VCE Chemistry Question Thread

[hums_student]

4) i was sure i was able to distinguish between the following types of bonding, but after doing a couple questions and getting my answers wrong, im somewhat confused between the differences between the following types of bonding, and how we know this type of bonding exists?( could anyone please help me with this, ive tried searching google, but im even more confused now)
-Dispersion
-Hydrogen
-Dipole-dipole

Dispersion: it's the weakest type of bonding and can only exist between non-polar molecules. Dispersion forces generally exist in all molecules.
Dipole-Dipole: Stronger than dispersion forces but weaker than hydrogen bonding. It exists between polar molecules.
Hydrogen: it's the strongest type of bonding, and exists when hydrogen bonds with Fluorine, Oxygen, or Nitrogen. Hydrogen bonding makes molecules soluble in water.

5) in part (c)  why does the marking scheme say 'partially' oppose the change. do we have to say partially? will this be the case for all other changes such as temp, volume etc? do we have to make a refernce to le chatliers principle?

Yes - you have to say partially. According to le Chatelier's principle it's always a partial opposition. And yes it applies to changes in temperature, volume, pressure, concentration etc.

6) also, when it says proton does it mean H+

Yes.

I'm not too sure on the other questions but maybe someone else can help out.

[Bri MT]

I wouldn't consider questions 1, 2 or 3  to be in this study design btw

[smamsmo22]

Hi, apologies if this question has come up before but I was struggling with understanding question 3 part ii from the 2018 Northern Hemisphere VCAA exam.

Question 3 (8 marks)
Nitrosyl chloride, NOCl, is a highly toxic gas used in the chemical industry as an oxidising agent.
The formation reaction of NOCl from nitrogen monoxide, NO, and chlorine, Cl2, is
2NO(g) + Cl2(g) ⇌ 2NOCl(g) H > 0 This reaction forms an equilibrium above 100 °C.
A scientist conducted two experiments on the equilibrium reaction of NOCl. The initial experiments were conducted in evacuated and sealed 4 L containers at 150 °C.
a. i.
Experiment 1 2 mol of NOCl was injected into a previously evacuated, sealed 4 L container.
Experiment 2 4 mol of NOCl was injected into another previously evacuated, sealed 4 L container.

ii.
If, for Experiment 1, the concentrations of NOCl and NO were equal at equilibrium, [NOCl] = [NO], then what conclusion could be made about the relative concentrations of NOCl and NO in Experiment 2 at equilibrium? Justify your answer.
2 marks

That's the question, I've read the answer guide obviously but if anyone could offer a clearer explanation that would be really helpful.


[minhalgill]

1) Why cant 10 be C?
2)for question 17, i dont understand how the dip around 3000 can be considered a peak, yet the one around 1550 isnt considered a peak?

[Lear]

Why cant 10 be C?

Side groups do not change in terms of ionisation in the zwitterion form. This only happens at high pH or low pH.

[minhalgill]

Side groups do not change in terms of ionisation in the zwitterion form. This only happens at high pH or low pH.

do the side groups never change at all or can they change to produce an overall neutral charge?

also, in the question 30 ive attatched, i understand how to solve the question using faradays constant etc, and finding the moles of electrons, but theyve done it another method (from where it says alternativeley) but i dont understand what theyve done?

also, in question 1 b ii, can we only say addition reaction or can we also say hydration reaction, as the marking scheme only has addition reaction?

[hums_student]

also, in the question 30 ive attatched, i understand how to solve the question using faradays constant etc, and finding the moles of electrons, but theyve done it another method (from where it says alternativeley) but i dont understand what theyve done?

The mol of elections stays constant for all four cases as all are set up under the same conditions: same time and same current (hence Q value will be the same, hence after dividing by Faraday's constant you end up with the same n(e-) value). As you know this stays constant you can jump straight to the step when the values differ between the options, which is the charges. You don't necessarily have to find the actual mass, dividing the molar mass by the charge will give you the ratio for the mass, which was what they did in the solutions.

also, in question 1 b ii, can we only say addition reaction or can we also say hydration reaction, as the marking scheme only has addition reaction?

As it's only one mark I would assume that saying addition is enough.

do the side groups never change at all or can they change to produce an overall neutral charge?

Not exactly sure what you're trying to say here. The side groups can change, as Lear said above, when it is in an extremely acidic or basic condition. The overall neutral charge thingo only happens when it's in a physiological pH, and in those cases side groups do not change.

[daniel.h]

In VCAA 2015 Q9ei) short answer, shouldn't the molar enthalpy of combustion be negative?

Also for VCAA 2015 9eii) short answer, how are we supposed to know that methyl palmitate is liquid not aqueous?

[sweetcheeks]

In VCAA 2015 Q9ei) short answer, shouldn't the molar enthalpy of combustion be negative?

Also for VCAA 2015 9eii) short answer, how are we supposed to know that methyl palmitate is liquid not aqueous?

Yes, the value should be negative.

It tells you higher up in the question that the non-aqueous layer contains the methyl esters, therefore the methyl palmitate is insoluble in water and must be a liquid (also, very rarely, if ever, would you use a bomb calorimeter to determine combustion of a dissolved substance).

[Azim.m]

Shouldn’t the value be positive? Because heat of combustion (molar enthalpy of combustion) would always give the amount of energy released. The heat of combustion is also positive in the data booklet.

[Lear]

Shouldn’t the value be positive? Because heat of combustion (molar enthalpy of combustion) would always give the amount of energy released. The heat of combustion is also positive in the data booklet.

Enthalpy is energy of products - energy of reactants. As it is an exothermic reactants the energy of products will be lower than energy of reactants and hence the enthalpy will be negative. Heat of combustion =/= Enthalpy.

[Lear]

Shouldn’t the value be positive? Because heat of combustion (molar enthalpy of combustion) would always give the amount of energy released. The heat of combustion is also positive in the data booklet.

Enthalpy is energy of products - energy of reactants. As it is an exothermic reactants the energy of products will be lower than energy of reactants and hence the enthalpy will be negative. Heat of combustion =/= Enthalpy.

[hums_student]

Hey guys, the screenshot below is from the examiners report for the 2018 VCAA NHT exam. Just wondering can this happen in the November exams? I've never seen this before, especially not in an official VCAA exam.


And it doesn't specify in the examiner's report, but if this was to happen would both options need to be selected in order to get the mark?

[TheAspiringDoc]

Hey guys, the screenshot below is from the examiners report for the 2018 VCAA NHT exam. Just wondering can this happen in the November exams? I've never seen this before, especially not in an official VCAA exam.
(Image removed from quote.)
It happened in a November PE exam. I think just selecting A would be totally fine
And it doesn't specify in the examiner's report, but if this was to happen would both options need to be selected in order to get the mark?

[Vaike]

Hey guys, the screenshot below is from the examiners report for the 2018 VCAA NHT exam. Just wondering can this happen in the November exams? I've never seen this before, especially not in an official VCAA exam.
*snip*
And it doesn't specify in the examiner's report, but if this was to happen would both options need to be selected in order to get the mark?

This only happens when the exam writers make a mistake; they originally intend for only one option to be correct but they haven't been careful enough or whatever and another option can be considered correct too. It's actually happened a couple times in VCAA chemistry exams (off memory, I'd say 2 or 3 times since 2002?). You'd get the mark for putting either down, so if you shaded A or C you'd get the mark. I'd strongly recommend against ever shading more than one bubble.