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March 29, 2024, 05:01:52 pm

Author Topic: VCE Chemistry Question Thread  (Read 2313606 times)  Share 

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sweetcheeks

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Re: VCE Chemistry Question Thread
« Reply #5985 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 #5986 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 #5987 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 #5988 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 #5989 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 #5990 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?

Syndicate

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Re: VCE Chemistry Question Thread
« Reply #5991 on: January 16, 2017, 08:42:29 pm »
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Isn't energy = n * heat of combustion. Don't you still need to divide by 12?

Yes, but the values listed on the table aren't the heat combustion values. They are the amount of energy released per gram of fuel.

So when you multiply 30 (Kj g^-1) by 573 x 10^6 g, g cancels out, leaving you with only kJ.   
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deStudent

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Re: VCE Chemistry Question Thread
« Reply #5992 on: January 16, 2017, 09:31:33 pm »
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Yes, but the values listed on the table aren't the heat combustion values. They are the amount of energy released per gram of fuel.

So when you multiply 30 (Kj g^-1) by 573 x 10^6 g, g cancels out, leaving you with only kJ.
Oh right, whoops didn't read the table properly.

Ty

cloudz99

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Re: VCE Chemistry Question Thread
« Reply #5993 on: January 19, 2017, 07:30:31 pm »
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I didn't do very well on this topic in the 1/2 exam so finding revision a little difficult!
Please help with this question, not sure what values to use in what equation :)

What volume of 0.460 M H2SO4 is required to neutralise 24.00 mL of 0.620 M NaOH?

Thanks in advance !!

Shadowxo

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Re: VCE Chemistry Question Thread
« Reply #5994 on: January 19, 2017, 09:54:50 pm »
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I didn't do very well on this topic in the 1/2 exam so finding revision a little difficult!
Please help with this question, not sure what values to use in what equation :)

What volume of 0.460 M H2SO4 is required to neutralise 24.00 mL of 0.620 M NaOH?

Thanks in advance !!

So if the H2SO4 is neutralising the NaOH, the n(H+) = n(OH-)
Since you're given both the volume and concentration, you can find out the number of NaOH
n=cV, n=0.620*0.02400 (as volume is in litres)
n=0.01488 = 0.0149 (3 sig figs)
n (OH-) = n (NaOH) = 0.0149
n (H+) = 2*n(H2SO4) as there are two hydrogen ions in each molecule
n(OH-)=n(H+) = 0.0149
n(H2SO4) = 1/2 n(H+) = 0.0744
V(H2SO4) = n/c = 0.0744/0.460 = 0.0162L = 16.2mL

Another route is n(H+) = n(OH-)
c(H+)*V(H+)=c(OH-)*V(OH-)
As the concentration of the H+ ions is twice that of H2SO4,
c(H2SO4)*2*V(H2SO4)=c(NaOH)*V(NaOH)
Then substitute values

So start with the part that you have the most information about, see what you need to do and go from there.
Bit rusty but hope this helps :)
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hodang

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Re: VCE Chemistry Question Thread
« Reply #5995 on: January 23, 2017, 03:07:45 pm »
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Can anyone tell me the correct conversions? (having it drawn would be easier with arrows like a chart)

MJ --> KJ (do you divide it by 10^3)
KJ ---> J , J --> KJ, g--> kG and so on, pretty confused

also anyone can help me with these questions and do working out (the questions are from the heinemann book 2) (i uploaded pics of the questions)

Syndicate

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Re: VCE Chemistry Question Thread
« Reply #5996 on: January 23, 2017, 03:24:17 pm »
+1
Can anyone tell me the correct conversions? (having it drawn would be easier with arrows like a chart)

MJ --> KJ (do you divide it by 10^3)
KJ ---> J , J --> KJ, g--> kG and so on, pretty confused

also anyone can help me with these questions and do working out (the questions are from the heinemann book 2) (i uploaded pics of the questions)

