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Mr Keshy

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Re: Chemistry 3/4 2013 Thread
« Reply #840 on: May 18, 2013, 11:41:20 pm »
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Don't take these as correct yet, I'm not entirely sure but I'll fix any errors as soon as someone points it out. :) I'm actually kinda confused about this too.. In some diagrams they put the states in there, in some, they don't. Some of this I found in my A+ notes book for Chem too.
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brightsky

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Re: Chemistry 3/4 2013 Thread
« Reply #841 on: May 18, 2013, 11:48:15 pm »
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Yeah ethanol is polar because of the hydroxyl functional group. But it is also non-polar. It serves a dual purpose.

Water is only polar. So in water you can only dissolve polar solutes. In ethanol, you can dissolve both.

Therefore, in the context of this question: when should ethanol be used instead of water? If the solute is non-polar.

And ethanol is not necessarily pure, for the same reasons as water. Yeah, it's pretty vague  :)

ahh i see what you mean now. i stand corrected.
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Re: Chemistry 3/4 2013 Thread
« Reply #842 on: May 19, 2013, 09:53:02 am »
+4
I'd say it's good practice to put states in :) And yeah, I do believe you need to have then written down.

I'm not too confident in this myself so please correct me if I'm wrong! But from my understanding thus far

So to go from an Alkane to a Chloroalkane you'd need Cl2(g) and the catalyst is light/heat. To go from an Alkene to a Chloroalkane you need HCl(g) and the catalyst is AlCl3(s).

To go from a Chloroalkane to an Alkanol, you need NaOH(aq) and again, there's no specified catalyst for that. To go from an Akene to a Chloroalkane you need H2O(g) where H3PO4(s) is the catalyst.

To go from an Alkanol to a Carboxylic acid you need either Cr2O72-(aq) or MnO4- and H+(aq) acts as the catalyst

Then to go from a Carboxylic acid to an ester, you need the Alcohol (ethanol/methanol.. etc) and your catalyst (H2SO4(l)

Strictly speaking, I wouldn't call heat a catalyst, as heat merely provides energy to overcome the activation energy barrier. A catalyst speeds up both the forwards and the backwards reaction equally and does not affect the equilibrium position of a reaction. Heat and light, as forms of energy, will affect equilibrium. Also, this heat and light are absorbed and are not emitted after the reaction. Therefore, they are consumed as part of the reaction; a catalyst isn't.

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Mr Keshy

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Re: Chemistry 3/4 2013 Thread
« Reply #843 on: May 19, 2013, 10:56:51 am »
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Strictly speaking, I wouldn't call heat a catalyst, as heat merely provides energy to overcome the activation energy barrier. A catalyst speeds up both the forwards and the backwards reaction equally and does not affect the equilibrium position of a reaction. Heat and light, as forms of energy, will affect equilibrium. Also, this heat and light are absorbed and are not emitted after the reaction. Therefore, they are consumed as part of the reaction; a catalyst isn't.

Oh okay. What is the heat/light classified as then? I thought light/heat sped the reaction, but what you said makes sense. So a catalyst merely speeds up a reaction, so it's not absolutely necessary right? It'll just take longer for the reaction to finish.

But for reactions like an Alkane to Chloroalkane where light/heat is required, you need the light/heat energy or else the reaction won't occur?
« Last Edit: May 19, 2013, 11:05:27 am by Mr Keshy »
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brightsky

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Re: Chemistry 3/4 2013 Thread
« Reply #844 on: May 19, 2013, 11:16:57 am »
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Technically, yes, but at times if you do not provide a suitable catalyst, the reaction will literally take hundreds of years to reach 'completion', and for obvious reasons, we don't have that amount of time to spare.

Yes, in the case of free radical halogenation, UV radiation is required or else the reaction will not take place. This is because UV radiation is required for the first step of the reaction mechanism involved. Without UV light, the chlorine molecule will not undergo homolysis, which is indispensable to this particular kind of reaction.
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Re: Chemistry 3/4 2013 Thread
« Reply #845 on: May 19, 2013, 11:34:59 am »
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Yes, catalysts are not necessary for the reaction to take place; they are only required to speed the reaction up. UV light here is an essential part of the reaction as it splits the chlorine molecule into chlorine radicals. However, sulfuric acid in the addition of water to ethene, for instance, is not absolutely necessary; it does, however, provide a reaction mechanism with a lower activation energy. A reaction IS possible without sulfuric acid, but I'm not sitting around waiting for the product.
« Last Edit: May 19, 2013, 11:36:49 am by nliu1995 »
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Re: Chemistry 3/4 2013 Thread
« Reply #846 on: May 20, 2013, 05:55:15 pm »
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Hello, I wanting to ask what sort of organic compounds we must know? i.e. we must know what kind of carboxyl derivatives? i.e. acyls, salt of carboxyls, nitriles, amides? or just esters? I'm revising over these, but there's a lot of info on their formation, reduction and oxidation, and I don't think I can remember all the processes and info of all these functional groups? My teacher saying they can put anything on there, but please if anyone has any info on this would be many thanks.

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Re: Chemistry 3/4 2013 Thread
« Reply #847 on: May 20, 2013, 06:06:51 pm »
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Hello i have a question:

Why is there a bold line at the bottom of carbohydrates? And why does the disaccharide composed of two fructose molecules comprise of one fructose that has been flipped?

Thanks in advance.

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Re: Chemistry 3/4 2013 Thread
« Reply #848 on: May 20, 2013, 06:12:36 pm »
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Hi guys,

Does anyone know what actually determines whether a substance is a strong acid or base? I mean, considering oxygen is more electronegative than chlorine, why is HCl a stronger acid than CH3COOH? A considering an OH bond is highly polarised, why are water and alcohols such weak acids?

