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April 18, 2024, 09:10:02 pm

Author Topic: VCE Chemistry Question Thread  (Read 2322653 times)  Share 

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TimewaveZero

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Re: VCE Chemistry Question Thread
« Reply #1155 on: July 12, 2014, 01:53:12 pm »
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Why is it undesirable to use a small volume of analyte (<2mL) in volumetric titration analysis?
« Last Edit: July 12, 2014, 01:56:45 pm by TimewaveZero »
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Yacoubb

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Re: VCE Chemistry Question Thread
« Reply #1156 on: July 12, 2014, 03:19:13 pm »
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Why is it undesirable to use a small volume of analyte (<2mL) in volumetric titration analysis?

Because this increases the chance of errors occuring, and thus renders your subsequent calculations regarding concentration, etc inaccurate and discreditable.

TimewaveZero

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Re: VCE Chemistry Question Thread
« Reply #1157 on: July 12, 2014, 04:11:14 pm »
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Because this increases the chance of errors occuring, and thus renders your subsequent calculations regarding concentration, etc inaccurate and discreditable.

Why does it increase the chance of errors occurring?
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TimewaveZero

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Re: VCE Chemistry Question Thread
« Reply #1158 on: July 12, 2014, 04:19:30 pm »
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A 1.862g sample of an organic compound undergoes combustion in excess oxygen to produce 3.321g of CO2 and 1.359g of H2O.

A) Calculate mass of carbon in the sample.
n(CO2) = 3.321 / 44 = 0.07548 mol
=> n(C) = n(CO2) => m(C) = 0.07548 * 12 = 0.9058g

B) Calculate mass of hydrogen in the sample.
n(H2O) = 1.359 / 18 = 0.07550 mol
n(H) = 2n(H2O) = 0.1510 mol => m(H) = 0.1510g

C) Calculate mass of oxygen in the sample
m(O) = m(sample) - m(C) + m(H) = 1.107g

D) Hence show the molecular formula of the compound is C3H6O2

Mol Ratio:     C    :      H    :    O
             0.07548: 0.1510: 0.06919
=              1.09  :    2.18  :      1

That ratio does not make any sense what so ever. What have I done wrong?
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Milkshake

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Re: VCE Chemistry Question Thread
« Reply #1159 on: July 12, 2014, 04:33:03 pm »
+1
A 1.862g sample of an organic compound undergoes combustion in excess oxygen to produce 3.321g of CO2 and 1.359g of H2O.

A) Calculate mass of carbon in the sample.
n(CO2) = 3.321 / 44 = 0.07548 mol
=> n(C) = n(CO2) => m(C) = 0.07548 * 12 = 0.9058g

B) Calculate mass of hydrogen in the sample.
n(H2O) = 1.359 / 18 = 0.07550 mol
n(H) = 2n(H2O) = 0.1510 mol => m(H) = 0.1510g

C) Calculate mass of oxygen in the sample
m(O) = m(sample) - (m(C) + m(H)) = 1.107g

D) Hence show the molecular formula of the compound is C3H6O2

Mol Ratio:     C    :      H    :    O
             0.07548: 0.1510: 0.06919
=              1.09  :    2.18  :      1

That ratio does not make any sense what so ever. What have I done wrong?

I think you forgot a bracket here

Edward21

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Re: VCE Chemistry Question Thread
« Reply #1160 on: July 12, 2014, 06:32:39 pm »
+3
Why is it undesirable to use a small volume of analyte (<2mL) in volumetric titration analysis?
I get the sneaking suspicion this is related to the latest NEAP trial exam..  ;)
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thushan

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Re: VCE Chemistry Question Thread
« Reply #1161 on: July 12, 2014, 10:00:31 pm »
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High percentage error. The burette has a fixed uncertainty (say +- 0.05 mL). If the titre volume were say 0.50 mL, the range would be 0.45 - 0.55 mL. Quite a significant percentage error.
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Yacoubb

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Re: VCE Chemistry Question Thread
« Reply #1162 on: July 13, 2014, 09:21:06 am »
+1
A 1.862g sample of an organic compound undergoes combustion in excess oxygen to produce 3.321g of CO2 and 1.359g of H2O.

A) Calculate mass of carbon in the sample.
n(CO2) = 3.321 / 44 = 0.07548 mol
=> n(C) = n(CO2) => m(C) = 0.07548 * 12 = 0.9058g

B) Calculate mass of hydrogen in the sample.
n(H2O) = 1.359 / 18 = 0.07550 mol
n(H) = 2n(H2O) = 0.1510 mol => m(H) = 0.1510g

C) Calculate mass of oxygen in the sample
m(O) = m(sample) - m(C) + m(H) = 1.107g

D) Hence show the molecular formula of the compound is C3H6O2

Mol Ratio:     C    :      H    :    O
             0.07548: 0.1510: 0.06919
=              1.09  :    2.18  :      1

That ratio does not make any sense what so ever. What have I done wrong?

