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Q26

Oh and the 2011 HSC question on titration is pretty good.

These should keep you busy for a while. I haven't decided on whether they're good or not, but difficulty is relative and i hope this practise is beneficial and will help you for your practical test. :) A standard solution was prepared by dissolving 1.432g of sodium carbonate in water. The...

goodnight

hai

Re: HSC 2012 Marathon :) cos(A + B - (A - B)) = cos(A + B - A + B) = cos2B @Carrot (your first post), i believe that's what happened to my thread lol.

Re: HSC 2012 Marathon :) Lol what? http://community.boredofstudies.org/showthread.php?t=274161

1. Heat capacity of water is 4.18J/gC 2. Yes you are correct that it is joules 3. You forgot to convert kilojoules to joules (ethanol molar heat)

I just tried it myself and it didn't work lol. Would attempt it again but i'm busy at the moment..

Okay thanks

Unsure if you'll need to take absolute value of the denominator but.. \int \frac{x^3 - 2}{x^3 + 1}dx = \int 1 - \frac{3}{x^3 + 1}dx = x - \int \frac{3}{(x + 1)(x^2 - x + 1)} \frac{3}{(x + 1)(x^2 - x + 1)} \equiv \frac{A}{x + 1} + \frac{Bx + C}{x^2 - x + 1} $And go from there$

Re: HSC 2012 Chemistry Marathon Question:

As x-> infinity, -1 <= cosx <= 1 and x -> infinity, so the value of cosx becomes negligible. So limit as x -> infinity, x + cosx = infinity

When you're taking limits of functions, what you're finding is what f(x) value the function is approaching. However, in these cases, e.g. f(x) = cosx, when taking the limit as x -> infinity, knowing that cosx is a periodic function that doesn't tend towards any f(x) value, it does not have a...

Re: HOW AM I GOING GUYS? Pretty well. What ATAR are you aiming for?

Should make your own thread and not bump something posted a long time ago lol. Use surface area formula, i.e. 600\pi = 2\pi r^2 + 2\pi r \times h Write volume formula, i.e. V = \pi r^2 \times h Use these two formulas to get something in terms of V and h or V and r and you can do the rest.

\int \sqrt{x^2 + a^2} dx = x\sqrt{x^2 + a^2} - \int \frac{2x}{2\sqrt{x^2 + a^2}} \times xdx = x\sqrt{x^2 + a^2} - \int \frac{x^2}{\sqrt{x^2 + a^2}}dx = x\sqrt{x^2 + a^2} - \int \frac{x^2 + a^2}{\sqrt{x^2 + a^2}} - \frac{a^2}{\sqrt{x^2 + a^2}}dx = x\sqrt{x^2 + a^2} - \int \sqrt{x^2 + a^2} dx...

I must have remembered the question wrongly.. Or i can't think of the method.. Ahh. Factorise bottom, partial fractions. Gets quite messy.... Too long actually..

Use: V = \frac{4}{3}\pi r^3 and SA = 4\pi r^2 \frac{d(SA)}{dt} = \frac{dV}{dt} \times \frac{d(SA)}{dV} \frac{dV}{dt} = 1.9 \times \frac{dV}{d(SA)} \frac{V}{SA} = \frac{1}{3}r V = \frac{rSA}{3} \frac{dV}{d(SA)} = \frac{r}{3} = 0.2 \therefore \frac{dV}{dt} = 1.9 \times 0.2 = 0.38mm^3/s

A bit tough for a year 9 student if you ask me lol. I think i just learnt simultaneous equations in year 9 let alone use it extensively like how i solved the 3rd question.