# Thread: HSC 2012-2015 Chemistry Marathon (archive)

1. ## HSC 2012-2015 Chemistry Marathon (archive)

The HSC Chemistry Marathon is an open chain of questions between students. It works by answering a question then posting another question and allowing the cycle to repeat itself.

Rules:
- After answering a question, always provide a new one - this is what keeps the thread alive.
- Allocate a number of marks for any question that you post.
- Do not cheat, if you cannot answer a question, do not search how to answer the question but rather, allow other students to answer the question.
- No copyrighted questions (eg CSSA and Independent) should be posted.

Tips:
- You may post more than one question.
- When possible, after questions have been answered, you can peer mark using the marking scheme.

EDIT:

This thread is an archive for the questions from older years.

Feel free to post in the marathon thread for the current HSC year about anything in this thread. Make sure to quote what you are inquiring about!

Originally Posted by nightweaver066
Hopefully people will participate.

Feel free to post long response questions.

I'll start off:

2. ## Re: HSC 2012 Chemistry Marathon

First question: 16g for a quarter of a mole, hence, molar mass is 64g

Only option D satisfies this --> answer is D.

I'm on my iPod so I can't make a good question.

Compare the penetrating power of alpha particles, beta particles and gamma rays.

3. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by someth1ng
First question: 16g for a quarter of a mole, hence, molar mass is 64g

Only option D satisfies this --> answer is D.

I'm on my iPod so I can't make a good question.

Compare the penetrating power of alpha particles, beta particles and gamma rays.
Alpha particles can only travel a few cm in the air and can be stopped by a sheet of paper.

Beta particles can travel even further, up to 2m, and can be stopped by a sheet of aluminium.

Gamma rays are highly penetrating, able to travel long distances which require thick sheets of lead or concrete to stop or reduce them significantly.

i.e. in terms of penetrating power, Gamma rays > Beta particles > Alpha particles

4. ## Re: HSC 2012 Chemistry Marathon

A- Citric acid

Analyse the relationship between the position of elements on the periodic table, and the acid-base behaviours of their oxides.

5. ## Re: HSC 2012 Chemistry Marathon

Question:

Concentration of acetic acid in vinegar = 0.34 M

Calculate the percentage (w/v) of acetic acid in the vinegar.

6. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by deswa1
A- Citric acid

Analyse the relationship between the position of elements on the periodic table, and the acid-base behaviours of their oxides.
On the left side of the periodic table, oxides are basic and as moved across the periodic table, oxides tend to become more acidic with few neutral oxides such as CO until you reach the noble gases that do not form oxides.

7. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by someth1ng
On the left side of the periodic table, oxides are basic and as moved across the periodic table, oxides tend to become more acidic with few neutral oxides such as CO until you reach the noble gases that do not form oxides.
Could include more.

Moving left to right across a period, we see a shift from basic oxides to amphoteric then to acidic oxides.

Moving down groups 1 and 2, we see increasing basicity of the oxides. Moving down groups 6 and 7, we see decreasing acidity of the oxides.

Noble gases do not form oxides.

8. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by Dylanamali
Question:

Concentration of acetic acid in vinegar = 0.34 M

Calculate the percentage (w/v) of acetic acid in the vinegar.
[acetic acid] = 0.34M = 0.34moles/litre
= (0.34 x (12.01 x 2 + 1.008 x 4+ 16 x 2))/litre
= 20.42g/litre
= 2% (g/ml) if you don't mind the units.

Qn:

9. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by nightweaver066
Could include more.

Moving left to right across a period, we see a shift from acidic oxides to amphoteric then to basic oxides.

Moving down groups 1 and 2, we see increasing basicity of the oxides. Moving down groups 6 and 7, we see decreasing acidity of the oxides.

Noble gases do not form oxides.
Also towards the top right of the non-metals (not including noble gases) there is increasing acidity of the oxides?

10. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by Kimyia
Also towards the top right of the non-metals (not including noble gases) there is increasing acidity of the oxides?
I made a mistake in my original post.

Meant to say shift from acidic oxides to amphoteric then to basic oxides across a period.

This includes your claim that towards the top right of the non-metals, there is increasing acidity of the oxides.

11. ## Re: HSC 2012 Chemistry Marathon

Identify the standard conditions required to produce a table of reduction potentials based on galvanic cells constructed using a hydrogen electrode.
An electrochemical cell is produced using two half cells. One cell consists of a solution of Fe2+ and Fe3+ with an inert platinum electrode, the other cell consists of a copper electrode in a solution of Cu2+. Write equations for each half cell and calculate the expected standard cell potential.

12. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by someth1ng
Identify the standard conditions required to produce a table of reduction potentials based on galvanic cells constructed using a hydrogen electrode.
An electrochemical cell is produced using two half cells. One cell consists of a solution of Fe2+ and Fe3+ with an inert platinum electrode, the other cell consists of a copper electrode in a solution of Cu2+. Write equations for each half cell and calculate the expected standard cell potential.
Standard conditions:
- 25 degrees celcius
- 1 atm
- Pure metals/liquids used
- 1 mol/L aqueous electrolytes used

$Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^- E(ox) = -0.34V$

$Fe^{3+}_{(aq)} + e^- \rightarrow Fe^{2+}_{(aq)} E(red) = 0.77V$

$E_{cell} = 0.77V - 0.34V = 0.43V$

Do my question above.

13. ## Re: HSC 2012 Chemistry Marathon

Also everyone in this thread is welcome to PM me for Chemistry papers. I have a lot.

14. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by nightweaver066
$E_{cell} = 0.77V - 0.34V = 0.43V$
Isn't it 1.11V?

(0.77V)-(-0.34)=1.11V

15. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by someth1ng
Isn't it 1.11V?

(0.77V)-(-0.34)=1.11V
I already reversed the reduction potential to get the oxidation potential.

Then i add both the reduction and the oxidation potentials to get the cell potential.

16. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by nightweaver066
I will just dot point this:

- Ethanol is the most used solvent because it has the ability to be a solvent for both polar and non-polar substances.
- The polar hydroxide functional group gives ethanol the ability to be miscible in polar substances.
- The non-polar alkyl part allows ethanol to be miscible in non-polar substances.
- Essentially, having the ability to be miscible in both polar and non-polar substances gives ethanol more effectiveness as a solvent.

17. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by nightweaver066
I already reversed the reduction potential to get the oxidation potential.

Then i add both the reduction and the oxidation potentials to get the cell potential.
Ah, okay. I got that question from the assessment that I'm currently doing but I just wanted to check my answer (SECRET!).

I'm tired now, gonna go sleep.

Good night.

18. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by someth1ng
I will just dot point this:

- Ethanol is the most used solvent because it has the ability to be a solvent for both polar and non-polar substances.
- The polar hydroxide functional group gives ethanol the ability to be miscible in polar substances.
- The non-polar alkyl part allows ethanol to be miscible in non-polar substances.
- Essentially, having the ability to be miscible in both polar and non-polar substances gives ethanol more effectiveness as a solvent.
Good answer. In a test, hopefully you'll draw a nice structural formula of ethanol showing both the hydroxide functional group and the alkyl chain.

19. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by nightweaver066
$Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^- E(ox) = -0.34V$

$Fe^{3+}_{(aq)} + e^- \rightarrow Fe^{2+}_{(aq)} E(red) = 0.77V$

$E_{cell} = 0.77V - 0.34V = 0.43V$
Thinking about this again, shouldn't the reduction potential of the copper be -0.34 since the reaction went the opposite direction as the one of the data sheet?

Hence, EMF=0.77--0.34=1.11V

OR

Is it always for any situation using the potential listed on the data sheet:
Ecell = Higher potential-Lower potential [negative is considered as lower than positive]

Higher potential=0.77
Lower potential=0.34
Ecell=0.77-0.34=0.43V

20. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by someth1ng
Thinking about this again, shouldn't the reduction potential of the copper be -0.34 since the reaction went the opposite direction as the one of the data sheet?

Hence, EMF=0.77--0.34=1.11V
Notice how i wrote E(ox). That means i already had reversed the reduction potential given on the data sheet to obtain the oxidation potential.

If you look on your sheet, it tells you that the reduction potential is 0.34V.

21. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by nightweaver066
Good answer. In a test, hopefully you'll draw a nice structural formula of ethanol showing both the hydroxide functional group and the alkyl chain.

Yeah, I normally draw pictures/diagrams of some sort when I can.

Originally Posted by nightweaver066
Notice how i wrote E(ox). That means i already had reversed the reduction potential given on the data sheet to obtain the oxidation potential.

If you look on your sheet, it tells you that the reduction potential is 0.34V.
Yeah, so it's ALWAYS:
Ecell=higher number-lower number ??

