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anat0my

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Re: VCE Chemistry Question Thread
« Reply #4350 on: November 03, 2015, 05:07:15 pm »
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Hi does anyone know what it takes to achieve a 30 SS? And what can I do to revise/learn the content most effectively? Sorry if this is in the wrong thread. Thanks.
« Last Edit: November 03, 2015, 05:10:40 pm by anat0my »

jyce

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Re: VCE Chemistry Question Thread
« Reply #4351 on: November 03, 2015, 06:14:55 pm »
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Hi does anyone know what it takes to achieve a 30 SS? And what can I do to revise/learn the content most effectively? Sorry if this is in the wrong thread. Thanks.

To achieve a study score of 30, I'd guess you need around 50% on the exam. When I say 'guess', I do mean guess though; I could be wrong.
As for revising, I recommend reading over the key knowledge dot points in the study design and making note of which concepts you feel least prepared for. I would then read up on those concepts and confirm your knowledge via practice questions, preferably VCAA ones. Practice exams are the key way to study Chemistry in my opinion, as the exam will be largely applications-based questions and so you need to expose yourself to a range of applications in the form of practice exams. Make sure you complete the VCAA 2013, 2014 and sample exams!

cosine

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Re: VCE Chemistry Question Thread
« Reply #4352 on: November 03, 2015, 06:22:48 pm »
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jyce how can I improve in chem?
These are my errors:

- pH questions - always get them wrong, dont even know what pH means. how can i fix this?
- I cant seem to get those fuel cells and like commercial cells. i am good/alright with galvanic and electrolysis but not industrial cells
- spectroscopy etc.
2016-2019: Bachelor of Biomedicine
2015: VCE (ATAR: 94.85)

cosine

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Re: VCE Chemistry Question Thread
« Reply #4353 on: November 03, 2015, 07:06:41 pm »
0
Please guys i need some help with 2011 Unit 4 exam MC:

Question 10: No idea how to do this one
Question 15 and 16: how can i get better at these ? and do you do this question?
Question 18: I know exactly what I should do but idk what faraday is.. Q = n(e)F is Q = 0.03 or is F = 0.03? How do you know? How do I do this question?

Thank you so much
2016-2019: Bachelor of Biomedicine
2015: VCE (ATAR: 94.85)

jyce

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Re: VCE Chemistry Question Thread
« Reply #4354 on: November 03, 2015, 07:09:45 pm »
+4
jyce how can I improve in chem?
These are my errors:

- pH questions - always get them wrong, dont even know what pH means. how can i fix this?
- I cant seem to get those fuel cells and like commercial cells. i am good/alright with galvanic and electrolysis but not industrial cells
- spectroscopy etc.

Hi cosine,

First of all pH is a measure of the power, or concentration, of hydrogen ions. As pH = -log10[H+], the greater the pH the lower the concentration of hydrogen ions, and vice versa. For strong acids, we assume that they have ionised in aqueous solution completely and therefore if, for example, we had a 0.1 M hydrochloric acid solution it will have ionised to produce 0.1 M of hydrogen ions and its pH will be -log10(0.1) = 1. Calculating the pH of a weak acid, on the other hand, is more complicated as there is not a complete ionisation; in fact, we often safely assume that a weak acid has not ionised at all. In this case, we need to write an expression for the equilibrium constant of the reaction, excluding the concentration of water, and go from there. If you had 0.1 M solution of a weak acid, it would ionise to produce less than 0.1 M of hydrogen ions and therefore have a pH greater than 1. This is by no means everything you need to know on pH, but it's some background information that might help you out.

On to your second point, industrial galvanic cells (i.e. primary, secondary and fuel cells) will most definitely be assessed. Here's some background information again:
- these three types of cells are all galvanic cells, and therefore: the cell reaction is spontaneous, with the oxidant being higher up on the electrochemical series than the reductant, with a positive E0 value for the cell reaction, and with the anode being negative and the cathode being positive
- primary cells cannot be recharged, meaning that the cell reaction cannot be reversed once it has reached equilibrium
- secondary cells on the other hand are able to be recharged, because the products of the discharge reaction remain in contact with the electrodes (i.e. the sites of reaction) and therefore remain available to become reactants in a reverse reaction, but an external power source is required as the reverse reaction is non-spontaneous
- when a secondary cell is recharging, it acts like an electrolytic cell: the recharge reaction is literally the reverse, so the anode becomes the site of reduction (i.e. the cathode) and the cathode becomes the anode but the polarities of the electrodes remain the same so the cathode is now the negative electrode and the anode is now positive
- fuel cells are not rechargeable but they also don't go flat like primary cells, and this is because they have a continuous supply of reactants.
Some tips regarding these commercial cells include to remember that they are galvanic cells and not electrolytic, to identify off the bat the anode and cathode and their polarities, the direction of electron flow and the direction of ion flow, to acknowledge that the electrolyte tends to participate in the cell reaction but with no net consumption or production (e.g. H+ ions might be produced at the anode, but then the same amount will be consumed at the cathode), and lastly to acknowledge that 'KOHES' does not always work to balance cell reactions as, for example, the environment may be alkaline rather than acidic.

