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@jimmysmith560, perhaps this thread belongs in the Chemistry forum?

Badly written questions are sadly common, as are questions with mistakes. You will find them from time to time, just hope it's not in an important assessment. The 2023 had three mistakes, IIRC, though hopefully none made any difference.

If the school listed in your profile is correct, there is no way that they are buying papers to use for assessments... though something weird did happen with one paper earlier this year...

Then it is likely that the answer they seek is the one you have - that despite its higher electronegativity, the much greater radius of chlorine results in a charge density that makes for a poorer H-bond acceptor when compared to nitrogen. There is material in modules 6 and 7 that will be...

Good, but that's not how it sounded when I read it. Also, there is a second factor here that none of the above has addressed. Atomic radius and reduced charge density explains why Cl atoms are generally ineffective as H bond acceptors, but what about the other side of the H bond? What I would...

Aluminium chloride is a covalent molecular substance, which actually exists primarily as a dimer, Al2Cl6. It reacts violently with water as the chloride ligands are displaced to produce its hydrate form, AlCl3.6H2O, which is much more accurately represented as [Al(H2O)6]Cl3, and which...

It's a badly written question If it means which of the options is the electron configuration of element X in its elemental form, then none of them is correct, but the best answer is C. If it means which of the options is the electron configuration of the form in which X exists in the compound...

The part about Cl's radius is correct, the parts about "nearly stealing" electrons from H is not. In an H-bonding situation, the H atom is interacting with two very electronegative atoms - one through a covalent bond, one through an electrostatic interaction (the H-bond). The electron from the...

This question was extremely easy under the old syllabus, as the following circle geometry theorem was part of the course. If AB is a chord of a circle, centred at O, then the angle at the centre of the circle standing on that arc is twice the angle at the circumference on the same arc. In...

Well, look on the bright side - at least we aren't Victoria! Though there were mistakes in the HSC Chemistry exam. No one should have been disadvantaged by accepting what the exam said, but it does have clear errors. :(

And the start of the alternative solution should refer to the projection of OA onto OC, rather than of AC onto OC... though they do correct themselves further on.

It is true that some aspects of vectors becomes problematic in complex numbers, which is an area that the HSC syllabus avoids, but the approach I have described is valid and the notation generally used.

Firstly, welcome @mandemindiguise, and yes, this is a suitable place for your question. Branching in most organic compounds leads to decreases in boiling point as the resulting molecular shapes are generally less efficient to pack together, increasing average intermolecular distances and thus...

This recursive definition yields a spiral of points in the Argand diagram... here is z_n for n \in \{1,\,2,\,3,\, ...,\,37\} starting from z_1 = 1.

There's a small error in the end of this. It should say: Since \sqrt{k} is real and i is imaginary, \left|\sqrt{k}+i\right|=\sqrt{\left(\sqrt{k}\right)^{2}+1^2}=\sqrt{k+1}=\text{RHS} as required

Express PQ and PR as vectors in terms of the given complex numbers. That is, for example, PQ = PO + OQ = OQ - OP = z2 - z1. Multiplication of a vector by i produces a vector equal in length to the original and rotated by 90 degrees in an anticlockwise direction. Thus, either PQ = iPR or PQ = iPR

Complex numbers, 3D vectors, and further methods of integration will all be relevant for Uni maths and engineering, as will some of mechanics (though more via physics and applications). They will also all be covered at Uni and won't be assumed knowledge, so looking at them in advance may be...

Absolutely, the intention is to approach the problem geometrically. However, getting the answer algebraically is better than not getting the answer at all. Further, understanding why the restriction exists and how it emerges from the algebra is a useful skill in constructing solid proofs.

The vector approach is definitely the way to go here, but we can derive it with algebra: \begin{align*} \arg{(z + i)} &= \arg{(z - 1)} \\ \arg{(z + i)} - \arg{(z - 1)} &= 0 \\ \arg{\left(\frac{z + i}{z - 1}\right)} &= 0 \qquad \qquad \text{(1)} \\ \\ \text{Taking $z = x + iy$, consider:}...

If you've drawn a ketone, written ketone, and spoken of carbonyl (correctly) in the justification that accompanies the table, I think there is a far argument to be made that you should be credited despite a minor mistake. However, if you've made the same mistake in the justification or if your...