this acronym SNAP
Symmetrical Non-polar Asymetrical Polar
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on April 13, 2014, 06:39:23 pm
If a molecule is symmetrical, it's non polar. I think you got a bit confused...
Remember this acronym SNAP
Symmetrical Non-polar Asymetrical Polar

Oh wow haha thanks :p
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 13, 2014, 08:03:58 pm
Sorry for the horrible drawing
Title: Re: VCE Chemistry Question Thread
Post by: drake on April 13, 2014, 08:06:59 pm
1-amino-3-methylbutane
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 13, 2014, 08:09:53 pm
1-amino-3-methylbutane

So does the amine group have to be on the smallest number of carbon or the methyl group?
Title: Re: VCE Chemistry Question Thread
Post by: saba.ay on April 13, 2014, 08:39:10 pm
So does the amine group have to be on the smallest number of carbon or the methyl group?

The amine has to be on the smallest Carbon. When naming, you work your way down the following list, giving the top one the smallest Carbon number:

• COOH group
• OH group
• NH2 group
• C=C double bond
• halo substituents (Cl, F, Br, I)
• alkyl groups (methyl, ethyl, etc)
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 13, 2014, 09:30:58 pm
The amine has to be on the smallest Carbon. When naming, you work your way down the following list, giving the top one the smallest Carbon number:

• COOH group
• OH group
• NH2 group
• C=C double bond
• halo substituents (Cl, F, Br, I)
• alkyl groups (methyl, ethyl, etc)

omg that's such a great tool. Thanks so much :)
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on April 14, 2014, 12:11:17 am
When converting an alkane to a carboxylic acid (eg ethane to ethanoic acid), why must we react it firstly with chlorine to form a chloroalkane(chloroethane), to then be reacted with h3po4 and oh- to form an alkanol(ethanol), to then be reacted with cr2o72- and h+ to form ethanoic acid?

cant we create ethanol from ethane in one reaction?
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on April 14, 2014, 04:36:44 am
When converting an alkane to a carboxylic acid (eg ethane to ethanoic acid), why must we react it firstly with chlorine to form a chloroalkane(chloroethane), to then be reacted with h3po4 and oh- to form an alkanol(ethanol), to then be reacted with cr2o72- and h+ to form ethanoic acid?

cant we create ethanol from ethane in one reaction?
Keep in mind that alkane is HIGHLY UNREACTIVE since it is a saturated hydrocarbon. Hence, there's really little chances for it to react with water to form alkanol like alkene does, the only way we can alter it is to use UV light to split the Cl2, forming a very high electronegativity ion that can attack and substitute one of the H+ on alkane. After that, we get a halo alkane, which can  undergo substitution reaction with NaOH for example to form alkanol, note that only then, the chlorine ion on the halo alkane creates a nucleophilic region (I.e possess a higher polarity degree) that the Cl- can be substituted with an OH-.
Hope this helps!
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on April 14, 2014, 09:54:12 am
Thank you so much :)
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 14, 2014, 11:07:37 am
What questions does VCAA typically ask about DNA?
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 14, 2014, 12:10:24 pm
Do we need to know the replication of DNA for chem? Its in ch 13 pg 209.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 14, 2014, 12:25:19 pm
Do we need to know the replication of DNA for chem? Its in ch 13 pg 209.

Nop!
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 14, 2014, 02:51:20 pm
Nop!

So we need to know nothing about protein synthesis?
eg:
DNA unzips
mRNA strand is synthesized to match DNA triplet code
mRNA travels outside nucleus to ribosomes in cytoplasm of cell.
tRNA matching the mRNA brings amino acids together to make the protein chain.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 14, 2014, 03:01:31 pm
Is the following statement important for vce?
- Two hydrogen bonds can form between the base pairs adenine and thymine. Three hydrogen bonds form betweem cytosine and guanine
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 14, 2014, 03:08:18 pm
So we need to know nothing about protein synthesis?
eg:
DNA unzips
mRNA strand is synthesized to match DNA triplet code
mRNA travels outside nucleus to ribosomes in cytoplasm of cell.
tRNA matching the mRNA brings amino acids together to make the protein chain.

Nup

Is the following statement important for vce?
- Two hydrogen bonds can form between the base pairs adenine and thymine. Three hydrogen bonds form betweem cytosine and guanine

Yup!
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 14, 2014, 04:53:48 pm
If ethanol is being made from chloroethane, can NaOH be used as a catalyst?
In the answers, it uses OH as the cataylst

Thanks
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 14, 2014, 05:32:41 pm
If ethanol is being made from chloroethane, can NaOH be used as a catalyst?
In the answers, it uses OH as the cataylst

Thanks

You could use NaOH or KOH.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 14, 2014, 06:47:46 pm
You could use NaOH or KOH.

ok thanks. But is the OH binds with the alkyl group, where does the Na go? It's never included in any of the equations from the textbook
Title: Re: VCE Chemistry Question Thread
Post by: vox nihili on April 14, 2014, 07:10:14 pm
ok thanks. But is the OH binds with the alkyl group, where does the Na go? It's never included in any of the equations from the textbook

The Na just floats around, the OH too. Remember, they're aqueous.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 14, 2014, 07:13:21 pm
The Na just floats around, the OH too. Remember, they're aqueous.

So we don't need to include Na in our equations?
Only if it asks for a balanced equation yeah
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on April 14, 2014, 07:38:27 pm
So we don't need to include Na in our equations?
Only if it asks for a balanced equation yeah
What happens in that reaction is the Cl- possesses a very high electronegativity, hence, it will attract to the Na+ in the NaOH and create NaCl. The OH- now left alone will come into substituting the place of Cl- and bond with the partial positive Carbon
Hence, the whole equation is for e.g.:
CH3Cl(g) + NaOH (aq) -> CH3OH + NaCl (aq)
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 14, 2014, 07:44:24 pm
What happens in that reaction is the Cl- possesses a very high electronegativity, hence, it will attract to the Na+ in the NaOH and create NaCl. The OH- now left alone will come into substituting the place of Cl- and bond with the partial positive Carbon
Hence, the whole equation is for e.g.:
CH3Cl(g) + NaOH (aq) -> CH3OH + NaCl (aq)

Oh that makes much more sense. Thanks :)
But writing the NaCl is not compulsory is it? Will I still get full marks without it?
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on April 14, 2014, 07:46:30 pm
Oh that makes much more sense. Thanks :)
But writing the NaCl is not compulsory is it? Will I still get full marks without it?
No worries. I don't know about the marking scheme, but if I write the reactant as NaOH, I will include NaCl. But if I write the reactant just as OH-, then I won't write NaCl.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 14, 2014, 08:40:45 pm
How would I draw a structural formula to show the formation of a disaccharide from 2 fructose molecules?
I'm confused all the parts in the fructose molecule where the bonding occurs, there is not going to be any glycosidic bond as there is nothing to bond.
I hope what I'm trying to say makes sense :P Probs doesn't but a diagram would really help
Thanks :D
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 14, 2014, 09:11:57 pm
When an acidic solution is diluted, does the pH increase or decrease? And pls explain why
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 14, 2014, 09:19:45 pm
If ethanol is being made from chloroethane, can NaOH be used as a catalyst?
In the answers, it uses OH as the cataylst

Thanks

What happens in that reaction is the Cl- possesses a very high electronegativity, hence, it will attract to the Na+ in the NaOH and create NaCl. The OH- now left alone will come into substituting the place of Cl- and bond with the partial positive Carbon
Hence, the whole equation is for e.g.:
CH3Cl(g) + NaOH (aq) -> CH3OH + NaCl (aq)

The electronegativity bit is correct. However, note that your product is NaCl (aq). In other words, your Cl- doesn't really bind to Na+.
What really happens is that the C-Cl bond is polar, with the carbon slightly positive. It therefore attracts a fully negative hydroxide ion. As Cl- is a weaker base than OH-, we can understand why Cl- is more stable than OH-, explaining why OH- substitutes for Cl- and not the other way around.
The OH- is actually consumed in the reaction; it's not a catalyst.

When an acidic solution is diluted, does the pH increase or decrease? And pls explain why

Acidic solution diluted => pH must increase. Sure, if it's a weak acid the percentage ionisation increases, but this increase in % ionisation predicted by Le Chatelier's principle cannot fully offset the volume increase and dilution, so the concentration of H+ decreases overall and the pH increases.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 15, 2014, 10:53:41 am
For q 4 SA in the 2011 exam 1, how do we know what the equation is?
Thanks
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 15, 2014, 11:58:10 am
For q 4 SA in the 2011 exam 1, how do we know what the equation is?
Thanks

You don't need an equation.

Mass of precipitate = 4.141g
Molar mass = 245.3g/mol
Therefore number of moles = 0.0168813697513252

There is one P in the precipitate, and two in P2O5

Thus, n(precipitate) x 1/2 = n(P2O5) = 0.008440684875662
Molar Mass of P2O5 = 142 g/mol
Thus, mass = 1.198577252344069

% Phosphorus =  (1.198577252344069/3.256) * 100 = 36.81% is your answer
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 15, 2014, 12:41:18 pm
You don't need an equation.

Mass of precipitate = 4.141g
Molar mass = 245.3g/mol
Therefore number of moles = 0.0168813697513252

There is one P in the precipitate, and two in P2O5

Thus, n(precipitate) x 1/2 = n(P2O5) = 0.008440684875662
Molar Mass of P2O5 = 142 g/mol
Thus, mass = 1.198577252344069

% Phosphorus =  (1.198577252344069/3.256) * 100 = 36.81% is your answer

Thanks heaps for that:) Just a clarification, in how did u know to multiply by 1/2 instead of 2? Isn't it unknown/known?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 15, 2014, 12:48:13 pm
Thanks heaps for that:) Just a clarification, in how did u know to multiply by 1/2 instead of 2? Isn't it unknown/known?

Because n(P2O5) x 2 = n(P)
Therefore, if we know the n(P) in the precipitate, dividing that by 2 (i.e. multiplying by 1/2) gives us n(P2O5).
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on April 15, 2014, 06:07:56 pm
when should i use the double bond equivalent formula?
Title: Re: VCE Chemistry Question Thread
Post by: katiesaliba on April 15, 2014, 07:32:40 pm
In gas-liquid chromatography, does the sample dissolve instead of adsorb? Does adsorption only occur between the high-boiling point hydrocarbon and the porous glass beads?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 15, 2014, 07:50:17 pm
In gas-liquid chromatography, does the sample dissolve instead of adsorb? Does adsorption only occur between the high-boiling point hydrocarbon and the porous glass beads?

Are we required to know this?? I mean, the difference between gas-liquid and gas-solid chromatography. I know it generally; so that the sample is vaporised, and that the components are swept through the column by the carrier gas.
Title: Re: VCE Chemistry Question Thread
Post by: katiesaliba on April 15, 2014, 08:01:28 pm
Are we required to know this?? I mean, the difference between gas-liquid and gas-solid chromatography. I know it generally; so that the sample is vaporised, and that the components are swept through the column by the carrier gas.

I'm not entirely sure. I just want to have a thorough understanding of the concept :)
Title: Re: VCE Chemistry Question Thread
Post by: thushan on April 15, 2014, 08:02:03 pm
Sorry for the horrible drawing

Careful drake - the amino group here is the principal functional group. The name of the compound is 3-methylbutan-1-amine. You only use the prefix 'amino-' when there is another principal functional group with a higher priority, that being COOH and OH groups.
Title: Re: VCE Chemistry Question Thread
Post by: thushan on April 15, 2014, 08:04:13 pm
In gas-liquid chromatography, does the sample dissolve instead of adsorb? Does adsorption only occur between the high-boiling point hydrocarbon and the porous glass beads?

Very good pickup Katie. Technically, in gas-liquid chromatography, the sample dissolves into the stationary phase, because the stationary phase is a liquid.

http://www.chemguide.co.uk/analysis/chromatography/gas.html

ChemGuide refers to dissolution of the sample in the stationary phase in gas-liquid chromatography.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 15, 2014, 08:05:01 pm
I'm not entirely sure. I just want to have a thorough understanding of the concept :)

Very good pickup Katie. Technically, in gas-liquid chromatography, the sample dissolves into the stationary phase, because the stationary phase is a liquid.

http://www.chemguide.co.uk/analysis/chromatography/gas.html

ChemGuide refers to dissolution of the sample in the stationary phase in gas-liquid chromatography.

Thanks thushan :)
Title: Re: VCE Chemistry Question Thread
Post by: katiesaliba on April 16, 2014, 12:58:02 am
In TLC, can the partition of a sample be measured in frequency? For example, if I were to compare the adsorption and desorption of a polar amino acid to a less polar amino acid, could I say that the polar amino acid partitions itself less frequently between both phases? Is partition even the right term? I've confused myself  :(

(not entirely sure if my example is right either...)
Title: Re: VCE Chemistry Question Thread
Post by: rhinwarr on April 16, 2014, 08:28:40 pm
In a redox titration using a coloured ion eg. permanganate, would the end point be exactly at the equivalence point?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 16, 2014, 08:33:25 pm
In TLC, can the partition of a sample be measured in frequency? For example, if I were to compare the adsorption and desorption of a polar amino acid to a less polar amino acid, could I say that the polar amino acid partitions itself less frequently between both phases? Is partition even the right term? I've confused myself  :(

(not entirely sure if my example is right either...)

You can say this :) my teacher has instructed me to use 'partitioning' between the stationary and mobile phase when describing and comparing two components of a sample and how frequently adsorption to and from the stationary phase occurs.

In a redox titration using a coloured ion eg. permanganate, would the end point be exactly at the equivalence point?

Many redox titrations are self-indicating. In theory, the end point is the equivalence point; that is, when the reactants are in their equivalent stoichiometric mole ratios, the indicator (or in this case the solution itself) changes colour simultaneously. In practice, the end point occurs right after the equivalence point.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 17, 2014, 09:35:53 pm
Do we need to know what 1,3 butadiene is for vce chem

Also, which carbon is 'carbon number on' on a benzene molecule? In terms of naming?
Title: Re: VCE Chemistry Question Thread
Post by: vox nihili on April 17, 2014, 09:53:27 pm
Do we need to know what 1,3 butadiene is for vce chem

Also, which carbon is 'carbon number on' on a benzene molecule? In terms of naming?

You should be able to draw 1,3-butadiene from your organic stuff. You don't have to know it per se, but you should be able to read that name and draw the structure.

Arbitrary. If it's attached to anything, normally the biggest group attached will be at C1.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 17, 2014, 10:41:17 pm
You should be able to draw 1,3-butadiene from your organic stuff. You don't have to know it per se, but you should be able to read that name and draw the structure.

Arbitrary. If it's attached to anything, normally the biggest group attached will be at C1.

what does 'diene' mean?

edit; means that there are two double bonds  :-)))
Title: Re: VCE Chemistry Question Thread
Post by: vox nihili on April 17, 2014, 10:59:37 pm
(http://i62.tinypic.com/2mgnm6r.jpg)

For some reason it's tilted, I have no idea why. Also apologies for UniMelb pad, it was the only one I had!
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 17, 2014, 11:05:34 pm
ohh makes sense. Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on April 18, 2014, 09:54:30 am
REDUCTION POTENTIALS AND GALVANIC CELL REACTIONS

Can anyone help with this, I'm super confused (this is uni Chem but I think it's the same stuff as VCE 3/4).

Question: Use the standard reduction potential table to predict the outcome of the following reactions in an aqueous solution.  If no reaction occurs, write NR.  If reaction does occur, write a balanced equation.

c) Ag+ + Fe -->

So, here's my logic/thought process.

From the table:
Ag+ + e- <--> Ag (s)  /  Eo = +0.8 V
Fe2+ + 2e- <--> Fe (s)  / Eo = -0.44 V

So far I see nothing wrong, looks like Ag+ is going to be the oxidant and undergo reduction?

A bit of swapsies and balancing:

2Ag+ + 2e- <--> 2Ag (s)  /  Eo = +0.8 V
Fe (s) <--> Fe2+ + 2e-   / Eo = -0.44 V

Eocell = Eooxidant - Eoreductant
Eocell = +0.8 - -0.44 = +1.24 V

This is a positive value and this the reaction will be spontaneous and will occur according to the following equation:

2Ag+ + Fe --> Fe2+ + 2Ag

So that's my thought process.... however, apparently there will be no reaction (according to the textbook).  Can anyone tell me what I'm doing wrong?
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on April 18, 2014, 11:05:34 am
Similar topic, I can't seem to get my head around these half-reactions when they get a bit more complicated.

Question: Write cell diagrams* using the following pairs of half-reactions.  For half-rections in which all the reactants are in solution, assume the use of inert platinum electrodes.  Identify the anode and the cathode.

f)
NO3- (aq) + 4H+ + 3e- <--> NO(g) + 2H2O (l)
MnO4- (aq) + 8H+ (aq) + 5e- <--> Mn2+ + 4H2O (l)

anode ---> Pt | NO (g), NO3- || MnO4-, Mn2+ | Pt <-- cathode

anode ---> Pt | H+ (aq), NO3- || MnO4-, Mn2+, H+ (aq) | Pt <-- cathode

I guess I'm just generally confused... something I'm missing... like where is the NO (g).... what part are the H+ ions playing here....?

*(Do you do cell diagrams in VCE?  The are written like: anode | electrocyte at anode || electrolyte at cathode | cathode
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 18, 2014, 04:51:03 pm
What is the name for H3C-O-CH3 and can you please explain how you found this answer?
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on April 18, 2014, 05:08:29 pm
What is the name for H3C-O-CH3 and can you please explain how you found this answer?
That would be dimethyl ether for the systematic name. Simply because you got 2 methyl groups sticking out from a -O- (which is an ether).
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 18, 2014, 05:14:52 pm
That would be dimethyl ether for the systematic name. Simply because you got 2 methyl groups sticking out from a -O- (which is an ether).

Do you know why it would also be called Methoxymethane?
Title: Re: VCE Chemistry Question Thread
Post by: thushan on April 18, 2014, 05:15:29 pm
What is the name for H3C-O-CH3 and can you please explain how you found this answer?

Methoxymethane. Dimethyl ether is also an acceptable name.

You aren't expected to name ethers in the VCE course.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 18, 2014, 05:19:03 pm
Methoxymethane. Dimethyl ether is also an acceptable name.

You aren't expected to name ethers in the VCE course.

Thanks :)

Do we need to know about the carbonyl functional group, aldehydes and kentones?
Title: Re: VCE Chemistry Question Thread
Post by: Aurelian on April 18, 2014, 05:32:54 pm
Thanks :)

Do we need to know about the carbonyl functional group, aldehydes and kentones?

You are not required to be able to name aldehydes and ketones, but you do need to know what they are, especially in the context of the oxidation of alcohols (which can generate aldehydes, ketones and/or carboxylic acids).
Title: Re: VCE Chemistry Question Thread
Post by: thushan on April 18, 2014, 07:09:31 pm
You are not required to be able to name aldehydes and ketones, but you do need to know what they are, especially in the context of the oxidation of alcohols (which can generate aldehydes, ketones and/or carboxylic acids).

I disagree - I think it's helpful to know what they are to get a better picture of what goes on, but that is quite peripheral knowledge, I highly doubt that that is assessable.
Title: Re: VCE Chemistry Question Thread
Post by: Aurelian on April 18, 2014, 07:11:57 pm
I disagree - I think it's helpful to know what they are to get a better picture of what goes on, but that is quite peripheral knowledge, I highly doubt that that is assessable.

? I also recommended knowing what they are? :S
Title: Re: VCE Chemistry Question Thread
Post by: Jawnle on April 19, 2014, 12:47:15 pm
What's the theory behind a non-polar solvent dissolving in a non-polar solute?

I understand how "like dissolves in like", and how polar solvents can dissolve in polar solutes because of their partial pos/neg charges but what about for non-polar, when all the charges are equal?
Thanks
Title: Re: VCE Chemistry Question Thread
Post by: thushan on April 19, 2014, 01:09:05 pm
Non-polar substances can dissolve in other non-polar solvents because the two substances exhibit dispersion forces with each other.
Title: Re: VCE Chemistry Question Thread
Post by: Jawnle on April 19, 2014, 01:15:52 pm
Non-polar substances can dissolve in other non-polar solvents because the two substances exhibit dispersion forces with each other.

Is it because of the instantaneous dipoles created from dispersion forces?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 19, 2014, 02:44:36 pm
What's the theory behind a non-polar solvent dissolving in a non-polar solute?

I understand how "like dissolves in like", and how polar solvents can dissolve in polar solutes because of their partial pos/neg charges but what about for non-polar, when all the charges are equal?
Thanks

Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on April 19, 2014, 02:46:05 pm
Is the base peak the same as the parent molecular ion in mass spectroscopy?
Also is the base peak/parent molecular ion the one with the highest intensity?

Thanks
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 19, 2014, 02:50:20 pm
Is the base peak the same as the parent molecular ion in mass spectroscopy?
Also is the base peak/parent molecular ion the one with the highest intensity?

Thanks

Base peak = peak with highest intensity
Which isn't always the parent molecular ion (it can be, no guarantee)
Parent molecular ion is the peak from the molecule you're analysing (i.e. the one before all of that fragmentation)
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 19, 2014, 05:42:37 pm
I don't understand the formula for finding the number of double bonds in a molecule. I would normally draw the molecule but this methods seems quicker.

Anyways, to find C3H5Cl, I would:

(7) - (6) + 0
-----------------
2

Even though it should have one double bond, not 1/2

What am I doing wrong?
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on April 19, 2014, 05:58:13 pm
I never got my head around dilution, so can someone clarify? :/

So I have a stock standard solution: 1000mg/100mL  of caffeine
I want to make concentrations of 5mg/100ml, 10mg/100ml and 20mg/100ml caffeine.

Say for the first concentration I want to obtain (5mg/100ml), do i simply get like 0.5mL of the standard (5mg/0.5mL), and then dilute it into 100mL?
Is that how it works? o.o
Or is my logic wrong. hahhaha

Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 19, 2014, 06:16:45 pm
I don't understand the formula for finding the number of double bonds in a molecule. I would normally draw the molecule but this methods seems quicker.

Anyways, to find C3H5Cl, I would:

(7) - (6) + 0
-----------------
2

Even though it should have one double bond, not 1/2

What am I doing wrong?

Hey

I have attached the actual formula. I think you were forgetting the a+1
a = number of Carbon atoms in the compound
b = number of Hydrogen atoms in the compound
c = number of Nitrogen atoms in the compound
f = number of Halogen atoms in the compound
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on April 19, 2014, 06:35:33 pm
I never got my head around dilution, so can someone clarify? :/

So I have a stock standard solution: 1000mg/100mL  of caffeine
I want to make concentrations of 5mg/100ml, 10mg/100ml and 20mg/100ml caffeine.

Say for the first concentration I want to obtain (5mg/100ml), do i simply get like 0.5mL of the standard (5mg/0.5mL), and then dilute it into 100mL?
Is that how it works? o.o
Or is my logic wrong. hahhaha

Thanks!
So initially, you have 1000mg/100mL, then after dilution, you have 5 mg/100 mL, you can see that 1000/5 = 200 . If concentration decrease by 200 folds, volume must have increased by 200 folds. Hence, 100 x 200 = 20 000 mL i.e if you fill the volume up to 20 00mL (adding 19 900 mL), you get a concentration of 5 mg/100mL.
Title: Re: VCE Chemistry Question Thread
Post by: rhinwarr on April 19, 2014, 06:58:17 pm
Do double bonds count as functional groups?
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 19, 2014, 07:00:32 pm
Do double bonds count as functional groups?

No  I believe they are just a characteristic that determines between alkanes and alkenes.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 19, 2014, 07:04:12 pm
No  I believe they are just a characteristic that determines between alkanes and alkenes.

I disagree - Double bonds between carbons atoms are funcitonal groups and are prioritised before halogens and alkyl branches when naming molecules :-)
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 19, 2014, 07:08:54 pm

I disagree - Double bonds between carbons atoms are funcitonal groups and are prioritised before halogens and alkyl branches when naming molecules :-)

Here's what the textbook said
A functional group is an atom or group of atoms which influences the chemical properties of an organic compound. Would you count a double bond as an atom?
Title: Re: VCE Chemistry Question Thread
Post by: Jawnle on April 19, 2014, 07:09:12 pm
No  I believe they are just a characteristic that determines between alkanes and alkenes.

I agree with yoshi97
Title: Re: VCE Chemistry Question Thread
Post by: Jawnle on April 19, 2014, 07:13:07 pm
Here's what the textbook said
A functional group is an atom or group of atoms which influences the chemical properties of an organic compound. Would you count a double bond as an atom?

I think an alkene is still considered a hydrocarbon (like an alkane) and hydrocarbons aren't functional groups...right?
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 19, 2014, 07:22:32 pm
I agree with yoshi97

yoshi97 ???
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 19, 2014, 07:24:48 pm
I think an alkene is still considered a hydrocarbon (like an alkane) and hydrocarbons aren't functional groups...right?

Yeah their not, hydrocarbons consist of functional groups.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 19, 2014, 07:38:33 pm
Because they are on the list of priorities (when naming) I always thought they were a functional group haha
Title: Re: VCE Chemistry Question Thread
Post by: Jawnle on April 19, 2014, 07:51:17 pm
for the reaction between butane and chlorine, I'd get a chlorobutane. How do I know the states of butane and chlorobutane?
Title: Re: VCE Chemistry Question Thread
Post by: Scooby on April 19, 2014, 07:52:46 pm
Yeah, it is a functional group. It's called an alkenyl group
Title: Re: VCE Chemistry Question Thread
Post by: Jawnle on April 19, 2014, 08:01:25 pm
Yeah, it is a functional group. It's called an alkenyl group

From the tsfx booklet
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 19, 2014, 08:03:00 pm
From the tsfx booklet

Alkyl and alkenyl are different lols
Title: Re: VCE Chemistry Question Thread
Post by: Scooby on April 19, 2014, 08:04:32 pm
From the tsfx booklet

Yeah but it's not an alkyl group
Title: Re: VCE Chemistry Question Thread
Post by: Scooby on April 19, 2014, 08:05:32 pm
And a carbonyl group is definitely a functional group so idk what they're on about
Title: Re: VCE Chemistry Question Thread
Post by: Jawnle on April 19, 2014, 08:10:02 pm
Yeah but it's not an alkyl group

Soz my bad haha didnt read properly
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on April 19, 2014, 08:41:14 pm
for the reaction between butane and chlorine, I'd get a chlorobutane. How do I know the states of butane and chlorobutane?
I don't think they assess states for organic chem
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 19, 2014, 09:08:46 pm
Finish this sentence (This is about DNA)

The encricled group is called a ___________________.
Title: Re: VCE Chemistry Question Thread
Post by: vox nihili on April 19, 2014, 09:24:36 pm
Finish this sentence (This is about DNA)

The encricled group is called a ___________________.

This sounds like there's a photo missing... hahaha
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 19, 2014, 10:26:22 pm
This sounds like there's a photo missing... hahaha

oh hahah I thought 'encricled' might be a chem term LOL
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 20, 2014, 10:17:50 am
The compound has the molecular formula C57H100O6 (Mr= 880 g/mol)
How many carbon-to-carbon bonds are present in this compound? Explain your answer
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 20, 2014, 11:10:55 am
C57H115OH

116-100 = 16
16/2 = 8

There are eight double bonds present. For every C=C double bond, two hydrogen atoms are lost. Therefore, because the general formula for 57 carbon atoms (given no double bonds) reveals we need 116 hydrogen atoms, and only 100 are present, means 16 atoms of hydrogen have been lost. As we mentioned, a C=C double bond exists for every two hydrogens lost. 16 divided by 2 gives 8. 8 double bonds are present in the compound.

The answer says 5  carbon-to-carbon double bonds.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 20, 2014, 11:16:43 am
The answer says 5  carbon-to-carbon double bonds.

Yeah sorry I misread the compound itself! Sorry!
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 20, 2014, 11:24:08 am
The answer says 5  carbon-to-carbon double bonds.

Okay got it:

The compound's molecular formula: C57H100O6

Consider the oxygen atoms as hydrogens
In this way, we have C57H106

Now, the general formula for an alkane is CnH2n+2

C57H116

Now, 116-106 = 10
10/2 = 5 C=C double bonds
Title: Re: VCE Chemistry Question Thread
Post by: thushan on April 20, 2014, 12:06:40 pm
Nah - when you work out the number of double bonds + rings in the compound, you ignore O atoms:

so C57H100O6 has the same number of double bonds + rings as C57H100.

Since C57H116 is the saturated alkane, it follows that, since you take of 2 H atoms for ever double bond you form,  there are (116-100)/2= 8 double bonds + rings in the structure.

Check the question again to see if there are any known double bonds and rings in the structure. Perhaps out of the 8 double bonds + rings, 5 would be C=C double bonds. There could be C=O double bonds too, remember.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 20, 2014, 12:18:51 pm
Nah - when you work out the number of double bonds + rings in the compound, you ignore O atoms:

so C57H100O6 has the same number of double bonds + rings as C57H100.

Since C57H116 is the saturated alkane, it follows that, since you take of 2 H atoms for ever double bond you form,  there are (116-100)/2= 8 double bonds + rings in the structure.

Check the question again to see if there are any known double bonds and rings in the structure. Perhaps out of the 8 double bonds + rings, 5 would be C=C double bonds. There could be C=O double bonds too, remember.

Thanks thushan! Do you know if there are any rules regarding this that we should be familiar of? Perhaps an explanation or even a link to one would be appreciated! :)
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 20, 2014, 06:55:34 pm
alkene + h20 ->

is the catalyst H+ or H3PO4?

And according to the text book, this is a hydrolysis reaction ?? Wouldn't it be a hydration reaction?

Can someone please clarify

Thanks :))
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 20, 2014, 07:11:50 pm
alkene + h20 ->

is the catalyst H+ or H3PO4?

And according to the text book, this is a hydrolysis reaction ?? Wouldn't it be a hydration reaction?

Can someone please clarify

Thanks :))

Reacting water with an alkene produces an alcohol. The catalyst is phosphoric acid, and this occcurs at 300oC.

Definitely not hydrolysis. This is an addition reaction.
Title: Re: VCE Chemistry Question Thread
Post by: Jawnle on April 20, 2014, 07:26:24 pm
Reacting water with an alkene produces an alcohol. The catalyst is phosphoric acid, and this occcurs at 300oC.

Definitely not hydrolysis. This is an addition reaction.

