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RKTR

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Re: VCE Chemistry Question Thread
« Reply #480 on: March 15, 2014, 10:03:25 pm »
+1
In chromatography, are the Rf values and the Rt value, inversely proportional?  ???
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grannysmith

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Re: VCE Chemistry Question Thread
« Reply #481 on: March 16, 2014, 03:49:46 pm »
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Do we need to show working out in chemistry?

HEN_iP

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Re: VCE Chemistry Question Thread
« Reply #482 on: March 16, 2014, 05:16:52 pm »
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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.
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ETTH96

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Re: VCE Chemistry Question Thread
« Reply #483 on: March 16, 2014, 09:06:13 pm »
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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

lzxnl

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Re: VCE Chemistry Question Thread
« Reply #484 on: March 16, 2014, 09:26:25 pm »
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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.
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rhinwarr

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Re: VCE Chemistry Question Thread
« Reply #485 on: March 17, 2014, 09:30:47 pm »
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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

soNasty

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Re: VCE Chemistry Question Thread
« Reply #486 on: March 18, 2014, 02:30:26 pm »
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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
« Last Edit: March 18, 2014, 02:32:25 pm by andrew2910 »

ETTH96

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Re: VCE Chemistry Question Thread
« Reply #487 on: March 18, 2014, 09:25:01 pm »
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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

lzxnl

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Re: VCE Chemistry Question Thread
« Reply #488 on: March 18, 2014, 09:30:22 pm »
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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
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Blondie21

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Re: VCE Chemistry Question Thread
« Reply #489 on: March 21, 2014, 05:29:13 pm »
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Is

CH3CH2CH(NH2)(CH2)2COOH

4-amino penatnoic acid?
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Aurelian

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Re: VCE Chemistry Question Thread
« Reply #490 on: March 21, 2014, 05:37:35 pm »
+1
Is

CH3CH2CH(NH2)(CH2)2COOH

4-amino penatnoic acid?

No, it's 4-aminohexanoic acid - there are six carbons in the chain.
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eagles

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Re: VCE Chemistry Question Thread
« Reply #491 on: March 21, 2014, 06:27:38 pm »
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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? :)

PB

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Re: VCE Chemistry Question Thread
« Reply #492 on: March 21, 2014, 07:11:04 pm »
+1
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
« Last Edit: March 21, 2014, 07:12:53 pm by PB »
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Re: VCE Chemistry Question Thread
« Reply #493 on: March 21, 2014, 07:25:56 pm »
+2
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

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Re: VCE Chemistry Question Thread
« Reply #494 on: March 21, 2014, 08:38:22 pm »
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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!