1 MJ = 1000kJ (which means you multiply MJ by 1000 or 10^3 to get kJ)
1kJ = 1000J or 1J = 1/1000 kJ
1kilogram (kg) = 1000g

Q4a/b/c) can't really help you with these, as you haven't posted the table up.
Q4d) 573 tonne = 573000000 g
energy released  = 573000000 x 30 =  1.7 x 10^10 kJ

Q5a) CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(l)   deltaH = -890 kJ mol^-1

Q5b)
Q = mcT(change in temperature)
Q = 500 x 4.18 x 80 =  167200 J or 167.2 kJ

This means that CH4 released 167.2  kJ of energy

Energy = n x Hc
-167.2= n x -890
therefore n = 0.187865 mol

M(CH4) = 16
m(CH4) = 0.187865  x 16 = 3.01 grams
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hodang

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Re: VCE Chemistry Question Thread
« Reply #5997 on: January 23, 2017, 08:49:13 pm »
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HEY thanks so much for answering Syndicate, the questions you've answered are the ones i needed

But for 4(d) - how did you get 573 tonnes to = 573 000 000 g  (we have to find mol right?, but how can you for dried brown coal or do we just have to convert the tonnes to g? if so did you use any formulas or?)
As for energy released, the formula is n x Hc, i know how you got 30, so was 573 000 000 your mol? (haha sorry for confusion,, my question is how did you get that as your mol)

P.S - ive always wanted to know, the heat of combustion for brown coal dried is -30, in every question, its a positive, and the minus sign is not included. Any reasons as to why this is the case?

Shadowxo

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Re: VCE Chemistry Question Thread
« Reply #5998 on: January 23, 2017, 09:29:18 pm »
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HEY thanks so much for answering Syndicate, the questions you've answered are the ones i needed

But for 4(d) - how did you get 573 tonnes to = 573 000 000 g  (we have to find mol right?, but how can you for dried brown coal or do we just have to convert the tonnes to g? if so did you use any formulas or?)
As for energy released, the formula is n x Hc, i know how you got 30, so was 573 000 000 your mol? (haha sorry for confusion,, my question is how did you get that as your mol)

P.S - ive always wanted to know, the heat of combustion for brown coal dried is -30, in every question, its a positive, and the minus sign is not included. Any reasons as to why this is the case?

1 tonne = 1000kg
1kg = 1000g
so 1 tonne = 1000*1000 = 1,000,000
573 tonnes = 573,000,000g
And in the table, it shows kJg-1 aka kJ per gram, not moles (often it will be mol though for these kinds of questions)
So 573,000,000g at 30kJ a gram = 1.7*1010kJ released

With the negative, it's probably showing the change in enthalpy, the difference in energy in the products vs reactants so if it's negative, the products have less energy than reactants, and the energy that is lost is lost as heat.
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Syndicate

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Re: VCE Chemistry Question Thread
« Reply #5999 on: January 23, 2017, 09:51:16 pm »
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HEY thanks so much for answering Syndicate, the questions you've answered are the ones i needed

But for 4(d) - how did you get 573 tonnes to = 573 000 000 g  (we have to find mol right?, but how can you for dried brown coal or do we just have to convert the tonnes to g? if so did you use any formulas or?)
 As for energy released, the formula is n x Hc, i know how you got 30, so was 573 000 000 your mol? (haha sorry for confusion,, my question is how did you get that as your mol)

P.S - ive always wanted to know, the heat of combustion for brown coal dried is -30, in every question, its a positive, and the minus sign is not included. Any reasons as to why this is the case?

Hi,

Shadowxo has basically covered it all  ;)

In addition to Shadowxo's answer,
All fuels undergo a combustion reaction to produce energy, which is a type of a exothermic reaction, meaning the energy is lost to the surroundings. A negative Hc indicates that energy is lost (the product(s) will have a lower amount of energy than the reactants).

(Btw I answered the same question at the top of this page  :P )
« Last Edit: January 23, 2017, 10:08:31 pm by Syndicate »
2017: Chemistry | Physics | English | Specialist Mathematics | Mathematics Methods
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