I think an addition reaction contains subsets of other various reactions.
Because an addition reaction could be
- Hydrogenation (addition of hydrogen)
- Hydration (addition of water)
- Halogenation (addition of a group 7 atom)

I think Blondie21 is confused about how the water is being added , but it's not referred to as a hydration reaction. A hydrolysis reaction is breaking something by adding H20. So if you can see, both hydration and hydrolysis reactions contain water as their reactants.

That's just the way I see it haha, I'm not too sure myself
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 20, 2014, 07:32:52 pm
I think an addition reaction contains subsets of other various reactions.
Because an addition reaction could be
- Hydrogenation (addition of hydrogen)
- Hydration (addition of water)
- Halogenation (addition of a group 7 atom)

I think Blondie21 is confused about how the water is being added , but it's not referred to as a hydration reaction. A hydrolysis reaction is breaking something by adding H20. So if you can see, both hydration and hydrolysis reactions contain water as their reactants.

That's just the way I see it haha, I'm not too sure myself

Well hydrolysis is when water reacts with a compound to break an existing covalent bond so I can see why it would say its hydrolysis. But I reckon its more suitable to call thisba hydration reaction (addition).
Title: Re: VCE Chemistry Question Thread
Post by: Jawnle on April 20, 2014, 07:44:58 pm
Well hydrolysis is when water reacts with a compound to break an existing covalent bond so I can see why it would say its hydrolysis. But I reckon its more suitable to call thisba hydration reaction (addition).

Yeah I agree it's more suitable to be called a hydration reaction (addition). Perhaps hydrolysis is more suited for esters and biochem you think?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 20, 2014, 09:50:08 pm
Yep. Hydrolysis normally results in the breaking up of a molecule into two. So for instance repeatedly hydrolysing urea (an amide) yields CO2 and two lots of NH3, hydrolysing ATP gives ADP and a phosphate ion and hydrolysing an ester gives an alcohol and a carboxylic acid. In all of these cases one of the reactants is broken up.
Title: Re: VCE Chemistry Question Thread
Post by: Jawnle on April 20, 2014, 09:57:42 pm
Yep. Hydrolysis normally results in the breaking up of a molecule into two. So for instance repeatedly hydrolysing urea (an amide) yields CO2 and two lots of NH3, hydrolysing ATP gives ADP and a phosphate ion and hydrolysing an ester gives an alcohol and a carboxylic acid. In all of these cases one of the reactants is broken up.

Would you agree that
Hydration: water molecules are added to the substance
Hydrolysis: Breaking-down process of molecule when it reacts with water, in other words, when it dissolves in water
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 20, 2014, 10:02:21 pm
We wouldn't consider the dissolution of an ionic salt as hydrolysis, for instance.
Title: Re: VCE Chemistry Question Thread
Post by: vox nihili on April 20, 2014, 10:10:50 pm
Would you agree that
Hydration: water molecules are added to the substance
Hydrolysis: Breaking-down process of molecule when it reacts with water, in other words, when it dissolves in water

Hydro-water
lysis-breaking

You're breaking a molecule with water.

As lzxnl pointed out for us though, hydrolysis has nothing to do with dissolution.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 21, 2014, 11:26:14 am
HOW do I find the monomers for the following polymers?
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 21, 2014, 11:36:01 am
BTW, do we need to memorise how each type of carbohydrate is formed i.e. suncrose is made from glucose + fructose?
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 21, 2014, 11:39:38 am
BTW, do we need to memorise how each type of carbohydrate is formed i.e. suncrose is made from glucose + fructose?

it would be good if you did know them since there aren't too many to remember. But it will definetly be helpful if you know the formation of sucrose because if you had to draw it, then you would draw a condensation reaction between glucose and fructose. I remember drawing sucrose in one of the questions from the textbook.
ps: this is a benefit of doing biol ;)
Title: Re: VCE Chemistry Question Thread
Post by: saba.ay on April 21, 2014, 11:53:12 am
HOW do I find the monomers for the following polymers?

You have to try and find the smallest repeating unit in each polymer chain. It's sometimes easier to try and find glycosidic or peptide, etc linkages to break them into monomers. One of the polymers is made from only one monomer while the other is made from 2 different monomers.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 21, 2014, 11:57:22 am
What structural similarities do the purine bases adenine and guanine share?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 21, 2014, 12:11:41 pm
What structural similarities do the purine bases adenine and guanine share?

Both are double-ringed structures.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 21, 2014, 12:16:52 pm
Both are double-ringed structures.

Tha answer says that adenine and guanine are derivatives of purine. What does this mean?
Title: Re: VCE Chemistry Question Thread
Post by: vox nihili on April 21, 2014, 12:22:25 pm
Tha answer says that adenine and guanine are derivatives of purine. What does this mean?

That they have a double ringed structure.

They're both synthesised from another chemical called purine, which gives them the double ring.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 21, 2014, 12:50:29 pm
Tha answer says that adenine and guanine are derivatives of purine. What does this mean?

Rishi have a read of pg 204
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 21, 2014, 01:28:51 pm
How would I draw the structure of 1,3-butadiene?
What is diene?
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on April 21, 2014, 01:36:21 pm
I never got my head around dilution, so can someone clarify? :/

So I have a stock standard solution: 1000mg/100mL  of caffeine
I want to make concentrations of 5mg/100ml, 10mg/100ml and 20mg/100ml caffeine.

Say for the first concentration I want to obtain (5mg/100ml), do i simply get like 0.5mL of the standard (5mg/0.5mL), and then dilute it into 100mL?
Is that how it works? o.o
Or is my logic wrong. hahhaha

Thanks!

So initially, you have 1000mg/100mL, then after dilution, you have 5 mg/100 mL, you can see that 1000/5 = 200 . If concentration decrease by 200 folds, volume must have increased by 200 folds. Hence, 100 x 200 = 20 000 mL i.e if you fill the volume up to 20 00mL (adding 19 900 mL), you get a concentration of 5 mg/100mL.

Does my method still work? o.o
Title: Re: VCE Chemistry Question Thread
Post by: Jawnle on April 21, 2014, 01:36:50 pm
How would I draw the structure of 1,3-butadiene?
What is diene?

When there are two double bonds

di - two
ene - double bond

there is a double bond on the first carbon and the third carbon of the mother chain
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 21, 2014, 01:40:32 pm
When there are two double bonds

di - two
ene - double bond

there is a double bond on the first carbon and the third carbon of the mother chain

Thanks heaps Jawnle :)
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 21, 2014, 01:49:19 pm
Is it possible to have a methyl group and a chlorine atom on the same carbon?
Cause I want to draw 2-chloro-2-methylbutane and I drew both functional groups on the same carbon but the teacher drew it differently.
Title: Re: VCE Chemistry Question Thread
Post by: rhinwarr on April 21, 2014, 01:51:51 pm
Yes, it should be:
Cl
|
C  -  C  -  C  -  C
|
C
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 21, 2014, 02:12:06 pm
Yes, it should be:
Cl
|
C  -  C  -  C  -  C
|
C

ok thanks. I was right then :)
It feels wonderful to prove teachers wrong sometimes. Agree?
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 21, 2014, 02:28:16 pm
Do we need to know about denaturation of enzymes? It's not in the study design
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 21, 2014, 02:32:03 pm
Do we need to know about denaturation of enzymes? It's not in the study design

Just know that enzymes change shape when the temp/ph increases or decreases above/below the optimum. The enzyme then experiences coagulation and can never bind with a substrate again. Poor enzyme
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 21, 2014, 02:33:50 pm
Poor enzyme

Hehehehe

Thanks Rishi :D
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 21, 2014, 03:09:43 pm
ok thanks. I was right then :)
It feels wonderful to prove teachers wrong sometimes. Agree?

Couldn't agree more. I did that heaps of times in year 12 chemistry hahaha.
Just know that enzymes change shape when the temp/ph increases or decreases above/below the optimum. The enzyme then experiences coagulation and can never bind with a substrate again. Poor enzyme

Sometimes, the denaturing isn't permanent and the enzyme can recover its previous shape upon reverting back to the original conditions. I'm not sure how common this is though.
If the temperature decreases too much, the enzyme may not have the energy to denature/denaturing may be too slow.

It IS helpful to have an idea of why temperature and pH changes denature enzymes.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 21, 2014, 03:57:41 pm
It IS helpful to have an idea of why temperature and pH changes denature enzymes.

Why does it? Do the bonds just break or something?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 21, 2014, 04:36:26 pm
Temperature increase => some bonds will break because the atoms have enough energy to break free
Temperature decrease => change the dissociation constant of water and the equilibrium constants of any acid-base equilibria in the protein, which can change the degree of ionisation of glutamate, aspartate, lysine or arginine residues and thus can affect ionic interactions in the tertiary structure of the protein (this occurs with a temperature increase too)

pH changes have similar effects on the degree of ionisation of the above amino acid residues.
Title: Re: VCE Chemistry Question Thread
Post by: Irving4Prez on April 21, 2014, 04:37:24 pm
Hey AN,

Is the stationary phase for both GC and GLC liquid? I thought the stationary phase consisted of inert solid particles. Secondly, what is the stationary phase for HPLC?

Thanks
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 21, 2014, 05:31:03 pm
Hey AN,

Is the stationary phase for both GC and GLC liquid? I thought the stationary phase consisted of inert solid particles. Secondly, what is the stationary phase for HPLC?

Thanks

1. Gas Chromatography can be either GSC or GLC (i.e. gas-solid chromatography and gas liquid-chromatography). Thus, the stationary phase is a solid in GSC, and a liquid in GLC.

2. Finely powdered alumina or silica gel packed into a column.
Title: Re: VCE Chemistry Question Thread
Post by: Irving4Prez on April 21, 2014, 05:38:58 pm
1. Gas Chromatography can be either GSC or GLC (i.e. gas-solid chromatography and gas liquid-chromatography). Thus, the stationary phase is a solid in GSC, and a liquid in GLC.

2. Finely powdered alumina or silica gel packed into a column.

So in GLC, the inert solid particles are coated in liquid whereas in GSC there is no liquid coating?

Okay, so it can either be solid/liquid for HPLC?
Title: Re: VCE Chemistry Question Thread
Post by: thushan on April 21, 2014, 06:46:40 pm
Thanks thushan! Do you know if there are any rules regarding this that we should be familiar of? Perhaps an explanation or even a link to one would be appreciated! :)

Hey - sure. So this is about working out the number of double bond equivalents (DBEs), which is basically the number of double bonds and rings.

DBE = 1/2 x (2(c+1) - h/cl/f/br/i + n)

The H/Cl/F/Br/I are just atoms that form one bond. Note that oxygen is not included in the equation.

Alternatively, another way to do it is to find a hydrocarbon with the same number of double bonds. Eg.

C3H6O will have the same number of double bonds as C3H6 (noting that O doesn't appear in the DBE formula), which is 1. Note here that we can effectively "ignore" oxygen atoms in looking for an equivalent hydrocarbon with the same number of double bonds.

C3H9N will have the same number of double bonds as C3H8, which is 0. To account for the N, add 1 to the number of H atoms.

C3H7Cl has the same number of double bonds as C3H8, which is 0. To account for the Cl/F/Br/I, subtract 1 from the number of H atoms.
Title: Re: VCE Chemistry Question Thread
Post by: katiesaliba on April 21, 2014, 09:30:10 pm
Is this correct: more protons then more energy released when an excited electron returns to its ground state
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 21, 2014, 10:02:06 pm
Is this correct: more protons then more energy released when an excited electron returns to its ground state

What's this for? It reminds me suspiciously of photosynthesis which isn't exactly chemistry.
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on April 21, 2014, 11:27:04 pm
Is this correct: more protons then more energy released when an excited electron returns to its ground state

Does this sentence even make sense?  I can't figure out what she's asking.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 21, 2014, 11:35:09 pm
Does this sentence even make sense?  I can't figure out what she's asking.

Me neither. Only sounds like photosynthesis because I hear "excited electrons", "more protons" and "energy released" in the same sentence.
Title: Re: VCE Chemistry Question Thread
Post by: thushan on April 22, 2014, 07:55:48 am
Is this correct: more protons then more energy released when an excited electron returns to its ground state

I think what Katie's asking is whether if the nucleus of an atom has more protons, then the energy difference of an electron in the ground state and in the excited state is greater.

To be honest, I don't actually know the answer to that conclusively. All I know is that a higher nuclear charge will tend to pull all the electron shells towards the nucleus (due to the attractive force), but I'm not sure whether as a result the electron shells will be closer to one another or further away. My educated guess is that the electron shells become closer to one another, which tends to decrease the difference between energy levels. On the flip side, the electron shells will both be closer to the nucleus, which may tend to increase the difference between energy levels of the valence shell and unoccupied shells. I don't know which effect is predominant.

On another note - lzxnl - I know you probably didn't mean it, but what you said came across to me as a little derisive and offensive. Just letting you know. Katie has every right to ask questions here without being derided.
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on April 22, 2014, 07:36:57 pm
What exactly is the chemical shift in NMR?
In a compound, how would I know where to put each peak (whether C-13/H1 NMR) along the horizontal axis with the chemical shift(ppm) on it?
I heard it was something to do with electronegativity or polarity of the different chemical environments of the compound..

help would be appreciated!

Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 22, 2014, 07:42:56 pm
How would I change 1,4-dichlorobutane to an alcohol? What would I add?
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 22, 2014, 08:18:45 pm
What effect does having a COOH group have on the physical and chemical properties of carbon compounds.
The only one I know it that it increases solubility.  :P
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on April 22, 2014, 08:40:47 pm
How would I change 1,4-dichlorobutane to an alcohol? What would I add?
You double the amount of NaOH that you normally because you have 2 Cl- there, hence, in the end, you have butane -1,4- diol, which has 2 OH- group on it.

What effect does having a COOH group have on the physical and chemical properties of carbon compounds.
The only one I know it that it increases solubility.  :P
Physical: you certainly increase boiling points, melting points, decrease volatility, increase solubility (as you increase the amount of both dispersion forces and polar bonds)
Chemical: it's an acid as the name suggests
Title: Re: VCE Chemistry Question Thread
Post by: Jawnle on April 22, 2014, 08:47:39 pm
You double the amount of NaOH that you normally because you have 2 Cl- there, hence, in the end, you have butane -1,4- diol, which has 2 OH- group on it.
Physical: you certainly increase boiling points, melting points, decrease volatility, increase solubility (as you increase the amount of both dispersion forces and polar bonds)
Chemical: it's an acid as the name suggests

I'm pre sure density increases aswell because the forces between molecules increase due to multiple hydrogen bonding/ increased dispersion forces, and this makes the molecules draw more closely to one another.
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on April 22, 2014, 09:06:16 pm
I'm pre sure density increases aswell because the forces between molecules increase due to multiple hydrogen bonding/ increased dispersion forces, and this makes the molecules draw more closely to one another.
Yeah, thanks for adding :D!
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 22, 2014, 09:26:11 pm
What exactly is the chemical shift in NMR?
In a compound, how would I know where to put each peak (whether C-13/H1 NMR) along the horizontal axis with the chemical shift(ppm) on it?
I heard it was something to do with electronegativity or polarity of the different chemical environments of the compound..

help would be appreciated!

Thanks!

Mmm well you have to have a pretty good understanding of the background information of NMR. However, I have never come across a questions which asks to literally place a peak onto an NMR spectrum! The questions you will normally recieve would have something to do with identifying what the molecule is.

You are right when you say it has something to do with electronegativity. BUT YOU DO NOT NEED TO KNOW THIS. THIS IS ONLY FOR YOUR UNDERSTANDING. From what I gather, we won't be asked to describe this background info at all.

The reason for the different chemical shifts depends on the 'shielding' of a nucleus and how 'exposed' it is. The energy from the radio waves cause the 'spinning' of electrons around each nucleus from the applied magnetic field. THe amount of nuclear shielding depends on other atoms surrounding the nucleus.

The chemical shift is basically the difference in energy needed to change the spin state compared to TMS.

Sooo back to electronegativity.. the more electronegative an atom is, the greater pull it will have from the other atoms .. decreasing the shielding.. leaving it more exposed and .:. allowing it to have a greater chemical shift

Like the hydrogen in -CH3- will absorb at a different rate than the hydrogen in -CH2- because of the different electronegiativity and thus the 'shielding'

CH2     CH3      TMS

That is normally the order ^^^^ (LMAO worst NMR ever). As you can see, the CH2 requires more energy to 'spin' because it is more 'exposed' as the electronegativity is greater :)

We add TMS as the reference sample because all the protons are in the same chemical enviro and will need the smallest chemical shift becuase it is the least 'exposed' ... because TMS is surrounded by CH3

Would love it if someone could please read over that and check my understanding. Thanks :)
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 22, 2014, 09:47:34 pm
What effect does having a COOH group have on the physical and chemical properties of carbon compounds.
The only one I know it that it increases solubility.  :P

The presence of a carboxyl (COOH) group in a compound gives it the ability to form hydrogen inter-particle bonds with water molecules. Thus, compounds with a carboxyl group will have increased solubility in water. With that, increases the melting and boiling point of the compound. The reason is because more energy is required to disturb the stronger hydrogen bonds that exist between the molecules of water and the carboxylic acid molecules, and thus, the temperature at which the compound boils/is melted increases. Also, decreased volatility. Because the hydrogen inter-particle bonds require more energy to disrupt, the compound cannot turn to a gaseous state as easily. Hence, the decreased volatility.

Chemically, it is an acid, and come all the chemical properties of acids with that, including pH<7, it being a proton donor, etc.
Title: Re: VCE Chemistry Question Thread
Post by: Jawnle on April 23, 2014, 04:25:45 pm
Hey guys, just want some opinions on whether it'd be better to go to the TSFX Chem Unit 3 revision and extension lecture, or the Unit 4 head start?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 23, 2014, 06:49:34 pm
Hey guys, just want some opinions on whether it'd be better to go to the TSFX Chem Unit 3 revision and extension lecture, or the Unit 4 head start?

If you can, both! I think U3 revision is excellent because you can consolidate everything + add any information to your knowledge bank that'll assist you for the exam. Unit 4 headstart is exactly as it sounds - it is essentially a head start to U4 and gives you a general outlook of the entire unit 4 chem course before you actually start it. I'd also go to Exam revision lectures at the end of the year. Any loose ends of your knowledge can be tied up and you'll be set for the exam.

If, however, you must choose, I'd go U3 revision and then U4 revision toward the end of the year.
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on April 23, 2014, 08:34:11 pm
Anyone got any particularly fantastic ways of remembering functional groups / nomenclature?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 23, 2014, 09:01:13 pm
alcohol => ol is at the end of the word, so if the alcohol group is the most important one, the name ends in ol
If it's a substituent, use 'hydroxy' as there is a 'hydrogen' and 'oxygen' making up 'hydroxy' and 'hydroxy' is like 'hydroxide' except shortened at the end, so it makes sense to put it at the front
oic acid you just have to remember
When naming esters, you have alkyl carboxylate. Example: methyl ethanoate can be thought of as a methyl group bonded to an ethanoate group, which is ethanoic acid minus a hydrogen, for instance
Ketones => 'one' at the back, like propanone; note how 'one' is also at the back of 'ketone'
Aldehydes have 'al', which is a bit weird given that the name begins with 'al'
The two above begin with 'oxo' if not the primary functional group, making sense as these functional groups only consist of oxygen atoms.
Halogen groups are easy xP
Methyl, ethyl etc you just have to know. It's only really up to butyl that you must rote learn as pentyl, hexyl are easier to work out.
Amides are fairly simple as they're named "ethanamide", for instance, with the only real difference being you name substituents attached to the nitrogen as N-methyl, for instance, instead of 2-methyl.

Nomenclature rules? Hmm. Longest hydrocarbon chain is the backbone. Your most important functional group is, well, the most distinguishing feature of your molecule. Substituents are like extra bits and the numbers tell you where they are. Then, order the substituents by alphabetical order. Anyone else you're unclear of?
Title: Re: VCE Chemistry Question Thread
Post by: DJA on April 23, 2014, 10:35:03 pm
How many moles of CH3COOH (acetic acid) are required to neutralise 0.14 moles of NaOH (sodium hydroxide)?

This question confuses me - what do you do when you have a weak acid like acetic acid?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 23, 2014, 11:18:21 pm
How many moles of CH3COOH (acetic acid) are required to neutralise 0.14 moles of NaOH (sodium hydroxide)?

This question confuses me - what do you do when you have a weak acid like acetic acid?

By definition (which is a pretty stupid one IMO) neutralisation is when n(acid) = n(base)
Why do I think it's stupid? Well, if you add 0.14 moles of NaOH to 0.14 moles of CH3COOH, you're left with 0.14 moles of CH3COONa (and some water), which is a weak base. If you're carrying out the experiment in water, you don't have a neutral pH at the end because all you have left is base. Therefore I think neutralisation is a misnomer :P
Title: Re: VCE Chemistry Question Thread
Post by: DJA on April 23, 2014, 11:42:24 pm
By definition (which is a pretty stupid one IMO) neutralisation is when n(acid) = n(base)
Why do I think it's stupid? Well, if you add 0.14 moles of NaOH to 0.14 moles of CH3COOH, you're left with 0.14 moles of CH3COONa (and some water), which is a weak base. If you're carrying out the experiment in water, you don't have a neutral pH at the end because all you have left is base. Therefore I think neutralisation is a misnomer :P

Thats LITERALLY exactly why I was so confused.

Thanks for clarifying!
Title: Re: VCE Chemistry Question Thread
Post by: darklight on April 24, 2014, 08:26:41 pm
Does anyone have any links for practice spectrums (IR/NMR) I can do: identifying bonds/molecules etc? Thanks :)
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on April 25, 2014, 11:02:16 am
Why are alkanols such as methanol and ethanol soluble in water, but octanol is not?
Title: Re: VCE Chemistry Question Thread
Post by: thushan on April 25, 2014, 11:09:51 am
Put simply, the hydrocarbon chain in octanol is very long, meaning lots of hydrogen bonds between water molecules have to be broken if dissolution were to occur. In return, only a small number of hydrogen bonds can be formed by the single OH group in the octanol with water. Hence, it is unfavourable for octanol to dissolve into water.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 25, 2014, 12:37:43 pm
Put simply, the hydrocarbon chain in octanol is very long, meaning lots of hydrogen bonds between water molecules have to be broken if dissolution were to occur. In return, only a small number of hydrogen bonds can be formed by the single OH group in the octanol with water. Hence, it is unfavourable for octanol to dissolve into water.

Thushan, could we say that because octanol is a longer molecule, there are more non-polar alkyl regions within the compound relative to say methanol or ethanol, and that as a result, shorter alcohols are more soluble than longer alcohols?
Title: Re: VCE Chemistry Question Thread
Post by: thushan on April 25, 2014, 12:52:15 pm
That's correct.
Title: Re: VCE Chemistry Question Thread
Post by: katiesaliba on April 26, 2014, 12:10:56 am
Concerning resonance, do nucleons that are already aligned against the external magnetic field flip, when exposed to radio waves, too? So, resonance doesn't just concern the alignment of nucleons with the external magnetic field?  :-\
Title: Re: VCE Chemistry Question Thread
Post by: vox nihili on April 26, 2014, 12:23:08 am
Concerning resonance, do nucleons that are already aligned against the external magnetic field flip, when exposed to radio waves, too? So, resonance doesn't just concern the alignment of nucleons with the external magnetic field?  :-\

Not in the slightest bit relevant to the VCE course.
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on April 26, 2014, 12:58:07 pm
what makes sulfuric acid such a good dehydrating agent to then be used as a catalyst in many reactions? eg esterification
Also, will the absence of sulfuric acid in a reaction of just say methanol and salicylic acid yield a higher percentage of water rather than methyl salicylate? if so, why?
Title: Re: VCE Chemistry Question Thread
Post by: thushan on April 26, 2014, 01:04:22 pm
Not essential to know for the exam - concentrated sulfuric acid is a very strong acid, and it also has a very high affinity for water. In other words, if exposed to hydrogen and oxygen atoms, it will often try to pull them out of the organic molecule in its attempt to find water to mix in (this is a highly simplified explanation). It uses its property as an acid to pull out O and H atoms from molecules.

If we mixed just methanol and salicylic acid without adding a catalyst, we'll get the same percentage yield of methyl salicylate, except it will take over 9000 years for the reaction to occur to get that yield of methyl salicylate. You'll learn why to an extent in Unit 4.
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on April 26, 2014, 01:10:38 pm
thanks thushan!
Title: Re: VCE Chemistry Question Thread
Post by: zhe0001 on April 26, 2014, 05:39:33 pm
How would an infra-red spectrum of butanoic acid differ from that of ethanoic acid?
Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on April 26, 2014, 07:16:27 pm
are we required to have knowledge of the principles of fractional distillation?
Title: Re: VCE Chemistry Question Thread
Post by: gohfish on April 26, 2014, 08:06:13 pm
are we required to have knowledge of the principles of fractional distillation?

Nope, as of the latest study design, fractional distillation is no longer part of the course
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on April 26, 2014, 11:37:46 pm
are we required to name ethers and if so, how to name?
Title: Re: VCE Chemistry Question Thread
Post by: Reus on April 26, 2014, 11:47:31 pm
In what part of the electromagnetic spectrum does benzene absorb?
Explain why humans see benzene as colourless.

Thanks.
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on April 27, 2014, 12:19:06 am
Explain why humans see benzene as colourless.

probably because the wavelength of benzene is outside of the visible spectrum that humans can see so it is colourless.

Also how do you name hydrocarbons in this form?

CH3(CH2)2CH2NH2
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 27, 2014, 09:18:51 am
In what part of the electromagnetic spectrum does benzene absorb?
Explain why humans see benzene as colourless.

Thanks.

Because benzene absorbs wavelengths of light that are not in the visible spectrum of the electromagnetic spectrum. Thus, it appears colourless.

Also how do you name hydrocarbons in this form?

CH3(CH2)2CH2NH2

1-Butylamine.

Title: Re: VCE Chemistry Question Thread
Post by: thushan on April 27, 2014, 10:33:55 am
Also how do you name hydrocarbons in this form?

CH3(CH2)2CH2NH2

butan-1-amine.

4 carbon compound => 'but'
no C=C double bonds => 'an'
an amino group in carbon 1 => '1-amine'

1-butylamine is an old name for that molecule, and would probably be penalised in the exam :/ - you need to use correct IUPAC terminology now.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 27, 2014, 10:39:31 am
butan-1-amine.

4 carbon compound => 'but'
no C=C double bonds => 'an'
an amino group in carbon 1 => '1-amine'

1-butylamine is an old name for that molecule, and would probably be penalised in the exam :/ - you need to use correct IUPAC terminology now.

Thanks for that thushan. My teacher taught us the way I said, but you're right. Ugh now I have to adjust for the SAC and then the exam.

So for instance:

CH3(CH2)3CH2NH2

Would be

Pentan-1-amine??
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 27, 2014, 10:51:29 am
are we required to name ethers and if so, how to name?

Nope, not needed for the exam
For simple ethers, there are two ways of naming them. Let's say you had CH3OCH3. You could call that dimethyl ether or methoxy methane. Similarly, CH3CH2OCH3 would be either ethyl methyl ether or methoxyethane, while CH3CH2CH2CH2OCH2CH3 would be butyl ethyl ether or ethoxybutane.

Thanks for that thushan. My teacher taught us the way I said, but you're right. Ugh now I have to adjust for the SAC and then the exam.

So for instance:

CH3(CH2)3CH2NH2

Would be

Pentan-1-amine??

Yes.
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on April 27, 2014, 10:57:04 am
thanks for the answers but I need to know what the brackets mean

e.g. (CH2)2 - is this saying a branch or just a chain of 2 more CH2 molecules
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 27, 2014, 11:10:35 am
The brackets can mean two things. If I use them to write CH3CH(CH3)CH2CH3, for instance, it means the CH3 in brackets is a sidegroup.
If I use them to write CH3(CH2)10 CH3, that means there are 10 CH2s in a chain.
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on April 27, 2014, 11:14:37 am
It represents a chain of 2 or more CH2 molecules.
Lets just say you have Butan-1-amine, it has a structural isomer of 2-methyl-propan-1-amine, which can be represented as CH2CH(CH3)CH2NH2. Notice that when there is an -methyl or -ethyl group, there is a -CH before it.

EDIT: Beaten lol :P
Title: Re: VCE Chemistry Question Thread
Post by: thushan on April 27, 2014, 11:48:06 am
Perfect. :)
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on April 27, 2014, 12:23:03 pm
ok so how would an ethyl group be represented

if (CH2)2 means just 2 CH2 chains would ethyl be (CH2CH3)?
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 27, 2014, 12:26:19 pm
ok so how would an ethyl group be represented

if (CH2)2 means just 2 CH2 chains would ethyl be (CH2CH3)?

Yes an ethyl group is CH2CH3
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 27, 2014, 12:28:50 pm
Could someone please give me a list of what catalysts to use in each reaction? I'm getting really confused what catalysts are meant to be used where
eg. H2SO4 is used during esterification

Greatly appreciated :)
Title: Re: VCE Chemistry Question Thread
Post by: katiesaliba on April 27, 2014, 12:32:26 pm
Atomic absorption spectroscopy (AAS) and UV–visible spectroscopy both involve absorption of light. Both can be used to determine the amount of copper in a solution.

b   Which technique would be simplest for the analysis of 0.5 M copper nitrate solution? Explain your answer.

What would be the answer to be and why, because don't both techniques require dilutions? Also, could questions of this calibre be on the exam?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 27, 2014, 12:34:35 pm
Could someone please give me a list of what catalysts to use in each reaction? I'm getting really confused what catalysts are meant to be used where
eg. H2SO4 is used during esterification

Greatly appreciated :)

• UV-light is required in converting alkanes to chloroalkanes.
• Converting alkenes to alcohols requires phosphoric acid + 300oC heat.
• Oxidation of primary alcohols to carboxylic acids requires acidified dichromate, or acidified permanganate.
• Esterification requires concentrated sulfuric acid.
• Chloroalkane --> Alcohol requires NaOH or KOH but that's a reagent!
• Ammonia is needed to produce an amine from a chloroalkane.
• Alkene to alkane (hydrogenation) requires a nickel catalyst + 200oC temperature.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on April 27, 2014, 12:36:08 pm
• UV-light is required in converting alkanes to chloroalkanes.
• Converting alkenes to alcohols requires phosphoric acid + 300oC heat.
• Oxidation of primary alcohols to carboxylic acids requires acidified dichromate, or acidified permanganate.
• Esterification requires concentrated sulfuric acid.
• Chloroalkane --> Alcohol requires NaOH or KOH but that's a reagent!
• Ammonia is needed to produce an amine from a chloroalkane.
• Alkene to alkane (hydrogenation) requires a nickel catalyst + 200oC temperature.