Lastly, with spectroscopy note that you do NOT need to know how the instrumentation (e.g. the monochromator) works, but rather for each spectroscopic technique you need to know which region of the electromagnetic spectrum it utilises, how this region of the spectrum interacts with the analyte, and how to interpret the data obtained:
- atomic absorption spectroscopy utilises visible light which promotes the valence electrons of atoms to higher energy levels; the data obtained is a series of absorbance measurements, one for the analyte and the rest for a series of standards, and the absorbances of the standards are graphed against the standard concentrations in what is known as a calibration graph, in order to relate absorbance and concentration in a mathematical relationship, AND then the concentration of the analyte is simply read off; be sure to account for dilution factors
- UV-visible spectroscopy utilises, in addition to visible light, ultraviolet light which has enough energy to promote electrons in atoms, ions and molecules; the data interpretation is basically the same as with AAS
- IR spectroscopy utilises infrared radiation which promotes molecules to higher vibrational energy levels (i.e. makes the covalent bonds in molecules more active); what comes out of IR spectroscopy is an IR spectrum and all you need to do is estimate the wavenumber ranges of major peaks and then compare these to the wavenumber ranges in the data book in order to identify the bonds responsible; particularly noteworthy peaks are those at about 1700 cm-1, indicative of a C=O bond, and a very broad band at around 2500-3300 cm-1, indicative of the O-H bond within a carboxyl functional group; IR spectroscopy is useful for identifying functional groups
- lastly, NMR spectroscopy utilises radio waves which change nuclei to a higher spin state; NMR spectra provide a number of useful pieces of information: the number of peaks/peak sets indicates the number of different hydrogen/carbon environments, the relative areas the peaks/peak sets indicates the relative numbers of hydrogen/carbon atoms in each different environment, the peak-splitting of a high-resolution 1H NMR spectrum indicates the number of hydrogens in adjacent environments according to the 'n + 1 rule' (e.g. in propane, the CH2 environment has six hydrogens in the adjacent environments and therefore the peak representing the CH2 group will be split into a 6 + 1 = septet), and lastly the chemical shifts, measured in ppm, could be useful (e.g. a peak set at 9-13 ppm of a 1H NMR spectrum is indicative of a carboxyl group); NMR spectroscopy is particularly useful for distinguishing between isomers.
Again, this is by no means an extensive explanation.

As for improving in these areas, I would recommend the same thing I said to anat0my: practice questions, preferably VCAA ones! It's really the only way to get better in Chemistry. You'll hopefully start to pick up on the main things VCAA draws from each of the above topics. It's hard for me to give you more specific help on these topics without more specific questions from yourself. So, if when doing practice questions on these topics you're having trouble, shoot me some more questions! Keep in mind there are many things in the course aside from pH, commercial galvanic cells and spectroscopy, and therefore obviously these three topics will only make up a fraction of the exam.
 
 
« Last Edit: November 06, 2015, 08:07:26 pm by jyce »

jyce

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Re: VCE Chemistry Question Thread
« Reply #4355 on: November 03, 2015, 07:47:11 pm »
+1
Please guys i need some help with 2011 Unit 4 exam MC:

Question 10: No idea how to do this one
Question 15 and 16: how can i get better at these ? and do you do this question?
Question 18: I know exactly what I should do but idk what faraday is.. Q = n(e)F is Q = 0.03 or is F = 0.03? How do you know? How do I do this question?

Thank you so much

Question 10
Let's take a look at the cell on the left first: the half-cell with ammonia has the positively charged electrode (i.e. the cathode, since this is a galvanic cell), and the half-cell with cyanate (CN-) has the negatively charged electrode (i.e. the anode). So, in this particular cell, the half-reaction with E01 (i.e. the one with ammonia) is the cathode reaction, and the half-reaction with E03 is the anode reaction. Therefore, E01 must be higher than E03. If you apply the same reasoning to the second galvanic cell, you should deduce that E02 is higher than E01. So, the answer is C (E02 > E01 > E03).

