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Bibliii

Bibliii
Make sure you have all the basics down pat, and understand what equations to ise when. Check if your library has any resources like the Oxford Study Buddy or ATARnotes topic tests. I found it really helped me. Understand your formula sheet, and write down any formulas that aren't on there, like the general equation of a sine or cosine function, things that are usueful. Use this when you are studying.

With math, the only way to actually study is by practicing. Practice heaps. Check out the QCAA exam examples. Recycle through different textbook questions - the cambridge one has really good ones. Talk to your school or local librarian and ask if they have year 12 methods resources - it always helps. You could check out the VCAA and HSC past exams, and pick out the ones that are relevant.

Hope this helped,
-jinx_58

Bibliii good luck homie! You’ve got this!

-jinx_58

Also, it would be great if you could download those files from that link ASAP because I think I need to keep the tab open for it to work and I can't promise that tab will stay open for that long haha.

Bibliii

Ka = the equilibirum constant of an acid dissociation reaction eg for the equilibrium system HCOOH + H2O <-> HCOO- + H3O+, Ka = [H3O+][HCOO-] / [HCOOH] (remember you exclude water for aqueous equilibria since its concentration change is negligible)

pKa = pH above which half of all acid molecules are deprotonated. Given by -log10(Ka), so higher pKa = weaker acid. If you inspect a titration curve, they're the half way points between equivalence points. They're really flat since when pH = pKa, an acid buffer system is most resistant to change according to the Henderson-Hasselbach equation (I don't know to what extent banana benders need to know this, I only learned Ka in uni chem in Victoria). You can memorise the property of pKa by the fact that when pH is above pKa, the equilibrium system is deficient in H+. According to Le Chat's principle, the system will want to partially oppose that, which it will by having your acid donate its proton.

Kb = the equilibrium constant of a base ionisation reaction eg the Kb for NH3 + H2O <--> NH4+ + OH-, Kb = [OH-][NH4+] / [NH3]. Kb and Ka can be generalised as [ionised species] / [unionised species]

pKb = the pH above which a base will be protonated. Similar story with pKa, when pH is above pKb, the system knows there's not enough H+. To compensate, you'll want less OH- in solution, forcing the base ionisation reaction to proceed in reverse

H3O+ = the protonated form of water, ie its conjugate acid. Abundant in acidic conditions since the acid has donated its H+ to the solvent molecule (H2O).

OH- = the deprotonated form of water ie its conjugate base. Abundant in basic conditions since the base has stolen H+s from H2O.

One small correction to jinx's excellent response, technically strong acids also have Ka and pKa values despite the fact that strong acid ionisation is pretty irreversible. For example, we are required to know in uni that pKa of HCl is -5 for purposes of comparing leaving group strength in organic reactions. Ditto for strong bases with regards to Kb and pKb

I don't know how QCE structures their data tests so someone else can help with that

Bibliii
Hey Bibliii,

Here are my answers to the theory questions in your first image. I skipped number 1 as I believe Lorentz force is a topic that is not in the syllabus and thus, you do not need to know (correct me if I'm wrong). However, if you like, I can take a look at the questions anyways.

3. If you take a look at either the centripetal force equation (F=mv²/r) or the centripetal acceleration (a=v²/r) equation, you will notice that if you double the velocity (i.e. 2v) then due to the square, it will become 4v². This means that the radius will increase by 4 times the original if you double the velocity.
4. Sin(theta) is used as this gives the perpendicular component (y-axis) whereas cos(theta) would give the parallel component (x-axis). The component that is needed to represent the force relative to the magnetic field is the perpendicular component which is why sin(theta) is used.
5. a) If the particle is positively charged, the direction of the force would be down. So, in this case, it would take path C.
   b) C (answered above as well)
   c) The kinetic energy should remain constant as the circular trajectory has a displacement of 0 and thus, there is no work done. No work done means no change in kinetic energy (according to the work-energy theorem).
   d) Negatively-charged particles would go up (path A)
   e) A neutral particle would travel in a straight line (path B) as magnetic fields only affect charged particles.

Hope this helps and answers your questions and let me know if you have any other issues!

• PP

Bibliii

Sorry Bibliii, I completely missed this! Would you still like help with these questions?

Bibliii
Hey Bibliii,

All good - we're all happy to help!