Perfect!!! Just what I wanted. Thanks Yacoubb :)
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 27, 2014, 12:38:06 pm
Atomic absorption spectroscopy (AAS) and UV–visible spectroscopy both involve absorption of light. Both can be used to determine the amount of copper in a solution.

b   Which technique would be simplest for the analysis of 0.5 M copper nitrate solution? Explain your answer.

What would be the answer to be and why, because don't both techniques require dilutions? Also, could questions of this calibre be on the exam?

I'll have a go, but I could be wrong.
I'd say not AAS. AAS is only reliable for very small concentrations, where the relationship between absorbance and concentration is linear at low concentrations (e.g. Units like ppm). In UV-visible spectroscopy, a coloured solution is required. Cu(NO3)2 is a blue solution, and so this is what makes UV-visible spectroscopy the better option.
Title: Re: VCE Chemistry Question Thread
Post by: darklight on April 27, 2014, 01:15:44 pm
I'll have a go, but I could be wrong.
I'd say not AAS. AAS is only reliable for very small concentrations, where the relationship between absorbance and concentration is linear at low concentrations (e.g. Units like ppm). In UV-visible spectroscopy, a coloured solution is required. Cu(NO3)2 is a blue solution, and so this is what makes UV-visible spectroscopy the better option.

I thought that colorimetry required coloured solutions, but you can use UV-visible for colourless solutions as well as this technique will utilise UV radiation..
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on April 27, 2014, 01:28:23 pm
I thought that colorimetry required coloured solutions, but you can use UV-visible for colourless solutions as well as this technique will utilise UV radiation..
yeah that's right.

Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 27, 2014, 01:33:50 pm
yeah that's right.

I thought that colorimetry required coloured solutions, but you can use UV-visible for colourless solutions as well as this technique will utilise UV radiation..

Oh yes my bad! What do you guys think though - AAS or UV-Vis?
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on April 27, 2014, 01:38:56 pm
Oh yes my bad! What do you guys think though - AAS or UV-Vis?
I reckon AAS, it's simpler than UV-Vis and it would get the job done.
just not sure how to answer the question  :P
Title: Re: VCE Chemistry Question Thread
Post by: katiesaliba on April 27, 2014, 01:43:33 pm
So, AAS requires smaller concentrations than UV-Vis for accuracy?
Thanks everyone! :)
Title: Re: VCE Chemistry Question Thread
Post by: Reus on April 27, 2014, 02:23:42 pm
So, AAS requires smaller concentrations than UV-Vis for accuracy?
Thanks everyone! :)

Yes it's AAS.
Btw guys thanks for the answer, however I noticed your questions are like the next few chapter or so, you guys individually ahead or school pace?
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on April 27, 2014, 03:06:54 pm
the answer's UV Vis
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 27, 2014, 03:22:34 pm
the answer's UV Vis

Yeah that's what I said. In Atomic Absorption Spectroscopy, you must dilute that copper nitrate solution to a very low concentration because only at very low concentrations is AAS actually a reliable instrumental technique. In UV-Visible Spectroscopy, those dilutions are not required.

So, the word 'simplest' refers to quickest/most efficient, and in this case, not having to make so many dilutions renders UV-Visible Spec the answer.
Title: Re: VCE Chemistry Question Thread
Post by: Reus on April 27, 2014, 03:28:17 pm
Anyone know anything about the Chemistry GTAC SAC? My school has it scheduled in 2 weeks or so, however we haven't discussed it, I'm guessing it's on spectroscopy. You guys done it yet? Any ideas?
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on April 27, 2014, 04:20:05 pm
how to do part b of the question. I'm trying to grasp the n+1 rule but I don't really know how to use it properly
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 28, 2014, 11:41:38 am
What is the name of this compound?

I always thought that the -OH group needs to be prioritised but apparently it is pronan-2-ol
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on April 28, 2014, 11:51:23 am
how to do part b of the question. I'm trying to grasp the n+1 rule but I don't really know how to use it properly
So your CH3 is next to a CH2, the number of hydrogen on its neighboring chemical environment will determine its number of split on the NMR,hence, 2+1=3 splits.
What is the name of this compound?

I always thought that the -OH group needs to be prioritised but apparently it is pronan-2-ol
You got a CH3 there, so there is 3 carbon in the compound, the -OH is bonded to the second one.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on April 28, 2014, 01:13:03 pm
Sooo even though there is a -OH group, it is not included in the 'mother chain'?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 28, 2014, 04:18:39 pm
Sooo even though there is a -OH group, it is not included in the 'mother chain'?

The hydroxyl group does not need to be at carbon 1. You must identify the longest chain, and the functional group must be present in this mother chain. So, you have CH3CHOHCH3, which is 2-propanol.

Hope this helped :)
Title: Re: VCE Chemistry Question Thread
Post by: Jawnle on April 28, 2014, 06:38:12 pm
Is it possible for C2H6 + OH- to produce ethanol? Or would I need to make it into a chloroalkane beforehand

The question is "Using the appropriate catalysts and other reaction conditions, which of the following set of reactants will produce ethanol as a product?
I) C2H6 +OH-
II) C2H4 + H2O
III) CH3CH2COOCH3 + H20

A) I only
B) II only
C) II and III only
D) I, II and III

For II I got ethanol, for III I got methanol, so it shouldn't be C or D. Since I'm certain that II produces ethanol, I'm guessing the answer is B?

What do you guys think??
Title: Re: VCE Chemistry Question Thread
Post by: katiesaliba on April 28, 2014, 07:30:41 pm
In gas chromatography, will more polar samples have a low retention time because the liquid hydrocarbon stationary phase is non-polar?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 28, 2014, 08:22:30 pm
Is it possible for C2H6 + OH- to produce ethanol? Or would I need to make it into a chloroalkane beforehand

The question is "Using the appropriate catalysts and other reaction conditions, which of the following set of reactants will produce ethanol as a product?
I) C2H6 +OH-
II) C2H4 + H2O
III) CH3CH2COOCH3 + H20

A) I only
B) II only
C) II and III only
D) I, II and III

For II I got ethanol, for III I got methanol, so it shouldn't be C or D. Since I'm certain that II produces ethanol, I'm guessing the answer is B?

What do you guys think??

Hydroxide isn't reactive enough to cleave a C-H bond. The reason why chlorine reacts with methane under UV light is because the chlorine molecule absorbs the UV light and dissociates to form individual Cl atoms. These only have 7 valence electrons and are thus VERY reactive. It takes something of that reactivity to react with a C-H bond.

From my own knowledge, only the hydration of ethene, in your list, produces ethanol, so you'd be correct.
Title: Re: VCE Chemistry Question Thread
Post by: Robert123 on April 28, 2014, 09:08:01 pm
How much do we need to know about galactose for chemistry?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 28, 2014, 10:18:05 pm
In gas chromatography, will more polar samples have a low retention time because the liquid hydrocarbon stationary phase is non-polar?

It depends on what stationary phase you ACTUALLY use. You could, for all we know, use some polar liquid that is still a liquid at 200 degrees.

How much do we need to know about galactose for chemistry?

Nothing at all I think. All I know is that it's an isomer of glucose and it was enough for me. The exam never asked anything on it.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on April 28, 2014, 10:19:46 pm
Do we need to know/state the state of phosphoric acid catalyst in hydration or the state of H2SO4 or any catalysts really in organic chemical pathways.
Title: Re: VCE Chemistry Question Thread
Post by: Reus on April 29, 2014, 05:25:15 pm
Why does oxygen not absorb IR radiation?
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on April 29, 2014, 06:09:14 pm
Why does oxygen not absorb IR radiation?
Because it's a diatomic molecule. Similar with N2.
This means they're not polarised
Title: Re: VCE Chemistry Question Thread
Post by: PsychoT on April 29, 2014, 06:30:18 pm
Why are Cyclohexane/Cyclohexene good organic solvents?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on April 29, 2014, 07:18:32 pm
Why does oxygen not absorb IR radiation?

It's not just because it's diatomic; it's homonuclear as well, which means any bond stretch wouldn't change its overall dipole moment. As a result, IR radiation is not absorbed.

Why are Cyclohexane/Cyclohexene good organic solvents?

They're non-polar much like other organic compounds, so there is little energy cost (if at all) for any organic solute to dissolve.
Title: Re: VCE Chemistry Question Thread
Post by: thushan on April 29, 2014, 08:01:51 pm
Do we need to know/state the state of phosphoric acid catalyst in hydration or the state of H2SO4 or any catalysts really in organic chemical pathways.

Not usually - you should know that the H2SO4 used as a catalyst in esterification must be H2SO4 (l) though - NOT H2SO4 (aq).
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on April 29, 2014, 11:42:30 pm
Question about naming amines
CH3CH2CH2NH2
It's okay to call this: Amino propane OR propan-1-amine? Just not 1-propyl-amine?
Is there a preferred way?

Thanks
Title: Re: VCE Chemistry Question Thread
Post by: PsychoT on April 30, 2014, 09:18:34 am
Question about naming amines
CH3CH2CH2NH2
It's okay to call this: Amino propane OR propan-1-amine? Just not 1-propyl-amine?
Is there a preferred way?

Thanks

I've always called in propan-1-amine, but that's just me.
Title: Re: VCE Chemistry Question Thread
Post by: Alwin on April 30, 2014, 02:20:04 pm
Question about naming amines
CH3CH2CH2NH2
It's okay to call this: Amino propane OR propan-1-amine? Just not 1-propyl-amine?
Is there a preferred way?

Thanks

Amino propane (iirc) is wrong because it could refer to CH3CH2CH2NH2 OR CH3CH(NH2)CH3

You should actually call it 1-aminopropane :) (see below)
Title: Re: VCE Chemistry Question Thread
Post by: thushan on April 30, 2014, 02:49:13 pm
Amino propane (iirc) is wrong because it could refer to CH3CH2CH2NH2 OR CH3CH(NH2)CH3

You should actually call it 1-aminopropane :)

propan-1-amine. 1-aminopropane is incorrect and is the old name, and those names were penalised in the 2008-2009 VCAA exam.
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on April 30, 2014, 05:11:50 pm
how would i make salicylic acid (2-hydroxybenzoic acid) from benzene?
Also, why is this called 2-hydroxybenzene phenol, why not "1-" ?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 01, 2014, 07:12:53 pm
how would i make salicylic acid (2-hydroxybenzoic acid) from benzene?
Also, why is this called 2-hydroxybenzene phenol, why not "1-" ?

Erm...it's just called hydroxybenzene. There shouldn't be a number at all.

Making salicylic acid is out of the course
As for making salicylic acid, you need phenol. Look up on Wiki how to make phenol. You can use N2O and form phenol and nitrogen gas, which is probably the best path (chlorobenzene is problematic as it's not easy to just make monosubstituted chlorobenzene by itself). React with sodium hydroxide to form sodium phenolate (conjugate base of phenol). Then react with CO2 at 100 atm and 390 K to make salicylic acid.
Title: Re: VCE Chemistry Question Thread
Post by: thushan on May 01, 2014, 08:46:28 pm
how would i make salicylic acid (2-hydroxybenzoic acid) from benzene?
Also, why is this called 2-hydroxybenzene phenol, why not "1-" ?

Hey man - it's quite unlikely that you'll be expected to know how to synthesise salicylic acid from benzene in the exam. :)
Title: Re: VCE Chemistry Question Thread
Post by: Alwin on May 01, 2014, 09:01:53 pm
propan-1-amine. 1-aminopropane is incorrect and is the old name, and those names were penalised in the 2008-2009 VCAA exam.
Cheers for the clarification!

Hey man - it's quite unlikely that you'll be expected to know how to synthesise salicylic acid from benzene in the exam. :)
I thought that naming aromatic compounds was not in the study design ever period. (?)

@thush I think I've been out of the system too long for chemistry as you can tell by my replies :P
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on May 01, 2014, 10:08:25 pm
thanks to both of you :)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 01, 2014, 10:09:45 pm
@thush I think I've been out of the system too long for chemistry as you can tell by my replies :P

Your nomenclature usage was penalised in 2008 apparently xP
Title: Re: VCE Chemistry Question Thread
Post by: Sanguinne on May 02, 2014, 07:01:34 pm
Why do non polar molecules dissolve in non polar solvents and why do polar molecules dissolve in polar solvents.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on May 02, 2014, 07:07:35 pm
Why do non polar molecules dissolve in non polar solvents and why do polar molecules dissolve in polar solvents.

Like dissolves like.
If a molecule is polar (has an OH- group) it will be able to form hydrogen bonds with the polar substance which is water.
If a molecule is non-polar, it forms dispersion forces with the non-polar solvent
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 02, 2014, 08:43:20 pm
Like dissolves like.
If a molecule is polar (has an OH- group) it will be able to form hydrogen bonds with the polar substance which is water.
If a molecule is non-polar, it forms dispersion forces with the non-polar solvent

Search the chem forum for my post on this. Polar molecules also form dispersion forces with polar solvents too; it's not just the fact that the non-polar molecules form dispersion forces that is relevant.
Polar molecules don't have to contain hydroxyl groups. Formaldehyde doesn't have a OH or NH but it still dissolves really well in water (400 g/L according to Wiki). Similarly, you can dissolve as much HCl as you want in water.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on May 02, 2014, 09:52:38 pm
Search the chem forum for my post on this. Polar molecules also form dispersion forces with polar solvents too; it's not just the fact that the non-polar molecules form dispersion forces that is relevant.
Polar molecules don't have to contain hydroxyl groups. Formaldehyde doesn't have a OH or NH but it still dissolves really well in water (400 g/L according to Wiki). Similarly, you can dissolve as much HCl as you want in water.

oh ok...thanks for the clarification :)
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on May 03, 2014, 01:14:19 pm
Begun to do Chem practice exam.
I know it's probably been asked a million times a year, but which papers are the best (for chem)?
Relative to the VCAA exams in terms of difficulty etc...

Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: DJA on May 03, 2014, 01:17:46 pm
With reaction pathways for Chloroalkanes to Alcohols what are the required conditions and reagents?

All I know is that it requires:
-H2O
-A Catalyst

I would like to know if 'heating' is also a required condition.

Cheers!
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on May 03, 2014, 01:23:54 pm
With reaction pathways for Chloroalkanes to Alcohols what are the required conditions and reagents?

All I know is that it requires:
-H2O
-A Catalyst

I would like to know if 'heating' is also a required condition.

Cheers!
nah, heating isn't required.
alternatively you could have either NaOH or KOH and no catalyst is required
Title: Re: VCE Chemistry Question Thread
Post by: DJA on May 03, 2014, 03:20:51 pm
In Chromatography, What are TLC, HPLC and GC all best used for in terms of analysing different kinds of substances?

TLC:

HPLC: Extremely sensitive analysis

GC: Volatile compounds

Could you guys help me complete this? Any examples of different substances under each type would be really helpful!

Thanks
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 03, 2014, 03:28:30 pm
With reaction pathways for Chloroalkanes to Alcohols what are the required conditions and reagents?

All I know is that it requires:
-H2O
-A Catalyst

I would like to know if 'heating' is also a required condition.

Cheers!

Your reaction will be quite slow; water isn't the best reactant to bring out this reaction. The oxygen doesn't have a high enough negative charge to quickly react with a chloroalkane. With hydroxide, it has a larger negative charge to attack the more positive C-Cl carbon. Best to stick with KOH or NaOH as mentioned.

In Chromatography, What are TLC, HPLC and GC all best used for in terms of analysing different kinds of substances?

TLC:

HPLC: Extremely sensitive analysis

GC: Volatile compounds

Could you guys help me complete this? Any examples of different substances under each type would be really helpful!

Thanks

TLC just costs less and is very fast
GC can't be used for compounds that undergo thermal decomposition at higher temperatures (like larger organic molecules for instance). HPLC is ideal for separating similar compounds like toluene and benzene.

Your textbook will have some for these; I've forgotten these to be honest.
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on May 03, 2014, 05:53:02 pm
primary standards should have a high molar mass (>110g/mol?) to minimise weight errors.
What does this mean exactly? and is specifying >110g/mol necessary?
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on May 03, 2014, 07:16:55 pm
Stoic calculations are made using moles.  Consider the following examples:

HCl  - low molar mass (36.45)
Real answer: 12grams, you got 13grams

12/36.5 = 0.32
13/36.5 = 0.35
difference = 0.03mol

Na2CO3  - higher molar mass (106)
Real answer: 12grams, you got 13grams

12/106 = 0.11
13/106 = 0.12
difference = 0.01

You can see how even though you have the same mass difference in each case, your number of mol difference is much less for a higher molar mass.  This reduces the amount of error during stoic calculations.

No real need to specify what molar mass is needed, unless your teacher has said to do so.

primary standards should have a high molar mass (>110g/mol?) to minimise weight errors.
What does this mean exactly? and is specifying >110g/mol necessary?
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on May 03, 2014, 07:29:05 pm
Why is it that an increase in temperature for an exothermic reaction causes K (eq constant) to decrease?
Conversely, why is does an increase in temperature for an endothermic reaction cause K to increase?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 04, 2014, 12:01:46 am
primary standards should have a high molar mass (>110g/mol?) to minimise weight errors.
What does this mean exactly? and is specifying >110g/mol necessary?

Let's pretend you used lithium fluoride as a primary standard. Its molar mass is 6.9 + 19 = 25.9 g/mol
Now let's say you had 259 mg of LiF, but your scale uncertainty was 5 mg. The uncertainty in the number of moles is then 5 mg/molar mass.
However, if your primary standard was KCl, for instance, with a molar mass of 39 + 35.5 = 74.5 g/mol, the numerical uncertainty in the number of moles would decrease given the same mass used.

Why is it that an increase in temperature for an exothermic reaction causes K (eq constant) to decrease?
Conversely, why is does an increase in temperature for an endothermic reaction cause K to increase?

There are a few ways to explain this.
1. Using the explicit dependence of the equilibrium constant on temperature. This is out of the course
2. Le Chatelier's principle (this IS in the course)

OK, so let's say you have A => B and this is exothermic. If we add heat, the system will attempt to partially negate this disturbance by reducing the system temperature. The backwards reaction does this, so the equilibrium shifts backwards, forming more reactant and less product and thus decreases K. Try use this logic for your endothermic reaction.

3. Considering the effects on the reaction rate. Heating a reaction flask actually increases the rate of the forward AND backward reaction. It can be proved that for an exothermic reaction, heating the reaction flask increases the forward reaction rate less than that of the backward reaction rate, and vice-versa for cooling. This is again out of the course.

Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on May 04, 2014, 04:16:47 pm
What does the study design mean by

addiction reactions of alkenes (addition of hydrogen halides and water limited to symmetrical alkenes)
Title: Re: VCE Chemistry Question Thread
Post by: OutstandingInDivination on May 04, 2014, 04:44:54 pm
What does the study design mean by

addiction reactions of alkenes (addition of hydrogen halides and water limited to symmetrical alkenes)

addition reactions involving water with an alkene

addition reactions involving a hydrogen halide (eg HCl) with an alkene
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on May 04, 2014, 04:46:03 pm
What does the study design mean by

addiction reactions of alkenes (addition of hydrogen halides and water limited to symmetrical alkenes)

Hydrogen halides:
When you react an alkene with H---X (where X is a halogen) to produce an alkyl halide.

Hydration:
Reacting an alkene with water, in the presence of phosphoric acid at 300 degrees Celcius, to produce an alcohol.

Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 04, 2014, 05:23:05 pm
Asymmetric alkene => addition reaction of water to propene can yield either isomer of propanol (reaction conditions will determine which isomer; there are many ways of effecting this reaction). VCAA avoids this complication by saying 'symmetric' alkenes only.
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on May 04, 2014, 09:49:25 pm
For the colorimetric determination of phosphate in fertiliser, why is it that the % phosphorous is greater than that of the manufacture's specified %?  So far all I can think of is ions (other than phosphate) absorbing the light.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 04, 2014, 10:48:37 pm
Just say you're reacting methanol and salicylic acid and form methyl salicylate and water yeah, and say that the yield of methyl salicylate is 80%.. What does the other 20% exist as?

Let's talk about molar quantities.
So you've formed 80% of the maximum possible amount of methyl salicylate. The 20% can be thought of as reacting back to form the original reactants. As the reaction here is 1 methanol + 1 salicylic acid => 1 water + 1 methyl salicylate, the number of moles of unreacted methanol is 1/4 (20%/80%) that of the methyl salicylate. Similar idea with the salicylic acid. Note that this calculation only gives the additional number of moles of methanol or methyl salicylate in conjunction with any excess reactant you may have had.
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on May 05, 2014, 12:25:44 am
Thank for that man!
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on May 05, 2014, 03:51:27 pm
Hi guys

Is it possible to do these problems without knowing the Kb values?

Q11.83
What amount of NH3 must be dissolved in water to give 500mL solution with a pH of 11.22?

Q11.84
Calculate the pH of 0.20 M NaCN.  What is the concentration of HCN in the solution?

Cheers.
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on May 05, 2014, 04:40:09 pm
Also, could anyone explain what exactly the difference is between Ka/Kb values and Kc?  Does Ka/Kb refer specifically and only to a weak acid or base in water reaction?  But Kc can refer to the equilibrium constant of any reaction?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 05, 2014, 08:16:09 pm
Hi guys

Is it possible to do these problems without knowing the Kb values?

Q11.83
What amount of NH3 must be dissolved in water to give 500mL solution with a pH of 11.22?

Q11.84
Calculate the pH of 0.20 M NaCN.  What is the concentration of HCN in the solution?

Cheers.

You definitely need acidity constants.

Also, could anyone explain what exactly the difference is between Ka/Kb values and Kc?  Does Ka/Kb refer specifically and only to a weak acid or base in water reaction?  But Kc can refer to the equilibrium constant of any reaction?

Kc is the equilibrium constant in terms of concentrations (generally used when referring to gas reactions). Ka is the acidity constant while Kb is the basicity constant; Ka * Kb = Kw for a conjugate acid base pair
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on May 06, 2014, 07:33:06 am
Does the addition reaction of ethene and water need a catalyst?

Aren't alkenes so reactive that a cataylst isn't needed?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on May 06, 2014, 09:12:44 am
Does the addition reaction of ethene and water need a catalyst?

Aren't alkenes so reactive that a cataylst isn't needed?

Ethene, when reacted with water, produces ethanol. In this hydration (addition) reaction, phosphoric acid catalyst in 300oC conditions is required.
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on May 06, 2014, 05:50:54 pm
does the benzene ring in acetylsalicylate count as a functional group?
Title: Re: VCE Chemistry Question Thread
Post by: Capristo on May 06, 2014, 07:30:05 pm
What is it that causes salicylic acid to have higher rf values than aspirin, considering salicylic acid is more polar?

Thanks  :)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 06, 2014, 08:36:13 pm
Does the addition reaction of ethene and water need a catalyst?

Aren't alkenes so reactive that a cataylst isn't needed?

You are confusing two types of reactivity. There is kinetic reactivity (lability), which describes how fast something reacts. There is also thermodynamic reactivity, in which physics predicts a reaction will occur to a great extent.  The difference can be shown in the following examples.
If you add an indicator to a solution and then add an acid, the acid will QUICKLY change the colour of the indicator (assuming you swirl the solution). Acid-base reactions are generally quite fast, even though they don't proceed to completion. So this is something that is fast kinetically but not favoured thermodynamically.

Now, let's look at the hydrogenation of ethene to form ethane. Thermodynamically, this reaction should proceed almost to completion. However it's not very fast. Catalysts and high temperatures are used if you need an acceptable reaction rate for some reasons, such as the hydration of ethene to form ethanol. Just because an alkene is reactive doesn't mean it'll react fast.

Another example. Lithium is the strongest reductant on your standard electrode potential table. It isn't, though, the fastest thing to react with water. Sodium will fizz, potassium will fizz violently and produce hydrogen gas so vigorously the hydrogen gas burns, while lithium does like nothing.

You'll learn more about the differences between reaction rate and extent of reaction.

What is it that causes salicylic acid to have higher rf values than aspirin, considering salicylic acid is more polar?

Thanks  :)

Rf values in what solvent and stationary phase?
Title: Re: VCE Chemistry Question Thread
Post by: Capristo on May 06, 2014, 10:23:51 pm

Rf values in what solvent and stationary phase?

Ethyl acetate and silica gel restrospectively. I'm not entirely sure if the results were correct. I was mainly researching some ways in which TLC could be used to qualitatively determine the purity of aspirin.
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on May 07, 2014, 10:46:45 am
Hi guys,

I don't understand whats happening here.

When people have a 'Barium Meal' i.e. BaSO4 (s) <--> Ba2+ + SO42-...
They make it less soluble (because Ba2+ is toxic to the body) by making it up in a solution with lots of extra SO4 ions.
This forces the backwards reaction to occur (Le Chateliers principle)...
But the Ksp doesn't change?
I can't get my head around this.....
Doesn't the Ksp have to change in order to reduce the solubility of Ba2+?
Title: Re: VCE Chemistry Question Thread
Post by: thushan on May 07, 2014, 10:59:11 am
Nup. So the solubility of barium sulfate is known as the maximal amount of barium sulfate that can be dissolved into a solution. It is not synonymous with Ksp.

Ksp = [Ba2+][SO4 2-], where the solution is already saturated.
Qsp = [Ba2+][SO4 2-], where the solution may or may not be saturated. Qsp < Ksp if the solution is not saturated, Qsp = Ksp if the solution is saturated.

Suppose we had a solution already containing sulfate ions (eg. K2SO4). The solubility of barium sulfate in this solution is the amount that can be dissolved into this solution to saturate it - i.e. bring Qsp up to Ksp. Since we already have sulfate ions in the solution, when we start adding BaSO4, it requires less BaSO4 to bring Qsp up to Ksp. In other words, less BaSO4 can be dissolved into this solution to achieve saturation.

Hence, the solubility of BaSO4 in a solution containing sulfate ions from other sources is less than that in water. In either case Ksp is the same.
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on May 07, 2014, 11:33:13 am
Hence, the solubility of BaSO4 in a solution containing sulfate ions from other sources is less than that in water. In either case Ksp is the same.

Thanks thushan.

Is it wrong for me to be thinking of Kw, Ka, Kb, Kc, Ksp etc as all basically the same thing, but the subscript just refers to the specific kind of reaction it is referring to (e.g. Kc is for gases only, Ka is for acid in water, Ksp is for salt in water)?

ie. are they all just an equilibrium constant for a reaction - which will only change with a change in temperature - and the 'formula' is:

Kwhatever = [Product 1]n[Product 2]m / [Reactant 1]p

Where the concentration of a solid or a liquid is always 1, and n, m, p are the stoiciometric constants for that species?

We have been kind of taught them all over the last couple of weeks and they haven't been terribly concise about it (IMO - but I am probably just not understanding it yet).
Title: Re: VCE Chemistry Question Thread
Post by: thushan on May 07, 2014, 05:55:59 pm
Thanks thushan.

Is it wrong for me to be thinking of Kw, Ka, Kb, Kc, Ksp etc as all basically the same thing, but the subscript just refers to the specific kind of reaction it is referring to (e.g. Kc is for gases only, Ka is for acid in water, Ksp is for salt in water)?

ie. are they all just an equilibrium constant for a reaction - which will only change with a change in temperature - and the 'formula' is:

Kwhatever = [Product 1]n[Product 2]m / [Reactant 1]p

Where the concentration of a solid or a liquid is always 1, and n, m, p are the stoiciometric constants for that species?

We have been kind of taught them all over the last couple of weeks and they haven't been terribly concise about it (IMO - but I am probably just not understanding it yet).

You're absolutely right.
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on May 07, 2014, 06:49:16 pm
You're absolutely right.

God that's so much less confusing. They could have taught it so much better haha. Thankyou!
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on May 07, 2014, 09:00:25 pm
Is this 3 or 4 chloroheptane?
CH3(CH2)2CHCl(CH2)2CH3
Book says it's 4, but going from right to left, the smallest chain is 3?
Title: Re: VCE Chemistry Question Thread
Post by: vox nihili on May 07, 2014, 09:05:46 pm
CH3(CH2)2CHCl(CH2)2CH3

Chlorine is attached to that carbon
Title: Re: VCE Chemistry Question Thread
Post by: Snorlax on May 07, 2014, 09:22:31 pm
thanks!

(CH3)2CHCH2CH3
Is the (CH3)2 just saying there's a methyl group on the second Carbon?
Title: Re: VCE Chemistry Question Thread
Post by: Bronzebottom64 on May 07, 2014, 09:31:56 pm
Sure is  :)
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on May 08, 2014, 01:09:34 am
thanks!

(CH3)2CHCH2CH3
Is the (CH3)2 just saying there's a methyl group on the second Carbon?

Another way to rewrite this is:

CH3CH2CH(CH3)2

It's actually saying that there are two methyl groups attached to the first carbon atom.

The name of this compound is methyl butane
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 08, 2014, 04:45:18 pm
Be careful. It's only methylbutane because the methyl group has no other carbon it could go on. Otherwise you'd need to specify the carbon number.
For instance, methylpentane isn't acceptable. 2-methylpentane and 3-methylpentane both exist and are distinct.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on May 08, 2014, 05:02:22 pm
Be careful. It's only methylbutane because the methyl group has no other carbon it could go on. Otherwise you'd need to specify the carbon number.
For instance, methylpentane isn't acceptable. 2-methylpentane and 3-methylpentane both exist and are distinct.

Yeah I'm aware of that :)
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on May 08, 2014, 07:50:00 pm
Need some help with these questions please
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 08, 2014, 08:15:24 pm
1 molecule of decane?
Decane is C10H22
Its molar mass is 12*10 + 22 = 142 g/mol
So in one mole, or 6.02*10^23 molecules, the mass is 142 g/mol
Try now work out the mass of one molecule.

5.0% w/v originally meant solute had 5% mass of the water, but as 5 g / 100g water is the same as 5 g/100 mL water. So really, you have 50 g / L solution
Ammonium hydroxide? What the. It's just a solution of ammonia, so the molar mass SHOULD just be 17 g/mol. However VCE somehow expects you to believe NH4OH actually exists (it doesn't), so the molar mass is 35 g/mol. A 50 g/L solution would then be pretty close to a 1.4 M solution.

