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HSC 2012-2015 Chemistry Marathon (archive) (10 Viewers)

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Immortalp00n

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re: HSC Chemistry Marathon Archive

I'm actually really sorry! I misinterpreted :(

Being so used to you hating on me I assumed it was negative/ you trying to tick me off.

Please accept the PDF through PM as an apology.
*need your e-mail...
Lol soz fo d rage.
Just a misinterpretation :)
 

nightweaver066

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re: HSC Chemistry Marathon Archive

Edit: Adding to the above.
Trend one:
Bp rises with increasing molecular mass. The MM is increasing due to increasing Carbon chain length. In turn, the BP rises because of ^ dispersion forces- temporarily induced polarity within the non-poplar molecules. This increased length --> increasing dispersion forces and thus more energy needed to break the intermolecular forces.

Trend 2:
Alkanoic acids have a higher BP than alkanols, which also have higher BPs than corresponding alkanes; alkanes having only dispersion forces involved is an attributable factor to this.
The other series have dispersion forces and dipole-dipole interactions. The dipole-dipole (2nd strongest intermolecular force) are greater in alkanoic acids than in alkanols, accounting for the greater BPs of this series.
In alkanoic acids and alkanols there Hydrogen bonding, the strongest of the intermolecular forces, holds the covalently bonded moleucles in structure more tightly than due to the strong force- making the intermolualar forces (all 3 involved, esp. H bonding) to be broken; this accounting for the significantly higher BPs of these two series.

The Hydrogen bonding is more marked in alkanoic acids because of the two O atoms and hence the 2 lone pairs of electrons present in the -COOH group in comparison to only the 1 O atom w/ 1 lone pair of electrons in the -OH alkanol group. The Hydrogen bonding is the strongest acting in the Alkanoic acid molecules thus the series has the highest BPs of the 3 shown in the graph.
Your explanation involving electrons is kinda weird..

Also seems like so much for a 4 marker lol. I'll try write a plan for how i would tackle this question:
- Explain why BP of alkanoic acids > alkanols > alkanes even though their similar MW
-> alkanoic acids have 2 polar groups
--> OH hydroxyl group (O highly electronegative inducing partial charges) & C=O bond slightly polar also due to O's electronegativity)
-> extensive hydrogen bonding & dipole dipole interactions
-> alkanol has 1 polar group (OH group)
-> less extensive hydrogen bonding
-> alkanes are non-polar, thus main IMF is dispersion forces which are much weaker than hydrogen bonds & dipole dip
-> BP proportional to strength/degree of IMF therefore BP of alkanoic acids > alkanols > alkanes when comparing ones of similar MW

- Explain why BP increases as molecular weight increases
-> dispersion forces is proportional to molecular weight
-> as molecular weight increases, dispersion forces between molecules increases
-> therefore IMF increases, and so all the BPs increase

I thought your answer seemed pretty long but turns out mine is probably equal in length haha.... but that's how I would do it.
 

Eg155

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re: HSC Chemistry Marathon Archive

Your explanation involving electrons is kinda weird..

Also seems like so much for a 4 marker lol. I'll try write a plan for how i would tackle this question:
- Explain why BP of alkanoic acids > alkanols > alkanes even though their similar MW
-> alkanoic acids have 2 polar groups
--> OH hydroxyl group (O highly electronegative inducing partial charges) & C=O bond slightly polar also due to O's electronegativity)
-> extensive hydrogen bonding & dipole dipole interactions
-> alkanol has 1 polar group (OH group)
-> less extensive hydrogen bonding
-> alkanes are non-polar, thus main IMF is dispersion forces which are much weaker than hydrogen bonds & dipole dip
-> BP proportional to strength/degree of IMF therefore BP of alkanoic acids > alkanols > alkanes when comparing ones of similar MW

- Explain why BP increases as molecular weight increases
-> dispersion forces is proportional to molecular weight
-> as molecular weight increases, dispersion forces between molecules increases
-> therefore IMF increases, and so all the BPs increase

I thought your answer seemed pretty long but turns out mine is probably equal in length haha.... but that's how I would do it.
Thankyou! Yeah, mine made sense in my head but I have a lot of trouble expressing what I mean lol

Your's is very concise and easy to understand. Exemplary example ! Tah.
 

