Answer:

Explanation:
Given that:

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

Multiplying (2) with equation (4) ; we have:

From equation (1) ; multiplying (-1) with equation (1); we have:

From equation (2); multiplying (3) with equation (2); we have:

Now; Adding up equation (5), (6) & (7) ; we get:



<u> </u>

<u> </u>
<u />
(According to Hess Law)


Answer:
The smell of a chocolate is from the presence of volatile compounds present in the chocolate bar which at room temperature readily changes phase from solid to liquid to vapor or gas
Explanation:
There are nearly 600 identified compounds present in a chocolate bar and out of these, there are volatile components which gives the chocolate bar its distinctive aroma.
These volatile chocolate contents readily change phase from solid to vapor, with very short duration liquid phase.
For example, 3 methylbutanal, vanillin, and several organic compounds which are known to be readily volatile.
Don't smoke, drive less, don't let toast burn, use an electrical fire instead of a wood one....
I can't think of anything else.