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re: HSC Chemistry Marathon Archive Answer should be A. By process of elimination, B is wrong because that is the electrolyte for the anode half-cell as the magnesium electrode is the anode. C is wrong as platinum doesn't take part in redox reactions because of its extremely low reactivity. D...

Anyone else getting really nervous for English?

re: HSC Chemistry Marathon Archive They gave the value of K, and they gave the initial concentrations of both reactants and products, and asked you to calculate the concentration at equilibrium. You had to calculate Q first with the initial concentrations (because the concentration of the...

re: HSC Chemistry Marathon Archive Our trial had a question about Q, so why not?

Never saw it in a trial or hsc exam as an essay question, but I have seen it been asked as a 4-5 marker in previous trials.

When can we expect the results?

On Wolfram, the graph looks like it intercepts the y-axis and x-axis. I did what you did, but wolfram graphed something weird.

I am asian, drsoccerball can confirm this.

Oh and also, to let everyone's expectations of me down, im not getting above 70. I counted my losses and it's not happening.

How was the graph question in question 12?

Perfect representation of me in the exam:

Gabriel Moussa

The question 16 one for 4u?

Unfortunately time was a major issue, in fact my working out for most questions was so messy and shocking, especially with the mechanics questions

That's correct, I had a feeling that I should of checked the values of p, well there goes a mark

Precisely this, the 3 values of p were 0,2,-2

I = \int \frac{\sqrt{\sin{x}}}{\cos{x}\sin{x}} dx $ Let $ u^2 = \sin{x} \therefore I = \int \frac{2u^2}{\cos^{2}{x} u^2} du = \int \frac{2u^2}{(1-u^4) u^2} du I = \int \frac{2du}{1-u^4} du = \int \frac{1}{1-u^2} + \frac{1}{1+u^2} du = \frac{1}{2}\int \frac{1}{1-u} + \frac{1}{1+u} du + \int...

for 4u or 3u?

Never knew pyramids had curved edges...

re: HSC Chemistry Marathon Archive I dont think you need to go into reaction quotient because when the system is at equilibrium, it still can favour one side. From what ive been taught: K <1 means the point of equilibrium is on the reactants side K >1 means the point of equilibrium is on the...