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Author Topic: VCE Chemistry Question Thread  (Read 2313312 times)  Share 

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keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #5970 on: January 06, 2017, 10:00:36 pm »
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Thanks EulerFan,

I'm still a bit confused. What should I be reading off the graph with the method you suggested? So I went n/(Mr*V) = 0.250/31 /(31*250E-3), then *1000 which gave me 0.104? If I had to read off the graph, I got C = 0.1M -> g/L, multiply by 31. This wouldn't be correct either?

I have a feeling that this is beyond what should be expected of VCE...

Reverse engineering the answer they got, I find that they saw a concentration of 0.41 mM, which is obviously wrong looking at the graph - in fact, this is four times (and three magnitudes) what you thought it should be. Now, assuming they accidentally put mg instead of g, this is only four times what you'd expect. But, it turns out that in a molybdenum phosphate complex, there are actually 4 phosphate ions, which would mean that the concentration of phosphate IS 4*0.10 (as the concentration is by complex, not by phosphate). You are not expected to know this, because nobody would know this without further information (or looking it up like I did/they just happen to know the compound beforehand).

tl;dr - I wouldn't worry, this seems beyond VCE level.

Gogo14

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Re: VCE Chemistry Question Thread
« Reply #5971 on: January 07, 2017, 04:34:03 pm »
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How do you do this, question? Cant seem to get the answer (c)
 
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Syndicate

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Re: VCE Chemistry Question Thread
« Reply #5972 on: January 07, 2017, 04:47:13 pm »
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How do you do this, question? Cant seem to get the answer (c)

« Last Edit: January 07, 2017, 04:48:53 pm by Syndicate »
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sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #5973 on: January 07, 2017, 04:55:19 pm »
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How do you do this, question? Cant seem to get the answer (c)
You need to find the OH- concentration of the solution first. We know that pH is a direct measurement of the H+ concentration of a solution. At 25 degrees the [H+][OH-] equals 10^-14.  10^-pH will give us the H+ concentration (10^-13). We can solve the [OH-]=10^-14/10^-13 which gives an OH- concentration of 0.1M. OH- and NaOH are in a 1:1 ratio, so the concentration of OH- will be equal to the concentration of NaOH. Now it is a matter of using C=n/V and n=m/Mw to determine the mass of sodium hydroxide.

Gogo14

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Re: VCE Chemistry Question Thread
« Reply #5974 on: January 07, 2017, 09:03:49 pm »
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Also, how can you have a negative pH value??
Brain is exploding, is this possible for a pH value over 14 as well?
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keltingmeith

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Re: VCE Chemistry Question Thread
« Reply #5975 on: January 07, 2017, 09:25:57 pm »
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Also, how can you have a negative pH value??
Brain is exploding, is this possible for a pH value over 14 as well?

Yup - remember the definition of pH:



If the concentration of protons is greater than 1 or less than 10^(-14), you get a pH below 0 or above 14.

sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #5976 on: January 07, 2017, 09:36:46 pm »
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Also, how can you have a negative pH value??
Brain is exploding, is this possible for a pH value over 14 as well?
Lets think about it mathematically. pH=-log(H+) and subsequently 10^-pH =H+ concentration. If the H+ concentration is 1.5M, -log(1.5) will be equal to -0.176. The negative sign in front of the log makes it confusing, but if you were to divide both sides by -1, the equation would become 0.176=log(1.5), it can now be seen easier that 10 to the power of what is equal to 1.5. Another way we can look at it is 10^-pH = H+, subbing in the pH as -0.176 will cancel the negative and the equation will become 10^0.176 = H+. It may be beneficial to ask your maths teacher when you get back, as logarithms and exponentials play a big role in methods and they may be able to explain it a lot better.

The same does apply in the higher end, you can have a basic solution that has a pH above 14.

Syndicate

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Re: VCE Chemistry Question Thread
« Reply #5977 on: January 10, 2017, 06:34:25 pm »
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Hey guys,

How much do we have to know about balancing overall redox equations under alkaline conditions? 
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sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #5978 on: January 10, 2017, 07:12:29 pm »
+1
Hey guys,

How much do we have to know about balancing overall redox equations under alkaline conditions?
You will need to know how to balance during alkaline conditions, acidic conditions are much more common but alkaline conditions have appeared on the exam a few times. Once you see a couple of examples they become quite easy If I remember correctly you use the KOHES method but after balancing the H+ you add the same amount of OH- to each side (enough to cancel out the H+). I'll see if I can dig up one of the examples my teacher showed me.

Gogo14

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Re: VCE Chemistry Question Thread
« Reply #5979 on: January 10, 2017, 10:19:23 pm »
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You will need to know how to balance during alkaline conditions, acidic conditions are much more common but alkaline conditions have appeared on the exam a few times. Once you see a couple of examples they become quite easy If I remember correctly you use the KOHES method but after balancing the H+ you add the same amount of OH- to each side (enough to cancel out the H+). I'll see if I can dig up one of the examples my teacher showed me.
What does KOHES stand for?
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sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #5980 on: January 10, 2017, 10:42:32 pm »
+1
What does KOHES stand for?
Key- balance key substances (everything but hydrogen and oxygen)
Oxygen- Balance oxygens with water
Hydrogen- balance hydrogens with H+
Electrons- balance the electrons (add electrons to once side to balance the charge
States- use states

Using the reduction of Cr2O72- (dichromate) to Cr3+ as an example
Cr2O72- --> Cr3+
K Cr2O72- --> 2Cr3+
O Cr2O72- --> 2Cr3+ + 7H2O
H Cr2O72- + 14H+ --> 2Cr3+ + 7H2O
E Cr2O72- + 14H+ + 6e- --> 2Cr3+ + 7H2O
S Cr2O72-(aq) + 14H+(aq) + 6e- --> 2Cr3+(aq) + 7H2O(l)

hodang

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Re: VCE Chemistry Question Thread
« Reply #5981 on: January 14, 2017, 07:28:18 pm »
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Anyone can help a friend out and give me the answers to
http://www.maribsc.vic.edu.au/sites/default/files/files/Unit%201%20Chemistry%202(1).pdf

So I can double check if what I've done is correct? Thanks heaps! xx

Gogo14

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Re: VCE Chemistry Question Thread
« Reply #5982 on: January 15, 2017, 08:24:25 pm »
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confused about anode and cathode terminology.
Which way do electrons travel?
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Syndicate

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Re: VCE Chemistry Question Thread
« Reply #5983 on: January 15, 2017, 08:26:27 pm »
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confused about anode and cathode terminology.
Which way do electrons travel?

Electrons travel from the anode (where oxidation occurs) to the cathode (where reduction occurs).
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Gogo14

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Re: VCE Chemistry Question Thread
« Reply #5984 on: January 15, 2017, 08:39:50 pm »
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Electrons travel from the anode (where oxidation occurs) to the cathode (where reduction occurs).
http://chemistry.about.com/od/electrochemistry/a/How-To-Define-Anode-And-Cathode.htm
this website says "The cathode is the source of electrons or an electron donor"
So......is it wrong?
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