2010 VCAA Chemistry U4 Suggested solutions

[Lighties]

They obviously weren't looking for that approach as it yields two answers..

The was I looked at it was A-factor of 2; B-factor of 2; C-no change; D - reduce by a factor of 4. Clearly the one which has undergone the biggest 'change' is D.

I don't really get how you looked at it =| But they asked for the 'percentage change in the concentration fraction', so wouldn't a change in the denominator of Q (as in D) affect it less than the same change in the numerator of Q (such as A)?

[a.zirek]

With question 2.2ai, when calculating Ka, do we include liquids? If not, then I don't think we include the molarity of the Lactic acid

[coletrain]

Thanks a lot for this Mao. 

[year12]

Thanks Mao for posting this up. Wasn't sure if I should go through and see how many marks I lost, but I just couldn't resist the temptation. Just a quick question.... The first Q of section 2, do we have to put M^2 next to the conc. fraction? (I was thinking.. maybe I should, but then, thought it's only a one mark question, so thought I'd get away with it!)

[d-ea-6]

With question 1.b.iii where methane and hydrogen are "burned", is that the same as combustion?

Because I assumed it was so I wrote down that

CH4 + 2 O2 → CO2 + 2 H2O and
2 H2 + O2 → 2 H2O

Therefore, the answer for the volume of hydrogen needed would be double the volume that Mao has given (12.4 L)

[year12]

With question 1.b.iii where methane and hydrogen are "burned", is that the same as combustion?

Because I assumed it was so I wrote down that

CH4 + 2 O2 → CO2 + 2 H2O and
2 H2 + O2 → 2 H2O

Therefore, the answer for the volume of hydrogen needed would be double the volume that Mao has given (12.4 L)

Yeah, burning = combustion but I don't get your logic. Are you saying that the volume needs to double because there is a coefficient of 2 in front of hydrogen...?

[d-ea-6]

Yes

[andy456]

I have a question too
for the lithium + water equation i mind blanked and wrote this

is this wrong??

[year12]

Yes

Oh, okay. Now I get you. But you don't need to worry about that. All you need to do is get the required energy that needs to be produced by burning hydrogen. Then, divide that by 286 which gives you the no. of mole of H2 you need which can be used to find the required volume (Just like Mao has shown in his solution) because 286 is for 1 mole of H2 burnt. Hope I make some sense!

[year12]

I have a question too
for the lithium + water equation i mind blanked and wrote this

is this wrong??

well.. it definitely isn't 'wrong'. It just isn't simplified as such. You just have the two ions combined to get 2LiOH! Doubt you will lost marks, the significance of this Q was to show that H2 is formed -> high pressure -> explosion.

[andy456]

I have a question too
for the lithium + water equation i mind blanked and wrote this

is this wrong??

well.. it definitely isn't 'wrong'. It just isn't simplified as such. You just have the two ions combined to get 2LiOH! Doubt you will lost marks, the significance of this Q was to show that H2 is formed -> high pressure -> explosion.

Ok so if my reaction is 'technically' right i should get the marks??

[Greggler]

one would hope so. I did the same thing, and just said gas is evolved very quickly, thus potential for kabooom

[andy456]

would that question be 2 for the equation and 1 for why it exploded, or the other way around

[m@tty]

would that question be 2 for the equation and 1 for why it exploded, or the other way around

Yeah, that sounds right. I dunno how they could give two for explosion..

[stonecold]

would that question be 2 for the equation and 1 for why it exploded, or the other way around

Yeah, that sounds right. I dunno how they could give two for explosion..

I said both of hydrogen gas is highly flammable/explosive and high pressure due to build up of hydrogen gas could lead to an explosion.