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December 06, 2021, 12:53:06 pm

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wingdings2791

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« Reply #9195 on: September 22, 2021, 11:00:16 am »
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"A student conducts an experiment to determine the ethanoic acid concentration in a commercial
brand of vinegar. An outline of her procedure and her measurements are provided below.
20.00 mL aliquots added to flasks and titrated against 0.150 M NaOH."

The solutions said that an error in this experimental design was that "20 mL added to 230 mL of water will not necessarily give 250 mL. Some liquids are miscible in each other. All concentration calculations will be subsequently affected," and suggested that "It should be made up to the mark and not have 20 mL added to 230 mL."

Would miscibility not result in an error regardless if we're filling the volumetric flask up to the mark or adding 20 to 230? Cheers

Hi saransh,
I think the main error described here is that the vinegar to the volumetric flask before the deionised water.

The analyte (vinegar) should always be added first to ensure that the correct concentration is achieved. When deionised water is added directly to the vinegar already in the flask, up to the $250 mL$ mark, you know that the volume totals as close to $250 mL$ as possible, and that the $n(vinegar)$ is as close to $20 mL$ as possible.

If the deionised water is added to the volumetric flask first, then the separately measured $20 mL$ of vinegar is added to the deionised water, you risk decreasing accuracy. As the uncertainty of the volumetric flask and the pipette used to measure the vinegar might result in inaccurate measurements, separately measuring the deionised water and vinegar and then adding them together combines the potential for error from two pieces of equipment. That is: when vinegar is added first, although inaccuracy can still come from the uncertainty of the $n(vinegar)$ measurement, the total $v$ wouldn't be affected as much.
If the deionised water is added first, both the vinegar and water would be more likely to be measured inaccurately, compounding the error. I'll list the possible errors from measurement in each case here:

Water after vinegar
High vinegar: less water added to achieve $v=250 mL$, inaccurately high $[CH_3COOH]$
Low vinegar: more water added to acheive $v=250 mL$, inaccurately low $[CH_3COOH]$

Vinegar after water
Low water/low vinegar: $v<250 mL$, inaccurately low $[CH_3COOH]$ calculated as $v=250 mL$ will be used
Low water/high vinegar: $v(solution)\approx\ 250 mL$, aliquots have higher $[CH_3COOH]$, high $[CH_3COOH]$ calculated
High water/low vinegar: $v(solution)\approx\ 250 mL$, aliquots have lower $[CH_3COOH]$, low $[CH_3COOH]$ calculated
High water/high vinegar: $v>250 mL$, high $[CH_3COOH]$ calculated as $v=250 mL$ will be used
High water/accurate vinegar: low $[CH_3COOH]$ from $v>250 mL$ and aliquot $[CH_3COOH]<$ actual
Low water/accurate vinegar: high $[CH_3COOH]$ from $v<250 mL$ and aliquot $[CH_3COOH]>$ actual
Accurate water/high vinegar: high $[CH_3COOH]$ from $v>250 mL$ and aliquot $[CH_3COOH]>$ actual
Accurate water/low vinegar: low $[CH_3COOH]$ from $v<250mL$ and aliquot $[CH_3COOH]<$ actual

As you can see, measuring the deionised water and vinegar separately, then adding the water first creates many more potential errors than adding the vinegar first and topping up the volume with water accordingly. This makes vinegar first, water second a more accurate approach.

Other errors
I don't see an error that would result from the miscibility of liquids. If anything, I would've thought that the analyte and titrant being immiscible would pose a problem, as this suggests that they would not react with each other and sit in the conical flask as a heterogenous solution.

Some other errors you could discuss, in case you need more ideas:
- No mention of equipment rinsing or swirling the solution to dissolve particles
- Indicator is not added to the analyte prior to titration
- The phrasing used is 'aliquots added to flasks and titrated against $0.150M NaOH$', which falsely implies that this is a back titration
- No mention of the titration process beyond 'titrate vinegar': should specify that three concordant titres must be achieved/results must be recorded (precision)

Anyways, that's what I think is going on in this question; it does seem a little strange. Hope I could help, let me know if there are any errors (see what I did there)
« Last Edit: September 22, 2021, 02:29:42 pm by wingdings2791 »
2019- Chinese SL [42]
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miyukiaura

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« Reply #9196 on: September 22, 2021, 02:19:01 pm »
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Hey guys,

I just had a question about determining the calibration factor by electrical calibration.
For a temperature-time graph (attached), how do you know if the calorimeter is well-insulated or poorly-insulated, and whether to extrapolate back to when the current was turned on, or to simply read off the graph to determine the change in temperature? I know that for poorly-insulated calorimeters, the graph should decrease and be less linear, however, if it isn't as obvious, how would you know which method to use?

