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June 07, 2025, 06:17:27 pm

Author Topic: VCE Chemistry Question Thread  (Read 2773725 times)  Share 

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Sine

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Re: VCE Chemistry Question Thread
« Reply #5910 on: November 13, 2016, 03:14:20 pm »
+1
http://m.imgur.com/LlQcYek

For this, is the reluctant O2- or H2O? Answer said H2O, don't get why..
The reductant causes reduction but itself is oxidised. F moves from oxidation number of 0 to neg 1. Thus water must be what is causing this, it is one of the reactants.Water is the reductant (O in water moves from oxidation number of -2 to 0)  Oxygen gas is a product hence will not be a reductant or oxidant in this reaction.

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Re: VCE Chemistry Question Thread
« Reply #5911 on: November 13, 2016, 03:42:44 pm »
+1
http://m.imgur.com/LlQcYek

For this, is the reluctant O2- or H2O? Answer said H2O, don't get why..

Reductant (aka reducing agent) refers to the compound/ element which has been oxidised. When an element/ compound undergoes oxidation, it means it has lost electrons, gained oxygen or lost hydrogen atoms. Only a reactant can be reduced or oxidised (O2 is a product like HF). H2O has lost its hydrogen atoms to the F element, therefore H2O is oxidised and is the reducing agent.
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Adequace

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Re: VCE Chemistry Question Thread
« Reply #5912 on: November 14, 2016, 03:07:14 pm »
0
Hey guys, I have another question sorry  :-X

http://m.imgur.com/s0p8ByO

For Q17) why isn't A correct?

Q18) how is the answer B? I don't get where the HCl comes from?

sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #5913 on: November 14, 2016, 03:27:45 pm »
+2
Hey guys, I have another question sorry  :-X

http://m.imgur.com/s0p8ByO

For Q17) why isn't A correct?

Q18) how is the answer B? I don't get where the HCl comes from?
Please keep the questions coming. I have nothing else to do.

Q17) The reaction is not an acid base reaction, rather a redox reaction. Sulfuric acid is a strong acid, but in this case that property is not being displayed, instead it is the ability to act as an oxidising agent.

Q18) this question is exploring the placement of substances on the electrochemical series. When metal X is added to Pb2+, X will be oxidised and Pb2+ reduced. This means that metal X is a stronger reductant than lead metal.

The electrochemical series shows us the line 2H+ + 2e- --> H2, which is above lead (Pb2+ + 2e- --> Pb(s)) , indicating that H+ is a stronger oxidant than Pb2+. From this, we can determine that Pb(s) is a stronger reductant than H2. Now, if we know that Metal X is a stronger reductant than lead, this means that metal X will react with H+ from an acid. HCl is used as a source of the H+.

keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #5914 on: November 14, 2016, 03:36:17 pm »
+2
Hey guys, I have another question sorry  :-X

http://m.imgur.com/s0p8ByO

For Q17) why isn't A correct?

A states that H2SO4 is ACTING (key word) as a strong acid. Yes, it IS a strong acid, but in this reaction, it isn't ACTING as a strong acid. The key is to notice the Magnesium - if you EVER have a metal go from M to MX (i.e., some metal salt), then it WILL have changed oxidation states, and the reaction WILL be redox. Redox reactions are NOT acid base reactions (at least, as far as VCE is concerned. The curious should google Lewis acids/bases, but don't confuse them with Bronsted and VCE!!). Since we know from looking at Mg (which goes from 0 -> +2 in oxidation state), we know that the reaction is redox, and hence NOT acid-base. If it's not an acid-base reaction, then H2SO4 CAN'T be acting as an acid. Instead, it's our oxidant.

Q18) how is the answer B? I don't get where the HCl comes from?

We use the electrochemical series. ;) Remember, spontaneous reactions happen with higher left to lower right compounds.

