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I think NaCl becomes bonded to the extremely polar H2O. As an ionic (Na+)(Cl-) is capable of making stronger ionic bonds than those of (2H+)(CO3 <2->) so this breaks down the bond. therefore this leads the reaction to move to the left as the unbonded H2O reduces from the reaction wit NaCl. I...

Thanks for the text. Can i print or r ua copyrite obsessed guy(jkz). Nweis thnk u v much for it. Gives me a huge clue. $@m

abt 13 peepz

Hey i can connect to ua server. Tnkz nwei.

4x^2 + 9y^2 = 36 x^2/9 + y^2/ 4 =1 b^2= a^2(1- e^2) 4=9(1- e^2) e= sq root 5 /3 foci( +/- ae, 0) (+/- sq root 5, O) P1= 5/9 + y^2/4 =1 y^2 =16/9 y=4/3 P1 x P2= 16/9 it doesnt come to 4 it comes to 16/9

make x=z^2 and then factorise: x^2 + bx + b^2/4 4x^2 + 4bx + b^2 then make b^2- 4ac>/= to 0 B=4b and a=4 and c=b^2 therefore delta =/ > 0 and b is an element of all real values.

Wat r u trying to say? Remeber that accordin to Einstein acheiving the "C" is almost impossible. E=mc^2+0.5mv^2 The energy required to acheive the speed of lite or 2.99999x10^8ms^-1 is also quite impossible. Elaborate on ua qsn more{do we really need to noe this 4 the HSC on relativistic space...

This rxn can give an answer- i.e: HCl + H2O<=>H3O+ + Cl- When temp increases two things may actualli happen: 1. The water evaporated and the equillibria shifts to the left in order to reduce the effects of the lost h20 and do the ph increases. 2. The eqn shifts to the left as...

find the general gradient of the point of intersection which must be: 2x/a^2+dy/dx(2y/b^2)=o dy/dx=-x(b^2)/y(a^2) if (x1,y1) is the xternal pt (y-y1)=-x(b^2)/y(a^2)[x-x1] x^2/a^2+y^2/b^2=x.x1/a^2+y.y1/b^2 [sub in the x1,y1] Although this will onli give u one eqn.

I saw the formula in Jacaranda Physics. Its really cool to use it in explanations.

congrats bro. U r a really smart kid. How the hell didnt u make any mistake in the three unit exam. Thats brilliant.

Thankz this is so gud. I will look for some of the things u posted when i study the text closer.

it is a long process of alpha emissionz. look at conquerin chem to see the diagriam of the decay.

Arg(z-2)+Arg(z-3+2i)=0 Arg(z-2) means that the line goes via the origin. so y2-y1/x2-x1 for the first case should equal the -ve the gradient of the second case. y/x-2=-(y+2)/x+3 y(x+3)=-(x-2)(y+2) xy+3y=-(xy+2x-2y-4) xy+3y+xy+2x-2y-4=0 y+2xy+2x=4 I think this is the answer but the graph...

Argz(z+3)=pi/2 z(z+3)=o z=0+0i ang -3+0i see the diagram in attachment.

Thankz mann. I waz readin the book n wrote a bit of info from there. But i will read it again to confirm.

I like this years geo. The onli thin i hate from this yr is the first module itz not as scientific as last yrs module 1. Nweiz i think the rest of the topicz r pretty interesting.

U can start by discussing the comparative heats of combustion. Then explaining the different processes in the manufacture of both fuels. U can also assess the environmental impacts of each fuel. Then compare all this in detail. go to conquering chemistry for the use of ethanol as a fuel. it will...

i guess this is too late 4 ua topic test but in future if u hav one u better check success one chemistry or take dot point chemistry as a test rather than a guide.

At orbital velocity the main force acting on the rocket is F=ma and centripetal force (xplainz y g-forces r zero durin orbit). U r rite but i just gave the general formula.