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hahaha, you oldie and stop stalking me >(

She's generally classified as 'new age' Though I like her songs too :)

Too much propoganda and 'dumbing down' imo. If the curriculum is to be reduced to absolute basics, then more opportunity should be available for students to pursue learning of their own based on their interests - it's the only benefit I can see from reducing the curriculum like this. Less...

Technically you're allowed to, but it's not a good idea - markers will just think you're lazy

YouTube - I will follow you into the dark (new version) "I will follow you into the dark" - Death cab for cutie

Impromptu No.4 in C sharp minor, Op.66 (Fantaisie-Impromptu) :)

No idea, maybe methods of learning such as group discussions? I have never looked at the English syllabus, lol

Some people are very accustomed to rote-learning, writing your own essay requires thinking :P

Complex, graphs, conics.. We're pretty slow :uhoh:

I don't think it's a major issue with subjects such as the sciences, as you're meant to display your understanding of concepts, etc, rather than individually analysing texts

about:blank ftw

Wow, was that difficult? You'd have missed most of the prelim course... Maybe I should've done that, though the school probably wouldn't have let me.

That makes a lot of sense, thanks! To clarify, is enthalpy change always measured per mol? I thought it was just heat change at a constant pressure.

From Jacaranda Chemistry 2: [Strong acid - strong base titration] The equivalence point is at the centre of the inflection (pH = 7). All three indicators change colour within this steep portion of the pH curve. Although the bromothymol blue end-point most closely matches the pH at the...

*joins bandwagon* Business studies was horrible.

That's what I was wondering :confused: Perhaps only the neutralisation reaction is relevant, the K and Cl are irrelevant? But I have no idea how to approach the question. (It's from the Conquering Chemistry blackline masters)

Does anyone know how to approach this question? When 50 mL of 1.0 mol/L potassium hydroxide solution is added to 50mL of a 1.0 mol/L hydrochloric acid solution, 2.8 kJ of heat energy is given out. Calculate the enthalpy change for the reaction: H+(aq) + OH-(aq) --> H2O(l)

What Aerath said. Plus for Module C, some classes do the Smithsonian September 11 website instead of The Fiftieth Gate. For Module B, we do the Gwen Harwood poems.

http://xkcd.com/435/