VCE Chemistry Question Thread

[vanilla and sunflower]

How did you know the Mg was in excess, the volume of the acid is not given...

haha..lol sorry about that

[paper-back]

When we are finding concordant results do we use 3 values that are within the range of 0.1ml(from max to min), or within the range of +/-0.1ml of the middle value?

When diluting a solution, why don't both the concentration of H3O+ and OH- decrease?(Answer - Q8 VCAA 2010 Exam 2 Examiners report)

If I had a 0.01M solution of HCl at 10ml, the H3O+ concentration is: 0.01, and [OH-] is 10^-12

Then if I were to dilute this solution to 1L
For the H3O soltution: (0.01 x 0.01)=10^-4
Therefore pH = 4

However when I do this dilution to the [OH] concentration:
(10^-12 x 0.01) = 10^-14
So the pH gained by doing this = 0?

Why does doing the dilution to the [H3O] and [OH] concentration yield different pH results?

[RazzMeTazz]

Are we required to know which range of wavelengths correspond to which part of the electromagnetic spectrum ? ( for e.g which wavelengths comprise the UV  region ) etc ? Thanks

[RazzMeTazz]

Are we required to know which range of wavelengths correspond to which part of the electromagnetic spectrum ? ( for e.g which wavelengths comprise the UV  region ) etc ? Thanks

[bonjour-sarah]

Are we required to know which range of wavelengths correspond to which part of the electromagnetic spectrum ? ( for e.g which wavelengths comprise the UV  region ) etc ? Thanks

No (according to study design) , but having a basic idea of which are higher/ lower in spectrum can be useful. i.e. Uv has higher frequency thus energy, and if we were to accidentally use it instead of IR, bonds would mostly likely break (at a very basic level of understanding) because it is too high for our purpose. I only say this because I have seen one question with this logic before

[Acid]

The production of chemicals (Sulfuric acid/ ammonia etc) is no longer examinable, right?

[coconut stripes]

How do you get the anode equation when given the full equation, for a galvanic cell?
E.g. Q10 VCAA Chemistry exam 2014:
the following reaction takes place as the cell discharges:
2Zn(s) + O2 (g)+2H2O (l) -> 2Zn(OH)2 (s).

The anode is Zn, Cathode is C.

Write a balanced half equation for the anode reaction?

[sparkyblossom]

Are we expected to know and write the state of certain substances at 25ºC regardless of whether they are in the data book or not? For example, that benzoic acid is a solid?

[bonjour-sarah]

The production of chemicals (Sulfuric acid/ ammonia etc) is no longer examinable, right?

No understanding required, if they do ask something, they have to give you the information, and then just applying normal equilibria principles

[bonjour-sarah]

In VCAA Exam 2 2009, Q3 cii)

It gives an equilibrium reaction and then asks to find mol t equilibrium. We are only given one value, and the assessors report shows the answer to make the assumption that [Ch3OCH3] = [H3O+] , which we would do if it was a weak acid, but this is just a normal equilibrium reaction and why can we do this here? Maybe I'm missing something really obvious but is it just that with no other option (lack of info?)

[cosine]

Need help with:

2009 Unit 4 chemistry exam:

Qestion 3cii). I dont know how to get the mol of dimethyl ether ?

Question 4e and f?

Question 5d: How can I learn more about fuel cells and such?

Question 2aii). why is the activation energy up to 120?

Thank you so much

[warya]

Guys I just looked at 2009 u4 and you guys are right...I just did it unassumingly as if it was a weak acid equilibrium and got it right, but it doesn't make sense as its just K???

[paper-back]

How do you get the anode equation when given the full equation, for a galvanic cell?
E.g. Q10 VCAA Chemistry exam 2014:
the following reaction takes place as the cell discharges:
2Zn(s) + O2 (g)+2H2O (l) -> 2Zn(OH)2 (s).

The anode is Zn, Cathode is C.

Write a balanced half equation for the anode reaction?

Zn in Zn(OH)2 has an oxidation number of 2+
The anode reaction is Zn -> Zn2+ + 2e-
Didn't see that Zn(OH)2 is in solid sate, sorry

In VCAA Exam 2 2009, Q3 cii)

It gives an equilibrium reaction and then asks to find mol t equilibrium. We are only given one value, and the assessors report shows the answer to make the assumption that [Ch3OCH3] = [H3O+] , which we would do if it was a weak acid, but this is just a normal equilibrium reaction and why can we do this here? Maybe I'm missing something really obvious but is it just that with no other option (lack of info?)

The reaction is:
2CH3OH(g) -> CH3OCH3(g) + H2O(g);

As CH3OCH3 and H2O are both products of the reaction, and they are both produced in a 2(CH3OH):1(CH3OCH3/H2O) molar ratio, we can assume that n(CH3OCH3)=n(H2O) - so concentrations are the same
We can do this regardless of if it is an acid/base equilibrium reaction or not



 

[jyce]

When we are finding concordant results do we use 3 values that are within the range of 0.1ml(from max to min), or within the range of +/-0.1ml of the middle value?

When diluting a solution, why don't both the concentration of H3O+ and OH- decrease?(Answer - Q8 VCAA 2010 Exam 2 Examiners report)

If I had a 0.01M solution of HCl at 10ml, the H3O+ concentration is: 0.01, and [OH-] is 10^-12

Then if I were to dilute this solution to 1L
For the H3O soltution: (0.01 x 0.01)=10^-4
Therefore pH = 4

However when I do this dilution to the [OH] concentration:
(10^-12 x 0.01) = 10^-14
So the pH gained by doing this = 0?

Why does doing the dilution to the [H3O] and [OH] concentration yield different pH results?

The answer to your first question is that concordant titres all have to be within 0.1 mL of each other, from the smallest to largest

In regards to the second question, the concentration of [OH^-] would not become 10^-14 M. Instead, if [H⁺] has decreased by a factor of 100, then [OH^-] has increased by a factor of 100.

[jyce]

GUYS, IN REGARDS TO VCAA 2009 EXAM 2 QUESTION 3CII:

You can make [CH₃OCH₃]equ. = [H₂O]equ. because:
1. initially, the concentrations of these products are the same as one another (i.e. they are both 0 M)
2. and the products are produced in a 1:1 ratio.

Don't think that this is exclusive to the ionisation of weak acids; you can do this whenever the above two conditions are satisfied.