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Author Topic: VCE Chemistry Question Thread  (Read 2313297 times)  Share 

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brightsky

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Re: VCE Chemistry Question Thread
« Reply #45 on: January 05, 2014, 12:55:39 pm »
+2
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.
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DJA

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Re: VCE Chemistry Question Thread
« Reply #46 on: January 05, 2014, 12:58:43 pm »
+1
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
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Professor_Oak

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Re: VCE Chemistry Question Thread
« Reply #47 on: January 05, 2014, 01:47:42 pm »
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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.
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DJA

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Re: VCE Chemistry Question Thread
« Reply #48 on: January 05, 2014, 04:12:09 pm »
0
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.
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Only Cheating Yourself

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Re: VCE Chemistry Question Thread
« Reply #49 on: January 05, 2014, 05:05:07 pm »
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How many hours a night do you guys plan on studying?
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psyxwar

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Re: VCE Chemistry Question Thread
« Reply #50 on: January 06, 2014, 01:45:47 pm »
+1
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.
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brightsky

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Re: VCE Chemistry Question Thread
« Reply #51 on: January 06, 2014, 01:53:48 pm »
+2
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.).
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psyxwar

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Re: VCE Chemistry Question Thread
« Reply #52 on: January 06, 2014, 01:56:23 pm »
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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
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DJA

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Re: VCE Chemistry Question Thread
« Reply #53 on: January 06, 2014, 02:11:52 pm »
0
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.  ::)
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #54 on: January 06, 2014, 03:18:30 pm »
+9
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.
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PsychoT

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Re: VCE Chemistry Question Thread
« Reply #55 on: January 07, 2014, 12:56:35 pm »
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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.
« Last Edit: January 07, 2014, 01:20:50 pm by tozer24 »
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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #56 on: January 07, 2014, 01:14:31 pm »
+2
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.
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PsychoT

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Re: VCE Chemistry Question Thread
« Reply #57 on: January 07, 2014, 01:18:53 pm »
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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.

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lzxnl

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Re: VCE Chemistry Question Thread
« Reply #58 on: January 07, 2014, 01:35:51 pm »
+2
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.
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Snorlax

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Re: VCE Chemistry Question Thread
« Reply #59 on: January 07, 2014, 01:59:31 pm »
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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
« Last Edit: January 07, 2014, 02:02:02 pm by Snorlax »
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