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March 28, 2024, 10:07:16 pm

Author Topic: VCE Chemistry Question Thread  (Read 2313054 times)  Share 

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Jawnle

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Re: VCE Chemistry Question Thread
« Reply #705 on: April 21, 2014, 01:36:50 pm »
+1
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

Rishi97

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Re: VCE Chemistry Question Thread
« Reply #706 on: April 21, 2014, 01:40:32 pm »
0
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 :)
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Rishi97

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Re: VCE Chemistry Question Thread
« Reply #707 on: April 21, 2014, 01:49:19 pm »
0
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.
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rhinwarr

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Re: VCE Chemistry Question Thread
« Reply #708 on: April 21, 2014, 01:51:51 pm »
0
Yes, it should be:
       Cl
        |
C  -  C  -  C  -  C
        |
       C

Rishi97

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Re: VCE Chemistry Question Thread
« Reply #709 on: April 21, 2014, 02:12:06 pm »
+1
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?
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Blondie21

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Re: VCE Chemistry Question Thread
« Reply #710 on: April 21, 2014, 02:28:16 pm »
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Do we need to know about denaturation of enzymes? It's not in the study design
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Rishi97

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Re: VCE Chemistry Question Thread
« Reply #711 on: April 21, 2014, 02:32:03 pm »
+1
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
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Blondie21

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Re: VCE Chemistry Question Thread
« Reply #712 on: April 21, 2014, 02:33:50 pm »
+1
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #713 on: April 21, 2014, 03:09:43 pm »
+2
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.
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Rishi97

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Re: VCE Chemistry Question Thread
« Reply #714 on: April 21, 2014, 03:57:41 pm »
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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?
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #715 on: April 21, 2014, 04:36:26 pm »
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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.
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Irving4Prez

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Re: VCE Chemistry Question Thread
« Reply #716 on: April 21, 2014, 04:37:24 pm »
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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

Yacoubb

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Re: VCE Chemistry Question Thread
« Reply #717 on: April 21, 2014, 05:31:03 pm »
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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.

Irving4Prez

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Re: VCE Chemistry Question Thread
« Reply #718 on: April 21, 2014, 05:38:58 pm »
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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?

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Re: VCE Chemistry Question Thread
« Reply #719 on: April 21, 2014, 06:46:40 pm »
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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.
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