For the 8.2 review questions 5 and 6, it is a bit tricky and I just realised I didn't really explain it that well. So the main thing you need to do is draw the magnetic field lines on both magnets. As soon as you do this, you will notice that since both are north poles, the lines of each one will come towards each other (--> <--). So if we take a look at point A, would the LHS or RHS magnetic field line be stronger? It would definitely be the LHS magnet because it is closer and the RHS magnet would have a much weaker magnetic field line. So, now you need to take both of those magnetic field lines as vectors and find the vector sum so like this: ------> + <--- = --->. And that final arrow is pointing towards East. Does that make sense?

Hope that clears it up and as always, let us know if you have any other questions.

• PP

Sorry guys, after 3 months off I’ve suddenly become completely inept at Chem

Thanks @Billzene and @PhytoPlankton

Much much better responses

Bibliii KOH as God notes is the strong base potassium hydroxide, potassium is denoted by the symbol K since it’s called kalium in Latin. To approach this question, I’d take -log10( ) of the concentration of KOH they gave you to get the pOH, and then rearrange the formula pOH + pH = pKw = 14 for pH

For 17, hydrocyanic acid is a very weak acid judging from its low Ka magnitude so set up an ICE table as follows

Initial [HCN] = 0.227 M, initial [CN-] = 0 M, initial [H+] = 0 M
Change [HCN] = -x M, change [CN-] = +x M, change [H+] = +x M
Equilibrium [HCN] = 0.227 - x M, equilibrium [CN-] = x M, equilibrium [H+] = x M

Then equate the magnitude of Ka with your Ka expression

6.2 x 10^-10 = (x M)² / (0.227 M)*

*In uni chem here in Victoria we’re always allowed to assume that the concentration change of an weak acid is 0, if QCE doesn’t allow it, you’ll have to use the quadratic formula.

Now solving for x should give you [H+], which you should then convert into pH

Ka isn’t the same as pKa, pKa is -log10(Ka) (whenever you see p in chem notation, it means -log10( ), such that pH = -log10(H+) and pOH = -log10(OH-))

18 is a titration question that involves a dilution, always start with a balanced equation and remember that your standard was diluted by a factor of 10.

So you know the original concentration of your standard was 31.8 g / 106 g/mol = 0.3 mol in 500 mL = 0.6 M. Diluting that by a factor 10 (use c1v1 = c2v2, where c1 = 0.6 M, v1 = 5 mL, c2 is your unknown and v2 = 50 mL). Your DILUTED concentration would be 0.06 M

Now you know 50 mL of a 0.06 M Na2CO3 solution reacted with 16.8 mL of H2SO4 solution. The balanced equation is pretty simple since everything is 1:1:1:1:1 (Na2CO3 + H2SO4 —> Na2SO4 + H2O + CO2). Find n(Na2CO3), which will be equivalent to n(H2SO4) according to the balanced molar ratio. Then use c = n/v to find [H2SO4], remember that v must be in L so 16.8 mL = 0.0168 L.

Bibliii

IKR!! everyone is so niceee. Honestly just being Indian is kinda stressful with all the aunties and family friends constantly talking about ATARs in Dec but I'm really lucky that my parents don't really force me and try their best to motivate me.

Howdy howdy!

I did QCE Chem and Phys so I understand the position you are in right now. Data tests usually consist of graphs or tables and identifying the relationship between two variables, using certain data points to find the value of a third variable etc. If your school library or any library has the Oxford Study Buddy for Chemistry and Physics, I highly suggest you get those - they saved my externals. If you cant find them, they’re usually $30 each.

For Physics, the data tests are usually based on different formulas. For Example, in a graph if you’re given how the current changes as the resistance changes across a circuit, you might be asked a) Find the voltage and b) Determine the mathematical relationship between the two.
Note: Always remember the mathematical relationship means the gradient. If you go to the QCAA website and look for the data tests there, it might be a better example.

For Chemistry, it’s very much based off experiments. There’s a few more explanations you need to give. For example, if it’s an equilibrium question, they’ll give you a graph and you’ll have to identify if it’s an endothermic or exothermic reaction, and give a sentence why. Other questions from this graph might be, that the concentration of the products were increased, how does it affect the equilibrium position, amount of reactants etc etc.

I hope this helped, and feel free to shoot more questions,
-jinx_58