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RE: The Steemit Crypto Challenge #11 - CC2k17.2

in #steemitcryptochallenge8 years ago (edited)

My previous count of 36 bonds is wrong because at first I thought the red color was a hint that there was a mistake in the molecule, so that I made another molecule out of it.
No, if the molecules are sketched correctly, we have altogether 34 bonds:

  • X = N2 and has 3 bonds (0 single bonds, 0 double bonds and 1 triple bond).
  • Y = 1,3-Cyclohexadiene (C6H8) and has 16 bonds (12 single bonds, 2 double bonds and 0 triple bonds).
  • Z = benzene (C6H6) and has 15 bonds (9 single bonds, 3 double bonds and 0 triple bonds).
  • That are altogether 34 bonds (21 single bonds, 5 double bonds and 1 triple bond).
  • I come to a higher number of bonds than @digicrypt because everywhere in Y, where you see a H2, there are 2 bonds, because every H atom has exactly 1 bond.
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well timsaid the password is in the counts furion has the "box" you have the key, any hints mate :) can either of you open the pass with either 1215 or 3415 dunno the exact idea behind the count, but he did say it was simple and 1215 sounds simple :D

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