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The 2021 BoS Chemistry Trial is now publicly available - Trebla posted a link in the thread in the general discussion forum: https://boredofstudies.org/threads/bos-trials-maths-and-chemistry-2021.397055/page-2#post-7399485 As the person who wrote it, I'm interested in any feedback, either...

There are plenty of students who are uncertain about what they want to study at uni, or whether they want to study at uni, or what they are interested in as a career. There will be plenty of students who change directions in the first semester / year of uni studies - transfers happen both to...

Remember, though, that sodium will also produce hydrogen with carboxylic acids.

Assume that chemical tests are done on purified products, not with possible unreacted materials (from the oxidation, say) still being around. If there is residual acid present then you are correct, added sodium carbonate would react with it regardless of whether the product formed is an acid or...

Places like T's textbooks are also good sources for reasonably-priced second hand books.

I think that some (many?) of the students facing the IB and HSC this year are experiencing feelings of relief that the exams are finally here, that after all the delays and disruption, there actually is an end to the high school journey in sight. Countless students in the past have quietly...

No problem, I see part of my role as clarifying for the sake of other readers. :) I think it helps everyone to be confident that information is reliable, and I hope it encourages students to help each other, knowing that any mistakes will be corrected and issues clarified so that they can...

Note that the statement is true only if the complex number has a modulus of 1, for the reasons noted in For a general complex number z, of any modulus, like z = Re^{i\theta}=R\big(\cos\theta + i\sin\theta\big), the result z + \overline{z} = 2\Re{(z)} is true for all z, but the result for z +...

I guess I should have anticipated that someone would be asked by a friend to ask! :) I thought about a few possible responses, but have settled on a serious one. For the sake of the BoS site, I think one great present for the Moderators would be to continue contributing after you've finished...

It could also be made needlessly complex by asking it in MX1 by vector methods

My point was, the question presents the idea that all analytical techniques are basically shine light through and measure absorbance, and asks students to evaluate it. The could certainly go for the differences in NMR, or fluorescence not being absorbance (not that fluorescence spectroscopy is...

Looking at the diagram in the question, the tripod and gauze on which the beaker is sitting is a long way above the flame of the burner. Reducing this gap with a shorter tripod would reduce heat loss, so the idea is sound, even if the implementation might be easier by placing the spirit burner...

It's worth noting that this thread has yet to consider the vastly more important topic of gifts for BoS Moderators...

No need to call me "Sir" :) Recycling is more environmental.

It's a probability, so 0 \leqslant P(A) \leqslant 1

Be environmentally friendly... use the fire to cook something nice, recycle notes into steak, perhaps?

Colourimetry is generally working at a singe wavelength, which is used to construct a calibration curve then measure unknowns. similar to AAS. UV/Vis spectroscopy generally involves scanning the spectrum to produce a graph of absorbance as a function of wavelength. UV/Vis can be carried out on...

@someth1ng, a 6 mark question that I have found from a 2020 Trial that you might like...

Put another way... If: I = \int_0^{\pi} x\sin{x}\ dx then I is the signed area enclosed by the curve y = x\sin{x}, the x-axis, and the boundaries x = 0 and x = \pi. Now, consider J = \int_0^{\pi} t\sin{t}\ dt J is the signed area enclosed by the curve y = t\sin{t}, the t-axis, and the...

\begin{align*} \int_0^{\pi} x\sin{x}\ dx &= \int_{x=0}^{x=\pi} (\pi - y)\sin{(\pi - y)} \times -1\ dy \qquad \text{putting $x = \pi - y \implies dx = -dy$} \\ &= -\int_{\pi}^0 (\pi - y)\sin{(\pi - y)}\ dy \qquad \text{as $x = \pi - y$, when $x = 0,\ y = \pi$ and when $x = \pi,\ y = 0$} \\ &=...