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what was your atar?

yeah i got this too xD but i was being careful with the significant figures and stuff cause considering this paper being too easy>.. markers can be mean and mark on significant figures

got this too! xD

i think b6 is over 85.. paper was easy... only worry would be weird marking criteria

ok since i got the same as him i'll might be able to help for 15 this is a simple projectile question use y=uyt+1/2at^2 with uy=0 since the table is horizontal and delta-y is -1 since the marble dropped off the tabke to the ground.... then you will end up with 4.9t^2 =1 and so t= squareroot of...

yes i believe you are all correct ! hope you are cause that means i am too xD

ok i honestly hate that exam ... it was too easy and will mean that cut off will be high assss!! would prefer it to be harder.... i hate super easy exams! everyone gonna end up getting really high marks which sucks11

Easy!! Some tricky questions but if your know your stuff you'll be fine!

Look guys they term ELECTROCHEMICAL ACTIVITY doesn't even exist in scientific dictionary and the exam this year have added a whole load of ambiguous terms like that... so i wont try to guess what the EXAM WRITERS meant cause they seriously need to RELEARN ENGLISH!!

ok If you haven't notice copper don't tend to form Cu+ in solution only cu2+ if you are not convinced then google copper nitrate ... the one that pops up would be copper (ii) nitrate ... and copper 2 nitrate is the one that is blue... copper (i) nitrate would most likely be red if it exists...

yeah cause the Bromine water would be in a kinda equilibrium where all reactants an products are present and any of them can react ==" oh well every text book and teachers said something different.. My teacher taught us to always think of bromine water as BrOh and HBr... thats why i chose B >< i...

It's A! O3 is soluble because it is polar and it is always polar regardless of which resonating state it is in. this is because of the BENT SHAPE of O3, somewhat like water... and this shape is due to presence of lone pairs of electrons on the central atom and this has nothing to do with its...

I was thinking of HBr reacting with the alkene instead of BrOH it is possible to produce A and B if HBr is in molecular form (gas) i wonder if the reaction still occur if it was HBr(aq) because if it does then you would get both A and B being correct... Logically it would be D but technically...

yeah it is more logical to be D but i was taught at my school to always think of Br2(aq) as an equilibrium with HBrO and HBr and the colour fade was due to a decrease in [HBrO] and [HBr] causing the equilibrium to shift right ... And hsc text books and study guides do this too so the question...

yes but you should realise the last step is not an equilibrium reaction and le chatelier's wont apply... and yield is always 100%

an experiment is not a model.. its a real thing... im thinking when they say models they meant like balls and sticks and computer simulations stuff.. well either way i skipped that question cause i can't write some BS about balls and sticks><

its around 0.31M you might've not calculated the [NaOH] correctly ..

Ok firstly methylethane does not exist.. it is still propane you just named it wrong... because you are meant to count the longest carbon chain in this case is 3 no matter if you draw it straight or bent.. it is still a chain... so it is still propane so all of your methylethane things are just...

ok heres mine with justifications for those hard ones... >< 1A 2D 3C 4D 5B 6A 7B 8C 9A Cause resonance and coordinate bond plays no role in solubility and since dissolving in water is not a reaction so O3 being reactive is irrelevant and O3 is bent so clearly polar no matter what resonance...

yes it is..