Eg155

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re: HSC Chemistry Marathon Archive

New Q: Explain how the structure and properties of polyethylene and polystyrene relate to the way each is used (4 marks).
 

Immortalp00n

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re: HSC Chemistry Marathon Archive

New Q: Explain how the structure and properties of polyethylene and polystyrene relate to the way each is used (4 marks).
Polystyrene is produced by the reaction of ethene and a benzene ring. It is used for many uses within the industry such as the production of disposable insulating cups and fast food containers because of the presence of the benzene ring making the polymers to be quite stiff, providing ethenyl benzene ( systematic name for polystyrene) with properties such as being both hard and rigid.
Polystyrene is also used in the manufacture of clear plastics, drinking glasses, audio cassettes, cd disks, computer and tv cabinets and the production of styrofoam by blowing gases through hot Polystyrene liquid and cooling them to shape.
Polyethylene is a very tough, flexible and durable polymer used for sandwich bags, cling wrap, car covers, squeeze
bottles, freezer bags, water pipes, wire and cable insulation.
The polyethylene used for the production of the uses mention before is a low density polymer, with properties including its very flexible and soft with a lower melting point than polystyrene.
Polystyrene on the other hand has a higher melting point and is much more stiff due to the presence of a benzene ring.


lol major mental diarrhea right there ^^^ pls feel free to rip through it and add extra points etc... coz im not 100% sure
 

deswa1

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re: HSC Chemistry Marathon Archive

About to go for a swim so don't have time to comment properly but:

- Easiest way to answer that would be in a table
- Distinguish between LDPE and HDPE (i.e. mention specifically which one you are talking about)
- Distinguish between crystal polystyrene and expanded polystyrene
- Draw the structural formulae

I'll have a proper read later
 

nerdasdasd

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re: HSC Chemistry Marathon Archive

FOr compare and contrast Q's, you can use tables.
 

Immortalp00n

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About to go for a swim so don't have time to comment properly but:

- Easiest way to answer that would be in a table
- Distinguish between LDPE and HDPE (i.e. mention specifically which one you are talking about)
- Distinguish between crystal polystyrene and expanded polystyrene
- Draw the structural formulae

I'll have a proper read later
derpity derp.
will do properly soon
 

Sy123

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deswa1

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Oh and while we're doing pH calculations, after someone does Sy's one, try this:

Q: What is the pH of a 10L solution of hydrochloric acid that contains 1x10^-8 moles of HCl in it?
 

Sy123

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Oh and while we're doing pH calculations, after someone does Sy's one, try this:

Q: What is the pH of a 10L solution of hydrochloric acid that contains 1x10^-8 moles of HCl in it?
Nice trick question, after normal calculation you would arrive at pH9, but HCl is acidic and hence we can never go above a 7 pH (assuming standard conditions). Therefore the pH is extremely close to 7.
 

HeroicPandas

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re: HSC Chemistry Marathon Archive

Havent done much calculations on acidic environment at school yet but i will have a try

Beaker 1:

Using pH = -log [H+] given pH=1

H+ = 0.1M

Now, using n = CV given V = 100mL and i just found C = 0.1M

n_1 = 0.1 x 100
n_1 = 10mol

Beaker 2:

Using n = CV given V= 1000 and since adding more liquid, it doesnt change the moles of final sol'n

10 = C x 1000
C= 0.01M

Now, using pH = -log[H+]

pH = -log[0.01]
pH = 2

Yes or no?
 

deswa1

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I haven't read the working but the answer is right. You could do it without calcs though- by adding 900ml of water, you made the concentration of H+ one tenth of what it was before and since ph is a logarithmic scale, the ph goes up 1
 

Sy123

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Havent done much calculations on acidic environment at school yet but i will have a try

Beaker 1:

Using pH = -log [H+] given pH=1

H+ = 0.1M

Now, using n = CV given V = 100mL and i just found C = 0.1M

n_1 = 0.1 x 100
n_1 = 10mol

Beaker 2:

Using n = CV given V= 1000 and since adding more liquid, it doesnt change the moles of final sol'n

10 = C x 1000
C= 0.01M

Now, using pH = -log[H+]

pH = -log[0.01]
pH = 2

Yes or no?
Correct, nice work.
 