Thanks  :)
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Re: Chemistry 3/4 2013 Thread
« Reply #849 on: May 20, 2013, 06:30:00 pm »
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acid dissociation constant (Ka), obtained via experimentation
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Alwin

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Re: Chemistry 3/4 2013 Thread
« Reply #850 on: May 20, 2013, 06:30:34 pm »
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Hello, I wanting to ask what sort of organic compounds we must know? i.e. we must know what kind of carboxyl derivatives? i.e. acyls, salt of carboxyls, nitriles, amides? or just esters?
My teacher saying they can put anything on there, but please if anyone has any info on this would be many thanks.
Check the study design. Personally, I just committed everything to memory

Why is there a bold line at the bottom of carbohydrates? And why does the disaccharide composed of two fructose molecules comprise of one fructose that has been flipped?
It's to show the 3D shape of the molecule, the solid black line comes 'out' of the page. So they carbon ring is on a perpendicular plane to the the hydrogen and other functional groups attached to the carbon back bone. eg the glucose molecule has hydrogen and -OH going up and down from the carbon ring, not hydrogens going toward the centre. Hope it makes sense. 3D modelling and angle of bonds is no longer in the course I think.

Hmmm, I've never heard of a double fructose molecule, only know sucrose is glucose and fructose molecule. Someone else may know about your disaccharide composed of two fructose molecules

Does anyone know what actually determines whether a substance is a strong acid or base? I mean, considering oxygen is more electronegative than chlorine, why is HCl a stronger acid than CH3COOH? A considering an OH bond is highly polarised, why are water and alcohols such weak acids?
Unit 4 chem. You'll get there when you get there ;) But if you still want a quick run down of it, I can give it a shot.
Expanding on what Brightsky has said, essentially (said simply) Ka tells you how likely the acid is to ionise. The table of values can be found in your data book, last page: entited 12. Acidity constants, Ka, of some weak acids at 25°C


EDIT: beaten to the last part. Added more detail.
« Last Edit: May 20, 2013, 06:40:23 pm by Alwin »
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Re: Chemistry 3/4 2013 Thread
« Reply #851 on: May 20, 2013, 06:33:04 pm »
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Unit 4 chem. You'll get there when you get there ;) But if you still want a quick run down of it, I can give it a shot

Yeah, that'd be good, if that's alright :P
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Re: Chemistry 3/4 2013 Thread
« Reply #852 on: May 20, 2013, 06:55:01 pm »
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Yeah, that'd be good, if that's alright :P

Okay, so the dissociation (ionisation) of an acid is an equilibrium reaction, ie it can go forwards and backwards.

where A represents the Acid.

For strong acids, such as HCl, the reaction goes forward a lot (for lack of a better word) so almost all of the acid dissociates.
For weak acids, such as CH3COOH, the reaction does not go forward "as much" so only some of the acid dissociates.

How do we measure this? With Ka
Skipping just a bit of equilibrium theory, I'll just jump straight to the equation


Using experimental data at 25°C (this is important), you can work out the concentration of H3O+, A- and HA in the resulting solution. You can see, that for strong acids they have high Ka since there is a very low concentration of HA (un-ionised) acid in the final solution. For weak acids, the majority is left not ionised, so there is a high concentration of HA hence low K

Hope that clears things up a bit!
« Last Edit: May 20, 2013, 06:59:34 pm by Alwin »
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Re: Chemistry 3/4 2013 Thread
« Reply #853 on: May 20, 2013, 07:50:29 pm »
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Hello, I wanting to ask what sort of organic compounds we must know? i.e. we must know what kind of carboxyl derivatives? i.e. acyls, salt of carboxyls, nitriles, amides? or just esters? I'm revising over these, but there's a lot of info on their formation, reduction and oxidation, and I don't think I can remember all the processes and info of all these functional groups? My teacher saying they can put anything on there, but please if anyone has any info on this would be many thanks.

From the study design:
Quote
• structure including molecular, structural and semi-structural formulae, and International Union
of Pure and Applied Chemistry (IUPAC) nomenclature of alkanes, alkenes, amines, haloalkanes,
alkanols (Cn
H2n+1OH), alkanoic acids (Cn
H2n+1COOH) and esters up to C10
• common reactions of organic compounds including equations: addition reactions of alkenes (addition
of hydrogen halides and water limited to symmetrical alkenes), substitution reactions of alkanes
and primary haloalkanes, oxidation of primary alkanols, and esterification
• chemical bonding:
– primary, secondary and tertiary structures of proteins
– the role of the tertiary structure of proteins in enzyme action
– denaturing of proteins: effect of changes in pH and temperature on bonding
– primary and secondary structure of DNA
• organic reaction pathways including appropriate equations and reagents:
– production of esters from alkenes
– condensation reactions that produce lipids (limited to triglycerides)
– condensation and polymerisation reactions that produce large biomolecules including
carbohydrates, proteins and DNA
– production of biochemical fuels including the fermentation of sugars to produce ethanol
– function of organic molecules in the design and synthesis of medicines including the production
of aspirin from salicylic acid.

So alkanes, alkenes, alkanols, carboxylic acids, haloalkanes, amines, amides and esters. Shouldn't need to know anything else I don't think - though benzene and cycloalkanes seem to come up even if they're not mentioned here. Plus carbohydrates, proteins and DNA stuff too I guess.

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Re: Chemistry 3/4 2013 Thread
« Reply #854 on: May 20, 2013, 07:59:09 pm »
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Are we required to be able to balance redox reactions in basic conditions?
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