Clearly D looks weird.
Title: Re: VCE Chemistry Question Thread
Post by: Reus on May 11, 2014, 07:36:18 pm
Struggling with this one tbh...
Anyone? Thanks in advance.
Title: Re: VCE Chemistry Question Thread
Post by: Alwin on May 11, 2014, 08:09:30 pm
Struggling with this one tbh...
Anyone? Thanks in advance.

hey :D It's perfectly normal to find some qs tough, that's what we're here for ;)

Part (a)
If we just read off the graph, the concentration of the copper in the solution is about 7.0 μg/ml

Part (b)
The volume of the solution was 200 ml (not 200+10=210 ml because it says it was "diluted to 200 ml")
So, the mass of copper is:  7.0 μg/ml x  200 ml = 1.4 mg

Part (c)
Assuming that the sample is indicative of all the ore, we can say the ratio of copper to ore is the same
ie copper:ore = 0.0014:2    (since we found 1.4 mg in the 2 g sample)
So, in 1 tonne: m(Cu) = 0.0014/2 x 1 = 0.0007 tonnes = 700g

Now is it commercially viable?
It must have a concentration of >0.4% w/w.
From the sample, %w/w = 0.0014/2 = 0.0007 = 0.07 %w/w so it is not commercially viable
Title: Re: VCE Chemistry Question Thread
Post by: Reus on May 11, 2014, 09:21:25 pm
hey :D It's perfectly normal to find some qs tough, that's what we're here for ;)

Part (a)
If we just read off the graph, the concentration of the copper in the solution is about 7.0 μg/ml

Part (b)
The volume of the solution was 200 ml (not 200+10=210 ml because it says it was "diluted to 200 ml")
So, the mass of copper is:  7.0 μg/ml x  200 ml = 1.4 mg

Part (c)
Assuming that the sample is indicative of all the ore, we can say the ratio of copper to ore is the same
ie copper:ore = 0.0014:2    (since we found 1.4 mg in the 2 g sample)
So, in 1 tonne: m(Cu) = 0.0014/2 x 1 = 0.0007 tonnes = 700g

Now is it commercially viable?
It must have a concentration of >0.4% w/w.
From the sample, %w/w = 0.0014/2 = 0.0007 = 0.07 %w/w so it is not commercially viable
Thanks heaps!
Title: Re: VCE Chemistry Question Thread
Post by: TimewaveZero on May 12, 2014, 10:59:20 am
Quick question... Alkanes will only undergo substitution reactions with halogens yes? also, will they only occur if IV light is applied or can it be heat as well?

also, will ALL halogenated alkanes react with water to form alkanols?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 12, 2014, 08:28:04 pm
Quick question... Alkanes will only undergo substitution reactions with halogens yes? also, will they only occur if IV light is applied or can it be heat as well?

also, will ALL halogenated alkanes react with water to form alkanols?

UV light is needed as UV light will break the halogen single bond (which is quite weak). Heat won't do the job.
Halogenated alkanes will react (slowly...not fast at all) with water to form alkanols; you're much better off reacting hydroxide with a halogenated alkane.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on May 13, 2014, 05:35:01 pm
Can someone please clarify my confusion.
Ohk, so a sodium ion is used to replace a cooh group in aspirin because it improves the soubility of the tablet. But, if sodium forms ion-dipole bonds with water, which is clearly stronger than hydrogen bonding, doesn't it mean that the aspirin would be less soluble with a Na group?  :-\
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 13, 2014, 06:16:11 pm
The problem is, if your sodium acetylsalicylate (called sodium AS from here on in) was to not dissolve, it would remain as crystals. Now, the crystals have strong ionic bonds between them, that is true.
However, when the crystal breaks up, there is a lot of energy released by the hydration of sodium cations (your ion-dipole bonds mentioned). Ion dipole bonds are only really meaningful when the crystal has dissolved in water (in the solid form, sure, the sodium cations may attract the oxygens, but the AS anion will repel the oxygens; it gets trippy) as then, each ion can form bonds to multiple water molecules. Sodium, for instance, bonds to six water molecules when dissolved. That's a lot of energy. I'm not exactly sure how many the AS forms.
Title: Re: VCE Chemistry Question Thread
Post by: ETTH96 on May 13, 2014, 08:41:25 pm
Hey guys can someone help me with this question?

The oxidation number of manganese changes as MnO2 (s) is converted to MnO42- how many electrons are gained or lost when one mole of MnO2 is completely converted into MnO2 2- (aq)?

And this one too:
when a polyester forms from reaction between the monomers HOCH2CH2OH and HOOCC6H4COOH the percentage, by mass, of carbon in the polyester is
a) greater than in the monomers
b)) less then in the monomers
c) the same as in the monomers
d) dependent on the length of the polymer chain

can you explain how to figure this out as well please!! thank you
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 13, 2014, 09:32:31 pm
Hey guys can someone help me with this question?

The oxidation number of manganese changes as MnO2 (s) is converted to MnO42- how many electrons are gained or lost when one mole of MnO2 is completely converted into MnO2 2- (aq)?

And this one too:
when a polyester forms from reaction between the monomers HOCH2CH2OH and HOOCC6H4COOH the percentage, by mass, of carbon in the polyester is
a) greater than in the monomers
b)) less then in the monomers
c) the same as in the monomers
d) dependent on the length of the polymer chain

can you explain how to figure this out as well please!! thank you

First one: manganese in manganese dioxide is in oxidation state +4, but in your manganate ion (lol interesting manganese species there), it is oxidation state +6, so it loses two electrons.
You made a typo I think; MnO22- does not exist from my knowledge.

As for the second one, think about it. In your polymerisation, the number of moles of carbon stays the same. However, you lose water when polymerising at each step. Therefore, you may have joined two pairs of monomers, meaning you have twice the amount of carbon as one pair of monomers (I'm considering the collection of diol and diacid as a pair of monomers), but as you've lost water, the total mass of the molecule is less than the mass of two pairs of monomers. Twice the carbon for less than twice the total mass => larger percentage by mass of carbon than in monomers.
Title: Re: VCE Chemistry Question Thread
Post by: ETTH96 on May 13, 2014, 09:42:25 pm
Sorry! i did make a typo! I was meant to type:
The oxidation number of manganese changes as MnO2 (s) is converted to MnO42- how many electrons are gained or lost when one mole of MnO2 is completely converted into MnO42- (aq)?

So, for MnO2, the oxidation number is +4, and for MnO4 2-, the oxidation number is +6.. so 2 electrons are lost?

This makes total sense but, its not one of the choices for the multiple choice :/

a) 3
b) 1.2 x 1024
c) 4
d) 1.8 x 1024

Could you help me again? thanks!
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 13, 2014, 10:14:56 pm
It's 2 electrons for every manganese atom. So one mole of manganese atoms...? You can do the rest.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on May 14, 2014, 11:44:41 pm
I have a few queries that I'd like to clarify:

• How can we calculate the m/e value of the base peak in the mass spectrum of a compound?
• Are we required to know different tests we can perform to determine one organic compound from another? If so, could someone give me a run down of the best tests to perform?
• What is the significance of TMS? Could I say that it's used in NMR spectroscopy as a point of reference? I know the answer, I just don't know how to properly phrase it.

Help would be greatly appreciated :)

Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on May 15, 2014, 07:40:40 am
1) The base peak is the one with the greatest intensity, the biggest peak. All you need to do is look at the spectrum and read off the m/e value.

2) Yes, you should know enough about organic reaction to explain tests which will be able to identify compounds:
- Alkenes - bromine water (is normally brown if an alkene is present it goes clear)
- Carboxylic acids - React with Na2CO3 to form carbon dioxide (acid base reaction)
- 1-alkanols - React with dichromate ions to form carboxylic acids (dichromate is orangish will change to greenish)

3) Yep TMS is added for a point of reference.

Hope this helps, pretty simplistic i know, if anyone else wants to add more detail go for it.

I have a few queries that I'd like to clarify:

• How can we calculate the m/e value of the base peak in the mass spectrum of a compound?
• Are we required to know different tests we can perform to determine one organic compound from another? If so, could someone give me a run down of the best tests to perform?
• What is the significance of TMS? Could I say that it's used in NMR spectroscopy as a point of reference? I know the answer, I just don't know how to properly phrase it.

Help would be greatly appreciated :)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 15, 2014, 08:03:15 pm
1) The base peak is the one with the greatest intensity, the biggest peak. All you need to do is look at the spectrum and read off the m/e value.

2) Yes, you should know enough about organic reaction to explain tests which will be able to identify compounds:
- Alkenes - bromine water (is normally brown if an alkene is present it goes clear)
- Carboxylic acids - React with Na2CO3 to form carbon dioxide (acid base reaction)
- 1-alkanols - React with dichromate ions to form carboxylic acids (dichromate is orangish will change to greenish)

3) Yep TMS is added for a point of reference.

Hope this helps, pretty simplistic i know, if anyone else wants to add more detail go for it.

You could also just use indicator paper for the water-soluble acids.

As for how TMS works, its purpose is to produce some number that doesn't change depending on your machine. The energy absorbed by a nucleus in NMR to change spin state is directly proportional to the external magnetic field. Now, the TMS and the proton in question experience slightly different magnetic fields due to this phenomenon called "shielding". Let's call the constant of proportionality k, the magnetic field experienced by TMS to be B(TMS), the magnetic field experienced by the proton in question be B(H) and E(TMS), E(H) to be the energy absorbed by the respective nuclei.
Then, the chemical shift is defined as $\frac{E(H) - E(TMS)}{E(TMS)} = \frac{kB(H) - kB(TMS)}{kB(TMS)}$
Clearly the k's cancel. This suggests that the chemical shift is really just the ratio of magnetic field strengths. If you can accept that doubling the external magnetic field will double the shielding effect, this last fraction really is constant. That is the effect of defining chemical shifts so weirdly.

Bit of a hand-wavy explanation but it hopefully shows what chemical shifts were designed for.
Title: Re: VCE Chemistry Question Thread
Post by: Einstein on May 15, 2014, 11:05:41 pm
Currently in year 11 near the end of organic chem, i want to make a start on unit 3/4 stuff, and was wondering best way to start? since half of the course will require stoich etc which we learn near term 3.

Really, what can i do in 3/4 book that can extend my knowledge of 1/2, for example do 3/4 organic chem to extend me 1/2 knowledge if you understand where im going,
Thanks
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on May 16, 2014, 10:59:24 am
I would say your best bet look at the organic reactions area of chemistry.  Chapter 9/10 in the Heinemann textbook is really just an extension of year 11, so start there, it that is okay have a look at biofuels and biomolecules.  It is all based around the same idea, how function groups react together.

Currently in year 11 near the end of organic chem, i want to make a start on unit 3/4 stuff, and was wondering best way to start? since half of the course will require stoich etc which we learn near term 3.

Really, what can i do in 3/4 book that can extend my knowledge of 1/2, for example do 3/4 organic chem to extend me 1/2 knowledge if you understand where im going,
Thanks
Title: Re: VCE Chemistry Question Thread
Post by: speedy on May 16, 2014, 03:47:56 pm
What differences would you expect to see in the HNMR and IR spectrum for aspirin vs. salicylic acid?
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on May 16, 2014, 04:50:56 pm
I have my aspirin sac on Monday and I have answered some revision questions
Could someone please check that I am correct?
- What's the purpose of adding Sulfuric acid?
The H2SO4 acts as a catalyst and speeds up the reaction
- What is the limiting reagent in the reaction?
The limiting reagent is salicylic acid as ethanoic anhydride is added in excess
- Why is water added at the end of the reaction?
Water is added at the end of the reaction to allow for crystallisation to occur and it cools down the reaction
-Why is cold water used?
Cold water is used to lower the solubility of aspirin. Water is also used to ensure that all other solutes and reactants are washed away.

Feel free to add anything
Much appreciated :)
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on May 17, 2014, 01:02:43 pm
- Why is water added at the end of the reaction?
Water is added at the end of the reaction to allow for crystallisation to occur and it cools down the reaction
The water also converts any excess acetic anhydride to acetic acid.
Title: Re: VCE Chemistry Question Thread
Post by: Eliaz on May 18, 2014, 11:02:20 am
How should i prepare for my EEI, i do have the results and my SAC is next monday . My cohort has performed our experiments on an energy drink

Title: Re: VCE Chemistry Question Thread
Post by: Einstein on May 18, 2014, 12:25:34 pm
can someone explain to me how the answer is C, much thanks :)
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on May 18, 2014, 12:53:46 pm
can someone explain to me how the answer is C, much thanks :)
start counting from COOH to find the longest carbon chain. methyl group at 6th carbon and propyl group at 5th carbon.
Hence 6 methyl 5 propyl octanoic acid
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on May 18, 2014, 12:58:32 pm
I have my aspirin sac on Monday and I have answered some revision questions
Could someone please check that I am correct?
- What's the purpose of adding Sulfuric acid?
The H2SO4 acts as a catalyst and speeds up the reaction
- What is the limiting reagent in the reaction?
The limiting reagent is salicylic acid as ethanoic anhydride is added in excess
- Why is water added at the end of the reaction?
Water is added at the end of the reaction to allow for crystallisation to occur and it cools down the reaction
-Why is cold water used?
Cold water is used to lower the solubility of aspirin. Water is also used to ensure that all other solutes and reactants are washed away.

Feel free to add anything
Much appreciated :)

For the sake of being meticulous, I'd probably specify that concentrated sulfuric acid is used to increase the rate of the condensation reaction (i.e. acts as a catalyst) by which acetylsalicylic acid - aspirin - is produced.

Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on May 18, 2014, 01:07:00 pm
can someone explain to me how the answer is C, much thanks :)

The first thing to do is count the longest (parent) chain. When a carboxyl group is present, that must be counted in the parent chain. The longest parent chain (including the carboxyl group) contains 8 carbon atoms. Hence, this is octanoic acid (a number is not required to specificy the location of the carboxyl group because it must be present at the terminal carbon atom). The next thing to do is notice that we have a methyl (CH3) group coming off one of the carbons, and a propyl (CH3CH2CH3) group coming off another carbon. The important thing to do now is note what number carbon they're coming off. Counting from the carbon of the carboxyl group (i.e. Carbon one), we can see that the propyl group is coming off carbon number 5, while the methyl group is coming off of carbon 6.

So, we know three things now:

Parent chain: octanoic acid
We also have 5-propyl and 6-methyl

So, our compound:

6-methyl-5-propyl octanoic acid, C.
Title: Re: VCE Chemistry Question Thread
Post by: Einstein on May 18, 2014, 06:32:43 pm
i got a different answer, and you can probably see where i sourced it from, can you tell me what i did wrong?

our teacher said you cant name the ethyls, methlys etc if they are a part of the chain.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on May 18, 2014, 07:09:26 pm
Relating to the aspirin practical, can anybody think of any sources of errors that could've resulted in a higher / lower percentage yield?
Thank you :)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 18, 2014, 07:11:04 pm
i got a different answer, and you can probably see where i sourced it from, can you tell me what i did wrong?

our teacher said you cant name the ethyls, methlys etc if they are a part of the chain.

Why can't you name the ethyl and methyl groups here? If you count your parent hydrocarbon chain as going up and then to the left (that one), you'll find that you can name it quite easily.
Title: Re: VCE Chemistry Question Thread
Post by: Einstein on May 18, 2014, 07:21:17 pm
Why can't you name the ethyl and methyl groups here? If you count your parent hydrocarbon chain as going up and then to the left (that one), you'll find that you can name it quite easily.

im not sure what you mean here , sorry
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on May 18, 2014, 07:23:12 pm
i got a different answer, and you can probably see where i sourced it from, can you tell me what i did wrong?

our teacher said you cant name the ethyls, methlys etc if they are a part of the chain.
start counting from the COOH...
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on May 18, 2014, 07:30:25 pm
apologies for shitty drawing but hope this helps
Title: Re: VCE Chemistry Question Thread
Post by: Einstein on May 18, 2014, 08:20:21 pm
start counting from the COOH...

ahh i got it, is that because the double bond takes precedence?

also how did you know to count from cooh then move to left and not right if that makes sense (in terms of determining longest chain) is this right?
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on May 18, 2014, 10:21:57 pm
do we need to worry about multiplicity for H1 NMR regarding peak splitting? ie. CH3-CH2-CH2-CH2-CH3, the hydrogen environment (first CH2) being split into 12 peaks.
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on May 18, 2014, 10:23:19 pm
ahh i got it, is that because the double bond takes precedence?

also how did you know to count from cooh then move to left and not right if that makes sense (in terms of determining longest chain) is this right?
he went left because counting left gives you a longer chain. With COOH the carbon there is always the carbon number 1 and you start counting from that.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on May 18, 2014, 10:40:59 pm
do we need to worry about multiplicity for H1 NMR regarding peak splitting? ie. CH3-CH2-CH2-CH2-CH3, the hydrogen environment (first CH2) being split into 12 peaks.

This may sound weird but did you do a STAV paper? I did one today and saw the exact same question lol.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 18, 2014, 10:59:32 pm
do we need to worry about multiplicity for H1 NMR regarding peak splitting? ie. CH3-CH2-CH2-CH2-CH3, the hydrogen environment (first CH2) being split into 12 peaks.

Most likely not

This may sound weird but did you do a STAV paper? I did one today and saw the exact same question lol.

So keen

ahh i got it, is that because the double bond takes precedence?

also how did you know to count from cooh then move to left and not right if that makes sense (in terms of determining longest chain) is this right?

It's not the double bond that takes precedence; the COOH functional group takes precedence.

In rare cases, it may be simpler to not treat the COOH as the parent functional group, but you won't come across these in VCE.
Title: Re: VCE Chemistry Question Thread
Post by: Einstein on May 19, 2014, 07:54:08 am

It's not the double bond that takes precedence; the COOH functional group takes precedence.

In rare cases, it may be simpler to not treat the COOH as the parent functional group, but you won't come across these in VCE.

Yeap, thats what i meant. Thanks for the help :)
Title: Re: VCE Chemistry Question Thread
Post by: Reus on May 19, 2014, 07:57:34 pm
I know we are required to know Chloroalkanes, but what about Chloroalkenes?
I presume it won't make a difference in theory except the alkyl group...
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on May 20, 2014, 10:21:11 pm
Hey can someone check if my understanding of the Collision Theory and how to increase reaction rates is accurate? Please feel free to edit/add/remove anything that is relevant to the VCAA course. Thanks!

Increasing concentration of reactants/ increasing pressure of gaseous reactant molecules:
By increasing the concentration of reactants, there is an increase in the number of collisions between reactant molecules. As a result of the increased collision rate, there is an increase in the rate of the overall reaction.

Catalysts lower the minimum quantum of energy required to disrupt bonds within reactant molecules. In the presence of a catalyst, the activation energy is lowered. As a result, a lower amount of energy must be absorbed by the bonds in the reactant molecules in order for the bonds to be broken. Once this has occured, the molecules which have absorbed the quantum of energy needed to activate the reaction collide with one another, increasing the rate at which the reaction occurs as collision rate increases.

Increasing temperature:
By increasing temperature, the amount of kinetic energy available also increases. As a result, molecules collide more frequently with one another. As the collision rate increases, the overall rate of the reaction also increases.

Increasing surface area of solids:
As surface area increases, the amount of the solid molecules exposed to other molecules' surfaces increases. As a result, there is a greater number of collisions between molecules at their surfaces. The increased collision rate leads to an increase in the overall reaction rate.

Help would be much appreciated. Thank you :)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 20, 2014, 10:50:09 pm
Hey can someone check if my understanding of the Collision Theory and how to increase reaction rates is accurate? Please feel free to edit/add/remove anything that is relevant to the VCAA course. Thanks!

Increasing concentration of reactants/ increasing pressure of gaseous reactant molecules:
By increasing the concentration of reactants, there is an increase in the number of collisions between reactant molecules. As a result of the increased collision rate, there is an increase in the rate of the overall reaction.

Catalysts lower the minimum quantum of energy required to disrupt bonds within reactant molecules. In the presence of a catalyst, the activation energy is lowered. As a result, a lower amount of energy must be absorbed by the bonds in the reactant molecules in order for the bonds to be broken. Once this has occured, the molecules which have absorbed the quantum of energy needed to activate the reaction collide with one another, increasing the rate at which the reaction occurs as collision rate increases.

Increasing temperature:
By increasing temperature, the amount of kinetic energy available also increases. As a result, molecules collide more frequently with one another. As the collision rate increases, the overall rate of the reaction also increases.

Increasing surface area of solids:
As surface area increases, the amount of the solid molecules exposed to other molecules' surfaces increases. As a result, there is a greater number of collisions between molecules at their surfaces. The increased collision rate leads to an increase in the overall reaction rate.

Help would be much appreciated. Thank you :)

Increase temperature => increase average kinetic energy of molecules => increase proportion of molecules that have more energy than activation energy to react => increase reaction rate. This is the second effect of temperature and is MUCH more important. In fact, the dependence on particle speed with temperature is v^2 proportional to T (for an ideal gas), but the relationship between the proportion of particles with more energy than the activation energy Ea is e^(-Ea/RT). See how important T is?

Fun fact: you can actually have a reaction that slows down if temperature is increased. Doesn't occur for many reactions but it happens :P

Also, solids aren't always molecules. Often, they're ions like in sodium chloride or even ammonium nitrate. Of course, some are, like iodine, ice and sulfur. I'm being picky here :P

Bear in mind that catalysts sometimes disrupt bonds themselves. Don't say quantum of energy as the use of quantum implies here that you can only absorb discrete multiples of a certain amount of energy. (ironic how the public uses "quantum" to mean large. Uninformed people)
You should then say that there are more particles with energy greater than the activation energy now that it has been lowered, instead of the bolded sentence.

I know we are required to know Chloroalkanes, but what about Chloroalkenes?
I presume it won't make a difference in theory except the alkyl group...

No difference really, except for iffy cases when you have isomerism about the double bond. There is a reason why alkenes aren't mentioned much in VCE :P
Title: Re: VCE Chemistry Question Thread
Post by: Saikyo on May 20, 2014, 11:00:02 pm
Can anyone help me with this question:

A student titrated an aliquot of standard sodium carbonate solution with hydrochloric acid in a burette. State whether the concentration determined for the hydrochloric acid would be likely to be higher than, lower than or unchanged from the actual value if the student had previously washed with water, but not dried, the following apparatus:

a) the pipette used to deliver the aliquot of sodium carbonate solution
b) the burette.

I'm confused why the answer says for a) would have a higher concentration than actual value?? To my understanding, wouldn't the sodium carbonate solution from pipette be diluted from the water and therefore having a less concentration??

Also, for b) if I dilute the acid in burette, wouldnt the concentration be also less therefore bigger titre is needed? But yet the answer says lower?

I'm not sure what I'm saying is correct, so could any of you clarify or point out any thing wrong I said. I would appreciate that :)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 20, 2014, 11:10:41 pm
Can anyone help me with this question:

A student titrated an aliquot of standard sodium carbonate solution with hydrochloric acid in a burette. State whether the concentration determined for the hydrochloric acid would be likely to be higher than, lower than or unchanged from the actual value if the student had previously washed with water, but not dried, the following apparatus:

a) the pipette used to deliver the aliquot of sodium carbonate solution
b) the burette.

I'm confused why the answer says for a) would have a higher concentration than actual value?? To my understanding, wouldn't the sodium carbonate solution from pipette be diluted from the water and therefore having a less concentration??

Also, for b) if I dilute the acid in burette, wouldnt the concentration be also less therefore bigger titre is needed? But yet the answer says lower?

I'm not sure what I'm saying is correct, so could any of you clarify or point out any thing wrong I said. I would appreciate that :)

So, in a), the pipette actually had less sodium carbonate than what the student thought. As a result, the student would have used less hydrochloric acid than they should have for the titration. As they would not have factored any issues in experimental technique into their calculations, they would assume the sodium carbonate concentration hadn't changed, so the number of moles of sodium carbonate, and thus hydrochloric acid, are both the correct values. However, the volume of hydrochloric acid has decreased, resulting in a higher calculated concentration than what it should have.

For b), a more dilute acid in the burette will require a higher titre volume than normal, yes. However, remember you're calculating the concentration of this HCl solution. You have a constant number of moles of sodium carbonate and thus HCl, yet your volume of HCl solution has increased due to this error. Therefore, the effect of this error is to decrease the calculated concentration of HCl.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on May 20, 2014, 11:51:18 pm
Increase temperature => increase average kinetic energy of molecules => increase proportion of molecules that have more energy than activation energy to react => increase reaction rate. This is the second effect of temperature and is MUCH more important. In fact, the dependence on particle speed with temperature is v^2 proportional to T (for an ideal gas), but the relationship between the proportion of particles with more energy than the activation energy Ea is e^(-Ea/RT). See how important T is?

Fun fact: you can actually have a reaction that slows down if temperature is increased. Doesn't occur for many reactions but it happens :P

Also, solids aren't always molecules. Often, they're ions like in sodium chloride or even ammonium nitrate. Of course, some are, like iodine, ice and sulfur. I'm being picky here :P

Bear in mind that catalysts sometimes disrupt bonds themselves. Don't say quantum of energy as the use of quantum implies here that you can only absorb discrete multiples of a certain amount of energy. (ironic how the public uses "quantum" to mean large. Uninformed people)
You should then say that there are more particles with energy greater than the activation energy now that it has been lowered, instead of the bolded sentence.

Thanks so much for that :)
Title: Re: VCE Chemistry Question Thread
Post by: Rod on May 21, 2014, 08:53:39 pm
Hey can someone check if my understanding of the Collision Theory and how to increase reaction rates is accurate? Please feel free to edit/add/remove anything that is relevant to the VCAA course. Thanks!

Increasing concentration of reactants/ increasing pressure of gaseous reactant molecules:
By increasing the concentration of reactants, there is an increase in the number of collisions between reactant molecules. As a result of the increased collision rate, there is an increase in the rate of the overall reaction.

Catalysts lower the minimum quantum of energy required to disrupt bonds within reactant molecules. In the presence of a catalyst, the activation energy is lowered. As a result, a lower amount of energy must be absorbed by the bonds in the reactant molecules in order for the bonds to be broken. Once this has occured, the molecules which have absorbed the quantum of energy needed to activate the reaction collide with one another, increasing the rate at which the reaction occurs as collision rate increases.

Increasing temperature:
By increasing temperature, the amount of kinetic energy available also increases. As a result, molecules collide more frequently with one another. As the collision rate increases, the overall rate of the reaction also increases.

Increasing surface area of solids:
As surface area increases, the amount of the solid molecules exposed to other molecules' surfaces increases. As a result, there is a greater number of collisions between molecules at their surfaces. The increased collision rate leads to an increase in the overall reaction rate.

Help would be much appreciated. Thank you :)
Your schools up to unit 4, aos 2 already ? :O
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on May 21, 2014, 08:54:43 pm
Your schools up to unit 4, aos 2 already ? :O
doubtful. knowing Yacoubb, pretty sure he is personally up to it  :P
Title: Re: VCE Chemistry Question Thread
Post by: Reus on May 21, 2014, 09:03:38 pm
Thanks so much for that :)

doubtful. knowing Yacoubb, pretty sure he is personally up to it  :P

How do you find the time Yacoubb! I struggle with the regulated AoS homework we're doing right now!  ::)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 21, 2014, 09:04:17 pm
Your schools up to unit 4, aos 2 already ? :O

That's aos 2? It's certainly listed first in my chemistry textbook before equilibrium. At least, I've run through/am running through these with my chem students before the equilibrium bit.
Title: Re: VCE Chemistry Question Thread
Post by: thushan on May 21, 2014, 09:15:32 pm
Your schools up to unit 4, aos 2 already ? :O

Nah, that's U4AOS1. U4AOS1 is reaction kinetics, equilibria and industrial chemistry.
Title: Re: VCE Chemistry Question Thread
Post by: TimewaveZero on May 21, 2014, 09:55:26 pm
Probably a silly question, but just wanted to clarify something that only occured to me then...

are haloalkanes (eg chloroalkane) part of the alkane homologous series or are haloalkanes a completely separate homologous series? Going by the chemistry definition of homologous series:having the same functional group but differing in composition by a fixed group of atoms. This would suggest that chloroalkanes are a homologous series itself, and then say, bromoalkanes are another one...
Title: Re: VCE Chemistry Question Thread
Post by: Rod on May 21, 2014, 10:07:06 pm
Nah, that's U4AOS1. U4AOS1 is reaction kinetics, equilibria and industrial chemistry.
Our school is only up to mass spec lol. But we did organics (aos 2) first.

Thushan how would you rate unit 4? Smaller in length? Larger? Harder? Easier? Than unit 3.

I have all my unit 4 sacs in the same term.
Title: Re: VCE Chemistry Question Thread
Post by: thushan on May 21, 2014, 10:13:59 pm
Smaller, more conceptual and less memory, and easier.

Check out the vTextbook videos for Unit 4 - I gave them :D
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on May 21, 2014, 10:19:04 pm
How do you find the time Yacoubb! I struggle with the regulated AoS homework we're doing right now!  ::)

Haha my school just finished unit 3. I'm up to area of study 2 now! V-textbook is AWESOME!

I really just find spare time (i.e. when I don't have SACs) to just go ahead. I'm aiming to finish the VCAA Chem course by the start of term 3, and dedicate all of term 3 to prac exams/revision of units 3 and 4 of Chem.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 21, 2014, 10:21:24 pm
Probably a silly question, but just wanted to clarify something that only occured to me then...

are haloalkanes (eg chloroalkane) part of the alkane homologous series or are haloalkanes a completely separate homologous series? Going by the chemistry definition of homologous series:having the same functional group but differing in composition by a fixed group of atoms. This would suggest that chloroalkanes are a homologous series itself, and then say, bromoalkanes are another one...

They would be, yes.

Our school is only up to mass spec lol. But we did organics (aos 2) first.

Thushan how would you rate unit 4? Smaller in length? Larger? Harder? Easier? Than unit 3.

I have all my unit 4 sacs in the same term.

Much less rote learning hahahaha
And much more fun for me

Haha my school just finished unit 3. I'm up to area of study 2 now! V-textbook is AWESOME!

I really just find spare time (i.e. when I don't have SACs) to just go ahead. I'm aiming to finish the VCAA Chem course by the start of term 3, and dedicate all of term 3 to prac exams/revision of units 3 and 4 of Chem.

