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  1. Holmes

    Chemistry: Post your doubts here!

    anyone help PLZ nov 2015 (13)
  2. Holmes

    Chemistry: Post your doubts here!

    To my inadequate knowledge the answer is "D" CH4 + 2O2 → CO2 + 2H2O Bond energy of reactants: =4(410)+2(496) = +2632 kj/mol now look at The bond energy of Products: = 2 (740)+ 4 (460) = +3320 kj/mol last step: enthalpy change = B.E of Reactants - B.E. of Products...
  3. Holmes

    Physics: Post your doubts here!

    Do u want me to show you how I got the answers? of course yes :)
  4. Holmes

    Physics: Post your doubts here!

  5. Holmes

    Physics: Post your doubts here!

  6. Holmes

    Physics: Post your doubts here!

    Here you go thanks in advance:cool:
  7. Holmes

    Physics: Post your doubts here!

    Thank you !Thelastmoment (y)
  8. Holmes

    Physics: Post your doubts here!

    Help !
  9. Holmes

    Chemistry: Post your doubts here!

    Oh sorry :oops: bytheway thanks for telling my mistake. Good luck
  10. Holmes

    Physics: Post your doubts here!

    explain?
  11. Holmes

    Mathematics: Post your doubts here!

    guide please.!!
  12. Holmes

    Physics: Post your doubts here!

    thank you bro....
  13. Holmes

    Physics: Post your doubts here!

    Explain why the answer is D Thanks in Advance :)
  14. Holmes

    Mathematics: Post your doubts here!

    Help ! kindly explain value of "common ratio" obtained Thanks in Advance. :)
  15. Holmes

    Chemistry: Post your doubts here!

    explain why the answer is B
  16. Holmes

    Chemistry: Post your doubts here!

    9701/12/O/N/10 (Chemistry) October 2010 paper 1 : Q8 ,Q11 , Q28, Q34, Q39. Help me please!! Thanks in advance :)
  17. Holmes

    Mathematics: Post your doubts here!

    Thank you!
  18. Holmes

    Mathematics: Post your doubts here!

    Thanks a lot. :)
  19. Holmes

    Mathematics: Post your doubts here!

    9709/12 June 2012 Pure Mathematics ! Help me please! thanks.
  20. Holmes

    Chemistry: Post your doubts here!

    KMnO4 is a famous oxidising agent. Using COLD KMnO4 will break "C=C" and add an -OH group on each carbon atom. Now count the total number of hydroxy groups present. Two are attached with each Carbon and one already attached to your absolute LEFT hand side (in the diagram ). So this narrows your...
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