So, check A - Pb MUST be more reactive than metal X. Not true - in fact, for Pb^2+ + X ----> Pb + X^2+, then Pb must be LESS reactive than X.
Now, B. The important part of HCl to H2 for electrochem is the H+. H+ is only slightly more reactive than Pb2+, so this very quickly seems like a good shot. This could be right, but for completions sake, let's check the others.
Next, C. Metal X must be able to react with Mg(NO3)2 to give Mg precipitate. Not true - fact, if X is Sn, Ni, Co (and so many others), then we will react with Pb^2+, but NOT Mg^2+.
Finally, D. Pb must be able to react with X(NO3)2. This is kinda true - if we supply an electric current, we will get electrolysis, by the nature of electrochemical reactions. However, this is less correct than B, because B is a spontaneous reaction, and it's implied that no current is used by the question.

So, A, C and D are all wrong, and the answer must be B. They try to throw you off by mentioning HCl, when it wasn't in the quesiton. BUT, by going to the electrochemical series, you can see why that might come into play.

EDIT: Beaten by sweetcheeks. :'( Leaving this here as I went through each option and why they're right/wrong.

Adequace

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Re: VCE Chemistry Question Thread
« Reply #5915 on: November 15, 2016, 12:08:46 pm »
0
Thanks for the help sweetcheeks and melting meithy  :)

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Re: VCE Chemistry Question Thread
« Reply #5916 on: November 15, 2016, 12:26:18 pm »
0
Can someone check my (unclear) working here http://imgur.com/i0hFk4g

I found the concentration of the initial solution and then found the concentration of the final solution. Since the mol ratios are all 1:1, I then directly solved for the mass of the OH- ions. I'm not sure if this is correct tho

sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #5917 on: November 15, 2016, 04:15:42 pm »
0
Can someone check my (unclear) working here http://imgur.com/i0hFk4g

I found the concentration of the initial solution and then found the concentration of the final solution. Since the mol ratios are all 1:1, I then directly solved for the mass of the OH- ions. I'm not sure if this is correct tho
You have used the C1V1=C2V2 formula, which in this case doesn't work, as the concentration is not being changing. You initially worked out the concentration of the solution to be 0.3125M. From there, you should use CxV=n 0.3125M x 0.750L to determine how many moles of sodium hydroxide are in the 750ml and then work out the mass of the hydroxide.

Edit: There is also another way to solve this question. You could determine what 1.25% of 750ml is, giving you the mass of NaOH in the solution. You can then calculate the mass fraction mOH-/mNaOH and multiply it by the mass of NaOH to find the mass of OH-.
« Last Edit: November 15, 2016, 04:17:36 pm by sweetcheeks »

Sine

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Re: VCE Chemistry Question Thread
« Reply #5918 on: November 15, 2016, 04:27:08 pm »
0
Can someone check my (unclear) working here http://imgur.com/i0hFk4g

I found the concentration of the initial solution and then found the concentration of the final solution. Since the mol ratios are all 1:1, I then directly solved for the mass of the OH- ions. I'm not sure if this is correct tho
Answer is B
1.25 g in 100mL so 7 x 1.25 = 9.375g in 750mL bottle







Closet is B

Use the C1V1=C2V2 formula only for dilutions where the concentration changes otherwise if you use the same concentration in the formula you will end up in the scenario where you get 0=0 for the equation.
« Last Edit: November 15, 2016, 04:32:57 pm by Sine »

Immune_Sushi

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Re: VCE Chemistry Question Thread
« Reply #5919 on: November 22, 2016, 10:21:58 pm »
0
Hi, im just wondering in what way that I was wrong:

Below is a (very) crude diagram of a cathode ray tube (picture the dots as electrons):
/---------------------------------------------------------\
|                               +            ..................          \
|  -   ......................................                         +     |
|                                _                                        /
|___________________________________/

The saying of AN OIL RIG CAT (Anode for Oxidation losing e- and Cathode for reduction gaining e-) contradicts the fact the the metal cathode at the negative end is gaining electrons since it is giving out electrons? Or does this apply to chemical reactions only?
Under what circumstances would you need AN OIL RIG CAT?
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keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #5920 on: November 28, 2016, 12:20:53 am »
0
Hi, im just wondering in what way that I was wrong:

Below is a (very) crude diagram of a cathode ray tube (picture the dots as electrons):
/---------------------------------------------------------\
|                               +            ..................          \
|  -   ......................................                         +     |
|                                _                                        /
|___________________________________/

The saying of AN OIL RIG CAT (Anode for Oxidation losing e- and Cathode for reduction gaining e-) contradicts the fact the the metal cathode at the negative end is gaining electrons since it is giving out electrons? Or does this apply to chemical reactions only?
Under what circumstances would you need AN OIL RIG CAT?