Finishing VCAA Chem course by start of term 3? Keen eh xP
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on May 21, 2014, 10:30:20 pm
Finishing VCAA Chem course by start of term 3? Keen eh xP

LOL I just want to do well in Chemistry. A lot of you guys are geniuses! I need LOTs of time to reach your standards :P
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on May 21, 2014, 10:31:39 pm
when performing the bromine test to see if the hydrocarbon is saturated or unsaturated, the Bromine is aqueous...I thought the first four alkanes were gas at room temperature? Could someone explain this?
Title: Re: VCE Chemistry Question Thread
Post by: darklight on May 21, 2014, 10:35:23 pm
Qn 2 should be an easy question.. but I keep getting none of the options  :-[
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on May 21, 2014, 10:39:00 pm
Qn 2 should be an easy question.. but I keep getting none of the options  :-[

Okay so at an absorbance of 24, the concentration is 13.7 ug/L. This, however, is not the concentration of copper ions in the original solution; this is the concentration of Cu2+ ions in the 'diluted' solution. So, we must calculate the concentration of the original solution by multiplying 13.7ug/L by the dilution factor, which is 100 (100mL/1mL), giving us 1370ug/L, which is C.
Title: Re: VCE Chemistry Question Thread
Post by: darklight on May 21, 2014, 10:42:55 pm
Okay so at an absorbance of 24, the concentration is 13.7 ug/L. This, however, is not the concentration of copper ions in the original solution; this is the concentration of Cu2+ ions in the 'diluted' solution. So, we must calculate the concentration of the original solution by multiplying 13.7ug/L by the dilution factor, which is 100 (100mL/1mL), giving us 1370ug/L, which is C.

Thanks Yacoubb, I understand that, but that's q1 not 2 :P
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on May 21, 2014, 10:47:38 pm
Thanks Yacoubb, I understand that, but that's q1 not 2 :P

is the answer C?
Title: Re: VCE Chemistry Question Thread
Post by: darklight on May 21, 2014, 10:48:42 pm

is the answer C?

Yes! How did you get that?  :D
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on May 21, 2014, 10:50:33 pm
mass of Cu2+ ions = 1370ug/L
in sample, mass of Cu2+ ions = 1.37 ug
n=m/m n=1.37x10-3 / 63.5
n(Cu2+) = 0.000022
n(Cu) = n(CuCl2)

n=m x M
m(CuCl2) = 0.002902g = 2.90 ug
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on May 21, 2014, 11:06:55 pm
Thanks Yacoubb, I understand that, but that's q1 not 2 :P

Oh woops, sorry! I thought you didn't get q1. swagsxcboi explained q2, so hopefully you got it !
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on May 21, 2014, 11:17:26 pm
when performing the bromine test to see if the hydrocarbon is saturated or unsaturated, the Bromine is aqueous...I thought the first four alkanes were gas at room temperature? Could someone explain this?
can't you bubble gas through solution?
Title: Re: VCE Chemistry Question Thread
Post by: Rod on May 22, 2014, 07:34:50 am
LOL I just want to do well in Chemistry. A lot of you guys are geniuses! I need LOTs of time to reach your standards :P
50 in chem right there :D

Hey Yacoubb, when you go ahead in chem, do you watch vtextbook, make detailed notes, and then finish of all the textbook questions? That's what I've been doing

Or do you do checkpoint/exam style questions instead after watching videos and making detailed notes?

Thanks :)
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on May 22, 2014, 09:25:13 am
50 in chem right there :D

Hey Yacoubb, when you go ahead in chem, do you watch vtextbook, make detailed notes, and then finish of all the textbook questions? That's what I've been doing

Or do you do checkpoint/exam style questions instead after watching videos and making detailed notes?

Thanks :)

What I do is read up on the concepts, watch V-textbook and basically write up my own notes. Once I've familiarised myself with the theory, I begin doing the textbook questions. After this, I do checkpoints for more complex questions. I find that this really worked for me.

Title: Re: VCE Chemistry Question Thread
Post by: Rod on May 22, 2014, 03:46:43 pm
What I do is read up on the concepts, watch V-textbook and basically write up my own notes. Once I've familiarised myself with the theory, I begin doing the textbook questions. After this, I do checkpoints for more complex questions. I find that this really worked for me.
I do the exact same thing except I tend to save the checkpoint questions for SAC prep.

Thanks Yacoubb :). Keep up the great work!
Title: Re: VCE Chemistry Question Thread
Post by: Reus on May 22, 2014, 05:46:48 pm
How does one write an equation illustrating fragmentation?
C5H12
72 m/e
Title: Re: VCE Chemistry Question Thread
Post by: jgoudie on May 23, 2014, 10:54:31 am
Fragmentation envolves the breaking up of an ion.  The 72 m/e is the mass of C5H12.  Was there a larger ion that this fragment broke up from?  Or a smaller ion that this breaks down into?  Other wise you would be looking at the formation of the parent molecular ion rather than fragmentation.

Seems the question is lacking some information.

How does one write an equation illustrating fragmentation?
C5H12
72 m/e
Title: Re: VCE Chemistry Question Thread
Post by: Einstein on May 23, 2014, 05:52:08 pm
what is delocalised electrons?

thanks
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 23, 2014, 06:57:41 pm
Delocalised electrons refer to electrons that don't reside in a bond between two atoms. For instance, in benzene each carbon has an electron that isn't confined to being between two carbons in a chemical bond, but can move around all of the carbon atoms.
Title: Re: VCE Chemistry Question Thread
Post by: Einstein on May 23, 2014, 08:55:29 pm
Delocalised electrons refer to electrons that don't reside in a bond between two atoms. For instance, in benzene each carbon has an electron that isn't confined to being between two carbons in a chemical bond, but can move around all of the carbon atoms.

thanks mate. And is that cyclobenzene?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 23, 2014, 09:06:12 pm
It's just called benzene
Title: Re: VCE Chemistry Question Thread
Post by: Reus on May 23, 2014, 09:20:06 pm
Just a question, for the unbalanced equation for the production of:
Methanol from chloromethane, I got the following:

CH3Cl(g)+ OH- --> CH3OH(aq) + Cl-

However the book's answers page has it as CH3Cl(g) -OH--> CH3OH(aq) + Cl-
(with the hydroxyl group on top of the arrow)
Does that really matter? Would mine be considered wrong? Or is it just a more technical way to express it?
Thanks.
Title: Re: VCE Chemistry Question Thread
Post by: thushan on May 23, 2014, 10:19:50 pm
Same thing. There are many ways of expressing it. The answers use a shorthand form. Your equation is also correct.
Title: Re: VCE Chemistry Question Thread
Post by: Einstein on May 25, 2014, 11:45:43 am
could someone please give me a definition of a molecule and compound, and any examples of each.

Thanks  ;D
Title: Re: VCE Chemistry Question Thread
Post by: Reus on May 25, 2014, 12:10:50 pm
could someone please give me a definition of a molecule and compound, and any examples of each.

Thanks  ;D
I'd define it as a molecule being formed when two or more atoms of the same element join together chemically (e.g. N2 or H2) whereas a compound is a molecule that contains at least two different elements (e.g. CO2 or NH3)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 25, 2014, 12:32:34 pm
Compounds include ionic compounds. This is a much broader category, but doesn't include allotropes of elements like ozone. Chemical compounds (that are pure) consist of atoms of different elements that can be separated by chemical reactions.
Molecules don't; we generally think of them as consisting of atoms joined by covalent bonds, and these do include things like O3, O2 and S8.
Title: Re: VCE Chemistry Question Thread
Post by: katiesaliba on May 25, 2014, 04:50:57 pm
For esters in year 12, do we only use alkanols with hydroxyl groups attached to their 1st carbon?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 25, 2014, 04:57:16 pm
They can give you other kinds of alcohols; not common though. Primary alcohols are most common in these reactions.
Title: Re: VCE Chemistry Question Thread
Post by: katiesaliba on May 25, 2014, 04:59:23 pm
They can give you other kinds of alcohols; not common though. Primary alcohols are most common in these reactions.
How would I write the semi-structural formula of an ester formed with a more complex alkanol? Could you please give me an example? Thank you :)
Title: Re: VCE Chemistry Question Thread
Post by: Scooby on May 25, 2014, 11:34:47 pm
A few questions to ask you geniuses!  ;D
• Acetaldehyde (CH3CHO) to Ethanoic Acid, what type of reaction would this be?
• How will this accumulation affect glycolysis and the pathways that produce CO2? Note that ethanol consumption leads to an accumulation of NADH.

It's oxidation. Ethanol is metabolised in the liver, which involves it first being converted to acetaldehyde, which is converted to ethanoic acid, which is converted to acetyl Co-A. As ethanol is metabolised, NAD+ is consumed and converted to NADH. Since NAD+ is required for glycolysis, the rate at which glycolysis occurs would reduce if NAD+ is depleted (NAD+ is essentially a limiting reactant).

Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 25, 2014, 11:59:45 pm
A few questions to ask you geniuses!  ;D
• Acetaldehyde (CH3CHO) to Ethanoic Acid, what type of reaction would this be?
• How will this accumulation affect glycolysis and the pathways that produce CO2? Note that ethanol consumption leads to an accumulation of NADH.

This isn't a question in the scope of the 3/4 chem course btw :P
but any interested bio students are free to read Scooby's answer above
Title: Re: VCE Chemistry Question Thread
Post by: darklight on May 26, 2014, 12:57:58 pm
Although benzene does not undergo the bromine test, will other cyclic molecules undergo addition reactions with bromine, causing the red solution to turn colourless? :)
Title: Re: VCE Chemistry Question Thread
Post by: Reus on May 26, 2014, 04:34:58 pm
Thanks scooby!
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on May 26, 2014, 07:14:22 pm
What are the functional groups of aspirin? Carboxyl group and...?
Title: Re: VCE Chemistry Question Thread
Post by: Reus on May 26, 2014, 07:23:38 pm
What are the functional groups of aspirin? Carboxyl group and...?
Is it not Ester?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on May 26, 2014, 07:27:24 pm
Is it not Ester?

Yeah just making sure! Thanks
Title: Re: VCE Chemistry Question Thread
Post by: eagles on May 26, 2014, 08:01:25 pm
Page 137 of Heinemann textbook, Table 9.4 Common functional groups in organic compounds.

Why do some of the formula for functional groups, eg. Chloro, hydroxyl and amino don't have the 'R' group whereas ester, amiss and ether do?

Thank you.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 26, 2014, 08:13:20 pm
Although benzene does not undergo the bromine test, will other cyclic molecules undergo addition reactions with bromine, causing the red solution to turn colourless? :)

Benzene...under certain conditions will react with bromine, although it's not an addition reaction; rather, if you have FeBr3 as a catalyst, you'll promote a substitution reaction where one of the bromine in Br2 substitutes for a H in benzene.
Title: Re: VCE Chemistry Question Thread
Post by: Reus on May 26, 2014, 08:27:06 pm
A few questions once again!
• Balanced equation for the conversion of CH3CHO to CH3COOH in the presence of NAD+
• The type of reaction of this would be?
• How does the omission of coenzymes and cofactors effect on the enzyme's activity?

Thanks!
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 26, 2014, 08:39:39 pm
A few questions once again!
• Balanced equation for the conversion of CH3CHO to CH3COOH in the presence of NAD+
• The type of reaction of this would be?
• How does the omission of coenzymes and cofactors effect on the enzyme's activity?

Thanks!

Ok first one.
CH3CHO to CH3COOH? NAD+ just goes to NADH, so the half reaction is NAD+ + H+ + 2e- => NADH for the reduction
As for the oxidation, CH3CHO => CH3COOH          Balance oxygens by adding water
CH3CHO + H2O => CH3COOH     Balance hydrogens
CH3CHO + H2O => CH3COOH + 2H+ + 2e-   See how I added electrons to balance the charge?
Put the two equations together to give NAD+ + CH3CHO => CH3COOH + H+ + H2O
This is a redox reaction

Omitting coenzymes and cofactors means the enzyme can't work. They're a crucial part of the enzyme's function. For example, the zinc cation in carbonic anhydrase is what facilitates the entire catalysis of H2CO3 formation.

Thanks so much! SO much clearer now. : ;D
However how does the accumulation affect the production of CO2?
I'm taking a wild guess that it increases it?

What happens is that by consuming ethanol and requiring the use of the oxidant NAD+,  your body has less NAD+. Remember how the Krebs/citric acid cycle requires 3 lots of NAD+ and 1 lot of FAD per molecule of acetyl CoA? If you run low on NAD+, the Krebs cycle can't run as fast and less CO2 is produced.
Title: Re: VCE Chemistry Question Thread
Post by: Reus on May 26, 2014, 08:56:21 pm
Ok first one.
Thank you!
Title: Re: VCE Chemistry Question Thread
Post by: danman.9 on May 26, 2014, 09:32:58 pm
Note that ethanol consumption leads to an accumulation of NADH. How will this accumulation affect
glycolysis and the pathways that produce CO2???
Title: Re: VCE Chemistry Question Thread
Post by: Reus on May 26, 2014, 09:37:37 pm
Note that ethanol consumption leads to an accumulation of NADH. How will this accumulation affect
glycolysis and the pathways that produce CO2???
Well well well...

It's oxidation. Ethanol is metabolised in the liver, which involves it first being converted to acetaldehyde, which is converted to ethanoic acid, which is converted to acetyl Co-A. As ethanol is metabolised, NAD+ is consumed and converted to NADH. Since NAD+ is required for glycolysis, the rate at which glycolysis occurs would reduce if NAD+ is depleted (NAD+ is essentially a limiting reactant).

What happens is that by consuming ethanol and requiring the use of the oxidant NAD+,  your body has less NAD+. Remember how the Krebs/citric acid cycle requires 3 lots of NAD+ and 1 lot of FAD per molecule of acetyl CoA? If you run low on NAD+, the Krebs cycle can't run as fast and less CO2 is produced.
Title: Re: VCE Chemistry Question Thread
Post by: danman.9 on May 26, 2014, 11:27:10 pm
thank you Reus  :)
Title: Re: VCE Chemistry Question Thread
Post by: Einstein on May 28, 2014, 05:51:05 pm
why exactly do shells further from the nucleus have higher energy?

also if metals have high melting and boiling points, and alkali (group 1 ) are metals, why do they have low melting and boiling points?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 28, 2014, 08:17:11 pm
why exactly do shells further from the nucleus have higher energy?

also if metals have high melting and boiling points, and alkali (group 1 ) are metals, why do they have low melting and boiling points?

These 'shells' can be thought of as electrons spending most of their time in a region of space. Now, electrostatic attraction, mathematically, behaves like gravitational attraction. They're both inverse square laws. Given this, do you expect a rock half a metre off the surface of the Earth or a rock 20 m above the surface of the Earth to have more potential energy? Similar idea for electrons.

why exactly do shells further from the nucleus have higher energy?

also if metals have high melting and boiling points, and alkali (group 1 ) are metals, why do they have low melting and boiling points?

Alkali metals only have 1 valence electron to contribute to the delocalised electron sea. The attractions between the cations and the electron sea are therefore much weaker. In addition, group 1 metals are the largest elements when it comes to ionic radii for their period, so the electrons are further away from the cations, weakening attractions.
Title: Re: VCE Chemistry Question Thread
Post by: Einstein on May 28, 2014, 11:04:26 pm
Thanks Izxnl :0 However i didnt grasp all the concepts their, i was wondering if you could dumben (if thats a word) it down a bit haha. :(
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on May 28, 2014, 11:18:45 pm
Which one didn't you get?

For the first one, can you imagine that a ball attracted to the Earth by gravity has more potential energy if you raise it higher and higher above the Earth's surface? A similar thing occurs with the attraction between the positive nucleus and the negative electrons. The further the electron is from the nucleus, the more potential energy it has.

As for the second one, alkali metals only have 1 electron in their valence shell to donate to the sea of delocalised electrons. Consider titanium with four valence electrons to donate. Its metallic bonding is going to be stronger as there are Ti 4+ cations with four times the number of electrons available to bond with as opposed to K+. Also the cation charge for Ti 4+ is bigger. Therefore, the attractions between the cations and the electrons are weaker for alkali metals so they melt at lower temperatures.
Title: Re: VCE Chemistry Question Thread
Post by: PsychoT on May 29, 2014, 01:00:06 pm
How do I determine the amount of peptide links in a protein? Or more more specifically, one molecule of insulin?
Title: Re: VCE Chemistry Question Thread
Post by: TimewaveZero on May 29, 2014, 07:11:47 pm
Name of this organic compound?

(CH3)2C(OH)CHClCH2CH3

I don't understand how you can have 2 consecutive CH3 groups in a molecule other than ethane?

EDIT: Nevermind, I realise now that it's 2 CH3 side chains from the following carbon.
Title: Re: VCE Chemistry Question Thread
Post by: Blondie21 on May 29, 2014, 08:26:54 pm
If
n(Na2SO4.H2O) = 0.094mol,
does
n(H2O) = 0.094mol
Title: Re: VCE Chemistry Question Thread
Post by: thushan on May 29, 2014, 08:38:09 pm
If
n(Na2SO4.H2O) = 0.094mol,
does
n(H2O) = 0.094mol
That's right.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on June 01, 2014, 08:59:13 am
Could someone please check if my answers to these questions are correct? Thanks :)

1. What are the differences between phenol and alcohol?
Phenol is an aromatic organic compound made up of a benzene ring attached to a hydroxyl group, whereas an alcohol is an organic compound containing a hydroxyl functional group.

2. How can we test a sample to prove that it is a phenol and not alcohol?
By reacting the sample with iron (III) salt. If a dark purple-coloured solution results, the sample contains phenol. However, if no reaction occurs and the colour of the solution does not become purple, the sample is an alcohol.

3a. What type of reaction is the synthesis of aspirin?
Esterification (condensation)

b. What does it mean when the reaction is written with a double arrow? And what is the effect of this on the yield of aspirin?
The double-arrow denotes this reaction as a reversible reaction. Because of this, the water produced as a by-product of this condensation reaction can react with the aspirin produced, hydrolysing the aspirin to produce salicylic acid and ethanoic acid. Hence, the yield of aspirin is quite low and this method of aspirin synthesis is inefficient.

4. What are the two functional groups of aspirin?
- Carboxyl group

5a. Phosphoric acid is a catalyst in this reaction. What is a catalyst?
A catalyst is a substance that increases the rate of chemical reactions by lowering the activation energy of that reaction.

b. How might a chemical prove that Phophoric acid acts as a catalyst, rather than as a reactant, in this reaction?
By repeating the experiment numerous times and each time using a different amount of H3PO4, while all other variables are kept constant. The mass of aspirin obtained is then measured. If there is minimal to no variation in the amount of aspirin obtained, this proves that because H3PO4 is a catalyst speeding up the reaction rate and not a reactant, the amount of phosphoric acid added does not affected the amount of aspirin obtained.

Title: Re: VCE Chemistry Question Thread
Post by: Limista on June 01, 2014, 09:40:01 am
Yacoubb, you could add something like this for part 5 in your own words:

A catalyst is not consumed in a reaction (it's crucial you mention this). This is why it is not a reactant.

The other two points you mentioned about how a catalyst speeds up a reaction and lowers activation energy by providing an alternative reaction pathway are correct.

And the rest of your answers look fine to me. Your method for checking H3PO4 acting as a catalyst for 5b) is also correct. Just bear in mind that as you increase the amount of H3PO4 that you react with x (where x = reagents), the mass of aspirin you calculate at the end would also increase. This is because H3PO4 will still be present amongst the products at the completion of the reaction, because as a catalyst, it is not consumed in the reaction.  :)
Title: Re: VCE Chemistry Question Thread
Post by: swagsxcboi on June 01, 2014, 09:43:56 am
4. What are the two functional groups of aspirin?
- Carboxyl group
not sure where your school stands on this, but my 'teacher' suggests that the benzene ring counts as a functional group.
Title: Re: VCE Chemistry Question Thread
Post by: Limista on June 01, 2014, 09:48:16 am
not sure where your school stands on this, but my 'teacher' suggests that the benzene ring counts as a functional group.

I don't recall the benzene ring (or aromatic compounds at all, actually!) being discussed to a great extent in the 3/4 course.

But the benzene is a functional group - that's true.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on June 01, 2014, 09:53:06 am
Yacoubb, you could add something like this for part 5 in your own words:

A catalyst is not consumed in a reaction (it's crucial you mention this). This is why it is not a reactant.

The other two points you mentioned about how a catalyst speeds up a reaction and lowers activation energy by providing an alternative reaction pathway are correct.

And the rest of your answers look fine to me. Your method for checking H3PO4 acting as a catalyst for 5b) is also correct. Just bear in mind that as you increase the amount of H3PO4 that you react with x (where x = reagents), the mass of aspirin you calculate at the end would also increase. This is because H3PO4 will still be present amongst the products at the completion of the reaction, because as a catalyst, it is not consumed in the reaction.  :)

Okay thanks for that. So what would be a better way to way to go with describing how H3PO4 is a catalyst and not a reactant? Like how should I word it?

not sure where your school stands on this, but my 'teacher' suggests that the benzene ring counts as a functional group.

I think that benzene is a functional group too, but I think the main ones my teacher was looking for were those two.

Title: Re: VCE Chemistry Question Thread
Post by: Limista on June 01, 2014, 10:04:59 am

5a. Phosphoric acid is a catalyst in this reaction. What is a catalyst?
A catalyst is a substance that increases the rate of chemical reactions by lowering the activation energy of that reaction, providing an alternative reaction pathway. It is also not consumed in a chemical reaction.

b. How might a chemical prove that Phophoric acid acts as a catalyst, rather than as a reactant, in this reaction?
By repeating the experiment numerous times and each time using a different amount of H3PO4, while all other variables (such as temperature, pressure, amount of reactant) are kept constant. The mass of aspirin obtained should increase proportionate to increase in amount of H3PO4 added. This proves that H3PO4 is acting as a catalyst, since it is not consumed in the reaction, and is still present in the reaction mixture at the completion of the reaction.

I've added to what you've written above.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 01, 2014, 11:02:13 am
Could someone please check if my answers to these questions are correct? Thanks :)

1. What are the differences between phenol and alcohol?
Phenol is an aromatic organic compound made up of a benzene ring attached to a hydroxyl group, whereas an alcohol is an organic compound containing a hydroxyl functional group.

2. How can we test a sample to prove that it is a phenol and not alcohol?
By reacting the sample with iron (III) salt. If a dark purple-coloured solution results, the sample contains phenol. However, if no reaction occurs and the colour of the solution does not become purple, the sample is an alcohol.

3a. What type of reaction is the synthesis of aspirin?
Esterification (condensation)

b. What does it mean when the reaction is written with a double arrow? And what is the effect of this on the yield of aspirin?
The double-arrow denotes this reaction as a reversible reaction. Because of this, the water produced as a by-product of this condensation reaction can react with the aspirin produced, hydrolysing the aspirin to produce salicylic acid and ethanoic acid. Hence, the yield of aspirin is quite low and this method of aspirin synthesis is inefficient.

4. What are the two functional groups of aspirin?
- Carboxyl group

5a. Phosphoric acid is a catalyst in this reaction. What is a catalyst?
A catalyst is a substance that increases the rate of chemical reactions by lowering the activation energy of that reaction.

b. How might a chemical prove that Phophoric acid acts as a catalyst, rather than as a reactant, in this reaction?
By repeating the experiment numerous times and each time using a different amount of H3PO4, while all other variables are kept constant. The mass of aspirin obtained is then measured. If there is minimal to no variation in the amount of aspirin obtained, this proves that because H3PO4 is a catalyst speeding up the reaction rate and not a reactant, the amount of phosphoric acid added does not affected the amount of aspirin obtained.

I think you also need to show that phosphoric acid actually speeds up the reaction too, although that's easy to prove xP

I've added to what you've written above.

Why should mass of aspirin increase? Assuming you've filtered and purified your product properly, the catalytic H3PO4 shouldn't be in the aspirin.
Title: Re: VCE Chemistry Question Thread
Post by: Limista on June 01, 2014, 11:25:09 am
I think you also need to show that phosphoric acid actually speeds up the reaction too, although that's easy to prove xP

Why should mass of aspirin increase? Assuming you've filtered and purified your product properly, the catalytic H3PO4 shouldn't be in the aspirin.

I didn't assume he would do this - as in filter and purify his product. But yes, once he's done this, then he should still find the same mass of aspirin. But the mass of product would have supposedly increased before purification. That's how he would measure that the increase in mass is due to the additional H3PO4.
Title: Re: VCE Chemistry Question Thread
Post by: Reus on June 01, 2014, 11:46:50 am
Anyone/school started Unit 4 already?
How is it, compared to Unit 3? More challenging? More interesting?
Starting next week  ;D
Title: Re: VCE Chemistry Question Thread
Post by: Scooby on June 01, 2014, 12:40:32 pm
Could someone please check if my answers to these questions are correct? Thanks :)

1. What are the differences between phenol and alcohol?
Phenol is an aromatic organic compound made up of a benzene ring attached to a hydroxyl group, whereas an alcohol is an organic compound containing a hydroxyl functional group.

Presence of a hydroxyl group in a compound doesn't prove that it's an alcohol. The hydroxyl group has to be the principal functional group
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on June 01, 2014, 01:12:46 pm
Presence of a hydroxyl group in a compound doesn't prove that it's an alcohol. The hydroxyl group has to be the principal functional group

So would I say that an alcohol is an organic with a hydroxyl group as the principal functional group?

I think you also need to show that phosphoric acid actually speeds up the reaction too, although that's easy to prove xP

Why should mass of aspirin increase? Assuming you've filtered and purified your product properly, the catalytic H3PO4 shouldn't be in the aspirin.

Thanks for that. Should I just say that if the mass of pure aspirin obtained has minimal to change as amount of phosphoric acid increases, and if the reaction occurs at a greater rate, this proves that H3PO4 is a catalyst and not in fact a reactant?

Thanks for the help guys!
Title: Re: VCE Chemistry Question Thread
Post by: Einstein on June 01, 2014, 09:07:45 pm
out of metallic, ionic and covalent what is the strongest type of bond and why?

thanks
Title: Re: VCE Chemistry Question Thread
Post by: IndefatigableLover on June 01, 2014, 09:10:01 pm
Hey AN,

When naming carbon compounds, I know we are required to place a ',' in between numbers (i.e 2,3 dimethyl) but I occasionally see them placed in between numbers and words. So when are we required to use a '-' and ','?

Hyphens must separate numbers and letters.
Commas must separate numbers.
Title: Re: VCE Chemistry Question Thread
Post by: Einstein on June 01, 2014, 09:30:22 pm
say ive got to elements (making up charges)

Cl^2- and (SO4)2^- , how does the cross over rule work?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 01, 2014, 10:43:28 pm
out of metallic, ionic and covalent what is the strongest type of bond and why?

thanks

That's actually a very dodgy kind of question as technically, all three are variants on the same theme. Ionic and covalent bonds only really differ in how polar the bond is (how large the electronegativity difference is) and metallic, covalent bonds both involve overlap of orbitals.
You don't need to know the above for VCE btw

In terms of the types of bonding commonly seen though, all three types of bonds are strong. It depends on the atoms really. Like, there are very weak covalent bonds (like F-F bonds, will react with almost anything), very weak metallic bonds (think caesium which melts at room temperature) and very weak ionic bonds (look up ionic liquids on Wiki for a list of these). In contrast, there are also strong covalent bonds (diamond), ionic bonds (sodium and lithium fluoride) and metallic bonds (titanium metal). All of these have melting points around or over 1000 degrees.

say ive got to elements (making up charges)

Cl^2- and (SO4)2^- , how does the cross over rule work?

What cross over rule?
Title: Re: VCE Chemistry Question Thread
Post by: Reus on June 01, 2014, 10:48:31 pm
say ive got to elements (making up charges)

Cl^2- and (SO4)2^- , how does the cross over rule work?
Cross over rule?
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on June 02, 2014, 05:04:54 pm
say ive got to elements (making up charges)

Cl^2- and (SO4)2^- , how does the cross over rule work?

You mean for e.g. BaCl2 ( Ba+2 and Cl-1 so swap over to get BaCl2) ???
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 02, 2014, 05:41:58 pm
That's more relying on the fact that the chemical formula for an ionic compound has to be electrically neutral, so you take the lowest integer multiple of the cations and anions and put them together so that the resulting compound is neutral

If this is the case, let's say you have Al 3+ and O 2-
If you have 2 Al 3+ and 3 O 2- you get an electrically neutral compound. Is that what you mean by 'crossing over'?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on June 02, 2014, 07:06:05 pm
Is 1:1 ethanol/dichloromethane a polar or non-polar solvent?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 02, 2014, 07:11:30 pm
Likely to be somewhere in-between. I'd say it has a more limited ability to dissolve polar things.
Title: Re: VCE Chemistry Question Thread
Post by: melbin123 on June 05, 2014, 04:49:26 pm
(http://img.tapatalk.com/d/14/06/05/y5u5yrev.jpg)
Help!! How do I do this question
Ans:C
Title: Re: VCE Chemistry Question Thread
Post by: Sense on June 05, 2014, 05:25:49 pm

(http://img.tapatalk.com/d/14/06/05/y5u5yrev.jpg)
Help!! How do I do this question
Ans:C

Remember that in polymerisation a water molecule is lost.

The weight of 4-hydroxyl butanoic acid is 104.

Each monomer will lose h2o so take away the weight of h2o = 104 - 18 = 86

86 x 450 = 38700
Title: Re: VCE Chemistry Question Thread
Post by: rinaheartsx3 on June 08, 2014, 06:03:40 am
Help with part d? My friend thinks the ratio when comparing the acid with bromine is 6:1 but I thought 2:1..explain?
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on June 08, 2014, 10:29:22 am
Chloroacetic acid is a weak monoprotic acid with a Ka of 1.3 × 10–3 M. For a 1.0 M solution of chloroacetic acid, calculate:
a) the pH
b) the percentage hydrolysis
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 08, 2014, 11:54:10 am
In shortcut VCE Chemistry style:

[H+] = [chloroacetate]
1.3 * 10^-3 = [H+][chloroacetate]/[chloroacetic acid]

The equilibrium concentration of chloroacetic acid is given by 1 - [H+]. Think about it. By mole ratios, the number of moles of chloroacetic acid reacting is equal to the number of moles of H+ formed
Now we're going to make an assumption (which I think doesn't work here) that the amount dissociated is negligible in comparison to the initial concentration. AKA 1 - [H+] is still close to 1
Then, we get 1.3*10^-3 = [H+][chloroacetate] = [H+]^2
[H+] = 0.036 M
You can find the others from this.

