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The "don't write in this area" section is where the booklets will be cut into individual sheets to feed into the scanner. This will be done by machine (take booklet, cut off margin and staple, feed sheet by sheet into scanner, produce image file of paper) so the scanned part will be as much of...

No, it doesn't. 1,2-dichloroprop-1-ene could not be made by addition to an alkene. Starting with propadiene (CH2=C=CH2) would yield 2,3-dichloroprop-1-ene assuming reaction with one equivalent (assuming that dienes like this react in this way), rather than to yield 1,2-dichloroprop-1-ene...

The addition of chlorine to an alkene produces a dichloroalkane. The fact that the mass spectrum appears to show only one is misleading. The product should be 1,2-dichloropropane. The mass spectrum of the compound can be seen at this link...

And there was Q31, which refers to a species that I'm doubtful exists. Copper(II) ions do react with lactate anions to make a compound known as copper(II) lactate, that can be thought of as a salt or as a coordination complex: [Cu(C3H5O3)2] - that is, [Cu(CH3CHOHCO2)2]. This year's HSC exam...

Well, the syllabus and the Department of Education / NESA / etc do things their own way... and trials often have questionable content... but functional groups and compound classes are not the same thing. And the HSC has a track record of being, shall we say, less than 100% on the accuracy...

PV = nRT is the way to go for Q27 (CO2 production from fermentation) because the gas volumes on the data sheet only apply to temperatures at 273 K and 298 K. Alternatively, if you used one of those (since the pressure matched) you would then need to adjust the volume using a Charles' Law...

You can use z=x+iy to show it rearranges to |y| < 1, and so is the region -1 < \Im{(z)} < 1. However, thinking of the expression as a vector, the interpretation of the statement is: z is positioned such that the distance from \bar{z} to z is less than 2 units Since \bar{z} is located at the...

The Cl- and Br- anions are 115% positively certainly definitely NOT acidic! How could they be? Acidity requires an ability to donate a proton (H+) under Lowry-Bronsted definitions or to be electron pair acceptors under Lewis definitions. Chloride and bromide salts are neutral, they do not...

The answer is right. The working is poor. If you set up an ICE table with the initial moles of the acid and conjugate base forms, then consider that the added HCl has the effect of converting the base form into the acid - so the acid moles goes up by n(H+) added and the base form goes down by...

Much wiser to start from ethanol than from ethene, hydrating ethene (a gas) to ethanol is not a school lab activity.

This is not a good question. Silver carbonate has a Ksp of 10-12, so it isn't soluble. Usually precipitating silver chloride is a good strategy but an ammonium salt (weak acid) will add ammonia molecules to the system, and the equilibrium constant for the formation of [Ag(NH3)2]+ is about 107...

7.5 % is correct

None of the answers given are correct. Anything based on the precipitation of PbCl2 being complete is wrong because the information given is that lead(II) chloride has just started to precipitate. Hence, it is a Ksp problem and not a titration calculation. You use the value of Ksp and the...

People like to see exams as a test of knowledge / understanding, but the true purpose is to produce a ranking order of candidates from first to last. If that order happens to correlate at least moderately well with knowledge or understanding, that's a bonus. The mark someone achieves on an...

Yes, every H environment should appear. Alcohol and amine signals can be broaden and look unlike a typical signal, but they should still appear. The reasons for these atypical features are beyond the syllabus.

It's still an H environment that will give rise to a signal, whether it is involved in splitting or not. In this monomer, none of the H environments are close enough for splitting to happen.

You need to put in light of a wavelength that will be absorbed (so 625 nm rather than 510 nm) and then measure the intensity the remains after passing through the sample.

The question is asking about the NMR of the monomer, which is 2-hydroxy-2-methylpropanoic acid... i.e. (CH3)2C(OH)COOH, which has 3 x 1H and 3 x 13C environments.

I don't see a typo in the question. I also don't see any uncertainty about what is being asked, nor what answer is correct. What caught my attention was when I heard that the students sitting the paper at a certain school were notified of an issue, because everyone who sat the exam state-wide...

Yes, so people know that (to take one example) the solutions of the BHHS 2019 Chemistry Trial score 16 / 20 on the MCQ...