That's... a phrase I've never heard before. However, you'll be pleased to know that cathode ray tubes do not feature on the study design at all! In fact, the reason that no one's answered you yet I believe is because nobody here knows what a cathode ray tube is. (well wishers, feel free to contradict me) However, I've done some research, and here's my understanding of the matter:

The phrase you're talking about is an electrochemical one - and, it is essentially purely for electrochemical cells. It's not a be all and end all - it's a mnemonic designed to help you remember what the anode and cathode are in a galvanic cell. So, be careful when using it for other things, because they won't make sense!

Now, in your galvanic cell, the cathode is the cell at which reduction occurs. So, what does a cathode ray tube do? It fires a beam of electrons at something. In this sense, you can think of it as creating a "ray of reduction" - hence why it's referred to as a "cathode ray tube".

Having said that, you've mentioned that since the negative part of the tube is donating electrons, that means it's gaining them? Seems a bit nonsensical to me! Remember, these electrons have to come from somewhere, and this is a closed system. So, if any part of the tube were to start with electrons and then donate them, would you expect that to be the negative terminal, or the positive terminal? The negative terminal, of course, so they must be starting from there. The fact that the negative terminal is giving out electrons has nothing to do with it gaining them, it's simply always had them to begin with.

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Re: VCE Chemistry Question Thread
« Reply #5921 on: November 28, 2016, 06:26:45 pm »
0
Thanks that was really helpful!
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Re: VCE Chemistry Question Thread
« Reply #5922 on: December 03, 2016, 08:15:33 pm »
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Hi guys, i'd appreciate any help on this question. . . .
Zinc Oxide powder and Zinc oxide nanoparticles both absorb UV light. What property of Zinc Oxide nanoparticles makes them more suitable than zinc oxide powder for used in sunscreen.
Can someone please explain it?



The answer is: zinc oxide nanoparticles are colourless.

Syndicate

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Re: VCE Chemistry Question Thread
« Reply #5923 on: December 03, 2016, 09:44:33 pm »
+1
Hi guys, i'd appreciate any help on this question. . . .
Zinc Oxide powder and Zinc oxide nanoparticles both absorb UV light. What property of Zinc Oxide nanoparticles makes them more suitable than zinc oxide powder for used in sunscreen.
Can someone please explain it?



The answer is: zinc oxide nanoparticles are colourless.

The question is alluding to how using ZnO nanoparticles in sunscreens can be more suitable for sunscreen users than applying sunscreens with ZnO powder to their skin. So basically ZnO powder particles are relatively larger than ZnO nanoparticles, which causes the sunscreen to have a thick, white texture leaving a white residue on the user's skin; whereas the nanaoparticles would prevent such build of this residue (as the ZnO nanoparticles are relatively small and have no colour), hence why the ZnO nanoparticles would be more suitable to use in sunscreens.

I would have also mentioned how a suncreen containing ZnO nanoparticles would have a larger concentration of ZnO, which would make it more "safer" to use than a sunscreen with ZnO powder. Although larger particles would reflect more UV rays, the size of the particle works inversely proportional to it's concentration in the sunscreen- the larger the size of the particles, the less the concentration of it (So there would be a certain range of the sizes of the ZnO particles that would be the most efficient to use in sunscreens).
« Last Edit: December 03, 2016, 09:59:51 pm by Syndicate »
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peanut

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Re: VCE Chemistry Question Thread
« Reply #5924 on: December 09, 2016, 12:23:58 pm »
0
When working through a question, do I round to appropriate sig figs at each step, or just for the final answer? In most cases, it shouldn't matter, but what is the correct way of doing it? Similarly, do I use a rounded value for the next step in my calculation, or just use the full (as many decimal places as possible) value on my calculator?

Thanks!