Now this is an example when you can JUST make the assumption 1- [H+] is still close to 1.
If you were to solve it properly, you would have 1.3 * 10^-3 = [H+]^2/(1-[H+]), which is a quadratic equation deemed by VCAA to be too hard for the average student to solve. Solving the quadratic gives [H+] = 0.0354 M, which is quite close.

In either case, in a VCE exam just do it the first way.
Title: Re: VCE Chemistry Question Thread
Post by: Reus on June 08, 2014, 02:37:25 pm
Find the % of Oxygen in:
Fe2O3

...49?
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on June 08, 2014, 02:41:07 pm
I think it's just

MM= 159.7g/mol
Mass of oxygen=48g

48/159.7 *100 = 30%
Title: Re: VCE Chemistry Question Thread
Post by: Reus on June 08, 2014, 04:36:20 pm
Ah God! How did that click in my head  :o ::)
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on June 08, 2014, 10:10:54 pm
ok so lets say i have an equilibrium reaction of like

$2SO_{2(g)}+ O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ yeah.. and its K value is 1.56M^-1.

Does this mean that it favours the exothermic (back) reaction?

Does the unit corresponding to M (molarity), whether it may be M^2, M^3, or M^-2 relate to whether or not a reaction is exo or endothermic? Or do i just look at the value of delta H..?

Because for some reason, all the exothermic reactions ive seen are like
$A \rightarrow B + C$

any clarification would be great :P just trying to get a grip on this new unit haha
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 08, 2014, 10:44:07 pm
ok so lets say i have an equilibrium reaction of like

$2SO_{2(g)}+ O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ yeah.. and its K value is 1.56M^-1.

Does this mean that it favours the exothermic (back) reaction?

Does the unit corresponding to M (molarity), whether it may be M^2, M^3, or M^-2 relate to whether or not a reaction is exo or endothermic? Or do i just look at the value of delta H..?

Because for some reason, all the exothermic reactions ive seen are like
$A \rightarrow B + C$

any clarification would be great :P just trying to get a grip on this new unit haha

Technically, equilibrium constants are meant to be expressed using a dimensionless quantity called a chemical activity, which is for most cases numerically equal to the concentration of the aqueous species in M. That's why we normally use molar concentrations in equilibrium constants. However, equilibrium constants aren't strictly meant to have units, but whatever.

In VCE chemistry, the unit comes from the fact that the equilibrium expression has [SO3]^2 on the top but [SO2]^2 [O2] on the bottom. As concentrations are in M, this means the overall units are M^2/M^3 = M^-1. Just means more gas particles on the left than on the right.

Exo/endothermic means delta H value negative/positive. It turns out that this reaction is exothermic. Really, you're meant to give a temperature when quoting equilibrium constants as these constants change with temperature only.
Exothermicity is not a very good indication of equilibrium position; dissolution of ammonium nitrate is endothermic but you'd hardly say it's disfavoured given the solubility rules that ammonium AND nitrate ions are soluble. Dissolution of salts is generally endothermic. Think about it; that's why you dissolve more at higher temperatures.
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on June 08, 2014, 10:47:52 pm
Thanks for the lzxnl!
Title: Re: VCE Chemistry Question Thread
Post by: melbin123 on June 09, 2014, 12:46:02 pm
@sense thx A lot buddy :)
Title: Re: VCE Chemistry Question Thread
Post by: Reus on June 10, 2014, 07:29:15 pm
Need help with this question, well I ended up working it out however clueless towards what the answer should be as we weren't provided an answer sheet.
So I'd love to see what you guys get :)

"Anglesite is a lead bearing ore found in large deposits in Australia. One particular deposit of Anglesite was found to contain 80% of lead (II) sulfate.
What mass of lead could be extracted from every one tonne (1000kg) of Anglesite processed?
"

Thanks guys  ;D
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on June 10, 2014, 08:01:12 pm
Need help with this question, well I ended up working it out however clueless towards what the answer should be as we weren't provided an answer sheet.
So I'd love to see what you guys get :)

"Anglesite is a lead bearing ore found in large deposits in Australia. One particular deposit of Anglesite was found to contain 80% of lead (II) sulfate.
What mass of lead could be extracted from every one tonne (1000kg) of Anglesite processed?
"

Thanks guys  ;D

Complete guess but did you get 800g?
Sorry if this is wrong :(
Title: Re: VCE Chemistry Question Thread
Post by: Reus on June 10, 2014, 08:20:50 pm
Complete guess but did you get 800g?
Sorry if this is wrong :(
Not quite...
I ended up with 683.24kg if it were 100% of PbSO4 in the extract.
However considering it is 80%; I went with 800kg instead of 1000kg which resulted in m(Pb)=546.59kg out of the 1000kg
Therefore each tonne of Anglesite contains 547kg of lead.

Yet, positive my calculations for the 80% of lead is incorrect, I have a feeling haha.
(Reason why I posted this :P to make sure if I'm right!)

So no need to be sorry! I'm probably wrong too  ::)
Title: Re: VCE Chemistry Question Thread
Post by: Bestie on June 11, 2014, 09:06:40 pm
what is the m/e of base peak of propan-1-ol in mass spectrometry after complete fragmentation?

thank you
Title: Re: VCE Chemistry Question Thread
Post by: Reus on June 11, 2014, 09:25:01 pm
M(propan-1-ol) = 60g/mol

We begin with C3H8O. Fragmentation of others will occur.
CH3O+ would be graphed with the highest point (Generally graphed for you). However given there is no data, O is most electronegative, compared to other fragmentations.(holds onto electrons the best and will form the base peak)

Molar mass of CH3O+ = 12+(1x3)+16
= 31.
Henceforth m/e base peak is 31.

Correct me if I'm wrong :)
Title: Re: VCE Chemistry Question Thread
Post by: Bestie on June 11, 2014, 09:28:02 pm
thank you but how do you know its CH3O+ and not something like OH+?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 12, 2014, 10:18:50 am
They give you a graph in which you can find the base peak. Then, you can work out an appropriate structure for it.
Title: Re: VCE Chemistry Question Thread
Post by: thushan on June 12, 2014, 11:18:35 am
It's unlikely that VCAA expects you to figure out to predict the structure of the ion that gives rise to the base peak from just the structure of the molecule itself, without giving you the m/z value of the base peak.
Title: Re: VCE Chemistry Question Thread
Post by: Reus on June 12, 2014, 03:46:35 pm
Haha yes, I knew it wasn't a requirement to do it without a graph.
Title: Re: VCE Chemistry Question Thread
Post by: Reus on June 12, 2014, 07:44:15 pm
Le Chatelier’s Principle? Can someone summarise this in the scope of this course  ;D
Thanks.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 12, 2014, 09:02:33 pm
Think negative feedback of biology. A system at equilibrium will PARTIALLY OPPOSE any changes to this equilibrium.

I say partially oppose for a reason. Let's say you initially had 10 moles of nitrogen gas in equilibrium with 5 moles of hydrogen gas and 10 moles of ammonia. If you then dump another 5 moles of ammonia gas into this system, the nitrogen concentration will increase, prompting the system to oppose that change by reducing the nitrogen concentration with a reaction of ammonia to reform nitrogen. However, no matter how the system tries to reduce the nitrogen concentration, the nitrogen concentration at the end will always be greater than the initial concentration of 10 moles/whatever volume. See why it's partially opposing?
Title: Re: VCE Chemistry Question Thread
Post by: Reus on June 12, 2014, 09:17:37 pm
Thanks!!
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on June 12, 2014, 09:33:52 pm
A 2.16g sample of an oxide of nitrogen is broken down to release nitrogen and oxygen gases. After the oxygen gas is removed, 0.56g of nitrogen remains. The oxide is likely to be:
a) NO
b) N4O
c) NO2
d) N2O5
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on June 12, 2014, 09:52:44 pm
Its d)

Why?

2N2O5----> 2N2 + 5O2

n(n2o5)= 2.16/108 = 0.02 mol

n(n2)=n(n2o5)=0.02 mol

m(n2)=0.02 x 28
m= 0.56g
Title: Re: VCE Chemistry Question Thread
Post by: Reus on June 14, 2014, 02:44:21 pm
Could someone explain the process/behaviour of particles when Iron anchors undergo corrosion in deep seas?
I know it's a low frequency of collisions between oxygen molecules and iron metal which results in corrosion, but why?
Any more detail required?
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on June 14, 2014, 04:56:49 pm
Is Kw temperature dependent???
Title: Re: VCE Chemistry Question Thread
Post by: keltingmeith on June 14, 2014, 05:06:03 pm
Yes. It's 10^-14 at 298 K, and if you work at anything else, they'll work you through it so you'll know what it should be.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 14, 2014, 08:53:42 pm
Is Kw temperature dependent???

Kw is the equilibrium constant for the auto-ionisation of water, i.e. H2O (l) <=> H+(aq) + OH-(aq)
It's an equilibrium constant so it must be temperature dependent
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on June 14, 2014, 09:47:21 pm
where to do you mark the + and - signs on a zwitterion when it is neutral, acidic and basic?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on June 14, 2014, 10:10:22 pm
where to do you mark the + and - signs on a zwitterion when it is neutral, acidic and basic?

In a zwitter ion, a + is placed at the NH3+ region, and a - is placed at the COO region.

When the amino acid is placed in an acid, it behaves as a base and thus accepts protons (becoming positively charged). A + is placed at the NH3 region. (i.e. where the proton has been gained at the amino group).

When the amino acid is placed in a base, it behaves as an acid and thus donates protons (becoming negatively charged). A - is placed at the COO region (i.e. where the proton has been lost from the carboxyl group).
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on June 14, 2014, 11:08:43 pm
how to join 2 nitrogenous bases together (ie. A-T or C-G)

- both are in data book but which bonds on the base join together and does dna produce any other small molecules like water after the equation like condensation reactions?
Title: Re: VCE Chemistry Question Thread
Post by: Reus on June 14, 2014, 11:14:30 pm
how to join 2 nitrogenous bases together (ie. A-T or C-G)

- both are in data book but which bonds on the base join together and does dna produce any other small molecules like water after the equation like condensation reactions?
Hyrdogen bonding between nitrogen bases. However keep in mind that adenine and thymine (purinse) have two bonds (double H bonds) and cytosine and guanine (pyrimidines) have three in between them (triple H bonds).

Well the nucleotides to form DNA, undergo a condensation reaction meaning H2O is produced.
4 nucleotides = 3 H2O's.
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on June 15, 2014, 04:16:13 pm
what does CO2 evolved mean?

when analysing an NMR spectra how do you know which chemical shift is right? e.g. 4.1 ppm can match different things in the data book. This concept is still confusing for me so can anyone explain
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on June 15, 2014, 04:28:00 pm
what does CO2 evolved mean?

when analysing an NMR spectra how do you know which chemical shift is right? e.g. 4.1 ppm can match different things in the data book. This concept is still confusing for me so can anyone explain

When a gas evolves, this means that a gas is produced from a reaction. So for instance, carbon dioxide evolves in the fermentation of glucose.

You're often given background information which should guide you into choosing the right chemical shift. On exams, you will find that questions on this topic give you background information on the compound in the form of it's chemical formula, an IR spectrum, a mass spectrum of the compound, and a proton/Carbon-13 NMR spectrum. Through this, you can properly identify the molecule.
Title: Re: VCE Chemistry Question Thread
Post by: Hannibal on June 15, 2014, 05:22:12 pm
Not sure if this is the right place to ask this, I hope so :)

What's the difference between a dipole and a polar molecule?
Title: Re: VCE Chemistry Question Thread
Post by: Reus on June 15, 2014, 05:33:10 pm
Not sure if this is the right place to ask this, I hope so :)

What's the difference between a dipole and a polar molecule?
Perfect place to ask!

A polar molecule is where there is separation of charge in the chemical bonds. One region of the molecule will have a slight positive charge and another with a slight negative charge. Think of it like the North and South Pole. A common example of a polar molecule is H2O. The O is negative yet the H2 is positive.
In comparison to dipole which also have a positive and a negative portion to the molecule. However these occur in cases of symmetry where dipole vectors cancel each other out. An example of this is all four bonds in carbon tetrachloride (CCl4) are polar because chlorine is very electronegative. This then results in four dipoles where all four vectors point from the central carbon to one of the four chlorines. Due to its tetrahedral shape, the vectors face in opposite directions hence cancel each other out.
Title: Re: VCE Chemistry Question Thread
Post by: keltingmeith on June 15, 2014, 05:42:46 pm
Not sure if this is the right place to ask this, I hope so :)

Look, if you can't get help with chemistry in here, I think a lot of us are in for some trouble, hahah.
Title: Re: VCE Chemistry Question Thread
Post by: Hannibal on June 15, 2014, 05:49:34 pm
Look, if you can't get help with chemistry in here, I think a lot of us are in for some trouble, hahah.
Thanks Reus! Haha yeah, I was just wondering because it looked like it was for Year 12's. I also wanted to ask if this was an accurate description of metallic bonding:

This bond requires large amounts of atoms, and occurs in most metals. This is because the atoms are structured in a way that the electrons are free to move, forming the electron sea model. This is the reason metals have high conductivity, ductility, malleability and low volatility.
Title: Re: VCE Chemistry Question Thread
Post by: Bestie on June 15, 2014, 05:51:43 pm
hello,
i was wondering what the answer to this question should be cause in the worked solution (funny enough didnt give any explaination of how they got their ans) got B but the back of the book said C and i said D???

thank you
Title: Re: VCE Chemistry Question Thread
Post by: keltingmeith on June 15, 2014, 05:55:46 pm
Hannibal - seems fairly accurate, minus the most metals part. The idea of "metallic bonding" is that it occurs in all metals, hahah.

Bestie - I'm with you there, I'd say D. Probably more correct to say B in an experimental sense, but given the data they've given you, certainly D is correct.
Title: Re: VCE Chemistry Question Thread
Post by: Bestie on June 15, 2014, 06:25:25 pm
yup thank you :)

i also found another one:
More energy is required to vibrate:
A C–C compared to C=C
B C–Cl compared to C–Br
C C–O compared to C–H
D C=C compared to C≡C

shouldn't the answer be B, the back of book says C?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 15, 2014, 07:40:18 pm
Thanks Reus! Haha yeah, I was just wondering because it looked like it was for Year 12's. I also wanted to ask if this was an accurate description of metallic bonding:

This bond requires large amounts of atoms, and occurs in most metals. This is because the atoms are structured in a way that the electrons are free to move, forming the electron sea model. This is the reason metals have high conductivity, ductility, malleability and low volatility.

One point of dispute: metallic bonding involves metal IONS, not atoms. Atoms are neutral (:

yup thank you :)

i also found another one:
More energy is required to vibrate:
A C–C compared to C=C
B C–Cl compared to C–Br
C C–O compared to C–H
D C=C compared to C≡C

shouldn't the answer be B, the back of book says C?

The problem with B is that although C-Cl has a stronger bond, Br is much larger.
Mass also affects the amount of energy required
Title: Re: VCE Chemistry Question Thread
Post by: Bestie on June 15, 2014, 07:55:13 pm
wasn't that with lower mass, more energy required?
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on June 15, 2014, 08:50:49 pm
wasn't that with lower mass, more energy required?
yeah I thought it was this too. The C-H band has a higher wavenumber (and thus energy) than the the C-O band, doesn't that mean more energy is needed to vibrate it?
Title: Re: VCE Chemistry Question Thread
Post by: Bestie on June 15, 2014, 08:54:57 pm
same here... but.. i don't know :(

can i ask another question then?
The amount of alcohol, CH3CH2OH, in your breath can be determined by
blowing into a tube containing acidified potassium dichromate, K2Cr2O7. Which
substance has been oxidised and which has been reduced?

thank you :)
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on June 15, 2014, 09:01:17 pm
same here... but.. i don't know :(

can i ask another question then?
The amount of alcohol, CH3CH2OH, in your breath can be determined by
blowing into a tube containing acidified potassium dichromate, K2Cr2O7. Which
substance has been oxidised and which has been reduced?

thank you :)
Alcohol gets oxidised to a carboxyllic acid (CH3COOH), meaning the K2Cr2O7 gets reduced (from unit 3 area of study 2).
Title: Re: VCE Chemistry Question Thread
Post by: Bestie on June 15, 2014, 09:45:32 pm
so alcohols are never reduced?
Title: Re: VCE Chemistry Question Thread
Post by: keltingmeith on June 15, 2014, 09:58:24 pm
It's not that they're never reduced - you need to know some of the common oxidation pathways of alcohols as a part of the area of study 2 in unit 3. That's one of them. Hate to say it Bestie, but you need to revise/read over the study design.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 16, 2014, 01:19:06 am
so alcohols are never reduced?

Well......I wouldn't say "never" reduced. Look up reactions out there that can convert alcohols to alkanes.
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on June 16, 2014, 11:46:07 pm
when drawing the structure of an ester should the alcohol (alkyl) part go on the left or right and what about the alkyanoate part?
Title: Re: VCE Chemistry Question Thread
Post by: keltingmeith on June 17, 2014, 12:06:31 am
It honestly doesn't matter - if it helps you remember to do it one way, do it that way. But there is no correct way of drawing esters as to which side you should the alkyl and which you should put the alkyanoate (which did mean a bunch of strife for me last year, let me tell you).
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on June 17, 2014, 11:17:57 am
Hi guys.
In general, I'm quite confused by some aspects of electrolysis.
In galvanic cells, using an ordered table of standard reduction potentials, we know that the strongest oxidant lies on the top left and the strongest reductant lies on the bottom right, so spontaneous reactions occur ‘along that diagonal’.  But in electrolysis, the reaction is not spontaneous, so is this principle switched?  Or do the reactions happen along the opposite diagonal (top right with bottom left)??
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on June 17, 2014, 11:20:43 am
Attached are two slides from our lectures (Fundamentals of Chemistry).

I can do the balancing/cancelling and adding together part, but I don’t know how to ‘get’ the initial half-equations from the worded answer.  What happens to the K?  And the O?  How do we know which half-reactions to kick off with?
Basically there is a missing link from my brain between reading the question and saying "Oh ok, I need to use Cr2O72- --> Cr3+ to kick off this question".

Can anyone help me with the thought process here??
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 17, 2014, 11:46:26 am
Hi guys.
In general, I'm quite confused by some aspects of electrolysis.
In galvanic cells, using an ordered table of standard reduction potentials, we know that the strongest oxidant lies on the top left and the strongest reductant lies on the bottom right, so spontaneous reactions occur ‘along that diagonal’.  But in electrolysis, the reaction is not spontaneous, so is this principle switched?  Or do the reactions happen along the opposite diagonal (top right with bottom left)??

Think of it this way. There is a quantity called the free energy that strives to reach a minimum at equilibrium. Free energy behaves a bit like potential energy in this regard. Now, a galvanic cell reaction is like having a ball roll down from the top of the hill; it's spontaneous as the energy decreases. An electrolytic cell reaction, however, is like having someone physically push the ball up the hill. Chemically, by applying a voltage, you're forcing the electrons to reduce a compound.

If you don't do the electrolysis properly and your products are in contact with each other, you're right; the top left bottom right principle will apply and the two products will react spontaneously. As an example, if you electrolyse molten sodium chloride, you'll get sodium metal and chlorine gas. The reason why they don't react with each other is they're physically separated from each other. If you let them make contact, they WILL react. Does this sort of help explain your queries?

Attached are two slides from our lectures (Fundamentals of Chemistry).

I can do the balancing/cancelling and adding together part, but I don’t know how to ‘get’ the initial half-equations from the worded answer.  What happens to the K?  And the O?  How do we know which half-reactions to kick off with?
Basically there is a missing link from my brain between reading the question and saying "Oh ok, I need to use Cr2O72- --> Cr3+ to kick off this question".

Can anyone help me with the thought process here??

OK. The potassium ions aren't affected at all. That's why you don't care about those.
I don't get your problem about the oxygen though.

You need to know 'dichromate => chromium(III)' as dichromate is what you start off with and chromium(III) ions are the end product. You just have to know that unfortunately. Just like you just have to know sodium metal is oxidised to Na+.
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on June 17, 2014, 12:01:45 pm
Think of it this way. There is a quantity called the free energy that strives to reach a minimum at equilibrium. Free energy behaves a bit like potential energy in this regard. Now, a galvanic cell reaction is like having a ball roll down from the top of the hill; it's spontaneous as the energy decreases. An electrolytic cell reaction, however, is like having someone physically push the ball up the hill. Chemically, by applying a voltage, you're forcing the electrons to reduce a compound.

Yep this makes sense (we haven't covered free energy but yeah)!

Quote
If you don't do the electrolysis properly and your products are in contact with each other, you're right; the top left bottom right principle will apply and the two products will react spontaneously. As an example, if you electrolyse molten sodium chloride, you'll get sodium metal and chlorine gas. The reason why they don't react with each other is they're physically separated from each other. If you let them make contact, they WILL react. Does this sort of help explain your queries?

Hmm yes I get it but that's not exactly what I was trying to ask (it's difficult to articulate what I'm finding confusing, sorry!).  I think there's just something I'm not thinking about in the right way.  Let me have a think about it and see if I can find some specific questions.

Quote
You need to know 'dichromate => chromium(III)' as dichromate is what you start off with and chromium(III) ions are the end product. You just have to know that unfortunately. Just like you just have to know sodium metal is oxidised to Na+.

Thanks, that explains that one!  :) Haha.  One of the weirdest things about chem is knowing what to 'rote learn' and what should be intuitive.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 17, 2014, 01:45:33 pm
Chem is, after all, based on experiment
By definition you have to rote learn some parts of it
Title: Re: VCE Chemistry Question Thread
Post by: Jason12 on June 17, 2014, 07:47:28 pm
when reacting an alkane with Cl2 to form a cholroalkane + HCL, why is UV light used?

how to calculate double bonds in a polyunsaturated fat e.g. linoleic acid
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on June 17, 2014, 08:02:00 pm
when reacting an alkane with Cl2 to form a cholroalkane + HCL, why is UV light used?
UV light provides energy to break the Cl-Cl bond to create free radicals (unpaired electrons). These radicals are very reactive and able to attack the C-H bond.
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on June 17, 2014, 08:07:42 pm
when reacting an alkane with Cl2 to form a cholroalkane + HCL, why is UV light used?

how to calculate double bonds in a polyunsaturated fat e.g. linoleic acid
If you are provided with the formula, try to work out from there. For linoleic fatty acids which is C17H31COOH. you can try to use DOB formula (which I'm too lazy to use :( ) or just replace the COOH with a Hydrogen, hence, you have C17H32, which is polyunsaturated (2 double bonds).
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on June 17, 2014, 10:08:59 pm
how to calculate double bonds in a polyunsaturated fat e.g. linoleic acid

A saturated fatty acid has the formula
$C_nH_{2n+1}COOH$
every double bond is 2H less than the saturated fatty acid

So for example, if we have $C_{17}H_{31}COOH$
the saturated version of that would be $C_{17}H_{35}COOH$  <--- 35 came from doubling 17, and adding 1.
Since its 31H rather than 35H, there are 2 double bonds.
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on June 18, 2014, 11:58:38 am
Okay lznxl and others... Here's a question (attached) on electrolysis that kind of outlines where I'm finding things confusing. Part b) is where I'm stuck.

I've also attached the answer for part b.

I thought reduction was happening at A and C, and oxidation happening at B and D but the answer indicates otherwise.

Just focussing on A and B, I wrote that the species in the beaker were Al3+, SO4, and H2O.
Then I wrote what I thought was the four relevant half reactions and their Eo voltage value:

O2 + 4H+ + 4e- <--> 2H2O (Eo = +1.23)
SO4 + 4H+ + 2e- <--> H2SO4 + H2O (Eo = +0.17)
2H2O + 2e- <--> H2 + 2OH- (Eo = -0.83)
Al3+ + 3e- <--> Al (Eo = -1.66)

But I don't know if these are the right half reactions... And how they use them to get the answer... Which two of the four are the ones that occur and why???

Question: (http://img.tapatalk.com/d/14/06/18/bupy4u2e.jpg)

(http://img.tapatalk.com/d/14/06/18/asuhynys.jpg)
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on June 18, 2014, 03:44:42 pm
The attached picture has been named in our notes as 3-ethyl-2methylhexane. Makes sense but couldn't it also be 3-isopropylhexane? There are two ways to get the six carbon backbone...(http://img.tapatalk.com/d/14/06/18/unujazas.jpg)
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on June 18, 2014, 07:00:40 pm
The attached picture has been named in our notes as 3-ethyl-2methylhexane. Makes sense but couldn't it also be 3-isopropylhexane? There are two ways to get the six carbon backbone...(http://img.tapatalk.com/d/14/06/18/unujazas.jpg)
Yeah you're correct.
Title: Re: VCE Chemistry Question Thread
Post by: vintagea on June 18, 2014, 07:54:55 pm
am i supposed to post my questions here?

A calorimeter containing 100 mL of water is calibrated by
passing a 3.00 A current through the instrument for 36.0 s
at a potential difference of 3.50 V. The temperature rises by
0.82°C.
Potassium hydroxide weighing 0.654 g is added to the
calorimeter and dissolved rapidly by stirring. The temperature
rises from 20.82°C to 22.23°C.
Determine the delta H

i know the calibration factor is 461J/celcius

i got -55.8 kj/mol but the answers say -55.8MJ/mol

did i get it wrong?

Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on June 19, 2014, 12:16:49 pm
Hi guys. See attached pic.
Just want to clarify a few things.
Even though we can make an 11-carbon long chain here (incl the substituent), we DONT do this because it's a ring, right? So this is cyclodecyne?
And then the numbering... Do you start at the triple bond, so the methyl group is on carbon 6?
So do we get 6-methylcyclodecyne?(http://img.tapatalk.com/d/14/06/19/u2y5a5eg.jpg)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 19, 2014, 09:14:09 pm
am i supposed to post my questions here?

A calorimeter containing 100 mL of water is calibrated by
passing a 3.00 A current through the instrument for 36.0 s
at a potential difference of 3.50 V. The temperature rises by
0.82°C.
Potassium hydroxide weighing 0.654 g is added to the
calorimeter and dissolved rapidly by stirring. The temperature
rises from 20.82°C to 22.23°C.
Determine the delta H

i know the calibration factor is 461J/celcius

i got -55.8 kj/mol but the answers say -55.8MJ/mol

did i get it wrong?

So...dH = heat / number of moles of reaction = 461 J/K * 1.39 K / (0.654/56.1 mol) = 55.0 kJ/mol
Exothermic => dH = -55.0 kJ/mol

Hi guys. See attached pic.
Just want to clarify a few things.
Even though we can make an 11-carbon long chain here (incl the substituent), we DONT do this because it's a ring, right? So this is cyclodecyne?
And then the numbering... Do you start at the triple bond, so the methyl group is on carbon 6?
So do we get 6-methylcyclodecyne?(http://img.tapatalk.com/d/14/06/19/u2y5a5eg.jpg)

I'm fairly sure you'd start at the triple bond. Triple bonds have higher priority than methyl groups. So this would be 6-methylcyclodecyne
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on June 20, 2014, 11:49:58 am
Thanks lznxl.

Question:
Which metals in the table will react with water? Write chemical reactions for these equations.

(http://www.mindset.co.za/assets/srptable.gif)

This is my main point of confusion when it comes to electrolysis/galvanic cells.  Water.  I don't even know where to start here...
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 20, 2014, 02:48:59 pm
Thanks lznxl.

Question:
Which metals in the table will react with water? Write chemical reactions for these equations.

(http://www.mindset.co.za/assets/srptable.gif)

This is my main point of confusion when it comes to electrolysis/galvanic cells.  Water.  I don't even know where to start here...

Look at where water is. It's at 1.23 V and -0.83 V. So anything above 1.23 V will oxidise water, while anything below -0.83 V will reduce water. So it looks like Au 3+ will oxidise water, while Al, Mg, Na, Ca, K, Li will all reduce water.
Title: Re: VCE Chemistry Question Thread
Post by: vintagea on June 21, 2014, 03:20:18 pm
thank you lzxnl :)

can you please help me: Suggest why salt is included in the contents of the bag.

Title: Re: VCE Chemistry Question Thread
Post by: darklight on June 21, 2014, 03:30:30 pm
Hi guys,
If a question asks for the combustion reaction of maltose, how do you know that maltose is (s)? Couldn't it also be (aq)? VCAA 2007,  exam 2, q4aii only allows for (s).
Title: Re: VCE Chemistry Question Thread
Post by: vintagea on June 21, 2014, 04:46:45 pm
Calculate the energy absorbed when 2.00 g of NH4Cl is
dissolved in:
i 100 mL water ii 1000 mL water
- find mols
-use delta H and the reaction equation...

i get how to do the question but the worked solutions says: The amount of energy absorbed does not depend on the volume of the solution?
why is that?
i know that changing the volume changes the concerntration of the solution because there is a constant 2g of NH4Cl is each solution... wouldn't the less concerntrated absorb less? i'm not sure... please clarify :)
Title: Re: VCE Chemistry Question Thread
Post by: Bestie on June 21, 2014, 06:16:59 pm
hello everyone !!!
i have a question:
why is it necessary to calibrate a calorimeter? don't you have to do it every time you are about to begin the experiment to calculate the heat of reaction, you can't just start it on one day and then finish it off on the next day using the same calibration factor... you have to do it all over again???
something about heat released in absorbed by instrument? how does that affect?
I'm doing the prac next week

thank you
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 21, 2014, 06:57:31 pm
thank you lzxnl :)

can you please help me: Suggest why salt is included in the contents of the bag.

It's a redox reaction => electrons need to flow. Dissolved salts increase the conductivity of the solution and thus speed up redox reactions.

hello everyone !!!
i have a question:
why is it necessary to calibrate a calorimeter? don't you have to do it every time you are about to begin the experiment to calculate the heat of reaction, you can't just start it on one day and then finish it off on the next day using the same calibration factor... you have to do it all over again???
something about heat released in absorbed by instrument? how does that affect?
I'm doing the prac next week

thank you

Well...you want an accurate result from your calorimeter and anything can affect its accuracy, including its temperature (like if it was used before), whether or not the instrument has been affected by heat etc. Basically you recalibrate it to be safe.

