MedVision ad

Search results

  1. H

    section 2 question

    WHY would u base it off something THEY will 100% know is plagiarism?
  2. H

    HSC 2008 question 25 help please

    Thanks for help guys i managed to work it out =]:)
  3. H

    HSC 2008 question 25 help please

    So i know reduction occurs at the cathode and oxidation at the annode but the question says Ni (electrode) is in Ni(NO3)2 aq and Pt (electrode) is in KCl aq (Cl2 gas is being put in) i went to the standard potentials but cant seem to work out which is the annode and which is the cathode? help...
  4. H

    chem question

    Small error in part (c) Therefore amount needed to release 22643789.474 kJ = (2643789.474)/285 = 9276.454294 moles (the 22643789.474 should be 2643789.474 there was just an extra 2 just a typo =])
  5. H

    chem question

    is the reason u divide heat of combustion by average molar mass is so u get kj per gram? hey soz haha but im really unsure why in part b you got: heat of combustion= delta H/n =5460 i dont understand how you got this?
Top