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May 08, 2024, 06:21:42 pm

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soccerboi

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Re: Soccerboi's unit 4 questions thread
« Reply #15 on: September 10, 2012, 06:24:07 pm »
0
Can someone please help me understand the reason for VCAA 2008 exam 2, Q3 b)? how do you know whether its the lowest or highest possible value?

Thanks in advance
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Lasercookie

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Re: Soccerboi's unit 4 questions thread
« Reply #16 on: September 10, 2012, 07:08:01 pm »
+2
"Which acid must have more than one acidic hydrogen per molecule? Give a reason for your answer"

The solutions are all 0.10M. We can see that the pH's vary, that is some donate their ion more so than others. For a monoprotic acid, at best they can completely donate a single ion completely. In this case, we'll have 



That's the lowest pH we could get out of a monoprotic acid with 0.10M of acid.

So in this case, if we were to have a pH less than 1, that would mean we'd need a . To get that we'd need more hydronium ions being donated -- in other words more than one acidic hydrogen per molecule.

Hence, Acid III, as that has a pH of 0.7.

Interestingly, if we solve in general:












So when pH is less than one, then for any volume. You could deduce that it's Acid III using the same reasoning as before - we only have 0.10M of acid, so a monoprotic acid could at best have a pH equal to 1.
« Last Edit: September 10, 2012, 07:09:39 pm by laseredd »

soccerboi

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Re: Soccerboi's unit 4 questions thread
« Reply #17 on: September 13, 2012, 08:35:19 pm »
0
VCAA 2008
Question 16
When comparing the electrolysis of molten NaF and that of a 1.0 M aqueous solution of NaF, which one of the
following statements is correct?
A. The product at the anodes is the same in both cells and the product at the cathodes is the same in both
cells.
B. The product at the anodes is the same in both cells but the products at the cathodes are different.
C. The product at the cathodes is the same in both cells but the products at the anodes are different.
D. The products at the cathodes of the cells are different and also the products at the anodes are different.

Can someone please explain this question to me, my teacher couldn't explain it =.=
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soccerboi

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Re: Soccerboi's unit 4 questions thread
« Reply #18 on: September 13, 2012, 08:53:14 pm »
0
Also, on the assessment report for VCAA 2008 Q 5, does anybody know what the colour coding on the table for sulfuric acid indicate?
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soccerboi

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Re: Soccerboi's unit 4 questions thread
« Reply #19 on: September 17, 2012, 07:21:04 am »
0
Which terminal of a rechargeable galvanic cell connects to the negative terminal of the power supply when its recharging? The positive terminal of the galvanic cell?
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joseph95

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Re: Soccerboi's unit 4 questions thread
« Reply #20 on: September 17, 2012, 09:34:13 pm »
+1
VCAA 2008
Question 16
When comparing the electrolysis of molten NaF and that of a 1.0 M aqueous solution of NaF, which one of the following statements is correct?
A. The product at the anodes is the same in both cells and the product at the cathodes is the same in both
cells.
B. The product at the anodes is the same in both cells but the products at the cathodes are different.
C. The product at the cathodes is the same in both cells but the products at the anodes are different.
D. The products at the cathodes of the cells are different and also the products at the anodes are different.

Can someone please explain this question to me, my teacher couldn't explain it =.=

For this question, you have to determine for each electrolysis what can possibly act as a reductant or an oxidant.

For molten NaF, the only oxidant present is Na+, and the only reductant present is F-.
Thus at the anode, F- ions will be oxidised to F2, and at the cathode Na+ ions will be reduced to Na.

Anode: F-(aq) --> F2(g) + 2e
Cathode: Na+ + e --> Na(s)

For an aqueous solution of NaF, Na+, F- and H2O are present.
H2O is a stronger oxidant than Na+, so water instead will be reduced at the cathode. H2O is also a stronger reductant than F-, and thus will be oxidised at the anode.

Anode: 2H2O(l) -->O2(g) + 4H+(aq) + 4e
Cathode: 2H2O(l) + 2e --> H2(g) + 2OH-(aq)

Thus the answer is D: the products at the cathodes and anodes are different.
 

charmanderp

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Re: Unit 4 questions thread
« Reply #21 on: September 17, 2012, 09:56:30 pm »
0
Which terminal of a rechargeable galvanic cell connects to the negative terminal of the power supply when its recharging? The positive terminal of the galvanic cell?
In recharging a cell, the negative terminal of the power supply is connected to the cathode. This is where reduction occurs. Since reduction is here the reverse process of oxidation, that means the cathode was previously the anode, which was the negative terminal of the galvanic cell.
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soccerboi

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Re: Unit 4 questions thread
« Reply #22 on: September 20, 2012, 07:03:23 pm »
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why isn't natural gas considered to be renewable? Can't we or cows just fart to make more? LOL
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charmanderp

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Re: Unit 4 questions thread
« Reply #23 on: September 20, 2012, 07:16:19 pm »
+2
notsureifsrs.jpg
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Lasercookie

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Re: Unit 4 questions thread
« Reply #24 on: September 20, 2012, 07:42:04 pm »
+1
why isn't natural gas considered to be renewable? Can't we or cows just fart to make more? LOL
I like this question. I'm not too sure, but I'll have a go at answering. Be aware that there might be inaccuracies here.

