VCE Chemistry Question Thread

[Syedali_]

Oh....
Hm, does the book have worked solutions??

Nope this what i done

[rachid.kam]

Nope this what i done

(Image removed from quote.)

Hey Syedali, I have just attached the relevant working out for your question. don't worry, I was in the same boat as you and it really bothered me that night, until my teacher went through it with us due to the ever-mounting complains from the students. And when we worked it out, my teacher concluded that they had done a calculation mistake, hence everybody was getting a different answer. Hope this helps. PM me for any more future questions, as I have actually encountered many questions like this until we finished the unit 3 course.

[Shadowxo]

Yep just clarifying
It should be 15,954 J, aka 1.6*10⁴ J/g, or 16 kJ/g
If using 1 sig fig it would go to 2*10⁴ J or 2*10kJ/g
Their answer is wrong - should be 1.6*10⁴ J not kJ (if they wanted it in J) or 16 kJ/g, and since the wood is to 1 sig fig, it should be rounded to 2*10⁴ J = 2*10 kJ/g

[rachid.kam]

Yep just clarifying
It should be 15,954 J, aka 1.6*10⁴ J,
If using 1 sig fig it would go to 2*10⁴ J
Their answer is wrong - should be 1.6*10⁴ J not kJ and since the wood is to 1 sig fig, it should be rounded to 2*10⁴ J

(To get to kJ just divide by 1,000 or 10³)

ShadowXO, just clarifying. Why did you use J instead of KJ, because the question explicitly states that it wants the first part of the answer in KJ per gram and the second part in MJ per tonne. I'm just a little confused.

[Shadowxo]

ShadowXO, just clarifying. Why did you use J instead of KJ, because the question explicitly states that it wants the first part of the answer in KJ per gram and the second part in MJ per tonne. I'm just a little confused.

Yes sorry I just use J out of habit - I was mainly explaining where the 1.6*10⁴ came from, to put it into kJ it would be 16 aka 2*10 kJ to 1 sig fig. I'll make it a bit clearer

[Syedali_]

ShadowXO, just clarifying. Why did you use J instead of KJ, because the question explicitly states that it wants the first part of the answer in KJ per gram and the second part in MJ per tonne. I'm just a little confused.

Thanks

[Gogo14]

So what does heating do to the pH?
Shouldnt the pH stay the same?

[sweetiepi]

So what does heating do to the pH?
Shouldnt the pH stay the same?

Ionisation occurs more with a rise in temperature. Therefore the equilibrium amount of H+ ions will be increased with temperature. Hence the neutral pH will be lower, as there is more H+ available. At room temperature, pH (neutral) is approximately 7.00 and at 100 degrees Celsius, the pH (neutral) is actually around 6.14.
Hope that clarifies this a little.

[Gogo14]

Ionisation occurs more with a rise in temperature. Therefore the equilibrium amount of H+ ions will be increased with temperature. Hence the neutral pH will be lower, as there is more H+ available. At room temperature, pH (neutral) is approximately 7.00 and at 100 degrees Celsius, the pH (neutral) is actually around 6.14.
Hope that clarifies this a little.

But doesnt OH increase as well?

[deStudent]

http://m.imgur.com/a/W2AQU
For part b) there's a small discrepancy in our answers. We didn't use the same method but we should be still get the same answer? Is there a flaw in my method?

Ty!

[peterpiper]

http://m.imgur.com/a/W2AQU
For part b) there's a small discrepancy in our answers. We didn't use the same method but we should be still get the same answer? Is there a flaw in my method?

Ty!

The solutions got it wrong. They divided 7000 with 2810 when it should've been 2803 kJmol^-1. So they should've ended up with your answer. What you did was fine I think.

[Shadowxo]

But doesnt OH increase as well?

It's a little tricky to wrap your head around.
pH depends on the concentration of H+.
At 25ºC, pH+pOH=14, and [H+][OH-]=10^-14.
As the temperature increases, the concentration of the H+ AND the OH- go up. Water is still "neutral" (same number of H+ and OH-) but it has a higher concentration of H+ and OH-, so it has a higher pH and higher pOH. So even though it's still neutral, the concentration of H+ (and OH-) has gone up and therefore the pH is higher.

http://m.imgur.com/a/W2AQU
For part b) there's a small discrepancy in our answers. We didn't use the same method but we should be still get the same answer? Is there a flaw in my method?

Ty!

It's a mistake in the book, they divided by 2810 instead of 2803, like peterpiper said. Both methods yield the same answer of 449.5 (rounded to 450) g/L.

[deStudent]

Thanks for the helps guys.

Edit:
http://m.imgur.com/a/tQ34l
The answer was D, but I chose B. I'm not sure how B is wrong? It seemed pretty confusing since O2 and H2O2 appear on both sides of the electrochemical series, but the question doesn't tell you which is the oxidant/reductant. A bit lost.

Thanks

[-273.15]

Hi guys,
Im super confused with electrolysis

first of all, I keep seeing electrolytic cells as two half cells like galvanic as well as just one half cell with both electrodes together. For sacs and the exam, which way should we draw it?

Secondly, with galvanic cells, using the electrochemical series you chose the strongest redundant and strongest oxidant and this is spontaneous. For electrolysis, it is not spontaneous and equations often show that the weakest oxidant and weakest redundant have been chosen. However I still see picking the strongest oxidant and redundant in e.g. electrolysis of aqueous ionic compounds. Im really confused with how to use the electrochemical series here

If someone could help id appreciate it heaps
thanks

[Syndicate]

Thanks for the helps guys.

Edit:
http://m.imgur.com/a/tQ34l
The answer was D, but I chose B. I'm not sure how B is wrong? It seemed pretty confusing since O2 and H2O2 appear on both sides of the electrochemical series, but the question doesn't tell you which is the oxidant/reductant. A bit lost.

Thanks

H2O2 is being oxidised (check the 9th row down on your electrochemical series). O2 cannot be oxidised, so H2O2 must be oxidised.
There aren't any H2O being oxidised below H2O2. So B is wrong.

A and C are obviously wrong. Hence D is the correct answer.

Hi guys,
Im super confused with electrolysis

first of all, I keep seeing electrolytic cells as two half cells like galvanic as well as just one half cell with both electrodes together. For sacs and the exam, which way should we draw it?

Secondly, with galvanic cells, using the electrochemical series you chose the strongest redundant and strongest oxidant and this is spontaneous. For electrolysis, it is not spontaneous and equations often show that the weakest oxidant and weakest redundant have been chosen. However I still see picking the strongest oxidant and redundant in e.g. electrolysis of aqueous ionic compounds. Im really confused with how to use the electrochemical series here

If someone could help id appreciate it heaps
thanks

The one with only one half cell with two electrodes is the electrolytic cell.

Weaker reductant = stronger oxidant
Weaker oxidant = stronger reductant

So you can still use the same method.