Hi guys,
If a question asks for the combustion reaction of maltose, how do you know that maltose is (s)? Couldn't it also be (aq)? VCAA 2007,  exam 2, q4aii only allows for (s).

You're burning it. I dunno why you'd be burning something aqueous as you'd be heating the water as well; not awfully efficient

Calculate the energy absorbed when 2.00 g of NH4Cl is
dissolved in:
i 100 mL water ii 1000 mL water
- find mols
-use delta H and the reaction equation...

i get how to do the question but the worked solutions says: The amount of energy absorbed does not depend on the volume of the solution?
why is that?
i know that changing the volume changes the concerntration of the solution because there is a constant 2g of NH4Cl is each solution... wouldn't the less concerntrated absorb less? i'm not sure... please clarify :)

The concentration of the solution has no bearing on how much energy is released. The amount of energy released is really only dependent on the number of moles of reactant, not the concentration of reactant.
What WILL change is the temperature change as then there's more water to absorb the energy released. The energy release doesn't change.
Title: Re: VCE Chemistry Question Thread
Post by: vintagea on June 21, 2014, 07:04:15 pm
thank you lzxnl, your help is very much appreciated :)

how do the salts increase the conductivity of the solution? they provide the electrons? how does it work?
Title: Re: VCE Chemistry Question Thread
Post by: thushan on June 21, 2014, 07:57:21 pm
Salt solutions contain mobile charged particles. An anion (eg. NO3 - ) travelling in a particular direction is effectively similar to an electron moving in the same direction, or a cation moving in the opposite direction.

It's the presence of mobile charged particles such as dissolved cations and anions, as well as mobile electrons (eg. delocalised electrons in metals), that cause a substance to be electrically conductive.
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on June 22, 2014, 08:55:41 am
Thanks lznxl for your solution yesterday.

I have another one, please see attached solution which also contains a diagram of the electrolysis setup (two cells in series).  There are four electrodes labelled A B C D and the ions in the aqueous solutions are shown underneath the beaker.  The question is to write what half-reactions are happening at each electrode.

My issue is that my answer for the half-reactions at B and D do not match the answers.  The provided answers say:
At B: 2H2O(l) + 2e– → H2(g) + 2OH–(aq)
At D: 2H+(aq) + 2e– → H2(g)

I think my working outlines my thought process fairly clearly so hopefully someone might be able to see where I have gone wrong.  I have been battling with the solutions that we have been provided, about 10% of them have been incorrect, so it is not outside the realms of possibility that the answer is in error.  I just want to know the truth!!  Thanks :)

NB.  The second last line should say "At D..." not "At B...".
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 22, 2014, 11:09:03 am
Thanks lznxl for your solution yesterday.

I have another one, please see attached solution which also contains a diagram of the electrolysis setup (two cells in series).  There are four electrodes labelled A B C D and the ions in the aqueous solutions are shown underneath the beaker.  The question is to write what half-reactions are happening at each electrode.

My issue is that my answer for the half-reactions at B and D do not match the answers.  The provided answers say:
At B: 2H2O(l) + 2e– → H2(g) + 2OH–(aq)
At D: 2H+(aq) + 2e– → H2(g)

I think my working outlines my thought process fairly clearly so hopefully someone might be able to see where I have gone wrong.  I have been battling with the solutions that we have been provided, about 10% of them have been incorrect, so it is not outside the realms of possibility that the answer is in error.  I just want to know the truth!!  Thanks :)

NB.  The second last line should say "At D..." not "At B...".

The reaction you quoted at B requires acid, whereas I only see aluminium sulfate in your solution (which doesn't have H+ as far as you're meant to know from the question; oddly enough aluminium sulfate is acidic which would actually change the entire question but you haven't been told that)
Even so, a bit of fiddling around with Nernst equations suggests that even at neutral pH, reduction of sulfate is more favourable than reduction of water. Sigh.

As for D, the only issue I can see is potentially whether or not you have 1 M sulfate ions there. Otherwise your answer should be ok.
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on June 22, 2014, 11:26:27 am

The reaction you quoted at B requires acid, whereas I only see aluminium sulfate in your solution (which doesn't have H+ as far as you're meant to know from the question; oddly enough aluminium sulfate is acidic which would actually change the entire question but you haven't been told that)
Even so, a bit of fiddling around with Nernst equations suggests that even at neutral pH, reduction of sulfate is more favourable than reduction of water. Sigh.

As for D, the only issue I can see is potentially whether or not you have 1 M sulfate ions there. Otherwise your answer should be ok.

Thankyou :)

Sorry yes the question did state that the concentration of both solutions was 1.0 M.

Okay I see what you mean, because H+ isn't present in the aluminium sulfate solution then the reduction reaction that I wrote for B can't occur. Or are you saying it still can? (We havent done Nernst equations).

Not quite sure if you are saying my answers for B and D are right or not....... Haha.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 22, 2014, 05:20:38 pm
Thankyou :)

Sorry yes the question did state that the concentration of both solutions was 1.0 M.

Okay I see what you mean, because H+ isn't present in the aluminium sulfate solution then the reduction reaction that I wrote for B can't occur. Or are you saying it still can? (We havent done Nernst equations).

Not quite sure if you are saying my answers for B and D are right or not....... Haha.

Yeah...I figured you hadn't look at the Nernst equation yet. Basically it allows you to find non-standard electrode cell potentials. Given that this is a Fundamentals subject, I'd say your answer to B is wrong because you don't have acid and that the given answer is correct. In reality, your reaction given in B would occur, but MUCH slower than the one given.

And I just realised the problem with D. 1 M sulfuric acid doesn't have 1 M sulfate ions; H2SO4 is a strong acid but HSO4- isn't. Hence your answer to D is wrong as well.
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on June 22, 2014, 05:25:15 pm

Yeah...I figured you hadn't look at the Nernst equation yet. Basically it allows you to find non-standard electrode cell potentials. Given that this is a Fundamentals subject, I'd say your answer to B is wrong because you don't have acid and that the given answer is correct. In reality, your reaction given in B would occur, but MUCH slower than the one given.

And I just realised the problem with D. 1 M sulfuric acid doesn't have 1 M sulfate ions; H2SO4 is a strong acid but HSO4- isn't. Hence your answer to D is wrong as well.

Cool man thanks. That all makes sense why now. I think that's way more complicated than anything we would get in an exam but that's cool. You learn more by doing problems beyond your level! Great - thanks again.
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on June 23, 2014, 02:32:10 pm
Hi everyone.

Could anyone help me with some MCQ questions?

1. Which of the following acids (together with their sodium salts) is the best choice to buffer a solution at pH 5.0?
A. chloroacetic acid (Ka = 1.4 x 10-3)
B. propanoic acid (Ka = 1.3 x 10-5)
C. benzoic acid (Ka = 6.4 x 10-5)
D. hypochlorous acid (Ka = 3.5 x 10-8)
E. carbonic acid (Ka = 4.3 x 10-7)

I've subsequently discovered that you solve this by saying pH = pKa, and pKa = -log10Ka, but I don't know where this comes from and if it always holds true?  So I can solve this problem, I just don't know why we use pH = pKa.

2. Calcium carbonate, CaCO3, has a solubility product, Ksp, of 2.8 × 10-9. What is the concentration of Ca2+ ions in a saturated solution of CaCO3?
A. 1.4x10-9 M
B. 2.8x10-9 M
C. 5.6x10-9 M
D. 5.3x10-5 M
E. 1.1x10-4 M

Is this just because at equilibrium, stoichiometry says that, [Ca2+] = [CO32-] = √Ksp ?  Or is there something more to it?

3. Iodide reacts with peroxodisulphate in aqueous solution to produce iodine and sulphate according to the equation:
2I-(aq) + S2O82-(aq) → I2(aq) + 2SO42-(aq)

If the rate at which I2 is produced is 2.0 x 10-6 Ms-1, then:

A. S2O82- is consumed at 1.0 x 10-6 Ms-1
B. I- is consumed at 2.0 x 10-6 Ms-1
C. SO42- is formed at 2.0 x 10-6 Ms-1
D. S2O82- is consumed at 2.0 x 10-6 Ms-1
E. SO42- is formed at 1.0 x 10-6 Ms-1

The provided answer says D, which makes sense, but I don't know why E isn't also true (by stoichiometric ratios)?  i.e. D is true because the stoichiometric ratios of I2 and S2O82- are 1:1, but isn't E also true because SO42- will form at half the rate of I2 because it is producing twice the amount?

4. Samples of chlorine, neon and nitrogen are contained in three identical 1litre flasks. All are at the same temperature. The mass of gas in each flask is the same. Which of the following is TRUE?

A. They are all at the same pressure.
B. The density of chlorine is higher than the other two.
C. The density of nitrogen is lower than the other two.
D. The number of Ne atoms is half the number of Cl atoms. The pressure is highest in the flask of neon.
E. The pressure is highest in the flask of neon.

I got the answer E, because of the following rationale, but the provided answer said B?
Atomic masses: N < Ne < Cl
Molecular mass of gas: Ne < N2 < Cl2
Therefore Ne flask has most number of molecules and Cl2 has the least (to make the same mass).

• A is false because n is different for each flask.
B is false because Cl2 has fewest molecules, thus is LEAST dense.
C is false because N2 is in the middle.
D is false because for 1 gram mass, n(Ne) = 0.05 mol and n(Cl) = 0.028 mol.
E is true because Ne contains the most molecules, maximizing PV=nRT.
So why is B the answer?

Thanks everybody :)
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 23, 2014, 04:15:57 pm
Hi everyone.

Could anyone help me with some MCQ questions?

1. Which of the following acids (together with their sodium salts) is the best choice to buffer a solution at pH 5.0?
A. chloroacetic acid (Ka = 1.4 x 10-3)
B. propanoic acid (Ka = 1.3 x 10-5)
C. benzoic acid (Ka = 6.4 x 10-5)
D. hypochlorous acid (Ka = 3.5 x 10-8)
E. carbonic acid (Ka = 4.3 x 10-7)

I've subsequently discovered that you solve this by saying pH = pKa, and pKa = -log10Ka, but I don't know where this comes from and if it always holds true?  So I can solve this problem, I just don't know why we use pH = pKa.

2. Calcium carbonate, CaCO3, has a solubility product, Ksp, of 2.8 × 10-9. What is the concentration of Ca2+ ions in a saturated solution of CaCO3?
A. 1.4x10-9 M
B. 2.8x10-9 M
C. 5.6x10-9 M
D. 5.3x10-5 M
E. 1.1x10-4 M

Is this just because at equilibrium, stoichiometry says that, [Ca2+] = [CO32-] = √Ksp ?  Or is there something more to it?

3. Iodide reacts with peroxodisulphate in aqueous solution to produce iodine and sulphate according to the equation:
2I-(aq) + S2O82-(aq) → I2(aq) + 2SO42-(aq)

If the rate at which I2 is produced is 2.0 x 10-6 Ms-1, then:

A. S2O82- is consumed at 1.0 x 10-6 Ms-1
B. I- is consumed at 2.0 x 10-6 Ms-1
C. SO42- is formed at 2.0 x 10-6 Ms-1
D. S2O82- is consumed at 2.0 x 10-6 Ms-1
E. SO42- is formed at 1.0 x 10-6 Ms-1

The provided answer says D, which makes sense, but I don't know why E isn't also true (by stoichiometric ratios)?  i.e. D is true because the stoichiometric ratios of I2 and S2O82- are 1:1, but isn't E also true because SO42- will form at half the rate of I2 because it is producing twice the amount?

4. Samples of chlorine, neon and nitrogen are contained in three identical 1litre flasks. All are at the same temperature. The mass of gas in each flask is the same. Which of the following is TRUE?

A. They are all at the same pressure.
B. The density of chlorine is higher than the other two.
C. The density of nitrogen is lower than the other two.
D. The number of Ne atoms is half the number of Cl atoms. The pressure is highest in the flask of neon.
E. The pressure is highest in the flask of neon.

I got the answer E, because of the following rationale, but the provided answer said B?
Atomic masses: N < Ne < Cl
Molecular mass of gas: Ne < N2 < Cl2
Therefore Ne flask has most number of molecules and Cl2 has the least (to make the same mass).

• A is false because n is different for each flask.
B is false because Cl2 has fewest molecules, thus is LEAST dense.
C is false because N2 is in the middle.
D is false because for 1 gram mass, n(Ne) = 0.05 mol and n(Cl) = 0.028 mol.
E is true because Ne contains the most molecules, maximizing PV=nRT.
So why is B the answer?

Thanks everybody :)

1. Buffer works best when [acid] = [conjugate base]
So if Ka = [H+][base]/[acid], then the [acid] and [conjugate base] cancel => Ka = [H+], pKa = pH

2. Exactly. There's nothing more to it.

3. Let's have a look at this carefully. For every mole of I2 formed, stoichiometric ratios suggest that 2 moles of sulfate are formed. Hence, if the rate of formation of I2 is 2 * 10^-6 M/s, the rate of formation of sulfate should be twice that, not half that.

4. I agree with your answer. The densities (mass/volume) are the same as the masses and volumes are all identical. I think the question is messed up.
Title: Re: VCE Chemistry Question Thread
Post by: hobbitle on June 23, 2014, 04:31:19 pm
Thanks lznxl!!!
Title: Re: VCE Chemistry Question Thread
Post by: Reus on June 24, 2014, 09:18:27 pm
Generally the arrows for a reversible reaction looks like this
$\LARGE \rightleftharpoons$

But is it okay if I wrote it like this?
$\LARGE \leftrightharpoons$

Just have a tendency to do it that way when I'm writing fast, just wondering if it matters haha?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 24, 2014, 09:28:53 pm
Well. You'll be unlucky if you lost a mark for that. I still wouldn't write them that way :P
Title: Re: VCE Chemistry Question Thread
Post by: darklight on June 25, 2014, 07:45:02 pm
Hi guys,
With amino acids in a neutral pH we would have NH3+ and COO- and we would call this amphiprotic as the carboy can donate a proton and amino can accept it. However, I was just wondering if the pH dropped to say 2 and we had NH3+ and COOH would this still be considered amphiprotic?  :)
Title: Re: VCE Chemistry Question Thread
Post by: LiquidPaperz on June 25, 2014, 08:15:09 pm
Could some please explain this to be simply. I get 0.01=mass of solvent (x) / 1000 ( because 1000ml in litre and density is 1g ml-1). I get an answer of 10 micro grams, but answer is D
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on June 27, 2014, 05:45:45 pm
is this an exothermic or endothermic reaction?
Title: Re: VCE Chemistry Question Thread
Post by: #J.Procrastinator on June 27, 2014, 06:15:21 pm
Could some please explain this to be simply. I get 0.01=mass of solvent (x) / 1000 ( because 1000ml in litre and density is 1g ml-1). I get an answer of 10 micro grams, but answer is D

1ppm = 1 mg/Litre (used for weight/volume questions)

0.01 ppm = 0.01 mg/ Litre.

Since it is asking for the mass of Cd in 1 Litre, you get a mass of 0.01 mg from direct conversion. For this question there is no need for working out. You just have to be able to recognise that 1 ppm = 1 mg/ L. Hence, D is the answer :)
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on June 27, 2014, 09:51:20 pm
is this an exothermic or endothermic reaction?

I'm going to say endothermic. But this is really vague; wouldn't you typically be given the thermochemical equation (with the enthalpy) for the reaction, from which you can decipher whether the reaction is endo- or exothermic by looking at whether delta H is + or -?
Title: Re: VCE Chemistry Question Thread
Post by: Reus on June 29, 2014, 02:18:10 pm
What do we call reactions that keep donating H+? Like when the left side keeps losing H
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on June 29, 2014, 02:22:37 pm
What do we call reactions that keep donating H+? Like when the left side keeps losing H

Are you talking about redox reactions?
If so, I like to look more-so at the loss/gain of electrons!!
Title: Re: VCE Chemistry Question Thread
Post by: Reus on June 29, 2014, 02:47:07 pm
Are you talking about redox reactions?
If so, I like to look more-so at the loss/gain of electrons!!
No no, uh I'm talking about acid and bases...
I remember there was a specific name for it, but in Units ½
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on June 29, 2014, 03:07:57 pm
Acid base reactions with diprotric/monoprotic/ acids?
I'm not sure whether or not they're called amphiprotic too
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on June 29, 2014, 03:18:47 pm
Acid base reactions with diprotric/monoprotic/ acids?
I'm not sure whether or not they're called amphiprotic too

Yeah I remember the names of species acting in that manner, but not actual reactions.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 29, 2014, 05:07:01 pm
Amphiprotic = can act as acid or base
So conjugate bases of polyprotic acids are amphiprotic. Try H2CO3's conjugate base of HCO3-, for instance
Title: Re: VCE Chemistry Question Thread
Post by: vintagea on June 29, 2014, 10:28:14 pm
hello
can someone please explain this? using the electrochemical series
Tin metal is added to solutions of tin(II) chloride to prevent oxidation of the tin(II) ions by oxygen in air.

thank you
Title: Re: VCE Chemistry Question Thread
Post by: #J.Procrastinator on June 30, 2014, 02:00:17 pm
How would I work this out?

The formation of hydrogen iodide from its elements is represented by the equation:
H2(g) + I2(g) → 2HI(g)
This endothermic reaction has an activation energy of 167 kJ mol–1 and the heat of
reaction, ΔH, is +28 kJ mol–1. What is the activation energy for the reverse reaction,
the decomposition of two mole of hydrogen iodide?

With the reverse equation, wouldn't the activation energy still be the same, with ΔH just being -28 kJ mol-1?

The answer is 139 kJ mol-1
Title: Re: VCE Chemistry Question Thread
Post by: keltingmeith on June 30, 2014, 03:16:46 pm
Let's consider the following two energy profile diagrams:

Endothermic
(http://1.bp.blogspot.com/-i1xurymmZss/UQAXw9P6rXI/AAAAAAAAAsA/EGj1XfFWY0w/s320/Endothermic.png)
Exothermic
(http://1.bp.blogspot.com/-oqZ9Sg_WU_o/UQAXrziNCaI/AAAAAAAAAr4/Be4ndltubaU/s1600/Exothermic.png)

Notice how the endothermic reaction has to overcome the energy of deltaH AS WELL AS a further little bit? Then, for the exothermic reaction, it only needs to get over that little bit?

So, let's call that little excess bit q, as just some random bit of energy. From the energy profile diagram, we can see that:

$E_a(endothermic) = q + |\Delta H|
\\ E_a(exothermic) = q$

And from this, we get:

$E_a(endothermic) = E_a(exothermic) + |\Delta H|$

Since you know the activation energy for the endothermic reaction, to find the activation energy, you either use the above formula to find the activation energy for the reverse, or exothermic, reaction. However, I do suggest logicing it from energy profile diagrams rather than trying to memorise this formula.
Title: Re: VCE Chemistry Question Thread
Post by: #J.Procrastinator on June 30, 2014, 04:23:07 pm
Let's consider the following two energy profile diagrams:

Endothermic
(http://1.bp.blogspot.com/-i1xurymmZss/UQAXw9P6rXI/AAAAAAAAAsA/EGj1XfFWY0w/s320/Endothermic.png)
Exothermic
(http://1.bp.blogspot.com/-oqZ9Sg_WU_o/UQAXrziNCaI/AAAAAAAAAr4/Be4ndltubaU/s1600/Exothermic.png)

Notice how the endothermic reaction has to overcome the energy of deltaH AS WELL AS a further little bit? Then, for the exothermic reaction, it only needs to get over that little bit?

So, let's call that little excess bit K, as just some random bit of energy. From the energy profile diagram, we can see that:

$E_a(endothermic) = q + |\Delta H|
\\ E_a(exothermic) = q$

And from this, we get:

$E_a(endothermic) = E_a(exothermic) + |\Delta H|$

Since you know the activation energy for the endothermic reaction, to find the activation energy, you either use the above formula to find the activation energy for the reverse, or exothermic, reaction. However, I do suggest logicing it from energy profile diagrams rather than trying to memorise this formula.

Oh I get it now, thank you very much! :)

Wait what does the q in the expression represent?
Title: Re: VCE Chemistry Question Thread
Post by: keltingmeith on June 30, 2014, 04:39:53 pm
Whoops - sorry, q is the excess bit (which I called k in my original post). Amended.
Title: Re: VCE Chemistry Question Thread
Post by: Lizzy7 on June 30, 2014, 05:11:31 pm
Hi everyone, if anyone could please give me explanations for the following... I've tried researching but I haven't figured it out :p

So regarding organic reaction pathways, why is it that I can't directly substitute an 'H' for a 'OH' (hydroxyl) functional group? or an 'H' for a 'NH2' (amine)?

Like it has to be a chlorine from a chloroalkane?

E.g. Chloroalkane + NaOH -> alkanol + NaCl
Title: Re: VCE Chemistry Question Thread
Post by: #J.Procrastinator on June 30, 2014, 06:15:54 pm
Whoops - sorry, q is the excess bit (which I called k in my original post). Amended.

Ahaha I figured, all goods. :)
Title: Re: VCE Chemistry Question Thread
Post by: brightsky on June 30, 2014, 08:23:21 pm
Hi everyone, if anyone could please give me explanations for the following... I've tried researching but I haven't figured it out :p

So regarding organic reaction pathways, why is it that I can't directly substitute an 'H' for a 'OH' (hydroxyl) functional group? or an 'H' for a 'NH2' (amine)?

Like it has to be a chlorine from a chloroalkane?

E.g. Chloroalkane + NaOH -> alkanol + NaCl

H- is a fairly poor leaving group. Cl-, however, is a fairly good one. hence, a nucleophile will readily displace Cl- but not H-. so to form an alcohol, you need to first convert alkane to haloalkane via free radical halogenation, and then convert haloalkane to alcohol in a nucleophilic substitution reaction.
Title: Re: VCE Chemistry Question Thread
Post by: Lizzy7 on June 30, 2014, 10:16:51 pm
H- is a fairly poor leaving group. Cl-, however, is a fairly good one. hence, a nucleophile will readily displace Cl- but not H-. so to form an alcohol, you need to first convert alkane to haloalkane via free radical halogenation, and then convert haloalkane to alcohol in a nucleophilic substitution reaction.

Thank you brightsky!! :D
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on June 30, 2014, 10:23:18 pm
By 'better leaving group', he means more stable. Chloride we know is quite stable (in water at least) because Cl- has no acid-base properties. H-, however, is a VERY small negative charge that is a ridiculously powerful base (the conjugate base of hydrogen gas and how strong an acid is H2?). Therefore H- is almost never formed.

Brightsky, there is one reaction I can imagine in which H- DOES act as a leaving group :P
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on July 01, 2014, 07:51:05 pm
Just to clarify something:

In NMR, is chemical shift proportional or inversely proportional to frequency of the radio waves?

I thought that shielded atoms experience the magnetic field of the NMR machine to a lesser extent, and hence require less energy for resonance to be induced (ie. chemical shift is proportional to frequency), but TSFX says the opposite, that is shielded atoms require more energy for resonance to be induced. Does the discrepancy arises due to there apparently being 2 kinds of NMR machines (ones where the magnetic field varies and ones where the radio wave frequency varies)?
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on July 01, 2014, 08:32:26 pm
Just to clarify something:

In NMR, is chemical shift proportional or inversely proportional to frequency of the radio waves?

I thought that shielded atoms experience the magnetic field of the NMR machine to a lesser extent, and hence require less energy for resonance to be induced (ie. chemical shift is proportional to frequency), but TSFX says the opposite, that is shielded atoms require more energy for resonance to be induced. Does the discrepancy arises due to there apparently being 2 kinds of NMR machines (ones where the magnetic field varies and ones where the radio wave frequency varies)?
Please correct me if any of these information is wrong. And I believe VCAA does not require too much details about these:

Radio waves with sufficient energy can break the magnetic field, causing the nuclei to “flip” and create a peak.
Normally in NMR, they might vary the magnetic field so that if you have a high magnetic field, radio waves strength will not be strong enough to cause flipping. And if you have a low magnetic field, radio waves will be strong enough to cause flipping.
However, since different nuclei experiences different chemical environment, different magnetic field is needed in order for that nuclei to flip. Electrons play a vital role in the chemical environment because they shield the nucleus from the magnetic field, forcing them to experience less energy, hence, larger magnetic field strength is needed.
In short, a highly “exposed” nucleus requires lower magnetic field to cause flipping (especially those that close to a high electronegativity atom because these atoms will draw the electrons from the nucleus away). You can called these being “deshielded”.
Now, in the NMR spectrum, on the LHS, we call that the “downfield” region and on the RHS, we call that the “upfield” region. Downfield is the region of nuclei that experiences the MOST magnetic field and upfield is the region of nuclei that experiences the LEAST magnetic field. Since chemical shift is measure in ppm, we can say that when a peak is 7ppm, it means the magnetic field required to cause flipping of the nucleus is 7 millionths less than that of the TMS. Hence, higher chemical shift (the more downfield it is), the more exposed a nucleus.
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on July 01, 2014, 08:42:51 pm
Please correct me if any of these information is wrong. And I believe VCAA does not require too much details about these:

Radio waves with sufficient energy can break the magnetic field, causing the nuclei to “flip” and create a peak.
Normally in NMR, they might vary the magnetic field so that if you have a high magnetic field, radio waves strength will not be strong enough to cause flipping. And if you have a low magnetic field, radio waves will be strong enough to cause flipping.
However, since different nuclei experiences different chemical environment, different magnetic field is needed in order for that nuclei to flip. Electrons play a vital role in the chemical environment because they shield the nucleus from the magnetic field, forcing them to experience less energy, hence, larger magnetic field strength is needed.
In short, a highly “exposed” nucleus requires lower magnetic field to cause flipping (especially those that close to a high electronegativity atom because these atoms will draw the electrons from the nucleus away). You can called these being “deshielded”.
Now, in the NMR spectrum, on the LHS, we call that the “downfield” region and on the RHS, we call that the “upfield” region. Downfield is the region of nuclei that experiences the MOST magnetic field and upfield is the region of nuclei that experiences the LEAST magnetic field. Since chemical shift is measure in ppm, we can say that when a peak is 7ppm, it means the magnetic field required to cause flipping of the nucleus is 7 millionths less than that of the TMS. Hence, higher chemical shift (the more downfield it is), the more exposed a nucleus.
Hmm is that what ppm means? Alright, thanks for the reply. So it's the magnetic field that's used to flip the nuclei and not the radio waves?

Another question:

Why does 1,2-dimethylbenzene have only 2 hydrogen environments? No matter how I look at it I get 3.
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on July 01, 2014, 09:18:47 pm
Hmm is that what ppm means? Alright, thanks for the reply. So it's the magnetic field that's used to flip the nuclei and not the radio waves?

Another question:

Why does 1,2-dimethylbenzene have only 2 hydrogen environments? No matter how I look at it I get 3.
Yeah, according to my knowledge, in NMR, we can either vary the magnetic field and keep the radio wave constant OR vary the radio wave and keep the magnetic field constant. If we keep a magnetic field constant and vary the radio wave frequency, different nuclei will resonate at different frequencies, then, we can measure this resonance and plot the graph. However, as long as I'm aware,in practice, it's it pretty easier to keep the radio wave constant and vary the magnetic field.

Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on July 01, 2014, 09:27:35 pm
Yeah, according to my knowledge, in NMR, we can either vary the magnetic field and keep the radio wave constant OR vary the radio wave and keep the magnetic field constant. If we keep a magnetic field constant and vary the radio wave frequency, different nuclei will resonate at different frequencies, then, we can measure this resonance and plot the graph. However, as long as I'm aware,in practice, it's it pretty easier to keep the radio wave constant and vary the magnetic field.

If you split them in half between the 4 and 5 C, you see that they are symmetrical from both sides and hence, it has 2 H environments.
Ok thanks!

Really? But aren't the 3 and 6 hydrogens, the 4 and 5 hydrogens and the methyl hydrogens exposed to different chemical environments?
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on July 01, 2014, 09:56:10 pm
Ok thanks!

Really? But aren't the 3 and 6 hydrogens, the 4 and 5 hydrogens and the methyl hydrogens exposed to different chemical environments?
Oops sorry, yeah I think there should be 3H environments too :( whilst if we have 1,4-dimethyl benzene, that can be 2.
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on July 01, 2014, 10:05:31 pm
Oops sorry, yeah I think there should be 3H environments too :( whilst if we have 1,4-dimethyl benzene, that can be 2.
tsfx y u do dis :(

http://www.chemicalbook.com/SpectrumEN_95-47-6_1HNMR.htm

The actual spectrum seems to show 2 peaks as well though. Maybe it's something beyond the VCE course?
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on July 01, 2014, 10:35:45 pm
tsfx y u do dis :(

http://www.chemicalbook.com/SpectrumEN_95-47-6_1HNMR.htm

The actual spectrum seems to show 2 peaks as well though. Maybe it's something beyond the VCE course?
I got a contradictory information here :(. Read page 231, scheme 4.6a
Title: Re: VCE Chemistry Question Thread
Post by: Aurelian on July 01, 2014, 10:49:10 pm
Hey guys,

There are three different environments formally, but in practice two of these environments have pretty much the same chemical shift, because they are very similar. As a result, two of the signals overlap with each other to give a messy looking multiplet at around 7.1 ppm. In light of this fact, in that spectrum from the Chemical Book this signal has been attributed jointly to all four of the aromatic hydrogens.

In short: formally, there are three environments, but in practice two of these environments behave as one.
Title: Re: VCE Chemistry Question Thread
Post by: Valyria on July 02, 2014, 11:29:00 am
Hey guys,

If a question involving calculations is split into multiple parts and the answer found in part (a) was adjusted to 3 significant figures, in part (b) do we use the adjusted figure or the figure on the calculator that has more than 3 significant figures? I thought the latter as it seems more accurate but a few questions from checkpoints uses the adjusted figure for future calculations.

Thanks :)
Title: Re: VCE Chemistry Question Thread
Post by: saba.ay on July 02, 2014, 12:06:40 pm
Hey guys,

If a question involving calculations is split into multiple parts and the answer found in part (a) was adjusted to 3 significant figures, in part (b) do we use the adjusted figure or the figure on the calculator that has more than 3 significant figures? I thought the latter as it seems more accurate but a few questions from checkpoints uses the adjusted figure for future calculations.