Question 15
The bottom reaction listed in the stem information has the higher voltage and therefore when the battery is DISCHARGING this is the cathode reaction, occurring in the forwards direction. When RECHARGING, this bottom reaction will occur in reverse, becoming an oxidation process and therefore now occurring at the ANODE. Therefore the answer to Question 15 is the bottom reaction written in reverse (i.e. C).

Question 16
When recharging a battery, the positive terminals are connected together and the negative terminals are connected together. Therefore, options C and D are already incorrect. When RECHARGING, the positive terminal of the battery is a site of OXIDATION, not reduction like you're used to with regular, discharging galvanic cells. Therefore, the answer is A.

Question 18
0.03 faradays does NOT = 0.03 C; rather, 0.03 faradays = 0.03 moles of electrons. Therefore, using stoichiometry and the relevant reduction half-reactions for Cr3+, Cu2+ and Ag+, you will have deposited 0.01 mol of chromium, 0.015 mole of copper and 0.03 mole of silver. Therefore, the answer is B. Why does 0.03 faradays = 0.03 moles of electrons, though? Think about it: one faraday would mean 96,500 C, and 96,500 C is the charge carried by one mole of electrons (1 faraday = 96,500 C = 1 mole electrons, so faradays = moles of electrons). Faradays are not to be confused with Faraday's constant, although they are very similar.

Hope this helps!
« Last Edit: April 11, 2016, 09:29:09 pm by jyce »

Acid

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Re: VCE Chemistry Question Thread
« Reply #4356 on: November 03, 2015, 07:56:26 pm »
0
Hey, does anyone have STAV 2015 trial exam solutions? My teacher only gave us the question booklet, and I have no way of contacting him atm  :-\

Also, could someone please tell me the answer for this question?

An aqueous solution containing ammonia, NH3, is titrated with 0.010 M hydrochloric acid. Assuming the process is carried out at 25c, at the equivalent point the pH is:
A. less than 7.0
B. more than 7.0 (I chose this because ammonia is a weak base, and HCl is a strong acid; pls correct if wrong)
C. 7.0
D. impossible to predict without being given the concentration of ammonia.

Thanks!
2015: English [50]
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jyce

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Re: VCE Chemistry Question Thread
« Reply #4357 on: November 03, 2015, 07:59:21 pm »
+2
Hey, does anyone have STAV 2015 trial exam solutions? My teacher only gave us the question booklet, and I have no way of contacting him atm  :-\

Also, could someone please tell me the answer for this question?

An aqueous solution containing ammonia, NH3, is titrated with 0.010 M hydrochloric acid. Assuming the process is carried out at 25c, at the equivalent point the pH is:
A. less than 7.0
B. more than 7.0 (I chose this because ammonia is a weak base, and HCl is a strong acid; pls correct if wrong)
C. 7.0
D. impossible to predict without being given the concentration of ammonia.

Thanks!

Hi, Acid.

As you yourself have stated, ammonia is a weak base and hydrochloric acid is a strong acid. At 25oC, a titration between a weak base and a strong acid has an equivalence point with a pH less than 7. In this particular scenario, the pH is less than 7 because one of the products of the titration reaction is a weak acid (NH4+). Similarly, the pH at the equivalence point of a titration between a strong base and a weak acid will be more than 7 (because there is some basic product), and a titration between a strong acid and a strong base has an equivalence point with a pH around if not at 7. A titration between a weak acid and a weak base does not yield a sharp or clear equivalence point, so a back titration is used instead.
The answer is 'A'.
« Last Edit: November 03, 2015, 08:35:50 pm by jyce »

Acid

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Re: VCE Chemistry Question Thread
« Reply #4358 on: November 03, 2015, 08:01:53 pm »
+1
Thank you so much!  :)
2015: English [50]
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bedigursimran

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Re: VCE Chemistry Question Thread
« Reply #4359 on: November 04, 2015, 05:51:49 pm »
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Hey guys, how do I get the oxidation numbers of compounds such as C4H6O5 and NaOH? Thanks!

jyce

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Re: VCE Chemistry Question Thread
« Reply #4360 on: November 04, 2015, 08:19:34 pm »
+1
Hey guys, how do I get the oxidation numbers of compounds such as C4H6O5 and NaOH? Thanks!

