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February 27, 2017, 05:46:35 pm

Author Topic: VCE Chemistry Question Thread  (Read 301504 times)  Share 

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sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #5985 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 #5986 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 #5987 on: January 10, 2017, 07:12:29 pm »
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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 #5988 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?

sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #5989 on: January 10, 2017, 10:42:32 pm »
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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 #5990 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 #5991 on: January 15, 2017, 08:24:25 pm »
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confused about anode and cathode terminology.
Which way do electrons travel?

Syndicate

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Re: VCE Chemistry Question Thread
« Reply #5992 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 #5993 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?

sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #5994 on: January 15, 2017, 08:42:48 pm »
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confused about anode and cathode terminology.
Which way do electrons travel?
Anode- site of oxidation. Cathode- site of reduction. In a galvanic cell (spontaneous redox) the anode is the negative electrode and the cathode is the positive. However in an electrolytic cell (non-spontaneous redox) the anode is the positive electrode. Electrons always travel from the anode to cathode. What confuses most people is the inverse polarities of galvanic and electrolytic cells.

sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #5995 on: January 15, 2017, 08:46:37 pm »
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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?
This specifically refers to electrolytic cells. In an electrolytic cell, a power supply is connected to two electrodes, the positive (anode) and negative (cathode). The cathode is the site of reduction, which requires electrons. The cathode has electrons available for ions to take.

If you had a solution of copper (ii) sulphate and you passed through a current, the Cu2+ ions would flow to the negative electrode (Cathode) and take some of the electrons.

deStudent

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Re: VCE Chemistry Question Thread
« Reply #5996 on: January 15, 2017, 10:43:48 pm »
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For this q http://m.imgur.com/MZw8xL9

4d:
Can someone confirm that the answer is -1.43E9 kJ, opposed to 1.7E10 kJ as the answer suggests?

Jakeybaby

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Re: VCE Chemistry Question Thread
« Reply #5997 on: January 16, 2017, 07:11:56 pm »
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For this q http://m.imgur.com/MZw8xL9

4d:
Can someone confirm that the answer is -1.43E9 kJ, opposed to 1.7E10 kJ as the answer suggests?
Do you have the tables that are listed in the question?
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Syndicate

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Re: VCE Chemistry Question Thread
« Reply #5998 on: January 16, 2017, 07:24:22 pm »
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For this q http://m.imgur.com/MZw8xL9

4d:
Can someone confirm that the answer is -1.43E9 kJ, opposed to 1.7E10 kJ as the answer suggests?

Dried brown coal releases 30 kJ per gram, which is the same as 30 GJ (\( 10^6 \) kJ) per tonne.

30 x 573 = 17190 GJ = 1.7 x \( 10^{10} \) kJ (2 sig figs)
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deStudent

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Re: VCE Chemistry Question Thread
« Reply #5999 on: January 16, 2017, 08:25:21 pm »
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Dried brown coal releases 30 kJ per gram, which is the same as 30 GJ (\( 10^6 \) kJ) per tonne.

30 x 573 = 17190 GJ = 1.7 x \( 10^{10} \) kJ (2 sig figs)
Isn't energy = n * heat of combustion. Don't you still need to divide by 12?