Natural gas is the mixture of hydrocarbon gases, it's extracted from natural gas reserves / oil reserves. "Oil and natural gas are produced by the same geological process according fossil fuel suggestion: anaerobic decay of organic matter deep under the Earth's surface". I believe when we say 'natural gas', I think that's what we refer to.

http://en.wikipedia.org/wiki/List_of_natural_gas_fields
http://en.wikipedia.org/wiki/Natural_gas_field
"In common usage, deposits rich in oil are known as oil fields, and deposits rich in natural gas are called natural gas fields."

http://en.wikipedia.org/wiki/Natural_gas#Sources

In fear of ruling out on what might be a revolutionary energy source that will fix the world's problems, I would be inclined to think that farts are not an efficient power source to harness. I'm not sure how you would go about it. http://en.wikipedia.org/wiki/Flatulence#Composition

Recall that "non-renewable" means that the rate of depletion is much greater than the rate of production.

I'm not entirely sure on this but stuff like Petroleum (http://en.wikipedia.org/wiki/Petroleum#Formation) is still being formed, just at a very very very slow rate (centuries, millions of years seem to be the figures thrown around from quick googling) If that implies that fossil fuels supply may one day - one day meaning a very very long time away - replenish themselves, I'm not sure. Either way, that won't be something that would happen in the next few centuries.

Also I just realised why fossil fuels are called fossil fuels. Now what was I talking about?

Oh yes, so there's Biogas, which is formed from from decaying "organic matter". That's a renewable source as you know.

There's also this http://en.wikipedia.org/wiki/Renewable_natural_gas
http://en.wikipedia.org/wiki/Substitute_natural_gas
« Last Edit: September 20, 2012, 07:43:54 pm by laseredd »

soccerboi

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Re: Unit 4 questions thread
« Reply #25 on: September 25, 2012, 09:32:19 pm »
0
The nickel–cadmium cell is a rechargeable cell used in mobile phones, laptop computers and many other portable electronic devices.
The two relevant half-equations for the discharge reaction are
Cd(s) + 2OH–(aq) → Cd(OH)2(s) + 2e–
NiOOH(s) + H2O(l) + e– → Ni(OH)2(s) + OH–(aq)

a. Will the pH in the region immediately surrounding the anode increase or decrease as the cell discharges? Give a reason for your answer.

Solution
Decrease because hydroxide ions are being consumed, meaning that the [OH–] is decreasing.

Isnt OH- also being produced in the Cd(OH)2(s)?
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Re: Unit 4 questions thread
« Reply #26 on: September 25, 2012, 10:09:33 pm »
+1
I think it has something to do with it asking for the region immediately surrounding the anode, where oxidation will occur, so only the top equation will happen immediately surrounding the anode.

soccerboi

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Re: Unit 4 questions thread
« Reply #27 on: September 26, 2012, 08:03:13 am »
0
A beaker containing 250 mL of 1.00 M CuSO4(aq) is electrolysed using carbon rods.
a. Write half-equations for the oxidation and reduction reactions.
i. oxidation reaction
ii. reduction reaction
Solution
i. 2H2O(l) → O2(g) + 4H+(aq) + 4e–
ii. Cu2+(aq) + 2e– → Cu(s)

i'm still confused about which water eqn to use? Why have they used that water eqn and not say the one at E value of 0.00?
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charmanderp

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Re: Unit 4 questions thread
« Reply #28 on: September 26, 2012, 05:24:09 pm »
+2
A beaker containing 250 mL of 1.00 M CuSO4(aq) is electrolysed using carbon rods.
a. Write half-equations for the oxidation and reduction reactions.
i. oxidation reaction
ii. reduction reaction
Solution
i. 2H2O(l) → O2(g) + 4H+(aq) + 4e–
ii. Cu2+(aq) + 2e– → Cu(s)

i'm still confused about which water eqn to use? Why have they used that water eqn and not say the one at E value of 0.00?
The equation at 0.00 does not involve water in any way...It's the reduction of H+ ions to form hydrogen gas (or oxidation of hydrogen gas yaddah yaddah).

Since you have Cu2+ acting as an oxidant and this is electrolysis, clearly you need water to act as a reductant.  The occasion where water acts as reductant with the strongest reduction potential is the equation at +1.23 which the solution has used.
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soccerboi

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Re: Unit 4 questions thread
« Reply #29 on: September 26, 2012, 05:45:15 pm »
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hey charmanderp, how did u know the Cu2+ is acting as an oxidant?
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