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October 19, 2025, 12:22:10 pm

Author Topic: Unit 4 Questions MEGATHREAD!  (Read 76450 times)  Share 

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Vincezor

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Re: Unit 4 Questions MEGATHREAD!
« Reply #285 on: October 31, 2011, 08:17:38 pm »
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um could someone help explain vcaa 2010 mc q8?

why the hydroxide ions increase when diluting HCl ...

I'm not sure if this is the best way to explain it, but this was my thought process:

When you dilute HCl with Water, the [H+] will decrease as a result of increased volume. Because it says 'distilled' water is added at a constant temp, that makes me think about the Kw value of water. Kw = [H+][OH-] = a constant

As temp is not changing, this value doesn't change. So if [H+] decreases, then [OH-] increases...

However, I wasn't very confident with the question; maybe you should wait for other responses just in case!
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Re: Unit 4 Questions MEGATHREAD!
« Reply #286 on: October 31, 2011, 08:43:02 pm »
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Hey guys, STAV 2008, MC 2
When a freshly cleaned piece of Mg is placed in de-oxygenated water
A. No reaction
B. Hydrogen bubble only would be formed
C. Hydroxide ions only would be formed
D. Both hydrogen gas and hydroxide ions are formed.

I said B, but the answer is D and I don't really get why. I'm guessing it has to do with me not knowing what de-oxygenated water is..
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mickeymouse

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Re: Unit 4 Questions MEGATHREAD!
« Reply #287 on: October 31, 2011, 09:16:35 pm »
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de-oxygenated shouldn't make a difference to the redox equations that will occur
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HenryP

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Re: Unit 4 Questions MEGATHREAD!
« Reply #288 on: October 31, 2011, 09:19:26 pm »
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I think it will make a difference.
Since there is no oxygen the reaction can't occur.
Instead will occur.
Therefore answer is D
« Last Edit: October 31, 2011, 09:22:15 pm by HenryP »
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mickeymouse

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Re: Unit 4 Questions MEGATHREAD!
« Reply #289 on: October 31, 2011, 09:29:06 pm »
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but I didn't think oxygen would be a reactant to consider.
if it didn't say "de-oxygenated" wouldn't we still only take into account magnesium metal and water in our databook

i'm confused lol ...
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Re: Unit 4 Questions MEGATHREAD!
« Reply #290 on: October 31, 2011, 09:33:06 pm »
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I think it will make a difference.
Since there is no oxygen the reaction can't occur.
Instead will occur.
Therefore answer is D
ahh ok that makes sense, thankyou!
i didnt really think about it haha, i thought maybe deoxygenated water had no O atoms at all, silly me
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Mao

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Re: Unit 4 Questions MEGATHREAD!
« Reply #291 on: November 01, 2011, 01:37:32 am »
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but I didn't think oxygen would be a reactant to consider.
if it didn't say "de-oxygenated" wouldn't we still only take into account magnesium metal and water in our databook

i'm confused lol ...

In theory, yes, you would only take into account magnesium metal and water.

In practice though, water has a certain amount of dissolved oxygen, and you will find the reaction doesn't proceed exactly as predicted.

Whilst it isn't something VCE students need to think about, the people writing the questions must make sure the questions are scientifically correct. Stating that the water is pure/deoxygenated is important.
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mickeymouse

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Re: Unit 4 Questions MEGATHREAD!
« Reply #292 on: November 01, 2011, 09:32:07 am »
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ahh okay thanks Mao, makes sense
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Re: Unit 4 Questions MEGATHREAD!
« Reply #293 on: November 01, 2011, 05:27:46 pm »
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Quick question:

It's given that metal solids will always act as a reductant. Do metal oxides always (or, if not always, usually) act as an oxidant?

(Namely in regards to electrochemistry)
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Re: Unit 4 Questions MEGATHREAD!
« Reply #294 on: November 01, 2011, 06:04:01 pm »
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Yup, because they can be reduced back to the metal.
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Re: Unit 4 Questions MEGATHREAD!
« Reply #295 on: November 01, 2011, 07:31:36 pm »
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In that case, how about something like, say, iron (II) oxide and iron (III) oxide?

I don't know if this is actually possible, but theoretically using both would allow you to obtain a metal oxide which is actually oxidised, right?
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Re: Unit 4 Questions MEGATHREAD!
« Reply #296 on: November 01, 2011, 08:24:39 pm »
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Quick question:

It's given that metal solids will always act as a reductant. Do metal oxides always (or, if not always, usually) act as an oxidant?

(Namely in regards to electrochemistry)

Fe2+ can act as either. It really depends on what reactions they participate in.

In that case, how about something like, say, iron (II) oxide and iron (III) oxide?

I don't know if this is actually possible, but theoretically using both would allow you to obtain a metal oxide which is actually oxidised, right?
Again, depends on the reductant. You can use Al(s), which will reduce both Fe2+ and Fe3+. Or you could use Pb(s) which will only reduce Fe3+ --> Fe2+, but will not reduce Fe2+ any further.

Just remember it's always the strongest oxidants reacting with strongest reductants. The roles they play can always be determined from the electrochemical series.
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Re: Unit 4 Questions MEGATHREAD!
« Reply #297 on: November 01, 2011, 08:46:46 pm »
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Thanks Mao and Thushan :D that clarifies a lot.

Oh, and with the strongest oxidants reacting with strongest reductants, I think that there are always possibilities for side reactions besides just the strongest oxidants reacting with strongest reductants, sheerly due to the fact that the reductant may simply collide with something which is a stronger oxidant but not necessarily the strongest; although obviously the former will be preferential.
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Re: Unit 4 Questions MEGATHREAD!
« Reply #298 on: November 03, 2011, 10:43:14 pm »
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D is wrong.

Liquids and solids are NEVER included in an equilibrium expression. The reasons are to do with something called 'activity.'

You're right HenryP it's B.

Sorry to bring back a discussion that happened a while ago.

I just did the VCAA 2010 exam and in question 2, they give you an equation that has two liquids as reactants. When using the Ka value in this situation, shouldn't the concentration of lactic acid (as it is a liquid) be omitted from Ka calculations?

P.S. That was likely a very noob question. Sorry in advance haha.

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Re: Unit 4 Questions MEGATHREAD!
« Reply #299 on: November 03, 2011, 10:49:09 pm »
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I see what you're getting at but I think thats just there to illustrate the reaction that takes place. The question tells you that the lactic acid is dissolved in water (4.5g in 500ml), this makes it an aqueous solution and will be included in the Ka calculation
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