Sy123

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I haven't read the working but the answer is right. You could do it without calcs though- by adding 900ml of water, you made the concentration of H+ one tenth of what it was before and since ph is a logarithmic scale, the ph goes up 1
Yeah this is what I was hoping people would do but it doesn't really matter.
 

nightweaver066

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Havent done much calculations on acidic environment at school yet but i will have a try

Beaker 1:

Using pH = -log [H+] given pH=1

H+ = 0.1M

Now, using n = CV given V = 100mL and i just found C = 0.1M

n_1 = 0.1 x 100
n_1 = 10mol

Beaker 2:

Using n = CV given V= 1000 and since adding more liquid, it doesnt change the moles of final sol'n

10 = C x 1000
C= 0.01M

Now, using pH = -log[H+]

pH = -log[0.01]
pH = 2

Yes or no?
Correct. One of the syllabus dot points requires you to explain that a change in concentration by a factor of 10 is a change in pH by a factor of 1 (due to base 10 logarithm).

So if you dilute it from 100mL to 1000mL, you've decreased the [H+] tenfold and so the pH will increase by 1 to become 2.

A typical MCQ.

Question
Explain how chemical changes and movement of charged species produce an electric current in a galvanic cell.
Use an example and include relevant half-equations with your answer. (4 marks)
 

HeroicPandas

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Question: Arrange the three polymers polyethylene, PVC and polystyrene in order of increasing flexibility. Explain your reasoning
 

HeroicPandas

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I haven't read the working but the answer is right. You could do it without calcs though- by adding 900ml of water, you made the concentration of H+ one tenth of what it was before and since ph is a logarithmic scale, the ph goes up 1

Correct. One of the syllabus dot points requires you to explain that a change in concentration by a factor of 10 is a change in pH by a factor of 1 (due to base 10 logarithm).

So if you dilute it from 100mL to 1000mL, you've decreased the [H+] tenfold and so the pH will increase by 1 to become 2.
ok
 

Eg155

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How far into the Acidic Environment module are you guys?
...we have not started, although I am trying to go ahead...
 

bleakarcher

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Correct. One of the syllabus dot points requires you to explain that a change in concentration by a factor of 10 is a change in pH by a factor of 1 (due to base 10 logarithm).

So if you dilute it from 100mL to 1000mL, you've decreased the [H+] tenfold and so the pH will increase by 1 to become 2.

A typical MCQ.

Question
Explain how chemical changes and movement of charged species produce an electric current in a galvanic cell.
Use an example and include relevant half-equations with your answer. (4 marks)
A galvanic cell is a battery consisting of two seperate half-cells both which consist of a metal (electrode) in an electrolyte solution, one at which a oxidation reaction occurs (anode-negative electrode) and one at which a reduction reaction occurs (cathode-positive electrode), a salt bridge (usually immersed in some electrolyte solution) which allows for the transportation of ions between the electrolyte solutions, and a wire connecting the anode to the cathode. A galvanic cell therefore runs on a redox reaction. This redox reaction will form a total EMF (electromotive force - a measure of the induced voltage at a negligibly small current) which allows for the movement of electrons through the wire. The direction of flow of electrons is such that they move from the anode (more reactive metal) to the cathode (less reactive metal). An example of a galvanic cell is the copper-zinc cell in which copper metal is immersed in copper sulfate solution and zinc metal is immersed in zinc sulfate solution with a neutral salt bridge. At the anode: Zn->Zn(2+)+2e- and at the cathode: Cu(2+)+2e->Cu. However, this means an increase in the Zn(2+) concentration in the aqueous ZnSO4 solution and an decrease in the Cu(2+) concentration in the aqueous CuSO4 solution. In order to maintain electrical neutrality in the half cells, a salt bridge is connected from from one electrolyte to the other so that Zn(2+) ions can flow to the CuSO4 solution (opposite to the direction of electron flow) and SO4(2-) can flow to the ZnSO4 solution (in the direction of the electrons). This creates a continuous current throughout the copper-zinc cell.

Please be as critical as you wish.
 
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