Chemistry: Discussion, Questions & Potential Solutions

[OldAnt]

No if it was percentage by mass of celloluse it would by 37% which is already given in the question

From memory, 144L at a density of 0.87 = 128 kg or so;
Source cellulose was 37% of 1000kg = 370 kg;
Percentage by mass = 128/370 = 34 or so percent..
...........
Re the systematic error question, I thought that the transformation of electrical --> heat energy might not be 100% efficient, which would be a systematic error in that calibration system.

[vector_graphics]

im pretty sure it was asking for the limitations of the chemical experiment conducted by student C. In that case would it not be the incomplete reaction of knh3 due to equilibrium?

i thought it was asking for the limitations of the chemical calibration?

[vector_graphics]

From memory, 144L at a density of 0.87 = 128 kg or so;
Source cellulose was 37% of 1000kg = 370 kg;
Percentage by mass = 128/370 = 34 or so percent..
...........
Re the systematic error question, I thought that the transformation of electrical --> heat energy might not be 100% efficient, which would be a systematic error in that calibration system.

density was 0.79gml

[DinWell]

From memory, 144L at a density of 0.87 = 128 kg or so;
Source cellulose was 37% of 1000kg = 370 kg;
Percentage by mass = 128/370 = 34 or so percent..
...........
Re the systematic error question, I thought that the transformation of electrical --> heat energy might not be 100% efficient, which would be a systematic error in that calibration system.

Electric heaters are basically 100% efficient. Or are at least close to the point where we can consider them to be so.

[matthewmwps]

im pretty sure it was asking for the limitations of the chemical experiment conducted by student C. In that case would it not be the incomplete reaction of knh3 due to equilibrium?

Yeah i kinda goofed on this one (or i think)

I wrote that the students didn't allow the calorimetry to cool before using the KNO3.
My reasoning was that in the method it said "after electrical configuration....". and at the last step of the electrical method it stated "allow to cool for 3 minutes" which wasn't necessarily enough time to cool back to room temperature.

My limitation for the electrical one was something like "students only measured final voltage and not initial voltage, which may have been inconsistent"

[matthewmwps]

Honestly, i'm just glad there was no H-NMR on this exam.

The practice one i did before this had a peak at like 2.5 or something close to it which was meant to be R-CH2-R despite the data book saying the range was close to 1.3-1.4. (the book does say "these aren't always the correct values" but still a 1.0 difference is too much to expect)

i was also surprised by the lack of thermo-chemical equations and Delta H stuff

[tiffanylps09]

Does anyone remember what they got for the IR and CNMR? Also what did the final molecule look like?

[petersenge]

For the systematic error for the electrical calibration, I said that the voltmeter may not give the exact voltage due to the possibility of it being faulty. Do you guys think that's a valid point?

[jondempsey17]

Does anyone remember what they got for the IR and CNMR? Also what did the final molecule look like?

I got 2 methyl propan-2-amine.
It looks like:
carbon centre, 3 methyls branching off with another amine branching off to make a

[meymanic]

Which one has the lower GI?

Idk if this is right but

I said smaller ones because the surface area of the smaller, thinner coconut shreds are greater, they can be packed together tightly and forms stronger forces in the body therefore, they are hydrolysed at a slower rate due to greater surface area hence lower GI.

The large coconut shreds have lower surface area and can’t be packed together tightly therefore forming weaker bonds and hence hydrolysed at a more rapid rate releasing more glucose in the blood levels hence higher GI for the large molecules


is this correct?! Maybe VCAA isn’t looking for a right answer but a justification of an answer.

[laws12]

I thought it was 2-methyl propan-1-amine as there were 3 carbon environments present, 2-methylpropan-2-amine only has 2. Looks like this

Does anyone remember what they got for the IR and CNMR? Also what did the final molecule look like?

I got 2 methyl propan-2-amine.
It looks like:
carbon centre, 3 methyls branching off with another amine branching off to make a

[G-Fr3sh]

From memory, 144L at a density of 0.87 = 128 kg or so;
Source cellulose was 37% of 1000kg = 370 kg;
Percentage by mass = 128/370 = 34 or so percent..
...........
Re the systematic error question, I thought that the transformation of electrical --> heat energy might not be 100% efficient, which would be a systematic error in that calibration system.

Density was grams per mL

[Lear]

-cut-
is this correct?! Maybe VCAA isn’t looking for a right answer but a justification of an answer.

I don’t think so. A greater surface area increases the frequency of collisions and by extension the rate of traction. This is one of the reasons we chew foods before swallowing.

This was quite a good question considering textbook’s predominantly focus on the levels of amylopectin and amylose on GI, rather than the actual physical properties of the food. To answer this question students actually needed to bring together content learnt from unit 3 about rates of reaction and surface area with content from unit 4 on GI.

For next year students reading this, make sure you try and build connections between information you learnt throughout the year and not consider them completely separate.

[not.yet.a.nerd]

I thought it was 2-methyl propan-1-amine as there were 3 carbon environments present, 2-methylpropan-2-amine only has 2. Looks like this

Yeah, I got the same thing. Because of the symmetrical end, that reduced the number of carbon environments to match the graph

[G-Fr3sh]

Yeah, so dividing by 370kg should yield ~30%?

I didn't do that