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I think the observation implicit in these questions / comments is important... why is this worth 4 marks? Is it just a mistake, or an indication that something is being missed? Start by considering the term "valency"... it is more than just the charges in @wizzkids' table, which tells us about...

The problem can be approached as an area between two curves, just relative to the y-axis. The curves that bound the region are x=e^{2 - y}, which is the "top" curve, and x = 1, which is the "bottom" curve. The area, against the y-axis, is then \begin{align*} \text{Shaded Area}\ &= \int_0^2...

\text{Since} \qquad \tan^{-1}{x} + \tan^{-1}{\left(\frac{1}{x}\right)} = \frac{\pi}{2} \qquad \text{for all $x > 0$} \text{And} \qquad \tan^{-1}{x} + \tan^{-1}{\left(\frac{1}{x}\right)} = -\frac{\pi}{2} \qquad \text{for all $x < 0$} \text{Your answer and the desired answer match but with...

Since the area is symmetric about the x-axis, its area can be found by doubling the integral of the function above the x-axis: \begin{align*} \text{The area of the property is given by } 2\int_0^1 y\,dx &= 2\int_0^1 \sqrt{9x}\,dx \qquad \text{as $y^2 = 9x \implies y = \sqrt{9x}$ above the...

Definitely positively certainly B and not D.

Question 1: \text{$y = \sin^{-1}{x}$ takes ratios between $-1 \le x \le 1$ and outputs angles between $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$.} \text{However, you are using this output as an input into a log function, which can only accept positive values. Hence, you need $\sin^{-1}{x} > 0$.}...

Sure... \begin{align*} \text{Going from Line 1:} \qquad \frac{n!\left[4!(n - 5)(n - 4) - 6!\right]}{(n -4)!6!4!} &= 0 \\ \text{to Line 2:} \qquad \qquad 4!(n - 5)(n - 4) - 6! &= 0 \end{align*} \\ \text{was done by either dividing both sides of the equation by $\frac{n!}{(n -4)!6!4!}$, or...

Your algebra is correct, you got to \begin{align*} \frac{n!\left[4!(n - 5)(n - 4) - 6!\right]}{(n -4)!6!4!} &= 0 \\ 4!(n - 5)(n - 4) - 6! &= 0 \qquad \text{as $\frac{n!}{(n -4)!6!4!} \neq 0$} \\ (n - 5)(n - 4) &= \frac{6!}{4!} = \frac{6 \times 5 \times 4!}{4!} \\ (n - 5)(n - 4) & = 30 \\ n &=...

Each row of Pascal's Triangle has terms that are distributed symmetrically. In binomial coefficient form, this symmetry property can be expressed as: \binom{n}{r} = \binom{n}{n-r} \begin{align*} \text{Applying the symmetry property to this problem yields:} \qquad \binom{n}{4} &= \binom{n}{n -...

Past papers

@wizzkids is correct that some metal hydroxide increase their solubility at high pH - aluminium hydroxide, for example - which the HSC covers in the form of an example of amphotericism. It pops up in modules 6 and 8, though is generally covered very briefly. Aluminium hydroxide is amphoteric...

Yes, they do. Hydroxide ions will react with hydronium / hydrogen ions and with unionised propanoic acid molecules, in each case producing water and (in the latter case) the propanoate anion. Which they react with principally will depend on relative concentrations as well as collision...

(a) is bounded above but not below as its a negative cubic, and so the region above y > 1 will be x \in (-\infty,\ a) or x \in (-\infty,\ a)\cup(b,\ c). (b) must be bounded below as \ln{x} \in \mathbb{R} \implies x > 0. It's also bounded above as x\ln{(x + 1) - (x + 1)\ln{x} < 0\ \forall x >...

This is backwards. Iron(II) hydroxide will not dissolve to any significant extent unless the hydroxide ion concentration is very low and so the solution needs to be acidic, not basic. Since its Ksp = 4.87 x 10-17 at 298 K and as [Fe2+] = Ksp / [OH-]2 (in a saturated solution): at 298 K and pH...

If the solution were prepared by dissolving iron(II) hydroxide into neutral water and the pH was 9.42 at equilibrium, your method would work. However, if the solution were prepared (say) by dissolving iron(II) chloride and then adding base to adjust the pH, then you will have no idea what the...

It should be potassium manganate(VII), indicating that the manganese atom is present in its +7 oxidation state... it is more commonly called potassium permanganate, KMnO4. The "per" prefix indicates a higher oxidation state than occurs with the simple -ate ending, typically by the presence of...

There's a lot more to redox than is addressed in the syllabus, and what is there isn't covered well. Oxidation states give you a good idea of what is likely to oxidise or reduce, by looking at what is present in unusual oxidation states. Potassium permanganate, KMnO4, has K+, Mn7+, and O2-...

You can look at the list if standard reduction potentials for typical reactions. This is an unusual combination as both hydrogen peroxide and permanganate are oxidising agents, but permanganate is much more powerful, so you will have oxidation of hydrogen peroxide to oxygen gas caused by...

For the first question, there is no solution to (x - 2)^2 < 0 because a real term squared can never be negative. For the second question, the answer is wrong. At x = 2, the expression (x - 3)(x - 1)(x - 2)^2 = -1 \times 1 \times 0^2 = 0 and so the inequality is NOT true at x = 2. The solution...

The "best" textbook depends a bit on the criteria used. For Advanced and MX1, I think Cambridge offers the best variety of questions and with suitably challenging material, but that doesn't make it the best starting place for everyone. Maths in Focus, for example, is overall too simple for...