Thanks so much.
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wingdings2791

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« Reply #9197 on: September 23, 2021, 01:34:05 pm »
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Hey guys,

I just had a question about determining the calibration factor by electrical calibration.
For a temperature-time graph (attached), how do you know if the calorimeter is well-insulated or poorly-insulated, and whether to extrapolate back to when the current was turned on, or to simply read off the graph to determine the change in temperature? I know that for poorly-insulated calorimeters, the graph should decrease and be less linear, however, if it isn't as obvious, how would you know which method to use?

Thanks so much.

Hello miyukiaura,
Determining the quality of insulation for a calorimeter largely depends on what happens to $T$ after the current is removed, as well as the ratio of $ΔT$(experimental) to $ΔT$(theoretical). For telling whether a calorimeter is well-insulated or poorly-insulated, I don't think you could within the scope of VCE, as it doesn't appear anywhere on the study design and there are no indications I can find for what is considered high/low (like how $K>M^4$ is a large equilibrium constant). Any determination of a calorimeter's insulation quality would probably have to be comparative.

Of course the better insulated a calorimeter system is, the less thermal energy will be lost to the environment, so just to reiterate:

Well-insulated:
- Less decline in $T$ after current is turned off (some is inevitable since no calorimeter is 100% insulated)
- Higher initial $ΔT$ (at instant current is removed), since less heat is lost during the reaction
- Lower CF (specific heat capacity of calorimeter is lower, less energy required to achieve same $ΔT$)

Poorly insulated:
- More decline in $T$ after current is removed (continues to lose thermal energy to environment)
- Lower initial $ΔT$, since more heat is lost to the environment during reaction
- Higher CF (more energy required to achieve same $ΔT$ as thermal waste is greater)

In general, I think that extrapolation is a safer approach (unless you're dealing with an ideal calorimeter). Some heat loss is always inevitable, so going with a line of best fit will probably produce the most reliable CF; if the calorimeter is close to ideal, the extrapolation will closely match anyway.

Hope this helps
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wingdings2791

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« Reply #9198 on: September 24, 2021, 10:57:25 pm »
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Hi everyone,
I'm doing my U4 AOS3 (prac) SAC on fuels, as my school has not assessed most of U3 AOS1. The prac focuses on producing a biodiesel from vegetable oil, then determining and comparing the density and energy content on each fuel.

Because of lockdown, we weren't able to complete the experiment ourselves, so the teachers provided the following sample data (to analyse as practice):

Spoiler
For the energy content determination (combusting fuels with a spirit burner to heat water, recording $\Delta T$ of water):
- $v(H_2O)=50\ mL$
- $v(fuel)=10\ mL$
- $\Delta m(fuel)=3.66g$ (for both fuels)
- $\Delta T=+28.5^\circ C, +31.0^\circ C$ (oil and biodiesel respectively)

Using this data, I determined the energy content of the oil and biodiesel each as follows:
$E=\frac{m(H_2O)\times 4.18Jg^{-1}K^{-1}\times \Delta T}{m(fuel)}$
$\therefore E(oil)=\frac{50mL\times 0.997gmL^{-1}\times 4.18Jg^{-1}K^{-1}\times 28.5K}{3.66g}\approx 1.62kJg^{-1}$
$E(biodiesel)=\frac{50mL\times 0.997gmL^{-1}\times 4.18Jg^{-1}K^{-1}\times 31.0K}{3.66g}\approx 1.76kJg^{-1}$
These values seem very inaccurate, given that fats/oils typically have a heat of combustion $\sim 37kJg^{-1}$. Am I forgetting something and doing a step wrong, or is this data just unrealistic? I've checked over the units, given data, and calculations a hundred times but still can't figure out what's wrong. Under the results column, the rubric does specify that 'any outliers should be identified (if appropriate)', but I'm hesitant to declare this a huge systematic error caused by a terribly calibrated thermometer or similar.

