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March 29, 2024, 11:30:58 pm

Author Topic: VCE Chemistry Question Thread  (Read 2313805 times)  Share 

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Rishi97

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Re: VCE Chemistry Question Thread
« Reply #615 on: April 14, 2014, 07:44:24 pm »
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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?
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nhmn0301

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Re: VCE Chemistry Question Thread
« Reply #616 on: April 14, 2014, 07:46:30 pm »
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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.
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Rishi97

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Re: VCE Chemistry Question Thread
« Reply #617 on: April 14, 2014, 08:40:45 pm »
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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
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Rishi97

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Re: VCE Chemistry Question Thread
« Reply #618 on: April 14, 2014, 09:11:57 pm »
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When an acidic solution is diluted, does the pH increase or decrease? And pls explain why
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #619 on: April 14, 2014, 09:19:45 pm »
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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.
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Rishi97

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Re: VCE Chemistry Question Thread
« Reply #620 on: April 15, 2014, 10:53:41 am »
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For q 4 SA in the 2011 exam 1, how do we know what the equation is?
Thanks
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Yacoubb

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Re: VCE Chemistry Question Thread
« Reply #621 on: April 15, 2014, 11:58:10 am »
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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

Rishi97

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Re: VCE Chemistry Question Thread
« Reply #622 on: April 15, 2014, 12:41:18 pm »
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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?
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Yacoubb

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Re: VCE Chemistry Question Thread
« Reply #623 on: April 15, 2014, 12:48:13 pm »
+1
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).

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Re: VCE Chemistry Question Thread
« Reply #624 on: April 15, 2014, 06:07:56 pm »
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when should i use the double bond equivalent formula?

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Re: VCE Chemistry Question Thread
« Reply #625 on: April 15, 2014, 07:32:40 pm »
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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? 
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Yacoubb

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Re: VCE Chemistry Question Thread
« Reply #626 on: April 15, 2014, 07:50:17 pm »
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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.

katiesaliba

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Re: VCE Chemistry Question Thread
« Reply #627 on: April 15, 2014, 08:01:28 pm »
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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 :)
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thushan

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Re: VCE Chemistry Question Thread
« Reply #628 on: April 15, 2014, 08:02:03 pm »
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Can someone please help me name the following compound?
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.
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thushan

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Re: VCE Chemistry Question Thread
« Reply #629 on: April 15, 2014, 08:04:13 pm »
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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.
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