Next question:
A student, attempting to obtain the equivalence point of a titration between acetic acid and sodium hydroxide, was unable to reach a consistent result using methyl orange as an indicator. Suggest one modification to the procedure which would give an acceptable result. Explain why sodium acetate solution is not pH neutral.

22. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by someth1ng
Yeah, I normally draw pictures/diagrams of some sort when I can.

Yeah, so it's ALWAYS:
Ecell=higher number-lower number ??
Yep.. The Fe3+ to Fe2- reduction potential is 0.77V.
The CU2+ to Cu reduction potential is 0.34V.

By following what you said, Ecell = 0.77 - 0.34 = 0.43V.

23. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by someth1ng
Yeah, I normally draw pictures/diagrams of some sort when I can.

Yeah, so it's ALWAYS:
Ecell=higher number-lower number ??

Next question:
A student, attempting to obtain the equivalence point of a titration between acetic acid and sodium hydroxide, was unable to reach a consistent result using methyl orange as an indicator. Suggest one modification to the procedure which would give an acceptable result. Explain why sodium acetate solution is not pH neutral.
Use phenolphthalein instead of methyl orange as the indicator.

$CH_3COONa_{(aq)} \rightarrow CH_3COO^-_{(aq)} + Na^+_{(aq)}$

$CH_3COO^-_{(aq)} + H_2O_{(l)} \rightarrow CH_3COOH_{(aq)} + OH^-_{(aq)}$

Therefore sodium acetate solution is not pH neutral but is basic, i.e. pH > 7.

24. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by nightweaver066
Use phenolphthalein instead of methyl orange as the indicator.

$CH_3COONa_{(aq)} \rightarrow CH_3COO^-_{(aq)} + Na^+_{(aq)}$

$CH_3COO^-_{(aq)} + H_2O_{(l)} \rightarrow CH_3COOH_{(aq)} + OH^-_{(aq)}$

Therefore sodium acetate solution is not pH neutral but is basic, i.e. pH > 7.
Oh, you use a basic indicator (phen...) when you are testing for basic substances and acidic indicators (methyl orange) for acidic substances?

Also, is the anion of the weak acid make basic solutions? eg CH3COO-

I don't know how to do that question:
How do you do this question?
A standard solution was prepared by dissolving 1.432g of sodium carbonate in water. The solution was made up to 250mL in a volumetric flask. Calculate the concentration of the sodium carbonate solution.The solution was used to determine the concentration of a solution of hydrochloric acid. Four 25.00mL samples of acid were titrated with the sodium carbonate solution. The average titration volume was 22.35mL. Why was it necessary to conduct four titrations? Calculate the concentration of the hydrochloric acid solution.

25. ## Re: HSC 2012 Chemistry Marathon

Originally Posted by someth1ng
I don't know how to do that question:
How do you do this question?
A standard solution was prepared by dissolving 1.432g of sodium carbonate in water. The solution was made up to 250mL in a volumetric flask. Calculate the concentration of the sodium carbonate solution.The solution was used to determine the concentration of a solution of hydrochloric acid. Four 25.00mL samples of acid were titrated with the sodium carbonate solution. The average titration volume was 22.35mL. Why was it necessary to conduct four titrations? Calculate the concentration of the hydrochloric acid solution.
Firstly, to calculate the concentration of Na2CO3 we will need to use the formula c=n/v.
Therefore from the given information v= 0.25L, and n will need to be calculated using n=m/M.
n(Na2CO3)= 1.432/2(22.99)+12.01+3(16) = 0.0135....
c(Na2CO3)= 0.0135...../0.25 = 0.0540......
Therefore the concentration of Na2CO3= 0.05404 (4 sig.figs)

Now, write a balanced equation:
Na2CO3(aq) + 2HCl(aq)2NaCl(aq) + H2O(l) + CO2(g)
Again, we need to use c=n/v to calculate the concentration of HCl but we are only give v=0.025L.
So using the concentration of Na2CO3 we found earlier and v=0.02235L we can calculate the number of mols that reacted.
c=n/v
n= c x v
n(Na2CO3)= 0.05404..... x 0.02235 = 0.001207......
From the equation we can see that there is a 1:2 ratio (Na2CO3: HCl)
Therefore HCl will have the twice the number of mols as Na2CO3.
Now, c=n/v
c(HCl)=0.002414..../0.025 = 0.09656 (4sig.figs.)

So the concentration of HCl is 0.09656

It is necessary to conduct the titration several times to obtain an average to ensure that your results are accurate.

Now, write a balanced equation:
Na2CO3(aq) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)
*facepalm*

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