The oxidation number of hydrogen in a compound is usually +1, oxygen in a compound is usually -2, and alkali metals in a compound (e.g. Na) are usually +1. As the two compounds you have mentioned are overall neutral, the oxidation numbers must add up to zero.

So, in C4H6O5, each H has an oxidation number of +1 and there are six of them, and each O is -2 and there are five of them. This gives +6 - 10 = -4. Therefore, the four carbon atoms need to contribute +4, to give a total of zero. This means that each C has an oxidation number of +4/4 = +1. With NaOH, we already know all the oxidation numbers: Na, being an alkali metal, is +1, O is -2 and H is +1. These numbers add up to give zero.

Most often, you will be given a compound where you already know all but one of the oxidation numbers. Another example would be H2SO4: we already know the oxidation numbers of H and O and we can use these, and the fact that the compound is neutral and therefore that the oxidation numbers must add up to zero, in order to calculate the oxidation number of S.

Note that in your question you referred to oxidation numbers as a property of compounds as a whole; this is not the case - oxidation numbers are properties of singular atoms, which add together in a compound.
« Last Edit: November 04, 2015, 08:31:52 pm by jyce »

paper-back

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Re: VCE Chemistry Question Thread
« Reply #4361 on: November 06, 2015, 12:24:50 pm »
0
Does Ae (activation energy) change when the molar ratios of a reaction is doubled?

sunshine98

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Re: VCE Chemistry Question Thread
« Reply #4362 on: November 06, 2015, 12:33:15 pm »
+2
Does Ae (activation energy) change when the molar ratios of a reaction is doubled?
Yep! I  got this  wrong in a SAC. It makes sense tho , because if you have more molecules you need more energy to break bonds between reactants than if you had less. 

sunshine98

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Re: VCE Chemistry Question Thread
« Reply #4363 on: November 06, 2015, 06:35:39 pm »
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I have tonnes of conceptual questions on electrochemistry, its been my weak point , so here goes :
- Why is it that by the simple act of separating half cells that we get electricity rather than heat energy?
-Though galvanic cells generate electricity , surely there is still some heat loss?
-Are all redox reactions exothermic?
-Why don't electrodes in secondary and primary cells act as catalysts? Why only in fuel cells?
-Do states matter in the electrochemical series? Like I was doing a q which had I2 as liquid but in the series it has it as solid?
-You know how theres that rule that if the concentration deviates from 1M Cl is a stronger reductant than water , does this only apply to concentrations greater than 1 or even lower than 1?
Thank you  :)

jyce

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Re: VCE Chemistry Question Thread
« Reply #4364 on: November 06, 2015, 06:48:00 pm »
+2
I have tonnes of conceptual questions on electrochemistry, its been my weak point , so here goes :
- Why is it that by the simple act of separating half cells that we get electricity rather than heat energy?
-Though galvanic cells generate electricity , surely there is still some heat loss?
-Are all redox reactions exothermic?
-Why don't electrodes in secondary and primary cells act as catalysts? Why only in fuel cells?
-Do states matter in the electrochemical series? Like I was doing a q which had I2 as liquid but in the series it has it as solid?
-You know how theres that rule that if the concentration deviates from 1M Cl is a stronger reductant than water , does this only apply to concentrations greater than 1 or even lower than 1?
Thank you  :)

1. By separating the two half-cells, the electrons are forced to travel along conducting wire connecting the half-cells. This flow of electrons in the external circuit is the electricity. Without separating the half-cells, the electrons would simply move through the solution, rather than travelling along wire.
2. Yes, galvanic cells always lose some energy as heat energy. One reason for this is that the wires connecting the half-cells have a resistance.
3. I believe some redox reactions are indeed endothermic, although many, including combustion reactions, are exothermic.
4. We only use catalytic electrodes in fuel cells because reactions in fuel cells tend to be particularly slow, whereas reactions in other types of cells are not so slow. We could use catalysts in other types of cells, but this is expensive and not necessary.
5. You should always use the states that are indicated in the particular question, NOT by the electrochemical series. A good example is the VCAA 2014 exam where magnesium and chloride ions were molten, not aqueous as in the series. States do matter.
6. Particularly high concentrations can change relative oxidising and reducing strengths, but I don't believe '1 M' is a magic number for this.

These are really excellent questions, by the way!
« Last Edit: November 06, 2015, 10:14:24 pm by jyce »