Search results

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...

Chlorine is a highly electronegative element A polar molecule has an uneven electron distribution A non-polar molecule has an even electron distribution In HCN the atoms have different electronegativity (the tendency to attract electrons) and so there will be an uneven electron...

Better to do it in person. Labs are an area you can't gain proper skills without doing it yourself. Watching a video isn't the same

It's probably not an actual lab then. Might just be like you watch some prerecorded videos and do the work in your own time. That's happened to me before for online labs at unsw

Because if you at the terms like [H2O]^2 which changes to [H2O] the only way to get rid of the power on the H2O is by taking a square root

For the 1st reaction: K = [O2] [HCL]^4 / [Cl2]^2 [H2O]^2 4.0 x 10^(-4) = [O2] [HCL]^4 / [Cl2]^2 [H2O]^2 For the 2nd reaction: K = [Cl2] [H2O] / [O2]^(1/2) [HCL]^2 We need to change the first expression into the 2nd one. This would involve flipping the fraction i.e. 1 / (4x10^-4) =...

It's a range of textbooks for subjects like biology and chemistry which put together past hsc and exam questions. It's not on this forum but you can buy these at bookshops like five senses