Okay, so we've established that this organic compound contains: carbon, hydrogen and oxygen. We know the masses of carbon dioxide and water produced.

n(CO2) = n(C) = 0.075477272 mol
m(C) = 0.905727272 g

n(H2O) * 2 = n(H) = 0.151 mol
m(H2O) = 0.151 g

m(O) = 1.862 - (0.151 + 0.9057) = 0.805272728 g
n(O) = 0.050329545 mol

C : H : O
0.075 : 0.151 : 0.05
1.5 : 3 : 1
Multiply by two to obtain simplest whole number ratio
3: 6 : 2

Thus, the empirical formula is C3H6O2, which is propanoic acid (CH3CH2COOH)

Blondie21

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Re: VCE Chemistry Question Thread
« Reply #1163 on: July 13, 2014, 10:44:42 am »
+1
For questions like "what can be deduced from the H NMR spectrum?," how specific do we need to be?
There are the things I am currently writing about:
 - Types of protons (from the data book) that each peak could be
 - Amount of H enviros
 - Amount of neighburing enviros

However, I feel like I am writing too much for a 2 mark Q.

What do you all write?


-----


Thus, the empirical formula is C3H6O2, which is propanoic acid (CH3CH2COOH)

Mmmm be careful.. we don't know this yet :P
Could also be ethyl methanonate.. or methyl ethanoate ..
« Last Edit: July 13, 2014, 10:52:54 am by Blondie21 »
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Rishi97

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Re: VCE Chemistry Question Thread
« Reply #1164 on: July 13, 2014, 11:07:41 am »
+1
For questions like "what can be deduced from the H NMR spectrum?," how specific do we need to be?
There are the things I am currently writing about:
 - Types of protons (from the data book) that each peak could be
 - Amount of H enviros
 - Amount of neighburing enviros

However, I feel like I am writing too much for a 2 mark Q.

What do you all write?

Hi Blondie

I believe that you are on the right track as this is what I would write as well. However, if there was a spectrum given, I would be more specific on the environments that created those peaks. Cause after doing some unit 3 practise exams, I've realised that VCAA like answers that relate to the context of the question not just vague answers. However, if what you've written is all that is given, I would also write something about chemical shift and electronegativity of elements.
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TimewaveZero

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Re: VCE Chemistry Question Thread
« Reply #1165 on: July 13, 2014, 11:30:49 am »
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I get the sneaking suspicion this is related to the latest NEAP trial exam..  ;)

I think your suspicion may be right ;)
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TimewaveZero

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Re: VCE Chemistry Question Thread
« Reply #1166 on: July 13, 2014, 11:32:33 am »
0
High percentage error. The burette has a fixed uncertainty (say +- 0.05 mL). If the titre volume were say 0.50 mL, the range would be 0.45 - 0.55 mL. Quite a significant percentage error.

The previous question was to work out how much analyte you would need for 20.0mL or reagent, thus the high percentage error wouldn't be the issue here no? Sorry, I probably should have put that info into the question  ::)
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Rishi97

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Re: VCE Chemistry Question Thread
« Reply #1167 on: July 13, 2014, 11:44:15 am »
0
A chemist uses fractional distillation to seperate a mixture of water, ethanol, propanol and ethylpropanoate. The first fraction collected will contain mostly:
a) Water
b) Ethanol
c) Propanol
d) ethylpropanoate
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Blondie21

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Re: VCE Chemistry Question Thread
« Reply #1168 on: July 13, 2014, 12:11:40 pm »
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A chemist uses fractional distillation to seperate a mixture of water, ethanol, propanol and ethylpropanoate. The first fraction collected will contain mostly:
a) Water
b) Ethanol
c) Propanol
d) ethylpropanoate

Fractional distillation is not in our study design

--

The number of ions in 3 mole of H3PO4 is closest to:
A) 1.81x1024
B) 7.22 x 1024
C) 3.61 x 1024
D) 1.44 x 1025

?? (Working out please!!!) <3

« Last Edit: July 13, 2014, 12:16:03 pm by Blondie21 »
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Rishi97

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Re: VCE Chemistry Question Thread
« Reply #1169 on: July 13, 2014, 12:12:46 pm »
+1
Fractional distillation is not in our study design

What a relief!!! I was freaking out after getting every fractional distillation question wriong. Phew... thanks :)
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