Search results

Yeah that last question honestly is actually very easy. But students who over analyse the question too much probably would've got stumped by it before even attempting it because it looks very complicated. Those students who just stuck to the basics such as soon as they see a K (equilibrium...

Solutions finished :D (bar the 33b which I will do). Hopefully these were useful to you after your exam. Keep in mind though the level of answer here may be higher than what is needed for a full mark since I am writing this from a perspective of someone who has tutored for 5 years and completed...

Yeah that was a bit odd they put that one in but I guess it's fair then since you can do it by elimination and they didn't ask for a reason. I think it would be problematic though if they asked to explain it and then I doubt 99% of students would know. Maybe a select few who did something like...

33) a) deltaG = deltaH - TdeltaS deltaH = -95 kJ/mol at 300 K TdeltaS = -80 kJ/mol at 300 K deltaG = -95 - (-80) deltag = -15 kJ/mol b) The deltaH for this reaction is negative and so therefore this reaction is exothermic (releases heat to the surroundings). The TdeltaS term is also less...

Short answer: 21) a) not the only possible answer - flammable substances as they are all fuels - address by keeping away open flames or ignition sources that may lead to a fire b) Flask 1: propanoic acid (polar so soluble in water and not able to oxidised) Flask 2: hex-1-ene (non-polar so...

Hi, here are my answers to this year's chemistry exam. I will post them progressively over today as I am not currently at home, so check back for updates later today. Comment by another user: just for convenience - this is the paper from the thread posted by @Life'sHard...

I might write solutions for the paper when I get home tonight will post it on here

Can some explain what the solution in part b is saying? I’m also confused with the absorbance graph and why it’s using volume instead of concentration | Bored Of Studies You can find it here

Swap numbers in the equation when writing the molar ratio. Will never go wrong For example, C6H12O6(aq) --> 2CO2(g) + 2C2H5OH(aq) Say we want to do molar ratio between carbon dioxide and glucose Write it without the number first: n(C6H12O6) = n(CO2) Swap the numbers from the equation...

There is some information on this website: English Extension 1 – HSC Raw Marks Database However, keep in mind the raw marks are not constant it depends on the difficulty of the exam each year and what is decided as the cut offs for each band

Yes you can learn module 7 before module 6. The content in module 6 isn't really linked to module 7, other than the reaction of amines and carboxylic acids with water. Aside from that rest of the content is different

Look at the unit it's not mol/L It is written as μmol/L μ = 10^-6 It's same thing as if you wrote 5 mg. You would need to do 5 x 10^-3 g. m = milli = 10^-3 μ = micro = 10^-6

By the way you don't need to memorise that explanation for HSC level but just be aware that some precipitates only form in certain types of pH

the precipitate dissolving with addition of HNO3 confirms it is phosphate. Phosphate is unable to form precipitates in a low pH because the PO4 3- reacts with H3O+ (due to the acidic environment). This makes HPO42- (PO43-(aq) + H3O+(aq) < -- > HPO4 2-(aq) + H2O(l)) and also can make H2PO4 - as...

Show the substitutions and full working. If you put in the wrong value how will the marker know what mistake you have made such as a substitution or calculator error.

One general way though to make questions more difficult from experience is by making questions that are linked to multiple areas of the syllabus across modules. I find that a lot of students have trouble for example being able to link module 7 organic chemistry reaction pathways with module 8...

Adding onto @CM_Tutor comments as I wrote the 2020 BOS trial paper last year. The main aspect that helps I would say in being able to write these type of questions is an understanding of the content at a level that is far beyond the scope of the syllabus. If I had just come out of Year 12 I...

K = [CH3OH] / [CO] [H2]^2 let [CH3OH] = x they say [CO] = [CH3OH] [CO] = x 2.25 = (x) / (x) [H2]^2 2.25 = 1/[H2]^2 1/2.25 = [H2]^2 take the square root and you have the asnwer

Normally you know what prac it is going to be or the pracs from which they are going to choose. To study for a practical exam it is slightly different from a normal exam. You want to know things like: - what are the common safety risks associated with the chemicals or equipment you are using...

Yeah there was that one but I mean in terms of the acid-base titrations