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Which one? I haven't got their solutions for it, unfortunately

At first, I approached the question as Jazz has, getting [Pb2+] = 0.0720 mol L-1 (3 sig fig). However, this was not an option provided, and I was suspicious as this has a final mixture with [Pb2+] = 0.02937 mol L-1 and [Cl-] = 0.05820... mol L-1 and Qsp = 0.02937 x 0.05820...2 = 9.949... x...

I came across this question as an MCQ and don't agree with any of the answers, so I'd like to hear what you think would be the answer. A student wishes to determine the concentration of a solution of lead(II) cations by using a precipitation titration. A 25.00 mL sample of the solution is...

Interesting, thanks for posting it. I have doubts about the red pathway. The pattern of dashes, from -1 to -0.8 to -0.6 along y = +0.6 and y = +0.4, looks consistent to me with a curve that is decreasing but concave up. I would, however, draw a curve similar to the red one if I started from x...

I've wondered the same... maybe @Trebla can explain?

This requires the conjugate root theorem. The result can be established without it, by proving that x^2 < 0. This can be done by noting that x^2 = \cfrac{-A \pm \sqrt{\Delta}}{2} \qquad \text{where} \qquad \Delta = A^2 - 4B and that A,\ B \in \mathbb{R}^+\ \implies\ \Delta = A^2 - 4B < A^2\...

Why is the quadratic positive definite?

Yes, you are right. Looking at the NESA description of Understanding Results, what I have described is moderating rather than aligning to the standards / performance bands.

Aligning is about making internal results match to external results. So, suppose a school had its Advanced Maths cohort with an average internal mark of 90% because they asked questions like "Solve for x: x + 2 = 3" and so artificially inflated the internal marks. Performance of the cohort on...

According to NESA: Further, the performance descriptor for Advanced Maths describes the requirements for Band 2 as: So, the question is, has your performance sufficiently met the descriptors for Band 2 or above, are have you not met enough of the outcomes for a report to be made? You...

The exam would have been finalised early in the year, certainly well before the major lockdown. Further, NESA tends to view issues like decreased opportunity for face-to-face teaching as a matter for special consideration rather than for exam setting.

It is certainly possible to do Physics with only Standard II Maths. It is certainly easier to do Physics with at least Advanced Maths as calculus is highly relevant to a significant amount of Physics concept, and Extension 1's handling of vectors is much more rigorous that the approach taken in...

You are correct, and well caught! Nash's solution to q13(a) is indeed incorrect. The solutions posted earlier answered this question correctly. Volumes rotated about the y-axis are indeed of the form \pi\int x^2\ dy which, in this case of one volume subtracted from another, gives...

Thank you. I'm hoping to get it finished in the next couple of days There was meant to be enough in the question to figure it out, based on knowledge of cation / anion testing from Module 1 and the question stating what was produced... but it did prove to be a challenge, based on my...

Nash's solution is missing the sketch in q12(d)(iii) and has the sane problem as the other solution in handling the absolute value in q12(b). It does use the vector approach that I believe was sought in q14(a), though.

Assuming that the slope field posted elsewhere in this thread is accurate, the path will have a minimum stationary point and always be above the x-axis

If helping others involves consolidating your own knowledge and understanding, then you can both benefit. Also, consider the perspective of your tutor - the best form of advertising is word of mouth, so refusing to tell others who you got help from is not doing any favours for the tutor, who...

Looking at the solutions provided above: there is a small error in Q5 where \theta \in \big[0,\ \pi\big] as \cos\theta < 0\ \implies\ \theta \in \left(\cfrac{\pi}{2},\ \pi\right], but the solution writer has stated that \theta \in \left[\cfrac{\pi}{2},\ \pi\right]. This is because \cos\theta =...

Yes, so long as you don't oversimply the question is so doing.

You are correct, so that means there should be open circles at z = -\pi and at z=0 and the boundary along the real axis between them should be dotted. Good catch!