VCE Chemistry Question Thread

[tashhhaaa]

is the industrial chemical eg. sulfuric acid that we studied examinable?

[Sine]

is the industrial chemical eg. sulfuric acid that we studied examinable?

As far as I know it will not be on exams from 2013+ but may be on VCAA exams before(and including) 2012

Will we be asked about the industrial chemicals?
For further confirmation^

[paper-back]

--> calibration factor for 1000 mL solution = 45.3125 / 10 = 4.53125 kJ K^-1
However, with ethanol the volume of water is double, meaning that now double the energy is required to raise the temperature by 1 K
--> calibration factor for 2000 mL solution = 4.53125 x 2 = 9.0625 kJ K^-1

Thanks Jyce!
Is it incorrect to do this?:
Calibration factor = 4.53KJ/K, as Energy(water) =4.18KJ/K, then CF of components = 0.35KJ/K (4.53KJ/K-4.18KJ/K) and then work from this?

[jyce]

Thanks Jyce!
Is it incorrect to do this?:
Calibration factor = 4.53KJ/K, as Energy(water) =4.18KJ/K, then CF of components = 0.35KJ/K (4.53KJ/K-4.18KJ/K) and then work from this?

I'm not sure what you mean, exactly. How about you try working it out the way you're thinking and see if you get the same answer I did?

[paper-back]

I'm not sure what you mean, exactly. How about you try working it out the way you're thinking and see if you get the same answer I did?

Sorry,

I did:
Calibration factor = 4.53KJ/K, as Energy(water) (SHC x 1000 x (change in)K) =4.18KJ/K, then CF of components = (4.53KJ/K-4.18KJ/K) 0.35KJ/K
The volume of water is increased to 2000ml so it'll be:
4.18 x 2000 x K = 8.36KJ/K
8.36KJ/K + (CF of components) 0.35KJ/K = (new CF) 8.71KJ/K
8.71KJ x 6.81K = 59.3KJ
59.3/n(Ethanol)= 1364KJ

Is this wrong?

[jyce]

Sorry,

I did:
Calibration factor = 4.53KJ/K, as Energy(water) (SHC x 1000 x (change in)K) =4.18KJ/K, then CF of components = (4.53KJ/K-4.18KJ/K) 0.35KJ/K
The volume of water is increased to 2000ml so it'll be:
4.18 x 2000 x K = 8.36KJ/K
8.36KJ/K + (CF of components) 0.35KJ/K = (new CF) 8.71KJ/K
8.71KJ x 6.81K = 59.3KJ
59.3/n(Ethanol)= 1364KJ

Is this wrong?

Huh, yes that would be correct. In my answer, I neglected to distinguish between the water and material components of the calibration factor.
I'm going to remove my answer, since yours is correct.
Well done

Note that this sort of question is unlikely to pop up in the VCAA exam. At least, I've never seen it pop up in a VCAA exam before, only in commercial papers.

[paper-back]

Huh, yes that would be correct. In my attempt, I neglected to distinguish between the water and material components of the calibration factor.
I'm going to remove my answer, since yours is correct.
Well done

Note that this sort of question is unlikely to pop up in the VCAA exam. At least, I've never seen it pop up in a VCAA exam before, only commercial papers.

Thanks for confirming! 

Does anything happen to the operation of a galvanic cell, if a negative terminal of a battery is connected to the positive terminal of the cathode of the galvanic cell, and the positive terminal is attached to the other electrode?

[jyce]

Thanks for confirming! 

Does anything happen a galvanic cell, if a negative terminal of a battery is connected to the positive terminal of the cathode of the galvanic cell, and the positive terminal is attached to the other electrode?

I think you'd just stuff up the cell

EDIT: Unless you actually mean an electrolytic cell, in which case the cell reaction would simply not occur.

[paper-back]

I think you'd just stuff up the cell

EDIT: Unless you actually mean an electrolytic cell, in which case the cell reaction would simply just stop occurring.

Would it increase the rate of reaction or something in the cathode if we're providing it also with electrons through the negative terminal of a battery?

[jyce]

Would it increase the rate of reaction or something in the cathode if we're providing it also with electrons through the negative terminal of a battery?

Well, you wouldn't want to connect the negative terminal of a battery to the positive electrode of a galvanic cell. You'd disrupt the flow of electrons, so no the rate would not increase. And if you connected the negative terminal of a battery to the negative electrode, and the positive to the positive, and if the products of the cell reaction have remained in contact with the electrodes then what you'd be doing is reversing the cell reaction (i.e. recharging the cell).

[jyce]

Hey guys,
I have an exam tomorrow afternoon and then I'm just chilling at home for the night, so tomorrow night I figure I'll keep this thread open and help you guys out with any last minute questions you have

[bonjour-sarah]

Hey guys,
I have an exam tomorrow afternoon and then I'm just chilling at home for the night, so tomorrow night I figure I'll keep this thread open and help you guys out with any last minute questions you have

Thanks for everything, you've already been an absolute incredible help to everyone!

[Mc47]

How do I input a number like 1x10^-2.1 into my calculator?

I can do 1x10^-2, but when i try to put '2.1' it comes up at '21'. Am I missing something?

[qwerty123456]

Why are enzymes ineffective at low temperatures?

[jyce]

How do I input a number like 1x10^-2.1 into my calculator?

I can do 1x10^-2, but when i try to put '2.1' it comes up at '21'. Am I missing something?

Uhh, are you forgetting to put in the '.'?? It could depend on your calculator, sorry.