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Phys/chem/bio you need usually like a 97+ hsc mark ext 1 maths 100 hsc mark unless it was a very hard exam ext 2 maths probably 98+ No clue about the ranking in the schools exactly but if there is only 20 places per subject then you need to really be like top 5 since the state rank is based on...

TDS = total dissolved solids. It's about understanding the difference between the parts of the method. The filtering is to remove the non-dissolved material like grass, dirt etc. that would be in a river/dam. After thats filtered then you can find out how much solid is dissolved by evaporating...

The reaction is N2(g) + 3H2(g) < -- > 2NH3(g) Liquefying the ammonia basically removes the gaseous form from the equation. Removing NH3(g) reduces [NH3] and so the equilibrium will shift right by Le chatelier's Principle to minimise the disturbance. Shifting to the right means more yield

Kind of depends how much are marks important for your degree and future. Most degrees having a high credit to distinction should be a good enough mark to get into some internships if you have other good skills and work experience that can give you a good resume and ability to discuss about...

I don't think most people would want to pay for a study space as stated by the other people above there are a lot of free places you can go to for studying such as local libraries. Have you tried maybe to put on some headphones. There are a lot of noise cancelling headphones these days that you...

This is mostly a memorise the formula and sub kind of question % means fraction x 100 W/v tells you what to put on the top and bottom of the fraction %w/v = mass / volume x 100 Sub the 1.5% and volume as 25 for Ethanol and solve for mass

If I was marking that above answer I wouldn't give it full marks. The H3A is coming from citric acid being triprotic. However, you can't just ignore the actual formula of the citric acid (C6H8O7) and make up your own formula of H3A. A mark would be lost in the writing of the equation since...

m(Pb2+) = 2.0 mg = 2 x 10^-3 g n(Pb2+) = m/MM = (2x10^-3) / 207.2 n(Pb2+) = 0.0000096525 moles n(Pb(NO3)2) = n(Pb2+) n(Pb(NO3)2) = 0.0000096525 moles m(Pb(NO3)2) = nMM = 0.0000096525 x (207.2 + 2(14.01 + 3x16)) = 0.00319... g Neither your or the textbook answer is correct

ppm is mg/L 2.0 ppm = 2.0 mg/L of Pb2+ 2.0 mg/ L x 1.0 L = 2.0 mg of Pb2+ find the moles of Pb2+ that equals moles of Pb(NO3)2 multiply that by molecular mass of Pb(NO3)2 to find the mass of Pb(NO3)2

Both First in mod 7 you learn the reactions Mod 8 emphasises the importance of physical observations such as colour changes or bubbles forming that can be used to confirm the chemical identity of a compound. In that you learn about changes in the colour of the reaction when an oxidation takes...

Not in the same kind of way. Year 11 is focused around using oxidation / reduction reactions to explain how electricity (voltage) is obtained in a galvanic cell. In Year 12 it's not asked in that same kind of way. It's more so about the general definition of what an oxidation / reduction...

Yeah but like this is unlikely for you to have an acid and base like that. Also weak acid and base titration graphs aren't very clear like those involving strong acid/base as you can see in the image above by the other user

With HSC knowledge how the questions are asked you can't do it This ties into the problem with the simple definitions of strong and weak acids learnt in HSC. A strong acid ionises fully while weak acid ionises partially. So even an acid that has 90% ionisation in this definition would be weak...

They most likely have written the equation for ethyne (C2H2) which has been mistaken for ethane (C2H6) Ethene combustion would be: C2H2(g) + 5/2 O2(g) --> 2CO2(g) + H2O(l) The answer for ethyne would be 2.5 moles of O2 For ethane, the answer would be 3.5 moles

Use n = m/MM to find a formula of aluminium chloride is AlCl3 n(Al) = n(AlCl3) since each aluminium chloride has 1 Al n(Cl) = 3xn(AlCl3) since each aluminium chloride has 3 Cl

Use the Avogadro's Law n = no. of particles / 6.022 x 10^23 that 6.022 x 10^23 is Avogadro's number and is found on the data sheet By combining that formula with n = m / MM, you can solve both of those questions 1) n = m / MM n = 0.213 / 32.07 n = 0.00664... moles n = no. of particles /...

Do questions from trial papers. The above with the olympiad although they may be harder than HSC, in my opinion it's not a good way to study for the subject in terms of the HSC. In the olympiad exams the questions are more problem focused, with not that much writing involved. HSC exams there is...

If you are majoring in the chem then it might be a good idea to do the specialist depending on how far your degree goes into chem. If you might do an honours year in the chemistry major, those extra lab skills and experience you get early on will be quite useful. If it's just a year 1 - 3...

I haven't done the exact courses since I went to UNSW, but we also have Chemistry 1A and Higher Chemistry 1A courses. The higher level of courses usually just means you do some extra content the normal level subject doesn't. Exam is a bit harder because of those extra questions but most of the...