You are viewing a single comment's thread from:

RE: #Chemistry Challenge 6

in #chemistrychallenge7 years ago

Let me try to give it a shot:
Q1
1 bee implies 0.1 gram; 4 kg bee implies 40000 bees
honey production rate = 9/20 gram/bee/day, implies 0.08 min per gram of honey.

  • To produce 500 g honey thus require 40 min


    500 gram honey = 150 gram glucose
  • = 0.8333 mole glucose
  • = 5.016667 x 1023 molecule of glucose


    Q2
    Density = mass / vol = 500 /(Pi(4)27.958)
  • = 1.25 gram/cm3


    Q3
    Enthalpy of reaction:
    C6H12O6 + 6 O2 --> 6CO2 + 6 H2O
    by completing the Hess cycle,
    The required enthalpy = (-393.5)(6) + (-285.9)(6) - (-1268)
  • = -2808.4 kJ/mol
Sort:  

Opus I forgot that was 1/10 of the honey. So I edit add a few line here:
Consumed honey = 50 gram = 15 gram glucose = 0.0833333 mole =
so ΔHR = -234 kJ

Right! So @jedigeiss 'stole' you this part of the solution because he was faster! :)
Well done nevertheless - hope you enjoyed the competition! (I will send you 3 STEEM.)

Haha !! That was really fun!!! This is probably the first time for me to work on a Hess cycle in the recent five years. Really took me some time back to the definition and make sure I didnt remember things wrongly XDD