The answer for this question is the smallest particle of matter.
Answer:

Explanation:
Hello,
In this case, for such formation of sulfur hexafluoride, the standard enthalpy of formation is -1220.47 kJ/mol (data extracted from NIST database). Next, we compute the moles in 10.0 grams of sulfur hexafluoride as shown below:

Next, for the given energy, we compute the total heat that is liberated:

Finally, we conclude such symbol has sense since negative heat is related with liberated heat.
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P and S
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441 g CaCO₃ would have to be decomposed to produce 247 g of CaO
<h3>Further explanation</h3>
Reaction
Decomposition of CaCO₃
CaCO₃ ⇒ CaO + CO₂
mass CaO = 247 g
mol of CaO(MW=56 g/mol) :

From equation, mol ratio CaCO₃ : CaO = 1 : 1, so mol CaO :

mass CaCO₃(MW=100 g/mol) :

The final temperature = 36 °C
<h3>Further explanation</h3>
The balanced combustion reaction for C₆H₆
2C₆H₆(l)+15O₂(g)⇒ 12CO₂(g)+6H₂O(l) +6542 kJ
MW C₆H₆ : 78.11 g/mol
mol C₆H₆ :

Heat released for 2 mol C₆H₆ =6542 kJ, so for 1 mol

Heat transferred to water :
Q=m.c.ΔT
