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I was trying to do an experiment called Pharaoh's Serpent and I read that the intermolecular forces between the $\ce{C}$ and the $\ce{Na2O}$ (formed in the thermal decomposition of the $\ce{NaHCO3}$) produced the "serpent" in the experiment, but I can't find what intermolecular force (if there's really one) that is.

What is the structure of this compound?

Here's a link to the experiment: https://www.youtube.com/watch?v=vQdK7gaZS0k

edited:

Here are the reactions that I can think of and I researched :

  1. Sodium bicarbonate decomposition ( with temperature ) :

$$\ce{2NaHCO3 -> Na2CO3 + CO2 + H2O}$$

$$\ce{Na2CO3 -> Na2O + CO2}$$

2.Fructose complete combustion:

$$\ce{C12H22O11 + 12O2 -> 12CO2 + 11H2O}$$

  1. Fructose incomplete combustion:

$$\ce{C12H22O11 -> 12C + 11H2O}$$

$$\ce{C12H22O11 +6O2 -> 11H2O + 12CO}$$

So , I found that the $\ce{C}$ ( produced in the fructose incomplete combustion) reacts with the $\ce{Na2O}$ ( produced in the sodium bicarbonate decomposition), composing the "body" of the "snake".

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closed as unclear what you're asking by pentavalentcarbon, airhuff, Todd Minehardt, Jan, Nilay Ghosh Jun 18 '17 at 4:01

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  • $\begingroup$ And where do you have Na2O molecules there, I wonder, cause not in solid... $\endgroup$ – Mithoron Jun 17 '17 at 16:02
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    $\begingroup$ en.wikipedia.org/wiki/… There's no Na2O or elemental carbon or bicarbonate in this reaction so I'm afraid you deeply confused here and question should be closed as unclear. $\endgroup$ – Mithoron Jun 17 '17 at 20:49
  • $\begingroup$ You can do the experiment using NaHCO3 or Mercury(II) thiocyanate. But since Mercury(II) thiocyanate is toxic , i decided to do using NaHCO3. Heres a link showing how to do the experiment melscience.com/en/experiments/sugar-snake $\endgroup$ – user158657 Jun 17 '17 at 20:54
  • $\begingroup$ Bicarbonates decompose to CO2 not C and still no idea what "intermolecular forces" you want. $\endgroup$ – Mithoron Jun 17 '17 at 21:07
  • $\begingroup$ It's sucrose not fructose and combusts completly. I have a feeling that editing won't help here :( I appreciate your responsiveness and additional effort, though. $\endgroup$ – Mithoron Jun 17 '17 at 22:17