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Google ‘large ceramic incense burner with lid’. Ceramic is a reasonably good insulator - that’s why we use it for hot beverages.

For Q9, K = [product] / [reactant] = [trans] / [cis] 0.17 = [trans] / 1 [trans] = 0.17 M

We agree on the answers to Q6 and Q8. I get 1.52 as the answer to Q7.

For Q6, I used an online calculator to get 0.39. What did you get?

I’m not sure what they are trying to test with question 5. carrotsss is correct about pressure per se not affecting the equilibrium when there are equal moles of gases on each side, but then the question mentions that the (forward) reaction is exothermic, so I don’t know if they are trying to...

Do they provide the answers to questions 6 & 7? @carrotsss has provided you with the correct approach, so if you provide their answers, we can determine if the textbook might be wrong (happens not uncommonly) and possibly reverse engineer from the answer.

TAFE does sound like a good option for now - allows you to get used to independent study, gain skills for the work force, work part time and have some breathing space to work out what you want to do in the longer term.

Your chem teacher is correct. The reasoning in your first statement is correct. Your second statement confuses cause and effect (& the wording “so less products have to be converted to products” is, itself, confusing as well). It isn’t that less reactants need to be converted to products to...

In theory you can do what you propose (see my previous post), but you’ll need your school’s support.

OP can do ext 2 in Year 11 next year if he completes preliminary extension this year. He can then study advanced, ext 1 and ext 2 next year and on the day that adv maths & ext 2 exams are held, he can choose to sit ext 2. This is from the combination of experiences of my elder daughter & my son.

If you are going to continue to invest in tutoring, I’d suggest consolidating everything you should know in Yr 11 (advanced & extension) before tackling Yr 12 content. For a start, make sure you fully understand every concept related to the marks you lost on last term’s 3U test, because you need...

For Q9, where the reaction is reversed, we take the reciprocal of the K of the forward reaction: K = 1/0.41 = 2.44 (to 2 d.p.)

Recall that the equation for calculating equilibrium constants for each reaction is: For question 8, the overall reaction is a combination of the forward reaction of the first reaction and the reverse reaction of the second, so we have to multiply K1 by the reciprocal of K2, or in otherwise...

I get that - it was more the general principle of communicating. That one should demonstrate understanding that when a proof is required, every step should be proven, so it’s unequivocal to state Bertrand‘s postulate (proven) or Bertrand’s postulate (now theorem). I’m not a maths marker (maths...

I guess technically, in an exam or competition, one would need to add that the postulate is now a theorem /has been proven, to show that you understand that a postulate can’t be used to prove another maths statement.

The symbols represent sets of numbers with specified conditions.

May also be of possible interest to some to note that there are Commonwealth funded courses and non funded courses. Non funded programs like Bond University’s are prohibitively expensive for the vast majority of students, however because most families can’t afford the fees...

I’m not a teacher or HSC marker, but at postgrad level, rounding to correct sig fig is for reporting outputs/results. For ongoing calculations, I would expect every step to be as accurate as possible to minimise the error of the final result.

I do agree with @Galactic Drama, but unfortunately this is a bad question and open to interpretation. The sig fig for an answer should only be to the least sig fig of any measurement performed, but it could be argued that this applies to measurements performed within the practical exercise and...

As you’ve said, they are structural isomers. Potassium permanganate doesn’t react with tertiary alcohols, therefore B must be the tertiary form of butanol, whereas C is likely to be primary butanol. Note: Given alkenes have the common molecular structure of CnH2n with one carbon and two...