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No, you can't. You know the mass of the whole solution (solute plus solvent) so you don't know the mass of NaOH or HCl present. You can only find the moles of each compound from n = CV.

Condensation polymerisation with one monomer: M(monomer) x n = M(polymer) + (n - 1) x M(water) Condensation polymerisation with two monomers: (M(monomer 1) + M(monomer 2)) x n = M(polymer) + (2n - 1) x M(water) Addition polymerisation: M(monomer) x n = M(polymer)

So, you calculate \Delta q in part (a), and then they give you a \Delta q to use in part (b) to find \Delta H. For part (b), it doesn't matter whether you use the moles of HCl or of NaOH as they are the same as the reaction is 1:1.

The masses are given for the solutions but you also have their concentrations and volumes, from which you can see that the reagents are present in their stoichiometric ratio and neither is present in excess... so, no limiting reagent calculation is needed.

I think the question means salt in the specific, non-chemical usage... ie. table salt, sodium chloride. The silver nitrate then precipitates the chloride ions as silver chloride solid, which is removed by filtration. The excess (remaining) silver cations are quantified by titration with...

Thanks for the advice, @Sam14113, I agree that incorrect questions and solutions are common. I was just surprised by this one as it is a paper that will have been done by many students and I wonder if no one else had noticed the mistake.

I have heard that some students were told that there was a typo in question 10, with students at at least one school warned about it during the exam. Does anyone know anything about this?

The solutions that @tywebb has provided a link to contain at least two mistakes... Q14 should be y = 6x + 1 rather than y = 54x + 1 (the solution uses m = 54 at x = 1 as the gradient of the tangent, but the point of interest was (0, 1) and so they should have found that m = 6 at x = 0). Q20...

I agree that the hint is awkward. I like the way Mok has used it, although the existence of cases is not treated. I kept thinking of how the perpendicular vector gas the direction of the altitude and trying to adjust its magnitude to match the height of the triangle. I suspect it will have...

Yes, and it should give you the answer 600\pi(2 - \ln{3}) cubic units.

I've figured out a non "otherwise" method for 14(c)(i) that doesn't require any difficult algebra, if anyone is interested: https://community.boredofstudies.org/threads/teacher-pov-2023-hsc-extension-1-solution-speedrun-send-help.405894/#post-7515536

I figured out what they wanted for 14(c)(i), the non-otherwise approach. Take vectors OA and OB as having lengths A and B, respectively. Without loss of generality, assume that OA has the larger angle between itself and the x-axis than has OB. Take OB' = b2i - b1j so (from the information...

Yes, I saw it, thanks. It's definitely the easy way so long as you have covered cross products of 3D vectors, which is beyond MX2 content and well beyond MX1. The approach that @epicmaths took, using ab\sin{C} / 2, and Lucid's solutions, are both "otherwise" methods too. I really wonder about...

To seat the students as one pair and three singles means many more possible arrangements, as it needs to be SS-X-S-X-S-X-S-X-, where X is not a student. Available to fill the four non-student positions are three teachers, so at least one of the teachers must be sub-divided. Now, a single...

Start from a teacher. The arrangement must be T-SS-T-SS-T-S-(and back to the original teacher), or one of two variations: T-SS-T-S-T-SS- and T-S-T-SS-T-SS-. Seat first teacher in 1 way Choose one of these 3 arrangements Seat the two remaining teachers in 2 ways Seat the students in 5! ways...

Everyone makes mistakes. They aren't going to say "Oh, you expanded a binomial expression incorrectly, you get 0 for the whole of Q12 plus a fifty mark penalty off your raw score!"

Indeed, and it is helpful to people to put out solutions... I also see it as helpful to note where they need corrections. Everyone makes mistakes. :)

Yes, I'm suggesting you may have made an error in the squaring, which would explain getting the 800pi term but not the 6 ln 3... and my earlier post should have y's rather than x's, fyi... I'll go and correct it.

Lucid's solution for 14(c)(ii) is wrong as the vectors given as OP and OQ are actually IP and JQ. Question 10 is wrong, too.

\begin{align*} \text{Applying the symmetry property } \qquad \binom{2023}{1943} &= \binom{2023}{2023-1943} \\ &= \binom{2023}{80} \qquad \text{(eqn 1)} \\ \\ \text{Applying the addition property (given)} \qquad \binom{2022}{80} + \binom{2022}{81} &= \binom{2023}{81} \qquad \text{(eqn 2)} \\ \\...