Any help would be greatly appreciated.
2019- Chinese SL [42]
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Mackenzie Aps

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« Reply #9199 on: September 28, 2021, 09:06:01 am »
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Does anyone have the NEAPS Chemistry 3/4 Trial exam and solutions

Mackenzie Aps

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« Reply #9200 on: September 28, 2021, 09:06:42 am »
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Does anyone have the NEAPS Chemistry 3/4 Trial exam and solutions for 2021

lm21074

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« Reply #9201 on: September 28, 2021, 12:51:11 pm »
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Does anyone have the NEAPS Chemistry 3/4 Trial exam and solutions for 2021
Hi Mackenzie,

It is against the forum rules to distribute copyrighted materials, including the NEAP trial exams and solutions.

Mackenzie Aps

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« Reply #9202 on: September 28, 2021, 03:26:54 pm »
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Apologies for that, they don't have this one. Not to worry, will use others.

Mackenzie Aps

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« Reply #9203 on: September 28, 2021, 05:51:07 pm »
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Hi Mackenzie,

It is against the forum rules to distribute copyrighted materials, including the NEAP trial exams and solutions.

Just wondering how other exam material requests and questions are posted but mine was not appropriate or against forum rules?

valjaybj

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« Reply #9204 on: October 04, 2021, 10:39:57 pm »
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hi can someone help me with U4 AOS3 (fuels prac) - my sac is in 2 days and I am extremely confused (as I haven't been formerly assessed on this part of the course and so I haven't bothered to study for it either) - what do I talk about in my discussion? The prac is about making biodiesel from a vegetable oil, determining the density and energy content of the biodiesel and comparing it to the original oil/ethanol. What limitations can I talk about? And how do I calculate the density and energy content? Oh and what would the DV and IV be? (I'm utterly lost so) any help would be GREATLY APPRECIATED!!!!!!!!!!!!!!

squid_2021

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« Reply #9205 on: October 06, 2021, 12:17:24 am »
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Just wondering, is this thread only for VCE chem 3&4? Looking for the thread (if there is one) for 1&2....

ArtyDreams

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« Reply #9206 on: October 06, 2021, 11:09:48 am »
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Just wondering, is this thread only for VCE chem 3&4? Looking for the thread (if there is one) for 1&2....

Feel free to post your 1/2 questions here as well! This thread is for both

Corey King

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« Reply #9207 on: October 12, 2021, 12:25:19 pm »
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Hey guys,
I'm struggle to see where I went wrong for a past question. I sent it to my teacher, but I think she gave up explaining it to me . Any help appreciated

https://gyazo.com/2cff1b72ae64054c76159fbb51cb3723
https://gyazo.com/b3c2ca680c5225f756c075d05e8ae6fe  (teacher response)

Thanks,
Corey

wingdings2791

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« Reply #9208 on: October 12, 2021, 04:26:23 pm »
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Hey guys,
I'm struggle to see where I went wrong for a past question. I sent it to my teacher, but I think she gave up explaining it to me . Any help appreciated

https://gyazo.com/2cff1b72ae64054c76159fbb51cb3723
https://gyazo.com/b3c2ca680c5225f756c075d05e8ae6fe  (teacher response)

Thanks,
Corey

Hey Corey,
I think the reason behind this incongruence of data might be this sentence in the data book:
'Typical 13C shift values relative to TMS = 0
These can differ slightly in different solvents'

Hence, I suspect that the disconnection between the C NMR and the lack of $R-CH_2-R$ is because of different range standards (and we should discount this by thinking of the potential for variation in the range). Here's the C NMR spectrum for methylpropan-1-ol, which indicates a similar chemical shift (~30 ppm) as the question indicates for a carbon bonded to two methyl groups and one hydrogen atom (which would be exactly the same environment)

Methylpropan-1-ol

ppm   Int.  Assign.

69.47   691      1
30.84   634      2
18.99   1000     3
Since this NMR spectra and the ones on VCAA exams seem to be from the same source, perhaps the standard range for $R_3-CH$ simply differs between VCAA and the labs that make these. I think VCAA possibly just wants students to account for slight variation in chemical shift ranges, focusing more on determining structure from the molecular formula and number of environments. Sorry for not being able to answer this better, but I think that may be what's going on
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