VCE Chemistry Question Thread

[K888]

The 2014 VCAA exam allowed up to 20 marks to be dropped. There are a lot of difficult/unseen concepts in this exam (Parts Per Million) and logical thinking questions.

I have found the NEAP to be the most difficult, I have dropped the most marks on them compared to other exams. However, I have noticed that there are some thing unrelated to the study design.

+1 for NEAP. Always find their stuff difficult.
And that 2014 exam...just did it as a practice...will agree that it wasn't easy by any stretch (for me at least)!

[Adequace]

Yep, your working is fine. Just out of curiosity, have you heard/used the KOHES method before?

Thanks, and no I haven't.

[HasibA]

can anyone guide me through this question? i seem to have forgot how to do it- thanks!

[sweetcheeks]

can anyone guide me through this question? i seem to have forgot how to do it- thanks!

The question wants to know the delta h of 1 mole of Calcium Chloride dissociating into water.
What we have is; the mass of the water (60.0g) (m), the change in temperature (+32.5C) (t) and the specific heat capacity (4.18J/C/G) (c)

The equation will be; c x m x t = joules
4.18 x 60.0g x 32.5C = 8151 joules of energy
Convert this into kilojoules 8.15kJ

This is how much energy is released when 11.1 grams of calcium chloride is added to water. We want to find how much energy one mole will produce. This can be achieved through simple ratios. First step is to find the mole of calcium chloride
n=m/M 11.1g/111gmol = 0.100 mole.

if 8.15kJ is released from 0.100 mole; 8.15kJ/0.100mole = -81.5kJ, which will be C.

[HasibA]

The question wants to know the delta h of 1 mole of Calcium Chloride dissociating into water.
What we have is; the mass of the water (60.0g) (m), the change in temperature (+32.5C) (t) and the specific heat capacity (4.18J/C/G) (c)

The equation will be; c x m x t = joules
4.18 x 60.0g x 32.5C = 8151 joules of energy
Convert this into kilojoules 8.15kJ

This is how much energy is released when 11.1 grams of calcium chloride is added to water. We want to find how much energy one mole will produce. This can be achieved through simple ratios. First step is to find the mole of calcium chloride
n=m/M 11.1g/111gmol = 0.100 mole.

if 8.15kJ is released from 0.100 mole; 8.15kJ/0.100mole = -81.5kJ, which will be C.

beautiful-thanks heaps!
btw- your 81.5 was negative because it was exothermic?

EDIT: can anyone explain the attachment part 2? thanks!

[sweetcheeks]

beautiful-thanks heaps!
btw- your 81.5 was negative because it was exothermic?

EDIT: can anyone explain the attachment part 2? thanks!

Yes, it is negative due to it being exothermic.

In relation to part two, lets establish what is happening. It appears that a solution of MnO4- with an unknown concentration is inside the burette. 25.00mL of a (C2O4)2- solution with a known concentration is added to an aliquot. The aliquot is used to standardise the MnO4-, the amount (moles) of C2O4 is being used as the method of determining the concentration.

The question asks the effect of rinsing with water of different glassware.
A) If water is used to rinse the 25.00mL pipette that is being used to deliver the C2O4-, some residual water will remain inside the pipette. When the sample is taken from the solution of C2O4-, it will be slightly diluted (less moles) than the concentration accounted for. The end point of the reaction will be achieved earlier than it should be (due to the MnO4- reacting with a lower amount of mole than expected), resulting in the calculation of a higher concentration.

B) When water is used to rinse the burette, residual amounts will remain. When the MnO4- solution is added to the burette it will be diluted, so a higher volume will be needed to be delivered than what it actually should be. This means that the solution will be calculated at a lower concentration than what it actually is.

C) rinsing the aliquot beaker with water will have no affect. The same amount of moles will be in the solution (as it is delivered by the 25.00mL pipette), regardless of how much water there is in there.

[Apink!]

Hello chemists,

I have a question that I would really appreciate if someone could answer me!

Methyl red is an acid/base indicator but it is also a weak acid.

A student has a 20.0 mL sample of NaOH solution of unknown concentration in a flask. She adds a few drops of methyl red solution of 0.0100 M concentration. The contents of the flask are yellow in colour. She continues to add methyl red until, after the addition of 15.0 mL, the flask and its contents turn red in colour.

The concentration of the NaOH

A. cannot be determined without the addition of another indicator
B. cannot be determined because methyl red is a weak acid
C. is 0.0075 M
D. 0.0133 M

I don't get this

[HasibA]

thanks sweetcheeks, solid explanation!
any tips for similar questions? i find i get stuck real quick on experimental errors etc.

also:
what assumed knowledge from yr 11 appears on the chem exams? i know solubility, common ions, common states of products/reactants, but what else that's not in the 3/4 course can appear on the exam, that's not explicitly taught in chem 3/4
ty guys!

[sweetcheeks]

thanks sweetcheeks, solid explanation!
any tips for similar questions? i find i get stuck real quick on experimental errors etc.

also:
what assumed knowledge from yr 11 appears on the chem exams? i know solubility, common ions, common states of products/reactants, but what else that's not in the 3/4 course can appear on the exam, that's not explicitly taught in chem 3/4
ty guys!

It can help to draw the setup when trying to identify experimental errors, it allows you to visualise what is going on. Practicing the questions is probably the best way to understand them better, as you can find common trends and issues that arise.

As for assumed knowledge;
Questions about polarity and intermolecular forces are often requried to explain chromatography questions. The Ideal Gas Equation (PV=nRT) is required, usually in the form of a multiple choice question.

[Sine]

How important are states of organic reactions, if you write g or l will you be fine?

[RuiAce]

How important are states of organic reactions, if you write g or l will you be fine?

They actually make states unimportant for certain cases in the VCE? Because I always included states.

(Note: For things such as combustion, use whatever would make sense. I can address examples.)

[HasibA]

How important are states of organic reactions, if you write g or l will you be fine?

i got a mark taken off for writing H2SO4 as gaseous, but thats moreso a catalyst.
that being said- on an exam, i got a mark taken off for writing Benzene as a solid :/ my chem teachers on hols so cant check if its right or not but yeah !

[RuiAce]

i got a mark taken off for writing H2SO4 as gaseous, but thats moreso a catalyst.
that being said- on an exam, i got a mark taken off for writing Benzene as a solid :/ my chem teachers on hols so cant check if its right or not but yeah !

Sulfuric acid catalyst is aqueous if you ever have to mention a state for it.

I didn't know that you could isolate benzene by itself but when I did a Google search just now I found out that it is a liquid.

[Sine]

C17H36(g /l) → C15H32(g/l) + C2H4(g)

from the 2007 VCAA Unit 3 exam - do you need to be consistent with states?

500th Post

[HasibA]

C17H36(g /l) → C15H32(g/l) + C2H4(g)

from the 2007 VCAA Unit 3 exam - do you need to be consistent with states?

500th Post

i reckon keep it all as gaseous!
and ahh i was told the catalyst for H2SO4 for esterification is concentrate, and had to be liquid :/ idk- will check it up once im back to school xD