Thanks :)

Always use the full figure for calculations, as it would be more accurate. But, when writing the final answer for part b) you have to consider the significant figures of that full figure as found in (a). Does that make sense ? Hope so. :P

Title: Re: VCE Chemistry Question Thread
Post by: Valyria on July 02, 2014, 01:16:49 pm
Always use the full figure for calculations, as it would be more accurate. But, when writing the final answer for part b) you have to consider the significant figures of that full figure as found in (a). Does that make sense ? Hope so. :P

So basically, the answer in part (b) should contain the same number of significant figures as the answer in part (a)?
Title: Re: VCE Chemistry Question Thread
Post by: saba.ay on July 02, 2014, 01:34:25 pm
So basically, the answer in part (b) should contain the same number of significant figures as the answer in part (a)?

Assuming that the smallest number of significant figures belongs to the value found in part (a), then yes. If the smallest number of significant figures belongs to another value introduced in part (b), then you'd obviously use that to determine the significant figures for the answer of part (b).
Title: Re: VCE Chemistry Question Thread
Post by: Valyria on July 02, 2014, 02:24:07 pm
Assuming that the smallest number of significant figures belongs to the value found in part (a), then yes. If the smallest number of significant figures belongs to another value introduced in part (b), then you'd obviously use that to determine the significant figures for the answer of part (b).

I thought the least precise piece of information provided within the stem of the question dictated how many significant figures we used for our answers :/
Title: Re: VCE Chemistry Question Thread
Post by: keltingmeith on July 02, 2014, 02:28:44 pm
It does, really. Think of it like this: if your least precise piece of information is to two significant figures, you're not going to round below that. That means in part a, the lowest amount of significant figures that you could have is two, because you're not going to go below that. So when you get to part b, you'll be using the lowest amount in the actual question, which is two.

However, consider this scenario: in part a, you round to two figures. But, in part b, you don't use the data from the question that's to two figures, but you do use your answer to part a. That means, the lowest amount of significant figures is coming from your answer to part a, which is two. So, you use that answer to guide your answer to part b.

Hope that all made sense. n.n;
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on July 02, 2014, 02:31:56 pm
So in gas chromatography the more volatile compounds are eluded first, right? Alkenes are more volatile than alkanes, does that mean they are eluded before alkanes? Asking cuz TSFX lists it the other way round
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on July 02, 2014, 02:38:26 pm
Also will we be penalised if we name redundantly?

for example, 1,1,1-tricholoro2,2-dimethylpropane, apparently we're supposed to omit the 2,2 since theres no other possible position for it?

Another example is 2-methylpropanoic acid, which is apparently just methylpropanoic acid. This doesnt seem to be an IUPAC convention either
Title: Re: VCE Chemistry Question Thread
Post by: Rod on July 02, 2014, 02:44:00 pm
Is energy from natural gases released via combustion?
Title: Re: VCE Chemistry Question Thread
Post by: alchemy on July 02, 2014, 06:36:51 pm
Five aspirin tablets, weighing 2.456g in total, are crushed and added to a 250mL beaker. 150mL of 0.252M NaOH is added to the beaker and the contents are heated gently. After heating, 20mL aliquots from the beaker are titrated against 0.15M HCl solution. The average titre is 14.88mL. Calculate the number of mole of aspirin and the % mass of aspirin in each tablet.

Sorry for the easy question, but something in my working out is preventing me from getting the correct answer.

Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on July 02, 2014, 08:28:52 pm
In mass spectrometry,
the relative intensities of the ions depend on:
- the energy of the bombarding electrons
- the stability of the ion fragments formed
- the ease at which ions can lose atoms

What exactly does this mean? What is it referring to when it says 'intensities"?

Thanks :-)
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on July 02, 2014, 08:49:46 pm
Five aspirin tablets, weighing 2.456g in total, are crushed and added to a 250mL beaker. 150mL of 0.252M NaOH is added to the beaker and the contents are heated gently. After heating, 20mL aliquots from the beaker are titrated against 0.15M HCl solution. The average titre is 14.88mL. Calculate the number of mole of aspirin and the % mass of aspirin in each tablet.

Sorry for the easy question, but something in my working out is preventing me from getting the correct answer.

I'm not the best at chemistry but I tried and got 81.43% as the percentage mass. I know this is defs wrong because I didn't use dilution factors but no harm in asking  :P
n(aspirin) = 0.002232 mol
Title: Re: VCE Chemistry Question Thread
Post by: alchemy on July 02, 2014, 10:08:29 pm
I'm not the best at chemistry but I tried and got 81.43% as the percentage mass. I know this is defs wrong because I didn't use dilution factors but no harm in asking  :P
n(aspirin) = 0.002232 mol

That's what I got the first time around, so it isn't correct. Btw, I fixed my mistake. Here's my working out:

n(NaOH)=0.252*0.15=0.0378
n(HCl)=0.15*0.0148=0.002232
NaOH(aq)+HCl(aq)-->NaCl(aq)+H20
n(NaOH)=n(HCl)=0.002232 in excess
n(NaOH) in excess in 250mL beaker: n(NaOH)*(150/20)=0.01674 .... this is the part that I initially made a mistake on. It's necessary to multiply the number of moles of NaOH by the number of aliquots that were added to the 250mL beaker.
n(NaOH) reacting with aspirin = 0.0378-0.01674=0.02106
n(Aspirin)=n(NaOH reacting with aspirin)/2 = 0.02106/2=0.0105
m(Aspirin)=0.0105*180=1.89g
Title: Re: VCE Chemistry Question Thread
Post by: Scooby on July 02, 2014, 10:13:57 pm
Also will we be penalised if we name redundantly?

for example, 1,1,1-tricholoro2,2-dimethylpropane, apparently we're supposed to omit the 2,2 since theres no other possible position for it?

Another example is 2-methylpropanoic acid, which is apparently just methylpropanoic acid. This doesnt seem to be an IUPAC convention either

Yeah, VCAA have penalised students for that in the past
Title: Re: VCE Chemistry Question Thread
Post by: thushan on July 02, 2014, 10:32:06 pm
Have VCAA actually penalised this in the past? Which year?
Title: Re: VCE Chemistry Question Thread
Post by: AngelWings on July 02, 2014, 10:54:17 pm
Have VCAA actually penalised this in the past? Which year?

Closest I've found is the Exam Assessment Report in the Unit 3 exam 2011, but it only states redundant numbers, not any penalising. (See Short Answer Q7aiv.)
Title: Re: VCE Chemistry Question Thread
Post by: Scooby on July 02, 2014, 11:15:57 pm
Closest I've found is the Exam Assessment Report in the Unit 3 exam 2011, but it only states redundant numbers, not any penalising. (See Short Answer Q7aiv.)

Yeah, I think that's what I'm remembering (I feel like it came up in some other exam as well but I can't remember where). Since the question asks for the systematic name, and then they've written that "ethan-1-ol is not the systematic name," it's probably safe to assume they didn't accept it. Either way, I'd just be wary of it
Title: Re: VCE Chemistry Question Thread
Post by: Valyria on July 04, 2014, 03:04:52 pm
What is the stationary & mobile phase for GLC and HPLC? And is there a difference between GC and GLC or are the terms used interchangeably?

Ty :)
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on July 04, 2014, 03:57:05 pm
What is the stationary & mobile phase for GLC and HPLC? And is there a difference between GC and GLC or are the terms used interchangeably?

Ty :)
GLC uses a porous solid coated with high boiling point liquid hydrocarbon (so it doesnt vaporise in the chromatography machine) as the stationary phase, and uses an inert carrier gas as the mobile phase, for example N2 or He. Inert doesnt necessarily mean noble gas btw, just one that will not reactive with the species under analysis.

HPLC uses small, high surface area solid particles as the stationary phase, eg. Alumina and uses some solvent as a mobile phase. Whats actually used depends on the setup and what you're trying to separate or identify

I dont know if they can be used interchangably, but I know there is also gas solid chromatography.
Title: Re: VCE Chemistry Question Thread
Post by: Penguuu on July 06, 2014, 10:08:28 am
Hi guys, quick question about properties of alkanes and alkenes. Which one has a higher boiling point? First I though alkenes because double bonds are harder to break, but apparently its alkanes? As alkenes have irregular shapes so the dispersion forces are weaker. Alkanes have regular structures that fit closely together allowing for stronger dispersion forces to exist between the molecules.

Title: Re: VCE Chemistry Question Thread
Post by: thushan on July 06, 2014, 10:11:19 am
To boil a substance, you need to break bonds BETWEEN molecules, which include:
- hydrogen bonds
- dipole dipole interactions
- dispersion forces (as for an alkene)

You don't need to break the covalent bonds within a molecule. Yes, double bonds are stronger (in an alkene), but boiling a substance does not require that double bonds be broken.

Dispersion forces are stronger when:
- the molecule is bigger
- the molecules are closely packed.
Title: Re: VCE Chemistry Question Thread
Post by: Penguuu on July 06, 2014, 11:16:29 am
To boil a substance, you need to break bonds BETWEEN molecules, which include:
- hydrogen bonds
- dipole dipole interactions
- dispersion forces (as for an alkene)

You don't need to break the covalent bonds within a molecule. Yes, double bonds are stronger (in an alkene), but boiling a substance does not require that double bonds be broken.

Dispersion forces are stronger when:
- the molecule is bigger
- the molecules are closely packed.

Thank you!
Title: Re: VCE Chemistry Question Thread
Post by: #J.Procrastinator on July 07, 2014, 10:58:19 am
With respect to finding to equilibrium constant, K, why is it that we don't include solids or liquids? For example, if we were finding the K constant for the photosynthesis reaction :

6CO2 (g) + 6H2O (l)  → C6H12O6 (aq) + 6O2 (g)

would we include the glucose or water molecule in the concentration fraction? What's the reason for this?

Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on July 07, 2014, 11:09:21 am
With respect to finding to equilibrium constant, K, why is it that we don't include solids or liquids? For example, if we were finding the K constant for the photosynthesis reaction :

6CO2 (g) + 6H2O (l)  → C6H12O6 (s) + 6O2 (g)

would we include the glucose or water molecule in the concentration fraction? What's the reason for this?
You don't include glucose or water in the formulation of K (edit: glucose is aqueous for photosynthesis so you do include it)

This is because the equilibrium constant actually uses the activity of the species rather than their molarity, and the activity of pure solids and liquids is defined to be one.

Further reading if you're keen, though it's not really relevant to VCE:

http://en.wikipedia.org/wiki/Thermodynamic_activity

Molarity approximates activity so we use it for simplicity's sake.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on July 07, 2014, 11:12:47 am
With respect to finding to equilibrium constant, K, why is it that we don't include solids or liquids? For example, if we were finding the K constant for the photosynthesis reaction :

6CO2 (g) + 6H2O (l)  → C6H12O6 (s) + 6O2 (g)

would we include the glucose or water molecule in the concentration fraction? What's the reason for this?

As you know, the equilibrium constant, takes account of concentrations. When a gas is dissolved in water, it's concentration will vary. However, when it is a solid or a liquid, it's concentration does not vary.
Also, I'm pretty sure that the glucose produced in photosynthesis, is in its aqueous state. Could someone please confirm this.
Thanks
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on July 07, 2014, 11:17:05 am
As you know, the equilibrium constant, takes account of concentrations. When a gas is dissolved in water, it's concentration will vary. However, when it is a solid or a liquid, it's concentration does not vary.
Also, I'm pretty sure that the glucose produced in photosynthesis, is in its aqueous state. Could someone please confirm this.
Thanks
Oh yeah my bad, yeah glucose is aqueous!
Title: Re: VCE Chemistry Question Thread
Post by: #J.Procrastinator on July 07, 2014, 11:21:58 am
You don't include glucose or water in the formulation of K (edit: glucose is aqueous for photosynthesis so you do include it)

This is because the equilibrium constant actually uses the activity of the species rather than their molarity, and the activity of pure solids and liquids is defined to be one.

Further reading if you're keen, though it's not really relevant to VCE:

http://en.wikipedia.org/wiki/Thermodynamic_activity

Molarity approximates activity so we use it for simplicity's sake.

As you know, the equilibrium constant, takes account of concentrations. When a gas is dissolved in water, it's concentration will vary. However, when it is a solid or a liquid, it's concentration does not vary.
Also, I'm pretty sure that the glucose produced in photosynthesis, is in its aqueous state. Could someone please confirm this.
Thanks

Oh thank you guys! :) Yes, whoops, glucose is in it's aqueuos state !
Title: Re: VCE Chemistry Question Thread
Post by: alchemy on July 07, 2014, 12:36:25 pm
Question attached.

Another question: if I was asked "what features might an infrared spectrum of propanoic acid have?" Without being provided any data, how would I answer it? The answers in the book actually refer to numerical values and I don't know how they got them.
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on July 07, 2014, 12:59:15 pm
Question attached.

Another question: if I was asked "what features might an infrared spectrum of propanoic acid have?" Without being provided any data, how would I answer it? The answers in the book actually refer to numerical values and I don't know how they got them.

Hi

The numbers are from the data book. Check the data book and based on your knowledge of what propanoic acid looks like, you will be able to find the wavelengths of the features such as a COOH group etc.
Title: Re: VCE Chemistry Question Thread
Post by: alchemy on July 07, 2014, 01:08:32 pm
Hi

The numbers are from the data book. Check the data book and based on your knowledge of what propanoic acid looks like, you will be able to find the wavelengths of the features such as a COOH group etc.

Oh ok. What about the question I attached? Were you able to get the answer to that as well?
Title: Re: VCE Chemistry Question Thread
Post by: keltingmeith on July 07, 2014, 01:14:50 pm
Yep, that's all in there.

Have you got a copy of the databook, alchemy? If not, I suggest printing off a copy from a VCAA exam, and keeping hold of it whenever you do any question from now on. If you come across something you don't have enough information for, see if there's something in the data book that can help you. Learning that book inside and out now can help you become very quick at pulling information out of it later in your exam. Last year, my teacher gave us all a copy, and we had to use it in all SACs, practice exams, practice questions, etc. and believe you and me - it certainly helped in the long run.
Title: Re: VCE Chemistry Question Thread
Post by: TimewaveZero on July 07, 2014, 01:54:20 pm
Hey guys, just wondering if anyone knows of a quick way to deduce weather a fatty acid is saturated/unsaturated without drawing it by looking at the number of carbon atoms compared to the number of hydrogen atoms?

I have come across a number of questions where your are required to decide which of two or more fatty acids would create a saturated /unsaturated fat.

For example;

C17H35-C=OOH

or

C16H31-C=OOH

Also, I just came across a question asking how many double bonds in a particular fatty acid. How can one devise this from a formula without drawing the entire molecule out?
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on July 07, 2014, 02:25:34 pm
A saturated fatty acid has the formula
$C_nH_{2n+1}COOH$
every double bond is 2H less than the saturated fatty acid

So for example, if we have $C_{17}H_{31}COOH$
the saturated version of that would be $C_{17}H_{35}COOH$  <--- 35 came from doubling 17, and adding 1.
Since its 31H rather than 35H, there are 2 double bonds.
Title: Re: VCE Chemistry Question Thread
Post by: TimewaveZero on July 07, 2014, 03:02:29 pm
Thankyou!! Much appreciation.
Title: Re: VCE Chemistry Question Thread
Post by: ETTH96 on July 07, 2014, 06:49:30 pm
Hey guys, can someone help me out with this?

300ml of 1.5 M HCl is mixed with 300 ml of 1.0 M NaOH in a thermally insulated container. the initial temperature of each solution before mixing was 18.
They react according to the equation:
H+ (aq) + OH- (aq) --> H2O(l)    Enthalpy=-56 kJmol-1

if 2.5 kJ is required to raise the temperature of 600ml of water by 1.0oC, then the final temperature, in degrees celsius, of the solution after mixing would be about:
a) 10
b) 25
c) 28
d) 40
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on July 07, 2014, 08:54:00 pm
Hey guys, can someone help me out with this?

300ml of 1.5 M HCl is mixed with 300 ml of 1.0 M NaOH in a thermally insulated container. the initial temperature of each solution before mixing was 18.
They react according to the equation:
H+ (aq) + OH- (aq) --> H2O(l)    Enthalpy=-56 kJmol-1

if 2.5 kJ is required to raise the temperature of 600ml of water by 1.0oC, then the final temperature, in degrees celsius, of the solution after mixing would be about:
a) 10
b) 25
c) 28
d) 40

Is the answer B? If it is, then here's the explanation.

Firstly, the energy released in this reaction:
n(NaOH) = 0.3 * 1 = 0.3 mol
0.3mol * 56kJ/mol = 16.8 kJ
So, 16.8kJ of energy will be released in this reaction (theoretically of course).

16800 J= 4.184 * 600 * change in temp
change in temp = 6.7 = 7

Change in Temp = Max Temp - Min Temp
7 = x - 18
x = 18 + 7
x = 25 degrees C, hence B.
Title: Re: VCE Chemistry Question Thread
Post by: hyunah on July 08, 2014, 11:37:28 am
hello
how would i do this question 14c) and 14d) i know 14 a and b are increase
for 14c) if both h30+ and OH- ions concerntration increases how does that equate to an overall ph decrease? cause i know the increase h30+ concerntration would account for the decreased ph, but wouldnt the increase oh- then counteract that?
for 14d) if 14c) is in fact ph decrease, wouldnt that mean the water becomes acidic, but then i know that h30+ concerntration = oh- concerntration therefore it has to be neutral, but then the ph decreased?

so confused????
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on July 08, 2014, 12:13:26 pm
[H3O+]=[OH-] in water at 25 degrees celcius.
When you increase the temperature, there is a higher amount of [H+] ions, therefore causing the pH of the water to decrease and hence become acidic.
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on July 08, 2014, 12:15:58 pm
hello
how would i do this question 14c) and 14d) i know 14 a and b are increase
for 14c) if both h30+ and OH- ions concerntration increases how does that equate to an overall ph decrease? cause i know the increase h30+ concerntration would account for the decreased ph, but wouldnt the increase oh- then counteract that?
for 14d) if 14c) is in fact ph decrease, wouldnt that mean the water becomes acidic, but then i know that h30+ concerntration = oh- concerntration therefore it has to be neutral, but then the ph decreased?

so confused????
A solution is considered as neutral when pH=7 at 25C . When you increase, or decrease the temperature outside 25C, a solution is considered neutral when [H+] = [OH-]. Hence, you can have a solution of pH = 6 at a given temperature but still called neutral.
Kw = [H30+] x [OH-] = 10^(-14) M at 25C
When you increase the temperature, forward reaction is favoured, hence, you obtained more of both [H30+] and [OH-]. This will change the value of Kw (that's why like all equilibrium constant, Kw is temperature dependent).
As mentioned above, because in a neutral solution, you have equal amount of hydronium and hydroxide, the Kw at any given temperature is
Kw= [H30+] x [OH-] = [H30+]^2
Since Kw is increasing when you apply heat, let's make up a larger Kw value and see how that changes the pH value.
New Kw = 51.3 x 10^(-14) M
[H3O+] = *sqrt* 51.3 x 10^(-14)
[H30+] = 7.16 x 10 ^ (-7)
Hence, pH = 6.14
In brief, even if you increase the temperature, the solution is still neutral despite the fact that it's pH is decreasing. pH 7 of a neutral solution only holds at 25C.
Correct me if I'm wrong.
[Beaten]
Title: Re: VCE Chemistry Question Thread
Post by: hyunah on July 08, 2014, 01:01:08 pm
thank you

can you please also explain 14c) why there a ph decrease?
if both h30+ and OH- ions concerntration increases how does that equate to an overall ph decrease? cause i know the increase h30+ concerntration would account for the decreased ph, but wouldnt the increase oh- then counteract that?
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on July 08, 2014, 02:04:28 pm
pH actually has no direct relationship with the concentration of hydroxide. It's JUST the H+ concentration; the corresponding increase in OH- with a higher temperature only means your solution is still neutral. As stated above, the increase in both H+ and OH- concentrations is a result of an increase in Kw. Normally we associate increasing [H+] with decreasing [OH-] because Kw is normally constant. It's not now.
Title: Re: VCE Chemistry Question Thread
Post by: hyunah on July 09, 2014, 09:51:08 am
thank you :)

can someone please explain the mistakes in my ans?

Sodium carbonate is readily soluble in water but calcium
carbonate is insoluble. A solution of sodium carbonate was
found to have a pH of 8. A quantity of calcium nitrate was
added to the solution. Explain the
effect the addition of calcium nitrate would have on the pH
of the solution.

na2co3 + ca(no3)2 (mixed with water ) --> caco3 + nano3
becasue caco3 are insoluble, the only ions left in the mixture is na+ and no3-
oh- ions from water react with na+ and H react with no3-
wheres the effect on ph?

Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on July 09, 2014, 11:47:41 am
thank you :)

can someone please explain the mistakes in my ans?

Sodium carbonate is readily soluble in water but calcium
carbonate is insoluble. A solution of sodium carbonate was
found to have a pH of 8. A quantity of calcium nitrate was
added to the solution. Explain the
effect the addition of calcium nitrate would have on the pH
of the solution.

na2co3 + ca(no3)2 (mixed with water ) --> caco3 + nano3
becasue caco3 are insoluble, the only ions left in the mixture is na+ and no3-
oh- ions from water react with na+ and H react with no3-
wheres the effect on ph?

I don't quite get what you're trying to do.

You have a few equilibria at play here.
Na2CO3(aq) + H2O(l) <=> NaHCO3(aq) + NaOH(aq)
Ca2+(aq) + CO32-(aq) <=> CaCO3(s)

Add calcium nitrate => increase [Ca2+] => second reaction goes more towards the right => decrease carbonate concentration => first reaction goes backwards to reform carbonate => hydroxide ions concentration drops => pH decreases
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on July 09, 2014, 05:39:04 pm
so i just did a practise exam and stumbled upon this question, which i didnt really know how to do. Could someone please give me insight as to how to tackle these sort of questions where you have an acid thats mono/di/triprotic etc and how mole calculations would differ?
(http://i62.tinypic.com/30tjg3q.png)

here's the answers, sorry for the pics being so big lol
(http://i60.tinypic.com/2hnmf51.jpg)
Title: Re: VCE Chemistry Question Thread
Post by: Reus on July 09, 2014, 06:06:42 pm
Would it be correct to say, that buffers ONLY occur with weak acids and its conjugate base, as if it were strong, it would ionise completely?

Thanks.
Title: Re: VCE Chemistry Question Thread
Post by: alchemy on July 09, 2014, 06:12:39 pm
so i just did a practise exam and stumbled upon this question, which i didnt really know how to do. Could someone please give me insight as to how to tackle these sort of questions where you have an acid thats mono/di/triprotic etc and how mole calculations would differ?

What I would do is pick any mono/di/tri protic acid that you know and calculate the concentration of them for the appropriate section in the table.
For example, for the first row of the table ("if the acid is monoprotic"), choose HCl to be the unknown acid in the question (since it's a monoprotic acid).
We can then write it's reaction with sodium carbonate in a balanced chemical equation as 2HCl + Na2CO3 --> 2NaCl + H2O + CO2
Therefore, the number of moles of the monoprotic acid (HCl) is 2*n(Na2CO3). From there, you can proceed to find the concentration of the diluted acid as well as the concentration of the original acid.
For the next row, pick any old diprotic acid you know (e.g. H2SO4) and repeat the above steps. Same applies for when it is a tri protic acid ;)
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on July 09, 2014, 06:17:53 pm
Thanks alchemy!!  ;D
Title: Re: VCE Chemistry Question Thread
Post by: hyunah on July 10, 2014, 03:46:47 pm
can someone please explain to me how to do part c) and e)?

thank you
Title: Re: VCE Chemistry Question Thread
Post by: BLACKCATT on July 10, 2014, 07:38:37 pm
Just wondering whether its better to finish all unit 3 practice exams now or save it for later? What do you guys think?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on July 10, 2014, 09:03:50 pm
Just wondering whether its better to finish all unit 3 practice exams now or save it for later? What do you guys think?

I'm doing a few now, here and there for a scope of questions. But, I'm saving the majority for later.
Title: Re: VCE Chemistry Question Thread
Post by: Reus on July 10, 2014, 11:25:13 pm
Could someone please explain why halving the volume (doubling the pressure) of an equilibrium will have the effect of a net forward reaction. Whereas when increasing pressure on the equilibrium reaction, there will be no effect?

Thanks :)
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on July 11, 2014, 12:18:11 am
When there is a reaction with more molecules on the LEFT as opposed to the right, increasing the pressure will cause a net forward reaction, to the side where there are less particles, to oppose that change in pressure.
eg. 2A+4B <===>2C+D

When there are equal numbers of particles on either side of a reaction, increasing the pressure will have no effect as the system cannot oppose that change.
eg. A+B<====>C+D
Title: Re: VCE Chemistry Question Thread
Post by: nhmn0301 on July 11, 2014, 12:22:23 am
Could someone please explain why halving the volume (doubling the pressure) of an equilibrium will have the effect of a net forward reaction. Whereas when increasing pressure on the equilibrium reaction, there will be no effect?

Thanks :)
Lol soNasty ninja'd :P
For equilibrium, it would be insufficient if you don't specify the equation you are referring to. I made a post a while ago here Re: Rod's Chemistry 3/4 Questions Thread so please have a look through it and let me know if you need any further clarifications :D!
Title: Re: VCE Chemistry Question Thread
Post by: Reus on July 11, 2014, 12:28:27 am
Thanks guys :)

Haha thought about posting the reaction too but cbf'd :P
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on July 11, 2014, 11:09:23 am
Could someone please explain why halving the volume (doubling the pressure) of an equilibrium will have the effect of a net forward reaction. Whereas when increasing pressure on the equilibrium reaction, there will be no effect?

Thanks :)

When the volume is halved, the pressure increases. In order to partially oppose this increase in pressure, the reaction that will be favoured will be the one that produces fewer molecules. In this way, pressure can be partially reduced. All you would then need to do is read your equation, and from that, look at which direction reaction produces FEWER molecules based on the mole ratios. The reaction that produces fewer particles will be favoured, and a net reaction would occur in that direction.
Title: Re: VCE Chemistry Question Thread
Post by: hyunah on July 11, 2014, 11:30:16 am

thank you
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on July 11, 2014, 02:40:42 pm
What exactly is the concentration fraction and how does it compare to an equilibrium constant?
Title: Re: VCE Chemistry Question Thread
Post by: Rishi97 on July 11, 2014, 02:46:53 pm
What exactly is the concentration fraction and how does it compare to an equilibrium constant?

I like to think of the concentration fraction like a ratio between the products and reactants. Only temperature effects the equilibirum constant but the concentration fraction can be effected by mostly everything.
eg. If I increase the concentration of a reactant, there will be more products produced so the concentration fraction will increase. BUT the equilibrium constant remains the same as there is no change is temperature.
Title: Re: VCE Chemistry Question Thread
Post by: soNasty on July 11, 2014, 02:56:04 pm
Oh ok. Thanks rish!!
Title: Re: VCE Chemistry Question Thread
Post by: alchemy on July 11, 2014, 03:13:23 pm
Which one of the following acts as a base when hydrofluoric acid is dissolved in an aqueous solution of sodium chloride?
a. Cl–
b. HF
c. H2O
d. NaCl
e. Na+

The answer is C. Can someone explain why.
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on July 11, 2014, 03:31:33 pm

thank you

When a galvanic cell is set up, you have two separate half-cells that are connected by connecting wires and a salt-bridge, in order to complete the circuit and enable its function. In a dry cell, a si-permeable membrane is used, that allows the two components to be separated, yet still connected by employing a membrane with a specific degree of permeability, which is how a dry cell ultimately functions.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on July 11, 2014, 10:02:50 pm
Which one of the following acts as a base when hydrofluoric acid is dissolved in an aqueous solution of sodium chloride?
a. Cl–
b. HF
c. H2O
d. NaCl
e. Na+

The answer is C. Can someone explain why.

NaCl, or Cl-, won't act as a base as Cl- is the conjugate base of an exceedingly strong acid (HCl), so it has no acid-base properties. Therefore neither a nor d is right. E is definitely wrong as Na+ ions don't do anything either. b is wrong as HF is the acid. c is correct as when you dissolve HF in water, a portion of the HF reacts by the following reaction
HF (aq) + H2O (l) <=> F- (aq) + H3O+ (aq)

Here the water accepts a proton and thus acts as a base
Title: Re: VCE Chemistry Question Thread
Post by: alchemy on July 12, 2014, 12:07:46 am
can someone please explain to me how to do part c) and e)?

thank you

Use the data book for part c. You will see that butanoic acid is weaker than ethanoic acid because butanoic acid has a lower Ka value than ethanoic acid. Therefore, for part e, the Ph of a similar solution of ethanoic acid would be lower.
Title: Re: VCE Chemistry Question Thread
Post by: thushan on July 12, 2014, 07:38:08 am
NaCl, or Cl-, won't act as a base as Cl- is the conjugate base of an exceedingly strong acid (HCl), so it has no acid-base properties. Therefore neither a nor d is right. E is definitely wrong as Na+ ions don't do anything either. b is wrong as HF is the acid. c is correct as when you dissolve HF in water, a portion of the HF reacts by the following reaction
HF (aq) + H2O (l) <=> F- (aq) + H3O+ (aq)

Here the water accepts a proton and thus acts as a base

Very solidly explained :D
Title: Re: VCE Chemistry Question Thread
Post by: psyxwar on July 12, 2014, 11:41:03 am
What exactly is the concentration fraction and how does it compare to an equilibrium constant?
I've actually never heard this term before. Isn't it Q, the reaction quotient?
Title: Re: VCE Chemistry Question Thread
Post by: Reus on July 12, 2014, 11:44:53 am
I've actually never heard this term before. Isn't it Q, the reaction quotient?
Same here, until I did that particular question haha.
Title: Re: VCE Chemistry Question Thread
Post by: lzxnl on July 12, 2014, 12:27:41 pm
I've actually never heard this term before. Isn't it Q, the reaction quotient?

Concentration fraction and reaction quotient refer to the same thing
Title: Re: VCE Chemistry Question Thread
Post by: TimewaveZero on July 12, 2014, 01:53:12 pm
Why is it undesirable to use a small volume of analyte (<2mL) in volumetric titration analysis?
Title: Re: VCE Chemistry Question Thread
Post by: Yacoubb on July 12, 2014, 03:19:13 pm
Why is it undesirable to use a small volume of analyte (<2mL) in volumetric titration analysis?

Because this increases the chance of errors occuring, and thus renders your subsequent calculations regarding concentration, etc inaccurate and discreditable.
Title: Re: VCE Chemistry Question Thread
Post by: TimewaveZero on July 12, 2014, 04:11:14 pm