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Re: HSC 2012 Chemistry Marathon Solution: n(BaSO_4) = \frac{1.8}{137.3+32.07+16 \times 4} = 7.71 \times 10^{-3} moles n(BaSO_4) = n(SO_4^{2-}) = 7.71 \times 10^{-3} moles m(SO_4^{2-}) = n(SO_4^{2-}) \times (32.07+16 \times 4) = 0.741g (SO_4^{2-})\%(mass) = \frac{0.741}{2} = 37\% Qn:

The moment you write the expression for the area. E.g. A = \int_0^3 f(x) dx + |\int_3^5 f(x) dx|

The area from -2 to 0 is exactly the same as the area from 0 to 2. It is an odd function with point symmetry about (0, 0) so the two sides of the graphs (x > 0 and x < 0) are essentially the same, just that one is flipped over.

Your answer tells me you attempted to find the area bounded between the two curves between x = -2 and x = 0. Since this area is below the x-axis, you must apply the absolute value sign as you cannot get negative areas. Now you need to consider the other region bounded by both curves, between x...

It is soluble. From wiki:

\int \frac{2}{\sqrt{2x + 1}} dx $Let $ u^2 = 2x + 1 u = \sqrt{2x + 1} du = \frac{1}{\sqrt{2x + 1}}dx \therefore \int 2du = 2u + c = 2\sqrt{2x + 1} + c

Thanks for that, your answer is correct. I assumed that they paid the first instalment in the same year they borrowed the money..

Re: HSC 2012 Chemistry Marathon Use phenolphthalein instead of methyl orange as the indicator. CH_3COONa_{(aq)} \rightarrow CH_3COO^-_{(aq)} + Na^+_{(aq)} CH_3COO^-_{(aq)} + H_2O_{(l)} \rightarrow CH_3COOH_{(aq)} + OH^-_{(aq)} Therefore sodium acetate solution is not pH neutral but is...

Re: HSC 2012 Chemistry Marathon Yep.. The Fe3+ to Fe2- reduction potential is 0.77V. The CU2+ to Cu reduction potential is 0.34V. By following what you said, Ecell = 0.77 - 0.34 = 0.43V.

Re: HSC 2012 Chemistry Marathon Notice how i wrote E(ox). That means i already had reversed the reduction potential given on the data sheet to obtain the oxidation potential. If you look on your sheet, it tells you that the reduction potential is 0.34V.

Re: HSC 2012 Chemistry Marathon Good answer. In a test, hopefully you'll draw a nice structural formula of ethanol showing both the hydroxide functional group and the alkyl chain. Post a new questions please!

I keep getting the answer wrong whilst doing superannuation and loan type questions.. Could someone please read over my working out and help me out? Thanks Qn: Jan borrows $8000 and agrees to repay it in equal instalments each year for 10 years. If interest is charged at 7 percent per annum on...

Re: HSC 2012 Chemistry Marathon I already reversed the reduction potential to get the oxidation potential. Then i add both the reduction and the oxidation potentials to get the cell potential.

Re: HSC 2012 Chemistry Marathon Standard conditions: - 25 degrees celcius - 1 atm - Pure metals/liquids used - 1 mol/L aqueous electrolytes used Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^- $ E(ox) = -0.34V Fe^{3+}_{(aq)} + e^- \rightarrow Fe^{2+}_{(aq)} $ E(red) = 0.77V E_{cell} = 0.77V -...

Repetition and memorising the methodical process is enough. Understanding is better though.

Only one as you requested. 1) A = \int^4_0 \sqrt{x}dx = [\frac{2}{3}x^{\frac{3}{2}}]^4_0 = \frac{2}{3}(4^{\frac{3}{2}} - 0) = \frac{16}{3}units^2

Re: HSC 2012 Chemistry Marathon I made a mistake in my original post. Meant to say shift from acidic oxides to amphoteric then to basic oxides across a period. This includes your claim that towards the top right of the non-metals, there is increasing acidity of the oxides.

Re: HSC 2012 Chemistry Marathon [acetic acid] = 0.34M = 0.34moles/litre = (0.34 x (12.01 x 2 + 1.008 x 4+ 16 x 2))/litre = 20.42g/litre = 2% (g/ml) if you don't mind the units. Qn:

3^{2n - 1} - 1 = \frac{3^{2n}}{3} - 1 = (\frac{3^n}{\sqrt{3}})^2 - 1 = (\frac{3^n}{\sqrt{3}} + 1)(\frac{3^n}{\sqrt{3}} - 1) Although i think you typed the LHS wrong.. so.. 3^{2n} - 1 = (3^n)^2 - 1 $, Can you see the two squares? Using indice laws, i can take out the squared like that to...

Re: HSC 2012 Chemistry Marathon Could include more. Moving left to right across a period, we see a shift from basic oxides to amphoteric then to acidic oxides. Moving down groups 1 and 2, we see increasing basicity of the oxides. Moving down groups 6 and 7, we see decreasing acidity of the...